collection of solutions to baby rudin

Real AnalysisMath 131AH

Rudin, Chapter #1

Dominique Abdi

1.1. If r is rational (r 6= 0) and x is irrational, prove that r + x and rx areirrational.

Solution. Assume the contrary, that r+x and rx are rational. Since the rationalnumbers form a field, axiom (A5) guarantees the existence of a rational number−r so that, by axioms (A4) and (A3), we have

x = 0 + x = (−r + r) + x = −r + (r + x).

Both −r and r + x are rational by assumption, so x is rational by axiom (A1),contradicting that x is irrational.

Similarly, axiom (M5) guarantees the existence of a rational number 1/r sothat, by axioms (M4) and (M3), we have

x = 1x =(

1r· r

)x =

1r

(rx).

Both 1/r and rx are rational by assumption, so x is rational by axiom (M1),contradicting that x is irrational.

1.2. Prove that there is no rational number whose square is 12.

Solution. Assume the contrary, that there is a rational number p such thatp2 = 12. Then there are integers m and n with p = m/n and for which 3 isnot a common factor. By assumption, m2 = 12n2, so m2 is divisible by 3, andtherefore m is divisible by 3. But then m2 is divisible by 9. This means that12n2 is divisible by 9, so 4n2 is divisible by 3. Thus n2 is divisible by 3, andtherefore n is divisible by 3. We have shown that 3 is a common factor of mand n, a contradiction.

1.3. Prove Proposition 1.15.

Solution. Suppose x, y, and z are elements of a field and x 6= 0. The axioms(M) imply

y = 1y =(

1x· x

)y =

1x

(xy) =1x

(xz) =(

1x· x

)z = 1z = z,

proving part (a). Part (b) follows from part (a) by setting z = 1, and part (c)follows from part (a) by setting z = 1/x. Since

x · 1x

= 1,

replacing x with 1/x in part (c) gives part (d).

1.4. Let E be a nonempty subset of an ordered set; suppose α is a lower boundof E and β is an upper bound of E. Prove that α ≤ β.

Solution. Since E is nonempty, there exists a point x ∈ E. Note that α is alower bound of E and β is an upper bound of E by the definition of supremumand infimum. Thus α ≤ x by the definition of lower bounds and x ≤ β by thedefinition of upper bounds, and therefore α ≤ β by Definition 1.5(ii).

1.5. Let A be a nonempty set of real numbers which is bounded below. Let−A be the set of all numbers −x, where x ∈ A. Prove that

inf A = − sup (−A) .

Solution. By the least-upper-bound property, α = inf A exists. By the definitionof infimum, α ≤ x for all x ∈ A, and if β > α, then there exists a y ∈ A suchthat β > y. Therefore Proposition 1.18(a) implies −α ≥ −x for all x ∈ A,and if γ < −α, then there exists a y ∈ A such that γ < −y. Thus −α is anupper bound of −A, and if γ < −α, then γ is not an upper bound of −A.By the definition of supremum, it follows that −α = sup (−A), and thereforeinf A = − sup (−A).

1.6. Fix b > 1.

(a) If m,n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, prove that

(bm)1/n = (bp)1/q.

Hence it makes sense to define br = (bm)1/n.

(b) Prove that br+s = brbs if r and s are rational.

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(c) If x is real, define B(x) to be the set of all numbers bt, where t is rationaland t ≤ x. Prove that

br = supB(r)

when r is rational. Hence it makes sense to define

bx = supB(x)

for every real x.

(d) Prove that bx+y = bxby for all real x and y.

Solution. Fix b > 1.

(a) If m/n = p/q, then mq/nq = np/nq, and therefore mq = np. For simplic-ity, let α = (bm)1/n and β = (bp)1/q. Then

α = (αnq)1/nq = (bmq)1/nq = (bnp)1/nq = (βnq)1/nq = β,

so (bm)1/n = (bp)1/q. Thus br = (bm)1/n is well-defined, since any tworepresentations of r yield the same value.

(b) Suppose r = m/n and s = p/q for some integers m,n, p, q with n > 0 andq > 0. Then r + s = (mq + np)/nq, and by part (a) and the corollary toTheorem 1.21, we have

brbs =(bm/n

) (bp/q

)=

(bmq/nq

) (bnp/nq

)= (bmq)1/nq (bnp)1/nq

=(bmq+np

)1/nq = b(mq+np)/nq = br+s.

(c) Suppose r = p/q for some integers p and q > 0, and fix bs ∈ B(r). Thens is rational and s ≤ r, so br−s ≥ 1 since b > 1. Multiplying through bybs yields br ≥ bs, so br is an upper bound of B(r).

Since r ≤ r, we have br ∈ B(r). Thus if γ < br, then γ is not an upperbound of B(r). Hence br = supB(r).

The definition bx = supB(x) for every real x makes sense because theequation br = supB(r) holds for every rational r by the argument above.

(d) Suppose x and y are real, r and s are rational, r ≤ x, and s ≤ y. Thenr + s ≤ x+ y, so

brbs = br+s ≤ bx+y.

Thus taking the supremum over all rationals r ≤ x and s ≤ y givesbxby ≤ bx+y.

Now suppose t is rational and t ≤ x+y. Then there exist rational numbersr and s such that t = r + s, r ≤ x, and s ≤ y. Thus

bt = br+s = brbs ≤ bxby,

and taking the supremum over all rationals t ≤ x+ y gives bx+y ≤ bxby.

Hence bx+y = bxby.

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1.8. Prove that no order can be defined in the complex field that turns it intoan ordered field.

Solution. By Definition 1.27, we have i 6= 0, and by Theorem 1.28 and Propo-sition 1.18(a,d), we have i2 = −1 < 0. This contradicts Proposition 1.18(d),and therefore no order can be defined in the complex field that turns it into anordered field.

1.9. Suppose z = a + bi, w = c + di. Define z < w if a < c, and also if a = cbut b < d. Prove that this turns the set of all complex numbers into an orderedset. Does this ordered set have the least-upper-bound property?

Solution. The real numbers form an ordered set, so one and only one of thefollowing holds: a < c, a = c, a > c. Similarly, either b < d, b = d, or b > d.There are 5 possible cases:

(1) If a < c, then z < w.

(2) If a > c, then z > w.

(3) If a = c and b < d, then z < w.

(4) If a = c and b > d, then z > w.

(5) If a = c and b = d, then z = w.

Thus one and only one of z < w, z = w, and z > w holds, so Definition 1.5(i)holds.

Let z1, z2, z3 be complex numbers satisfying z1 < z2 and z2 < z3. Then oneand only one of the following holds,

(1) Re(z1) < Re(z2),

(2) Re(z1) = Re(z2) and Im(z1) < Im(z2),

and similarly, one and only one of the following holds,

(1) Re(z2) < Re(z3),

(2) Re(z2) = Re(z3) and Im(z2) < Im(z3).

This yields 2 possible cases:

(1) If Re(z1) < Re(z2) or if Re(z2) < Re(z3), then Re(z1) < Re(z3).

(2) If Re(z1) = Re(z2) = Re(z3), then Im(z1) < Im(z2) < Im(z3).

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Thus z1 < z3 in all cases, so Definition 1.5(ii) holds.Hence the set of all complex numbers is an ordered set.

Now let E = {in : n ∈ Z}. Then E is a nonempty subset of C which isbounded above by 1. Suppose α+ iβ = supE. Then α ≥ 0 since α is an upperbound. But if α > 0, then α/2 is a smaller upper bound of E, contradictingthat α + iβ = supE. Thus α = 0, and so iβ = supE. But for any β, thereexists an integer n > β, contradicting that iβ is an upper bound of E. HencesupE does not exist, so the least-upper-bound property does not hold.

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Rudin, Chapter #2

Dominique Abdi

2.1. Prove that the empty set is a subset of every set.

Solution. Assume the contrary, that there is a set E such that the empty set isnot a subset of E. Then there is an element x ∈ ∅ such that x /∈ E, but thiscontradicts that the empty set is empty. Hence ∅ ⊂ E.

2.2. A complex number z is said to be algebraic if there are integers a0, . . . , an,not all zero, such that

a0zn + a1z

n−1 + · · ·+ an−1z + an = 0.

Prove that the set of all algebraic numbers is countable.

Solution. Let p be a polynomial of degree n with integer coefficients and let

Zp := {z ∈ C : p(z) = 0} .

By the fundamental theorem of algebra, Zp has at most n elements.Now define the set

Pn(Z) :=

{n∑k=0

akzk : ak ∈ Z, an 6= 0

},

consisting of all polynomials of degree n with integer coefficients, and let

En := {(a0, a1, . . . , an) : ak ∈ Z, an 6= 0} .

Example 2.5 can be easily modified to show that the set of non-zero integers iscountable. Thus by Example 2.5 and Theorem 2.13, En is countable. Define

f : En → Pn(Z), f(a0, a1, . . . , an) :=n∑k=0

akzk.

Every polynomial of degree n is uniquely determined by its n + 1 coefficients,so f is one-to-one and onto. Hence Pn(Z) is countable, and by Theorem 2.12,the set

P (Z) :=∞⋃n=0

Pn(Z),

consisting of all polynomials with integer coefficients, is countable.Since the algebraic numbers are defined to be the roots of polynomials with

integer coefficients, the set A of all algebraic numbers can be written as

A =⋃

p∈P (Z)

Zp,

a union of a countable collection of finite (hence at most countable) sets. Bythe corollary to Theorem 2.12, A is at most countable. But for any n ∈ Z,

p(z) = z − n

is a polynomial with integer coefficients for which z = n is a root, so Z ⊂ A.We have shown that A is an at most countable set which contains an infinite

subset. Hence A is infinite and at most countable, and thus A is countable.

2.3. Prove that there exist real numbers which are not algebraic.

Solution. Assume that every real number is algebraic. Then R is countable byExercise 2. This contradicts the corollary to Theorem 2.43, which states that Ris uncountable.

2.4. Is the set of all irrational real numbers countable?

Solution. Assume that the set Qc of all irrational real numbers is countable.Since R = Q ∪ Qc and Q is countable by the corollary to Theorem 2.13, itfollows that R is countable by the corollary to Theorem 2.12. This contradictsthe corollary to Theorem 2.43, which states that R is uncountable.

2.5. Construct a bounded set of real numbers with exactly three limit points.

Solution. Define the sets

E0 :={

1n

: n ∈ N}, E1 :=

{n− 1n

: n ∈ Z}, E2 :=

{2n− 1n

: n ∈ N},

and E := E0 ∪ E1 ∪ E2. Then E is bounded since E ⊂ (0, 2).We now show that E′ = {0, 1, 2}. Fix ε > 0 and choose N such that N > 1/ε.

Then 0 < 1/n < ε for all n ≥ N , and in particular,

1n∈ (−ε, ε), n− 1

n∈ (1− ε, 1 + ε),

2n− 1n

∈ (2− ε, 2 + ε)

for all n ≥ N . This shows that an open interval of arbitrarily small radius ε > 0centered at 0, 1, or 2 intersects E at infinitely many points, so {0, 1, 2} ⊂ E′.

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Now suppose x /∈ {0, 1, 2}. If x < 0, then (2x, 0) ∩ E = ∅, so x /∈ E′. Ifx > 0, then there are three possible cases:

Case 1 : If 0 < x < 1, then choose k ∈ N such that

1k + 1

< x ≤ 1k,

and define δ to be the smallest non-zero element of the set{|x| ,

∣∣∣∣x− 1k + 1

∣∣∣∣ , ∣∣∣∣x− 1k

∣∣∣∣ , |x− 1|}.

Case 2 : If 1 < x < 2, then choose k ∈ N such that

k − 1k

< x ≤ k

k + 1,

and define δ to be the smallest non-zero element of the set{|x− 1| ,

∣∣∣∣x− k − 1k

∣∣∣∣ , ∣∣∣∣x− k

k + 1

∣∣∣∣ , |x− 2|}.

Case 3 : If x > 2, then define δ := x− 2.In all three cases, we have (x−δ, x+δ)∩E = ∅ by the choice of δ, so x /∈ E′.

Hence E′ = {0, 1, 2}.

2.6. Let E′ be the set of all limit points of a set E. Prove that E′ is closed.Prove that E and E have the same limit points. (Recall that E = E ∪E′.) DoE and E′ always have the same limit points?

Solution. By Definition 2.26 and Theorem 2.27(b), to prove that E′ is closed,it suffices to show that E′ contains its limit points. Let x be a limit point of E′

and let Nr(x) be a neighborhood of x. Then there exists a point y ∈ Nr(x)∩E′with x 6= y. Let r′ := r − d(x, y) and note that Nr′(y) ∩ E contains a pointz ∈ E since y is a limit point of E. Moreover, x 6= z since d(x, z) > 0. Thetriangle inequality gives

d(x, z) ≤ d(x, y) + d(y, z) < d(x, y) + r′ = r,

so z ∈ Nr(x). Thus x is a limit point of E, so therefore (E′)′ ⊂ E′ and henceE′ is closed.

We now show that E and E have the same limit points. If x ∈ E′ and r > 0,then there exists a point y ∈ Nr(x) ∩ E with x 6= y. Since E ⊂ E, we havey ∈ Nr(x) ∩ E, and therefore x ∈ (E)′. Conversely, suppose that x ∈ (E)′ andr > 0. Then there exists a point y ∈ Nr(x) ∩ E with x 6= y. If y /∈ E, theny ∈ E′ since E = E ∪ E′. Setting r′ := r − d(x, y), then there exists a z ∈ Ewith z 6= y and z 6= x (since d(x, z) > 0). In either case, Nr(x) contains a pointof E distinct from x, so x ∈ E′. Thus E′ = (E)′.

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The sets E and E′ need not have the same limit points. Consider the setE defined in the solution to Exercise 2.5. We showed that E′ = {0, 1, 2}. Ifx ∈ E′, then (x− 1, x+ 1) ∩ E′ = {x}, and if x /∈ E′, then for

δ := min {|x− y| : y ∈ E′} ,

we have (x− δ, x+ δ) ∩ E′ = ∅. Thus (E′)′ = ∅, and therefore E′ 6= (E′)′.

2.7. Let A1, A2, A3, . . . be subsets of a metric space.

(a) If Bn =⋃ni=1Ai, prove that Bn =

⋃ni=1Ai, for n = 1, 2, 3, . . ..

(b) If B =⋃∞i=1Ai, prove that B ⊃

⋃∞i=1Ai.

Show, by an example, that the inclusion can be proper.

Solution.

(a) Fix x ∈⋃ni=1Ai. Then x ∈ Ai = Ai ∪ A′i for some i. If x ∈ Ai, then

x ∈ Bn since Ai ⊂ Bn ⊂ Bn. If x ∈ A′i, then for every r > 0, there is apoint y ∈ Ai ⊂ Bn with x 6= y such that d(x, y) < r. Thus x ∈ B′n ⊂ Bn,so⋃ni=1Ai ⊂ Bn.

Fix x /∈⋃ni=1Ai. Then

x ∈

(n⋃i=1

Ai

)c=

n⋂i=1

(Ai)c.

Thus there exist r1, . . . , rn > 0 such that Nri(x) ⊂ (Ai)c for each i. Let

r := min {r1, . . . , rn}. Then Nr(x) ⊂⋂ni=1Nri

(x), so Nr(x) ⊂⋂ni=1(Ai)c.

In particular, Nr(x) contains no points of any Ai, and therefore no pointsof Bn. If Nr(x) contains a point y ∈ B′n, then letting r′ := r− d(x, y), wesee that Nr′(y) contains a point z ∈ Bn by the definition of limit points,and z ∈ Nr′(y) ⊂ Nr(x) by the triangle inequality, contradicting thatNr(x) ∩Bn = ∅. Thus

x ∈ Bcn ∩ (B′n)c = (Bn ∪B′n)c = (Bn)c,

so therefore Bn ⊂⋃ni=1Ai, and by the result proved above, Bn =

⋃ni=1Ai.

(b) Fix x ∈⋃∞i=1Ai. Then x ∈ Ai = Ai ∪ A′i for some i. If x ∈ Ai, then

x ∈ B since Ai ⊂ B ⊂ B. If x ∈ A′i, then for every r > 0, there is apoint y ∈ Ai ⊂ B with x 6= y such that d(x, y) < r. Thus x ∈ B′ ⊂ B, soB ⊃

⋃∞i=1Ai.

For all i ∈ N, let Ai := {1/i}. Then Ai = Ai since finite sets are closed bythe corollary to Theorem 2.20. However, B = {1/i : i ∈ N}, and therefore

B ={

1i

: i ∈ N}∪ {0} ⊃ {1/i : i ∈ N} =

∞⋃i=1

Ai,

and the inclusion is proper since 0 /∈⋃∞i=1Ai.

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2.8. Is every point of every open set E ⊂ R2 a limit point of E? Answer thesame question for closed sets in R2.

Solution. Yes, every point of every open set E ⊂ R2 a limit point of E. LetE ⊂ R2 be open and fix x ∈ E. Then x is an interior point of E, so thereexists an r > 0 such that Nr(x) ⊂ E. Let Ns(x) be an arbitrary neighborhoodof x. If s ≥ r, then Nr(x) ⊂ Ns(x), so Nr(x) ⊂ Ns(x) ∩ E. If s < r, thenNs(x) ⊂ Nr(x) ⊂ E, so Ns(x) = Ns(x)∩E. In either case, Ns(x)∩E is itself aneighborhood of x in R2 and therefore contains infinitely many points, and inparticular, some point y 6= x. Thus x is a limit point of E.

The same result does not hold for closed sets in R2. A set consisting of asingle point x ∈ R2 is closed since its complement is open, but finite sets haveno limit points by the corollary to Theorem 2.20.

2.9. Let E◦ denote the set of all interior points of a set E.

(a) Prove that E◦ is always open.

(b) Prove that E is open if and only if E = E◦.

(c) If G ⊂ E and G is open, prove that G ⊂ E◦.

(d) Prove that the complement of E◦ is the closure of the complement of E.

(e) Do E and E always have the same interiors?

(f) Do E and E◦ always have the same closures?

Solution.

(a) Fix x ∈ E◦. Then there exists an r > 0 such thatNr(x) ⊂ E. If d(x, y) < rand r′ = r − d(x, y), then d(y, z) < r′ implies

d(x, z) ≤ d(x, y) + d(y, z) < r,

so y ∈ E◦. This shows that Nr(x) ⊂ E◦, so x is an interior point of E◦.Thus E◦ is open.

(b) If E = E◦, then E is open by part (a). If E is open, then every point ofE is an interior point, so E ⊂ E◦. Since every interior point of a set mustbe contained in the set, the reverse inclusion holds as well, so E = E◦.

(c) Suppose G ⊂ E and G is open, and fix x ∈ G. Then x is an interior pointof G, so there exists an r > 0 such that Nr(x) ⊂ G ⊂ E. Thus x is aninterior point of E, so G ⊂ E◦.

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(d) Fix x ∈ (E◦)c. Then x is not an interior point of E, so every neighborhoodNr(x) intersects Ec. Thus x is in Ec or x is a limit point of Ec, so

x ∈ Ec ∪ (E′)c = E.

Now suppose x /∈ (E◦)c. Then x ∈ E◦, so there exists an r > 0 such thatNr(x) ⊂ E. Thus x ∈ E and x is not a limit point of Ec, so

x ∈ E ∩ (E′)c = (Ec)c ∩ (E′)c = (E ∪ E′)c = (E)c.

(e) No, E and E do not always have the same interiors. For example, notethat Q◦ = ∅ because every neighborhood of a rational number containsirrational numbers. However, Q = R, so (Q)◦ = R since R is open.

(f) No, E and E◦ do not always have the same closures. For example, notethat Q◦ = ∅ because every neighborhood of a rational number containsirrational numbers. Therefore Q◦ = ∅ = ∅ since the empty set is closed,but Q = R.

2.10. Let X be an infinite set. For p ∈ X and q ∈ X, define

d(p, q) =

{1 (if p 6= q)0 (if p = q).

Prove that this is a metric. Which subsets of the resulting metric space areopen? Which are closed? Which are compact?

Solution. Parts (a) and (b) of Definition 2.15 follow immediately from the def-inition of d. If p = q, then part (c) is trivial. If p 6= q, then for every r ∈ X,either p 6= r or q 6= r. Thus d(p, r) + d(q, r) ≥ 1, and since d(p, q) ≤ 1, part (c)holds in this case as well. Thus d is a metric.

Fix E ⊂ X and x ∈ E. Then

N1/2(x) = {y ∈ X : d(x, y) < 1/2} = {x} ,

so every point is a neighborhood of itself. By Theorem 2.19, every point is open,and since

E =⋃x∈E{x} ,

every set is open by Theorem 2.24(a).If E ⊂ X, then Ec is open by what we just proved, so therefore every set is

closed by the corollary to Theorem 2.23.Suppose K ⊂ X is compact. Then the open cover {{x} : x ∈ K} has a finite

subcover, so K must be finite. Conversely, if K is finite, say K = {x1, . . . , xn},then given any open cover {Gα}α∈A of K, there are indices αi such that xi ∈ Gαi

for all i = 1, . . . , n. Thus {Gα1 , . . . , Gαn} is a finite subcover of K, so K is

compact. Hence the compact subsets of X are the finite subsets.

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2.11. For x ∈ R1 and y ∈ R1, define

d1(x, y) = (x− y)2,

d2(x, y) =√|x− y|,

d3(x, y) =∣∣x2 − y2

∣∣ ,d4(x, y) = |x− 2y| ,

d5(x, y) =|x− y|

1 + |x− y|.

Determine, for each of these, whether it is a metric or not.

Solution.

(a) Since

d1(2, 0) = (2− 0)2 = 4,

d1(2, 1) = (2− 1)2 = 1,

d1(1, 0) = (1− 0)2 = 1,

it follows that

d1(2, 0) = 4 > 2 = d1(2, 1) + d1(1, 0),

so d1(x, y) is not a metric by Definition 2.15(c).

(b) Since |x− y| > 0 if x 6= y, we have d2(x, y) =√|x− y| > 0 if x 6= y. But

if x = y, then d2(x, y) = 0. Thus Definition 2.15(a) holds.

Since |x− y| = |y − x|, we have

d2(x, y) =√|x− y| =

√|y − x| = d2(y, x),

so Definition 2.15(b) holds.

By the triangle inequality, we have

|x− z| ≤ |x− y|+ |y − z|

≤ |x− y|+ 2√|x− y|

√|y − z|+ |y − z|

=(√|x− y|+

√|y − z|

)2

,

and taking square roots gives d2(x, z) ≤ d2(x, y) + d2(y, z), so Defini-tion 2.15(c) holds.

Hence d2(x, y) is a metric.

(c) Sinced3(1,−1) =

∣∣12 − (−1)2∣∣ = |1− 1| = 0,

Definition 2.15(a) fails, so d3(x, y) is not a metric.

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(d) Sinced4(2, 1) = |2− 2(1)| = |2− 2| = 0,

Definition 2.15(a) fails, so d4(x, y) is not a metric.

(e) Since |x− y| defines a metric, we have

d5(x, y) =|x− y|

1 + |x− y|≥ 0,

with equality holding if and only if x = y. Thus Definition 2.15(a) holds.

Since |x− y| = |y − x|, we have

d5(x, y) =|x− y|

1 + |x− y|=

|y − x|1 + |y − x|

= d5(y, x),

so Definition 2.15(b) holds.

Definition 2.15(c) is satisfied if and only if

d5(x, y) +d5(y, z)−d5(x, z) =|x− y|

1 + |x− y|+|y − z|

1 + |y − z|− |x− z|

1 + |x− z|≥ 0.

Let a := |x− y|, b := |y − z|, and z := |x− z|. Then taking

(1 + a)(1 + b)(1 + c)

as the common denominator in the fraction above, the numerator is

a+ ab+ ac+ abc+ b+ ba+ bc+ abc− c− ac− cb− abc= a+ b− c+ 2ab+ abc ≥ (2 + c) ab ≥ 0.

Hence d5(x, y) is a metric.

2.12. Let K ⊂ R1 consist of 0 and the numbers 1/n, for n = 1, 2, 3, . . .. Provethat K is compact directly from the definition (without using the Heine-Boreltheorem).

Solution. Let {Gα}α∈A be an open cover of K. Then there is an index α0 ∈ Asuch that 0 ∈ Gα0 . Since Gα0 is an open set containing 0, it contains anopen interval (−ε, ε) for some ε > 0. Let N be the largest integer such thatN ≤ 1/ε. Note that for all n > N , we have 0 < 1/n < ε, so 1/n ∈ Gα0 , andfor all positive n ≤ N , there exist indices αn ∈ A such that 1/n ∈ Gαn . Then{Gα0 , Gα1 , . . . , GαN

} is a finite subcover of K, so K is compact.

2.13. Construct a compact set of real numbers whose limit points form a count-able set.

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Solution. Define a set

E := {1/m+ 1/n : m,n ∈ N} ∪ {1/n : n ∈ N} ∪ {0} .

Then E′ = {1/n : n ∈ N} ∪ {0}, so E′ is countable and E′ ⊂ E, so E is closed.Moreover, E ⊂ [0, 2], so E is bounded. By the Heine-Borel theorem, it followsthat E is compact.

2.14. Give an example of an open cover of the segment (0, 1) which has nofinite subcover.

Solution. For every n ∈ N, define Gn := (1/n, 1). Then each Gn is an openinterval in R, and

⋃∞n=1Gn = (0, 1), so {Gn}∞1 is an open cover of (0, 1).

Suppose {Gn}∞1 has a finite subcover C of (0, 1). Then there is a largest integerN such that GN ∈ C. But then

⋃Gn∈C Gn = (1/N, 1), so every ε satisfying

0 < ε < 1/N is an element of (0, 1) but is not covered by C. This contradictsthat C is a cover, so therefore {Gn}∞1 has no finite subcover of (0, 1).

2.15. Show that Theorem 2.36 and its Corollary become false (in R1, for ex-ample) if the word “compact” is replaced by “closed” or by “bounded.”

Solution. For every n ∈ N, define En := [n,∞) ⊂ R. Then En is closed, andfor any finite subcollection C, let N be the largest integer such that EN ∈ C.Then

⋂En∈C En = [N,∞) 6= ∅, but

⋂∞n=1En = ∅. Thus Theorem 2.36 and its

corollary are false if “compact” is replaced by “closed.”For every n ∈ N, define En := (0, 1/n) ⊂ R. Then En is bounded, and

for any finite subcollection C, let N be the largest integer such that EN ∈ C.Then

⋂En∈C En = (0, 1/N) 6= ∅, but

⋂∞n=1En = ∅. Thus Theorem 2.36 and its

corollary are false if “compact” is replaced by “bounded.”

2.16. Regard Q, the set of all rational numbers, as a metric space, withd(p, q) = |p− q|. Let E be the set of all p ∈ Q such that 2 < p2 < 3. Show thatE is closed and bounded in Q, but that E is not compact. Is E open in Q?

Solution. Fix x ∈ Ec. Since√

2 and√

3 are irrational, either x2 > 3 or x2 < 2.Assume without loss of generality that x2 > 3 (the proof of the other case issimilar). Let r := x−

√3 and note that

Nr(x) ={p ∈ Q : 3 < p2 < (x+ r)2

}⊂ Ec,

proving that x is an interior point of Ec. Thus Ec is open, and by the corollaryto Theorem 2.23, E is closed.

9

Since E ⊂ [0, 2], it follows that E is bounded.For every integer n ≥ 10, define Gn := (

√2 + 1/n,

√3) ∩ Q. If p ∈ E, then

2 < p2 < 3, so there is an integer N such that (√

2 + 1/N)2 < p. Thus p ∈ GN ,and therefore {Gn}∞10 is an open cover of E. If C is a finite subcollection of{Gn}∞10, then there is a largest integer k such that Gk ∈ C. Choose p ∈ Q suchthat

√2 < p <

√2 + 1/k. Then p /∈

⋃Gn∈C Gn = (

√2 + 1/k,

√3) ∩ Q, so C is

not a cover of E. Thus {Gn}∞10 contains no finite subcover of E, and thereforeE is not compact.

Yes, E is open in Q. Since E = (√

2,√

3)∩Q, E is open by Theorem 2.30.

2.19.

(a) If A and B are disjoint closed sets in some metric space X, prove thatthey are separated.

(b) Prove the same for disjoint open sets.

(c) Fix p ∈ X, δ > 0, define A to be the set of all q ∈ X for which d(p, q) < δ,define B similarly, with > in place of <. Prove that A and B are separated.

(d) Prove that every connected metric space with at least two points is un-countable. Hint : Use (c).

Solution.

(a) If A and B are closed, then A = A and B = B, so

A ∩B = A ∩B = ∅ and A ∩B = A ∩B = ∅.

(b) If A and B are open, then Ac and Bc are closed. Since A ⊂ Bc, we haveA ⊂ Bc by Theorem 2.27(c), and similarly, B ⊂ Ac. Thus

A ∩B = ∅ = A ∩B.

(c) The set A is open since A = Nδ(p). Fix x ∈ B and let r := δ − d(p, x).Then r > 0 and Nr(x) ⊂ B by the triangle inequality, so B is open.Moreover, A ∩B = ∅ since d(p, q) cannot be both greater than δ and lessthan δ. By part (b), A and B are separated.

(d) Suppose X is a metric space which is at most countable. Fix x ∈ X. Thenthere exists an r > 0 such that {y ∈ X : d(x, y) = r} = ∅, for otherwise,we would have an injective function from (0,∞) into X. The sets

A := {y ∈ X : d(x, y) < r} and B := {y ∈ X : d(x, y) > r}

separate X by part (c).

10

2.20. Are closures and interiors of connected sets always connected? (Look atsubsets of R2.)

Solution. The closure of a connected set is always connected. Suppose E isconnected and E ⊂ A∪B, where A∩B = ∅ = A∩B. Then E ∩B = ∅ withoutloss of generality, for otherwise, A and B separate E. Thus E ⊂ A, and byTheorem 2.27(c), E ⊂ A. Since A ∩ B = ∅, it follows that E ∩ B = ∅. Thus Eis not a union of two nonempty separated sets.

The interior of a connected set need not be connected. Define

E :={x ∈ R2 : d(x,±1) ≤ 1

}.

Then E is connected, but

E◦ ={x ∈ R2 : d(x,−1) < 1

}∪{x ∈ R2 : d(x, 1) < 1

},

which is a union of two nonempty separated sets by Exercise 2.19(b).

2.21. Let A and B be separated subsets of some Rk, suppose a ∈ A, b ∈ B,and define

p(t) = (1− t)a + tb

for t ∈ R1. Put A0 = p−1(A), B0 = p−1(B). [Thus t ∈ A0 if and only ifp(t) ∈ A.]

(a) Prove that A0 and B0 are separated subsets of R1.

(b) Prove that there exists t0 ∈ (0, 1) such that p(t0) /∈ A ∪B.

(c) Prove that every convex subset of Rk is connected.

Solution.

(a) Fix t ∈ A0. Then p(t) ∈ A, and since A ∩ B = ∅, p(t) /∈ B. Thus thereis an r > 0 such that Nr(p(t)) ⊂ B

c. In particular, if s ∈ Nr(t), then

s /∈ B0. Assume that there exists an s ∈ Nr(t) such that s ∈ B0. Ifr′ := r − |s− t|, then there exists a point s′ ∈ Nr′(s) ∩B0. But

|t− s′| ≤ |t− s|+ r′ < r

by the triangle inequality, so therefore s′ ∈ Nr(t), contradicting thatNr(t) ⊂ Bc0. We have thus shown that Nr(t) ⊂ (B0)c. The same ar-gument applied to a point t′ ∈ B0 shows that there is a neighborhood oft′ which is contained in (A0)c. Thus A ∩B = ∅ = A ∩B.

11

(b) Assume the contrary, that p(t) ∈ A ∪ B for all t ∈ (0, 1). Then we have[0, 1] ⊂ A0 ∪B0, and by part (a),(

A0 ∩ [0, 1])∩(B0 ∩ [0, 1]

)= (A0 ∩B0) ∩ [0, 1] = ∅ ∩ [0, 1] = ∅,(

A0 ∩ [0, 1])∩(B0 ∩ [0, 1]

)= (A0 ∩B0) ∩ [0, 1] = ∅ ∩ [0, 1] = ∅.

The sets A0 ∩ [0, 1] and B0 ∩ [0, 1] are nonempty since 0 ∈ A0 ∩ [0, 1]and 1 ∈ B0 ∩ [0, 1], and they are disjoint since A0 and B0 are disjointby part (a). Thus A0 ∩ [0, 1] and B0 ∩ [0, 1] separate [0, 1], contradictingTheorem 2.47.

(c) Assume that there exists a convex set E ⊂ Rk such that E = A ∪ B fortwo nonempty separated sets A and B. Fix a ∈ A and b ∈ B. By thedefinition of convexity,

p(t) := (1− t)a + tb ∈ E = A ∪B

for every t ∈ (0, 1). This contradicts part (b). Hence every convex set inRk is connected.

12

Rudin, Chapter 3

Dominique Abdi

3.1. Prove that the convergence of {sn} implies convergence of {|sn|}. Is theconverse true?

Solution. We first prove the reverse triangle inequality from Exercise 1.13. Ifa, b ∈ C (or Rk), then

|a| = |a− b+ b| ≤ |a− b|+ |b| ,

so |a| − |b| ≤ |a− b|. Interchanging a and b, we see that |b| − |a| ≤ |a− b|, andtherefore ∣∣ |a| − |b| ∣∣ ≤ |a− b| .

Now suppose sn → s. Fix ε > 0 and N such that |sn − s| < ε for all n ≥ N .The reverse triangle inequality then yields∣∣ |sn| − |s|

∣∣ ≤ |sn − s| < ε

for all n ≥ N , so |sn| → |s|.The converse is not true. Consider the sequence sn := (−1)n. This sequence

diverges, but since |sn| =∣∣(−1)n

∣∣ = 1 for all n ∈ N, we have |sn| → 1.

3.2. Calculate limn→∞

(√n2 + n− n

).

Solution. By elementary algebra, we have(√n2 + n− n

)=

n√n2 + n+ n

=1√

1 + 1/n+ 1.

Since 1/n → 0, Theorem 3.3 and the continuity of the square root function onthe positive real numbers yields

limn→∞

(√n2 + n− n

)= lim

n→∞

1√1 + 1/n+ 1

=12.

3.3. If s1 =√

2, and

sn+1 =√

2 +√sn (n = 1, 2, 3, . . .),

prove that {sn} converges, and that sn < 2 for n = 1, 2, 3, . . ..

Solution. First, note that

s1 =√

2 <√

2 +√

2 = s2 and s2 =√

2 +√

2 < 2.

Suppose s1 < s2 < · · · < sn < 2. Then

sn =√

2 +√sn−1 <

√2 +√sn = sn+1.

Moreover, sn < 2 implies that√sn < 2, and therefore 2 +

√sn < 4. Taking

square roots, we have

sn+1 =√

2 +√sn < 2,

and therefore s1 < · · · < sn < sn+1 < 2. By induction, {sn} is a monotoni-cally increasing sequence which is bounded above by 2, so {sn} converges byTheorem 3.14.

3.4. Find the upper and lower limits of the sequence {sn} defined by

s1 = 0; s2m =s2m−1

2; s2m+1 =

12

+ s2m.

Solution. We can prove by induction that

s2m =12− 1

2mand s2m+1 = 1− 1

2m−1

for all m ∈ N. Since the subsequence of even terms converges to 1/2 and thesubsequence of odd terms converges to 1, both 1/2 and 1 are subsequentiallimits of {sn}. If x /∈ {1/2, 1}, then let r := 1

2 min {|x− 1| , |x− 1/2|} and notethat sn ∈ Nr(x) for at most finitely many n ∈ N. Thus 1/2 and 1 are the onlysubsequential limits of {sn}, so therefore

lim supn→∞

sn = 1 and lim infn→∞

sn =12.

3.5. For any two real sequences {an}, {bn}, prove that

lim supn→∞

(an + bn) ≤ lim supn→∞

an + lim supn→∞

bn,

provided the sum on the right is not of the form ∞−∞.

2

Solution. Assume that the sum on the right is not of the form ∞−∞.Suppose one of the terms on the right is −∞, say lim sup an = −∞. Then

an → −∞, and for any M > 0, there are only finitely many n ∈ N such thatbn > M . Thus an + bn → −∞, so lim sup(an + bn) = −∞.

Now suppose one of the terms on the right is ∞, say lim sup an =∞. Thenfor any M > 0, there are infinitely many n ∈ N such that an > M , but onlyfinitely many n ∈ N such that bn < −M . Thus there are infinitely many n ∈ Nsuch that an + bn > M , so lim sup(an + bn) =∞.

Finally, suppose α := lim sup an and β := lim sup bn are both finite and fixε > 0. By Theorem 3.17(b), there exist N1 and N2 such that n ≥ N1 impliesan < α+ ε/2, and n ≥ N2 implies bn < β + ε/2. Let N := max {N1, N2}. Thenn ≥ N implies

an + bn < α+ β + ε.

But ε > 0 is arbitrary, so therefore

an + bn ≤ α+ β = lim supn→∞

an + lim supn→∞

bn

for all but finitely many n. Thus every subsequential limit of {an + bn} isbounded above by lim sup an + lim sup bn, and in particular,

lim supn→∞


an + lim supn→∞

bn

by Theorem 3.17(a).

3.8. If∑an converges, and if {bn} is monotonic and bounded, prove that∑

anbn converges.

Solution. By Theorem 3.14, {bn} converges to some number b. Define

cn :=

{b− bn if {bn} is monotonically increasing,bn − b if {bn} is monotonically decreasing.

Then {cn} is monotonically decreasing and lim cn = 0. Moreover, for all N ∈ N,

N∑n=1

anbn =N∑

n=1

ancn + b

N∑n=1

an

if {bn} is decreasing, for example, and similarly if {bn} is increasing. By Theo-rem 3.42,

∑ancn converges, and therefore

∑anbn converges.

3.20. Suppose {pn} is a Cauchy sequence in a metric space X, and some sub-sequence {pni

} converges to a point p ∈ X. Prove that the full sequence {pn}converges to p.

3

Solution. Fix ε > 0. Choose N1 such that ni ≥ N1 implies d(pni , p) < ε/2,and choose N2 such that m,n ≥ N2 implies d(pm, pn) < ε/2. Now defineN := max {N1, N2}. Then n ≥ N implies

d(pn, p) ≤ d(pn, pni) + d(pni , p) < ε

for some ni ≥ N . Thus lim pn = p.

3.21. Prove the following analogue of Theorem 3.10(b): If {En} is a sequence ofclosed nonempty and bounded sets in a complete metric space X, if En ⊃ En+1,and if

limn→∞

diamEn = 0,

then⋂∞

1 En consists of exactly one point.

Solution. For every n ∈ N, choose a point xn ∈ En. Fix ε > 0 and choose Nsuch that diamEN < ε. Then m,n ≥ N implies xm, xn ∈ EN since En ⊃ En+1

for all n, so therefored(xm, xn) ≤ diamEN < ε.

Thus {xn} is a Cauchy sequence, and since X is complete, {xn} converges tosome point x ∈ X. For every N ∈ N, {xn}∞N ⊂ EN and EN is closed, sotherefore x ∈ EN . Hence x ∈

⋂∞1 En.

If x, y ∈⋂∞

1 En and x 6= y, then d(x, y) > 0. Then diamEn ≥ d(x, y) > 0for all n, contradicting that lim diamEn = 0.

4


Rudin, Chapter 4

Dominique Abdi

4.1. Suppose f is a real function defined on R1 which satisfies

limh→0

[f(x+ h)− f(x− h)] = 0

for every x ∈ R1. Does this imply that f is continuous?

Solution. No. Define f : R→ R by

f(x) :=

{0 if x 6= 0,1 if x = 0.

.

Thenlimh→0

[f(h)− f(−h)] = 0,

but f is not continuous at x = 0.

4.2. If f is a continuous mapping of a metric space X into a metric space Y ,prove that

f(E) ⊂ f(E)

for every set E ⊂ X. Show, by an example, that f(E) can be a proper subsetof f(E).

Solution. Fix x ∈ E and a sequence {xn}∞1 ⊂ E such that limxn = x. Then{f(xn)}∞1 is a sequence in f(E), and since f is continuous, lim f(xn) = f(x).If f(x) /∈ f(E) ∪ f(E)′ = f(E), then there is a neighborhood Nr(f(x)) whichdoes not intersect f(E). By Theorem 4.8, f−1(Nr(f(x))) is open, so thereis a neighborhood Ns(x) which does not intersect E. This contradicts thatlimxn = x, so f(x) ∈ f(E) and therefore f(E) ⊂ f(E).

Define f : N→ R by f(n) = 1/n. Since N = N, we have

f(N) ={

1n

: n ∈ N}⊂{

1n

: n ∈ N}∪ {0} = f(N).

4.3. Let f be a continuous real function on a metric space X. Let Z(f) (thezero set of f) be the set of all p ∈ X at which f(p) = 0. Prove that Z(f) isclosed.

Solution. Let x be a limit point of Z(f) and let {xn}∞1 ⊂ Z(f) be a sequencesuch that limxn = x. Then

f(x) = limn→∞

f(xn) = limn→∞

0 = 0,

so x ∈ Z(f). Thus Z(f) contains its limit points, so Z(f) is closed.

4.4. Let f and g be continuous mappings of a metric space X into a metricspace Y , and let E be a dense subset of X. Prove that f(E) is dense in f(X).If g(p) = f(p) for all p ∈ E, prove that g(p) = f(p) for all p ∈ X. (In otherwords, a continuous mapping is determined by its values on a dense subset ofits domain.)

Solution. Fix y ∈ f(X) and choose x ∈ X such that f(x) = y. Since E isdense in X, there is a sequence {xn}∞1 ⊂ X such that limxn = x. Consider thesequence {f(xn)}∞1 ⊂ f(E). Since f is continuous, lim f(xn) = f(x) = y, soy ∈ f(E) ∪ f(E)′ = f(E). Thus f(X) ⊂ f(E), so f(E) is dense in f(X).

Fix p ∈ X and choose a sequence {pn} ⊂ E such that lim pn = p. Since fand g are continuous and f(pn) = g(pn), we have

f(p) = limn→∞

f(pn) = limn→∞

g(pn) = g(p),

proving that f(p) = g(p) for all p ∈ X.

4.5. If f is a real continuous function defined on a closed set E ⊂ R1, provethat there exist continuous real functions g on R1 such that g(x) = f(x) forall x ∈ E. (Such functions g are called continuous extensions of f from E toR1.) Show that the result becomes false if the word “closed” is omitted. Extendthe result to vector-valued functions. Hint : Let the graph of g be a straightline on each of the segments which constitute the complement of E (compareExercise 29, Chap. 2). The result remains true if R1 is replaced by any metricspace, but the proof is not so simple.

Solution. Since E is closed, Ec is the union of an at most countable collectionof disjoint open intervals C = {In} by Exercise 2.29. If (−∞, c) is one of theseintervals for some c ∈ R, then define g(x) := f(c) for all x ∈ (−∞, c), andsimilarly if (c,∞) is one of the intervals. On the remaining intervals In =(an, bn), define

g(x) := f(an) +(f(bn)− f(an)

bn − an

)(x− an) .

2

Finally, define g(x) := f(x) for all x ∈ E. Then clearly g extends f to R and gis continuous on E◦ and on Ec. It remains to show that g is continuous at theendpoints of the intervals In ∈ C.

Let x be the endpoint of some interval In ∈ C. If {xj} is a sequence inE converging to x, then lim g(xj) = g(x) = f(x). If {xj} is a sequence in Ec

converging to x = an, then

limj→∞

g(xj) = limj→∞

[f(an) +

(f(bn)− f(an)

bn − an

)(xj − an)

]= f(an),

and similarly if x = bn. Thus g is continuous on R.If the word “closed” is omitted, let E := (0, 1), and define f(x) := 1/x for all

x ∈ E. If f has a continuous extension g to R, then g is continuous on [−1, 1],and therefore bounded by the intermediate value theorem. However, f is notbounded in any neighborhood of x = 0, so therefore g is not bounded either, acontradiction.

If f : E → Rk is continuous and E ⊂ R is closed, then f has a continuousextension g : R→ Rk. Each component function fj : R→ R of f = (f1, . . . , fk)extends to a continuous function gj : R→ R by the proof above, and it followsthat g = (g1, . . . , gk) is a continuous extension of f to R.

4.6. If f is defined on E, the graph of f is the set of points (x, f(x)), for x ∈ E.In particular, if E is a set of real numbers, and f is real-valued, the graph of fis a subset of the plane.

Suppose E is compact, and prove that f is continuous on E if and only ifits graph is compact.

Solution. Suppose f is continuous and let {Uα × Vβ} be an open cover of thegraph of f . Then {Uα} is an open cover of E, so there exist α1, . . . , αm suchthat {Uαi

}m1 is a finite subcover of E. Similarly, {Vβ} is an open cover of f(E),and since f(E) is compact by Theorem 4.14, there exist β1, . . . , βn such that{Vβj}n1 is a finite subcover of f(E). Thus{

Uαi× Vβj

: 1 ≤ i ≤ m, 1 ≤ j ≤ n}

is a finite subcover of {Uα × Vβ}, proving that the graph of f is compact.Now suppose f is not continuous. Then there is a sequence {xn}∞1 ⊂ E such

that limxn = x ∈ E, but f(xn) does not converge to f(x). Thus there is anε > 0 such that d(f(xn), f(x)) ≥ ε for infinitely many n, and by restricting to asubsequence, we can assume without loss of generality that d(f(xn), f(x)) ≥ εfor all n. If the graph of f is compact, then the sequence {(xn, f(xn))}∞1 containsa convergent subsequence by Theorem 3.6(a). That is, (xnk

, f(xnk))→ (x, f(x))

for some x ∈ E, contradicting that d(f(xn), f(x)) ≥ ε for all n. The graph of fis therefore not compact.

3

4.7. If E ⊂ X and if f is a defined on X, the restriction of f to E is thefunction g whose domain of definition is E, such that g(p) = f(p) for p ∈ E.Define f and g on R2 by: f(0, 0) = g(0, 0) = 0, f(x, y) = xy2/(x2 + y4),g(x, y) = xy2/(x2 + y6) if (x, y) 6= (0, 0). Prove that f is bounded on R2, thatg is unbounded in every neighborhood of (0, 0), and that f is not continuous at(0, 0); nevertheless, the restrictions of both f and g to every straight line in R2

are continuous!

Solution. First, note that

0 ≤ (x− y2)2 = x2 − 2xy2 + y4

implies x2 + y4 ≥ 2xy2. Thus

|f(x, y)| =∣∣∣∣ xy2

x2 + y4

∣∣∣∣ ≤ ∣∣∣∣ xy2

2xy2

∣∣∣∣ =12,

so f is bounded on R2.Along the curve y = x1/3, we have

limx→0

∣∣∣g(x, x1/3)∣∣∣ = lim

x→0

∣∣∣∣x5/3

2x2

∣∣∣∣ = limx→0

∣∣∣∣ 12x1/3

∣∣∣∣ =∞,

proving that g is unbounded in every neighborhood of (0, 0).Along the curve y = x1/2, we have

limx→0

f(x, x1/2) = limx→0

x2

2x2=

126= 0 = f(0, 0),

so f is not continuous at (0, 0).It is clear that f and g are continuous away from (0, 0) and along the axes,

so it suffices to consider lines of the form y = cx for some fixed constant c ∈ R.Since

limx→0

f(x, cx) = limx→0

cx3

x2 + c4x4= limx→0

cx

1 + c4x2= 0,

limx→0

g(x, cx) = limx→0

cx3

x2 + c6x6= limx→0

cx

1 + c6x4= 0,

the restrictions of both f and g to every straight line in R2 are continuous.

4.8. Let f be a real uniformly continuous function on the bounded set E inR1. Prove that f is bounded on E.

Show that the conclusion is false if boundedness of E is omitted from thehypothesis.

4

Solution. Fix ε > 0 and choose δ > 0 such that |f(x), f(y)| < ε whenever|x− y| < δ. The collection {Nδ(x) : x ∈ E} is an open cover of E. In fact, thisis also an open cover of E, for otherwise, there is a limit point p of E withthe property that p /∈ Nδ(x) for all x ∈ E, a contradiction. Since E is closedand bounded, it is compact by the Heine-Borel theorem, so there exist pointsx1, . . . , xn ∈ E such that {Nδ(xj)}n1 is a finite subcover of E. But then for everyx ∈ E, there is a j such that |x− xj | < δ, and therefore |f(x)− f(xj)| < ε.That is, {Nε(f(xj))}n1 is a finite cover of f(E), so diam f(E) ≤ nε. Thus f isbounded on E.

If E = R and f(x) := x on E, then f is uniformly continuous since forevery ε > 0, |x− y| < ε implies that |f(x)− f(y)| < ε. However, f is notbounded.

4.9. Show that the requirement in the definition of uniform continuity can berephrased as follows, in terms of diameters of sets: To every ε > 0 there existsa δ > 0 such that diam f(E) < ε for all E ⊂ X with diamE < δ.

Solution. Suppose f is uniformly continuous, fix ε > 0, and choose δ > 0 suchthat d(x, y) < δ implies d(f(x), f(y)) < ε. Given a set E with diamE < δ, wehave d(x, y) < δ for all x, y ∈ E. Thus d(f(x), f(y)) < ε for all x, y ∈ E, sotherefore diam f(E) < ε.

Conversely, fix ε > 0 and choose δ > 0 such that diam f(E) < ε for allE ⊂ X with diamE < δ. If d(x, y) < δ, let E := {x, y}. Then diamE < δ andf(E) = {f(x), f(y)}. By assumption, diam f(E) < ε, that is, d(f(x), f(y)) < ε,proving that f is uniformly continuous.

4.10. Complete the details of the following alternate proof of Theorem 4.19: Iff is not uniformly continuous, then for some ε > 0 there are sequences {pn}, {qn}in X such that dX(pn, qn)→ 0 but dY (f(pn), f(qn)) > ε. Use Theorem 2.37 toobtain a contradiction.

Solution. Suppose f is not uniformly continuous. Then there exists an ε > 0such that for any n ∈ N, there are points pn, qn ∈ X such that dX(pn, qn) < 1/n,but dY (f(pn), f(qn)) > ε. Thus {pn} and {qn} are sequences in X such thatdX(pn, qn)→ 0, but dY (f(pn), f(qn)) > ε.

By Theorem 4.14, f(X) is compact. If the sequences {pn} and {qn} are bothfinite, then pn = qn for all n sufficiently large, contradicting that

dY (f(pn), f(qn)) > ε

for all n. Therefore the set E consisting of the sequences {pn} and {qn} isinfinite, so it has a limit point x by Theorem 2.37. It follows that there are

5

increasing sequences of positive integers {nk} and {n′k} such that lim pnk= x

and lim qn′k

= x. Then

limk→∞

f(pnk) = f(x) = lim

k→∞f(qn′

k)

since f is continuous, and dX(pnk, qn′

k)→ 0. Choose k sufficiently large so that

dY (f(pnk), f(x)) <

ε

2and dY (f(qn′

k), f(x)) <

ε

2.

Then

dY (f(pnk), f(qn′

k)) ≤ dY (f(pnk

), f(x)) + dY (f(qn′k), f(x)) < ε,

contradicting thatdY (f(pnk

), f(qn′k)) > ε

for all n.

4.11. Suppose f is a uniformly continuous mapping of a metric space X intoa metric space Y and prove that {f(xn)} is a Cauchy sequence in Y for everyCauchy sequence {xn} is in X. Use this result to give an alternative proof ofthe theorem stated in Exercise 13.

Solution. Fix ε > 0 and choose δ > 0 such that d(f(x), f(y)) < ε wheneverd(x, y) < δ. Now choose N such that d(xm, xn) < δ for all m,n ≥ N . Thend(f(xm), f(xn)) < ε for all m,n ≥ N , so {f(xn)} is a Cauchy sequence.

Let E be a dense subset of a metric space X, and let f be a uniformlycontinuous real function defined on E. For every x ∈ X, there is a sequence{xn}∞1 ⊂ E such that limxn = x. Since every convergent sequence is Cauchy,the result above shows that {f(xn)}∞1 is a Cauchy sequence in R, and thusconverges. If limxn = x and lim yn = x, then

limn→∞

f(xn) = f(x) = limn→∞

f(yn)

by the continuity of f , so we can extend f to X by defining f(x) = lim f(xn)for any sequence {xn}∞1 ⊂ E converging to x. Moreover, by Theorem 4.2, theextension of f is continuous on X.

4.12. A uniformly continuous function of a uniformly continuous function isuniformly continuous.

State this more precisely and prove it.

6

Solution. Let X,Y, Z be metric spaces and suppose f : X → Y and g : Y → Zare uniformly continuous. Then g ◦ f is uniformly continuous.

Proof. Fix ε > 0 and choose η > 0 such that dY (y1, y2) < η implies

dZ(g(y1), g(y2)) < ε.

Now choose δ > 0 such that dX(x1, x2) < δ implies dY (f(x1), f(x2)) < η. ThendX(x1, x2) < δ implies dY (f(x1), f(x2)) < η, which in turn implies

dZ(g(f(x1), g(f(x2))) < ε.

Thus g ◦ f is uniformly continuous.

4.14. Let I = [0, 1] be the closed unit interval. Suppose f is a continuousmapping of I into I. Prove that f(x) = x for at least one x ∈ I.

Solution. The function g : I → I defined by g(x) := f(x) − x is continuous byTheorem 4.4. It suffices to show that g(x) = 0 for some x ∈ I. If g(0) = 0 org(1) = 1, then there is nothing to prove. Otherwise, g(0) > 0 since 0 < f(0) ≤ 1,and g(1) < 0 since 0 ≤ f(1) < 1. By the intermediate value theorem, there is apoint x ∈ I such that g(x) = 0.

7

Rudin, Chapter 5

Dominique Abdi

5.1. Let f be defined for all real x, and suppose that

|f(x)− f(y)| ≤ (x− y)2

for all real x and y. Prove that f is constant.

Solution. If x 6= y, dividing by |x− y| and taking the limit as y → x gives

limy→x

∣∣∣∣f(x)− f(y)

x− y

∣∣∣∣ ≤ limy→x|x− y| = 0.

Thus f ′(x) exists and f ′(x) = 0 for every x ∈ R. By Theorem 5.11(b), f isconstant.

5.2. Suppose f ′(x) > 0 in (a, b). Prove that f is strictly increasing in (a, b),and let g be its inverse function. Prove that g is differentiable, and that

g′(f(x)) =1

f ′(x)(a < x < b).

Solution. Fix x, y ∈ (a, b) with x < y. By the mean value theorem, there is anx0 ∈ (a, b) with

f(y)− f(x) = f ′(x0)(y − x) > 0.

Thus f is strictly increasing.Fix p = f(x) in the range of f , let {pn} be a sequence in the range of f

converging to p, and choose xn ∈ (a, b) such that f(xn) = pn. Then

limn→∞

g(pn)− g(p)

pn − p= lim

n→∞

g(f(xn))− g(f(x))

f(xn)− f(x)= lim

n→∞

xn − xf(xn)− f(x)

=1

f ′(x).

5.3. Suppose g is a real function on R1, with bounded derivative (say |g′| ≤M).Fix ε > 0, and define f(x) = x+ εg(x). Prove that f is one-to-one if ε is smallenough. (A set of admissible values of ε can be determined which depends onlyon M .)

Solution. First, note that f is differentiable by Theorem 5.3 because g is differ-entiable. Moreover,

f ′(x) = 1 + εg′(x).

If ε < 1/M , then

|εg′(x)| < |g′(x)|M

≤ 1,

so f ′(x) > 0. By the mean value theorem, if x < y, then there is an x0 ∈ (x, y)such that

f(y)− f(x) = f ′(x0)(y − x) > 0,

so f is strictly increasing, and therefore one-to-one.

5.4. If

C0 +C1

2+ · · ·+ Cn−1

n+

Cn

n+ 1= 0,

where C0, . . . , Cn are real constants, prove that the equation

C0 + C1x+ · · ·+ Cn−1xn−1 + Cnx

n = 0

has at least one real root between 0 and 1.

Solution. Define

f(x) := C0x+C1

2x2 + · · ·+ Cn

n+ 1xn+1.

Then f is a polynomial, and is therefore differentiable, and

f ′(x) = C0 + C1x+ · · ·+ Cn−1xn−1 + Cnx

n.

Since f(0) = 0 and

f(1) = C0 +C1

2+ · · ·+ Cn−1

n+

Cn

n+ 1= 0

by assumption, by the mean value theorem, there is an x ∈ (0, 1) such that

0 = f(1)− f(0) = f ′(x).

5.5. Suppose f is defined and differentiable for every x > 0, and f ′(x)→ 0 asx→ +∞. Put g(x) = f(x+ 1)− f(x). Prove that g(x)→ 0 as x→ +∞.

Solution. Fix ε > 0 and choose N such that x > N implies |f ′(x)| < ε. Thenfor any x ≥ N , by the mean value theorem, there is an x0 ∈ (x, x+ 1) such that

g(x) = f(x+ 1)− f(x) = f ′(x0)(x+ 1− x) = f ′(x0) < ε.

Thus g(x)→ 0 as x→ +∞.

2

5.6. Suppose

(a) f is continuous for x ≥ 0,

(b) f ′(x) exists for x > 0,

(c) f(0) = 0,

(d) f ′ is monotonically increasing.

Put

g(x) =f(x)

x(x > 0)

and prove that g is monotonically increasing.

Solution. Since f is differentiable and x > 0, g is differentiable by Theorem 5.3.It suffices to prove that g′ is non-negative by Theorem 5.11(a). Moreover, since

g′(x) =xf ′(x)− f(x)

x2,

it suffices to prove that xf ′(x) ≥ f(x).Fix x > 0. By the mean value theorem, there is an x0 ∈ (0, x) such that

f(x) = f(x)− f(0) = f ′(x0)(x− 0) = xf ′(x0) ≤ xf ′(x),

where the first equality follows by (c), and the last inequality follows by (d).

5.7. Suppose f ′(x), g′(x) exist, g′(x) 6= 0, and f(x) = g(x) = 0. Prove that

limt→x

f(t)

g(t)=f ′(x)

g′(x).

(This also holds for complex functions.)

Solution.

f ′(x)

g′(x)= lim

t→x

f(t)− f(x)

t− x· t− xg(t)− g(x)

= limt→x

f(t)− f(x)

g(t)− g(x)= lim

t→x

f(t)

g(t)=f(x)

g(x).

5.8. Suppose f ′ is continuous on [a, b] and ε > 0. Prove that there exists δ > 0such that ∣∣∣∣f(t)− f(x)

t− x− f ′(x)

∣∣∣∣ < ε

whenever 0 < |t− x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. (This could be expressed bysaying that f is uniformly differentiable on [a, b] if f ′ is continuous on [a, b].)Does this hold for vector-valued functions too?

3

Solution. Fix ε > 0. By Theorem 4.19, f ′ is uniformly continuous on [a, b],so there exists a δ > 0 such that |x− y| < δ implies |f ′(x)− f ′(y)| < ε. Fixx ∈ [a, b], and without loss of generality, fix t ∈ [a, b] with 0 < t−x < δ. By themean value theorem, there is an x0 ∈ (x, t) such that

f(t)− f(x)

t− x= f ′(x0).

Then 0 < |x− x0| < δ, so∣∣∣∣f(t)− f(x)

t− x− f ′(x)

∣∣∣∣ = |f ′(x0)− f ′(x)| < ε.

Now suppose f′ is continuous on [a, b] for some function f : Rn → R, f =(f1, . . . , fn). Then each component fj is differentiable, f′ = (f ′1, . . . , f

′n), and

each f ′j is continuous on [a, b]. Fix ε > 0, and by the result above, choose δj > 0such that ∣∣∣∣fj(t)− fj(x)

t− x− f ′j(x)

∣∣∣∣ < ε√n

whenever 0 < |t− x| < δj , a ≤ x ≤ b, a ≤ t ≤ b, for j = 1, . . . , n. Then∣∣∣∣ f(t)− f(x)

t− x− f′(x)

∣∣∣∣ =

∣∣∣∣(f1(t)− f1(x)

t− x− f ′1(x), . . . ,

fn(t)− fn(x)

t− x− f ′n(x)

)∣∣∣∣=

n∑j=1

(fj(t)− fj(x)

t− x− f ′j(x)

)21/2

<

n∑j=1

(ε√n

)21/2

= ε

whenever 0 < |t− x| < δj , a ≤ x ≤ b, a ≤ t ≤ b.

5.9. Let f be a continuous real function on R1, of which it is known that f ′(x)exists for all x 6= 0 and that f ′(x) → 3 as x → 0. Does it follow that f ′(0)exists?

Solution. Fix ε > 0 and choose δ > 0 such that |f ′(x)− 3| < ε whenever0 < |x| < δ. Now fix t ∈ R such that 0 < |t| < δ. By the mean value theorem,there is an x0 ∈ (0, t) (or x0 ∈ (t, 0) if t < 0) such that

f(t)− f(0)

t= f ′(x0).

Then 0 < |x0| < δ, so therefore∣∣∣∣f(t)− f(0)

t− 3

∣∣∣∣ = |f ′(x0)− 3| < ε.

4

This proves that

limt→0

f(t)− f(0)

t= 3,

so therefore f ′(0) exists.

5.10. Suppose f and g are complex differentiable functions on (0, 1), f(x)→ 0,g(x)→ 0, f ′(x)→ A, g′(x)→ B as x→ 0, where A andB are complex numbers,B 6= 0. Prove that

limx→0

f(x)

g(x)=A

B.

Compare with Example 5.18. Hint :

f(x)

g(x)=

{f(x)

x−A

}· x

g(x)+A · x

g(x).

Apply Theorem 5.13 to the real and imaginary parts of f(x)/x and g(x)/x.

Solution. Define f1 := Re(f), f2 := Im(f), g1 := Re(g), and g2 := Im(g). Then

f ′1 = Re(f ′), f ′2 = Im(f ′), g′1 = Re(g′), g′2 = Im(g′),

so therefore

f ′1(x)→ Re(A), f ′2 → Im(A), g′1 → Re(B), g′2 → Im(B)

as x→ 0. Moreover, fi → 0 and gi → 0 for i = 1, 2 as x→ 0. By Theorem 5.13,

limx→0

f(x)

x= lim

x→0

(f1(x)

x+ i

f2(x)

x

)= lim

x→0

f1(x)

x+ i lim

x→0

f2(x)

x

= limx→0

f ′1(x) + i limx→0

f ′2(x) = Re(A) + i Im(A) = A,

and similarly, g(x)/x → B as x → 0. Since B 6= 0, it follows from Theo-rem 4.4(c) that

limx→0

f(x)

g(x)= lim

x→0

f(x)

x· x

g(x)=A

B.

5.11. Suppose f is defined in a neighborhood of x, and suppose f ′′(x) exists.Show that

limh→0

f(x+ h) + f(x− h)− 2f(x)

h2= f ′′(x).

Show by an example that the limit may exist even if f ′′(x) does not.Hint : Use Theorem 5.13.

5

Solution. Since f ′′(x) exists, so does f ′(x), and by Theorem 5.2, f is continuousat x. Therefore f(x+h)→ f(x) and f(x−h)→ f(x) as h→ 0, so the numeratorand denominator of the fraction

f(x+ h) + f(x− h)− 2f(x)

h2

both converge to 0 as h→ 0. By Theorem 5.13,

limh→0

f(x+ h) + f(x− h)− 2f(x)

h2= lim

h→0

f ′(x+ h)− f ′(x− h)

2h

= limh→0

f ′(x+ h)− f ′(x) + f ′(x)− f ′(x− h)

2h

=1

2limh→0

f ′(x+ h)− f ′(x)

h+

1

2limh→0

f ′(x)− f ′(x− h)

h

= f ′′(x).

Define f : R→ R by

f(x) :=

{1/x if x 6= 0,

0 if x = 0.

Then f is not continuous at x = 0, so f ′(0) and f ′′(0) do not exist by Theo-rem 5.2. However, for any h 6= 0, we have

f(h) + f(−h)− 2f(0)

h2= 0,

so the fraction above converges to 0 as h→ 0.

5.12. If f(x) = |x|3, compute f ′(x), f ′′(x) for all real x, and show that f (3)(0)does not exist.

Solution. If x > 0, then f(x) = x3, so f ′(x) = 3x2 and f ′′(x) = 6x. If x < 0,then f(x) = −x3, so f ′(x) = −3x2 and f ′′(x) = −6x. Since

limt→0

f(t)

t= lim

t→0

|t|3

sgn(t) |t|= lim

t→0sgn(t) |t|2 = 0,

we have f ′(0) = 0, and since

limt→0

f ′(t)

t= lim

t→0

3 |t|2

sgn(t) |t|= lim

t→0sgn(t)3 |t| = 0,

we have f ′′(0) = 0. However, f (3)(0) does not exist because

limt→0+

f ′′(t)

t= lim

t→0

6t

t= 6, lim

t→0−

f ′′(t)

t= lim

t→0

−6t

t= −6,

shows that the limit of f ′′(t)/t as t→ 0 does not exist.

6

Math 131C: Homework 1 Solutions

(From Rudin, Chapter 7)

Problem 14:The continuity of Φ follows from the continuity of x and y, which follows from the Weierstrass M -test

(Theorem 7.10, Rudin) and the continuity and boundedness of f .To show that Φ maps the Cantor set onto I2, we follow the hint. Note that to as given belongs to

the Cantor set (its ternary expansion contains no 1’s). We have

3kt0 =∞∑

i=1

3k−i−1(2ai)

= 2k−1∑

i=1

3k−i−1ai +∞∑

j=0

3−j−1(2aj+k)

≡∞∑

j=0

3−j−1(2aj+k) (mod 2),

so that

f(3kt0) = f

( ∞∑

j=0

3−j−1(2aj+k))

by the 2-periodicity of f . Now suppose ak = 0; then

0 ≤∞∑

j=0

3−j−1(2aj+k) ≤ 13

=⇒ f

( ∞∑

j=0

3−j−1(2aj+k))

= f(3kt0) = 0 = ak.

(The lower bound follows from setting ai = 0 for i > k, and the upper bound follows from setting ai = 1for i > k.) Similarly, if ak = 1, we have

23≤

∞∑

j=0

3−j−1(2aj+k) ≤ 1 =⇒ f

( ∞∑

j=0

3−j−1(2aj+k))

= f(3kt0) = 1 = ak.

The hint and surjectivity of Φ follow immediately. �

Problem 15:Such an f must be constant on [0,∞). Fix arbitrary nonnegative numbers x and y. Let ε > 0 be

arbitrary, and choose δ > 0 s.t.

|x− y| < δ ⇒ |fn(x)− fn(y)| < ε ∀x, y ∈ [0, 1], ∀n ∈ N.

Now choose n ∈ N s.t. |x−y|n < δ and x

n , yn ≤ 1 . Then | xn − y

n | < δ, so

|f(x)− f(y)| =∣∣∣∣ fn

(x

n

)− fn

(y

n

)∣∣∣∣ < ε.

1

Since ε was arbitrary, we have f(x) = f(y). But x and y were arbitrary, so f is constant. �

Problem 16:Take ε > 0. Choose δ > 0 as in the definition of equicontinuity for the family {fn}. The collection

{Bδ(x) |x ∈ K} (where Bδ(x) is the open ball of radius δ centered at x) is an open cover of K. Bycompactness of K, there exists a finite subcover, so there exist x1, . . . , xm ∈ K s.t. K =

⋃n1 Bδ(xi). Say

fn → f pointwise, and choose N ∈ N s.t. n > N ⇒ |fn(xi)− f(xi)| < ε, 1 ≤ i ≤ m.Now take any x ∈ K. x ∈ Bδ(xi) for some i, so |fn(x) − fn(xi)| < ε for all n by equicontinuity.

This in turn implies |f(x)− f(xi)| ≤ ε (by continuity of the absolute value and pointwise convergence of{fn}). Finally, we have

n > N =⇒ |fn(x)− f(x)| ≤ |fn(x)− fn(xi)|+ |fn(xi)− f(xi)|+ |f(xi)− f(x)|< ε + ε + ε.

Since x was arbitrary, fn → f uniformly on K. �

Problem 18:By the Fundamental Theorem of Calculus and the uniform boundedness of {fn}, the sequence {Fn}

has uniformly bounded derivatives. By the Mean Value Theorem, this implies that {Fn} is uniformlyLipschitz; i.e., there exists M s.t.

|Fn(x)− Fn(y)||x− y| ≤ M

for all x, y ∈ [a, b] and all n. But then |Fn(x) − Fn(y)| ≤ M |x − y| for all n, x, and y, so the family{Fn} is equicontinuous. Furthermore, {Fn} is pointwise (in fact, uniformly) bounded by the uniformboundedness of {fn}. By the Arzela-Ascoli Theorem (Theorem 7.25, Rudin), {Fn} contains a uniformlyconvergent subsequence. �

Problem 20:Since f is continuous on a compact set it is bounded; say |f(x)| ≤ M ∀x. The complex conjugate

f(x) is continuous, so by the Weierstrass Approximation Theorem, there is a sequence of complex poly-nomials Pn s.t. Pn → f uniformly on [0, 1]. Since f is bounded, fPn → ff = |f |2 uniformly as well(since ‖fPn − ff‖∞ ≤ M‖Pn − f‖∞ → 0). By Theorem 7.16 in Rudin,

0 = limn→∞

∫ 1

0

f(x)Pn(x) dx =∫ 1

0

limn→∞

f(x)Pn(x) dx =∫ 1

0

|f(x)|2 dx.

By continuity of f , this implies f ≡ 0 on [0,1]. �

Problem 21:To see that A vanishes at no point of K and separates points, simply notice that f(eiθ) = eiθ is in

A . Proving the hint is a pretty straightforward exercise in integration and application of Theorem 7.16.Finally, consider g(eiθ) = e−iθ, which is a continuous function on K. We have

∫ 2π

0

g(eiθ)eiθ dθ = 2π 6= 0,

which shows that g is not in the uniform closure of A by the hint. �

2

Problem 3:Suppose Ax = Ay. Then A(x− y) = 0 ⇒ x = y. �

Problem 4:These results follow immediately from the definition of a vector space and linearity. You don’t have

to verify all the vector space axioms; since the range and kernel are subsets of vector spaces, it sufficesto show they are subspaces (i.e. that they are closed under addition and scalar multiplication).

Problem 5:Given A ∈ L(Rn,R), let yA = (Ae1, Ae2, . . . , Aen) (where {e1, . . . , en} is the standard basis for Rn).

By linearity, Ax = x · yA for all x ∈ Rn. The uniqueness of yA is immediate.

Now by Cauchy-Schwarz, we have

‖A‖ = sup|x|≤1

x · yA ≤ sup|x|≤1

|x||yA| = |yA|.

But if x = yA

|yA| , we have |x| = 1 and Ax = |yA|, so ‖A‖ ≥ |yA|. So ‖A‖ = |yA|. �

Problem 6:For (x, y) 6= (0, 0), we can compute (Djf)(x, y) via the usual differentiation formulas just as in Math

32A or the like. At (0, 0), we consider the definition of partial derivatives:

(D1f)(0, 0) = limh→0

f(0 + h, 0)− f(0, 0)h

= limh→0

0− 0h

= 0.

Similarly, (D2f)(0, 0) = 0. So the partial derivatives of f exist everywhere in R2.

However, f is not continuous at (0, 0), since f(a, a) = 12 for all a ∈ R. (So there is no neighbor-

hood of the origin on which |f(x, y)− f(0, 0)| = |f(x, y)| < 12 ). �

Problem 7:Fix x ∈ E and ε > 0. Since the partial derivatives of f are bounded in E (say by M), the Mean

Value Theorem implies that |f(a+hej)− f(a)| < hM for all a ∈ E and all h ∈ R such that a+hej ∈ E(1 ≤ j ≤ n). Now choose δ < ε

nM s.t. Bδ(x) ⊂ E (we can choose such a δ since E is open).

Now pick an arbitrary y in Bδ(x). Write y − x =∑n

1 hjej , v0 = 0, and vk =∑k

1 hjej . (Notevn = y − x). It’s pretty easy to see that hj < δ for all j. Now we have

1

|f(y)− f(x)| = |f(x + (y − x))− f(x)|

= |n∑

j=1

[f(x + vj)− f(x + vj−1)]|

≤n∑

j=1

|f(x + vj)− f(x + vj−1)|

≤n∑

j=1

hjM

<

n∑

j=1

δM = ε,

where we’ve used the Mean Value Theorem result from the first paragraph. So |y − x| < δ⇒ |f(y)− f(x)| < ε, and f is continuous at x. �

Problem 8:Let x = (x1, . . . , xn), and let Ej be the j-cross-section of E through x; i.e.,

Ej = {x ∈ R | (x1, . . . , xj−1, x, xj+1, . . . , xn) ∈ E}.Now define fj : Ej → R by fj(x) = f(x1, . . . , xj−1, x, xj+1, . . . , xn). Since f has a local maximum at x,fj has a local maximum at xj ∈ Ej for all j. By single-variable calculus, this implies f ′j(xj) = 0 for allj. But since f ′j(xj) = Djf(x) (VERIFY!), this implies f ′(x) = 0 by Theorem 9.17 in Rudin. �

Problem 9:To show f is constant, fix some x0 ∈ E and consider the set A = {x ∈ E | f(x) = f(x0)}. We want

to show that A = E, but since E is connected and A is nonempty, it suffices to show that A is both openand closed in the relative topology on E. Since E is open, the relative topology coincides with the usualtopology on Rn.

First we show A is open. Take any x ∈ A. Since E is open, there is an open ball Br(x) ⊂ E.Now Br(x) is convex, so Theorem 9.19 (or its corollary) in Rudin implies that f is constant on Br(x).So, since x ∈ A, this implies Br(x) ⊂ A. So A is open.

To show A is closed, use an exactly analogous argument to show that Ac is open. So A is bothclosed and open, which implies that A = E. �

Problem 10:(This whole problem is easier to think about if you draw a picture in two dimensions.) Fix any

x0 = (x(0)1 , . . . , x

(0)n ) ∈ E. We want to show that for all x = (x1, x

(0)2 , . . . , x

(0)n ) ∈ E, f(x) = f(x0). As

in Problem 8, define f1(x) = f(x, x(0)2 , . . . , x

(0)n ). Note that f ′1(x) = (D1f)(x, x

(0)2 , . . . , x

(0)n ), so f ′1(x) = 0

for all x at which f1 is defined. Now take any x1 s.t. (x1, x(0)2 , . . . , x

(0)n ) ∈ E. By convexity, we have

(x, x(0)2 , . . . , x

(0)n ) ∈ E for all x

(0)1 ≤ x ≤ x1, so f1 is defined for all such x. Since f ′1(x) = 0 for all these

2

x, the Mean Value Theorem shows that f1(x(0)1 ) = f1(x1). So f(x0) = f(x1, x

(0)2 , . . . , x

(0)n ), as desired. �

Note that the proof just given only required that E be “convex in the first variable”; i.e., we canrelax the convexity condition to

(a, x2, . . . , xn) ∈ E and (b, x2, . . . , xn) ∈ E =⇒ (x, x2, . . . , xn) ∈ E ∀ a ≤ x ≤ b.

But, as noted, the proof doesn’t work for arbitrary connected regions. Draw the union of the followingthree regions in R2:

A = {(x, y) | − 2 < x < 2, 0 < y < 1}B = {(x, y) | − 2 < x < −1, −2 < y ≤ 0}

C = {(x, y) | 1 < x < 2, −2 < y ≤ 0}.Take

f(x, y) =

0, (x, y) ∈ A−y, (x, y) ∈ B

y, (x, y) ∈ C

Then clearly ∂f∂x ≡ 0 on the union of these regions, but f does not depend only on y. The problem is

that we can’t apply the Mean Value Theorem because of the “gap” between B and C.

3

Problem 18:(a) If we define f(x, y) = (u(x, y), v(x, y)), then the range of f is R2. The slickest way to see this

is to note that if z = x + iy, then u = <(z2) and v = =(z2) (where < and = denote real and imagi-nary parts, respectively). Then since the map z 7→ z2 maps C onto C, it follows that the range of f isall of R2. If you don’t want to appeal to complex variables, you can also show this by brute-force algebra.

(b) The Jacobian of f is in fact 0 at the origin. However, it is nonzero at all other points of R2, so f isone-to-one on a neighborhood of every point of R2\{(0, 0)}. But it’s easily seen that f(x, y) = f(−x,−y)for all (x, y), so f is not injective on R2\{(0, 0)}. (This is the same as the fact that z 7→ z2 is not injectiveon C \ {0}.)

(c) Once again, this is much easier if you appeal to the complex-variable interpretation of f , soidentify R2 with C. Since u + iv = (x + iy)2, we can write

x + iy =√

u + iv,

where √ denotes a branch of the square root function defined in a neighborhood of u+ iv (note that wecan’t find such a branch if u = v = 0). In a neighborhood of (0, π/3), we can take the standard branchof the square root, which is represented most easily in polar form (recall eiθ = cos θ + i sin θ):

√reiθ =

√rei θ

2 .

Using some trig identities (viz., the half-angle formulas), we can rewrite this in the upper half-plane as:

√u + iv =

√√u2 + v2 + u

2+ i

√√u2 + v2 − u

2,

where the square roots on the right hand side are just the usual square roots of positive real numbers.Rewriting this in R2-notation:

f−1(u, v) =(√√

u2 + v2 + u

2,

√√u2 + v2 − u

2

).

This is a good way to brush up on playing with complex numbers. I’ll let you compute the derivativesin question; it’s just calculus.

(d ) The images under f of lines parallel to the coordinate axes are parabolas, unless of coursewe’re considering the coordinate axes themselves, in which case we get we get the nonnegative u-axis asthe image of the x-axis and the nonpositive u-axis as the image of the y-axis.

Problem 19:First subtract the second equation from the first and compare with the third to see that u must be

1

0 or 3. To show that we can solve for x, y, u, in terms of z, apply the Implicit Function Theorem:Let f(x, y, z, u) = (3x + y − z + u2, x− y + 2z + u, 2x + 2y − 3z + 2u) = (f1, f2, f3). Then define

Ax,y,u =

∂f1∂x

∂f1∂y

∂f1∂u

∂f2∂x

∂f2∂y

∂f2∂u

∂f3∂x

∂f3∂y

∂f3∂u

=

3 1 2u1 −1 12 2 2

Then detAx,y,u = 8u− 12 6= 0 since u = 0 or u = 3, so the Implicit Function Theorem gives the desiredresult. The next two claims follow by analogous reasoning.

Finally, to show that we cannot solve for x, y, z in terms of u, note that the system can be expressedas

A

xyz

=

−u2

−u−2u

,

where

A =

3 1 −11 −1 22 2 −3

.

But detA = 0, so the system cannot be solved.

Problem 21:(a) Compute ∇f = 6(x2 − x, y2 + y). Then clearly the gradient is zero at precisely the points (0, 0),

(1, 0), (0,−1), (1,−1). (1, 0) is a local minimum and (0,−1) is a local maximum, while the other two aresaddle points. This can easily be seen either by examining the Hessian of f or by examining f directlynear the points in question.

(b) We’ll first try to describe the set S. With some cleverness, you can notice that f(x,−x) = 0 forall x ∈ R. This should suggest that we can factor f as

f(x, y) = (x + y)P (x, y),

where P is a polynomial. P turns out to be

P (x, y) = 2x2 − 3x− 2xy + 3y + 2y2.

The set of zeroes of P is an ellipse; so that S is the union of this ellipse and the line y=−x. (Try plottingthis in Mathematica or some other such program.)

Now by the Implicit Function Theorem, the only candidates for points that have no neighborhoodsin which f(x, y) = 0 can be solved for y in terms of x (or vice versa) will be points at which D1f = 0and D2f = 0, i.e. ∇f = 0. (This is a NOT(p OR q) ⇐⇒ (NOT p) AND (NOT q) thing). Using part(a), it’s easy to see that the only points of S at which ∇f = 0 are (0, 0) and (1,−1). These both lie onthe line y =−x and the ellipse P (x, y) = 0, so the curves “cross” at these points. If you examine a plotof the points in S, it’s pretty clear that it is neither a graph of y as a function of x nor a graph of x asa function of y. So these are the points we’re looking for. As an example of how to be more rigorousabout this, we’ll consider the point (0, 0). Pick a positive ε ¿ 1. Then for any x0 ∈ (−ε, ε), the point(x0,−x0) satisfies f(x, y) = 0. But there is also a y0 s.t. P (x0, y0) = 0 ⇒ f(x0, y0) = 0. It’s pretty easy

2

to see that this y0 must have the same sign as x0, so y0 6= −x0. So, in short, we can’t solve for y interms of x in some small neighborhood of the (0, 0). An analogous argument shows we can’t solve forx in terms of y, and the point (1,−1) is treated similarly. (I realize this is probably confusing; it reallyhelps to draw a picture).

Problem 24:Compute the Jacobian matrix of f :

[f ′(x, y)] =1

(x2 + y2)2

[4xy2 −4x2y

y3 − x2y x3 − y2x

].

Noticing that the first column is −xy times the second, we see that f ′(x, y) has rank 1. The range of f is

an ellipse, as you can see from the relation

f21 + 4f2

2 = 1.

Problem 25:(a) By the hint, SA is a projection in Rn. The zj used to define S are linearly independent (VER-

IFY!), so S is injective. Then it’s immediate that N (SA) = N (A). Finally, SAzj = zj so thatR(SA) = R(S), since clearly R(SA) ⊆ R(S).

(b) This is immediate from the Rank-Nullity Theorem, but we can do the problem as Rudin suggestswithout using this result. We’ll use the following result from the proof of 9.31 (b) in Rudin:

P a projection in Rn =⇒ dimN (P ) + dimR(P ) = n.

Since we noted in part (a) that the zj are linearly independent, we know that rank(S) = r = rank(A).The desired result now follows immediately from part (a).

Problem 28:To show ϕ is continuous, it suffices to consider points (x, t) with t ≥ 0 since ϕ is odd in t. The

following is just a sketch of the continuity argument; I expect a bit more detail on the actual homeworks.First consider a point (x0, t0) with 0 < x0 <

√t0. Argue that you can find an ε > 0 such that 0 < x <

√t

for all (x, y) ∈ Bε((x0, y0)). Then ϕ(x, t) = x on this ball, and the continuity follows immediately. Treatthe cases

√t0 < x < 2

√t0 and 2

√t0 < x similarly. Now consider a point of the form (

√t0, t0) (i.e.

x =√

t0). Since the functions ϕ1(x, t) = x and ϕ2(x, t) = −x + 2√

t are both continuous, choose δ > 0so that both |ϕ1(x, t) − ϕ1(

√t0, t0)| < ε and |ϕ2(x, t) − ϕ2(

√t0, t0)| < ε for a given ε > 0. This δ will

“work” for ϕ in the usual sense. Treat the other “boundary” cases similarly.Showing that (D2ϕ)(x, 0) = 0 for all x is routine; just use the definition of partial derivatives.Now |t| < 1

4 ⇒ 2√|t| < 1, so we have

f(t) =∫ 1

−1

ϕ(x, t) dx =∫ 1

0

ϕ(x, t) dx = sgn(t)[ ∫ √

|t|

0

x dx +∫ 2√|t|

√|t|

(−x + 2√|t|) dx

]= t,

where

sgn(t) =

−1, t < 0

0, t = 01, t > 0.

3

So this implies f ′(t) = 1 for |t| < 14 , and hence

1 = f ′(0) 6=∫ 1

−1

(D2ϕ)(x, 0) dx = 0.

This suggests you should take care when differentiating under the integral sign.

4

PRINCIPLES OF MATHEMATICAL ANALYSIS.

WALTER RUDIN

Disclaimer: these solutions were typed at warp speed, often with little orno preparation. And very little proofreading. Consequently, there are bound to beloads of mistakes. Consider it part of the challenge of the course to find these errors.

(And when you do find some, please let me know so I can fix them!)

7. Sequences and Series of Functions

1. Prove that every uniformly convergent sequence of bounded functions is uni-formly bounded.{fn} is uniformly Cauchy, so |fn(x)−fm(x)| < 1, for n,m bigger than some

large N . Then∣∣∣|fn(x)| − |fN(x)|

∣∣∣ ≤ |fn(x)− fN(x)| < 1, so

|fn(x)| < |fN(x)|+ 1.

Let M be a bound for fN . Then

|fn(x)| < M + 1,

so {fn}∞n=N is uniformly bounded by M + 1. Define

K := max{sup |f1|, sup |f2|, . . . , sup |fN−1|,M + 1}.Then {fn}∞n=1 is uniformly bounded by K.

2. Show that {fn}, {gn} converge uniformly on E implies {fn + gn} convergesuniformly on E. If, in addition, {fn}, {gn} are sequences of bounded functions,prove that {fngn} converges uniformly.

supx∈E

{|(fn + gn)(x)− (f + g)(x)|} = supx∈E

{|fn(x)− f(x) + gn(x)− g(x)|}≤ sup

x∈E{|fn(x)− f(x)|+ |gn(x)− g(x)|}

≤ supx∈E

|fn(x)− f(x)|+ supx∈E

|gn(x)− g(x)|n→∞−−−−−→ 0 + 0.

When {fn}, {gn} are bounded sequences,

supx∈E

{|(fngn)(x)− (fg)(x)|}

March 23, 2006. Solutions by Erin P. J. Pearse.

1

Principles of Mathematical Analysis — Walter Rudin

= supx∈E

{|fngn(x)− fgn(x) + fgn(x)− fg(x)|}≤ sup

x∈E{|fn(x)− f(x)||gn(x)|+ |f(x)||gn(x)− g(x)|}

≤ supx∈E

|fn(x)− f(x)||gn(x)|+ supx∈E

|f(x)||gn(x)− g(x)|≤ sup

x∈E|fn(x)− f(x)| ·Mg + sup

x∈EMf · |gn(x)− g(x)|.

We can introduce the uniform bounds Mf and Mg by problem 1 and theadditional hypothesis. Then it is clear that the last line goes to 0 as n →∞.

3. Construct sequences {fn}, {gn} which converge uniformly on a set E, but suchthat {fngn} does not converge uniformly.

Work on I = (0, 1). Let fn(x) = 1x

+ xn

and gn(x) = − 11−x

− 1−xn

so thatgn is the horizontal reflection of fn, translated 1 to the right. (Graph them tosee it.)

fn(x) = 1x

+ xn

n→∞−−−−−→ 1x

= f(x) sup0≤x≤1

|fn − f | = supx∈I

xn

= 1n→ 0

gn(x) = − 11−x

− 1−xn

n→∞−−−−−→ − 11−x

= g(x) sup0≤x≤1

|gn − g| = supx∈I

1−xn

= 1n→ 0

fngn(x) = −(n + (x− 1)2)(n + x2)

n2x(x− 1)

n→∞−−−−−→ 1

x− x2= fg(x)

sup0≤x≤1

|fngn − fg| = supx∈I

∣∣∣∣x2(x− 1)2 + n + 2nx(x− 1)

n2x(x− 1)

∣∣∣∣ = ∞, for any n.

To see the sup is infinite, check x = 0, 1.

4. Consider the sum

f(x) =∞∑

n=1

1

1 + n2x.

For what x does the series converge absolutely? On what intervals does itconverge uniformly? On what intervals does it fail to converge uniformly? Isf bounded?

For xk = − 1k2 , we get

f(xk) =∞∑

n=1

1

1− (nk

)2 .

The kth term of the sum is undefined, so f is undefined at xk = − 1k2 , k =

1, 2, . . .To be completed.

2

Solutions by Erin P. J. Pearse

5. Define a sequence of functions by

fn(x) =

0(x < 1

n+1

),

sin2 πx

(1

n+1< x ≤ 1

n

),

0(

1n

< x).

Show that the series {fn} converges to a continuous function, but not uni-formly. Use the series

∑fn to show that absolute convergence, even for all x,

does not imply uniform convergence.For x ≤ 0, fn(x) = 0 for every n, so lim fn(x) = 0. For x > 0,

n > N :=[

1x

]=⇒ 1

n< x =⇒ fn(x) = 0.

Thus fn(x)pw−−−→ f(x) := 0 for x ∈ R.

To see that the convergence is not uniform, consider

f ′n(x) = 2(sin π

x

) (cos π

x

) (− πx2

).

f ′n(x) = 0 when• sin π

x= 0, in which case x = 1

kfor some k ∈ Z, or

• cos πx

= 0, in which case x = 22k+1

for some k ∈ Z.For each fn, only a few of these values occur where fn is not defined to be 0,so checking these values of x,

fn

(1n

)= sin(nπ) = 0

fn

(1

n+1

)= sin((n + 1)π) = 0

fn

(2

2n+1

)= sin

(2n+1

2π)

= 1.

So Mn = supx{|fn(x)− f(x)|} = 1 for each n, and clearly Mn → 1 6= 0.The series

∑∞n=1 fn(x) converges absolutely for all x ∈ R: for any fixed x,

there is only one nonzero term in the sum.The series

∑∞n=1 fn(x) does not converge uniformly: check partial sums.

supx

∣∣∣∣∣N∑

n=1

fn(x)−N+1∑n=1

fn(x)

∣∣∣∣∣ = supx|fN+1(x)| = 1,

so the sequence of partial sums is not Cauchy (in the topology of uniformconvergence), hence cannot converge.

6. Prove that the series ∞∑n=1

(−1)n x2 + n

n2

converges uniformly in every bounded interval, but does not converge abso-lutely for any value of x.

To see uniform convergence on a bounded interval,

supa<x<b

{|fN(x)− f(x)|} = supa<x<b

{∣∣∣∑∞

n=N+1(−1)n x2+n

n2

∣∣∣}

3


={∣∣∣

∑∞n=N+1

(−1)n c2+nn2

∣∣∣}

(7.1)

where c := max{|a|, |b|}. We will use the alternating series test to show∑∞n=1(−1)n c2+n

n2 converges. From this, it will follow that (7.1) goes to 0 asN →∞.(i) For all x ∈ R, x2+n

n2 > 0. So the sum alternates.

(ii) For fixed x, lim x2+nn2 = lim 1

2n= 0. (L’Hop)

(iii) To check monotonicity, prove

x2 + n + 1

(n + 1)2≤ x2 + n

n2

by cross-multiplying.

7. For n = 1, 2, 3, . . . , and x ∈ R, put fn(x) = x1+nx2 . Show that {fn} converges

uniformly to a function f , and that the equation f ′(x) = limn→∞ f ′n(x) iscorrect if x 6= 0 but false if x = 0.

8. If

I(x) =

{0 (x ≤ 0),

1 (x > 0),

if {xn} is a sequence of distinct points in (a, b), and if∑ |cn| converges, then

prove that the series

f(x) =∞∑

n=1

cnI(x− xn) (a ≤ x ≤ b)

converges uniformly, and that f is continuous for every x 6= xn.

4

9. {fn} are continuous and fnunif−−−→ f on E. Show lim fn(xn) = f(x) for every

sequence {xn} ⊆ E with xn → x.Pick N1 such that

n ≥ N1 =⇒ supx∈E

|fn(x)− f(x)| < ε/2.

Then surely |fn(xn) − f(xn)| < ε/2 holds for each n ≥ N1. By Thm. 7.12, fis continuous, so pick N2 such that

n ≥ N2 =⇒ |f(xn)− f(x)| < ε/2.

Then we are done because

|fn(xn)− f(x)| ≤ |fn(xn)− f(xn)|+ |f(xn)− f(x)|.

11. {fn}, {gn} are defined on E with:

(a) {∑Nn=1 fn} uniformly bounded, (b) gn

unif−−−→ g on E, and (c) gn(x) ≤gn−1(x) ∀x ∈ E, ∀n. Prove

∑fngn converges uniformly on E.

Define a := supx∈E |fn(x)| and bn := supx∈E |gn(x)|. Then

supx∈E

∣∣∣∣∣∞∑

N+1

fn(x)gn(x)

∣∣∣∣∣ ≤∞∑

N+1

supx∈E

|fn(x)gn(x)| ≤∞∑

N+1

anbn → 0,

by Thm. 3.42.

13. {fn} is monotonically increasing on R, and 0 ≤ fn(x) ≤ 1

(a) Show ∃f, {nk} such that f(x) = limk→∞ fnk(x),∀x ∈ R.

By Thm. 7.23, we can find a subsequence {fni} such that {fni

(r)} con-verges for every rational r. Thus we may define f(x) := supr≤x f(r),where the supremum is taken over r ∈ Q. It is clear that f is monotone,because

x < y =⇒ {r ≤ x} ⊆ {r ≤ y}and the supremum can only increase on a larger set. Thus, f has at mosta countable set of discontinuities, by Thm. 4.30, pick x such that f iscontinuous at x.We want to show fni

(x) → f(x). Fix ε > 0. Since f is continuous at x,choose δ such that |x − y| < δ =⇒ |f(x) − f(y)| < ε/3. Now pick arational number r ∈ [x− δ

3, x]. Then

|fni(x)− f(x)| ≤ |fni

(x)− fni(r)|+ |fni

(r)− f(r)|+ |f(r)− f(x)|. (7.2)

On the RHS of (7.2): the last term is less than ε/3 by the choice of δ;and the middle term is less than ε/3 whenever i ≥ N1 for some large N1,because the subsequence converges on the rationals. It remains to showthe first term gets small.

5

Pick some rational s ∈ [x, x + δ/3]. Then r ≤ x ≤ s and the continuityof f at x shows

|r − s| < δ =⇒ |f(r)− f(s)| < ε/3.

Also, since the fniare monotone,

fni(r) ≤ fni

(x) ≤ fni(s). (7.3)

With

|fni(r)− fni

(s)| ≤ |fni(r)− f(r)|+ |f(r)− f(s)|+ |f(s)− fni

(s)|,some large N2, i ≥ N2 gives |fni

(r) − fni(s)| < ε. By (7.3), this shows

|fni(r)− fni

(x)| < ε.

(b) If f is continuous, show fnk→ f uniformly on compact sets.

Let K be compact. Fix ε > 0. Since f is uniformly continuous onK, pick δ such that |x − y| < δ =⇒ |f(x) − f(y)| < ε/3. Since K

is compact, we can find {x1, . . . xJ} such that K ⊆ ⋃Jj=1 Bδ(xj), where

Bδ(xj) := (xj − δ, xj + δ).

16. {fn} is equicontinuous on a compact set K, fnpw−−−→ f on K. Prove {fn}

converges uniformly on K.Define f by f(x) := lim fn(x). Fix ε. From equicontinuity, find δ such that

|x− y| < δ =⇒ |fn(x)− fn(y)| < ε/3, ∀n, x, y.

Letting n →∞, this gives

|x− y| < δ =⇒ |f(x)− f(y)| < ε/3, ∀x, y.

Since K is compact, we can choose a finite set {x1, . . . , xJ} such that

K ⊆⋃J

j=1Bδ(xj)

where Bδ(xj) := (xj − δ, xj + δ). For each xj, we know fn(xj) → f(xj), sopick N big enough that

n ≥ N =⇒ |fn(xj)− f(xj)| < ε/3, ∀j = 1, . . . , J.

For any x ∈ K, x ∈ Bδ(xj) for some j. Thus for all x ∈ K,

|fn(x)− f(x)| ≤ |fn(x)− fn(xj)|+ |fn(xj)− f(xj)|+ |f(xj)− f(x)|.

6

18. Let {fn} be uniformly bounded and Fn(x) :=∫ x

af(t) dt for x ∈ [a, b]. Prove

∃{Fnk} which converges uniformly on [a, b]

We need to show {Fn} is equicontinuous. Then by Thm. 6.20, each Fn iscontinuous; and by Thm. 7.25(b), we’re done. So fix ε > 0, let x < y, and letM be the uniform bound on the {fn}.

|Fn(x)− Fn(y)| =∣∣∣∣∫ y

x

fn(t) dt

∣∣∣∣ ≤∫ y

x

|fn(t)| dt ≤ M(y − x)

Then pick any δ < ε/M and

|x− y| < δ =⇒ |Fn(x)− Fn(y)| ∀n,

so {Fn} is equicontinuous.

7


20. f is continuous on [0, 1] and∫ 1

0f(x)xn dx = 0, n = 0, 1, 2, . . . . Prove that

f(x) ≡ 0.Let g be any polynomial. Then g(x) = a0 + a1x + a2x

2 + · · · + aKxK . Bylinearity of the integral,

∫ 1

0

f(x)g(x) dx =K∑

k=0

ak

∫ 1

0

f(x)xk dx =K∑

k=0

0 = 0.

By the Weierstrass theorem, let {fn}∞n=1 be a sequence of polynomials whichconverge uniformly to f on [0, 1]. Then

∫ 1

0

f 2(x) dx = limn→∞

∫ 1

0

f(x)fn(x) dx = limn→∞

0 = 0.

Then by Chap. 6, Exercise 2, f 2(x) ≡ 0. Thus f(x) ≡ 0.

21. Let K be the unit circle in C and define

A :=

{f(eiθ) =

N∑n=0

cneinθ ... cn ∈ C, θ ∈ R}

.

To see that A separates points and vanishes at no point, note that A con-tains the identity function f(eiθ) = eiθ.

To see that there are continuous functions on K that are not in the uniformclosure of A, note that

∫ 2π

0

f(eiθ)eiθ dθ = 0 ∀f ∈ A, (7.4)

and hence for g = lim gn (uniform limit) with gn ∈ A,

∫ 2π

0

g(eiθ)eiθ dθ = limn→∞

∫ 2π

0

gn(eiθ)eiθ dθ = 0,

by Thm. 7.16. Thus, all functions in the closure of A satisfy (7.4). However,if we choose an h which is not in A, like

h(eiθ) = e−iθ,

then h is clearly continuous on K, and

∫ 2π

0

h(eiθ)eiθ dθ =

∫ 2π

0

1 dθ = 2π.

Thus h is not in the uniform closure of A.

8

22. Assume f ∈ R(α) on [a, b] and prove that there are polynomials Pn such that

limn→∞

∫ b

a

|f − Pn|2dα = 0.

We need to find {Pn} such that ‖f−Pn‖2n→∞−−−−−→ 0. Fix ε > 0. By Chap. 6,

Exercise 12, we can find g ∈ C[a, b] such that ‖f − g‖2 < ε/2. Note that∫ b

a

|g − P |2 ≤∫ b

a

sup |g − P |2 = sup |g − P |2(b− a).

Then by the Weierstrass theorem, we can find a polynomial P such that

‖g − P‖2 ≤ sup |g − P |(b− a) < ε/2.

By Chap. 6, Exercise 11, this gives ‖f − P‖2 < ε.

23. Put P0 = 0 and define Pn+1(x) := Pn(x) + (x2 − P 2n(x)) /2 for n = 0, 1, 2, . . . .

Prove that limn→∞ Pn(x) = |x| uniformly on [−1, 1].Note that if Pn is even, the definition will force Pn+1 to be even, also. Now

P1 = x2

2, so assume 0 ≤ Pn−1 ≤ 1 for |x| ≤ 1. Then

Pn = Pn−1 +x2

2− P 2

n−1

2=

x2

2+ Pn−1

(1− Pn−1

2

).

By elementary calculus,

0 ≤ y ≤ 1 =⇒ f(y) = y(1− y2) takes values in [0, 1

2]. (7.5)

Since |x| ≤ 1 also implies x2

2∈ [0, 1], this gives 0 ≤ Pn ≤ 1. Then

0 ≤ |x|+ Pn(x)

2≤ 1

0 ≤ 1− |x|+ Pn(x)

2≤ 1. (7.6)

To see Pn(x) ≤ |x|, consider that for x ≥ 0, the inequality

x− P1(x) = x− x2/2 ≥ 0

holds in virtue of the positivity of f in (7.5). Then by the symmetry of evenfunctions, this is true for |x| ≤ 1. Now suppose Pn−1 ≤ |x|, i.e., |x|−Pn−1(x) ≥0. Then the given identity, and (7.6), give

|x| − Pn = (|x| − Pn−1)(1− 1

2(|x|+ Pn−1)

) ≥ 0.

We have established

0 ≤ Pn(x) ≤ Pn+1(x) ≤ |x| for− 1 ≤ x ≤ 1. (7.7)

Now, for n = 1, it is clear that

|x| − P1(x) = |x| − |x|22≤ |x|

(1− |x|

2

)1

.

9

So suppose |x| − Pn−1(x) ≤ |x|(1− |x|

2

)n−1

. Then, multiplying each side by(1− |x|

2− Pn−1(x)

2

)and using the identity, we get

(|x| − Pn−1(x))(1− |x|

2− Pn−1(x)

2

)≤ |x|

(1− |x|

2

)n−1 (1− |x|

2− Pn−1(x)

2

)

|x| − Pn(x) ≤ |x|(1− |x|

2

)n−1 (1− |x|

2

)

|x| − Pn(x) ≤ |x|(1− |x|

2

)n

.

Now consider gn(x) = x(1− x

2

)non [0, 1]. gn(x) =

(1− x

2

)n−1(1− (n+1)x

2

)

shows that gn has extrema at 2 and 2n+1

; only the latter is in [0, 1]. Then for

fn(x) = |x|(1− |x|

2

)n

, fn has extrema

fn

(± 2n+1

)= 2

n+1

(n

n+1

)n< 2

n+1.

We have established

|x| − Pn(x) ≤ |x|(1− |x|2

x

)n

< 2n+1

. (7.8)

Now sup∣∣∣|x| − Pn(x)

∣∣∣ < 2n+1

n→∞−−−−−→ 0 and Pnunif−−−→ |x|.

10

8. Some Special Functions

7. If 0 < x < π2, prove that 2

π< sin x

x< 1.

To see the first inequality, suppose ∃x0 ∈ (0, π2) with 2

π≥ sin x0

x 0. Since

limx→0sin x

x= 1, IVT gives an y with 2

π= sin y

y. Define g(x) = sin x − 2

πx so

g(y) = 0. Then g′(x) = cos(x) − 2π

and g′′(x) = − sin x < 0 for x in theinterval. By IVT again, there is a point z ∈ (0, y) with g′(z) = 0, so g′(x) < 0

for x ∈ (y, π2). Then g(y) = 0 implies that g(π

2) < 0. <↙

To see the second inequality, put f(x) = x− sin x. Then f ′(x) = 1− cos xshows f ′(0) = 0 and f ′(x) > 0 for 0 < x < π

2.

8. For n = 0, 1, 2, . . . and x ∈ R, prove | sin nx| ≤ n| sin x|.This is clearly true (with equality) for n = 0, 1, so induct on n; suppose

| sin nx| ≤ n| sin x|. We could use this assumption, to say

| sin(n + 1)x| = | sin(nx + x)| ≤ | sin nx|+ | sin x| = (n + 1)| sin x|,if we could prove the central inequality. From the definitions of C(x) and S(x)given in (46),

sin(x + y) = sin x cos y + cos x sin y.

Applying this,

| sin(nx + x)| ≤ | sin nx cos x|+ | cos nx sin x|≤ | sin nx|+ | sin x|,

since | cos x| ≤ 1 for all x.

9. (a) Put sN = 1 + 12

+ · · ·+ 1N

. Prove that γ = limN→∞(sN − log N) exists.Note that

γ = limN→∞

N∑

k=1

1

k− log N =

∫ ∞

1

(1

[x]− 1

x

)dx.

The following sum telescopes:

N∑

k=1

[1

k− (log(k + 1)− log k)

]= sN − log(N + 1),

so rewrite the summand as

ak :=1

k− log

(1 +

1

k

).

We will bound ak by 12x−2. Note that for

f(x) =x2

2− x + log(1 + x),

11

we have f ′(x) = x− 1+ 11+x

and f ′′(x) = 1 = 1(1+x)2

. This gives f ′(0) = 0

and f ′′(x) > 0 for all x > 0, so that f is always positive for x > 0 and

0 ≤ x− log(1 + x) ≤ x2

2.

In particular, this is true for x = 1k. Since

∑1k2 converges, this shows

limN→∞

N∑

k=1

ak = limN→∞

(sN − log(N + 1)

)

exists. Finally,

log(N + 1)− log N = log

(1 +

1

N

)N→∞−−−−−→ log 1 = 0

shows that limN→∞ (sN − log N) = 0.

(b) Roughly how large must m be such that N = 10m satisfies sN > 100?From the above, and problem #13, we have

0 ≤ sN − log N ≤ π2

6=⇒ log N ≤ sN ≤ log N +

π2

6.

Then m ≈ 43.43 will ensure log 10m > 100.

12. Suppose that f is periodic with f(x) = f(x + 2π), and for δ ∈ (0, π),

f(x) =

{1, |x| ≤ δ,

0, δ < |x| < π.

(a) Compute the Fourier coefficients of f .

cn =1

2π

∫ δ

−δ

e−inx dx =−1

2πin

(e−inδ − einδ

)=

sin nδ

πn.

(b) Conclude that

∞∑n=1

sin nδ

n=

π − δ

2, (0 < δ < π).

∑

n∈Z=

∑

n∈Z

sin nδ

πn

2∞∑

n=1

cn =1

π

∑

n∈Z

sin nδ

πn.

12

(c) Deduce from Parseval’s theorem that∞∑

n=1

sin2 nδ

n2δ=

π − δ

2.

Parseval’s theorem gives equality between

1

2π

∫ π

−π

|f(x)|2 dx =1

2π

∫ δ

−δ

1 dx =δ

π

and ∑

n∈Z|cn|2 =

∑

n∈Z

sin2 nδ

π2n2=

δ

π2

∑

n∈Z

sin2 nδ

n2δ.

Then ∞∑n=1

|cn|2 =1

2

(δ

π− c0

)=

Note that c0 = δπ

also.

(d) Let δ → 0 and prove that∫ ∞

0

(sin x

x

)2

dx =π

2.

(e) Put δ = π/2 in (c). What do you get?

2

π

∞∑n=0

sin2(π2n)

n2=

π

4=⇒

∞∑

k=0

1

(2k + 1)2=

π2

8.

13. Put f(x) = x for 0 ≤ x < 2π, and apply Parseval’s Theorem to conclude∞∑

n=1

1

n2=

π2

6.

Apply it to the 2π-periodic function f(x) = x on (−π, π) instead. Integra-tion by parts gives

cn =1

2π

∫ π

−π

xeinx dx =(−1)n

in,

so with Parseval’s theorem,

∑

n∈Z

1

n2=

∑

n∈Z|cn|2 =

1

2π

∫ π

−π

|f(x)|2 dx =π2

3.

Since we have |cn| = |c−n|, we find the desired series by subtracting

c0 =1

2π

∫ π

−π

x dx = 12[x2]π−π = 0

13


from each side, and dividing the remainder by 2.

∞∑n=1

1

n2=

1

2

(∑

n∈Z|cn|2 − c0

)=

1

2

(π2

3− 0

)=

π2

6.

9. Functions of Several Variables

5. Prove that to every A ∈ L(Rn,R) there corresponds a unique y ∈ Rn suchthat Ax = x · y.

For N = {x ∈ Rn ... Ax = 0}, N is a closed subspace by Problem 4. IfN = Rn, then Ax = 0 · x ∀x, and we’re done. So suppose N 6= Rn. Thenthere is a nonzero vector

x0 ∈ N⊥ = {u ∈ Rn ... u · v = 0,∀v ∈ N}.We check Ax = yA · x for yA defined by

yA =Ax0

‖x0‖2x0.

First, if x ∈ N , then Ax = 0 = yA · x. Next, if x = αx0, then

Ax = A(αx0) = αAx0 =Ax0

‖x0‖2x0 · αx0 = yA · αx0.

Since the functions A(x) and yA · x are linear functions of x and agree on Nand x0, they must agree on the space spanned by N and x0. But N and x0

span Rn, since every element y ∈ Rn can be written

y =

(y − Ay

Ax0

x0

)+

Ay

Ax0

x0.

Thus Ax = yA · x for all x ∈ Rn. If Ax = y′ · x also, then

‖y′ − yA‖2 = A(y′ − yA)− A(y′ − yA) = 0.

So y′ = yA is unique. Equality of the norms comes from the inequalities:

‖A‖ = sup‖x‖≤1

|Ax| = sup‖x‖≤1

|yA · x| ≤ sup‖x‖≤1

‖yA‖‖x‖ = ‖yA‖, and

‖A‖ = sup‖x‖≤1

|Ax| ≥∣∣∣∣A

(yA

‖yA‖)∣∣∣∣ = yA · yA

‖yA‖ = ‖yA‖.

6. If f(0, 0) = 0 and

f(x1, x2) =x1x2

x21 + x2

2

, for (x1, x2) 6= 0,

prove that D1f(x1, x2) and D2f(x1, x2) exist for every point (x1, x2) ∈ R2,although f is not continuous at (0, 0).

14

First, the derivatives are

D1f(x1, x2) =x2(x

22 − x2

1)

(x21 + x2

2)2

, and D2f(x1, x2) =x1(x

21 − x2

2)

(x21 + x2

2)2

,

and so clearly exist wherever (x, y) 6= (0, 0). To check the origin, consider thatalong the axes,

D1f(x1, 0) =0

x41

= 0, and D2f(0, x2) =0

x42

= 0.

However, for f to be continuous at (0, 0), we must have

f(0, 0) = lim(x,y)→(0,0)

f(x, y),

no matter how (x, y) → (0, 0)! Define γ(t) : R→ R2 by

γ(t) = (x(t), y(t)) = (t, t),

so that we approach the origin along the diagonal. Then

limt→0

f(x(t), y(t)) = limt→0

f(t, t) = limt→0

t2

2t2= lim

t→0

1

2=

1

26= 0.

7. Suppose that f is a R-valued defined in an open set E ⊆ Rn, and that thepartial derivatives D1f, . . . , Dnf are bounded in E. Prove that f is continuousin E.

Take m = 1 as in 9.21. Pick x ∈ E and consider the ball of radius rB(x, r) ⊆ E. Choose h ∈ Rn such that ‖h‖ < r. Define vk ∈ Rn by

vk = (h1, . . . , hk, 0, . . . , 0) for k = 1, . . . , n.

Then as in (42),

f(x + h)− f(x) =n∑

j=1

[f(x + vj)− f(x + vj−1)] .

Since |vk| < r for 1 ≤ k ≤ n, and since B = B(x, r) is convex, the segmentswith endpoints x + vj−1 and x + vj lie in B. Then vj = vj−1 + hjej, where

hjej = (0, . . . , 0, hj, 0, . . . , 0).

Then the Mean Value Theorem applies to the partials and shows that the jth

summand ishj(Djf)(x + vj−1 + tjhjej)

for some tj ∈ (0, 1). By hypothesis, we have M such that

|(Djf)(x)| ≤ M j = 1, . . . , n, ∀x ∈ E.

Applying this to the absolute value of the above difference,

|f(x + h)− f(x)| ≤n∑

j=1

|hj| ·M ≤ nM max{|hj|} ≤ nM‖h‖.

15


8. Suppose that f is a differentiable real function in an open set E ⊆ Rn, andthat f has a local maximum at a point x ∈ E. Prove that f ′(x) = 0.

Let {ej} be the standard basis vectors of Rn, and define

ϕj(t) = f(x0 + tej), for t ∈ R.

Then ϕj : R→ R is differentiable, so by Thm. 5.8,

ϕ′j(0) = (Djf)(x0) = 0.

But the partial derivatives (Djf)(x0) are precisely the columns of the matrix(Df)(x0) = f ′(x).

9. If f is a differentiable mapping of a connected open set E ⊆ Rn into Rm, andin f ′(x) = 0 for every x ∈ E, prove f is constant in E.

First, suppose m = 1. Pick some x0 ∈ E and define

C = {x ∈ E ... f(x) = f(x0)},D = {x ∈ E ... f(x) 6= f(x0)} = E \ C.

Clearly, C ∪ D = E and C ∩ D = ∅. If we can show each of C, D is open,then D must be empty, or else we’d have a disconnection of the connected setE.

E is open, so for any x ∈ C, pick ε > 0 such that B(x, ε) ⊆ E. Thenfor any y ∈ B(x, ε), the segment [x, y] ⊆ B(x, ε). Since f ′(x) = 0, we havef(x) = f(y) by Thm. 5.11(b). Also, we chose x ∈ C, so f(x) = f(y) = f(x0).This puts B(x, ε) ⊆ C and shows C is open.

Now suppose we have some x ∈ E \C. Since E is open, we can again chooseε > 0 such that B(x, ε) ⊆ E. By the exact same argument, we get

y ∈ B(x, ε) =⇒ f(x) = f(y) 6= f(x0),

so that B(x, ε) ⊆ D and D is open.Finally, if m > 1, then apply this argument to each component of f .

10. Suppose f is a differentiable mapping of R into R3 such that |f(t)| = 1 forevery t. Prove f ′(t) · f(t) = 0. Interpret this geometrically.

Since ‖f(t)‖ = 1, we have

(f1(t))2 + (f2(t))

2 + (f3(t))2 = 1.

Differentiating both sides,

2f1(t)f′1(t) + 2f2(t)f

′2(t) + 2f3(t)f

′3(t) = 0

f(t) · f ′(t) = 0.

This means that for any curve on the unit sphere, the tangent at p ∈ S1 isorthogonal to p, i.e., the surface of a sphere is orthogonal to the radius at anypoint.

16


14. Define f(0, 0) = 0 and f(x, y) = x3/(x2 + y2) for (x, y) 6= (0, 0).

-2-1 0 1 2

-2-1

012

-2

-1

0

1

2

-1 0 1 2

-2-1

012

(a) Prove that D1f and D2f are bounded functions in R2 so that f is con-tinuous.We have the partial derivatives

D1f =x4 + 3x2y2

x4 + 2x2y2 + y4, and

D2f =−2x3y

x4 + 2x2y2 + y4.

Boundedness at ±∞:

|D1f | ≤ 3(x4 + 3x2y2)

x4 + 2x2y2= 3, as |x| → ∞, and

|D2f | = |2x3||x4/y + 2x2y + y3|

|y|→∞−−−−−→ 0.

Boundedness away from ∞: only need to check zeroes of the denomina-tors, and there is only (0, 0).

D1f(0, 0) = limx→0

f(x, 0)− f(0, 0)

xlimx→0

1 = 1, and

D2f(0, 0) = limy→0

f(0, y)− f(0, 0)

ylimy→0

0 = 0.

Thus f is continuous by Prob. 7.

(b) Let u ∈ R2, |u| = 1. Show (Duf)(0, 0) exists and |(Duf)(0, 0)| ≤ 1.The directional derivative is

Duf(0, 0) = limt→0

f [(0, 0) + tu]− f(0, 0)

t= lim

t→0

1

t· t3x3

t2(x2 + y2)

=x3

x2 + y2,

17


for u = (x, y). Since |u| = x2 + y2 = 1, this implies |x| ≤ 1. Hence wehave

Duf(0, 0) = x3 and |Duf(0, 0)| = |x3| ≤ 1.

(c) Let γ be a differentiable mapping of R into R2 with γ(0) = (0, 0) and|γ′(0)| > 0. Put g(t) = f(γ(t)) and prove that g is differentiable for everyt ∈ R.Put γ(t) = (x(t), y(t)) so that

g(t) =x(t)3

x(t)2 + y(t)2, for t 6= 0.

Since γ is differentiable for t 6= 0, we have x(t), y(t) differentiable for t 6= 0,and hence g(t) is differentiable for t 6= 0, by the chain rule, Thm. 9.15.1

It remains to check that g is differentiable at the origin.

g′(0) = limh→0

g(h)− g(0)

h

= limh→0

1

h· x(h)3

x(h)2 + y(h)2

= limh→0

x(h)3

h3· h2

x(h)2 + y(h)2

=

(limh→0

x(h)

h

)3 (limh→0

h2

x(h)2 + y(h)2

)

= (x′(0))3 · 1

|γ′(0)|2 .

The last two lines are easiest to see by working backwards.

Also show γ ∈ C ′ =⇒ g ∈ C ′.Note that g′(t) = f ′(γ(t))γ′(t). Since the additional hypothesis is thatγ′(t) is continuous (which is equivalent to saying x(t), y(t) are continuous,by Thm. 4.10), we just need that f ′(γ(t)) is continuous. Since the chainrule gives

g′(t) =x(t)2 (x(t)2x′(t) + 3y(t)2x′(t)− 2x(t)y(t)y′(t))

(x(t)2 + y(t)2)2 ,

it is clear that g′(t) is continuous whenever x, y are not simultaneously 0.For t = 0 (or other t0 s.t. x(t0) = y(t0) = 0), replace x(t) by x′(0)t + o(t)(using Taylor’s Thm.) and similarly replace y(t) by y′(0)t + o(t). Thus

g′(t) ≈ (x′(0)t+o(t))2((x′(0)t+o(t))2x′(t)+3(y′(0)t+o(t))2x′(t)−2(x′(0)t+o(t))(y′(0)t+o(t))y′(t))((x′(0)t+o(t))2+(y′(0)t+o(t))2)2

,

1By (a), the partials of f exist and are continuous in an open nbd of (0, 0). Hence, f is continuouslydifferentiable at (0, 0) by Thm. 9.21.

18


so that g′(t) t→0−−−−→ g′(0).

(d) In spite of this, f is not differentiable at (0, 0).Let u = (x, y) be a unit vector. By Rudin’s (40),

Du = D1f(0, 0)x + D2f(0, 0)y = 1 · x + 0 · y = x.

But from (b) we have Du = x3 6= x for any x other than −1, 0, 1.

(Bonus problem) This problem isn’t in Rudin.Define f : R2 → R by

f(x, y) =

{0, (x, y) = (0, 0),

x2yx4+y2 , otherwise.

Let γa be any straight line through the origin with slope a and γa(0) = (0, 0). Thismeans that (up to parametrization) γa(t) = (t, at) for some a ∈ R, or γ∞(t) = (0, t).Show that

limt→0

f(γa(t)) = f(0, 0) = 0,

but that f is not continuous at the origin.

-1

-0.5

0

0.5

1-1

0

1-0.5-0.2500.250.5

-1

-0.5

0

0.5

1

-0.5-0.2500.25

Continuity of the restriction follows by a basic computation:

limt→0

f(γa(t)) = limt→0

at3

t4 + (at)2= lim

t→0

at

t2 + a2= 0, and

limt→0

f(γ∞(t)) = limt→0

0

0 + t2= 0,

To see the other part, approach the origin along a parabolic curve: let γ(t) =(x(t), y(t)) = (t, t2). Then

limt→0

f(γ(t)) = limt→0

t2t2

t4 + (t2)2= lim

t→0

t4

2t4= lim

t→0

1

2=

1

26= 0.

19

15. Define f(x, y) = x2 + y2 − 2x2y − 4x6y2

(x4+y2)2.

-2-101

2

-2-1

01

2

-10

0

10

20

-10

0

10

(a) Prove that 4x4y2 ≤ (x4 + y2)2, so that f ∈ C0.

(x4 + y2)2 − 4x4y2 = x8 + 2x4y2 + y4 − 4x4y2

= (x4 − y2)2 ≥ 0

Now f(x, y) = x2 + y2 − 2x2y − x2ϕ(x, y), where 0 ≤ ϕ(x, y) ≤ 1. Thisgives f continuous at (0, 0).

(b) For 0 ≤ θ < π, t ∈ R, let gθ(t) = f(t cos θ, t sin θ). Show that gθ(0) = 0,g′θ(0) = 0, g′′θ (0) = 2. Each gθ thus has a strict local minimum at t = 0,i.e., on each line through (0, 0), f has a strict local minimum at (0, 0).

Since sin θ, cos θ are bounded, gθ(0) = f(0, 0) = 0 by (a). By somenightmarishly tortuous (but elementary) calculation, get the other tworesults.

(c) Show that (0, 0) is not a local minimum of f , since f(x, x2) = −x4.

Substituting in y = x2:

f(x, x2) = x2 + x4 − 2x4 − 4x10

(2x4)2

= x2 − x4 − x2

= −x4.

16. Define f(t) = t+2t2 sin(

1t

), f(0) = 0. Then f ′(0) = 1, f ′ bounded in (−1, 1),

but f is not injective in any nbd of 0.

20


-0.2 -0.1 0.1 0.2

-0.2

-0.1

0.1

0.2

First note that f ′(t) = 1 − 2 cos(

1t

)+ 4t sin

(1t

)for t 6= 0. Since sin t, cos t

are bounded by 1, we have

|f ′(t)| ≤ 1 + 4 + 2 = 7, for t ∈ (−1, 1) \ {0}.At t = 0, we have that

f(t)− f(0)

t=

t + 2t2 sin 1t

t= 1 + 2t sin 1

t

shows f ′(0) = 1 by the Sandwich theorem applied to t + 2t2 and t − 2t2. Sof ′ is bounded in all of (−1, 1).

Define sn = 12nπ

for n = 1, 2, . . . , so that

f ′(sn) = 1− 2 cos(2nπ) + 2sin 2nπ

nπ= 1− 2 = −1.

Then define tn = 2(4n+3)π

for n = 1, 2, . . . , so that

f ′(tn) = 1− 2 cos (4n+3)π2

+ 8(4n+3)π

sin (4n+3)π2

= 1 + 8(4n+3)π

> 0.

Since f ′ changes sign between each successive sn and tn, and since sn, tn → 0,f fails to be injective in every neighbourhood of 0.

17. Let f = (f1, f2) : R2 → R2 be given by

f1(x, y) = ex cos y, f2(x, y) = ex sin y.

(a) What is the range of f?

Note that if we identify (0, 1) =√−1 = i, then

f(z) = f(x + iy) = ez = ex+iy = ex cos y + iex sin y.

But you only really need to see that (ex cos y, ex sin y) is polar coordinatesfor a point with radius ex and argument y, to see that the range is anypoint of R2 with radius r > 0, i.e., R2 \ (0, 0).

21


(b) Show that the Jacobian of f is nonzero. Thus, every point of R2 has aneighbourhood in which f is injective. However, f is not injective.

The Jacobian is the determinant of partials:∣∣∣∣ex cos y −ex sin yex sin y ex cos x

∣∣∣∣ = e2x(cos2 x + sin2 x) = e2x 6= 0.

But f is not injective, since f(x, y) = f(x, y + 2nπ), for n ∈ Z.

(c) Put a = (0, π3) and b = f(a). Let g be the continuous inverse of f ,

defined in a neighbourhood of b, such that g(b) = a. Find an explicitformula for g, compute f ′(a) and g′(b), and verify Rudin’s (52).

For u = ex cos y, v = ex sin y, one verifies that

x = 12log(u2 + v2) = log r, for r = ex, and

y = tan−1 vu

= arg θ, for θ = vu.

For the derivatives,

f ′(a) =

[12

−√

32√

32

12

],

and

g′(u, v) = 1u2+v2

[u v

−v u

]=⇒ g′(b) =

[12

√3

2

−√

32

12

],

Finally,

f ′(g(u, v)) = f ′(log

√u2 + v2, tan−1 v

u

)

=√

u2 + v2

[cos

(tan−1 v

u

) − sin(tan−1 v

u

)

sin(tan−1 v

u

)cos

(tan−1 v

u

)]

=√

u2 + v2

[ u√u2+v2 − v√

u2+v2

v√u2+v2 − u√

u2+v2

]

=

[u −vv u

]

= [g′(u, v)]−1

(d) What are the images under f of lines parallel to the coordinate axes?

Lines parallel to the x-axis are mapped to straight lines through theorigin, parameterized exponentially. Lines parallel to the y-axis are cir-cles about the origin of radius ex, parameterized with constant speed.Diagonal lines γ(t) = (at + b, btc) will get mapped to (et cos t, et sin t),counterclockwise logarithmic spirals emanating from the origin.

22

PRINCIPLES OF MATHEMATICAL ANALYSIS.

WALTER RUDIN

10. Integration of Differential Forms

1. Let H be a compact convex set in Rk with nonempty interior. Let f ∈ C(H), putf(x) = 0 in the complement of H, and define

∫

Hf as in Definition 10.3.

Prove that∫

Hf is independent of the order in which the k integrations are carried

out. Hint: Approximate f by functions that are continuous on Rk and whose supportsare in H, as was done in Example 10.4.

2. For i = 1, 2, 3, . . . , let ϕi ∈ C(R1) have support in (2−i, 21−i), such that∫ϕi = 1. Put

f(x, y) =

∞∑

i=1

[ϕi(x) − ϕi+1(x)]ϕi(y).

Then f has compact support in R2, f is continuous except at (0, 0), and∫

f(x, y) dx dy = 0 but

∫

f(x, y) dy dx = 1.

Observe that f is unbounded in every neighbourhood of (0, 0).

3. (a) If F is as in Theorem 10.7, put A = F′(0), F1(x) = A−1F(x). Then F′1(0) = I.

Show that

F1(x) = Gn◦Gn−1◦. . .◦G1(x)

in some neighbourhood of 0, for certain primitive mappings G1, . . . ,Gn. Thisgives another version of Theorem 10.7:

F(x) = F′(0)Gn◦Gn−1◦. . .◦G1(x).

(b) Prove that the mapping (x, y) 7→ (y, x) of R2 onto R2 is not the composition ofany two primitive mappings, in any neighbourhood of the origin. (This showsthat the flips Bi cannot be omitted from the statement of Theorem 10.7.)

4. For (x, y) ∈ R2, define

F(x, y) = (ex cos y − 1, ex sin y).

May 20, 2005. Solutions by Erin P. J. Pearse.

1


� Prove that F = G2◦G1, where

G1(x, y) = (ex cos y − 1, y)

G2(u, v) = (u, (1 + u) tan v)

are primitive in some neighbourhood of (0, 0).

G2◦G1(x, y) = G2(ex cos y − 1, y)

=

(

ex cos y − 1, (ex cos y)sin y

cos y

)

= (ex cos y − 1, ex sin y)

= F(x, y).

G1(x, y) = (g1(x, y), y) and G2(u, v) = (u, g2(u, v) are clearly primitive.

� Compute the Jacobians of G1,G2,F at (0, 0).

JG1(0) = detG′

1(x)∣∣∣x=0

=

∣∣∣∣

ex cos y −ex sin y0 1

∣∣∣∣(x,y)=(0,0)

= ex cos y∣∣∣(x,y)=(0,0)

= 1 · 1 = 1.

JG2(0) =

∣∣∣∣

1 0tan v (1 + u) sec2 v

∣∣∣∣(x,y)=(0,0)

= (1 + u) sec2 v∣∣∣(x,y)=(0,0)

= 1 · 12 = 1.

JF(0) =

∣∣∣∣

ex cos y −ex sin yex sin y ex cos y

∣∣∣∣(x,y)=(0,0)

= e2x cos2 y + e2x sin2 y∣∣∣(x,y)=(0,0)

= e2x(cos2 y + sin2 y

)∣∣∣(x,y)=(0,0)

= e2x∣∣∣(x,y)=(0,0)

= 1.

� DefineH2(x, y) = (x, ex sin y)

and findH1(u, v) = (h(u, v), v)

2


so that F = H1◦H2 in some neighbourhood of (0, 0).We have

H1◦H2(x, y) = H1(x, ex sin y)

= (h(x, ex sin y), ex sin y)

= (ex cos y − 1, ex sin y)

for h(u, v) = eu cos (arcsin(e−uv)) − 1, valid in the neighbourhood R ×(−π

2, π

2

)

of (0, 0). Indeed, for (u, v) = (x, ex sin y),

eu cos(arcsin(e−uv)

)− 1 = ex cos

(arcsin(e−xex sin y)

)− 1

= ex cos (arcsin(sin y)) − 1

= ex cos y − 1.

5. Formulate and prove an analogue of Theorem 10.8, in which K is a compact subsetof an arbitrary metric space. (Replace the functions ϕi that occur in the proof ofTheorem 10.8 by functions of the type constructed in Exercise 22 of Chap. 4.)

3


6. Strengthen the conclusion of Theorem 10.8 by showing that the functions ψi can bemade differentiable, even smooth (infinitely differentiable). (Use Exercise 1 of Chap.8 in the construction of the auxiliary functions ϕi.)

7. (a) Show that the simplex Qk is the smallest convex subset of Rk that contains thepoints 0 = e0, e1, . . . , ek.

We first show that Qk is the convex hull of the points 0, e1, . . . , ek, i.e.,

x ∈ Qk ⇐⇒ x =

k∑

i=0

ciei, with 0 ≤ ci ≤ 1,

k∑

k=0

ci = 1.

(⇒) Consider x = (x1, . . . , xk) ∈ Qk. Let ci = xi for 1 = 1, . . . , k and define

c0 = 1 −∑k

i=1 xi. Then ci ∈ [0, 1] for i = 0, 1, . . . , k and∑k

i=0 ci = 1. Moreover,

k∑

i=0

ciei =

(

1 −k∑

i=1

xi

)

e0 +

k∑

i=1

ciei

=

(

1 −k∑

i=1

xi

)

0 +k∑

i=1

xiei = (x1, . . . , xk) = x.

(⇐) Now given x as a convex combination x =∑k

i=0 ciei, write x = (c1, . . . , ck).Then 0 ≤ ci ≤ 1 and

k∑

i=0

ci = 1 =⇒k∑

i=1

ci ≤ 1,

so x ∈ Qk.Now if K is a convex set containing the points 0, e1, . . . , ek, convexity implies itmust also contain all points x =

∑ki=0 ciei, i.e., it must contain all of Qk.

(b) Show that affine mappings take convex to convex sets.

An affine mapping T may always be represented as the composition of a linearmapping S followed by a translation B. Translations (as congruences) obviouslypreserve convexity, so it suffices to show that linear mappings do.Let S be a linear mapping. If we have a convex set K with u,v ∈ K, then anypoint between u,v is (1 − λ)u + λv for some λ ∈ [0, 1], and in the image of S,the linearity of S gives

S ((1 − λ)u + λv) = (1 − λ)S(u) + λS(v),

which shows that any point between u and v gets mapped to a point betweenS(u) and S(v), i.e., convexity is preserved.

4


8. Let H be the parallelogram in R2 whose vertices are (1, 1), (3, 2), (4, 5), (2, 4).

� Find the affine map T which sends (0, 0) to (1, 1), (1, 0) to (3, 2), and (0, 1) to(2, 4).

Since T = B◦S, where S is linear and B is a translation, let B(x) = x + (1, 1)and find S which takes I2 to (0, 0), (2, 1), (3, 4), (1, 3). Once we fix a basis (let’suse the standard basis in Rk), then S corresponds to a unique matrix A. Thenwe can define S in terms of its action on the basis of R2:

A =

| |S(e1) S(e2)

| |

=

[2 11 3

]

.

Then S(e1) = S(1, 0) = (2, 1)T , and S(e2) = S(0, 1) = (1, 3)T , as you can check.Now for b = (1, 1), we have

T (x) = B◦S(x) = Ax + b =

[2 11 3

] [x1

x2

]

+

[11

]

=

[2x1 + x2 + 1x1 + 3x2 + 1

]

.

� Show that JT = 5.

Let T1(x1, x2) = 2x1 + x2 + 1 and T2(x1, x2) = x1 + 3x2 + 1.

JT =

∣∣∣∣∣∣

ddx1T1

ddx2T1

ddx1T2

ddx2T2

∣∣∣∣∣∣

=

∣∣∣∣

2 11 3

∣∣∣∣= 6 − 1 = 5.

� Use T to convert the integral

α =

∫

H

ex−y dx dy

to an integral over I2 and thus compute α.

T satisfies all requirements of Thm 10.9, so we have

α =

∫

H

ex−y dy =

∫

I2

e(2x1+x2+1)−(x1+3x2+1)|5| dx

= 5

∫ 1

0

∫ 1

0

ex1−2x2 dx1 dx2

= 5

(∫ 1

0

ex1 dx1

)(∫ 1

0

e−2x2 dx2

)

= 5 [ex1 ]10[−1

2e−2x2

]1

0

= 5(e1 − 1)(−12e−2 + 1

2)

= 52(e1 − 1)(1 − e−2)

= 52(e1 − 1 − e−1 + e−2).

5


9. Define (x, y) = T (r, θ) on the rectangle

0 ≤ r ≤ a, 0 ≤ θ ≤ 2π

by the equationsx = r cos θ, y = r sin θ.

Show that T maps this rectangle onto the closed disc D with center at (0, 0) andradius a, that T is one-to-one in the interior of the rectangle, and that JT (r, θ) = r.If f ∈ C(D), prove the formula for integration in polar coordinates:

∫

D

f(x, y) dx dy =

∫ a

0

∫ 2π

0

f(T (r, θ)) rdr dθ.

Hint: Let D0 be the interior of D, minus the segment from (0, 0) to (0, a). As itstands, Theorem 10.9 applies to continuous functions f whose support lies in D0. Toremove this restriction, proceed as in Example 10.4.

6

12. Let Ik be the unit cube and Qk be the standard simplex in Rk; i.e.,

Ik = {u = (u1, . . . , uk) ... 0 ≤ ui ≤ 1, ∀i}, and

Qk = {x = (x1, . . . , xk) ... xi ≥ 0,∑xi ≤ 1.}

Define x = T (u) by

x1 = u1

x2 = (1 − u1)u2

...

xk = (1 − u1) . . . (1 − uk−1)uk.

The unit cube Ik The standard simplex Qk

Figure 1. The transform T , “in slow motion”.

� Show thatk∑

i=1

xi = 1 −k∏

i=1

(1 − ui).

This computation is similar to (28)–(30), in Partitions of Unity, p.251.By induction, we begin with the (given) basis step x1 = u1, then assume

j∑

i=1

xi = 1 −j∏

i=1

(1 − ui), for some j < k.

Using this assumption immediately, we can write∑j+1

i=1xi =∑j

i=1xi + xj+1

=(

1 −∏ji=1(1 − ui)

)

+(

(1 − u1) . . . (1 − uj−1)uj

)

= 1 − (1 − u1) . . . (1 − uj) + (1 − u1) . . . (1 − uj)uj+1

= 1 −(

(1 − u1) . . . (1 − uj) − (1 − u1) . . . (1 − uj)uj+1

)

= 1 − (1 − u1) . . . (1 − uj)(1 − uj+1)

= 1 −∏j+1

i=1 (1 − ui).

7


� Show that T maps Ik onto Qk, that T is 1-1 in the interior of Ik, and that itsinverse S is defined in the interior of Qk by u1 = x1 and

ui =xi

1 − x1 − · · · − xi−1

for i = 1, 2, . . . , k.

Note that T is a continuous map (actually, smooth) because each of its compo-nent functions is a polynomial.

First, we show that T (∂Ik) = ∂Qk, so that the boundary of Ik is mapped ontothe boundary of Qk. Injectivity and surjectivity for the interior will be given bythe existence of the inverse function.

Let u ∈ ∂Ik. Then for at least one uj, either uj = 0 or uj = 1. uj = 0 impliesxj = 0 so that x ∈ Ej ⊆ ∂Qk. Meanwhile, uj = 1 implies

∑xi = 1−

∏(1−ui) =

1 − 0 = 1, which puts x ∈ E0 ⊆ ∂Qk. Therefore, T (∂Ik) ⊆ ∂Qk.

Let x ∈ ∂Qk. Then either xj = 0 for some 1 ≤ j ≤ k (because x ∈ Ej) or else∑xj = 1 (because x ∈ E0), or both.

case (i) Let xj = 0 so x = (x1, . . . , 0, . . . , xk), and take xε = (x1, . . . , ε, . . . , xk),so xε ∈ (Qk)◦. We have

T−1(xε) =(

x1,x2

1−x1, x3

1−x1−x2, . . . ,

xj

1−x1−···−xj−1,

xj+1

1−x1−···−xj, . . . , xk

1−x1−···−xk−1

)

=(

x1,x2

1−x1, x3

1−x1−x2, . . . , ε

1−x1−···−xj−1,

xj+1

1−x1−···−xj−1−ε, . . . , xk

1−x1−···−xk−1

)

ε→0−−−−→(

x1,x2

1−x1, x3

1−x1−x2, . . . , 0,

xj+1

1−x1−···−xj−1, . . . , xk

1−x1−···−xk−1

)

where the last step uses the continuity guaranteed by the Inverse FunctionTheorem. Now check that for

u =(

x1,x2

1−x1, x3

1−x1−x2, . . . , 0,

xj+1

1−x1−···−xj−1, . . . , xk

1−x1−···−xk−1

)

,

we do in fact have T (u) = x:

T(

x1,x2

1−x1, x3

1−x1−x2, . . . , 0,

xj+1

1−x1−···−xj−1, . . . , xk

1−x1−···−xk−1

)

=(

x1, (1 − x1)x2

1−x1, (1 − x1)(1 − x2

1−x1) x3

1−x1−x2,

. . . , (1 − x1)(1 − x2

1−x1) . . . (1 − xj−2

1−x1−···−xj−3)

xj−1

1−x1−···−xj−2,

(1 − x1)(1 − x2

1−x1) . . . (1 − xj−1

1−x1−···−xj−2)

xj

1−x1−···−xj−1,

(1 − x1)(1 − x2

1−x1) . . . (1 − xj

1−x1−···−xj−1)

xj+1

1−x1−···−xj,

. . . , (1 − x1)(1 − x2

1−x1) . . . (1 − xk−1

1−x1−···−xk−2) xk

1−x1−···−xk−1

)

= (x1, x2, x3, . . . xj−1, 0, (1 − x1 − x2 − · · · − xj−1)xj+1

1−x1−···−xj, . . . , xk)

= (x1, x2, x3, . . . xj−1, 0, xj+1, . . . , xk)

8

case (ii)∑k

i=1 xi = 1.

Then by the previous part, 1 −∏k

i=1(1 − ui) = 1, so∏k

i=1(1 − ui) = 0.This can only happen if one of the factors is 0, i.e., if one of the ui is 1.This puts u ∈ ∂Ik.

For the interior, we have an inverse.

S◦T (u) = S(u1, (1 − u1)u2, . . . , (1 − u1)(1 − u2) . . . uk)

=

(

u1,(1 − u1)u2

1 − u1

, . . . ,(1 − u1)(1 − u2) . . . (1 − uk−1)uk

1 − u1 − (1 − u1)u2 − · · · − (1 − u1)(1 − u2) . . . uk−1

)

= (u1, u2, . . . , uk)

= u.

where the third equality follows by successive distributions against the last factor:

[(1 − u1)(1 − u2) . . . (1 − uk−2)] (1 − uk−1)

= (1 − u1)(1 − u2) . . . (1 − uk−2) − (1 − u1)(1 − u2) . . . (1 − uk−2)uk−1.

Injectivity may also be shown directly as follows: let u and v be distinct pointsof (Ik)◦, with T (u) = x and T (v) = y. We want to show x 6= y. From theprevious part, we have T−1(∂Qk) ⊆ ∂Ik, so x,y ∈ (Qk)◦. Let the jth coordinatebe the first one in which u differs from v, so that ui = vi for i < j, but uj 6= vj.Then

xj = (1 − u1) . . . (1 − uj−1)uj

= (1 − v1) . . . (1 − vj−1)uj

6= (1 − v1) . . . (1 − vj−1)vj = yj.

So x 6= y.

� Show that

JT (u) = (1 − u1)k−1(1 − u2)

k−2 . . . (1 − uk−1),

and

JS(x) = [(1 − x1)(1 − x1 − x2) . . . (1 − x1 − · · · − xk−1)]−1 .

JT (u) =

∣∣∣∣∣∣∣∣∣∣

1 0 0 . . . 0−u2 1 − u1 0 . . . 0

−(1 − u2)u3 −(1 − u1)u3 (1 − u1)(1 − u2) . . . 0...

......

...−(1 − u2) . . . uk . . . . . . . . . (1 − u1) . . . (1 − uk−1)

∣∣∣∣∣∣∣∣∣∣

= 1 · (1 − u1) · (1 − u1)(1 − u2) · · · · · (1 − u1)(1 − u2) . . . (1 − uk−1)

= (1 − u1)k−1(1 − u2)

k−2 . . . (1 − uk−1).

9


JS(u) =

∣∣∣∣∣∣∣∣∣∣∣

1 0 0 . . . 0x2

(1−x1)−2

11−x1

0 . . . 0x3

(1−x1−x2)−2

x3

(1−x1−x2)−2

11−x1−x2

. . . 0...

......

...xk

(1−x1−···−xk−1)−2

xk

(1−x1−···−xk−1)−2

xk

(1−x1−···−xk−1)−2 . . . 11−x1−···−xk−1

∣∣∣∣∣∣∣∣∣∣∣

= 1 · 11−x1

· 11−x1−x2

· · · · · 11−x1−···−xk−1

= [(1 − x1)(1 − x1 − x2) . . . (1 − x1 − · · · − xk−1)]−1

10


13. Let r1, . . . rk be nonnegative integers, and prove that∫

Qk

xr1

1 . . . xrk

k dx =r1! . . . rk!

(k + r1 + · · · + rk)!.

Hint: Use Exercise 12, Theorems 10.9 and 8.20. Note that the special case r1 = · · · =rk = 0 shows that the volume of Qk is 1/k!.

Theorem 8.20 gives that for α, β > 0 we have∫ 1

0

tα−1(1 − t)β−1 dt =Γ(α)Γ(β)

Γ(α + β).

Theorem 10.9 says when T is a 1-1 mapping of an open set E ⊆ Rk into Rk withinvertible Jacobian, then

∫

Rk

f(y) dy =

∫

Rk

f(T (x))|JT (x)| dx

for any f ∈ C(E). Since we always have∫

cl(E)f =

∫

int(E)f, we can apply this theorem

to the T given in the previous exercise:∫

Qk

xr1

1 . . . xrk

k dx =

∫

Ik

T1(u)r1T2(u)r2 . . . Tk(u)rk |JT (u)|, du

=

∫

Ik

ur1

1 [(1 − u1)u2]r2 . . . [(1 − u1) . . . (1 − uk−1)uk]

rk

·[(1 − u1)

k(1 − u2)k−1 . . . (1 − uk−1)

]du

=

∫

Ik

ur1

1 ur2

2 . . . urk

k (1 − u1)r2+···+rk+k−1(1 − u2)

r3+···+rk+k−2

. . . (1 − uk−2)rk−1+rk+2(1 − uk−1)

rk+1 du

=

(∫

I

ur1

1 (1 − u1)r2+···+rk+k−1 dx1

)(∫

I

ur2

2 (1 − u2)r3+···+rk+k−2 dx1

)

. . .

(∫

I

urk−1

k−1 (1 − uk−1)rk+1 dxk

)(∫

I

urk

k dxk

)

.

Now we can use Theorem 8.20 to evaluate each of these, using the respective values

αj = 1 + rj, β = rj+1 + · · · + rk + k − j + 1,

so that, except for the final factor∫

I

urk

k dxk =

[urk+1

k

rk + 1

]1

0

=1

rk + 1,

the above product of integrals becomes a product of gamma functions:

= Γ(1+r1)Γ(r2+...rk+k)Γ(1+r1+r2+···+rk+k)

· Γ(1+r2)Γ(r3+...rk+k−1)Γ(r2+r3+···+rk+k)

· Γ(1+r3)Γ(r4+...rk+k−2)Γ(r3+r4+···+rk+k−1)

· · · · Γ(1+rk−1)Γ(rk+2)

Γ(1+2+rk−1+rk)· 1

rk+1

then making liberal use of Γ(1 + rj) = rj! from Theorem 8.18(b), we have

=r1!...rk−1!

(r1+...rk+k)!· Γ(r2+···+rk+k)

Γ(r2+···+rk+k)· Γ(r3+···+rk+k−1)

Γ(r3+···+rk+k−1). . . Γ(rk+2)

rk+1.

11


Now making use of xΓ(x) = Γ(x + 1) from Theorem 8.19(a), to simplify the finalfactor, we have

Γ(rk+2)rk+1

= Γ((rk+1)+1)rk+1

= (rk+1)Γ(rk+1)rk+1

= Γ(rk + 1) = rk!,

so that the final formula becomes∫

Qk

xr1

1 . . . xrk

k dx = r1!...rk!(r1+...rk+k)!

.

Note: if we take rj = 0, ∀j, then this formula becomes

∫

Qk

1dx =0! . . . 0!

(0 + . . . 0 + k)!=

1

k!,

showing that the volume of Qk is 1k!

.

15. If ω and λ are k- and m-forms, respectively, prove that ω ∧ λ = (−1)kmλ ∧ ω.

Write the given forms as

ω =∑

ai1,...,ik(x) dxi1 ∧ . . . ∧ dxik =∑

IaIdxI

and

λ =∑

bj1,...,jk(x) dxj1 ∧ . . . ∧ dxjm

=∑

JaJdxJ .

Then we have

ω ∧ λ =∑

I,J

aI(x) bJ(x) dxI ∧ dxJ and λ ∧ ω =∑

I,J

aI(x) bJ(x) dxJ ∧ dxI ,

so it suffices to show that for every summand,

dxI ∧ dxJ = (−1)kmdxJ ∧ dxI .

Making excessive use of anticommutativity (Eq. (46) on p.256),

dxI ∧ dxJ = dxi1 ∧ . . . ∧ dxik ∧ dxj1 ∧ . . . ∧ dxjm

= (−1)kdxj1 ∧ dxi1 ∧ . . . ∧ dxik ∧ dxj2 ∧ . . . ∧ dxjm

= (−1)2kdxj1 ∧ dxj2 ∧ dxi1 ∧ . . . ∧ dxik ∧ dxj3 ∧ . . . ∧ dxjm

...

= (−1)mkdxj1 ∧ . . . ∧ dxjm∧ dxi1 ∧ . . . ∧ dxik

= (−1)kmdxJ ∧ dxI .

12

16. If k ≥ 2 and σ = [p0,p1, . . . ,pk] is an oriented affine k-simplex, prove that ∂2σ = 0,directly from the definition of the boundary operator ∂. Deduce from this that∂2Ψ = 0 for every chain Ψ.

Denote σi = [p0, . . . ,pi−1,pi+1, . . . ,pk] and for i < j, let σij be the (k−2)-simplexobtained by deleting pi and pj from σ.

Now we use Eq. (85) to compute

∂σ =k∑

j=0

(−1)jσj.

Now we apply (85) again to the above result and get

∂2σ = ∂(∂σ) = ∂

(k∑

j=0

(−1)jσj

)

=

k∑

j=0

(−1)j∂σj

=

k∑

j=0

(−1)j

k−1∑

i=0

(−1)iσij

=k∑

j=0

k−1∑

i=0

(−1)i+jσij

Lemma. For i < j, σij = σj−1,i.

Proof. These both correspond to removing the same points pi, pj, just in differentorders (because pj is the (j − 1)th vertex of σi).

σ = [p0, ...,pi, ...,pj, ...,pk] 7→ σj = [p0, ...,pi, ...,pk] 7→ σij = [p0, ...,pk]

σ = [p0, ...,pi, ...,pj, ...,pk] 7→ σi = [p0, ...,pj, ...,pk] 7→ σj−1,i = [p0, ...,pk].

�

Thus we split the previous sum along the “diagonal”:

∂2σ =k−1∑

i≥j=0

(−1)i+jσij

︸︷︷︸

A

+k∑

i<j=1

(−1)i+jσij

︸︷︷︸

B

.

Continuing on with B for a bit, we have

B =k∑

i<j=1

(−1)i+jσj−1,i by lemma

=k∑

j=1

j−1∑

i=0

(−1)i+jσj−1,i rewriting the sum

=

k∑

i=1

i−1∑

j=0

(−1)i+jσi−1,j swap dummies i↔ j

13

...

...

...

i0

0

1

1

2

2

k-1

k-1

j

Figure 2. Reindexing the double sum over i, j.

=

k−1∑

i=0

i∑

j=0

(−1)i+j+1σi,j reindex i 7→ i+ 1

= (−1)

k−1∑

j=1

k−1∑

i=j

(−1)i+jσi,j see Figure 2

= −A.Whence

∂2σ = A+B = A− A = 0.

In order to see how this works with the actual faces, define F ik : Qk−1 7→ Qk for

i = 0, . . . , k − 1 to be the affine map F ik = (e0, . . . , ei, . . . ek), where ei means omit

ei, i.e.,

F ik(ej) =

{

ej, j < i

ej+1, j ≥ i.

Then define the ith face of σ to be σi = σ◦F ik. Then

e0

e0 p

0

e1

e1

p1

e2 p

2

F21

F20

F22

Q1 Q2σ(Q2)

σ

σ2

σ0

σ1

Figure 3. The mappings F ik and σ, for k = 2.

∂2σ = ∂(∂σ) = ∂(σ0 − σ1 + σ2)

= ∂(σ0) − ∂(σ1) + ∂(σ2)

= (p2 − p1) − (p2 − p0) + (p1 − p0)

= 0

14


17. Put J2 = τ1 + τ2, where

τ1 = [0, e1, e1 + e2], τ2 = −[0, e2, e2 + e1].

� Explain why it is reasonable to call J2 the positively oriented unit square in R2.Show that ∂J2 is the sum of 4 oriented affine 1-simplices. Find these.

τ1 has p0 = 0,p1 = e1,p2 = e1 + p2, with arrows pointing toward higher index.τ2 has p0 = 0,p1 = e1,p2 = e1 + p2, with arrows pointing toward lower index.

e0

e1

e2

p0

p1 = e

1

p2 = e

1 + e

2

τ1

e0

e1

e2

p0

p1 = e

2p

2 = e

1 + e

2

−τ2

Figure 4. τ1 and −τ2.

e0

e1

e2

p0

p1 = e

2p

2 = e

1 + e

2

τ2

τ1 + τ

1

Figure 5. τ2 and τ1 + τ2.

So

∂(τ1 + τ2) = ∂[0, e1, e1 + e2] − ∂[0, e2, e2 + e1]

= [e1, e1 + e2] − [0, e1 + e2] + [0, e1] − [e2, e2 + e1] + [0, e2 + e1] − [0, e2]

= [0, e1] + [e1, e1 + e2] + [e2 + e1, e2] + [e2, 0].

So the image of τ1 + τ2 is the unit square, and it is oriented counterclockwise,i.e., positively.

� What is ∂(τ1 − τ2)?

∂(τ1 − τ2) = ∂[0, e1, e1 + e2] + ∂[0, e2, e2 + e1]

= [e1, e1 + e2] − [0, e1 + e2] + [0, e1] + [e2, e2 + e1] − [0, e2 + e1] + [0, e2]

= [0, e1] + [e1, e1 + e2] − [e2 + e1, e2] − [0, e2] − 2[0, e2 + e1].

15


20. State conditions under which the formula∫

Φ

f dω =

∫

∂Φ

fω −∫

Φ

(df) ∧ ω

is valid, and show that it generalizes the formula for integration by parts.Hint: d(fω) = (df) ∧ ω + f dω.

By the definition of d, we have

d(fω) = d(f ∧ ω) = (df ∧ ω) + (−1)0(f ∧ dω) = (df) ∧ ω + f dω.

Integrating both sides yields∫

Φ

d(fω) =

∫

Φ

(df) ∧ ω +

∫

Φ

f dω.

If the conditions of Stokes’ Theorem are met, i.e., if Φ is a k-chain of class C ′′ in anopen set V ⊆ Rn and ω is a (k − 1)-chain of class C ′ in V , then we may apply it tothe left side and get

∫

∂Φ

fω =

∫

Φ

(df) ∧ ω +

∫

Φ

f dω,

which is equivalent to the given formula.

21. Consider the 1-form

η =x dy − y dx

x2 + y2in R2\{0}.

(a) Show∫

γη = 2π 6= 0 but dη = 0.

Put gg(t) = (r cos t, r sin t) as in the example. Then

x2 + y2 7→ r2, dx 7→ −r sin t dt, anddy 7→ r cos t dt

and the integral becomes∫

γ

η =

∫ 2π

0

−r sin t

r2(−r sin t)dt+

∫ 2π

0

r cos t

r2(r cos t)dt

=

∫ 2π

0

sin2 t+ cos2 tdt

=

∫ 2π

0

dt

= 2π 6= 0.

However, using df =∑n

i=1∂f∂x1dx1, we compute

dη = d

(x dy

x2 + y2− y dx

x2 + y2

)

= d

(x dy

x2 + y2

)

− d

(y dx

x2 + y2

)

16


1

u

t2π

Γ

Φ

γ

Figure 6. The homotopy from Γ to γ.

= d

(x

x2 + y2

)

∧ dy +x

x2 + y2∧ d(dy)

− d

(y

x2 + y2

)

∧ dx− y

x2 + y2∧ d(dx)

=

[(x2 + y2) · 1 − x(2x)

(x2 + y2)2dx+

(x2 + y2) · 0 − x(2y)

(x2 + y2)2dy

]

∧ dy

−[(x2 + y2) · 0 − y(2x)

(x2 + y2)2dx+

(x2 + y2) · 1 − y(2y)

(x2 + y2)2dy

]

∧ dx

= 1(x2+y2)2

(

(y2 − x2)dx ∧ dy − 2xy dy ∧ dy

+ 2xy dx ∧ dx− (x2 − y2)dy ∧ dx)

= 1(x2+y2)2

(

(y2 − x2)dx ∧ dy + (x2 − y2)dx ∧ dy)

= 1(x2+y2)2

(y2 − x2 + x2 − y2)dx ∧ dy

= 0 dx ∧ dy = 0.

(b) With γ as in (a), let Γ be a C ′′ curve in R2\{0}, with Γ(0) = Γ(2π), such that theintervals [γ(t),Γ(t)] do not contain 0 for any t ∈ [0, 2π]. Prove that

∫

Γη = 2π.

For 0 ≤ t ≤ 2π and 0 ≤ u ≤ 1, define the straight-line homotopy between γ andΓ:

Φ(t, u) = (1 − u)Γ(t) + uγ(t), Φ(t, 0) = Γ(t), Φ(t, 1) = γ(t).

Then Φ is a 2-surface in R2\{0} whose parameter domain is the rectangle R =[0, 2π] × [0, 1] and whose image does not contain 0. Now define s(x) : I → R2

by s(x) = (1− x)γ(0) + xΓ(x), so s(I) is the segment connecting the base pointof γ to the base point of Γ. Also, denote the boundary of R by

∂R = R1 +R2 +R3 +R4,

17


where R1 = [0, 2πe1], R2 = [2πe1, 2πe1 + e2], R3 = [2πe1 + e2, e2], R4 = [e2, 0].

∂Φ = Φ(∂R)

= Φ(R1 +R2 +R3 +R4)

= Φ(R1) + Φ(R2) + Φ(R3) + Φ(R4)

= Γ + s+ γ + s

= Γ + s− γ − s

= Γ − γ,

where γ denotes the reverse of the path γ (i.e., traversed backwards). Then∫

Γ

η −∫

γ

η =

∫

Γ−γ

η =

∫

∂Φ

η ∂Φ = Γ − γ

=

∫

Φ

dη by Stokes

=

∫

Φ

0 by (a)

= 0

shows that ∫

Γ

η =

∫

γ

η.

(c) Take Γ(t) = (a cos t, b sin t) where a, b > 0 are fixed. Use (b) to show∫ 2π

0

ab

a2 cos2 t + b2 sin2 tdt = 2π.

We have

x(t) = Γ1(t) = a cos t and y(t) = Γ2(t) = b sin t

and

dx = ∂x∂tdt = −a sin t dt and dy = ∂y

∂tdt = b cos t dt,

so

x dy − y dx

x2 + y2=ab sin2 t+ ab cos2 t

a2 cos2 t + b2 sin2 t=

ab

a2 cos2 t + b2 sin2 t,

and the integral becomes

2π =

∫

Γ

η =

∫ 2π

0

ab cos2 t dt+ ab sin2 t dt

a2 cos2 t + b2 sin2 t

=

∫ 2π

0

ab

a2 cos2 t+ b2 sin2 tdt.

18

(d) Show that

η = d(arctan y

x

)

in any convex open set in which x 6= 0 and that

η = d(

− arctan xy

)

in any convex open set in which y 6= 0. Why does this justify the notationη = dθ, even though η isn’t exact in R2\{0}?

θ(x, y) = arctan(

yx

)is well-defined on

D+ = {(x, y) ... x > 0}.On D+ we have

dθ = d(arctan y

x

)

= ∂θ∂xdx + ∂θ

∂ydy

=∂∂x

(yx

)

1 +(

yx

)2dx+

∂∂y

(yx

)

1 +(

yx

)2dy

=−yx−2 dx

1 +(

yx

)2 +1xdy

1 +(

yx

)2

=−y dxx2 + y2

+x dy

x2 + y2.

So dθ = η on D+ (and similarly on D− = {(x, y) ... x < 0}), which is everywhereθ is defined. If we take

B+ = {(x, y) ... y > 0},then

d(

− arctan xy

)

= − 1/y dx

1 + (x/y)2+

(x/y)2 dy

1 + (x/y)2= − y dx

x2 + y2+

x dy

x2 + y2,

and similarly on B− = {(x, y) ... y < 0}.We say η = dθ because θ gives arg(x, y); if we let γ(t) = (cos t, sin t) as above,then this becomes

arctan(y

x

)

= arctan

(sin t

cos t

)

= t,

where the inverse is defined, i.e., for 0 ≤ t ≤ 2π.Suppose η were exact: ∃λ, a 0-form on R2

0 such that dλ = η. Then

dθ − dλ = d(θ − λ) = 0 =⇒ θ − λ = c

=⇒ θ = λ+ c

=⇒ θ = λ

19

but λ is continuous and θ isn’t. Alternatively, if η were exact,

∫

γ

η =

∫

γ

dλ =

∫

∂γ

λ =

∫

∅

λ = 0,

contradicting (b). <↙

(e) Show (b) can be derived from (d).

Taking η = dθ,

∫

Γ

η =

∫

Γ

dθ =

[

θ

]2π

0

= 2π − 0 = 2π.

(f) Γ is a closed C ′ curve in R20 implies that

1

2π

∫

Γ

η = Ind(Γ) :=1

2πi

∫ b

a

Γ′(t)

Γ(t)dt.

We use the following basic facts from complex analysis:

log z = log |z| + i arg z, so arg z = −i(log z − log |z|).

First, fix |Γ| = c, so Γ is one or more circles around the origin in whateverdirection, but of constant radius. Then

∫

Γ

η =

∫

Γ

dθ =

∫ b

a

d(θ(Γ(t)))

= −i∫ b

a

d(log Γ(t) − log |Γ(t)|)

= −i∫ b

a

d(log Γ(t))

= −i∫ b

a

Γ′(t)

Γ(t)dt.

Now use (b) for arbitrary Γ.

20


24. Let ω =∑ai(x) dxi be a 1-form of class C ′′ in a convex open set E ⊆ Rn. Assume

dω = 0 and prove that ω is exact in E.

� Fix p ∈ E. Define

f(x) =

∫

[p,x]

ω, x ∈ E.

Apply Stokes’ theorem to the 2-form dω = 0 on the affine-oriented 2-simplex[p,x,y] in E.

By Stokes’ Thm,

0 =

∫

[p,x,y]

dω =

∫

∂[p,x,y]

ω Stokes’ Thm

=

∫

[x,y]−[p,y]+[p,x]

ω def of ∂

=

∫

[x,y]

ω −∫

[p,y]

ω +

∫

[p,x]

ω Rem 10.26, p.267

=

∫

[x,y]

ω − f(y) + f(x) def of f

f(y) − f(x) =

∫

[x,y]

ω.

� Deduce that

f(y) − f(x) =

n∑

i=1

(yi − xi)

∫ 1

0

ai((1 − t)x + ty) dt

for x,y ∈ E. Hence (Dif)(x) = ai(x).

Note that γ = [x,y] is the straight-line path from x to y, i.e., γ(t) = (1−t)x+ty.Then the right-hand integral from the last line of the previous derivation becomes

f(y) − f(x) =

∫

[x,y]

ω

=

∫

[x,y]

n∑

i=1

ai(u) dui

=

∫ 1

0

n∑

i=1

ai ((1 − t)x + ty) ∂∂t

(

(1 − t)xi + tyi

)

dt

=

n∑

i=1

(yi − xi)

∫ 1

0

ai ((1 − t)x + ty) dt.

Putting y = x + sei, we plug this into the formula for the ith partial derivative.Note that all terms other than j = i are constant with respect to xi, and hence

21


vanish. This leaves

(Dif)(x) = lims→0

f(x + sei) − f(x)

s

= lims→0

(xi+s)−xi

s

∫ 1

0

ai ((1 − t)x + t(x + sei)) dt

= lims→0

1s· s∫ 1

0

ai (x + tsei)) dt

= lims→0

∫ 1

0

ai (x + tsei)) dt

=

∫ 1

0

lims→0

ai (x + tsei)) dt

=

∫ 1

0

ai (x) dt

= ai(x)

∫ 1

0

dt

= ai(x).

We can pass the limit through the integral using uniform convergence.

27. Let E be an open 3-cell in R3, with edges parallel to the coordinate axes. Suppose(a, b, c) ∈ E, fi ∈ C ′(E) for i = 1, 2, 3

ω = f1dy ∧ dz + f2dz ∧ dx+ f3dx ∧ dy,and assume that dω = 0 in E. Define

λ = g1dx+ g2dy

where

g1(x, y, z) =

∫ z

c

f2(x, y, s) ds−∫ y

b

f3(x, t, c) dt

g2(x, y, z) = −∫ z

c

f1(x, y, s) ds,

for x, y, z) ∈ E.

� Prove that dλ = ω in E.

We compute the exterior derivative:

dλ = d(g1dx+ g2dy)

= dg1 ∧ dx+ g1 ∧ d(dx) + dg2 ∧ dy + g2 ∧ d(dy)

= d

[∫ z

c

f2(x, y, s) ds−∫ y

b

f3(x, t, c) dt

]

∧ dx+ d

[

−∫ z

c

f1(x, y, s) ds

]

∧ dy

22


= ∂∂y

(∫ z

c

f2(x, y, s) ds

)

dy ∧ dx+ ∂∂z

(∫ z

c

f2(x, y, s) ds

)

dz ∧ dx

− ∂∂y

(∫ y

b

f3(x, t, c) dt

)

dy ∧ dx− ∂∂z

(∫ y

b

f3(x, t, c) dt

)

dz ∧ dx

− ∂∂x

(∫ z

c

f1(x, y, s) ds

)

dx ∧ dy − ∂∂z

(∫ z

c

f1(x, y, s) ds

)

dz ∧ dy

= ∂∂z

(∫ z

c

f1(x, y, s) ds

)

dy ∧ dz + ∂∂z

(∫ z

c

f2(x, y, s) ds−∫ y

b

f3(x, t, c) dt

)

dz ∧ dx

∂∂y

(∫ y

b

f3(x, t, c) dt−∫ z

c

f2(x, y, s) ds

)

dx ∧ dy

− ∂∂x

(∫ z

c

f1(x, y, s) ds

)

dx ∧ dy

Note that the third integral above is constant with respect to z, and hence vanishes.

= f1(x, y, z)dy ∧ dz + f2(x, y, z)dz ∧ dx+ f3(x, y, c)dx ∧ dy

−[∫ z

c

(∂∂yf2(x, y, s) + ∂

∂xf1(x, y, s)

)

ds

]

dx ∧ dy

= f1(x, y, z)dy ∧ dz + f2(x, y, z)dz ∧ dx

+

[

f3(x, y, c) −∫ z

c

(∂∂yf2(x, y, s) + ∂

∂xf1(x, y, s)

)

ds

]

dx ∧ dy. (10.1)

Since ω is closed,

0 = dω = d(f1dy ∧ dz) + d(f2dz ∧ dx) + d(f3dx ∧ dy)= df1 ∧ dy ∧ dz + f1 ∧ d(dy ∧ dz) + df2 ∧ dz ∧ dx+ f2 ∧ d(dz ∧ dx)

+ df3 ∧ dx ∧ dy + f3 ∧ d(dx ∧ dy)= ∂

∂xf1dx ∧ dy ∧ dz + ∂

∂yf2dy ∧ dz ∧ dx + ∂

∂zf3dz ∧ dx ∧ dy

=(

∂∂xf1 + ∂

∂yf2 + ∂

∂zf3

)

dx ∧ dy ∧ dz

=⇒ ∂∂zf3 = −

(∂∂xf1 + ∂

∂yf2

)

,

so the last line of the previous derivation, (10.1), becomes[

f3(x, y, c) +

∫ z

c

(∂∂zf3

)ds

]

dx ∧ dy

=[

f3(x, y, c) +(

f3(x, y, z) − f3(x, y, c))]

dx ∧ dy

= f3(x, y, z) dx ∧ dyand we finally obtain

dλ = f1(x, y, z)dy ∧ dz + f2(x, y, z)dz ∧ dx+ f3(x, y, z)dx ∧ dy= ω.

23


� Evaluate these integrals when ω = ζ and thus find the form λ = −(z/r)η, where

η = x dy−y dxx2+y2 .

Now the form is

ω = ζ = 1r3 (x dy ∧ dz + y dz ∧ dx + z dx ∧ dy) ,

sof1 =

x

r3, f2 =

y

r3, f3 =

z

r3.

The integrals are∫ z

c

f2(x, y, s) ds =

∫ z

c

y(x2 + y2 + s2)−3/2 ds = yx2+y2

(

zr− c√

x2+y2+c2

)

,

∫ y

b

f3(x, t, c) dt =

∫ y

b

c(x2 + t2 + c2)−3/2 dt = cx2+c2

(

y√x2+y2+c2

− b√x2+b2+c2

)

,

∫ z

c

f1(x, y, s) ds =

∫ z

c

x(x2 + y2 + s2)−3/2 ds = xx2+y2

(

zr− c√

x2+y2+c2

)

.

Thus λ = g1 dx+ g2 dy is

λ = yx2+y2

(

zr− c√

x2+y2+c2

)

dx− cx2+c2

(

y√x2+y2+c2

− b√x2+b2+c2

)

dx

− xx2+y2

(

zr− c√

x2+y2+c2

)

dy

=(− z

r

) (

− y dxx2+y2 + xdy

x2+y2

)

+

[

cx2+c2

(

b√x2+b2+c2

− y√x2+y2+c2

)

− yc

(x2+y2)√

x2+y2+c2

]

dx

+ xc

(x2+y2)√

x2+y2+c2dy

=(− z

r

)η,

due to some magic cancellations at the end.

24


28. Fix b > a > 0 and define

Φ(r, θ) = (r cos θ, r sin θ)

for a ≤ r ≤ b, 0 ≤ θ ≤ 2π. The range of Φ is thus an annulus in R2. Put ω = x3dy,and show ∫

Φ

dω =

∫

∂Φ

ω

by computing each integral separately.

Starting with the left-hand side,

dω = ∂∂xx3dx ∧ dy + ∂

∂yx3dy ∧ dy = 3x2dx ∧ dy + 0 = 3x2dx ∧ dy,

so ∣∣∣∣

cos θ −r sin θsin θ r cos θ

∣∣∣∣= r cos2 θ + r sin2 θ = r,

whence

3x2dx ∧ dy = 3r2(cos2 θ) r dr dθ,

and thus∫

Φ

dω =

∫ 2π

0

∫ b

a

3r3 cos2 θ dr dθ

=

(∫ 2π

0

cos2 θ dθ

)(∫ b

a

3r3 dr

)

=[

12θ + 1

4sin 2θ

]2π

0

[34r4]b

a

=(π + 1

4(sin 4π − sin 0)

)· 3

4

(b4 − a4

)

= 3π4

(b4 − a4).

Meanwhile, on the right-hand side, note that

∂Φ = Γ − γ,

as in Example 10.32 or Exercise 21(b). Thus,∫

Φ

ω =

∫

Γ−γ

ω =

∫

Γ

x3 dy −∫

γ

x3 dy

=

∫ 2π

0

b3 cos3 θ(b cos θ) dθ −∫ 2π

0

a3 cos3 θ(a cos θ) dθ

= b4∫ 2π

0

cos4 θ dθ − a4

∫ 2π

0

cos4 θ dθ

=(b4 − a4

)∫ 2π

0

cos4 θ dθ

=(b4 − a4

)· 3π

4.

25

11. Lebesgue Theory

1. If f ≥ 0 and∫

Ef dµ = 0, prove that f(x) = 0 almost everywhere on E.

First define

En := {f > 1n} and A :=

⋃

En = {f 6= 0}.We need to show that µA = 0, so suppose not. Then

µA > 0 =⇒∑

En > 0

=⇒ µEn > 0 for some n.

But then we’d have∫

E

f dµ ≥∫

En

f dµ > 1n· µ(En) > 0. <↙

2. If∫

Af dµ = 0 for every measurable subset A of a measurable set E, then f(x) =ae 0

on E.

First, suppose f ≥ 0. Then f =ae 0 on every A ⊆ E by the previous exercise. Inparticular, E ⊆ E, so f =ae 0 on E.

Now let f � 0. Then f = f+ − f−, and each of f+, f− are 0 on E by the aboveremark, so f =ae 0 on E, too.

3. If {fn} is a sequence of measurable functions, prove that the set of points x at which{fn(x)} converges is measurable.

Define f(x) = lim fn(x), so f is measurable by (11.17) and |fn(x) − f(x)| is mea-surable by (11.16),(11.18). Then we can write the given set as

{x ... ∀ε > 0, ∃N s.t. n ≥ N =⇒ |fn(x) − f(x)| < ε}= {x ... ∀k ∈ N, ∃N s.t. n ≥ N =⇒ |fn(x) − f(x)| < 1/k}

=∞⋂

k=1

{x ... ∃N s.t. n ≥ N =⇒ |fn(x) − f(x)| < 1/k}

=

∞⋂

k=1

∞⋃

n=1

{x ... |fn(x) − f(x)| < 1/k}

4. If f ∈ L(µ) on E and g is bounded and measurable on E, then fg ∈ L(µ) on E.

Since g is bounded, ∃M such that |g| ≤M . Then∣∣∣∣

∫

E

fg dµ

∣∣∣∣≤M

∫

E

|f | dµ <∞

by (11.23(d)) and (11.26).

26

MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1

Generally, a “solution” is something that would be acceptable if turned in in theform presented here, although the solutions given are often close to minimal in thisrespect. A “solution (sketch)” is too sketchy to be considered a complete solutionif turned in; varying amounts of detail would need to be filled in.

Problem 1.1: If r ∈ Q \ {0} and x ∈ R \ Q, prove that r + x, rx 6∈ Q.

Solution: We prove this by contradiction. Let r ∈ Q\{0}, and suppose that r+x ∈Q. Then, using the field properties of both R and Q, we have x = (r +x)− r ∈ Q.Thus x 6∈ Q implies r + x 6∈ Q.

Similarly, if rx ∈ Q, then x = (rx)/r ∈ Q. (Here, in addition to the fieldproperties of R and Q, we use r 6= 0.) Thus x 6∈ Q implies rx 6∈ Q.

Problem 1.2: Prove that there is no x ∈ Q such that x2 = 12.

Solution: We prove this by contradiction. Suppose there is x ∈ Q such thatx2 = 12. Write x = m

n in lowest terms. Then x2 = 12 implies that m2 = 12n2.Since 3 divides 12n2, it follows that 3 divides m2. Since 3 is prime (and by uniquefactorization in Z), it follows that 3 divides m. Therefore 32 divides m2 = 12n2.Since 32 does not divide 12, using again unique factorization in Z and the fact that3 is prime, it follows that 3 divides n. We have proved that 3 divides both m andn, contradicting the assumption that the fraction m

n is in lowest terms.

Alternate solution (Sketch): If x ∈ Q satisfies x2 = 12, then x2 is in Q and satisfies(

x2

)2 = 3. Now prove that there is no y ∈ Q such that y2 = 3 by repeating theproof that

√2 6∈ Q.

Problem 1.5: Let A ⊂ R be nonempty and bounded below. Set −A = {−a : a ∈A}. Prove that inf(A) = − sup(−A).

Solution: First note that −A is nonempty and bounded above. Indeed, A containssome element x, and then −x ∈ A; moreover, A has a lower bound m, and −m isan upper bound for −A.

We now know that b = sup(−A) exists. We show that −b = inf(A). That −b isa lower bound for A is immediate from the fact that b is an upper bound for −A.To show that −b is the greatest lower bound, we let c > −b and prove that c is nota lower bound for A. Now −c < b, so −c is not an upper bound for −A. So thereexists x ∈ −A such that x > −c. Then −x ∈ A and −x < c. So c is not a lowerbound for A.

Problem 1.6: Let b ∈ R with b > 1, fixed throughout the problem.

Comment: We will assume known that the function n 7→ bn, from Z to R, isstrictly increasing, that is, that for m, n ∈ Z, we have bm < bn if and only ifm < n. Similarly, we take as known that x 7→ xn is strictly increasing when n is

Date: 1 October 2001.

1

2 MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1

an integer with n > 0. We will also assume that the usual laws of exponents areknown to hold when the exponents are integers. We can’t assume anything aboutfractional exponents, except for Theorem 1.21 of the book and its corollary, becausethe context makes it clear that we are to assume fractional powers have not yetbeen defined.

(a) Let m, n, p, q ∈ Z, with n > 0 and q > 0. Prove that if mn = p

q , then(bm)1/n = (bp)1/q.

Solution: By the uniqueness part of Theorem 1.21 of the book, applied to thepositive integer nq, it suffices to show that[

(bm)1/n]nq

=[(bp)1/q

]nq

.

Now the definition in Theorem 1.21 implies that[(bm)1/n

]n

= bm and[(bp)1/q

]q

= bp.

Therefore, using the laws of integer exponents and the equation mq = np, we get[(bm)1/n

]nq

=[[

(bm)1/n]n]q

= (bm)q = bmq

= bnp = (bp)n =[[

(bp)1/q]q]n

=[(bp)1/q

]nq

,

as desired.

By Part (a), it makes sense to define bm/n = (bm)1/n for m, n ∈ Z with n > 0.This defines br for all r ∈ Q.

(b) Prove that br+s = brbs for r, s ∈ Q.

Solution: Choose m, n, p, q ∈ Z, with n > 0 and q > 0, such that r = mn and

s = pq . Then r + s = mq+np

nq . By the uniqueness part of Theorem 1.21 of the book,applied to the positive integer nq, it suffices to show that[

b(mq+np)/(nq)]nq

=[(bm)1/n(bp)1/q

]nq

.

Directly from the definitions, we can write[b(mq+np)/(nq)

]nq

=[[

b(mq+np)]1/(nq)

]nq

= b(mq+np).

Using the laws of integer exponents and the definitions for rational exponents, wecan rewrite the right hand side as[

(bm)1/n(bp)1/q]nq

=[[

(bm)1/n]n]q [[

(bp)1/q]q]n

= (bm)q(bp)n = b(mq+np).

This proves the required equation, and hence the result.

(c) For x ∈ R, define

B(x) = {br : r ∈ Q ∩ (−∞, x]}.Prove that if r ∈ Q, then br = sup(B(r)).

Solution: The main point is to show that if r, s ∈ Q with r < s, then br < bs.Choose m, n, p, q ∈ Z, with n > 0 and q > 0, such that r = m

n and s = pq . Then

MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1 3

also r = mqnq and s = np

nq , with nq > 0, so

br = (bmq)1/(nq) and bs = (bnp)1/(nq).

Now mq < np because r < s. Therefore, using the definition of c1/(nq),

(br)nq = bmq < bnp = (bs)nq.

Since x 7→ xnq is strictly increasing, this implies that br < bs.Now we can prove that if r ∈ Q then br = sup(B(r)). By the above, if s ∈ Q

and s ≤ r, then bs ≤ br. This implies that br is an upper bound for B(r). Sincebr ∈ B(r), obviously no number smaller than br can be an upper bound for B(r).So br = sup(B(r)).

We now define bx = sup(B(x)) for every x ∈ R. We need to show that B(x)is nonempty and bounded above. To show it is nonempty, choose (using theArchimedean property) some k ∈ Z with k < x; then bk ∈ B(x). To show itis bounded above, similarly choose some k ∈ Z with k > x. If r ∈ Q ∩ (−∞, x],then br ∈ B(k) so that br ≤ bk by Part (c). Thus bk is an upper bound for B(x).This shows that the definition makes sense, and Part (c) shows it is consistent withour earlier definition when r ∈ Q.

(d) Prove that bx+y = bxby for all x, y ∈ R.

Solution:In order to do this, we are going to need to replace the set B(x) above by the

set

B0(x) = {br : r ∈ Q ∩ (−∞, x)}(that is, we require r < x rather than r ≤ x) in the definition of bx. (If you areskeptical, read the main part of the solution first to see how this is used.)

We show that the replacement is possible via some lemmas.

Lemma 1. If x ∈ [0,∞) and n ∈ Z satisfies n ≥ 0, then (1 + x)n ≥ 1 + nx.

Proof: The proof is by induction on n. The statement is obvious for n = 0. Soassume it holds for some n. Then, since x ≥ 0,

(1 + x)n+1 = (1 + x)n(1 + x) ≥ (1 + nx)(1 + x)

= 1 + (n + 1)x + nx2 ≥ 1 + (n + 1)x.

This proves the result for n + 1.

Lemma 2. inf{b1/n : n ∈ N} = 1. (Recall that b > 1 and N = {1, 2, 3, . . . }.)Proof: Clearly 1 is a lower bound. (Indeed, (b1/n)n = b > 1 = 1n, so b1/n > 1.) Weshow that 1+x is not a lower bound when x > 0. If 1+x were a lower bound, then1 + x ≤ b1/n would imply (1 + x)n ≤ (b1/n)n = b for all n ∈ N. By Lemma 1, wewould get 1 + nx ≤ b for all n ∈ N, which contradicts the Archimedean propertywhen x > 0.

Lemma 3. sup{b−1/n : n ∈ N} = 1.

Proof: Part (b) shows that b−1/nb1/n = b0 = 1, whence b−1/n = (b1/n)−1. Sinceall numbers b−1/n are strictly positive, it now follows from Lemma 2 that 1 is anupper bound. Suppose x < 1 is an upper bound. Then x−1 is a lower bound for

{b1/n : n ∈ N}. Since x−1 > 1, this contradicts Lemma 2. Thus sup{b−1/n : n ∈N} = 1, as claimed.

Lemma 4. bx = sup(B0(x)) for x ∈ R.

Proof: If x 6∈ Q, then B0(x) = B(x), so there is nothing to prove. If x ∈ Q,then at least B0(x) ⊂ B(x), so bx ≥ sup(B0(x)). Moreover, Part (b) shows thatbx−1/n = bxb−1/n for n ∈ N. The numbers bx−1/n are all in B0(x), and

sup{bxb−1/n : n ∈ N} = bx sup{b−1/n : n ∈ N}because bx > 0, so using Lemma 3 in the last step gives

sup(B0(x)) ≥ sup{bx−1/n : n ∈ N} = bx sup{b−1/n : n ∈ N} = bx.

Now we can prove the formula bx+y = bxby. We start by showing that bx+y ≤bxby, which we do by showing that bxby is an upper bound for B0(x + y). Thus letr ∈ Q satisfy r < x+y. Then there are s0, t0 ∈ R such that r = s0 + t0 and s0 < x,t0 < y. Choose s, t ∈ Q such that s0 < s < x and t0 < t < y. Then r < s + t, sobr < bs+t = bsbt ≤ bxby. This shows that bxby is an upper bound for B0(x + y).

(Note that this does not work using B(x + y). If x + y ∈ Q but x, y 6∈ Q, thenbx+y ∈ B(x + y), but it is not possible to find s and t with bs ∈ B(x), bt ∈ B(y),and bsbt = bx+y.)

We now prove the reverse inequality. Suppose it fails, that is, bx+y < bxby. Thenbx+y

by< bx.

The left hand side is thus not an upper bound for B0(x), so there exists s ∈ Q withs < x and

bx+y

by< bs.

It follows thatbx+y

bs< by.

Repeating the argument, there is t ∈ Q with t < y such thatbx+y

bs< bt.

Therefore

bx+y < bsbt = bs+t

(using Part (b)). But bs+t ∈ B0(x + y) because s + t ∈ Q and s + t < x + y, so thisis a contradiction. Therefore bx+y ≤ bxby.

Problem 1.9: Define a relation on C by w < z if and only if either Re(w) < Re(z)or both Re(w) = Re(z) and Im(w) < Im(z). (For z ∈ C, the expressions Re(z)and Im(z) denote the real and imaginary parts of z.) Prove that this makes C anordered set. Does this order have the least upper bound property?

Solution: We verify the two conditions in the definition of an order. For the first,let w, z ∈ C. There are three cases.

Case 1: Re(w) < Re(z). Then w < z, but w = z and w > z are both false.Case 2: Re(w) > Re(z). Then w > z, but w = z and w < z are both false.

Case 3: Re(w) = Re(z). This case has three subcases.Case 3.1: Im(w) < Im(z). Then w < z, but w = z and w > z are both false.Case 3.2: Im(w) > Im(z). Then w > z, but w = z and w < z are both false.Case 3.3: Im(w) = Im(z). Then w = z, but w > z and w < z are both false.These cases exhaust all possibilities, and in each of them exactly one of w < z,

w = z, and w > z is true, as desired.Now we prove transitivity. Let s < w and w < z. If either Re(s) < Re(w)

or Re(w) < Re(z), then clearly Re(s) < Re(z), so s < z. If Re(s) = Re(w)and Re(w) = Re(z), then the definition of the order requires Im(s) < Im(w) andIm(w) < Im(z). We thus have Re(s) = Re(z) and Im(s) < Im(z), so s < z bydefinition.

It remains to answer the last question. We show that this order does not havethe least upper bound property. Let S = {z ∈ C : Re(z) < 0}. Then S 6= ∅ because−1 ∈ S, and S is bounded above because 1 is an upper bound for S.

We show that S does not have a least upper bound by showing that if w is anupper bound for S, then there is a smaller upper bound. First, by the definition ofthe order it is clear that Re(w) is an upper bound for

{Re(z) : z ∈ S} = (−∞, 0).

Therefore Re(w) ≥ 0. Moreover, every u ∈ C with Re(u) ≥ 0 is in fact an upperbound for S. In particular, if w is an upper bound for S, then w − i < w and hasthe same real part, so is a smaller upper bound.

Note: A related argument shows that the set T = {z ∈ C : Re(z) ≤ 0} also hasno least upper bound. One shows that w is an upper bound for T if and only ifRe(w) > 0.

Problem 1.13: Prove that if x, y ∈ C, then ||x| − |y|| ≤ |x − y|.Solution: The desired inequality is equivalent to

|x| − |y| ≤ |x − y| and |y| − |x| ≤ |x − y|.We prove the first; the second follows by exchanging x and y.

Set z = x − y. Then x = y + z. The triangle inequality gives |x| ≤ |y| + |z|.Substituting the definition of z and subtracting |y| from both sides gives the result.

Problem 1.17: Prove that if x, y ∈ Rn, then

‖x + y‖2 + ‖x − y‖2 = 2‖x‖2 + 2‖y‖2.

Interpret this result geometrically in terms of parallelograms.

Solution: Using the definition of the norm in terms of scalar products, we have:

‖x + y‖2 + ‖x − y‖2 = 〈x + y, x + y〉 + 〈x − y, x − y〉= 〈x, x〉 + 〈x, y〉 + 〈y, x〉 + 〈y, y〉+ 〈x, x〉 − 〈x, y〉 − 〈y, x〉 + 〈y, y〉= 2〈x, x〉 + 2〈y, y〉 = 2‖x‖2 + 2‖y‖2.

The interpretation is that 0, x, y, x + y are the vertices of a parallelogram, andthat ‖x+y‖ and ‖x−y‖ are the lengths of its diagonals while ‖x‖ and ‖y‖ are each


the lengths of two opposite sides. Therefore the sum of the squares of the lengthsof the diagonals is equal to the sum of the squares of the lengths of the sides.

Note: One can do the proof directly in terms of the formula ‖x‖2 =∑n

k=1 |xk|2.The steps are all the same, but it is more complicated to write. It is also lessgeneral, since the argument above applies to any norm that comes from a scalarproduct.



Problem 2.2: Prove that the set of algebraic numbers is countable.

Solution (sketch): For each fixed integer n ≥ 0, the set Pn of all polynomialswith integer coefficients and degree at most n is countable, since it has the samecardinality as the set {(a0, . . . , an) : ai ∈ N} = Nn+1. The set of all polynomialswith integer coefficients is

⋃∞n=0 Pn, which is a countable union of countable sets

and so countable. Each polynomial has only finitely many roots (at most n fordegree n), so the set of all possible roots of all polynomials with integer coefficientsis a countable union of finite sets, hence countable.

Problem 2.3: Prove that there exist real numbers which are not algebraic.

Solution (Sketch): This follows from Problem 2.2, since R is not countable.

Problem 2.4: Is R \ Q countable?

Solution (Sketch): No. Q is countable and R is not countable.

Problem 2.5: Construct a bounded subset of R with exactly 3 limit points.

Solution (Sketch): For example, use{1n : n ∈ N

} ∪ {1 + 1

n : n ∈ N} ∪ {

2 + 1n : n ∈ N

}.

Problem 2.6: Let E′ denote the set of limit points of E. Prove that E′ is closed.Prove that E

′= E′. Is (E′)′ always equal to E′?

Solution (Sketch): Proving that E′ is closed is equivalent to proving that (E′)′ ⊂E′. So let x ∈ (E′)′ and let ε > 0. Choose y ∈ E′ ∩ (Nε(x) \ {x}). Chooseδ = min(d(x, y), ε − d(x, y)) > 0. Choose z ∈ E ∩ (Nδ(y) \ {y}). The triangleinequality ensures z 6= x and z ∈ Nε(x). This shows x is a limit point of E.

Here is a different way to prove that (E′)′ ⊂ E′. Let x ∈ (E′)′ and ε > 0.Choose y ∈ E′ ∩ (Nε/2(x) \ {x}). By Theorem 2.20 of Rudin, there are infinitelymany points in E ∩ (Nε/2(y) \ {y}). In particular there is z ∈ E ∩ (Nε/2(y) \ {y})with z 6= x. Now z ∈ E ∩ (Nε(x) \ {x}).

To prove E′= E′, it suffices to prove E

′ ⊂ E′. We first claim that if A and Bare any subsets of X, then (A ∪ B)′ ⊂ A′ ∪ B′. The fastest way to do this is toassume that x ∈ (A ∪ B)′ but x 6∈ A′, and to show that x ∈ B′. Accordingly, letx ∈ (A ∪ B)′ \ A′. Since x 6∈ A′, there is ε0 > 0 such that Nε0(x) ∩ A contains no


1


points except possibly x itself. Now let ε > 0; we show that Nε(x) ∩ B contains atleast one point different from x. Let r = min(ε, ε0) > 0. Because x ∈ (A ∪ B)′,there is y ∈ Nr(x)∩ (A∪B) with y 6= x. Then y 6∈ A because r ≤ ε0. So necessarilyy ∈ B, and thus y is a point different from x and in Nr(x) ∩ B. This shows thatx ∈ B′, and completes the proof that (A ∪ B)′ ⊂ A′ ∪ B′.

To prove E′ ⊂ E′, we now observe that

E′= (E ∪ E′)′ ⊂ E′ ∪ (E′)′ ⊂ E′ ∪ E′ = E′.

An alternate proof that E′ ⊂ E′ can be obtained by slightly modifying either of

the proofs above that (E′)′ ⊂ E′.For the third part, the answer is no. Take

E = {0} ∪ {1n : n ∈ N

}.

Then E′ = {0} and (E′)′ = ∅. (Of course, you must prove these facts.)

Problem 2.8: If E ⊂ R2 is open, is every point of E a limit point of E? What ifE is closed instead of open?

Solution (Sketch): Every point of an open set E ⊂ R2 is a limit point of E. Indeed,if x ∈ E, then there is ε > 0 such that Nε(x) ⊂ E, and it is easy to show that x isa limit point of Nε(x).

(Warning: This is not true in a general metric space.)Not every point of a closed set need be a limit point. Take E = {(0, 0)}, which

has no limit points.

Problem 2.9: Let E◦ denote the set of interior points of a set E, that is, theinterior of E.

(a) Prove that E◦ is open.

Solution (sketch): If x ∈ E◦, then there is ε > 0 such that Nε(x) ⊂ E. Since Nε(x)is open, every point in Nε(x) is an interior point of Nε(x), hence of the bigger setE. So Nε(x) ∈ E◦.

(b) Prove that E is open if and only if E◦ = E.

Solution: If E is open, then E = E◦ by the definition of E◦. If E = E◦, then E isopen by Part (a).

(c) If G is open and G ⊂ E, prove that G ⊂ E◦.

Solution (sketch): If x ∈ G ⊂ E and G is open, then x is an interior point of G.Therefore x is an interior point of the bigger set E. So x ∈ E◦.

(d) Prove that X \ E◦ = X \ E.

Solution (sketch): First show that X \ E◦ ⊂ X \ E. If x 6∈ E, then clearly x ∈X \ E. Otherwise, consider x ∈ E \ E◦. Rearranging the statement that x fails tobe an interior point of E, and noting that x itself is not in X \E, one gets exactlythe statement that x is a limit point of X \ E.

Now show that X \ E ⊂ X \E◦. If x ∈ X \E, then clearly x 6∈ E◦. If x 6∈ X \Ebut x is a limit point of X \E, then one simply rearranges the definition of a limitpoint to show that x is not an interior point of E.

(e) Prove or disprove:(E

)◦= E.


Solution (sketch): This is false. Example: take E = (0, 1)∪(1, 2). We have E◦ = E,E = [0, 2], and

(E

)◦= (0, 2).

Another example is Q.

(f) Prove or disprove: E◦ = E.

Solution (sketch): This is false. Example: take E = (0, 1) ∪ {2}. Then E =[0, 1] ∪ {2}, E◦ = (0, 1), and E

◦= [0, 1].

The sets Q and {0} are also examples: in both cases, E◦ = ∅.

Problem 2.11: Which of the following are metrics on R?

(a) d1(x, y) = (x − y)2.

Solution (Sketch): No. The triangle inequality fails with x = 0, y = 2, and z = 4.

(b) d2(x, y) =√|x − y|.

Solution (Sketch): Yes. Some work is needed to check the triangle inequality.

(c) d3(x, y) = |x2 − y2|.Solution (Sketch): No. d3(1,−1) = 0.

(d) d4(x, y) = |x − 2y|.Solution (Sketch): No. d4(1, 1) 6= 0. Also, d4(1, 6) 6= d4(6, 1).

(e) d5(x, y) =|x − y|

1 + |x − y| .

Solution (Sketch): Yes. Some work is needed to check the triangle inequality. Youneed to know that t 7→ t

1+t is nondecreasing on [0,∞), and that a, b ≥ 0 implies

a + b

1 + a + b≤ a

1 + a+

b

1 + b.

Do the first by algebraic manipulation. The second isa + b

1 + a + b=

a

1 + a + b+

b

1 + a + b≤ a

1 + a+

b

1 + b.

(This is easier than what most people did the last time I assigned this problem.)

Problem 2.14: Give an example of an open cover of the interval (0, 1) ⊂ R whichhas no finite subcover.

Solution (sketch): {(1/n, 1) : n ∈ N}. (Note that you must show that this works.)

Problem 2.16: Regard Q as a metric space with the usual metric. Let E = {x ∈Q : 2 < x2 < 3}. Prove that E is a closed and bounded subset of Q which is notcompact. Is E an open subset of Q?

Solution (sketch): Clearly E is bounded.We prove E is closed. The fast way to do this is to note that

Q \ E = Q ∩[(

−∞,−√

3)∪

(−√

2,√

2)∪

(√3,∞

)],

and so is open by Theorem 2.30. To do it directly, suppose x ∈ Q is a limit pointof E which is not in E. Since we can’t have x2 = 2 or x2 = 3, we must havex2 < 2 or x2 > 3. Assume x2 > 3. (The other case is handled similarly.) Letr = |x| − √

3 > 0. Then every z ∈ Nr(x) satisfies

|z| ≥ |x| − |x − z| > |x| − r > 0,

which implies that z2 > (|x| − r)2 = 3. This shows that z 6∈ E, which contradictsthe assumption that x is a limit point of E.

The fast way to see that E is not compact is to note that it is a subset of R,but is not closed in R. (See Theorem 2.23.) To prove this directly, show that, forexample, the sets {

y ∈ Q : 2 + 1n < y2 < 3 − 1

n

}form an open cover of E which has no finite subcover.

To see that E is open in Q, the fast way is to write

E = Q ∩[(

−√

3,−√

2)∪

(√2,−

√3)]

,

which is open by Theorem 2.30. It can also be proved directly.

Problem 2.19: Let X be a metric space, fixed throughout this problem.(a) If A and B are disjoint closed subsets of X, prove that they are separated.

Solution (Sketch): We have A ∩ B = A ∩ B = A ∩ B = ∅ because A and B areclosed.


1

(b) If A and B are disjoint open subsets of X, prove that they are separated.

Solution (Sketch): X \ A is a closed subset containing B, and hence containing B.Thus A ∩ B = ∅. Interchanging A and B, it follows that A ∩ B = ∅.

(c) Fix x0 ∈ X and δ > 0. Set

A = {x ∈ X : d(x, x0) < δ} and B = {x ∈ X : d(x, x0) > δ}.Prove that A and B are separated.

Solution (Sketch): Both A and B are open sets (proof!), and they are disjoint. Sothis follows from Part (b).

(d) Prove that if X is connected and contains at least two points, then X isuncountable.

Solution: Let x and y be distinct points of X. Let R = d(x, y) > 0. For eachr ∈ (0, R), consider the sets

Ar = {z ∈ X : d(z, x) < r} and Br = {z ∈ X : d(z, x) > r}.They are separated by Part (c). They are not empty, since x ∈ Ar and y ∈ Br.Since X is connected, there must be a point zr ∈ X \ (Ar ∪Br). Then d(x, zr) = r.

Note that if r 6= s, then d(x, zr) 6= d(x, zs), so zr 6= zs. Thus r 7→ zr defines aninjective map from (0, R) to X. Since (0, R) is not countable, X can’t be countableeither.

Problem 2.20: Let X be a metric space, and let E ⊂ X be a connected subset.Is E necessarily connected? Is int(E) necessarily connected?

Solution to the first question (sketch): The set int(E) need not be connected. Theeasiest example to write down is to take X = R2 and

E = {x ∈ R2 : ‖x − (1, 0)‖ ≤ 1} ∪ {x ∈ R2 : ‖x − (−1, 0)‖ ≤ 1}.Then

int(E) = {x ∈ R2 : ‖x − (1, 0)‖ < 1} ∪ {x ∈ R2 : ‖x − (−1, 0)‖ < 1}.This set fails to be connected because the point (0, 0) is missing. A more dramaticexample is two closed disks joined by a line, say

E = {x ∈ R2 : ‖x − (2, 0)‖ ≤ 1} ∪ {x ∈ R2 : ‖x − (−2, 0)‖ ≤ 1}(1)

∪ {(α, 0) ∈ R2 : − 3 ≤ α ≤ 3}.(2)

Then

int(E) = {x ∈ R2 : ‖x − (2, 0)‖ < 1} ∪ {x ∈ R2 : ‖x − (−2, 0)‖ < 1}.

Solution to the second question: If E is connected, then E is necessarily connected.To prove this using Rudin’s definition, assume E = A∪B for separated sets A andB; we prove that one of A and B is empty. The sets A0 = A ∩ E and B0 = B ∩ Eare separated sets such that E = A0∪B0. (They are separated because A0 ⊂ A andB0 ⊂ B.) Because E is connected, one of A0 and B0 must be empty; without lossof generality, A0 = ∅. Then A ⊂ E \ E. Therefore E ⊂ B. But then A ⊂ E ⊂ B.Because A and B are separated, this can only happen if A = ∅.

Alternate solution to the second question: If E is connected, we prove that E isnecessarily connected, using the traditional definition. Thus, assume that E = A∪Bfor disjoint relatively open sets A and B; we prove that one of A and B is empty.The sets A0 = A ∩ E and B0 = B ∩ E are disjoint relatively open sets in E suchthat E = A0 ∪ B0. Because E is connected, one of A0 and B0 must be empty;without loss of generality, A0 = ∅. Then A ⊂ E \ E and is relatively open in E.

Now let x ∈ A. Then there is ε > 0 such that Nε(x) ∩ E ⊂ A. So Nε(x) ∩ E ⊂E \ E, which implies that Nε(x) ∩ E = ∅. This contradicts the fact that x ∈ E.Thus A = ∅.

Problem 2.22: Prove that Rn is separable.

Solution (sketch): The subset Qn is countable by Theorem 2.13. To show that Qn

is dense, let x = (x1, . . . , xn) ∈ Rn and let ε > 0. Choose y1, . . . , yn ∈ Q such that|yk−xk| < ε

n for all k. (Why is this possible?) Then y = (y1, . . . , yn) ∈ Qn∩Nε(x).

Problem 2.23: Prove that every separable metric space has a countable base.

Solution: Let X be a separable metric space. Let S ⊂ X be a countable densesubset of X. Let

B = {N1/n(s) : s ∈ S, n ∈ N}.Since N and S are countable, B is a countable collection of open subsets of X.

Now let U ⊂ X be open and let x ∈ U . Choose ε > 0 such that Nε(x) ⊂ U .Choose n ∈ N such that 1

n < ε2 . Since S is dense in X, there is s ∈ S ∩ N1/n(x),

that is, s ∈ S and d(s, x) < 1n . Then x ∈ N1/n(s) and N1/n(s) ∈ B. It remains to

show that N1/n(s) ⊂ U . So let y ∈ N1/n(s). Then

d(x, y) ≤ d(x, s) + d(s, y) < 1n + 1

n < ε,

so y ∈ Nε(x) ⊂ U .

Problem 2.25: Let K be a compact metric space. Prove that K has a countablebase, and that K is separable.

The easiest way to do this is actually to prove first that K is separable, andthen to use Problem 2.23. However, the direct proof that K has a countable baseis not very different, so we give it here. We actually give two versions of the proof,which differ primarily in how the indexing is done. The first version is easier towrite down correctly, but the second has the advantage of eliminating some of thesubscripts, which can be important in more complicated situations. Note that thesecond proof is shorter, even after the parenthetical remarks about indexing aredeleted from the first proof. Afterwards, we give a proof that every metric spacewith a countable base is separable.

Solution 1: We prove that K has a countable base. For each n ∈ N, the open setsN1/n(x), for x ∈ K, form an open cover of K. Since K is compact, this open coverhas a finite subcover, say

{N1/n(xn,1), N1/n(xn,2), . . . , N1/n(xn,kn)}

for suitable xn,1, xn,2, . . . , xn,kn∈ K. (Note: For each n, the collection of x’s is

different; therefore, they must be labelled independently by both n and a second

parameter. The number of them also depends on n, so must be called kn, k(n), orsomething similar.)

Now let

B = {N1/n(xn,j) : n ∈ N, 1 ≤ j ≤ kn}.(Note that both subscripts are used here.) Then B is a countable union of finitesets, hence countable. We show that B is a base for K.

Let U ⊂ K be open and let x ∈ U . Choose ε > 0 such that Nε(x) ⊂ U . Choosen ∈ N such that 1

n < ε2 . Since the sets

N1/n(xn,1), N1/n(xn,2), . . . , N1/n(xn,kn)

cover K, there is j with 1 ≤ j ≤ kn such that x ∈ N1/n(xn,j). (Here we see whythe double indexing is necessary: the list of centers to choose from depends on n,and therefore their names must also depend on n.) By definition, N1/n(xn,j) ∈ B.It remains to show that N1/n(xn,j) ⊂ U . So let y ∈ N1/n(xn,j). Since x and y areboth in N1/n(xn,j), we have

d(x, y) ≤ d(x, xn,j) + d(xn,j , y) < 1n + 1

n < ε,

so y ∈ Nε(x) ⊂ U .

Solution 2: We again prove that K has a countable base. For each n ∈ N, the opensets N1/n(x), for x ∈ K, form an open cover of K. Since K is compact, this opencover has a finite subcover. That is, there is a finite set Fn ⊂ K such that the setsN1/n(x), for x ∈ Fn, still cover K. Now let

B = {N1/n(x) : n ∈ N, x ∈ Fn}.Then B is a countable union of finite sets, hence countable. We show that B is abase for K.

Let U ⊂ K be open and let x ∈ U . Choose ε > 0 such that Nε(x) ⊂ U . Choosen ∈ N such that 1

n < ε2 . Since the sets N1/n(y), for y ∈ Fn, cover K, there is

y ∈ Fn such that x ∈ N1/n(y). By definition, N1/n(y) ∈ B. It remains to show thatN1/n(y) ⊂ U . So let z ∈ N1/n(y). Since x and z are both in N1/n(y), we have

d(x, z) ≤ d(x, y) + d(y, z) < 1n + 1

n < ε,

so z ∈ Nε(x) ⊂ U .

It remains to prove the following lemma.

Lemma: Let X be a metric space with a countable base. Then X is separable.

Proof: Let B be a countable base for X. Without loss of generality, we may assume∅ 6∈ B. For each U ∈ B, choose an element xU ∈ U . Let S = {xU : U ∈ B}. ClearlyS is (at most) countable. We show it is dense. So let x ∈ X and let ε > 0. Ifx ∈ S, there is nothing to prove. Otherwise Nε(x) is an open set in X, so thereexists U ∈ B such that x ∈ U ⊂ Nε(x). In particular, xU ∈ Nε(x). Since xU 6= xand since ε > 0 is arbitrary, this shows that x is a limit point of S.

Problem 3.1: Prove that if (sn) converges, then (|sn|) converges. Is the conversetrue?

Solution (sketch): Use the inequality∣∣|sn| − |s|∣∣ ≤ |sn − s| and the definition of the

limit. The converse is false. Take sn = (−1)n. (This requires proof, of course.)

Problem 3.2: Calculate limn→∞(√

n2 + 1 − n).

Solution (sketch):√

n2 + 1 − n =n√

n2 + 1 + n=

1√1 + 1

n2 + 1→ 1

2.

(Of course, the last step requires proof.)

Problem 3.3: Let s1 =√

2, and recursively define

sn+1 =√

2 +√

sn

for n ∈ N. Prove that (sn) converges, and that sn < 2 for all n ∈ N.

Solution (sketch): By induction, it is immediate that sn > 0 for all n, so that sn+1

is always defined.Next, we show by induction that sn < 2 for all n. This is clear for n = 1. The

computation for the induction step is

sn+1 =√

2 +√

sn ≤√

2 +√

2 < 2.

To prove convergence, it now suffices to show that (sn) is nondecreasing. (SeeTheorem 3.14.) This is also done by induction. To start, observe that

s2 =√

2 +√

s1 =

√2 +

√√2 >

√2.

The computation for the induction step is:

sn+1 − sn =√

2 +√

sn −√

2 +√

sn−1 =√

sn −√sn−1√

2 +√

sn +√

2 + √sn−1

> 0.


1

Problem 3.4: Let s1 = 0, and recursively define

sn+1 ={

12 + sn n is even12sn n is odd .

for n ∈ N. Find lim supn→∞ sn and lim infn→∞ sn.

Solution (sketch): Use induction to show that

s2m =2m−1 − 1

2mand s2m+1 =

2m − 12m

.

It follows that

lim supn→∞

sn = 1 and lim infn→∞ sn = 1

2 .

Problem 3.5: Let (an) and (bn) be sequences in R. Prove that

lim supn→∞


an + lim supn→∞

bn,

provided that the right hand side is defined, that is, not of the form ∞ − ∞ or−∞ + ∞.

Four solutions are presented or sketched. The first is what I presume to be thesolution Rudin intended. The second is a variation of the first, which minimizesthe amount of work that must be done in different cases. The third shows whatmust be done if one wants to work directly from Rudin’s definition. The fourth isthe “traditional” proof of the result, and proceeds via the traditional definition.

Solution 1 (sketch): We give a complete solution for the case

lim supn→∞

an ∈ (−∞,∞) and lim supn→∞

bn ∈ (−∞,∞).

One needs to consider several other cases, but the basic method is the same.Define

a = lim supn→∞

an and b = lim supn→∞

bn.

Let c > a + b. We show that c is not a subsequential limit of (an + bn).Let ε = 1

3 (c − a − b) > 0. Use Theorem 3.17 (b) of Rudin to choose N1 ∈ Nsuch that n ≥ N1 implies an < a + ε, and also to choose N2 ∈ N such that n ≥ N2

implies bn a+ b+2ε, it follows that c is not a subsequential limit of (an + bn).

We conclude that a + b is an upper bound for the set of subsequential limits of(an + bn). Therefore lim supn→∞(an + bn) ≤ a + b.

Solution 2: This solution is a variation of Solution 1, designed to handle all casesat once. (You will see, though, that the case breakdown can’t be avoided entirely.)

As in Solution 1, define

a = lim supn→∞

an and b = lim supn→∞

bn,

and let c > a + b. We show that c is not a subsequential limit of (an + bn).We first find r, s, t ∈ R such that

a < r, b < s, c > 0, and r + s + t ≤ c.

If a = ∞ or b = ∞, this is vacuous, since no such c can exist. Next, suppose a andb are finite. If c = ∞, then

r = a + 1, s = b + 1, and c = 1

will do. Otherwise, let ε = 13 (c − a − b) > 0, and take

r = a + ε, s = b + ε, and t = ε.

Finally, suppose at least one of a and b is −∞, but neither is ∞. Exchanging thesequences if necessary, assume that a = −∞. Choose any s > b, choose any t > 0,and set r = c − s − t, which is certainly greater than −∞.

Having r, s, and t, use Theorem 3.17 (b) of Rudin to choose N1 ∈ N such thatn ≥ N1 implies an < r, and also to choose N2 ∈ N such that n ≥ N2 impliesbn < s. For n ≥ max(N1, N2), we then have an + bn < r + s. It follows that everysubsequential limit l of (an + bn) satisfies l ≤ r + s. Since c = r + s + t > r + s, itfollows that c is not a subsequential limit of (an + bn).

We conclude that a + b is an upper bound for the set of subsequential limits of(an + bn). Therefore lim supn→∞(an + bn) ≤ a + b.

Solution 3 (sketch): We only consider the case that both a = lim supn→∞ an andb = lim supn→∞ bn are finite. Let s = lim supn→∞(an + bn). Then there is asubsequence (ak(n) + bk(n)) of (an + bn) which converges to s. Further, (ak(n))is bounded. (This sequence is bounded above by assumption, and it is boundedbelow because (ak(n) + bk(n)) is bounded and (bk(n)) is bounded above.) So thereis a subsequence (al(n)) of (ak(n)) which converges. (That is, there is a strictlyincreasing function n → r(n) such that the sequence (ak◦r(n)) converges, and we letl = k ◦r : N → N. Note that if we used traditional subsequence notation, we wouldhave the subsequence j 7→ ankj

at this point.) Let c = limn→∞ al(n). By similarreasoning to that given above, the sequence (bl(n)) is bounded. Therefore it has aconvergent subsequence, say (bm(n)). (With traditional subsequence notation, wewould now have the subsequence i 7→ ankji

. You can see why I don’t like traditionalnotation.) Let d = limn→∞ bm(n). Since (am(n)) is a subsequence of (al(n)), we stillhave limn→∞ am(n) = c. So limn→∞(am(n) + bm(n)) = c+d. But also am(n) + bm(n)

is a subsequence of (ak(n) + bk(n)), and so converges to s. Therefore s = c + d. Wehave c ≤ a and d ≤ b by the definition of lim supn→∞ an and lim supn→∞ bn, givingthe result.

Solution 4 (sketch): First prove that

lim supn→∞

xn = limn→∞ sup

k≥nxk.

(We will probably prove this result in class; otherwise, see Problem A in Home-work 6. This formula is closer to the usual definition of lim supn→∞ xn, whichis

lim supn→∞

xn = infn∈N

supk≥n

xk,

using a limit instead of an infimum.)Then prove that

supk≥n

(ak + bk) ≤ supk≥n

ak + supk≥n

bk,

provided that the right hand side is defined. (For example, if both terms on the rightare finite, then the right hand side is clearly an upper bound for {ak + bk : k ≥ n}.)Now take limits to get the result.

Remark: It is quite possible to have

lim supn→∞

(an + bn) < lim supn→∞

an + lim supn→∞

bn.

Problem 3.21: Prove the following analog of Theorem 3.10(b): If

E1 ⊃ E2 ⊃ E3 ⊃ · · ·are closed bounded nonempty subsets of a complete metric space X, and if

limn→∞diam(En) = 0,

then⋂∞

n=1 En consists of exactly one point.

Solution (sketch): It is clear that⋂∞

n=1 En can contain no more than one point, sowe need to prove that

⋂∞n=1 En 6= ∅.

For each n, choose some xn ∈ En. Then, for each n, we have

{xn, xn+1, . . . } ⊂ En,

whence

diam({xn, xn+1, . . . }) ≤ diam(En).

Therefore (xn) is a Cauchy sequence. Since X is complete, x = limn→∞ xn existsin X. Since En is closed, we have x ∈ En for all n. So x ∈ ⋂∞

n=1 En.

Problem 3.22: Prove the Baire Category Theorem: If X is a complete metricspace, and if (Un) is a sequence of dense open subsets of X, then

⋂∞n=1 Un is dense

in X.

Note: In this formulation, the statement is true even if X = ∅.

Solution (sketch): Let x ∈ X and let ε > 0. We recursively construct points xn ∈ Xand numbers εn > 0 such that

d(x, x1) <ε

3, ε1 <

ε

3, εn → 0,

and

Nεn+1(xn+1) ⊂ Un+1 ∩ Nεn(xn)

for all n. Problem 3.21 will then imply that∞⋂

n=1

Nεn(xn) 6= ∅.

(Note that diam(Nεn

(xn))≤ 2εn.) One easily checks that

∞⋂n=1

Nεn(xn) ⊂ Nε(x) ∩

∞⋂n=1

Un.

Thus, we will have shown that⋂∞

n=1 Un contains points arbitrarily close to x,proving density.

Since U1 is dense in X, there is x1 ∈ U1 such that d(x, x1) < ε3 . Choose ε1 > 0

so small that

ε1 < 1, ε1 <ε

3, and N2ε1(x1) ⊂ U1.

Then also

Nε1(x1) ⊂ U1.

Given εn and xn, use the density of Un+1 in X to choose

xn+1 ∈ Un+1 ∩ Nεn/2(xn).

Choose εn+1 > 0 so small that

εn+1 <1

n + 1, εn+1 <

εn

2, and N2εn+1(xn+1) ⊂ Un+1.

Then also

Nεn+1(xn+1) ⊂ Un+1.

This gives all the required properties. (We have εn → 0 since εn < 1n for all n.)

Note: We don’t really need to use Problem 3.21 here. If we always requireεn < 2−n in the argument above, we will get d(xn, xn+1) < 2−n−1 for all n. Thisinequality implies that (xn) is a Cauchy sequence.



Problem 3.6: Investigate the convergence or divergence of the following series.(Note: I have supplied lower limits of summation, which I have chosen for max-

imum convenience. Of course, convergence is independent of the lower limit, pro-vided none of the individual terms is infinite.)

(a)∞∑

n=0

(√n + 1 −√

n).

Solution: The n-th partial sum is(√

n + 1 −√n)

+(√

n −√n − 1

)+ · · · +

(√1 −

√0)

=√

n + 1.

We have limn→∞√

n = ∞, so limn→∞√

n + 1 = ∞. Therefore the series diverges.

Remark: This sort of series is known as a telescoping series. The more interestingcases of telescoping series are the ones that converge.

Alternate solution: We calculate:

√n + 1 −√

n =(√

n + 1 −√n) ·

(√n + 1 +

√n√

n + 1 +√

n

)

=1√

n + 1 +√

n≥ 1√

n + 1 +√

n + 1=

12√

n + 1.

Now∑∞

n=11√n

diverges by Theorem 3.28 of Rudin. Therefore∑∞

n=11

2√

ndiverges,

and hence so does∑∞

n=01

2√

n+1. So the comparison test implies that

∞∑n=0

(√n + 1 −√

n)

diverges.

(b)∞∑

n=1

√n + 1 −√

n

n.


1

Solution (sketch):√

n + 1 −√n

n=

1n

(√n + 1 +

√n) ≤ 1

n3/2.

Therefore the series converges by the comparison test.

(c)∞∑

n=1

(n√

n − 1)n

.

Solution: We use the root test (Theorem 3.33 of Rudin). With an = ( n√

n − 1)n,we have n

√an = n

√n− 1. Theorem 3.20 (c) of Rudin implies that limn→∞ n

√n = 1.

Therefore limn→∞ n√

an = 0. Since limn→∞ n√

an < 1, convergence follows.

Alternate solution (sketch): Let xn = n√

n − 1, so that(n√

n − 1)n = xn

n and (1 + xn)n = n.

The binomial formula implies that

n = (1 + xn)n = 1 + nxn +n(n − 1)

2· x2

n + · · · ≥ n(n − 1)2

· x2n,

from which it follows that

0 ≤ xn ≤√

2n − 1

for n ≥ 2. Hence, for n ≥ 4,

(n√

n − 1)n = xn

n ≤(

2n − 1

)n/2

≤((

23

)1/2)n

.

Since(

23

)1/2< 1, the series converges by the comparison test.

(d)∞∑

n=0

11 + zn

,

for z ∈ C arbitrary.

Solution: We show that the series converges if and only if |z| > 1.If z = exp(2πir), with r = k/l, with k an odd integer and l an even integer, then

zn = −1 for infinitely many values of n, so that infinitely many of the terms of theseries are undefined. Convergence is therefore clearly impossible.

In all other cases with |z| ≤ 1, we have

|1 + zn| ≤ 1 + |zn| ≤ 1 + 1 = 2,

which implies that ∣∣∣∣ 11 + zn

∣∣∣∣ ≥ 12.

The terms thus don’t converge to 0, and again the series diverges.

Now let |z| > 1. Then |1 + zn| ≥ |zn| − 1 = |z|n − 1. Choose N such that ifn ≥ N then |z|n > 2. For such n, we have

|z|n2

> 1,

whence

|1 + zn| ≥ |z|n − 1 =|z|n2

+( |z|n

2− 1

)>

|z|n2

.

So ∣∣∣∣ 11 + zn

∣∣∣∣ ≤ 2 · 1|z|n

for all n ≥ N . Since |z| > 1, the comparison test implies that∞∑

n=0

11 + zn

converges.

Problem 3.7: Let an ≥ 0 for n ∈ N. Suppose∑∞

n=1 an converges. Show that∑∞n=1

1n

√an converges.

Solution (sketch): Using the inequality 2ab ≤ a2 + b2, we get√

an

n≤ 1

2

(an +

1n2

).

Since both∑∞

n=1 an and∑∞

n=11

n2 converge,∑∞

n=11n

√an converges by the com-

parison test.

Problem 3.8: Let (bn) be a bounded monotone sequence in R, and let an ∈ C besuch that

∑∞n=1 an converges. Prove that

∑∞n=1 anbn converges.

Solution (sketch): We first reduce to the case limn→∞ bn = 0. Since (bn) is abounded monotone sequence, it follows that b = limn→∞ bn exists. Set cn = bn − b.Then (cn) is a bounded monotone sequence with limn→∞ cn = 0. Since anbn =ancn +anb and

∑∞n=1 anb converges, it suffices to prove that

∑∞n=1 ancn converges.

That is, we may assume that limn→∞ bn = 0.With this assumption, if b1 ≥ 0, then b1 ≥ b2 ≥ · · · ≥ 0, so

∑∞n=1 anbn converges

by Theorem 3.42 in the book. Otherwise, replace bn by −bn.

Problem 3.9: Find the radius of convergence of each of the following power series:

(a)∞∑

n=0

n3zn.

Solution 1: Use Theorem 3.20 (c) of Rudin in the second step to get

lim supn→∞

n√

n3 =(

limn→∞

n√

n)3

= 1.

It now follows from Theorem 3.39 of Rudin that the radius of convergence is 1.

Solution 2: We show that the series converges for |z| < 1 and diverges for |z| > 1.For |z| = 0, convergence is trivial.

For 0 < |z| < 1, we use the ratio test (Theorem 3.34 of Rudin). We have

limn→∞

|(n + 1)3zn+1||n3zn| = lim

n→∞ |z|(

n + 1n

)3

= |z| limn→∞

(1 +

1n

)3

= |z|(

1 + limn→∞

1n

)3

= |z|.

For |z| < 1 the hypotheses of Theorem 3.34 (a) of Rudin are therefore satisfied, sothat the series converges.

For |z| > 1, we use the ratio test. The same calculation as in the case 0 < |z| < 1gives

limn→∞

|(n + 1)3zn+1|n3zn| = |z|.

Since |z| > 1, it follows that there is N such that for all n ≥ N we have∣∣∣∣ |(n + 1)3zn+1||n3zn| − |z|

∣∣∣∣ < 12 (|z| − 1).

In particular,

|(n + 1)3zn+1||n3zn| > 1

for n ≥ N . The hypotheses of Theorem 3.34 (b) of Rudin are therefore satisfied,so that the series diverges.

Theorem 3.39 of Rudin implies that there is some number R ∈ [0,∞] such thatthe series converges for |z| < R and diverges for |z| > R. We have therefore shownthat R = 1.

Solution 3: We calculate

limn→∞

|(n + 1)3||n3| = lim

n→∞

(1 +

1n

)3

=(

1 + limn→∞

1n

)3

= 1.

According to Theorem 3.37 of Rudin, we have

lim infn→∞

|(n + 1)3||n3| ≤ lim inf

n→∞n√|n3| ≤ lim sup

n→∞n√|n3| ≤ lim sup

n→∞|(n + 1)3|

|n3| .

Therefore limn→∞ n√|n3| exists and is equal to 1. It now follows from Theorem 3.39

of Rudin that the radius of convergence is 1.

(b)∞∑

n=0

2n

n!· zn.

Solution (sketch): Use the ratio test to show that the series converges for all z.(See Solution 2 to Part (a).) So the radius of convergence is ∞.

Remark: Note that∞∑

n=0

2n

n!· zn = e2z.

(c)∞∑

n=0

2n

n2· zn.

Solution (sketch): Either the root or ratio test gives radius of convergence equal to12 . (Use the methods of any of the three solutions to Part (a).)

(d)∞∑

n=0

n3

3n· zn.

Solution (sketch): Either the root or ratio test gives radius of convergence equal to3. (Use the methods of any of the three solutions to Part (a).)

Problem 3.10: Suppose the coefficients of the power series∑∞

n=0 anzn are inte-gers, infinitely many of which are nonzero. Prove that the radius of convergence isat most 1.

Solution 1: For infinitely many n, the numbers an are nonzero integers, and there-fore satisfy |an| ≥ 1. So, if |z| ≥ 1, then infinitely many of the terms anzn haveabsolute value |anzn| ≥ |an| ≥ 1, and the terms of the series

∑∞n=0 anzn don’t

approach zero. This shows that∑∞

n=0 anzn diverges for |z| ≥ 1, and therefore thatits radius of convergence is at most 1.

Solution 2 (Sketch): There is a subsequence (ak(n)) of (an) such that |ak(n)| ≥ 1for all n. So k(n)

√|ak(n)| ≥ 1 for all n, whence lim supn→∞n√|an| ≥ 1.

Remarks: (1) It is quite possible that infinitely many of the an are zero, so thatlim infn→∞ n

√|an| could be zero. For example, we could have an = 0 for all odd n.(2) In Solution 2, the expression n

√|ak(n)|, and its possible limit as n → ∞, haveno relation to the radius of convergence.

Problem 3.16: Fix α > 0. Choose x1 >√

α, and recursively define

xn+1 =12

(xn +

α

xn

).

(a) Prove that (xn) is nonincreasing and limn→∞ xn =√

α.

Solution (sketch): Using the inequality a2 + b2 ≥ 2ab, and assuming xn > 0, we get

xn+1 =12

(xn +

α

xn

)≥ √

xn ·√

α

xn=

√α.

This shows (using induction) that xn >√

α for all n. Next,

xn − xn+1 =x2

n − α

2xn> 0.

Thus (xn) is nonincreasing. We already know that this sequence is bounded below(by α), so x = limn→∞ xn exists. Letting n → ∞ in the formula

xn+1 =12

(xn +

α

xn

).

gives

x =12

(x +

α

x

).

This equation implies x = ±√α, and we must have x =

√α because (xn) is bounded

below by√

α > 0.

(b) Set εn = xn −√α, and show that

εn+1 =ε2

n

2xn<

ε2n

2√

α.

Further show that, with β = 2√

α,

εn+1 < β

(ε1

β

)2n

.

Solution (sketch): Prove all the relations at once by induction on n, together with

the statement εn+1 > 0. For n = 1, the relation εn+1 =ε2

n

2xnis just algebra, the

inequalityε2

n

2xn<

ε2n

2√

αfollows from x1 >

√α, and the inequality εn+1 < β

(ε1

β

)2n

is just a rewritten form ofε2

n

2xn<

ε2n

2√

α. The statement εn+1 > 0 is clear from

εn+1 =ε2

n

2xnand x1 > 0.

Now assume all this is known for some value of n. As before, the relation

εn+1 =ε2

n

2xnis just algebra, and implies that εn+1 > 0. (We know that xn =

√α+εn >

√α > 0.) Since εn > 0, we have xn >

√α, so the inequality

ε2n

2xn<

ε2n

2√

αfollows. To get the other inequality, write

εn+1 <ε2

n

2√

α=

ε2n

β<

(1β

)·[β ·

(ε1

β

)2n−1]2

= β

(ε1

β

)2n

.

(c) Specifically take α = 3 and x1 = 2. show thatε1

β<

110

, ε5 < 4 · 10−16, and ε6 < 4 · 10−32.

Solution (sketch): This is just calculation.

Problem 3.23: Let X be a metric space, and let (xn)n∈N and (yn)n∈N be Cauchysequences in X. Prove that limn→∞ d(xn, yn) exists.

Solution (sketch): Since R is complete, it suffices to show that (d(xn, yn))n∈N isa Cauchy sequence. Let ε > 0. Choose N so large that if m, n ≥ N , then bothd(xm, xn) < ε

2 and d(ym, yn) < ε2 . Then check that, for such m and n,

|d(xm, ym) − d(xn, yn)| ≤ d(xm, xn) + d(ym, yn) < ε.

Problem 3.24: Let X be a metric space.(a) Let (xn)n∈N and (yn)n∈N be Cauchy sequences in X. We say they are

equivalent, and write (xn)n∈N ∼ (yn)n∈N, if limn→∞ d(xn, yn) = 0. Prove thatthis is an equivalence relation.

Solution (sketch): That (xn)n∈N ∼ (xn)n∈N, and that (xn)n∈N ∼ (yn)n∈N implies(yn)n∈N ∼ (xn)n∈N, are obvious. For transitivity, assume (xn)n∈N ∼ (yn)n∈N

and (yn)n∈N ∼ (zn)n∈N. Then (xn)n∈N ∼ (zn)n∈N follows by taking limits in theinequality

0 ≤ d(xn, zn) ≤ d(xn, yn) + d(yn, zn).

(b) Let X∗ be the set of equivalence classes from Part (a). Denote by [(xn)n∈N]the equivalence class in X∗ of the Cauchy sequence (xn)n∈N. If (xn)n∈N and(yn)n∈N are Cauchy sequences in X, set

∆0((xn)n∈N, (yn)n∈N) = limn→∞ d(xn, yn).

Prove that ∆0((xn)n∈N, (yn)n∈N) only depends on [(xn)n∈N] and [(yn)n∈N]. More-over, show that the formula

∆([(xn)n∈N], [(yn)n∈N]) = ∆0((xn)n∈N, (yn)n∈N)

defines a metric on X∗.

Solution (sketch): It is easy to check that ∆0 is a semimetric, that is, it satisfies allthe conditions for a metric except that possibly

∆0((xn)n∈N, (yn)n∈N) = 0

Date: 5 November 2001.

1

without having (xn)n∈N = (yn)n∈N. (For example,

∆0((xn)n∈N, (zn)n∈N) = limn→∞ d(xn, zn) ≤ lim

n→∞ d(xn, yn) + limn→∞ d(yn, zn)

= ∆0((xn)n∈N, (yn)n∈N) + ∆0((yn)n∈N, (zn)n∈N)

because d(xn, zn) ≤ d(xn, yn) + d(yn, zn); the other properties are proved simi-larly.) We further note that, by definition, (xn)n∈N ∼ (yn)n∈N if and only if∆0((xn)n∈N, (yn)n∈N) = 0.

Now we prove that ∆0((xn)n∈N, (yn)n∈N) only depends on

[(xn)n∈N] and [(yn)n∈N].

Let (xn)n∈N ∼ (rn)n∈N and (yn)n∈N ∼ (sn)n∈N. Then, by the previous paragraph,

∆0((xn)n∈N, (yn)n∈N)

≤ ∆0((xn)n∈N, (rn)n∈N) + ∆0((rn)n∈N, (sn)n∈N) + ∆0((sn)n∈N, (yn)n∈N)

= 0 + ∆0((rn)n∈N, (sn)n∈N) + 0 = ∆0((rn)n∈N, (sn)n∈N);

similarly

∆0((rn)n∈N, (sn)n∈N) ≤ ∆0((xn)n∈N, (yn)n∈N).

Thus

∆0((rn)n∈N, (sn)n∈N) = ∆0((xn)n∈N, (yn)n∈N).

The previous paragraph implies that ∆ is well defined. It is now easy to checkthat ∆ satisfies all the conditions for a metric except that possibly

∆([(xn)n∈N], [(yn)n∈N]) = 0

without having [(xn)n∈N] = [(yn)n∈N]. (For example,

∆([(xn)n∈N], [(zn)n∈N]) = ∆0((xn)n∈N, (zn)n∈N)

≤ ∆0((xn)n∈N, (yn)n∈N) + ∆0((yn)n∈N, (zn)n∈N)

= ∆([(xn)n∈N], [(yn)n∈N]) + ∆0([(yn)n∈N], [(zn)n∈N]);

the other properties are proved similarly.)Finally, if ∆([(xn)n∈N], [(yn)n∈N]) = 0 then it follows from the definition of

(xn)n∈N ∼ (yn)n∈N that we actually do have [(xn)n∈N] = [(yn)n∈N]. So ∆ is ametric.

(c) Prove that X∗ is complete in the metric ∆.

The basic idea is as follows. We start with a Cauchy sequence in X∗, which is asequence of (equivalence classes of) Cauchy sequences in X. The limit is supposedto be (the equivalence class of) another Cauchy sequence in X. This sequence isconstructed by taking suitable terms from the given sequences. The choices get alittle messy. Afterwards, we will give a different proof.

Solution (sketch): Let (ak)k∈N be a Cauchy sequence in X∗; we show that itconverges. Each ak is an equivalence class of Cauchy sequences in X. We maytherefore write ak = [(x(k)

n )n∈N], where each (x(k)n )n∈N is a Cauchy sequence in

X. The limit we construct in X∗ will have the form a = [(x(f(n))n )] for a suitable

function f : N → N.We recursively construct

M(1) < M(2) < M(3) < · · · and N(1) < N(2) < N(3) < · · ·

such that(1) ∆(ak, al) < 2−r−2 for k, l ≥ M(r).(2) For all k, l ≤ M(r) and m ≥ N(r), we have d(x(k)

m , x(l)m ) < ∆(ak, al) +

2−r−2.(3) For all k ≤ M(r) and m, n ≥ N(r), we have d(x(k)

m , x(k)n ) < 2−r−2.

To do this, first use the fact that (ak)k∈N is a Cauchy sequence to find M(1).Then choose N(1) large enough to satisfy (2) and (3) for r = 1; this can be donebecause limm→∞ d(x(k)

m , x(l)m ) = ∆(ak, al) (for (2)) and because (x(k)

n )n∈N is Cauchy(for (3)), and using the fact that there are only finitely many pairs (k, l) to considerin (2) and only finitely many k to consider in (3). Next, use the fact that (ak)k∈N

is a Cauchy sequence to find M(2), and also require M(2) > M(1). Choose N(2) >N(1) by the same reasoning as used to get N(1). Proceed recursively.

We now take a to be the equivalence class of the sequence

x(1)1 , . . . , x

(1)N(1)−1, x

(M(1))N(1) , . . . , x

(M(1))N(2)−1, x

(M(2))N(2) , . . . , x

(M(2))N(3)−1, x

(M(3))N(3) , . . . .

That is, the function f above is given by f(n) = M(r) for N(r) ≤ n ≤ N(r+1)−1.We show that (x(f(n))

n )n∈N is Cauchy. First, estimate:

d(x

(M(r))N(r) , x

(M(r+1))N(r+1)

)≤ d

(x

(M(r))N(r) , x

(M(r))N(r+1)

)+ d

(x

(M(r))N(r+1), x

(M(r+1))N(r+1)

)

< 2−r−2 + ∆(aM(r), aM(r+1)) + 2−r−3

< 2−r−2 + 2−r−2 + 2−r−3 < 3 · 2−r−2.

The first term on the second line is gotten from (2) above, because M(r) ≤ M(r)and N(r), N(r + 1) ≥ N(r). The other two terms on the second line are gottenfrom (3) above (for r + 1)), because M(r), M(r + 1) ≤ M(r + 1) and N(r + 1) ≥N(r + 1). The estimate used to get the third line comes from (1) above. Then useinduction to show that s ≥ r implies

d(x

(M(r))N(r) , x

(M(s))N(s)

)≤ 3[2−r−2 + 2−r−3 + · · · + 2−s−1].

Now let n ≥ N(r) be arbitrary. Choose s ≥ r such that N(s) ≤ n ≤ N(s+1)−1.Then f(n) = M(s), so

d(x(f(n))

n , x(M(s))N(s)

)= d

(x(M(s))

n , x(M(s))N(s)

)< 3 · 2−s−2,

using (3) above with r = s and k = M(s). Therefore

d(x

(M(r))N(r) , x(f(n))

n

)< 3[2−r−2 + 2−r−3 + · · · + 2−s−1 + 2−s−2] < 3 · 2−r−1.

Finally, if m, n ≥ N(r) are arbitrary, then

d(x(f(m))

m , x(f(n))n

)≤ d

(x(f(m))

m , x(M(r))N(r)

)+ d

(x

(M(r))N(r) , x(f(n))

n

)< 3 · 2−r.

This is enough to prove that (x(f(n))n )n∈N is Cauchy.

It remains to show that ∆(ak, a) → 0. Fix k, choose r with M(r−1) < k ≤ M(r),and let n ≥ N(r). From the previous paragraph we have

d(x

(M(r))N(r) , x(f(n))

n

)< 3 · 2−r−1.

Since n, N(r) ≥ N(r) and k ≤ M(r), condition (3) above gives

d(x(k)

n , x(k)N(r)

)< 2−r−2.

Furthermore,

d(x

(k)N(r), x

(M(r))N(r)

)< ∆(ak, aM(r)) + 2−r−2 < 2−r−1 + 2−r−2,

where the first step uses (2) above and the inequalities k, M(r) ≤ M(r) and N(r) ≥N(r), while the second step uses (1) above and the inequality r ≥ M(r − 1).Combining these estimates using the triangle inequality, we get

d(x(k)n , x(f(n))

n ) < 2−r−2 + [2−r−1 + 2−r−2] + 3 · 2−r−1 < 2−r+2.

Therefore

∆(ak, a) = limn→∞ d(x(k)

n , x(f(n))n ) ≤ 2−r+2

for M(r − 1) < k ≤ M(r). Since M(r) → ∞, this implies that ∆(ak, a) → 0.

Here is a perhaps slicker way to do the same thing, although it isn’t any shorter.Essentially, by passing to suitable subsequences, we can take the representative ofthe limit to be the diagonal sequence, that is, f(n) = n in the proof above. Theconstruction requires the following lemmas.

Lemma 1. Let (xn)n∈N be a Cauchy sequence in a metric space Y . Then there isa subsequence (xk(n))n∈N of (xn)n∈N such that d(xk(n+1), xk(n)) < 2−n for all n.

Proof (sketch): Choose k(n) recursively to satisfy k(n + 1) > k(n) and d(xl, xm) <2−n for all l, m ≥ k(n).

Lemma 2. Let (xn)n∈N be a sequence in a metric space Y . Suppose∞∑

k=1

d(xk, xk+1)

converges. Then (xn)n∈N is Cauchy. Moreover, if n > m then

d(xm, xn) ≤n−1∑k=m

d(xk, xk+1).

Proof (sketch): The Cauchy criterion for convergence of a series implies that forall ε > 0, there is N such that if n > m ≥ N , then

∑n−1k=m d(xk, xk+1) < ε. But

the triangle inequality gives d(xm, xn) ≤ ∑n−1k=m d(xk, xk+1). Thus if n > m ≥ N

then d(xm, xn) < ε. The case m > n ≥ N is handled by symmetry, and the casen = m ≥ N is trivial.

Lemma 3. Let (xn)n∈N be a Cauchy sequence in a metric space Y , and let(xk(n))n∈N be a subsequence. Then limn→∞ d(xn, xk(n)) = 0.

Proof (sketch): Let ε > 0. Choose N such that if m, n ≥ N then d(xm, xn) < ε. Ifn ≥ N , then k(n) ≥ n ≥ N , so d(xn, xk(n)) < ε.

Lemma 4. Let (xn)n∈N be a Cauchy sequence in a metric space Y . If (xn)n∈N

has a convergent subsequence, then (xn)n∈N converges.

Proof (sketch): Let (xk(n))n∈N be a subsequence with limit x. Then, usingLemma 3,

d(xn, x) ≤ d(xn, xk(n)) + d(xk(n), x) → 0.

Proof of the result: Let (ak) be a Cauchy sequence in X∗; we show that it converges.Use Lemma 1 to choose a subsequence (ar(k))k∈N such that

∆(ar(k), ar(k+1)) < 2−k

for all k. By Lemma 4, it suffices to show that (ar(k))k∈N converges. Withoutloss of generality, therefore, we may assume the original sequence (ak)k∈N satisfies∆(ak, ak+1) < 2−k for all k.

Each ak is an equivalence class of Cauchy sequences in X. By Lemmas 1 and 3,we may write ak = [(x(k)

n )n∈N], with d(x(k)n , x

(k)n+1) < 2−n for all n.

We now estimate d(x(k)n , x

(k+1)n ). For ε > 0, we can find m > n such that

d(x(k)m , x(k+1)

m ) < ∆([(x(k)n )n∈N], [(x(k+1)

n )n∈N]) + ε

= ∆(ak, ak+1) + ε < 2−k + ε.

Now, using Lemma 2,

d(x(k)n , x(k)

m ) < 2−n + 2−n−1 + · · · + 2−m+1 < 2−n+1.

The same estimate holds for d(x(k+1)n , x

(k+1)m ). Therefore

d(x(k)n , x(k+1)

n ) ≤ d(x(k)n , x(k)

m ) + d(x(k)m , x(k+1)

m ) + d(x(k+1)m , x(k+1)

n )

< 2−n+1 + ε + 2−n+1.

Since ε > 0 is arbitrary, this gives

d(x(k)n , x(k+1)

n ) ≤ 2−n+2

for all n and k.Now define yn = x

(n)n . First, observe that

d(yn, yn+1) ≤ d(x(n)n , x

(n)n+1) + d(x(n)

n+1, x(n+1)n+1 )

≤ 2−n + 2−n+1 < 2−n+2.

Therefore (yn)n∈N is Cauchy, by Lemma 2. So a = [(yn)n∈N] ∈ X∗.It remains to show that ∆(an, a) → 0. If m > n, then we use the estimates

d(x(n)n , x

(n)m ) < 2−n+1 (as above) and d(yn, ym) < 2−n+3 (obtained similarly, using

Lemma 2 again) to get

d(x(n)m , ym) ≤ d(x(n)

m , x(n)n ) + d(yn, ym) < 2−n+4.

In particular,

∆(an, a) = limm→∞ d(x(n)

m , ym) ≤ 2−n+4.

Thus ∆(an, a) → 0, as desired.

(d) Define f : X → X∗ by f(x) = [(x, x, x, . . . )]. Prove that f is isometric, thatis, that ∆(f(x), f(y)) = d(x, y) for all x, y ∈ X.

Solution (sketch): This is immediate.

(e) Prove that f(X) is dense in X∗, and that f(X) = X∗ if X is complete.

Solution (sketch): To prove density, let [(xn)n∈N] ∈ X∗, and let ε > 0. Choose Nsuch that if m, n ≥ N then d(xm, xn) < ε

2 . Then

∆(f(xN ), [(xn)n∈N]) = limn→∞ d(xN , xn),

which is at most ε2 because d(xN , xn) < ε

2 for n ≥ N . (Note that the limit existsby Problem 3.23.) In particular, ∆(f(xN ), [(xn)n∈N]) < ε.

Now assume X is complete. Then f(X) is complete, because f is isometric.Therefore it suffices to prove that a complete subset of a metric space is closed. (Asubset of a metric space which is both closed and dense must be equal to the wholespace.)

Accordingly, let Y be a metric space, and let E ⊂ Y be a complete subset. Let(xn)n∈N be a sequence in E which converges to some point y ∈ Y ; we show y ∈ E.(By a theorem proved in class, this is sufficient to verify that E is closed.) Now(xn)n∈N converges, and is therefore Cauchy. Since E is complete, there is x ∈ Esuch that xn → x. By uniqueness of limits, we have x = y. Thus y ∈ E, as desired.

Problem A. Prove the equivalence of four definitions of the lim sup of a sequence.That is, prove the following theorem.

Theorem. Let (an) be a sequence in R. Let E be the set of all subsequentiallimits of (an) in [−∞,∞]. Define numbers r, s, t, and u ∈ [−∞,∞] as follows:

(1) r = sup(E).(2) s ∈ E and for every x > s, there is N ∈ N such that n ≥ N implies an < x.(3) t = infn∈N supk≥n ak.(4) u = limn→∞ supk≥n ak.

Prove that s is uniquely determined by (2), that the limit in (4) exists in [−∞,∞],and that r = s = t = u.

Note: You do not need to repeat the part that is done in the book (Theo-rem 3.17).

Solution: Theorem 3.17 of Rudin implies that s is uniquely determined by (2) andthat s = r.

Define bn = supk≥n ak, which exists in (−∞,∞]. We clearly have

{ak : k ≥ n + 1} ⊂ {ak : k ≥ n},so that supk≥n+1 ak ≤ supk≥n ak. This shows that the sequence in (4), which hasvalues in (−∞,∞], is nonincreasing. Therefore it has a limit u ∈ [−∞,∞], andmoreover

u = infn∈N

bn = infn∈N

supk≥n

ak = t.

We finish the proof by showing that s ≤ t and t ≤ r.To show that s ≤ t, let x > s; we show that x > t. Choose y with x > y > s. By

the definition of s, there is N ∈ N such that n ≥ N implies an < y. This impliesthat y ≥ supn≥N an, so that x > supn≥N an. It follows that x is not a lower boundfor {supn≥N an : N ∈ N}. So x > t by the definition of a greatest lower bound.

To show that t ≥ r, let x > t; we show that x ≥ r. Since x > t, it followsthat x is not a lower bound for the set {supn≥N an : N ∈ N}. Accordingly, there isN0 ∈ N such that supn≥N0

an < x. In particular, n ≥ N0 implies an < x. Now let(ak(n)) be any convergent subsequence of (an). Choose N ∈ N such that n ≥ Nimplies k(n) ≥ N0. Then n ≥ N implies ak(n) < x, from which it follows thatlimn→∞ ak(n) ≤ x. This shows that x is an upper bound for the set E, so thatx ≥ sup(E) = r.

Generally, a “solution” is something that would be acceptable if turned in inthe form presented here, although the solutions given are usually close to minimalin this respect. A “solution (sketch)” is too sketchy to be considered a completesolution if turned in; varying amounts of detail would need to be filled in.

Problem 4.1: Let f : R → R satisfy limh→0[f(x+h)−f(x−h)] = 0 for all x ∈ R.Is f necessarily continuous?

Solution (Sketch): No. The simplest counterexample is

f(x) ={

0 x 6= 01 x = 0 .

More generally, let f0 : R → R be continuous. Fix x0 ∈ R, and fix y0 ∈ R withy0 6= f0(x0). Then the function given by

f(x) ={

f0(x) x 6= x0

y0 x = x0

is a counterexample. There are even examples with a nonremovable discontinuity,such as

f(x) ={ 1

|x| x 6= 00 x = 0

.

Problem 4.3: Let X be a metric space, and let f : X → R be continuous. LetZ(f) = {x ∈ X : f(x) = 0}. Prove that Z(f) is closed.

Solution 1: The set Z(f) is equal to f−1({0}). Since {0} is a closed subset of Rand f is continuous, it follows from the Corollary to Theorem 4.8 of Rudin thatZ(f) is closed in X.

Solution 2: We show that X \ Z(f) is open. Let x ∈ X \ Z(f). Then f(x) 6= 0.Set ε = 1

2 |f(x)| > 0. Choose δ > 0 such that y ∈ X and d(x, y) < δ imply|f(x) − f(y)| < ε. Then f(y) 6= 0 for y ∈ Nδ(x). Thus Nδ(x) ⊂ X \ Z(f) withδ > 0. This shows that X \ Z(f) is open.

Note that we really could have taken ε = |f(x)|. Also, there is no need to doanything special if Z(f) is empty, or even to mention the that case separately: theargument works (vacuously) just as well in that case.

Solution 3 (sketch): We show Z(f) contains all its limit points. Let x be a limitpoint of Z(f). Then there is a sequence (xn) in Z(f) such that xn → x. Since f iscontinuous and f(xn) = 0 for all n, we have

f(x) = limn→∞ f(xn) = 0.

So x ∈ Z(f).

Date: 12 Nov. 2001.

1

Again, there is no need to treat separately the case in which Z(f) has no limitpoints.

Problem 4.4: Let X and Y be metric spaces, and let f, g : X → Y be continuousfunctions. Let E ⊂ X be dense. Prove that f(E) is dense in f(X). Prove that iff(x) = g(x) for all x ∈ E, then f = g.

Solution: We first show that f(E) is dense in f(X). Let y ∈ f(X). Choose x ∈ Xsuch that f(x) = y. Since E is dense in X, there is a sequence (xn) in E such thatxn → x. Since f is continuous, it follows that f(xn) → f(x). Since f(xn) ∈ f(E)for all n, this shows that x ∈ f(E).

Now assume that f(x) = g(x) for all x ∈ E; we prove that f = g. It suffices toprove that F = {x ∈ X : f(x) = g(x)} is closed in X, and we prove this by showingthat X \ F is open. Thus, let x0 ∈ F . Set ε = 1

2d(f(x0), g(x0)) > 0. Chooseδ1 > 0 such that if x ∈ X satisfies d(x, x0) < δ, then d(f(x), f(x0)) < ε. Chooseδ2 > 0 such that if x ∈ X satisfies d(x, x0) < δ, then d(g(x), g(x0)) < ε. Setδ = min(δ1, δ2). If d(x, x0) < δ, then (using the triangle inequality several times)

d(f(x), g(x)) ≥ d(f(x0), g(x0)) − d(f(x), f(x0)) − d(g(x), g(x0))

> d(f(x0), g(x0)) − ε − ε = 0.

So f(x) 6= g(x). This shows that Nδ(x0) ⊂ X \ F , so that X \ F is open.

The second part is closely related to Problem 4.3. If Y = R (or Cn, or . . . ),then {x ∈ X : f(x) = g(x)} = Z(f − g), and f − g is continuous when f andg are. For general Y , however, this solution fails, since f − g won’t be defined.The argument given is the analog of Solution 2 to Problem 4.3. The analog ofSolution 3 to Problem 4.3 also works the same way: F is closed because if xn → xand f(xn) = g(xn) for all n, then limn→∞ f(xn) = limn→∞ g(xn). The analog ofSolution 1 can actually be patched in the following way (using the fact that theproduct of two metric spaces is again a metric space): Define h : X → Y × Y byh(x) = (f(x), g(x)). Then h is continuous and D = {(y, y) : y ∈ Y } ⊂ Y × Y isclosed, so {x ∈ X : f(x) = g(x)} = h−1(D) is closed.

Problem 4.6: Let E ⊂ R be compact, and let f : E → R be a function. Provethat f is continuous if and only if the graph G(f) = {(x, f(x)) : x ∈ E} ⊂ R2 iscompact.

Remark: This statement is my interpretation of what was intended. Normallyone would assume that E is supposed to be a compact subset of an arbitrary metricspace X, and that f is supposed to be a function from E to some other metric spaceY . (In fact, one might as well assume E = X.) The proofs are all the same (withone exception, noted below), but require the notion of the product of two metricspaces. We make X × Y into a metric space via the metric

d((x1, y1), (x2, y2)) =√

d(x1, x2)2 + d(y1, y2)2;

there are other choices which are easier to deal with and work just as well.

We give several solutions for each direction. We first show that if f is continuousthen G(f) is compact.


Solution 1 (Sketch): The map x 7→ (x, f(x)) is easily checked to be continuous,and G(f) is the image of the compact set E under this map, so G(f) is compactby Theorem 4.14 of Rudin.

Solution 2 (Sketch): The graph of a continuous function is closed, as can be verifiedby arguments similar to those of Solutions 2 and 3 to Problem 4.3. The graph is asubset of E×f(E). This set is bounded (clear) and closed (check this!) in R2, and istherefore compact. (Note: This does not work for general metric spaces. However,it is true in general that the product of two compact sets, with the product metric,is compact.) Therefore the closed subset G(f) is compact.

Now we show that if G(f) is compact then f is continuous.

Solution 1 (Sketch): We know that the function g0 : E×R → E, given by g0(x, y) =x, is continuous. (See Example 4.11 of Rudin.) Therefore g = g0|G(f) : G(f) → Eis continuous. Also g is bijective (because f is a function). Since G(f) is compact,it follows (Theorem 4.17 of Rudin) that g−1 : E → G(f) is continuous. Further-more, the function h : E × R → R, given by h(x, y) = y, is continuous, again byExample 4.11 of Rudin. Therefore f = h ◦ g−1 is continuous.

Solution 2 (Sketch): Let (xn) be a sequence in E with xn → x. We show thatf(xn) → f(x). We do this by showing that every subsequence of (f(xn)) has inturn a subsubsequence which converges to f(x). (To see that this is sufficient, let(yn) be a sequence in some metric space Y , let y ∈ Y , and suppose that (yn) does notconverge to y. Find a subsequence (yk(n)) of (yn) such that infn∈N d(yk(n), y) > 0.Then no subsequence of (yk(n)) can converge to y.)

Accordingly, let (f(xk(n))) be a subsequence of (f(xn)). Let (xk(n)) be the cor-responding subsequence of (xn). If (xk(n)) is eventually constant, then alreadyf(xk(n)) → f(x). Otherwise, {xk(n) : n ∈ N} is an infinite set, whence so is{(xk(n), f(xk(n))) : n ∈ N} ⊂ G(f). Since G(f) is compact, this set has a limitpoint, say (a, b). It is easy to check that a must equal x. Since G(f) is compact, itis closed, so b = f(x). Since (a, b) is a limit point of G(f), there is a subsequenceof

((xk(n), f(xk(n)))

)which converges to (a, b). Using continuity of projection onto

the second coordinate, we get a subsequence of(f(xk(n))

)which converges to f(x).

Solution 3: We first observe that the range Y = {f(x) : x ∈ E} of f is compact.Indeed, Y is the image of G(f) under the map (x, y) 7→ y, which is continuous byExample 4.11 of Rudin. So Y is compact by Theorem 4.14 of Rudin. It suffices toprove that f is continuous as a function from E to Y , as can be seen, for example,from the sequential criterion for limits (Theorem 4.2 of Rudin).

Now let x0 ∈ E and let V ⊂ Y be an open set containing f(x0). We must findan open set U ⊂ E containing x0 such that f(U) ⊂ V . For each y ∈ Y \ V , thepoint (x0, y) is not in the closed set G ⊂ E × R. Therefore there exist open setsRy ⊂ E containing x0 and Sy ⊂ Y containing y such that (Ry × Sy) ∩ G = ∅.Since Y \ V is compact, there are n and y(1), . . . , y(n) ∈ Y \ V such that the setsSy(1), . . . , Sy(n) cover Y \ V . Set U = Ry(1) ∩ · · · ∩ Ry(n) to obtain f(U) ⊂ V .


Problem 4.7: Define f, g : R2 → R by

f(x, y) =

xy2

x2 + y4(x, y) 6= (0, 0)

0 (x, y) = (0, 0)

and

g(x, y) =

xy2

x2 + y6(x, y) 6= (0, 0)

0 (x, y) = (0, 0).

Prove:(1) f is bounded.(2) f is not continuous at (0, 0).(3) The restriction of f to every straight line in R2 is continuous.(4) g is not bounded on any neighborhood of (0, 0).(5) g is not continuous at (0, 0).(6) The restriction of g to every straight line in R2 is continuous.

Solution (Sketch): (1) We use the inequality 2ab ≤ a2 + b2 (which follows froma2 + b2 − 2ab = (a − b)2 ≥ 0). Taking a = |x| and b = y2, we get 2|x|y2 ≤ x2 + y4,which implies |f(x, y)| ≤ 1

2 for all (x, y) ∈ R2.(2) Set xn = 1

n2 and yn = 1n . Then (xn, yn) → (0, 0), but f(xn, yn) = 1

2 6→ 0 =f(0, 0).

(3) Clearly f is continuous on R2 \ {(0, 0)}, so the restriction of f to everystraight line in R2 not going through (0, 0) is clearly continuous. Furthermore, therestriction of f to the y-axis is given by (0, y) 7→ 0, which is clearly continuous.

Every other line has the form y = ax for some a ∈ R. We have

f(x, ax) =a2x

1 + a4x2

for all x ∈ R, so the restriction of f to this line is given by the continuous function

(x, y) 7→ a2x

1 + a4x2.

(4) Set xn = 1n3 and yn = 1

n . Then (xn, yn) → (0, 0), but g(xn, yn) = n → ∞.(5) This is immediate from (4).(6) Clearly g is continuous on R2 \ {(0, 0)}, so the restriction of g to every

straight line in R2 not going through (0, 0) is clearly continuous. Furthermore, therestriction of g to the y-axis is given by (0, y) 7→ 0, which is clearly continuous.

Every other line has the form y = ax for some a ∈ R. We have

g(x, ax) =a3x

1 + a6x4

for all x ∈ R, so the restriction of g to this line is given by the continuous function

(x, y) 7→ a3x

1 + a6x4.

Problem 4.8: Let E ⊂ R be bounded, and let f : E → R be uniformly continu-ous. Prove that f is bounded. Show that a uniformly continuous function on anunbounded subset of R need not be bounded.

Solution (Sketch): Choose δ > 0 such that if x, y ∈ E satisfy |x − y| < δ, then|f(x) − f(y)| < 1. Choose n ∈ N such that 1

n < δ. Since E is a bounded subsetof R, there are finitely many closed intervals

[ak, ak + 1

n

]whose union contains

the closed interval [inf(E), sup(E)] and hence also E. Let S be the finite set ofthose k for which E ∩ [

ak, ak + 1n

] 6= ∅. Thus E ⊂ ⋃k∈S

[ak, ak + 1

n

]. Choose

bk ∈ E ∩ [ak, ak + 1

n

]. Set M = 1 + maxk∈S |f(bk)|.

We show that |f(x)| ≤ M for all x ∈ E. For such x, choose k ∈ S suchthat x ∈ [

ak, ak + 1n

]. Then |x − bk| ≤ 1

n < δ, so |f(x) − f(bk)| < 1. Thus|f(x)| ≤ |f(x) − f(bk)| + |f(bk)| < 1 + M .

As a counterexample with E unbounded, take E = R and f(x) = x for all x.

Problem 4.9: Let X and Y be metric spaces, and let f : X → Y be a function.Prove that f is uniformly continuous if and only if for every ε > 0 there is δ > 0such that whenever E ⊂ X satisfies diam(E) < δ, then diam(f(E)) < ε.

Solution: Let f be uniformly continuous, and let ε > 0. Choose δ > 0 suchthat if x1, x2 ∈ X satisfy d(x1, x2) < δ, then d(f(x1), f(x2)) < 1

2ε. Let E ⊂X satisfy diam(E) < δ. We show that diam(f(E)) < ε. Let y1, y2 ∈ f(E).Choose x1, x2 ∈ E such that f(x1) = y1 and f(x2) = y2. Then d(x1, x2) < δ,so d(f(x1), f(x2)) < 1

2ε. This shows that d(y1, y2) < 12ε for all y1, y2 ∈ f(E).

Therefore

diam(f(E)) = supy1,y2∈E

d(y1, y2) ≤ 12ε < ε.

Now assume that for every ε > 0 there is δ > 0 such that whenever E ⊂X satisfies diam(E) < δ, then diam(f(E)) < ε. We prove that f is uniformlycontinuous. Let ε > 0. Choose δ > 0 as in the hypotheses. Let x1, x2 ∈ Xsatisfy d(x1, x2) < δ. Set E = {x1, x2}. Then diam(E) < δ. So d(f(x1), f(x2)) =diam(f(E)) < ε.

Problem 4.10: Use the fact that infinite subsets of compact sets have limit pointsto give an alternate proof that if X and Z are metric spaces with X compact, andf : X → Z is continuous, then f is uniformly continuous.

Date: 19 Nov. 2001.

1

Solution: Assume that f is not uniformly continuous. Choose ε > 0 for which thedefinition of uniform continuity fails. Then for every n ∈ N there are xn, yn ∈ Xsuch that d(xn, yn) < 1

n and d(f(xn), f(yn)) ≥ ε. Since X is compact, the sequence(xn) has a convergent subsequence. (See Theorem 3.6 (a) of Rudin.) Let x =limn→∞ xk(n). Since d(xk(n), yk(n)) < 1

k(n) ≤ 1n , we also have limn→∞ yk(n) = x.

If f were continuous at x, we would have

limn→∞ f(xk(n)) = lim

n→∞ f(yk(n)) = f(x).

This contradicts d(f(xn), f(yn)) ≥ ε for all n. To see this, choose N ∈ N such thatn ≥ N implies

d(f(xk(n)), f(x)) < 13ε and d(f(yk(n)), f(x)) < 1

3ε.

Then

d(f(xk(N)), f(yk(N)) ≤ d(f(xk(n)), f(x)) + d(f(x), f(yk(n))) < 13ε + 1

3ε = 23ε,

but by construction d(f(xk(N)), f(yk(N)) ≥ ε.

Remark: It is not correct to simply claim that the sequence (xn) has a convergentsubsequence (xk(n)) and the sequence (yn) has a convergent subsequence (yk(n)).If one chooses convergent subsequences of (xn) and (yn), they must be called, say,(xk(n)) and (yl(n)) for different functions k, l : N → N.

It is nevertheless possible to carry out a proof by passing to convergent subse-quences of (xn) and (yn). The following solution shows how it can be done. Thissolution is not recommended here, but in other situations it may be the only wayto proceed.

Alternate solution: Assume that f is not uniformly continuous. Choose ε > 0 forwhich the definition of uniform continuity fails. Then for every n ∈ N there arexn, yn ∈ X such that d(xn, yn) < 1

n and d(f(xn), f(yn)) ≥ ε. Since X is compact,the sequence (xn) has a convergent subsequence (xk(n)). (See Theorem 3.6 (a)of Rudin.) Then (yk(n)) is a sequence in a compact metric space, and therefore,again by Theorem 3.6 (a) of Rudin, it has a convergent subsequence (yk(r(n))). Letl = k ◦ r. Then (xl(n)) is a subsequence of the convergent sequence (xk(n)), andtherefore converges.

Let x = limn→∞ xl(n) and y = limn→∞ yl(n). Now d(xl(n), yl(n)) < 1l(n) ≤ 1

n . Itfollows that d(x, y) = 0. (To see this, let ε > 0, and choose N1, N2, N3 ∈ N so largethat n ≥ N1 implies d(xl(n), x) < 1

3ε, so large that n ≥ N2 implies d(yl(n), y) < 13ε,

and so large that n ≥ N3 implies 1n < 1

3ε. Then with n = max(N1, N2, N3), we get

d(x, y) ≤ d(x, xl(n)) + d(xl(n), yl(n)) + d(yl(n), y) < 13ε + 1

n + 13ε < ε.

Since this is true for all ε > 0, it follows that d(x, y) = 0.)We now know that limn→∞ yl(n) = x = limn→∞ xl(n).If f were continuous at x, we would have

limn→∞ f(xl(n)) = lim

n→∞ f(yl(n)) = f(x).

This contradicts d(f(xn), f(yn)) ≥ ε for all n. To see this, choose N ∈ N such thatn ≥ N implies

d(f(xl(n)), f(x)) < 13ε and d(f(yl(n)), f(x)) < 1

3ε.

Then

d(f(xl(N)), f(yl(N)) ≤ d(f(xl(n)), f(x)) + d(f(x), f(yl(n))) < 13ε + 1

3ε = 23ε,

but by construction d(f(xl(N)), f(yl(N)) ≥ ε.

Problem 4.11: Let X and Y be metric spaces, and let f : X → Y be uniformlycontinuous. Prove that if (xn) is a Cauchy sequence in X, then (f(xn)) is a Cauchysequence in Y . Use this result to prove that if Y is complete, E ⊂ X is dense,and f0 : E → Y is uniformly continuous, then there is a unique continuous functionf : X → Y such that f |E = f0.

Solution: We prove the first statement. Let (xn) be a Cauchy sequence in X.Let ε > 0. Choose δ > 0 such that if x1, x2 ∈ X satisfy d(s1, s2) < δ, thend(f(s1), f(s2)) < ε. Choose N ∈ N such that if m, n ∈ N satisfy m, n ≥ N , thend(xm, xn) < δ. Then whenever m, n ∈ N satisfy m, n ≥ N , we have d(xm, xn) < δ,so that d(f(xm), f(xn)) < ε. This shows that (f(xn)) is a Cauchy sequence.

Now we prove the second statement. The neatest arrangement I can think of isto prove the following lemmas first.

Lemma 1. Let X and Y be metric spaces, with Y complete, let E ⊂ X, and letf : E → Y be uniformly continuous. Let (xn) be a sequence in E which convergesto some point in X. Then limn→∞ f(xn) exists in Y .

Proof: We know that convergent sequences are Cauchy. This is therefore immediatefrom the first part of the problem and the definition of completeness.

Lemma 2. Let X and Y be metric spaces, with Y complete, let E ⊂ X, and letf : E → Y be uniformly continuous. Then for every ε > 0 there is δ > 0 suchthat whenever x1, x2 ∈ X satisfy d(x1, x2) < δ, and whenever (rn) and (sn) aresequences in E such that rn → x1 and sn → x2, then

d(

limn→∞ f(rn), lim

n→∞ f(sn))

< ε.

Note that the limits exist by Lemma 1.

Proof of Lemma 2: Let ε > 0. Choose ρ > 0 such that whenever x1, x2 ∈ Esatisfy d(x1, x2) < ρ, then d(f(x1), f(x2)) < 1

2ε. Set δ = 12ρ > 0. Let x1, x2 ∈ X

satisfy d(x1, x2) < δ, and let (rn) and (sn) be sequences in E such that rn → x1

and sn → x2. Let y1 = limn→∞ f(rn) and y2 = limn→∞ f(sn). (These exist byLemma 1.) Choose N so large that for all n ∈ N with n ≥ N , the following fourconditions are all satisfied:

• d(rn, x1) < 14ρ.

• d(sn, x2) < 14ρ.

• d(f(rn), y1) < 14ε.

• d(f(sn), y2) < 14ε.

We then have

d(rN , sN ) ≤ d(rN , x1) + d(x1, x2) + d(x2, sN ) < 14ρ + 1

2ρ + 14ρ = ρ.

Therefore d(f(rN ), f(sN )) < 12ε. So

d(y1, y2) ≤ d(y1, f(rN )) + d(f(rN ), f(sN )) + d(f(sN ), y2) < 14ε + 1

2ε + 14ε = ε,

as desired.

Theorem. Let X and Y be metric spaces, with Y complete, let E ⊂ X, and letf : E → Y be uniformly continuous. Then there is a unique continuous functionf : X → Y such that f |E = f0.

Proof: If f exists, then it is unique by Problem 4.4, which was in the previousassignment. So we prove existence. For x ∈ X, we want to define f(x) by choosinga sequence (rn) in E with limn→∞ rn = x and then setting f(x) = limn→∞ f0(rn).We know that such a sequence exists because E is dense in X. We know thatlimn→∞ f0(rn) exists, by Lemma 1. However, we must show that limn→∞ f0(rn)only depends on x, not on the sequence (rn).

To prove this, let (rn) and (sn) be sequences in E with

limn→∞ rn = lim

n→∞ sn = x.

Let ε > 0; we show that

d(

limn→∞ f0(rn), lim

n→∞ f0(sn))

< ε.

(Since ε is arbitrary, this will give limn→∞ f0(rn) = limn→∞ f0(sn).) To do this,choose δ > 0 according to Lemma 2. We certainly have d(x, x) < δ. Therefore theconclusion of Lemma 2 gives

d(

limn→∞ f0(rn), lim

n→∞ f0(sn))

< ε,

as desired.We now get a well defined function f : X → Y by setting f(x) = limn→∞ f0(rn),

where (rn) is any sequence in E with limn→∞ rn = x. By considering the constantsequence xn = x for all n, we see immediately that f(x) = f0(x) for x ∈ E.We show that f is continuous, in fact uniformly continuous. Let ε > 0. Chooseδ > 0 according to Lemma 2. For x1, x2 ∈ X with d(x1, x2) < δ, choose (bydensity of E, as above) sequences (rn) and (sn) in E such that rn → x1 andsn → x2. Then d (limn→∞ f0(rn), limn→∞ f0(sn)) < ε. By construction, we havef(x1) = limn→∞ f0(rn) and f(x2) = limn→∞ f0(sn). Therefore we have shown thatd(f(x1), f(x2)) < ε, as desired.

The point of stating Lemma 2 separately is that the proof that f is well defined,and the proof that f is continuous, use essentially the same argument. By puttingthat argument in a lemma, we avoid repeating it.

Problem 4.12: State precisely and prove the following: “A uniformly continuousfunction of a uniformly continuous function is uniformly continuous.”

Solution: Here is the precise statement:

Proposition. Let X, Y , and Z be metric spaces. Let f : X → Y and g : Y → Zbe uniformly continuous functions. Then g ◦ f is uniformly continuous.

Proof: Let ε > 0. Choose ρ > 0 such that if y1, y2 ∈ Y satisfy d(y1, y2) < ρ, thend(g(y1), g(y2)) < ε. Choose δ > 0 such that if x1, x2 ∈ X satisfy d(x1, x2) < δ,then d(f(x1), f(x2)) < ρ. Then whenever x1, x2 ∈ X satisfy d(x1, x2) < δ, we haved(f(x1), f(x2)) < ρ, so that d(g(f(x1)), g(f(x2))) < ε.

Problem 4.14: Let f : [0, 1] → [0, 1] be continuous. Prove that there is x ∈ [0, 1]such that f(x) = x.

Solution: Define g : [0, 1] → R by g(x) = x − f(x). Then g is continuous. Sincef(0) ∈ [0, 1], we have g(0) = −f(0) ≤ 0, while since g(1) ∈ [0, 1], we have g(1) =1 − f(1) ≥ 0. If g(0) = 0 then x = 0 satisfies the conclusion, while if g(1) = 0 thenx = 1 satisfies the conclusion. Otherwise, g(0) < 0 and g(1) > 0, so Theorem 4.23of Rudin provides x ∈ (0, 1) such that g(x) = 0. This x satisfies f(x) = x.

Something much more general is true, namely the Brouwer Fixed Point Theorem:

Theorem. Let n ≥ 1, and let B = {x ∈ Rn : ‖x‖ ≤ 1}. Let f : B → B becontinuous. Then there is x ∈ B such that f(x) = x.

The proof requires higher orders of connectedness, and is best done with algebraictopology.

Problem 4.16: For x ∈ R, define [x] by the relations [x] ∈ Z and x − 1 < [x] ≤ x(this is called the “integer part of x” or the “greatest integer function”), and define(x) = x − [x] (this is called the “fractional part of x”, but the notation (x) is notstandard). What discontinuities do the functions x 7→ [x] and x 7→ (x) have?

Solution (Sketch): Both functions are continuous at all noninteger points, sincex ∈ (n, n + 1) implies [x] = n and (x) = x− n; both expressions are continuous onthe interval (n, n + 1).

Both functions have jump discontinuities at all integers: for n ∈ Z, we have

limx→n+

[x] = limx→n+

n = n = f(n) and limx→n−

[x] = limx→n−

(n − 1) = n − 1 6= f(n),

and also

limx→n+

(x) = limx→n+

(x − n) = 0 = f(n)

and

limx→n−

(x) = limx→n−

[x − (n − 1)] = 1 6= f(n).

Problem 4.18: Define f : R → R by

f(x) ={

0 x ∈ R \ Q1q x = p

q in lowest terms .

(By definition, we require q > 0. If x = 0 we take p = 0 and q = 1.) Prove thatf is continuous at each x ∈ R \ Q, and that f has a simple discontinuity at eachx ∈ Q.

Solution: We show that limx→0 f(x) = 0 for all x ∈ R. This immediately im-plies that f is continuous at all points x for which f(x) = 0 and has a removablediscontinuity at every x for which f(x) 6= 0.

Let x ∈ R, and let ε > 0. Choose N ∈ N such that 1N < ε. For 1 ≤ n ≤ N , let

Sn ={a

n: a ∈ Z and 0 <

∣∣∣an− x

∣∣∣ < 1}

.

Then Sn is finite; in fact, card(Sn) ≤ 2n. Set

S =N⋃

n=1

Sn,

which is a finite union of finite sets and hence finite. Note that x 6∈ S. Set

δ = min(

1, miny∈S

|y − x|)

.

Then δ > 0 because x 6∈ S and S is finite.Let 0 < |y−x| < δ. If y 6∈ Q, then |f(y)−0| = 0 < ε. Otherwise, because y 6∈ S,

|y − x| < 1, and y 6= x, it is not possible to write y = pq with q ≤ N . Thus, when

we write y = pq in lowest terms, we have q > N , so f(y) = 1

q < 1N < ε. This shows

that |f(y) − 0| < ε in this case also.

Problem 4.15: Prove that every continuous open map f : R → R is monotone.

Sketches of two solutions are presented. The second is what I expect peopleto have done. The first is essentially a careful rearrangement of the ideas of thesecond, done so as to minimize the number of cases. (You will see when readingthe second solution why this is desirable.)

Solution (Sketch):

Lemma 1. Let f : R → R be continuous and open. Let a, b ∈ R satisfy a < b.Then f(a) 6= f(b).

Proof (sketch): Suppose f(a) = f(b). Let m1 and m2 be the minimum and maxi-mum values of f on [a, b]. (These exist because f is continuous and [a, b] is compact.)If m1 = m2, then m1 = m2 = f(a) = f(b), and f((a, b)) = {m1} is not an open set.Since (a, b) is open, this is a contradiction. So suppose m1 < m2. If m1 6= f(a),choose c ∈ [a, b] such that f(c) = m1. Then actually c ∈ (a, b). So f((a, b)) con-tains f(c) but contains no real numbers smaller than f(c). This is easily seen tocontradict the assumption that f((a, b)) is open. The case m2 6= f(a) is handledsimilarly, or by considering −f in place of f .

Note: The last part of this proof is the only place where I would expect asubmitted solution to be more complete than what I have provided.

Lemma 2. Let f : R → R be continuous and open. Let a, b ∈ R satisfy a < b. Iff(a) < f(b), then whenever x ∈ R satisfies x < a, we have f(x) < f(a).

Proof: We can’t have f(x) = f(a), by Lemma 1. If f(x) = f(b), we again have acontradiction by Lemma 1. If f(x) > f(b), then the Intermediate Value Theoremprovides z ∈ (x, a) such that f(z) = f(b). Since z < b, this contradicts Lemma 1.If f(a) < f(x) < f(b), then the Intermediate Value Theorem provides z ∈ (a, b)such that f(z) = f(x). Since x < z, this again contradicts Lemma 1. The onlyremaining possibility is f(x) < f(a).

Lemma 3. Let f : R → R be continuous and open. Let a, b ∈ R satisfy a < b. Iff(a) < f(b), then whenever x ∈ R satisfies b < x, we have f(b) < f(x).

Proof: Apply Lemma 2 to the function x 7→ −f(−x).

Lemma 4. Let f : R → R be continuous and open. Let a, b ∈ R satisfy a < b. Iff(a) < f(b), then whenever x ∈ R satisfies a < x < b, we have f(a) < f(x) < f(b).

Date: 30 Nov. 2001.

1

Proof: Lemma 1 implies that f(x) is equal to neither f(a) nor f(b). If f(x) < f(a),we apply Lemma 3 to −f , with b and x interchanged, to get f(b) < f(x). Thisimplies f(b) < f(a), which contradicts the hypotheses. If f(x) > f(b), we applyLemma 2 to −f , with a and x interchanged, to get f(a) > f(x). This againcontradicts the hypotheses.

Corollary 5. Let f : R → R be continuous and open. Let a, b ∈ R satisfy a < b,and suppose that f(a) < f(b). Let x ∈ R. Then:

(1) If x ≤ a then f(x) ≤ f(a).(2) If x ≤ b then f(x) ≤ f(b).(3) If x ≥ a then f(x) ≥ f(a).(4) If x ≥ b then f(x) ≥ f(b).

Proof: If we have equality (x = a or x = b), the conclusion is obvious. With strictinequality, Part (1) follows from Lemma 2, and Part (4) follows from Lemma 3.Part (2) follows from Lemma 4 if x > a, from Lemma 2 if x < a, and is trivial ifx = a. Part (3) follows from Lemma 4 if x < b, from Lemma 3 if x > b, and istrivial if x = b.

I won’t actually use Part (2); it is included for symmetry.

Now we prove the result. Choose arbitrary c, d ∈ R with c < d. We havef(c) 6= f(d) by Lemma 1. Suppose first that f(c) < f(d). Let r, s ∈ R satisfyr ≤ s. We prove that f(r) ≤ f(s), and there are several cases. I will try to arrangethis to keep the number of cases as small as possible.

Case 1: r ≤ c ≤ s. Then f(r) ≤ f(c) ≤ f(s) by Parts (1) and (3) of Corollary 5,taking a = c and b = d.

Case 2: r ≤ s ≤ a. Then f(s) ≤ f(a) by Part (1) of Corollary 5, taking a = cand b = d. Further, f(r) ≤ f(s) by Part (1) of Corollary 5, taking a = s and b = d.

Case 3: a ≤ r ≤ s. Then f(a) ≤ f(r) by Part (3) of Corollary 5, taking a = cand b = d. Further, f(r) ≤ f(s) by Part (4) of Corollary 5, taking a = c and b = r.

The case f(c) > f(d) follows by applying the preceding argument to −f .

Alternate solution (Brief sketch):Suppose f is not monotone; we prove that f is not open. Since f isn’t nonde-

creasing, there exist a, b ∈ R such that a f(b); and since f isn’tnonincreasing, there exist c, d ∈ R such that c < d and f(c) < f(d). Now there arevarious cases depending on how a, b, c, and d are arranged in R, and depending onhow f(a) and f(b) relate to f(c) and f(d). Specifically, there are 13 possible waysfor a, b, c, and d to be arranged in R, namely:

a < b < c < d(1)

a < b = c < d(2)

a < c < b < d(3)

a < c < b = d(4)

a < c < d < b(5)

a = c < b < d(6)

a = c < b = d(7)

a = c < d < b(8)

c < a < b < d(9)

c < a < b = d(10)

c < a < d < b(11)

c < a = d < b(12)

c < d < a < b(13)

Of these, the arrangement (7) gives an immediate contradiction. For each of theothers, we find x < y < z such that f(y) < f(x), f(z) (so that f is not openby Lemma 2 of the previous solution), or such that f(y) > f(x), f(z) (so that fis not open by Lemma 3 of the previous solution). Many cases break down intosubcases depending on how the values of f are arranged. We illustrate by treatingthe arrangement (1).

Suppose a < b < c < d and f(b) < f(c). Set

x = a, y = b, and z = d.

Then x < y < z and f(y) < f(x), f(z), so Lemma 2 applies. Suppose, on the otherhand, that a < b < c < d and f(b) ≥ f(c). Set

x = a, y = c, and z = d.

Then again x < y < z and f(y) < f(x), f(z), so Lemma 2 applies.

Problem 5.1: Let f : R → R be a function such that

|f(x) − f(y)| ≤ (x − y)2

for all x, y ∈ R. Prove that f is constant.

Solution: We first prove that f ′(x) = 0 for all x ∈ R. For h 6= 0,∣∣∣∣f(x + h) − f(x)h

∣∣∣∣ =|f(x + h) − f(x)|

|h| ≤ |h2||h| = |h|.

It follows immediately that

f ′(x) = limh→0

f(x + h) − f(x)h

= 0.

It now follows that f is constant. (See Theorem 5.11 (b) of Rudin’s book.)

The solution above is the intended solution. However, there is another solutionwhich is nearly as easy and does not use calculus.

Alternate solution: Let x, y ∈ R and let ε > 0; we prove that |f(x) − f(y)| < ε. Itwill clearly follow that f is constant.

Choose N ∈ N with N > ε−1(x − y)2. The hypothesis implies that, for any k,we have∣∣f (

x + (k − 1) · 1N (y − x)

) − f(x + k · 1

N (y − x))∣∣ ≤ (

1N (y − x)

)2

=(

1N

) ((y − x)2

N

)<

1N

· ε.

Therefore

|f(x) − f(y)| ≤N∑

k=1

∣∣f (x + (k − 1) · 1

N (y − x)) − f

(x + k · 1

N (y − x))∣∣

< N · 1N

· ε = ε.

Problem 5.2: Let f : (a, b) → R satisfy f ′(x) > 0 for all x ∈ (a, b). Prove that fis strictly increasing, that its inverse function g is differentiable, and that

g′(f(x)) =1

f ′(x)

for all x ∈ (a, b).

Solution: That f is strictly increasing on (a, b) follows from the Mean Value The-orem and the fact that f ′(x) > 0 for all x ∈ (a, b).

Define

c = infx∈(a,b)

f(x) and d = supx∈(a,b)

f(x).

(Note that c could be −∞ and d could be ∞.) Our next step is to prove that fis a bijection from (a, b) to (c, d). Clearly f is injective, and has range containedin [c, d]. If c = f(x) for some x ∈ (a, b), then there is q ∈ (a, b) with q < x. Thiswould imply f(q) < c, contradicting the definition of c. So c is not in the range off . Similarly d is not in the range of f . So the range of f is contained in (c, d). Forsurjectivity, let y0 ∈ (c, d). By the definitions of inf and sup, there are r, s ∈ (a, b)such that f(r) < y0 < f(s). Clearly r < s. The Intermediate Value Theoremprovides x0 ∈ (r, s) such that f(x0) = y0. This shows that the range of f is all of(c, d), and completes the proof that f is a bijection from (a, b) to (c, d).

Now we show that g : (c, d) → (a, b) is continuous. Again, let y0 ∈ (c, d), andchoose r and s as in the previous paragraph. Since f is strictly increasing, andagain using the Intermediate Value Theorem, we see that f |[r,s] is a continuousbijection from [r, s] to [f(r), f(s)]. Since [r, s] is compact, the function

(f |[r,s]

)−1 =g|[f(r), f(s)] is continuous. Since y0 ∈ (f(r), f(s)), it follows that g is continuous aty0. Thus g is continuous.

Now we find g′. Fix x0 ∈ (a, b), and set y0 = f(x0). For y ∈ (c, d) \ {y0}, wewrite

g(y) − g(y0)y − y0

=(

f(g(y)) − f(x0)g(y) − x0

)−1

.

(Note that g(y) 6= x0 because g is injective.) Since g is continuous, we havelimy→y0 g(y) = x0. Therefore

limy→y0

f(g(y)) − f(x0)g(y) − x0

= f ′(x0).

Hence

limy→y0

g(y) − g(y0)y − y0

=1

f ′(x0).

That is, g′(y0) exists and is equal to1

f ′(x0), as desired.

Note: I believe, but have not checked, that further use of the IntermediateValue Theorem can be substituted for the use of compactness in the proof that gis continuous.

Problem 5.3: Let g : R → R be a differentiable function such that g′ is bounded.Prove that there is r > 0 such that the function f(x) = x + εg(x) is injectivewhenever 0 < ε < r.

Solution: Set M = max(0, supx∈R(−g′(x)). Set r = 1M . (Take r = ∞ if M = 0.)

Suppose 0 < ε < r, and define f(x) = x + εg(x) for x ∈ R. For x ∈ R, we have

f ′(x) = 1 + εg′(x) = 1 − ε(−g′(x)) ≥ 1 − εM > 1 − rM = 0

(except that 1 − rM = 1 if M = 0). Thus f ′(x) > 0 for all x, so the Mean ValueTheorem implies that f is strictly increasing. In particular, f is injective.

Note: The problem as stated in Rudin’s book is slightly ambiguous: it could beinterpreted as asking that f(x) = x + εg(x) be injective whenever −r < ε < r. Toprove this version, take M = supx∈R |g′(x)|, and estimate

f ′(x) = 1 + εg′(x) ≥ 1 − |ε|M > 1 − rM = 0.

Problem 5.4: Let C0, C1, . . . , Cn ∈ R. Suppose

C0 +C1

2+ · · · + Cn−1

n+

Cn

n + 1= 0.

Prove that the equation

C0 + C1x + C2x2 + · · · + Cn−1x

n−1 + Cnxn = 0

has at least one real solution in (0, 1).

Solution (Sketch): Define f : R → R by

f(x) = C0x +C1x

2

2+ · · · + Cn−1x

n

n+

Cnxn+1

n + 1for x ∈ R. Then f(0) = 0 (this is trivial) and f(1) = 0 (this follows from thehypothesis). Since f is differentiable on all of R, the Mean Value Theorem providesx ∈ (0, 1) such that f ′(x) = 0. Since

f ′(x) = C0 + C1x + C2x2 + · · · + Cn−1x

n−1 + Cnxn,

this is the desired conclusion.

Problem 5.5: Let f : (0,∞) → R be differentiable and satisfy limx→∞ f ′(x) = 0.Prove that limx→∞[f(x + 1) − f(x)] = 0.

Solution: Let ε > 0. Choose M ∈ R such that x > M implies |f ′(x)| < ε.Let x > M . By the Mean Value Theorem, there is z ∈ (x, x + 1) such thatf(x + 1) − f(x) = f ′(z). Then |f(x + 1) − f(x)| = |f ′(z)| < ε. This shows thatlimx→∞[f(x + 1) − f(x)] = 0.

Problem 5.9: Let f : R → R be continuous. Assume that f ′(x) exists for allx 6= 0, and that limx→0 f ′(x) = 3. Does it follow that f ′(0) exists?

Solution: We prove that f ′(0) = 3. Define g(x) = x. Then

limx→0

f ′(x)g′(x)

= 3

by assumption. Therefore Theorem 5.13 of Rudin (L’Hospital’s rule) applies to thelimit

limx→0

f(x) − f(0)x

= limx→0

f(x) − f(0)g(x)

(because f − f(0) vanishes at 0 and has derivative f ′). Thus

f ′(0) = limx→0

f(x) − f(0)x

= limx→0

f ′(x)g′(x)

= 3.

In particular, f ′(0) exists.

Note: It is mathematically bad practice (although it is tolerated in freshmancalculus courses) to write

limx→0

f(x) − f(0)g(x)

= limx→0

f ′(x)g′(x)

= 3

before checking that

limx→0

f ′(x)g′(x)

exists, because the equality

limx→0

f(x) − f(0)g(x)

= limx→0

f ′(x)g′(x)

is only known to hold when the second limit exists.

Problem 5.11: Let f be a real valued function defined on a neighborhood ofx ∈ R. Suppose that f ′′(x) exists. Prove that

limh→0

f(x + h) + f(x − h) − 2f(x)h2

= f ′′(x).

Show by example that the limit might exist even if f ′′(x) does not exist.

Solution (Sketch): Check using algebra that

limh→0

f ′(x + h) − f ′(x − h)2h

= limh→0

(f ′(x + h) − f ′(x)

2h+

f ′(x) − f ′(x − h)2h

)

= f ′′(x).

Now use Theorem 5.13 of Rudin (L’Hospital’s rule) to show that

limh→0

f(x + h) + f(x − h) − 2f(x)h2

= limh→0

f ′(x + h) − f ′(x − h)2h

= f ′′(x).

For the counterexample, take

f(t) =

1 t > x0 t = x−1 t < x

.

Then

f(x + h) + f(x − h) − 2f(x)h2

= 0

for all h 6= 0. This shows that the limit can exist even if f isn’t continuous at x.

Note 1: I gave a counterexample for an arbitrary value of x, but it suffices togive one at a single value of x, such as x = 0.

Note 2: A legitimate counterexample must be defined at x, since it must satisfyall the hypotheses except for the existence of f ′′(x).

Note 3: Another choice for the counterexample is

f(t) ={

(t − x)2 t ≥ x−(t − x)2 t < x

.

This function is continuous at x, and even has a continuous derivative on R, butf ′′(x) doesn’t exist. One can also construct examples which are continuous nowhereon R.

Note 4: It is tempting to use L’Hospital’s rule a second time, to get

limh→0

f ′(x + h) − f ′(x − h)2h

= limh→0

f ′′(x + h) + f ′′(x − h)2

.

This reasoning is not valid, since the second limit need not exist. (We do notassume that f ′′ is continuous.)

Problem 5.13: Let a and c be fixed real numbers, with c > 0, and define f =fa,c : [−1, 1] → R by

f(x) ={ |x|a sin (|x|−c) x 6= 0

0 x = 0

Prove the following statements. (You may use the standard facts about the func-tions sin(x) and cos(x).)

Note: The book has

f(x) ={

xa sin (|x|−c) x 6= 00 x = 0 .

However, unless a is a rational number with odd denominator, this function willnot be defined for x < 0.

(a) f is continuous if and only if a > 0.

Solution (Sketch): Since x 7→ sin(x) is continuous, we need only consider continuityat 0. If a > 0, then limx→0 f(x) = 0 since |f(x)| ≤ |x|a and limx→0 |x|a = 0.

Now define sequences (xn) and (yn) by

xn =1[(

2n + 12

)π]1/c

and yn =1[(

2n + 32

)π]1/c

.

Note that

limn→∞xn = lim

n→∞ yn = 0

and

sin(|xn|−c

)= 1 and sin

(|yn|−c)

= −1

for all n. (We will use these sequences in other parts of the problem.)If now a = 0, then

limn→∞ f(xn) = 1 and lim

n→∞ f(yn) = −1,

so limx→0 f(x) does not exist, and f is not continuous at 0. If a < 0, then

limn→∞ f(xn) = ∞ and lim

n→∞ f(yn) = −∞,

with the same result.

Note: Since f(0) is defined to be 0, we actually need only consider limn→∞ f(xn).The conclusion limx→0 f(x) does not exist is stronger, and will be useful later.

(b) f ′(0) exists if and only if a > 1.

Solution (Sketch): We test for existence of

f ′(0) = limh→0

f(h) − f(0)h

= limh→0

f(h)h

= limh→0

fa−1, c(h),

which we saw in Part (a) exists if and only if a − 1 > 0. Moreover (for use below),note that if the limit does exist then it is equal to 0.

(c) f ′ is bounded if and only if a ≥ 1 + c.

Solution (Sketch): Boundedness does not depend on f ′(0) (or even on whetherf ′(0) exists). So we use the formula

f ′(x) = axa−1 sin(x−c

)+ cxa−c−1 cos

(x−c

)for x > 0, and for x < 0 we use

f ′(x) = −f ′(−x) = −f ′(|x|) = −a|x|a−1 sin(|x|−c

) − c|x|a−c−1 cos(|x|−c

).

If a − c − 1 ≥ 0, then also a − 1 ≥ 0 (recall that c > 0), and f ′ is bounded (byc + a).

Otherwise, we consider the sequences (wn) and (zn) given by

wn =1

[2nπ]1/cand yn =

1

[(2n + 1) π]1/c.

Since

sin(w−c

n

)= sin

(z−c

n

)= 0

and

cos(w−c

n

)= 1 and cos

(z−c

n

)= −1,

arguments as in Part (a) show that

limn→∞ f ′(wn) = −∞ and lim

n→∞ f ′(zn) = ∞,

so f ′ is not bounded.

(d) f ′ is continuous on [−1, 1] if and only if a > 1 + c.

Solution (Sketch): If a < 1 + c, then f ′ is not bounded on [−1, 1] \ {0} by Part (c),and therefore can’t be the restriction of a continuous function on [−1, 1]. If a = 1+c,then the sequences of Part (c) satisfy

limn→∞ f ′(wn) = −c and lim

n→∞ f ′(zn) = c,

so again f ′ can’t be the restriction of a continuous function on [−1, 1]. If a > 1+ c,then also a > 1, and limx→0 f ′(x) = 0 by reasoning similar to that of Part (a).Moreover f ′(0) = 0 by the extra conclusion in the proof of Part (b). So f ′ iscontinuous at 0, hence continuous.

f ′′(0) exists if and only if a > 2 + c.

Solution (Sketch): This is reduced to Part (d) in the same way Part (b) was reducedto Part (a). As there, note also that f ′′(0) = 0 if it exists.

(f) f ′′ is bounded if and only if a ≥ 2 + 2c.

Solution (Sketch): For x 6= 0, we have

f ′′(x) = a(a − 1)|x|a−2 sin(|x|−c

)+ (2ac − c2 − c)xa−c−2 cos

(|x|−c)

− c2|x|a−2c−2 sin(|x|−c

).

(One handles the cases x > 0 and x < 0 separately, as in Part (c), but this time theresulting formula is the same for both cases.) Since c > 0, if a−2c−2 ≥ 0 then alsoa − c − 2 ≥ 0 and a − 2 ≥ 0, so f ′′ is bounded on [−1, 1] \ {0}. For a − 2c − 2 < 0,consider

f ′′(xn) = a(a − 1)xa−2n − c2xa−2c−2

n .

Since xn → 0 and a− 2c− 2 < min(0, a− 2), one checks that the term −c2xa−2c−2n

dominates and f ′′(xn) → −∞. So f ′′ is not bounded.

(g) f ′′ is continuous on [−1, 1] if and only if a > 2 + 2c.

Solution (Sketch): Recall from the extra conclusion in Part (e) that f ′′(0) = 0 if itexists. If a− 2c− 2 > 0, then also a− c− 1 > 0 and a− 2 > 0, so limx→0 f ′′(x) = 0by a more complicated version of the arguments used in Parts (a) and (d). Ifa − 2c − 2 < 0, then f ′′ isn’t bounded on [−1, 1] \ {0}, so f ′′ can’t be continuouson [−1, 1]. If a − 2c − 2 = 0, then a − 2 > 0. Therefore

f ′′(xn) = a(a − 1)xa−2n − c2xa−2c−2

n = a(a − 1)xa−2n − c2 → c2 6= 0

as n → ∞. So f ′′ is not continuous at 0.

Problem 5.12: Define f : R → R by f(x) = |x|3. Compute f ′(x) and f ′′(x) forall real x. Prove that f ′′′(0) does not exist.

Solution: To make clear exactly what is being done, we first prove a lemma.

Lemma. Let (a, b) ⊂ R be an open interval, and let f, g : R → R be functions.Let c ∈ (a, b), and suppose f ′(c) exists. Suppose that there is ε > 0 such that(c − ε, c + ε) ⊂ (a, b) and g(x) = f(x) for all x ∈ (c − ε, c + ε). Then g′(c) existsand g′(c) = f ′(c).

Proof: We have

g(c + h) − g(c)h

=f(c + h) − f(c)

h

for all h with 0 < |h| < ε. Therefore

limh→0

g(c + h) − g(c)h

= limh→0

f(c + h) − f(c)h

.

Now we start the calculation. For x > 0, we have f(x) = x3. Therefore, by thelemma, f ′(x) = 3x2 and f ′′(x) = 6x. Similarly, for x < 0, we have f(x) = −x3, sof ′(x) = −3x2 and f ′′(x) = −6x.

The lemma is not useful for x = 0. So we calculate directly. For h 6= 0, we have∣∣∣∣f(h) − f(0)h

∣∣∣∣ =∣∣∣∣ |h|

3

h

∣∣∣∣ = |h|2.

Therefore

f ′(0) = limh→0

f(h) − f(0)h

= 0.

With this result in hand, we can calculate f ′′(0). For h 6= 0, we have |f ′(h)| =3|h|2 (regardless of whether h is positive or negative), so∣∣∣∣f

′(h) − f ′(0)h

∣∣∣∣ =∣∣∣∣f

′(h)h

∣∣∣∣ =∣∣∣∣3|h|

2

h

∣∣∣∣ = 3|h|.

Therefore

f ′′(0) = limh→0

f ′(h) − f ′(0)h

= 0.

Date: 14 January 2002.

1

Finally, we consider f ′′′(0). For h > 0, we have

f ′′(h) − f ′′(0)h

=6h

h= 6.

For h < 0, we have

f ′′(h) − f ′′(0)h

=−6h

h= −6.

So

f ′′′(0) = limh→0

f ′′(h) − f ′′(0)h

does not exist.

Problem 5.22: Let f : R → R be a function. We say that x ∈ R is a fixed pointof f if f(x) = x.

(a) Suppose that f is differentiable and f ′(t) 6= 1 for all t ∈ R. Prove that f hasat most one fixed point.

Solution: Suppose f has two distinct fixed points r and s. Without loss of generalityr < s. Apply the Mean Value Theorem on the interval [r, s], to find c ∈ (r, s) suchthat f(s) − f(r) = f ′(c)(s − r). Since f(r) = r and f(s) = s, and since s − r 6= 0,this implies that f ′(c) = 1. This contradicts the assumption that f ′(t) 6= 1 for allt ∈ R.

(b) Define f by f(t) = t + (1 + et)−1 for t ∈ R. Prove that 0 < f ′(t) < 1 for allt ∈ R, but that f has no fixed points. (You may use the standard properties of theexponential function from elementary calculus.)

Solution: If x is a fixed point for f , then

x = f(x) = x +1

1 + ex,

whence1

1 + ex= 0.

This is obviously impossible.Using the fact from elementary calculus that the derivative of et is et, and using

the differentiation rules proved in Chapter 5 of Rudin’s book, we get

f ′(t) = 1 − et

(1 + et)2.

Since 0 < et < 1 + et < (1 + et)2, we have

0 <et

(1 + et)2< 1

for all t, from which it is clear that 0 < f ′(t) < 1 for all t.

(c) Suppose there is a constant A < 1 such that |f ′(t)| ≤ A for all t ∈ R. Provethat f has a fixed point. Prove that if x0 ∈ R, and that if the sequence (xn)n∈N isdefined recursively by xn+1 = f(xn), then (xn)n∈N converges to a fixed point of f .

Solution: It suffices to prove the last statement. First, observe that the version ofthe Mean Value Theorem in Theorem 5.19 of Rudin’s book implies that |f(s) −f(t)| ≤ A|s − t| for all s, t ∈ R.

Now let x0 ∈ R be arbitrary, and define the sequence (xn)n∈N recursively as inthe statement. Using induction and the estimate above, we get

|xn+1 − xn| ≤ An|x1 − x0|for all n. Using the triangle inequality in the first step and the formula for the sumof a geometric series at the second step, we get, for n ∈ N and m ∈ N,

|xn+m − xn| ≤m−1∑k=0

An+k|x1 − x0| =An − An+m

1 − A· |x1 − x0| ≤ An

1 − A· |x1 − x0|.

With this estimate, we can prove that (xn)n∈N is a Cauchy sequence. Let ε > 0.Since 0 ≤ A < 1, we have limN→∞ AN = 0, so we can choose N so large that

AN

1 − A· |x1 − x0| < ε.

For m, n ≥ N , we then have

|xm − xn| ≤ Amin(m,n)

1 − A· |x1 − x0| < ε.

This shows that (xn)n∈N is a Cauchy sequence.Since R is complete, x = limn→∞ xn exists. It is trivial that limn→∞ xn+1 = x

as well. Using the continuity of f at x in the first step, we get

f(x) = limn→∞ f(xn) = lim

n→∞xn+1 = x,

that is, x is a fixed point for f .

We can give an alternate proof of the existence of a fixed point, which is ofinterest, even though I do not see how to use it to show that the fixed point is thelimit of the sequence described in the problem without essentially redoing the othersolution.

Partial alternate solution (Sketch): Without loss of generality f(0) 6= 0. Weconsider only the case f(0) > 0; the proof for f(0) < 0 is similar. Define b =f(0)(1−A)−1 > 0, and define g(x) = f(x)−x. Then g is continuous and g(0) > 0.The version of the Mean Value Theorem in Theorem 5.19 of Rudin’s book impliesthat |f(b) − f(0)| ≤ Ab. Therefore f(b) ≤ f(0) + Ab, whence

g(b) = f(b) − b ≤ f(0) + Ab − b = f(0) + (A − 1)f(0)(1 − A)−1 = 0.

So the Intermediate Value Theorem provides x ∈ [0, b] such that g(x) = 0, that is,f(x) = x.

Problem 5.26: Let f : [a, b] → R be differentiable, and suppose that f(a) = 0 andthere is a real number A such that |f ′(x)| ≤ A|f(x)| for all x ∈ [a, b]. Prove thatf(x) = 0 for all x ∈ [a, b].

Hint: Fix x0 ∈ [a, b], and define

M0 = supx∈[a,b]

|f(x)| and M1 = supx∈[a,b]

|f ′(x)|.

For x ∈ [a, b], we then have

|f(x)| ≤ M1(x0 − a) ≤ A(x0 − a)M0.

If A(x0 − a) < 1, it follows that M0 = 0. That is, f(x) = 0 for all x ∈ [a, x0].Proceed.

Solution (Sketch): Choose numbers xk with

a = x0 < x1 < · · · < xn = b

and such that A(xk − xk−1) < 1 for 1 ≤ k ≤ n. We prove by induction on k thatf(x) = 0 for all x ∈ [a, xk]. This is immediate for k = 0. So suppose it is knownfor some k; we prove it for k + 1.

Define

M0 = supx∈[xk−1, xk]

|f(x)| and M1 = supx∈[xk−1, xk]

|f ′(x)|.

The hypotheses imply that M1 ≤ AM0. For x ∈ [xk−1, xk], we have (using theversion of the Mean Value Theorem in Theorem 5.19 of Rudin’s book at the secondstep)

|f(x)| = |f(x) − f(xk)| ≤ M1(x − xk) ≤ AM0(x − xk).

In this inequality, take the supremum over all x ∈ [xk−1, xk], getting

M0 = supx∈[xk−1, xk]

|f(x)| ≤ AM0 supx∈[xk−1, xk]

(x − xk) = A(xk+1 − xk)M0.

Since 0 ≤ A(xk+1−xk) < 1 and M0 ≥ 0, this can only happen if M0 = 0. Thereforef(x) = 0 for all x ∈ [xk−1, xk], and hence for all x ∈ [a, xk]. This completes theinduction step, and the proof.

Problem 6.2: Let f : [a, b] → R be continuous and nonnegative. Assume that∫ b

af = 0. Prove that f = 0.

Solution (Sketch): Assume that f 6= 0. Choose x0 ∈ [a, b] such that f(x0) > 0. Bycontinuity, there is δ > 0 such that f(x) > 1

2f(x0) for |x − x0| < δ. Let

I = [a, b] ∩ [x0 − 1

2δ, x0 + 12δ

],

which is an interval of positive length, say l. It is now easy to construct a partitionP such that L(P, f) ≥ l · 1

2f(x0) > 0.

Alternate solution (Sketch): Let x0, δ, I, and l be as above. Let χI be the char-acteristic function of I. Check that χI is integrable, and

∫ b

aχI = l, by choosing

a partition P of [a, b] such that L(P, χI) = U(P, χI) = l. Then g = 12f(x0)χI is

integrable, and∫ b

ag = 1

2f(x0)l. Since f ≥ χI , we have∫ b

af ≥ ∫ b

aχI > 0.

Problem 6.4: Let a, b ∈ R with a < b. Define f : [a, b] → R by

f(x) ={

0 x ∈ R \ Q1 x ∈ Q .

Prove that f is not Riemann integrable on [a, b].

Solution (Sketch): For every partition P = (x0, x1, . . . , xn) of [a, b], every subinter-val [xj−1, xj ] contains both rational and irrational numbers. Therefore L(P, f) = 0and U(P, f) = b − a for every P .

Problem A: Let X be a complete metric space, and let f : X → X be a function.Suppose that there is a constant k such that d(f(x), f(y)) ≤ kd(x, y) for all x, y ∈X.

(1) Prove that f is uniformly continuous.

Solution (Sketch): For ε > 0, take δ = k−1ε.

(2) Suppose that k < 1. Prove that f has a unique fixed point, that is, there isa unique x ∈ X such that f(x) = x.

Solution (Sketch): If r, s ∈ X are fixed points, then d(r, s) = d(f(r), f(s)) ≤kd(r, s). Since 0 ≤ k < 1, this implies that d(r, s) = 0, that is, r = s. So f has atmost one fixed point.

The proof that f has a fixed point is essentially the same as the proof of Prob-lem 5.22 (c). Choose any x0 ∈ X. Define a sequence (xn)n∈N recursively byxn+1 = f(xn) for n ≥ 1. The same calculations as there show that

d(xn, xn+1) ≤ knd(x0, x1)

for all n, that

d(xn, xn+m) ≤ kn

1 − k· d(x0, x1)

for n ∈ N and m ∈ N, and hence (using k < 1) that (xn)n∈N is a Cauchy sequence.Since X is assumed to be complete, x = limn→∞ xn exists. Using the continuity off at x, it then follows, as there, that x is a fixed point for f .

(3) Show that the conclusion in Part (2) need not hold if X is not complete.

Solution: Take X = R \ {0}, with the restriction of the usual metric on R, anddefine f : X → X by f(x) = 1

2x for all x ∈ X. Then d(f(x), f(y)) = 12d(x, y) for

all x, y ∈ X, but clearly f has no fixed point.It follows from Part (2) that X is not complete. (This is also easy to check

directly:(

1n

)n∈N\{0} is a Cauchy sequence which does not converge.)


Generally, a “solution” is something that would be acceptable if turned in inthe form presented here, although the solutions given are usually close to minimalin this respect. A “solution (sketch)” is too sketchy to be considered a completesolution if turned in; varying amounts of detail would need to be filled in. A“solution (nearly complete)” is missing the details in just a few places; it would beconsidered a not quite complete solution if turned in.

Problem 6.7: Let f : (0, 1] → R be a function, and suppose that f |[c,1] is Riemannintegrable for every c ∈ (0, 1). Define

∫ 1

0

f = limc→0+

∫ 1

c

f

if this limit exists and is finite.

(a) If f is the restriction to (0, 1] of a Riemann integrable function on [0, 1], showthat the new definition agrees with the old one.

Solution: The function F (x) =∫ x

0f is continuous, so that

limc→0+

∫ 1

c

f = limc→0+

(∫ 1

0

f − F (c))

=∫ 1

0

f − F (0) =∫ 1

0

f.

(b) Give an example of a function f : (0, 1] → R such that limc→0+

∫ 1

cf exists

but limc→0+

∫ 1

c|f | does not exist.

Solution (nearly complete): We will use the fact that the series∑∞

n=1(−1)n 1n con-

verges but does not converge absolutely.Define f : (0, 1] → R by setting f(x) = 1

n (−1)n · 2n for x ∈ (12n , 1

2n−1

]and

n ∈ N. Then h(c) =∫ 1

c|f | increases as c decreases to 0, and is is easy to check

that

h

(12n

)=

n∑k=1

2k

k· 12k

=n∑

k=1

1k

.

Therefore limc→0+

∫ 1

c|f | = ∞.

Now we prove that

limc→0+

∫ 1

c

f =∞∑

n=1

(−1)n

n.

Date: 23 Jan. 2002.

1

(Note that it is clear that this should be true, but that proving it requires a littlecare.) First, one checks that ∫ 1

12n

f =n∑

k=1

(−1)k

k.

Second, note that the series converges by the Alternating Series Test. Let its sumbe a. Let ε > 0, and choose N such that if n ≥ N than∣∣∣∣∣a −

n∑k=1

(−1)k

k

∣∣∣∣∣ < ε.

Take δ = 12N . Let 0 < c < δ, and choose n such that c ∈ [

12n , 1

2n−1

). Note that

n − 1 ≥ N . If n is even, then∫ 1

12n

f ≥∫ 1

c

f ≥∫ 1

12n−1

f.

Since n − 1 ≥ N , we have∣∣∣∣∣a −∫ 1

12n

f

∣∣∣∣∣ =

∣∣∣∣∣a −n∑

k=1

(−1)k

k

∣∣∣∣∣ < ε and

∣∣∣∣∣a −∫ 1

12n−1

f

∣∣∣∣∣ =

∣∣∣∣∣a −n−1∑k=1

(−1)k

k

∣∣∣∣∣ < ε.

Therefore∣∣∣a − ∫ 1

cf∣∣∣ < ε. The case n odd is handled similarly. Thus, whenever

0 < c < δ we have∣∣∣a − ∫ 1

cf∣∣∣ < ε. This shows that limc→0+

∫ 1

cf = a, as desired.

Problem 6.8: Let f : [a,∞) → R be a function, and suppose that f |[a,b] is Rie-mann integrable for every b > a. Define∫ ∞

a

f = limb→∞

∫ b

a

f

if this limit exists and is finite. In this case, we say that∫ ∞

af converges.

Assume that f is nonnegative and nonincreasing on [1,∞). Prove that∫ ∞1

f

converges if and only if∑∞

n=1 f(n) converges.

Solution (nearly complete): First assume that∑∞

n=1 f(n) converges. Since f isnonnegative, b 7→ ∫ b

af is nondecreasing. It is therefore easy to check (as with

sequences) that limb→∞∫ b

1f exists if and only if b 7→ ∫ b

1f is bounded.

Define g : [1,∞) → R by g(x) = f(n) for x ∈ [n, n + 1) and n ∈ N. Since fis nondecreasing, it follows that g ≥ f . So if b ≥ 1 we can choose n ∈ N withn + 1 ≥ b, giving ∫ b

1

f ≤∫ n

1

f ≤∫ n+1

1

g =n∑

k=1

f(k) ≤∞∑

k=1

f(k).

This shows that b 7→ ∫ b

1f is bounded.

Now assume that∫ ∞1

f converges. This clearly implies that the sequence n 7→∫ n

1f is bounded (in fact, converges). Define h : [1,∞) → R by h(x) = f(n + 1) for

x ∈ [n, n+1) and n ∈ N. Since f is nondecreasing, it follows that h ≤ f . Thereforen∑

k=1

f(k) = f(1) +n∑

k=2

f(k) = f(1) +∫ n

1

h ≤ f(1) +∫ n

1

f.

This shows that the partial sums∑n

k=1 f(k) form a bounded sequence. Since theterms are nonnegative, it follows that

∑∞n=1 f(n) converges.

Problem 6.10: Let p, q ∈ (1,∞) satisfy 1p + 1

q = 1.

(a) For u, v ∈ [0,∞) prove that

uv ≤ up

p+

vq

q,

with equality if and only if up = vq.

Comment: The solution below was found by first letting x = up and y = vq, so thatthe inequality is equivalent to x1/py1/q = x

p + yq for x, y ∈ [0,∞); then dividing by

y and noting that 1q − 1 = − 1

p , so that the inequality is equivalent to

(x

y

)1/p

≤(

1p

) (x

y

)+

1q;

then letting α = xy .

Solution: We first claim that α1/p ≤ 1p · α + 1

q for all α ≥ 0, with equality if andonly if α = 1. That we have equality for α = 1 is clear. Set

f(α) = α1/p − 1p· α − 1

q

for α ≥ 0. Then

f ′(α) =1p· α1/p−1 − 1

p=

1p

(α1/p−1 − 1

)

for all α > 0. Fix α > 1. The Mean Value Theorem provides β ∈ (1, α) such that

f(α) − f(1) = f ′(β)(α − 1).

Since β > 1 and 1p − 1 < 0, we have f ′(β) < 0. Since f(1) = 0, we therefore get

f(α) < 0. This proves the claim for α > 1. For 0 ≤ α < 1, a similar argumentworks, using the fact that f ′(β) > 0 for 0 < β < 1.

Now we prove the statement of the problem. If vq = 0, the inequality reduces to0 ≤ 1

pup. Clearly this is true for all u ≥ 0, and equality holds if and only if up = 0.Otherwise, set α = up · v−q. Applying the claim, we get

(up

vq

)1/p

≤(

1p

) (up

vq

)+

1q

for all u ≥ 0 and v > 0, with equality if and only if up = vq. Since vq > 0, we canmultiply by vq, getting

uvq−q/p ≤ up

p+

vq

q

for all u ≥ 0 and v > 0, with equality if and only if up = vq. Now the relationship1p + 1

q = 1 implies that q − qp = 1, completing the proof.

(b) (with the function α taken to be α(x) = x). Let α : [a, b] → R be a nonde-creasing function. Let f, g : [a, b] → R be nonnegative functions which are Riemann


integrable, and such that∫ b

a

fp = 1 and∫ b

a

gq = 1.

Prove that ∫ b

a

fg ≤ 1.

Solution: The function fg is Riemann integrable, by Theorem 6.13 (a) of Rudin.The inequality in Part (a) implies that fg ≤ 1

pfp + 1q gq on [a, b]. Therefore

∫ b

a

fg ≤∫ b

a

(1pfp + 1

q gq)

= 1p

∫ b

a

fp + 1q

∫ b

a

gq = 1p + 1

q = 1.

(c) (with the function α taken to be α(x) = x). Let f, g : [a, b] → C be Riemannintegrable complex valued functions. Prove that

∣∣∣∣∣∫ b

a

fg

∣∣∣∣∣ ≤(∫ b

a

|f |p)1/p (∫ b

a

|g|q)1/q

.

Solution: First, we note that the hypotheses imply that fg, |f |p, and |g|q areRiemann integrable. For the second two, use Theorems 6.25 and 6.11 of Rudin.For the first, use the decompositions of f and g into real and imaginary parts, andthen use Theorems 6.13 (a) and 6.12 (a).

Next observe that ∣∣∣∣∣∫ b

a

fg

∣∣∣∣∣ ≤∫ b

a

|f | · |g|.

Therefore it is enough to show that

∫ b

a

|f | · |g| ≤(∫ b

a

|f |p)1/p (∫ b

a

|g|q)1/q

.

Let

α =

(∫ b

a

|f |p)1/p

and β =

(∫ b

a

|g|q)1/q

,

and apply Part (b) to the functions f0 = α−1|f | and g0 = β−1|g|.(d) Prove that the result in Part (c) also holds for the improper Riemann integrals

defined in Problems 6.7 and 6.8. (Only actually do the case of the improper integraldefined in Problems 6.7.)

Solution (nearly complete): We assume that∫ 1

0

|f |p and∫ 1

0

|g|q

exist in the extended sense defined in Problem 6.7, and we need to prove that∫ 1

0fg

exists and that ∣∣∣∣∫ 1

0

fg

∣∣∣∣ ≤(∫ 1

0

|f |p)1/p (∫ 1

0

|g|q)1/q

.

Once existence is proved, the inequality follows immediately from Part (c) by takingthe limit as c → 0+.

Define

I(c) =∫ 1

c

fg

for c ∈ (0, 1). We claim that for every ε > 0 there is δ > 0 such that wheneverc1, c2 ∈ (0, δ), then |I(c1) − I(c2)| < ε. This will imply that limc→0+ I(c) exists.Indeed, restricting to c = 1

n with n ∈ N ∩ [2,∞) gives a Cauchy sequence in C,which necessarily has a limit α, and it is easy to show that limc→0+ I(c) = α as cruns through arbitrary values too.

To prove the claim, let ε > 0. Choose δ > 0 so small that if 0 < c < δ then∣∣∣∣∫ 1

0

|f |p −∫ 1

c

|f |p∣∣∣∣ <

(√ε)p

and ∣∣∣∣∫ 1

0

|g|q −∫ 1

c

|g|q∣∣∣∣ <

(√ε)q

.

Since |f |p and |g|q are nonnegative, it follows that whenever 0 < c1 < c2 < δ, then∫ c2

c1

|f |p <(√

ε)p and

∫ c2

c1

|g|q <(√

ε)q

.

Now apply Part (c) to get

|I(c2) − I(c1)| =∣∣∣∣∫ c2

c1

fg

∣∣∣∣ ≤(∫ c2

c1

|f |p)1/p (∫ c2

c1

|g|q)1/q

<(√

ε) (√

ε)

= ε.

This takes care of the case c1 < c2. The reverse case is obtained easily from thisone, and the case c1 = c2 is trivial.

Problem 6.11 (with the function α taken to be α(x) = x): For any Riemannintegrable function u, define

‖u‖2 =

(∫ b

a

|u|2)1/2

.

Prove that if f , g, and h are all Riemann integrable, then

‖f − h‖2 ≤ ‖f − g‖2 + ‖g − h‖2.

Hint: Use the Schwarz inequality, as in the proof of Theorem 1.37 of Rudin.

Comment: As I read the problem, it is intended that the functions be real. However,the complex case is actually more important, so I will give the proof in that case.

Also, one can use Problem 6.10 (c) at an appropriate stage, but that isn’t actuallynecessary.


Solution (nearly complete): It is enough to prove that

‖f + g‖2 ≤ ‖f‖2 + ‖g‖2.

Start by defining, for f , g Riemann integrable,

〈f, g〉 =∫ b

a

fg.

Note that, by an argument similar to the one given in the proof of Problem 6.10 (c),fg is in fact Riemann integrable. It is now immediate to verify the following facts:

(1) The set R([a, b]) of Riemann integrable complex functions is a vector space(with the pointwise operations).

(2) For every g ∈ R([a, b]), the function f 7→ 〈f, g〉 is linear.(3) 〈g, f〉 = 〈f, g〉 for all f, g ∈ R([a, b]).(4) 〈f, f〉 ≥ 0 for all f ∈ R([a, b]).(5) ‖f‖2

2 = 〈f, f〉 for all f ∈ R([a, b]).We next show that

|〈f, g〉| ≤ ‖f‖2‖g‖2

for all f, g ∈ R([a, b]). Note that the proof uses only properties (1) through (5)above.

Choose α ∈ C with |af | = 1 and α〈f, g〉 = |〈f, g〉|. It is clear that

‖αf‖2 = ‖f‖2 and 〈αf, g〉 = |〈f, g〉|.Therefore it suffices to prove the inequality with αf in place of f . That is, withoutloss of generality we may assume 〈f, g〉 ≥ 0.

For t ∈ R, define

p(t) = 〈f + tg, f + tg〉.Using Properties (2) and (3), and the fact that 〈f, g〉 is real, we may expand thisas

p(t) = 〈f, f〉 + 2t〈f, g〉 + t2〈g, g〉.Thus, p is a quadratic polynomial. Property (4) implies that its discriminant isnonpositive, that is,

4〈f, g〉2 − 4〈f, f〉〈g, g〉 ≤ 0.

This implies the desired inequality.Now we prove the inequality

‖f + g‖2 ≤ ‖f‖2 + ‖g‖2.

Again, the proof depends only on properties (1) through (5) above, and uses nothingabout integrals except what went into proving those properties.

We write (using Properties (2) and (3) to expand at the second step, and thepreviously proved inequality at the third step):

‖f + g‖22 = 〈f + g, f + g〉 = 〈f, f〉 + 2Re(〈f, g〉) + 〈g, g〉≤ ‖f‖2

2 + ‖f‖2‖g‖2 + ‖g‖22 = (‖f‖2 + ‖g‖2)2.

Now take square roots.



Problem 6.15: Let f : [a, b] → R be a continuously differentiable function satis-fying f(a) = f(b) = 0 and

∫ b

af(x)2 dx = 1. Prove that

∫ b

a

xf(x)f ′(x) dx = − 12 and

(∫ b

a

[f ′(x)]2 dx

) (∫ b

a

x2f(x)2 dx

)> 1

4 .

Comments: (1) I do not use the notation “f2(x)”, because it could reasonably beinterpreted as f(f(x)).

(2) This result is related to the Heisenberg Uncertainly Principle in quantummechanics.

Solution (Sketch): For the equation, use integration by parts (Theorem 6.22 ofRudin), ∫ b

a

u(x)v′(x) dx = u(b)v(b) − u(a)v(a) −∫ b

a

u′(x)v(x) dx,

with

v′(x) = f(x)f ′(x), v(x) = 12f(x)2, u(x) = x, and u′(x) = 1.

The inequality (∫ b

a

[f ′(x)]2 dx

) (∫ b

a

x2f(x)2 dx

)≥ 1

4

now follows from the Holder inequality (Problem 6.10 of Rudin) with p = q = 2.(Also see the solution given for Problem 6.11.)

To get strict inequality requires more work. First, we need to know that if(∫ b

a

f(x)g(x) dx

)2

=

(∫ b

a

f(x)2 dx

)(∫ b

a

g(x)2 dx

),

which in the notation used in the solution to Problem 6.11 is

〈f, g〉2 = 〈f, f〉 · 〈g, g〉(for real valued f and g), then f and g are linearly dependent, that is, either one ofthem is zero or there is a constant λ such that f = λg. This is proved below. Giventhis, and taking g(x) = xf ′(x), the hypotheses and the first equation proved above

Date: 28 Jan. 2002.

1

immediately rule out f = 0 or g = 0. Thus, if the strict inequality fails above, thereis a nonzero constant λ ∈ R such that

f ′(x) = λxf(x)

for all x ∈ [a, b].The next step is motivated by the observation that the last equation is a differ-

ential equation for f whose solutions have the form

f(x) = 0 and f(x) = exp(

12λx2 + γ

)for all x (for some constant γ). Let

S = {x ∈ [a, b] : f(x) 6= 0} ⊂ [a, b].

Then S is open in [a, b] because f is continuous. We know that S 6= ∅, because∫ b

af(x)2 dx = 1. So choose x0 ∈ S. Let

c = inf{t ∈ [a, x0] : [t, x0] ⊂ S}.Then c < x0 and (c, x0] ⊂ S. We show that [c, x0] ⊂ S. On the interval (c, x0],since f(x) 6= 0, we can rewrite the differential equation above as

f ′(x)f(x)

= λx.

Integrating, we get that there is a constant γ ∈ R such that

log(f(x)) = 12λx2 + γ

for x ∈ (c, x0]. (We have implicitly used Theorem 5.11 (b) of Rudin here: if f ′ = 0on an interval, then f is constant.) Rewrite the above equation as

f(x) = exp(

12λx2 + γ

)for x ∈ (c, x0]. Since f is continuous at c, we let x approach zero from above to get

f(c) = exp(

12λc2 + γ

) 6= 0.

So c ∈ S, whence [c, x0] ⊂ S.Since S is open in [a, b], it is easy to see that this implies that c = a. (Otherwise,

c is a limit point of [a, b]\S but c 6∈ [a, b]\S.) We saw above that f(c) 6= 0. However,this now contradicts the assumption that f(a) = 0, thus proving the required strictinequality.

It remains to prove the criterion for 〈f, g〉2 = 〈f, f〉 · 〈g, g〉. Referring to thesolution to Problem 6.11 of Rudin, and taking α(x) = x there, we see that equalityimplies that the polynomial

p(t) = 〈f, f〉 + 2t〈f, g〉 + t2〈g, g〉used there has a real root, which implies the existence of t such that

0 = 〈f + tg, f + tg〉 =∫ b

a

(f + tg)2.

Since f and g are assumed real and continuous here, and since we are using theRiemann integral, this implies that f + tg = 0.

Problem 7.1: Prove that every uniformly convergent sequence of bounded func-tions is uniformly bounded.

Solution: Let E be a set, let fn : E → C be bounded functions, and assume thatfn → f uniformly. By the definition of uniform convergence, there is N such thatif n ≥ N then ‖fn − f‖∞ < 1. Set

M = max (‖f1‖∞, ‖f2‖∞, . . . , ‖fN−1‖∞, ‖fN‖∞ + 1) .

Then obviously ‖fn‖∞ ≤ M for 1 ≤ n < N . For n ≥ N , we have

‖fn‖∞ ≤ ‖fN‖∞ + ‖fn − f‖∞ < ‖fN‖∞ + 1 ≤ M.

This shows that supn∈N ‖fn‖∞ ≤ M , that is, that (fn) is uniformly bounded.

Problem 7.2: Let fn : E → C and gn : E → C be uniformly convergent sequencesof functions. Prove that fn + gn converges uniformly. If, in addition, the functionsfn and gn are all bounded, prove that fngn converges uniformly.

Solution: Let f and g be the functions (assumed to exist) such that fn → funiformly and gn → g uniformly.

For the sum, we show that fn +gn → f +g uniformly. Let ε > 0, choose N1 ∈ Nsuch that n ≥ N1 implies

supx∈E

|fn(x) − f(x)| < 12ε,

and choose N2 ∈ N such that n ≥ N2 implies

supx∈E

|gn(x) − g(x)| < 12ε.

Set N = max(N1, N2). Let n ≥ N and let x ∈ E. Then

|(fn + gn)(x) − (f + g)(x)| ≤ |fn(x) − f(x)| + |gn(x) − g(x)|≤ sup

y∈E|fn(y) − f(y)| + sup

x∈E|gn(y) − g(y)|.

Therefore n ≥ N implies

supx∈E

|(fn + gn)(x) − (f + g)(x)|

≤ supy∈E

|fn(y) − f(y)| + supy∈E

|gn(y) − g(y)| < 12ε + 1

2ε = ε.

This shows that fn + gn → f + g uniformly.(Here is an alternate way of presenting the computation. Having chosen N as

above, for any n ≥ N we have

supx∈E

|(fn + gn)(x) − (f + g)(x)| ≤ supx∈E

[|fn(x) − f(x)| + |gn(x) − g(x)|]≤ sup

x∈E|fn(x) − f(x)| + sup

x∈E|gn(x) − g(x)|

< 12ε + 1

2ε = ε.

Note that the second inequality in this computation in general is not an equality.)Now consider the uniform convergence of the products. Since the functions are

bounded, we will use norm notation.First note that, for any bounded functions f and g, we have ‖fg‖∞ ≤ ‖f‖∞‖g‖∞.

Indeed, for any x ∈ E we have

|f(x)g(x)| = |f(x)| · |g(x)| ≤ ‖fg‖∞ ≤ ‖f‖∞‖g‖∞,

and we take the supremum over x ∈ E to get the result.

Now let ε > 0. Choose N1 ∈ N such that n ≥ N1 implies

‖fn(x) − f(x)‖∞ <ε

2(‖g‖∞ + 1),

and choose N2 ∈ N such that n ≥ N2 implies

‖gn(x) − g(x)‖∞ < min(

1,ε

2‖f‖∞ + 1

).

Set N = max(N1, N2). First, note that n ≥ N implies

‖gn‖∞ ≤ ‖g‖∞ + ‖gn − g‖∞ < ‖g‖∞ + 1.

Therefore, using the inequality in the previous paragraph and the triangle inequalityfor ‖ · ‖∞ (proved in class), we see that n ≥ N implies

‖fngn − fg‖∞ ≤ ‖fngn − fgn‖∞ + ‖fgn − fg‖∞≤ ‖fn − f‖∞‖gn‖∞ + ‖f‖∞‖gn − f‖∞≤ ε

2(‖g‖∞ + 1)· (‖g‖∞ + 1) + ‖f‖∞ · ε

2‖f‖∞ + 1< ε.

This shows that fngn → fg uniformly.

Problem 7.3: Construct a set E and two uniformly convergent sequences fn : E →C and gn : E → C of functions such that fngn does not converge uniformly. (Ofcourse, fngn converges pointwise.)

Solution (Sketch): Set E = R, let fn be the constant function fn(x) = 1n for all

x ∈ R, and let f(x) = 0 for all x. Define g by g(x) = x for all x ∈ R, and letgn = g for all n. It is obvious that fn → f uniformly and gn → g uniformly. Alsofngn → 0 pointwise. However, fn(n)gn(n) = 1 for all n.

Problem 7.4: Consider the series∞∑

n=1

11 + n2x

.

For which values of x does the series converge absolutely? On which closed intervalsdoes the series converge uniformly? On which closed intervals does the series fail toconverge uniformly? Is the sum continuous wherever the series converges? Is thesum bounded?

Comment: The problem was written in the book with the word “interval”. Becauseof Rudin’s strange terminology, I believe “closed interval” is what was meant.

Solution (Sketch):The situation will be clear with two lemmas. Define

fn(x) =1

1 + n2x

for all x and n.

Lemma 1. For every ε > 0, we have∞∑

n=1

‖fn|[ε,∞)‖∞ < ∞.

Proof: Observe that if x ≥ ε, then

0 < fn(x) =1

1 + n2x≤ 1

1 + n2ε<

1n2ε

,

so that

‖fn|[ε,∞)‖∞ ≤ 1n2ε

,

and∞∑

n=1

‖fn|[ε,∞)‖∞ ≤ 1ε

∞∑n=1

1n2

< ∞.

Lemma 2. For every ε > 0, there is N ∈ N such that∞∑

n=N

‖fn|(−∞,−ε]‖∞ < ∞.

Proof: Choose N ∈ N such that N2 > 2ε−1. For x ≥ ε and n ≥ N we then have

n2x − 1 ≥ n2ε − 1 ≥ 2n2

N2− 1 ≥ n2

N2.

Thus

fn(−x) = − 1n2x − 1

satisfies

0 > fn(−x) > −N2

n2.

So

‖fn|(−∞,−ε]‖∞ ≤ N2

n2,

and∞∑

n=N

‖fn|(−∞,−ε]‖∞ ≤ N2∞∑

n=N

1n2

< ∞.

It is easy to check that∑∞

n=1 fn(x) fails to converge at x = 0, and the seriesdoesn’t converge at x = − 1

n2 for any n ∈ N because one of the functions is notdefined there. Thus, the series does not converge even pointwise on any closedinterval containing any point of the set

S = {0} ∪ {−1, − 14 , − 1

9 , . . .}

.

For any closed interval not containing any points of S (even an infinite closedinterval), the two lemmas above make it clear that the series

∑∞n=1 fn(x) converges

uniformly to a bounded continuous function. In particular, the sum is continuousoff S.

The sum of the series is, however, not bounded on its domain. From the above,the domain is clearly R \ S. We show carefully that if U is any neighborhood ofany point x0 ∈ S, then the sum is not bounded on U ∩ (R \ S).

We first consider the case x0 = − 1n2 for some n. Choose r with

0 < r <1n2

− 1(n + 1)2

,

and set

E = U ∩[− 1

n2− r,

1n2

+ r

]and E0 = E ∩ (R \ S) = E \

{− 1

n2

}.

Since E0 ⊂ U ∩ (R \S), it suffices to show that the sum is not bounded on E0. Setε = 1

(n+1)2 , and choose N as in Lemma 2 for this value of ε. We alse require thatN > n. Then

∑∞k=N fk converges to a bounded function on E, by the conclusion

of Lemma 2. Moreover, each fk, for 1 ≤ k < N but k 6= n, is continuous on thecompact set

[− 1n2 − r, 1

n2 + r], hence bounded on its subset E. It follows that∑

k 6=n fk is bounded on E, hence on E0. However, fn is not bounded on E0, since− 1

n2 is a limit point of E0 and

limx→(− 1

n2 )−fn(x) = −∞ and lim

x→(− 1n2 )+

fn(x) = ∞.

It follows that∞∑

k=1

fk = fn +∑k 6=n

fk

is the sum of a bounded function on E0 and an unbounded function on E0, henceis not bounded on E0.

Now consider the case x0 = 0. Then U also contains − 1n2 for some n. Therefore

the sum is not bounded on U ∩ (R \ S) by the previous case.

We point out that, when showing that∑∞

k=1 fk is not bounded on E0, it is notsufficient to simply show that fn is unbounded there while all other fk are boundedthere. Here is an example of a series

∑∞k=0 gk which converges on R \ {0} to a

bounded function, with g0 unbounded but with gk bounded for all k 6= 0. Set

g0(x) = − 1x2

and, for k ≥ 1,

hk(x) = min(

k − 1,1x2

)and gk(x) = hk+1(x) − hk(x).

Then g0 is not bounded, |gk(x)| ≤ 1 for all k ≥ 1 and all x ∈ R \ {0}, but∑∞k=0 gk(x) = 0 for all x ∈ R \ {0}.

Problem 6.12 (with α(x) = x): Let ‖ ·‖2 be as in Problem 6.11. Let f : [a, b] → Rbe Riemann integrable, and let ε > 0. Prove that there is a continuous functiong : [a, b] → R such that ‖f − g‖2 < ε.

Hint: For a suitable partition P = (x0, x1, . . . , xn) of [a, b], define g on [xj−1, xj ]by

g(t) =xj − t

xj − xj−1f(xj−1) +

t − xj−1

xj − xj−1f(xj).

Solution (nearly complete): Fix f as in the problem. For any partition P of [a, b],let gP be the function defined in the hint. Check that gP is well defined andcontinuous.

We now choose a suitable partition. Let M = supx∈[a,b] |f(x)|. Choose ρ > 0 sosmall that

ρ2(b − a) < 12ε2.

Choose δ > 0 so small that

4M2ρ−1δ < 12ε2.

Choose a partition P as in the hint such that

U(P, f) − L(P, f) < δ.

Let

Mj = supx∈[xj−1, xj ]

f(x) and mj = infx∈[xj−1, xj ]

f(x).

On [xj−1, xj ], we then have both

mj ≤ f ≤ Mj and mj ≤ gP ≤ Mj .

Therefore

‖f − gP ‖22 =

∫ b

a

|f − gP |2 ≤n∑

j=1

(Mj − mj)2(xj − xj−1).

Let

S = {j ∈ {1, 2, . . . , n} : Mj − mj ≥ ρ}.

Date: 17 Feb. 2002.

1

We have

δ > U(P, f) − L(P, f) =n∑

j=1

(Mj − mj)(xj − xj−1)

≥∑j∈S

(Mj − mj)(xj − xj−1) ≥ ρ∑j∈S

(xj − xj−1),

whence ∑j∈S

(xj − xj−1) ≤ ρ−1δ

and ∑j∈S

(Mj − mj)2(xj − xj−1) ≤ (2M)2ρ−1δ < 12ε2.

For j 6∈ S, we have Mj − mj < ρ. Therefore∑j 6∈S

(Mj − mj)2(xj − xj−1) ≤ ρ2∑j 6∈S

(xj − xj−1)

≤ ρ2n∑

j=1

(xj − xj−1) = ρ2(b − a) < 12ε2.

So

‖f − gP ‖22 ≤

n∑j=1

(Mj − mj)2(xj − xj−1) < 12ε2 + 1

2ε2 = ε2,

whence ‖f − gP ‖2 < ε.

Problem 7.5: For x ∈ (0, 1) and n ∈ N, define

fn(x) =

0 0 < x < 1n+1

sin2(

πx

)1

n+1 ≤ x ≤ 1n

0 1n < x < 1

.

Prove that there is a continuous function f such that fn → f pointwise, but thatfn does not converge uniformly to f . Use the series

∑∞n=1 fn to show that absolute

convergence of a series at every point does not imply uniform convergence of theseries.

Solution (nearly complete): All functions appearing here are bounded, so we maywork in the space Cb((0, 1)) with the metric obtained from the norm ‖ · ‖∞.

It is easy to see that fn(x) → 0 for all x. However, ‖fn‖∞ = 1 for all n, so fn

does not converge uniformly to 0.The series

∑∞n=1 fn(x) converges absolutely for all x, because for each fixed x

only one term is nonzero. The series∑∞

n=1 fn does not converge uniformly: lettingsn be the n-th partial sum sn =

∑nk=1 fk, we get ‖sn − sn−1‖∞ = ‖fn‖∞ = 1 for

all n, so the sequence (sn) is not Cauchy in the metric which determines uniformconvergence.

Problem 7.6: Prove that the series∞∑

n=1

(−1)n · x2 + n

n2

converges uniformly on every bounded interval in R, but does not converge abso-lutely for any value of x.

Solution (nearly complete): Rewrite:∞∑

n=1

(−1)n · x2 + n

n2= x2

∞∑n=1

(−1)n · 1n2

+∞∑

n=1

(−1)n · 1n

.

For each fixed x, everything on the right converges, so this is legitimate; moreover,it follows that the original series converges for all x.

For each fixed x, the first series on the right converges absolutely but the sec-ond does not. Since the difference of absolutely convergent series is absolutelyconvergent, the series on the left does not converge absolutely.

Now fix a bounded interval I ⊂ R. We want to show that the original seriesconverges uniformly on I. For x ∈ I and n ∈ N, set

fn(x) =n∑

k=1

(−1)k · x2 + k

k2, gn(x) =

n∑k=1

(−1)k · x2

k2, and hn(x) =

n∑k=1

(−1)k · 1k

.

All functions appearing here are bounded, so we may work in the space Cb(I) withthe metric obtained from the norm ‖ · ‖∞. It is easy to see that the sequence(hn) converges in this metric to the function h with constant value

∑∞n=1(−1)n · 1

n ,and (because I is bounded) that the sequence (gn) converges in this metric to thefunction g(x) = x2

∑∞n=1(−1)n · 1

n2 . Therefore fn = gn+hn converges in this metricto the function f = g + h. (The proof is the same as the proof of the convergenceof the sum of convergent sequences of complex numbers.)

Alternate proof of the failure of absolute convergence: Fix x ∈ R. Then∣∣∣∣(−1)n · x2 + n

n2

∣∣∣∣ =x2 + n

n2≥ n

n2=

1n

.

Since∑∞

n=11n diverges, the series

∞∑n=1

∣∣∣∣(−1)n · x2 + n

n2

∣∣∣∣diverges by the comparison test.

Alternate proof of uniform convergence: Define fn, gn, hn, f , g, and h as above.Let ε > 0. Choose M so large that I ⊂ [−M, M ]. Using the convergence of∑∞

n=1(−1)n · 1n2 , choose N1 such that if n ≥ N1 then∣∣∣∣∣

∞∑k=1

(−1)k · 1k2

−n∑

k=1

(−1)k · 1k2

∣∣∣∣∣ <ε

2M.

Using the convergence of∑∞

n=1(−1)n · 1n , choose N2 such that if n ≥ N2 then∣∣∣∣∣

∞∑k=1

(−1)k · 1k−

n∑k=1

(−1)k · 1k

∣∣∣∣∣ <ε

2.

Then n ≥ max(N1, N2) implies

‖gn − g‖∞ ≤ 12ε and ‖hn − h‖∞ ≤ 1

2ε,

from which it is easy to conclude that

‖fn − f‖∞ ≤ ε.

Second alternate proof of uniform convergence: We are going to use the AlternatingSeries Test on a series of functions, but of course this needs justification. Webegin with an observation on the Alternating Series Test for a series of numbers∑∞

k=1(−1)kak, where

a1 > a2 > a3 > · · · > 0 and limn→∞ an = 0.

We claim (see below) that for any N we have∣∣∣∣∣∞∑

k=1

(−1)kak −N∑

k=1

(−1)kak

∣∣∣∣∣ ≤ aN+1.

Assuming this for the moment, consider an interval of the form [−M,M ] forM ∈ (0,∞). (All bounded intervals are contained in intervals of this form.) Foreach x ∈ [−M,M ], we have

x2

12>

x2

22>

x2

32> · · · > 0

and112

>222

>232

> · · · > 0,

whencex2 + 1

12>

x2 + 222

>x2 + 3

32> · · · > 0.

Also it is easy to see that

limn→∞

x2 + n

n2= 0.

The Alternating Series Test therefore gives pointwise convergence of the series.Moreover, applying the claim at the first step, we get, for x ∈ [−M,M ] and N ∈ N,∣∣∣∣∣

∞∑k=1

(−1)k · x2 + k

k2−

N∑k=1

(−1)k · x2 + k

k2

∣∣∣∣∣ ≤x2 + N + 1(N + 1)2

≤ M2 + N + 1(N + 1)2

.

The right hand side is independent of x ∈ [−M,M ] and has limit 0 as n → ∞, sothat the series converges uniformly on [−M,M ].

It remains to prove the claim above. Let sn =∑n

k=1(−1)kak be the partialsums. We can write an even partial sum s2n as

s2n = (a2 − a1) + (a4 − a3) + · · · + (a2n − a2n−1).

Using the analogous formula for s2n+2 and the fact that a2n+2 − a2n+1 < 0, we gets2n+2 < s2n. That is,

s0 > s2 > s4 > · · · .

Therefore∞∑

k=1

(−1)kak < s2n

for all n. A similar argument on the odd partial sums, using a2n −a2n+1 > 0, gives

s1 < s3 < s5 < · · · and∞∑

k=1

(−1)kak > s2n+1

In particular, if N = 2n is even then∣∣∣∣∣sN −∞∑

k=1

(−1)kak

∣∣∣∣∣ = s2n −∞∑

k=1

(−1)kak < s2n − s2n+1 = a2n+1 = aN+1,

while if N = 2n is odd then∣∣∣∣∣sN −∞∑

k=1

(−1)kak

∣∣∣∣∣ =∞∑

k=1

(−1)kak − s2n+1 < s2n+2 − s2n+1 = a2n+2 = aN+1.

This proves the claim, and completes the proof of uniform convergence.

Problem 7.7: For x ∈ (0, 1) and n ∈ N, define

fn(x) =x

1 + nx2.

Show that there is a function f such that fn → f uniformly, and that the equationf ′(x) = limn→∞ f ′

n(x) holds for x 6= 0 but not for x = 0.

Solution: We show that ‖fn‖∞ ≤ n−1/2, which will imply that fn → 0 uniformly.For |x| ≤ n−1/2, we have 1 + nx2 ≥ 1, so

|fn(x)| ≤ |x| ≤ n−1/2.

For |x| ≥ n−1/2, we can write

fn(x) =x−1

x−2 + n.

Since x−2 + n ≥ n, this gives

|fn(x)| ≤ x−1

n≤ n1/2

n= n−1/2.

So |fn(x)| ≤ n−1/2 for all x, proving the claim. This proves the first statement withf = 0.

Now we consider the second part. We calculate:

f ′n(x) =

1 − nx2

(1 + nx2)2.

Clearly f ′n(0) = 1 for all n, so limn→∞ f ′

n(0) = 1 6= f ′(0). Otherwise, we can rewrite

f ′n(x) =

1n2 − 1

nx2

1n2 + 1

n · 2x2 + x4,

and for each fixed x 6= 0 we clearly have limn→∞ f ′n(x) = 0 = f ′(0).

Alternate solution for the first part (Sketch): Using the formula for f ′n in the solu-

tion to the second part above, and standard calculus methods, show that fn(x) hasa global maximum at n−1/2, with value 1

2n−1/2, and a global minimum at −n−1/2,with value − 1

2n−1/2. (Note that this requires consideration of limx→∞ fn(x) andlimx→−∞ fn(x) as well as those numbers x for which f ′

n(x) = 0. Forgetting to do

so is already a blunder in elementary calculus.) This shows that ‖fn‖∞ = 12n−1/2,

so clearly limn→∞ ‖fn‖∞ = 0.

Second alternate solution for the first part (Sketch): The inequality (a − b)2 ≥ 0implies that 2ab ≤ a2 + b2 for all a, b ∈ R. Apply this result with a = 1 andb = |x|√n to get

2|x|√n ≤ 1 +(|x|√n

)2 = 1 + nx2

for all n ∈ N and x ∈ R. It follows that

|fn(x)| =|x|

1 + nx2≤ 1

2√

n

for all n ∈ N and x ∈ R. Therefore fn → 0 uniformly on R.

Problem 7.9: Let X be a metric space, let (fn) be a sequence of continuousfunctions from X to C, and let f be a function from X to C. Show that if fn

converges uniformly to f , then for every sequence (xn) in X with limit x, one haslimn→∞ fn(xn) = f(x). Is the converse true?

Solution: We prove the direct statement. Let (xn) be a sequence in X with xn → x,and let ε > 0. Note that f is continuous because it is the uniform limit of continuousfunctions. So there is δ > 0 such that whenever y ∈ X satisfies d(y, x) < δ, then|f(y) − f(x)| < 1

2ε. Choose N1 ∈ N so that if n ≥ N1 then for all x ∈ X we have|fn(x) − f(x)| < 1

2ε. Choose N2 ∈ N so that if n ≥ N2 then d(xn, x) < δ. Taken = max(N1, N2). Then for n ≥ N we have d(xn, x) < δ so that |f(xn)−f(x)| < 1

2ε.Therefore

|fn(xn) − f(x)| ≤ |fn(xn) − f(xn)| + |f(xn) − f(x)| < 12ε + 1

2ε = ε.

The converse is false in general. Take X = R, fn(x) = 1nx, and f = 0. Then

fn(x) → f(x) for all x, but the convergence is not uniform. Let (xn) be a sequencein R such that limn→∞ xn = x. Then (xn) is bounded, so there is M such that(xn) and x are all contained in [−M, M ]. Clearly fn|[−M, M ] converges uniformlyto f |[−M, M ]. The direct statement above therefore gives limn→∞ fn(xn) = f(x).(This can also be checked directly:

|fn(xn) − f(x)| =∣∣ 1nxn

∣∣ ≤ 1nM,

whence limn→∞ fn(xn) = f(x).)The converse is true if X is compact. (This was, strictly speaking, not asked

for.) We first show that the hypotheses imply that f is continuous. Let x ∈ X, andlet (xn) be a sequence in X with xn → x; we show that f(xn) → f(x). Inductivelychoose k(1) < k(2) < · · · such that |fk(n)(xn)−f(xn)| < 1

n . Define a new sequence(yn) in X by taking yk(n) = xn and yn = x for n 6∈ {k(1), k(2), . . . }. Clearlylimn→∞ yn = x. By hypothesis, we therefore have limn→∞ fn(yn) = f(x). Now(xn) is a subsequence of (yn), and (fk(n)(xn)) is the corresponding subsequence of(fn(yn)). Therefore fk(n)(xn) → f(x). Since |fk(n)(xn) − f(xn)| < 1

n , it followseasily that f(xn) → f(x). So f is continuous.

Now we prove that the convergence is uniform. Suppose not. Then there is ε > 0such that for all N ∈ N there is n ≥ N with ‖fn − f‖∞ ≥ ε. Accordingly, thereare k(1) < k(2) < · · · and xn ∈ X such that |fk(n)(xn)− f(xn)| > 1

2ε. Passing to asubsequence, we may assume that there is x ∈ X such that xn → x. (This is wherewe use compactness.) As before, define a new sequence (yn) in X by taking yk(n) =

xn and yn = x for n 6∈ {k(1), k(2), . . . }. Clearly limn→∞ yn = x. By hypothesis,we therefore have limn→∞ fn(yn) = f(x). Also, limn→∞ f(yn) = f(x) because fis continuous. However, since yk(n) = xn, we have |fk(n)(yk(n)) − f(yk(n))| > 1

2εfor all n. So (fk(n)(yk(n))) and (f(yk(n))) can’t have the same limit. This is acontradiction, and therefore we must have fn → f uniformly.

Problem 7.10: For x ∈ R, let (x) denote the fractional part of x, that is, (x) =x − n for x ∈ [n, n + 1). For x ∈ R, define

f(x) =∞∑

n=1

(nx)n2

.

Show that the set of points at which f is discontinuous is a countable dense subsetof R. Show that nevertheless f is Riemann integrable on every closed boundedinterval in R.

Solution (nearly complete): Set

fn(x) =(nx)n2

.

First observe that ‖fn‖∞ = 1n2 , so that

∑∞n=1 fn converges uniformly by the Weier-

strass test.

Lemma. Let X be a metric space, let fn : X → C be functions, and suppose thatfn → f uniformly. Let x0 ∈ X. If all fn are continuous at x0, then f is continuousat x0.

The proof can be obtained from Theorem 7.11 of Rudin, but we can give adirect proof which imitates the standard proof that the uniform limit of continuousfunctions is continuous.

Proof: Let ε > 0. Choose N ∈ N so that if n ≥ N then for all x ∈ X we have|fn(x) − f(x)| < 1

3ε. Use the continuity of fN at x0 to choose δ > 0 such thatwhenever x ∈ X satisfies d(x, x0) < δ, then |fN (x) − fN (x0)| < 1

3ε. For all x ∈ Xsuch that d(x, x0) < δ, we then have

|f(x) − f(x0)| ≤ |f(x) − fN (x)| + |fN (x) − fN (x0)| + |fN (x0) − f(x0)|< 1

3ε + 13ε + 1

3ε = ε.

Our particular functions fn are continuous at every point of R\Q, so the lemmaimplies that f is continuous on R \ Q.

We now show that f is not continuous at any point of Q. So let x ∈ Q. Writex = p

q in lowest terms. (If x ∈ Z, take q = 1.) One checks that limt→x+ fq(t) <

limt→x− fq(t). So define

α = limt→x−

fq(t) − limt→x+

fq(t) > 0.

Choose N ∈ N so large that n ≥ N implies∞∑

k=n

1k2

< 12α.

For any k, we clearly have limt→x+ fk(t) ≤ limt→x− fk(t). Set n = max(N, q).Then

limt→x−

n∑k=1

fk(t) − limt→x+

n∑k=1

fk(t) ≥ α.

Moreover,

0 ≤∞∑

k=n+1

fk(t) ≤∞∑

k=n+1

1k2

< 12α

for all t, whence

lim inft→x−

∞∑k=n+1

fk(t) − lim supt→x+

∞∑k=n+1

fk(t) ≥ − 12α.

(We use the obvious definitions of

lim inft→x−

g(t) and lim supt→x+

g(t)

here. Note, however, that these have not been formally defined, so if this is usedin a solution that is turned in, it needs justification. To avoid this, simply evaluateeverything at particular sequences converging to x from below and above, such assm = x − 1

m and tm = x + 1m .) Therefore

lim inft→x−

f(t) − lim supt→x+

f(t)

=

(lim

t→x−

n∑k=1

fk(t) − limt→x+

n∑k=1

fk(t)

)

+

(lim inft→x−

∞∑k=n+1

fk(t) − lim supt→x+

∞∑k=n+1

fk(t)

)

≥ 12α > 0.

This is clearly incompatible with continuity of f at x.The Riemann integrability of f on closed bounded intervals is immediate from

Theorem 7.16 of Rudin.

Problem 7.16: Let K be a compact metric space, and let (fn) be a uniformlyequicontinuous sequence of functions in C(X). Suppose that (fn) converges point-wise. Prove that (fn) converges uniformly.

Solution: We show that (fn) is a Cauchy sequence in C(K). Since C(K) is complete(Theorem 7.15 of Rudin), this will imply that (fn) converges uniformly.

Let ε > 0. Choose δ > 0 such that for all n ∈ N, and for all x, y ∈ K suchthat d(x, y) < δ, we have |fn(x) − fn(y)| < 1

4ε. Since K is compact, there existx1, x2, . . . , xk ∈ K such that the open δ-balls Nδ(x1), Nδ(x2), . . . , Nδ(xk) coverK. Since each sequence (fn(xj)) converges, there is N ∈ N such that wheneverm, n ≥ N and 1 ≤ j ≤ k we have |fm(xj) − fn(xj)| < 1

4ε. Now let x ∈ K bearbitrary. Choose j such that x ∈ Nδ(xj). For m, n ≥ N we then have

|fm(x) − fn(x)| ≤ |fm(x) − fm(xj)| + |fm(xj) − fn(xj)| + |fn(xj) − fn(x)|< 1

4ε + 14ε + 1

4ε = 34ε.

Since x is arbitrary, this shows that ‖fm − fn‖∞ ≤ 34ε < ε.

Alternate solution: Let f(x) = limn→∞ fn(x). We show that (fn) converges uni-formly to f .

Let ε > 0. Choose δ > 0 such that for all n ∈ N, and for all x, y ∈ Ksuch that d(x, y) < δ, we have |fn(x) − fn(y)| < 1

4ε. Letting n → ∞, we find that|f(x)−f(y)| ≤ 1

4ε whenever x, y ∈ K satisfy d(x, y) < δ. Since K is compact, thereexist x1, x2, . . . , xk ∈ K such that the open δ-balls Nδ(x1), Nδ(x2), . . . , Nδ(xk)cover K. Choose N ∈ N such that whenever n ≥ N and 1 ≤ j ≤ k, then|fn(xj)−f(xj)| < 1

4ε. Now let x ∈ K be arbitrary. Choose j such that x ∈ Nδ(xj).For n ≥ N we then have

|fn(x) − f(x)| ≤ |fn(x) − fn(xj)| + |fn(xj) − f(xj)| + |f(xj) − f(x)|< 1

4ε + 14ε + 1

4ε = 34ε.

Since x is arbitrary, this shows that ‖fn − f‖∞ ≤ 34ε < ε.

Problem 7.20: Let f ∈ C([0, 1]), and suppose that∫ 1

0

f(x)xn dx = 0

Date: 11 Feb. 2002.

1


for all n ∈ N. Prove that f = 0.Hint: The hypotheses imply that∫ 1

0

f(x)p(x) dx = 0

for every polynomial p. Use the Weierstrass Theorem to show that∫ 1

0

|f(x)|2 dx = 0.

Note: As is clear from Rudin’s hint, which suggested showing that∫ 1

0

f(x)2 dx = 0,

Rudin tacitly assumed that f is real, while I have assumed that f is complex.

Solution (Sketch): It is clear that the hypotheses imply that∫ 1

0

f(x)p(x) dx = 0

for every polynomial p. Use the Weierstrass Theorem to choose polynomials pn

such that limn→∞ ‖pn − f‖∞ = 0. Using Theorem 7.16 of Rudin, we get∫ 1

0

|f(x)|2 dx =∫ 1

0

ff = limn→∞

∫ 1

0

fpn = 0.

Since f is continuous, it is easy to show that this implies f = 0. (See Problem 6.2of Rudin, solved in an earlier solution set.)

Problem 7.21: Let S1 = {z ∈ C : |z| = 1} be the unit circle in the complex plane.Let A ⊂ C(S1) be the subalgebra consisting of all functions of the form

f(eiθ) =N∑

n=0

cneinθ

for θ ∈ R, with arbitrary N ∈ N and c0, c1, . . . , cN ∈ C. Then A separates thepoints of S1 and vanishes at no point of S1, but A is not dense in C(S1).

Hint: For every f ∈ A we have∫ 2π

0

f(eiθ)eiθ dθ = 0,

and this is also true for every f ∈ A.

Solution (Sketch): It is easy to check that A is a subalgebra. (Anyway, the problemis worded in such a way that you are to assume this is true.) The subalgebra Aseparates the points of S1 because it contains the function f(z) = z, and vanishesnowhere because it contains the constant function 1. The verification of the hintfor f ∈ A is direct from the computation, valid for any n ≥ 0,∫ 2π

0

einθ · eiθ dθ =∫ 2π

0

ei(n+1)θ dθ =1

i(n + 1)

(ei(n+1)·2π − ei(n+1)·0

)= 0.

The case f ∈ A now follows from Theorem 7.16 of Rudin.

To complete the proof, it suffices to find f ∈ C(S1) such that∫ 2π

0

f(eiθ)eiθ dθ 6= 0.

The function f(z) = z (that is, f(eiθ) = e−iθ) will do.

Problem 7.22 (with α(x) = x for all x): Let f : [a, b] → C be bounded andRiemann integrable on [a, b]. Prove that there are polynomials pn such that

limn→∞

∫ b

a

|f − pn|2 = 0.

(Compare with Problem 6.12 of Rudin.)

Solution (Sketch): In the notation of Problem 6.11 of Rudin (see an earlier solu-tion set), the conclusion is that limn→∞ ‖f − pn‖2

2 = 0. We prove the equivalentstatement limn→∞ ‖f − pn‖2 = 0.

Taking real and imaginary parts, without loss of generality f is real. (Thisreduction uses the triangle inequality for ‖ · ‖2, Problem 6.11 of Rudin, which issolved in an earlier solution set.) Further, it is enough to find, for every ε > 0, apolynomial p such that ‖f − p‖2 < ε.

Use Problem 6.12 of Rudin (solved in an earlier solution set) to find g ∈ CR([0, 1])such that ‖f − g‖2 < 1

2ε. Use the Weierstrass theorem to find a polynomial p suchthat

‖g − p‖∞ <ε√

2[b − a].

Then check that ‖g − p‖2 < 12ε, so that triangle inequality for ‖ · ‖2, Problem 6.11

of Rudin, implies that ‖f − p‖2 < ε.

Problem 7.24: Let X be a metric space, with metric d. Fix a ∈ X. For eachp ∈ X, define fp : X → C by

fp(x) = d(x, p) − d(x, a)

for x ∈ X. Prove that |fp(x)| ≤ d(a, p) for all x ∈ X, that f ∈ Cb(X), and that‖fp − fq‖ = d(p, q) for all p, q ∈ X.

Define Φ: X → Cb(X) by Φ(p) = fp for p ∈ X. Then Φ is an isometry, that is,a distance preserving function, from X to a subset of Cb(X).

Let Y be the closure of Φ(X) in Cb(X). Prove that Y is complete. Concludethat X is isometric to a dense subset of a complete metric space.

Solution (Sketch): That |fp(x)| ≤ d(a, p) follows from two applications of thetriangle inequality for d, namely

d(x, p) ≤ d(x, a) + d(a, p) and d(x, a) ≤ d(x, p) + d(a, p).

This shows that f is bounded.To prove continuity of fp, we observe that the map x 7→ d(x,w) is continuous

for every w ∈ X. Indeed, an argument using two applications of the triangleinequality and very similar to the above shows that |d(x,w)− d(y, w)| ≤ d(x, y) forall x, y ∈ X. This implies that x 7→ d(x,w) is continuous (in fact, Lipschitz withconstant 1).


The inequality |d(x,w) − d(y, w)| ≤ d(x, y) for all w, x, y ∈ X gives

|fx(w) − fy(w)| = |d(w, x) − d(w, y)| ≤ d(x, y)

for all w ∈ X, whence ‖fx − fy‖ ≤ d(x, y). Also,

fx(y) − fy(y) = d(y, x) − d(y, a) − [d(y, y) − d(y, a)] = d(x, y),

whence ‖fx − fy‖ ≥ d(x, y). It follows that ‖fx − fy‖ = d(x, y), which says exactlythat Φ is isometric.

Define Y = Φ(X). The space Cb(X) is complete by Theorem 7.15 of Rudin, soY is complete by the discussion after Definition 3.12 of Rudin. It follows that X isisometric to a dense subset of the complete metric space Y .

Problem 8.2: Define

ajk =

0 j < k−1 j = k2k−j j > k

That is, ajk is the number in the j-th row and k-th column of the array:

−1 0 0 0 . . .12 −1 0 0 . . .14

12 −1 0 . . .

18

14

12 −1 . . .

......

......

. . .

Prove that∞∑

j=1

∞∑k=1

ajk = −2 and∞∑

k=1

∞∑j=1

ajk = 0.

Solution (Sketch): It is immediate from the formula for the sum of a geometricseries that the column sums (which are clearly all the same) are all 0. Using theformula for the sum of a finite portion of a geometric series, one sees that the rowsums are −1, − 1

2 , − 14 , − 1

8 , . . . , and the sum of these is −2.

Problem 8.3: Prove that∞∑

j=1

∞∑k=1

ajk =∞∑

k=1

∞∑j=1

ajk

if aj,k ≥ 0 for all j and k. (The case +∞ = +∞ may occur.)

Comment: We give a solution which combines the finite and infinite cases. This isshorter than a solution using a case breakdown.

Solution: We show that∞∑

j=1

∞∑k=1

ajk = sup

∑(j,k)∈S

ajk : S ⊂ N×N finite

.

Since the right hand side is unchanged when j and k are interchanged, this willprove the result.

Date: 18 Feb. 2002.

1

It is clear that if S ⊂ N×N is finite, then∞∑

j=1

∞∑k=1

ajk ≥∑

(j,k)∈S

ajk.

This implies

∞∑j=1

∞∑k=1

ajk ≥ sup

∑(j,k)∈S


.

For the reverse inequality, let s ∈ R be an arbitrary number satisfying

s <∞∑

j=1

∞∑k=1

ajk;

we prove that

sup

∑(j,k)∈S


> s.

Choose ε > 0 such that

s + ε <∞∑

j=1

∞∑k=1

ajk.

Choose m ∈ N such thatm∑

j=1

∞∑k=1

ajk > s + ε.

If for some j0 with 1 ≤ j ≤ m, the sum∑∞

k=1 aj0k is infinite, choose n such that∑nk=1 aj0k > s. Then clearly S = {j0} × {1, 2, . . . , n} is a subset of N ×N such

that∑

(j,k)∈S ajk > s, and we are done. Otherwise, for 1 ≤ j ≤ m choose nj ∈ Nsuch that

nj∑k=1

ajk >∞∑

k=1

ajk −ε

m.

Setn = max(n1, n2, . . . , nm) and S = {1, 2, . . . ,m} × {1, 2, . . . , n}.

Then ∑(j,k)∈S

ajk >m∑

j=1

( ∞∑k=1

ajk −ε

m

)=

m∑j=1

∞∑k=1

ajk − ε > s.

This proves the desired inequality.

Note: We can even avoid the case breakdown in the last paragraph, as follows.Choose b1, b2, . . . , bm such that b1 + b2 + · · · + bm > s and bj <

∑∞k=1 ajk for

1 ≤ j ≤ m. Then choose nj ∈ N such that∑nj

k=1 ajk > bj . However, at this pointthe case breakdown seems easier.

Problem 8.4: Prove the following limit relations:

(a) limx→0

bx − 1x

= log(b) for b > 0.

Solution (Sketch): Set f(x) = bx = exp(x log(b)). The desired limit is by definitionf ′(0), which can be gotten from the second expression for f using the chain rule.


(b) limx→0

log(1 + x)x

= 1.

Solution (Sketch): This limit is f ′(0) with f(x) = log(1 + x).

(c) limx→0

(1 + x)1/x = e.

Solution (Sketch): By Part (b), we have

limx→0

log((1 + x)1/x

)= 1.

Apply exp to both sides, using continuity of exp.

(d) limn→∞

(1 +

x

n

)n

= ex.

Solution (Sketch): Write (1 +

x

n

)n

=[(

1 +x

n

)n/x]x

,

note that xn → 0 as n →∞, and apply Part (c) (using continuity at the appropriate

places).

Problem 8.5: Find the following limits:

(a) limx→0

e− (1 + x)1/x

x.

Comment: We first do a calculation, which shows what the answer is, and thengive a sketch of a solution in the correct logical order.

Calculation (Sketch): Rewrite and then use L’Hospital’s Rule to get

limx→0

e− (1 + x)1/x

x= lim

x→0

e− exp(

1x log(1 + x)

)x

= − limx→0

(x + 1)1/x(

1x(x+1) −

log(x+1)x2

)1

.

Rewrite:

− limx→0

(x + 1)1/x

(1

x(x + 1)− log(x + 1)

x2

)= −

(limx→0

(x + 1)1/x)(

limx→0

xx+1 − log(x + 1)

x2

)= −e lim

x→0

xx+1 − log(x + 1)

x2.

Use L’Hospital’s rule to get

−e limx→0

xx+1 − log(x + 1)

x2= e lim

x→0

1x+1 −

1(x+1)2

2x= e lim

x→0

12(x + 1)2

=e

2.

Solution (Sketch): We observe that

limx→0

1x+1 −

1(x+1)2

2x= lim

x→0

12(x + 1)2


exists (and is equal to 12 ). Since the other hypotheses of L’Hospital’s Rule are also

verified, we can use it to show that

− limx→0

xx+1 − log(x + 1)

x2= lim

x→0

1x+1 −

1(x+1)2

2x=

12.

Therefore

− limx→0

(x + 1)1/x

(1

x(x + 1)− log(x + 1)

x2

)=

e

2.

Since the other hypotheses of L’Hospital’s Rule are also verified, we can use it againto show that

limx→0

e− (1 + x)1/x

x= − lim

x→0

(x + 1)1/x(

1x(x+1) −

log(x+1)x2

)1

=e

2.

(b) limn→∞

n

log(n)

(n1/n − 1

).

Solution: Rewrite the limit as

limn→∞

n

log(n)

(n1/n − 1

)= lim

n→∞

exp(

1n log(n)

)− 1

1n log(n)

Now limn→∞1n log(n) = 0, so

exp(

1n log(n)

)− 1

1n log(n)

= limh→0

exp(h)− 1h

= exp′(0) = 1.

(c) limx→0

tan(x)− x

x[1− cos(x)].

Solution 1 (Sketch): Write

tan(x)− x

x[1− cos(x)]=(

1cos(x)

)(sin(x)− x cos(x)

x− x cos(x)

).

Since limx→0 cos(x) = 1, we get

limx→0

tan(x)− x

x[1− cos(x)]= lim

x→0

sin(x)− x cos(x)x− x cos(x)

.

Now apply L’Hospital’s rule three times (being sure to check that the hypothesesare satisfied!):

limx→0


= limx→0

x sin(x)1− cos(x) + x sin(x)

= limx→0

x cos(x) + sin(x)x cos(x) + 2 sin(x)

= limx→0

2 cos(x)− x sin(x)3 cos(x)− x sin(x)

=23.

(See the solution to Part (a) for the logically correct way to write this.)


Solution 2 (Sketch; not recommended): Apply L’Hospital’s rule three times to theoriginal expression (being sure to check that the hypotheses are satisfied!):

limx→0

tan(x)− x

x[1− cos(x)]= lim

x→0

[sec(x)]2 − 11− cos(x) + x sin(x)

= limx→0

2[sec(x)]2 tan(x)x cos(x) + 2 sin(x)

= limx→0

2[sec(x)]4 + 4[sec(x)]2[tan(x)]2

3 cos(x)− x sin(x)=

23.

(See the solution to Part (a) for the logically correct way to write this.)This solution is not recommended because of the messiness of the differentiation.

(The results given here were obtained using Mathematica.)

Solution 3 (Sketch; the mathematical justification for the power series manipula-tions is easy but is not provided here): We compute the limit by substitution ofpower series. We check easily that the usual power series

sin(x) = x− 13!x

3 + 15!x

5 − 17!x

7 + · · · and cos(x) = 1− 12!x

2 + 14!x

4 − 16!x

6 + · · ·

follow from the definitions of sin(x) and cos(x) in terms of exp(ix) and from thedefinition of exp(z). Start from the equivalent limit given in Solution 1. We have


=

(x− 1

3!x3 + 1

5!x5 − · · ·

)− x

(1− 1

2!x2 + 1

4!x4 − · · ·

)x− x

(1− 1

2!x2 + 1

4!x4 − · · ·

)=

(12! −

13!

)x3 +

(15! −

14!

)x5 + · · ·

12!x

3 − 14!x

5 + · · ·

=

(12! −

13!

)+(

15! −

14!

)x2 + · · ·

12! −

14!x

2 + · · ·.

The last expression defines a function of x which is continuous at 0, so

limx→0

tan(x)− x

x[1− cos(x)]= lim

x→0


=

(12! −

13!

)(12!

) =23.

(d) limx→0

x− sin(x)tan(x)− x

.

Solution 1 (Sketch): Write


= cos(x)(

x− sin(x)sin(x)− x cos(x)

).

Since limx→0 cos(x) = 1, we get

limx→0


= limx→0


.

Now apply L’Hospital’s rule three times (being sure to check that the hypothesesare satisfied!):

limx→0


= limx→0

1− cos(x)x sin(x)

= limx→0

sin(x)x cos(x) + sin(x)

= limx→0

cos(x)2 cos(x)− x sin(x)

=12.


Solution 2 (Sketch): Apply L’Hospital’s rule twice to the original expression (beingsure to check that the hypotheses are satisfied!):

limx→0


= limx→0

1− cos(x)[sec(x)]2 − 1

= limx→0

sin(x)2[sec(x)]2 tan(x)

= limx→0

12[sec(x)]3

=12.


Solution 3 (Sketch): Use the same method as Solution 3 to Part (c). Details areomitted.

Problem 8.6: Let f : R → R be a nonzero function satisfying f(x+y) = f(x)f(y)for all x, y ∈ R.

(a) Suppose f is differentiable. Prove that there is c ∈ R such that f(x) =exp(cx) for all x ∈ R.

Solution (nearly complete): Since f(0)f(x) = f(x) for all x, and since there is somex such that f(x) 6= 0, it follows that f(0) = 1. The computation

limh→0

f(x + h)− f(x)h

= f(x) limh→0

f(h)− f(0)h

= f(x)f ′(0)

shows that f ′(x) = f(x)f ′(0) for all x ∈ R. Now let g(x) = f(x) exp(−xf ′(0))for x ∈ R. Differentiate g using the product rule and the formula above, gettingg′(x) = 0 for all x. So g is constant. Since f(0) = 1, we get g(0) = 1. Thereforef(x) = exp(xf ′(0)) for all x ∈ R.

Alternate solution: Use the solution to Part (b).

(b) Suppose f is continuous. Prove that there is c ∈ R such that f(x) = exp(cx)for all x ∈ R.

Solution (nearly complete): We have f(0) = 1 for the same reason as in the firstsolution to Part (a). By continuity, there is a > 0 such that f(a) > 0. Next,define g(x) = f(x) exp(−a−1x log(f(a))) for x ∈ R. Then g is continuous, satisfiesg(x + y) = g(x)g(y) for all x, y ∈ R, and g(a) = 1. Set

S = inf{x > 0: g(x) = 1} and x0 = inf S.

We have S 6= ∅ because a ∈ S.First suppose that x0 > 0. Then g(x0) = 1 by continuity. We have g

(12x0

)2 =g(x0) = 1 but g

(12x0

)6= 1, so g

(12x0

)= −1. Then g

(14x0

)2 = g(

12x0

)= −1,

contradicting the assumption that g(

14x0

)is real. So x0 = 0.

We can now show that g(x) = 1 for all x ∈ R. Let ε > 0. Choose δ > 0 suchthat whenever y ∈ R satisfies |y − x| < δ, then |g(y)− g(x)| < ε. By the definition

of S and because x0 = 0, there is z ∈ S such that 0 < z < δ. Choose n ∈ Z suchthat |nz − x| < δ. Then g(nz) = g(z)n = 1, so |g(x) − 1| < ε. This shows thatg(x) = 1.

So f(x) = exp(cx) with c = a−1 log(f(a)).

Alternate solution (Sketch): We have f(0) = 1 for the same reason as in the firstsolution to Part (a). Furthermore, x ∈ R implies

f(x) = f(

12x)2 ≥ 0

since f(

12x)∈ R. Moreover, if f(x) = 0 then f

(12x)

= 0, and by inductionf (2−nx) = 0 for all n. Since f is continuous and limn→∞ 2−nx = 0, this contradictsf(0) = 1. Therefore f(x) > 0 for all x.

Define g(x) = exp(x log(f(1))) for x ∈ R. For n ∈ N,

f(

1n

)n = f(1) and g(

1n

)n = g(1) = f(1).

Since both f(

1n

)and g

(1n

)are positive, and positive n-th roots are unique, it

follows that f(

1n

)= g

(1n

)for all n ∈ N. An easy argument now shows that

f(x) = g(x) for all x ∈ Q. Since f and g are continuous, and since Q is dense inR, it follows that f = g. So f(x) = exp(cx) with c = log(f(1)).

Problem 8.7: Prove that2π

<sin(x)

x< 1

for 0 < x < 12π.

Solution (Sketch): The inequality sin(x)x < 1 is the same as sin(x) < x. This is

proved by noting that sin(0) = 0 and that the derivative 1− cos(x) of x− sin(x) isstrictly positive for 0 < x < 1

2π.This also implies that sin(x)

x < 1 for x = 12π, so that 2

π < 1.For the other inequality, suppose there is x0 with 0 < x0 < 1

2π such that

2π

≥ sin(x0)x0

.

Using the Intermediate Value Theorem and limx→0sin(x)

x = 1, there is x0 with0 < x0 < 1

2π such that

2π

=sin(x0)

x0.

Set f(x) = sin(x)− 2π ·x. For 0 < x ≤ 1

2π, we have f ′′(x) = − sin(x) < 0, so that f ′

is strictly decreasing on[0, 1

2π]. The Mean Value Theorem gives z ∈ (0, x0) such

that f ′(z) = 0, so f ′(x) < 0 for x0 ≤ x ≤ 12π. Since f(x0) = 0, we get f

(12π

)< 0,

a contradiction.

Alternate solution 1 (Sketch): Suppose f : [0, a] → R is a continuous function suchthat:

(1) f(0) = 0.(2) f ′(x) exists for x ∈ (0, a).(3) f ′ is strictly decreasing on (0, a).

We claim that the function g(x) = x−1f(x) is strictly decreasing on (0, a].To see that the claim implies the result, take f(x) = sin(x) and a = 1

2π. Weconclude that g is strictly decreasing on

(0, 1

2π]. We get the result by observing

that

limx→0+

sin(x)x

= sin′(0) = cos(0) = 1.

Date: 27 Feb. 2002.

1

To prove the claim, let 0 < x1 < x2 ≤ b. Use the Mean Value Theorem to choosec1 ∈ (0, x1) and c2 ∈ (x1, x2) such that

f(x1)x1

=f(x1) − f(0)

x1 − 0= f ′(c1) and

f(x2) − f(x1)x2 − x1

= f ′(c2).

Then c1 < c2, so f ′(c1) > f ′(c2), whence

f(x1)x1

>f(x2) − f(x1)

x2 − x1.

Multiply both sides of this inequality by x1(x2 − x1) > 0 to get

(x2 − x1)f(x1) > x1(f(x2) − f(x1)).

Multiply out and cancel −x1f(x1) to get x2f(x1) > x1f(x2), and divide by x1x2

to get g(x1) > g(x2).

Alternate solution 2 (Sketch): Since

sin(

12π

)12π

=2π

and limx→0+

sin(x)x

= sin′(0) = cos(0) = 1,

it suffices to prove that the function g(x) = x−1 sin(x) is strictly decreasing on(0, 1

2π). We do this by showing that g′(x) < 0 on this interval.

Begin by observing that the function h(x) = x cos(x) − sin(x) satisfies h(0) = 0and h′(x) = − sin(x) for all x. Therefore h′(x) < 0 for x ∈ (

0, 12π

], and from

h(0) = 0 we get h(x) < 0 for x ∈ (0, 1

2π]. Now calculate, for x ∈ (

0, 12π

),

g′(x) =x cos(x) − sin(x)

x2=

h(x)x2

< 0.

This is the required estimate.

Alternate solution 3 (Sketch): As in the second alternate solution, we set g(x) =x−1 sin(x) and show that g′(x) < 0 for x ∈ (

0, 12π

). Define

q(x) =sin(x)cos(x)

− x

for x ∈ [0, 1

2π). Then the quotient rule and the relation sin2(x) + cos2(x) = 1 give

q′(x) =1

cos2(x)− 1.

For x ∈ (0, 1

2π)

we have 0 < cos(x) < 1, from which it follows that q′(x) > 0.Since q is continuous on

[0, 1

2π)

and q(0) = 0, we get q(x) > 0 for x ∈ (0, 1

2π). It

follows that sin(x) > x cos(x) for x ∈ (0, 1

2π). As in the second alternate solution,

this implies that g′(x) < 0 for x ∈ (0, 1

2π).

Problem 8.8: For n ∈ N ∪ {0} and x ∈ R, prove that

| sin(nx)| ≤ n| sin(x)|.Note that this inequality may be false for n not an integer. For example,∣∣sin (

12xm

)∣∣ > 12 |sin(x)| .

Solution (Sketch): Combining the formula exp(i(x + y)) = exp(ix) exp(iy) witheither the result

cos(x) = Re(exp(ix)) and sin(x) = Im(exp(ix))

(for x real) or the definitions

cos(x) = 12 [exp(ix) + exp(−ix)] and sin(x) = 1

2i [exp(ix) − exp(−ix)],

prove the addition formula

sin(x + y) = sin(x) cos(y) + cos(x) sin(y).

(Using the first suggestion gives this only for real x and y, but that is all that isneeded here.)

Now prove the result by induction on n. For n = 0 the desired inequality says0 ≤ 0 for all x, which is certainly true. Assuming it is true for n, we have (using theaddition formula in the first step and the induction hypothesis and the inequality| cos(a)| ≤ 1 for all real a in the second step)

| sin((n + 1)x)| ≤ | sin(x)| · | cos(nx)| + | cos(x)| · | sin(nx)|≤ | sin(x)| + n| sin(x)| = (n + 1)| sin(x)|

for all x ∈ R.

Alternate solution (Sketch): We first prove the inequality for 0 ≤ x ≤ 12π. If

x ∈ [0, 1

2π]

satisfies sin(x) ≥ 1n , then since | sin(nx)| ≤ 1 there is nothing to prove.

Otherwise, x < π2n by Problem 8.7. Since t 7→ cos(t) is nonincreasing on

[0, 1

2π]

(itsderivative − sin(t) is nonpositive there), we get cos(nx) ≤ cos(x) for x ∈ [

0, π2n

].

With f(x) = n sin(x) − sin(nx), we therefore have f(0) = 0 and

f ′(x) = n[cos(x) − cos(nx)] ≤ 0

for x ∈ [0, π

2n

]. Since also sin(nx) ≥ 0 on this interval, the inequality is proved for

0 < x ≤ π2n and hence 0 ≤ x ≤ 1

2π.For − 1

2π ≤ x ≤ 0, the inequality follows from the fact that t 7→ sin(t) is an oddfunction. (This is easily seen from the definition.) For 1

2π ≤ x ≤ 32π, reduce to

− 12π ≤ x ≤ 1

2π using the identity sin(k(x + π)) = (−1)k sin(kx) (which is easilyderived from the definition and exp(iπ) = −1). The inequality now follows for allx by periodicity.

Problem 8.9: (a) For n ∈ N set

sn = 1 + 12 + 1

3 + · · · + 1n .

Prove that

γ = limn→∞[sn − log(n)]

exists. (This limit is called Euler’s constant. Numerically, γ ≈ 0.5772. It is notknown whether γ is rational or not.)

Solution: We have

1n − [log(n + 1) − log(n)] = 1

n −∫ n+1

n

1t dt =

∫ n+1

n

(1n − 1

t

)dt.

for n ∈ N. Since the integrand is between 0 and 1n − 1

n+1 , it follows that

0 ≤ 1n − [log(n + 1) − log(n)] ≤ 1

n − 1n+1 .

Sincen∑

k=1

(1k − [log(k + 1) − log(k)]

)= sn − log(n + 1),

we get by adding up terms

sn − log(n + 1) ≤ 1 − 1n+1 < 1

for all n; also,

sn − log(n + 1) = sn−1 − log(n) + 1n − [log(n + 1) − log(n)] ≥ sn−1 − log(n).

Therefore (sn − log(n + 1)) is a bounded nondecreasing sequence, hence converges.Next observe that

log(n + 1) − log(n) =∫ n+1

n

1t dt,

so that

0 ≤ log(n + 1) − log(n) ≤ 1n .

It follows that limn→∞[log(n + 1) − log(n)] = 0, whence

limn→∞[sn − log(n)] = lim

n→∞[sn − log(n + 1)].

Alternate solution 1: Let Pn be the partition Pn = (1, 2, . . . , n) of [1, n]. Setf(x) = x−1. Since f is nonincreasing, we have

L(Pn, f) =n∑

k=2

1k

and U(Pn, f) =n−1∑k=1

1k

.

Therefore

sn − log(n) = 1 + L(Pn, f) −∫ n

1

f = 1n + U(Pn, f) −

∫ n

1

f.

We prove that (sn − log(n)) is nonincreasing. We have

[sn − log(n)] − [sn+1 − log(n + 1)]

=[1 + L(Pn, f) −

∫ n

1

f

]−

[1 + L(Pn+1, f) −

∫ n+1

1

f

]

=∫ n+1

n

f − [L(Pn+1, f) − L(Pn, f)] =∫ n+1

n

f − 1n+1 .

The last expression is nonnegative because f(x) ≥ 1n+1 for all x ∈ [n, n + 1]. So

(sn − log(n)) is nonincreasing.We prove that (sn − log(n)) is bounded below (by 0). Indeed,

0 ≤ U(Pn, f) −∫ n

1

f ≤ 1n + U(Pn, f) −

∫ n

1

f = sn − log(n)

for all n.Since (sn − log(n)) is nonincreasing and bounded below, limn→∞(sn − log(n))

exists.

Alternate solution 2: First prove the following lemma.

Lemma. For every x ≥ 0, we have 0 ≤ x − log(1 + x) ≤ 12x2.

Proof: Set g(x) = x − log(1 + x) and h(x) = 12x2 for x ∈ (−1, ∞). Then g(0) =

h(0) = 0. Furthermore, for all x ≥ 0 we have

g′(x) = 1 − 11 + x

=x

1 + x≥ 0 and h′(x) − g′(x) = x − x

1 + x=

x2

1 + x≥ 0.

Therefore 0 ≤ g(x) ≤ h(x) for all x ≥ 0.

Now define

bk = 1k − [log(k) − log(k + 1)] = 1

k − log(1 + 1

k

)for k ≥ 1. (That the two expressions are equal follows from the algebric properties ofthe function log. See Equation (40) on of Rudin’s book.) Since log(1) = 0,we have

sn − log(n + 1) =n∑

k=1

bk.

By the lemma, we have 0 ≤ bk ≤ 12k−2 for k ≥ 1. Since

∑∞k=1 k−2 converges, the

Comparison Test shows that∑∞

k=1 bk converges, whence limn→∞[sn − log(n + 1)]exists.

Now show that limn→∞[log(n + 1) − log(n)] = 0 as in the first solution, andconclude as there that limn→∞[sn − log(n)] exists.

Alternate solution 3 (Outline): This solution differs from the previous one only inthe method of proof of the lemma. Instead of comparing derivatives, we use thederivative form of the remainder in Taylor’s Theorem (see Theorem 5.15 of Rudin’sbook) to compare log(1 + x) with the Taylor polynomials of degrees 1 and 2.

(b) Roughly how large must m be so that n = 10m satisfies sn > 100?

Solution (Sketch): The proof above gives 0 < sn−log(n+1) < 1 for all n. Thereforesn ∈ (log(n + 1), 1 + log(n + 1)). We have log(n + 1) ≥ 100 if and only if n >exp(100) − 1. So it suffices to take

m = log10(exp(100)) = 100 log10(e) ≈ 43.43.

Problem 8.10: Prove that ∑p prime

1p

diverges.Hint: Given N , let p1, p2, . . . , pk be those primes that divide at least one integer

in {1, 2, . . . , N}. Then

N∑n=1

1n≤

k∏j=1

(1 +

1pj

+1p2

j

+ · · ·)

=k∏

j=1

(1 − 1

pj

)−1

≤ exp

k∑

j=1

1pj

.

The last inequality holds because (1 − x)−1 ≤ exp(2x) for 0 ≤ x ≤ 12 .

Solution (Sketch): It suffices to verify the inequalities because∑∞

n=11n diverges.

For the first inequality, let m be the largest power of any prime appearing in the

prime factorization of any integer in {1, 2, . . . , N}. Then one checks thatk∏

j=1

(1 +

1pj

+1p2

j

+ · · · + 1pm

j

)=

∑n∈S

1n

,

where S is the set of all integers whose prime factorization involves only the primesp1, p2, . . . , pk, and in which no prime appears with multiplicity greater than m.Moreover, {1, 2, . . . , N} ⊂ S. For the other inequality, since 0 ≤ 1

p ≤ 12 for all

primes p, it suffices to check that (1 − x)−1 ≤ exp(2x) for 0 ≤ x ≤ 12 . Now

1 + 2x ≤ exp(2x) for all x ≥ 0 by any of a number of arguments. The inequality(1 − x)−1 ≤ 1 + 2x for 0 ≤ x ≤ 1

2 is easily verified by multiplying both sides by1 − x.

Problem 8.11: Let f : [0,∞) → R be a function such that limx→∞ f(x) = 1 andf is Riemann integrable on every interval [0, a] for a > 0. Prove that

limt→0+

t

∫ ∞

0

e−txf(x) dx = 1.

Solution (Sketch): Let ε > 0. Choose a > 0 such that |f(x) − 1| < 12ε for x ≥ a.

Since f is Riemann integrable on [0, a], it is bounded there. Choose M such that|f(x)| ≤ M for all x ∈ [0, a]. Choose δ so small that δ(M +1) < 1

2ε. Then 0 < t < δimplies

t

∫ ∞

0

e−tx · 1 dx = 1.

and

t

∫ ∞

0

e−tx|f(x) − 1| dx ≤ δ

∫ a

0

e−tx(M + 1) dx + t

∫ ∞

a

e−tx 12ε dx

≤ δ(M + 1) + 12εe−at < ε.



Problem 5.27: Let R = [a, b] × [α, β] ⊂ R2 be a rectangle in the plane, and letϕ : R → R be a function. A solution of the initial value problem y′ = ϕ(x, y) andy(a) = c (with c ∈ [α, β]) is, by definition, a differentiable function f : [a, b] → [α, β]such that f(a) = c and such that f ′(t) = ϕ(t, f(t)) for all t ∈ [a, b].

Assume that there is a constant A such that

|ϕ(x, y2) − ϕ(x, y1)| ≤ A|y2 − y1|for all x ∈ [a, b] and y1, y2 ∈ [α, β]. Prove that such a problem has at most onesolution.

Hint: Apply Problem 5.26 to the difference of two solutions.Note that this uniqueness theorem does not hold for the initial value problem

y′ = y1/2 and y(0) = 0, which has two solutions f1(x) = 0 for all x and f2(x) = 14x20

for all x. Find all other solutions to this initial value problem.

Solution (Sketch): Let f1 and f2 be two solutions to the initial value problemy′ = ϕ(x, y) and y(a) = c. Then for all t ∈ [a, b] we have

|f ′2(t) − f ′

1(t)| = |ϕ(t, f(t)) − ϕ(t, f(t))| ≤ A|f ′2(t) − f ′

1(t)|,and also f2(c) − f1(c) = 0, so Problem 5.26 shows that f2 − f1 = 0.

Now we investigate the solutions to y′ = y1/2 and y(0) = 0. The intervals arenot specified, but it seems reasonable to assume that [0,∞) × [0,∞) is intended.

First, for r ∈ [0,∞) define gr : [0,∞) → [0,∞) by

gr(t) ={

0 0 ≤ t ≤ r14 (t − r)2 t > r

,

and define g∞(t) = 0 for all t. Check that gr is in fact a solution. (This istrivial everywhere except at t = r, where one must calculate g′r(t) directly from thedefinition.) We are going to show that these are all solutions.

We claim that if t0 ∈ R and c > 0, then the initial value problem y′ = y1/2

and y(t0) = c has the solution f(t) = 14 (t − r)2 for t ≥ t0, where r is chosen to be

r = t0 − 2√

c. Applying the uniqueness result in the first part of the problem on[t0, b]× [α, β], with α ≤ c ≤ 1

4 (b− r)2 ≤ β, and letting b → ∞, α → 0, and β → ∞,we find that there are no other solutions f : [t0,∞) → (0,∞). (Be sure to checkthat the hypotheses hold!)

Date: 5 March 2002.

1

Now suppose g : [0,∞) → [0,∞) is a solution satisfying g(0) = 0. We claim thatthere is r ∈ [0,∞] such that g = gr. Without loss of generality g 6= g∞, so thereexists s > 0 such that g(s) > 0. Let

r = inf({s ∈ [0,∞) : g(s) > 0}).By continuity, we must have g(r) = 0. We will show that g = gr.

Suppose not. There are three cases to consider. First, suppose there is t ≤ rsuch that g(t) > 0. Apply the claim above with t0 = t to obtain

g(r) = 14

(r − t + 2

√g(t)

)2

6= 0,

a contradiction. Next, suppose there is t > r such that g(t) = 0. Choose s ∈ (r, t)such that g(s) 6= 0, and apply the claim above with t0 = t to obtain

g(t) = 14

(t − s + 2

√g(s)

)2

6= 0,

a contradiction. Finally, suppose there is t > r such that g(t) 6= gr(t) and g(t) 6= 0.Repeated use of the claim, with t0 running through a sequence decreasing monoton-

ically to t− 2√

g(t), shows that g(s) = 14

(s − t + 2

√g(t)

)2

for all s > t− 2√

g(t).

If t − 2√

g(t) > r, then continuity gives g(t − 2

√g(t)

)= 0, and we obtain a con-

tradiction as in the second case. If t − 2√

g(t) < r, then we obtain a contradictionas in the first case. If t − 2

√g(t) = r, then one checks that g = gr.

Comment: The techniques for finding solutions found in nonrigorous differentialequations courses (such as Math 256 at the University of Oregon) are not proofs ofanything, and therefore have no place in a formal proof in this course. (They can,of course, be used to find solutions in scratchwork.) One sees this from the factthat these methods do not find most of the solutions gr given above. Such methodsmay be used in scratchwork to find solutions, but the functions found this way mustbe verified to be solutions, and can only be expected to be the only solutions whenthe hypotheses of the first part of the problem are satisfied.

Problem 8.20: The following simple computation yields a good approximation toStirling’s formula.

Define f, g : [1,∞) → R by

f(x) = (m + 1 − x) log(m) + (x − m) log(m + 1)

for m = 1, 2, . . . and x ∈ [m, m + 1], and

g(x) =x

m− 1 + log(m)

for m = 1, 2, . . . and x ∈ [m − 1

2 , m + 12

)(x ∈ [

1, m + 12

)if m = 1). Draw the

graphs of f and g. Prove that f(x) ≤ log(x) ≤ g(x) for x ≥ 1, and that∫ n

1

f(x) dx = log(n!) − 12 log(n) > − 1

8 +∫ n

1

g(x) dx.

Integrate log(x) over [1, n]. Conclude that78 < log(n!) − (

n + 12

)log(n) + n < 1

for n = 2, 3, 4, . . . . Thus

e7/8 <n!

(n/e)n√

n< e.

Solution (Sketch): Note: No graphs are included here.We need the following lemma, which is essentially a concavity result.

Lemma. Let [a, b] ⊂ R be a closed interval, and let f : [a, b] → R be a continuousfunction such that f ′′(x) exists for all x ∈ (a, b) and f ′′(x) ≤ 0 on (a, b). Then

f(x) ≥ f(a) +f(b) − f(a)

b − a· (x − a)

for all x ∈ [a, b].

Proof (Sketch): Let

q(x) = f(x) − f(a) +f(b) − f(a)

b − a· (x − a)

for x ∈ [a, b]. Then q satisfies the same hypotheses as f (note that q′′ = f ′′), andq(a) = q(b) = 0. It suffices to prove that q(x) ≥ 0 for all x ∈ (a, b).

Suppose x0 ∈ (a, b) and q(x0) < 0. Use the Mean Value Theorem to choosec1 ∈ (a, x0) and c2 ∈ (x0, b) such that q′(c1) < 0 and q′(c2) > 0. Then c1 < c2,so the Mean Value Theorem, applied to q′, gives d ∈ (c1, c2) such that q′′(d) > 0.This is a contradiction.

Alternate Proof (Sketch): As in the first proof, let

q(x) = f(x) − f(a) +f(b) − f(a)

b − a· (x − a),

which satisfies q′′(x) ≤ 0 on (a, b) and q(a) = q(b) = 0; it suffices to prove thatq(x) ≥ 0 for all x ∈ (a, b).

Use the Mean Value Theorem to choose c ∈ (a, b) such that q′(c) = 0. Sinceq′′(x) ≤ 0 for all x ∈ (a, b), it follows that q′ is nonincreasing. Therefore q′(x) ≥ 0for x ∈ [a, c] and q′(x) ≤ 0 for x ∈ [c, b]. It follows that q is nondecreasing on [a, c],so for x ∈ [a, c] we have q(x) ≥ q(a) = 0. It also follows that q is nonincreasing on[c, b], so for x ∈ [c, b] we have q(x) ≥ q(b) = 0.

The lemma can be used directly to show that f(x) ≤ log(x).To show g(x) ≥ log(x), show that g(m) = log(m), that g′(x) ≤ log′(x) for

m − 12 ≤ x ≤ m, and that g′(x) ≥ log′(x) for m ≤ x ≤ m + 1

2 .Next, use the inequality f(x) ≤ log(x) ≤ g(x) for x ≥ 1 to get∫ n

1

f(x) dx ≤∫ n

1

log(x) dx ≤∫ n

1

g(x) dx.

This inequality is actually strict (as is required to solve the problem). To see this,use the fact that there are x1 and x2 such that f is continuous at x1, such that g iscontinuous at x2, such that f(x1) < log(x1), and such that log(x2) < g(x2). (Thisneeds proof!)

To calculate∫ n

1f(x) dx, calculate

∫ m+1

m

f(x) dx = 12 [log(m + 1) − log(m)],

and add these up. To estimate∫ n

1g(x) dx, calculate

∫ m+ 12

m− 12

g(x) dx = log(m) and∫ 3/2

1

g(x) dx = 18 ,

and estimate ∫ n

n− 12

g(x) dx ≤ 12g(n) = 1

2 log(n).

(The exact answer for the last one is 12 log(n)− 1

8n .) Then add these up. Substitutingthe resulting values in the strict version of the inequality above, and rearranging,yields

78 < log(n!) − (

n + 12

)log(n) + n < 1,

as desired. Then exponentiate.

Problem 8.23: Let γ : [a, b] → C \ {0} be a continuously differentiable closedcurve. Define the index of γ to be

Ind(γ) =1

2πi

∫ b

a

γ′(t)γ(t)

dt.

Prove that Ind(γ) ∈ Z.Compute Ind(γ) for γ(t) = exp(int) and [a, b] = [0, 2π].Explain why Ind(γ) is often called the winding number of γ about 0.Hint for the first part: Find ϕ : [a, b] → C such that

ϕ′ =γ′

γand ϕ(a) = 0.

Show that ϕ(b) = 2πiInd(γ). Show that γ exp(−ϕ) is constant, so that γ(a) = γ(b)implies exp(ϕ(b)) = exp(ϕ(a)) = 1.

Solution (Sketch): The function ϕ of the hint exists by the Fundamental Theoremof Calculus (Theorem 6.20 of Rudin; not Theorem 6.21 of Rudin). That ϕ(b) =2πiInd(γ) is clear from the construction of ϕ and the definitions. That γ exp(−ϕ)is constant follows by differentiating it and using the formula for ϕ′ to simplify thederivative to zero. That exp(ϕ(b)) = exp(ϕ(a)) = 1 is now clear. So ϕ(b) ∈ 2πiZ,whence Ind(γ) ∈ Z.

The second statement is a simple computation; the answer is n.Why Ind(γ) is often called the winding number: In the case considered in the

second statement, it counts the number of times the curve goes around 0 in thepositive sense. (This is true in general. See the comment in the solution to Prob-lem 8.24.)

Problem 8.24: Let γ and Ind(γ) be as in Problem 8.23. Suppose the range of γdoes not intersect the negative real axis. Prove that Ind(γ) = 0.

Hint: The function c 7→ Ind(γ + c), defined on [0,∞), is a continuous integervalued function such that limc→∞ Ind(γ + c) = 0.

Solution (Sketch): One checks that

c 7→ (γ + c)′

γ + c=

γ′

γ + c

is a continuous function from [0,∞) to C([a, b]), for example, by checking thatcn → c implies

(γ + cn)′

γ + cn→ (γ + c)′

γ + c

uniformly. Theorem 7.16 of Rudin and the sequential criterion for continuity thenimply that c 7→ Ind(γ + c) is continuous. Furthermore,

(γ + c)′

γ + c→ 0

uniformly as c → ∞. Therefore limc→∞ Ind(γ + c) = 0. Since Ind(γ + c) ∈ Z for allc and since [0,∞) is connected, we must have Ind(γ + c) = 0 for all c, in particularfor c = 0.

Alternate solution (Sketch): We can give a direct proof of continuity, which is inprinciple nicer than what was done above.

The first step is to show that

r = inft∈[a,b], c∈[0,I)

|γ(t) + c| > 0.

Since [a, b] is compact, M = supt∈[a,b] |γ(t)| < ∞. Therefore c ≥ 2M implies

inft∈[a,b]

|γ(t) + c| ≥ M.

Define f : [a, b] × [0, 2M ] → C by f(t, c) = |γ(t) + c|. Then f is continuous andnever vanishes on [a, b] × [0, 2M ], so compactness of [a, b] × [0, 2M ] implies that

inft∈[a,b], c∈[0, 2M ]

|γ(t) + c| > 0.

So

inft∈[a,b], c∈[0,I)

|γ(t) + c| ≥ min(

M, inft∈[a,b], c∈[0, 2M ]

|γ(t) + c|)

> 0.

Now let ε > 0. Set

δ =2πr2ε

1 + (b − a) sups∈[a,b] |γ′(s)| > 0.

Note that sups∈[a,b] |γ′(s)| is finite, because γ′ is continuous and [a, b] is compact.Let c, d ∈ [0,∞) satisfy |c − d| < δ. Then t ∈ [a, b] implies∣∣∣∣ γ′(t)

γ(t) + c− γ′(t)

γ(t) + c

∣∣∣∣ =∣∣∣∣ γ′(t)(d − c)(γ(t) + c)(γ(t) + d)

∣∣∣∣≤ δ sups∈[a,b] |γ′(s)|

r2<

2πε

b − a.

Therefore

|Ind(γ + d) − Ind(γ + c)| ≤ 12πi

∫ b

a

∣∣∣∣ γ′(t)γ(t) + c

− γ′(t)γ(t) + c

∣∣∣∣ dt < ε.

We give a solution which uses a different method to get a lower bound on |γ(t)+c|and also makes explicit the use of the fact that the composite of two continuousfunctions is continuous. It is possible to combine parts of this solution with partsof the previous solution to obtain two further arrangements of the proof.

Second alternate solution (Sketch): Define F : [0,∞) → C([a, b]) by

F (c)(t) =γ′(t)

γ(t) + c,

that is, F (c) is the function

t 7→ γ′(t)γ(t) + c

which is in C([a, b]). Further define I : C([a, b]) → C by

I(f) =1

2πi

∫ b

a

f.

Then Ind(γ + c) = I ◦F (c). We prove that c 7→ Ind(γ + c) is continuous by provingthat I and F are continuous.

We prove that F is continuous. Let c0 ∈ [0,∞) and let ε > 0. Set M =sups∈[a,b] |γ′(s)|, which is finite because γ′ is continuous and [a, b] is compact. Setr = infs∈[a,b] |γ(s) + c0|, which is strictly positive because [a, b] is compact, γ iscontinuous, and γ(s) + c0 6= 0 for all s ∈ [a, b]. Choose

δ = min(

r

2,

r2ε

2(1 + M)

)> 0.

Suppose c ∈ [0,∞) satisfies |c − c0| < δ. Since |γ(s) + c0| > r for all s ∈ [a, b],and since |c − c0| < 1

2r, it follows that |γ(s) + c0| > 12r for all s ∈ [a, b]. Therefore

1[γ(s) + c0][γ(s) + c]

<2r2

for all s ∈ [a, b]. So

‖F (c) − F (c0)‖∞ = sups∈[a,b]

∣∣∣∣ γ′(s)γ(s) + c

− γ′(s)γ(s) + c0

∣∣∣∣ = sups∈[a,b]

∣∣∣∣ γ′(s)(c0 − c)[γ(s) + c0][γ(s) + c]

∣∣∣∣≤

(sup

s∈[a,b]

|γ′(s)|)|c − c0|

(sup

s∈[a,b]

1[γ(s) + c0][γ(s) + c]

)

≤ M |c − c0| · 2r2

< ε.

This proves continuity of F at c0.Now we prove continuity of I. Let ε > 0. Choose

δ =2πε

b − a> 0.

Let f, g ∈ C([a, b]) satisfy ‖f − g‖∞ < δ. Then (using Theorem 6.25 of Rudin atthe first step)

|I(f) − I(g)| ≤ 12π

∫ b

a

|f − g| ≤ ‖f − g‖∞(b − a)2π

<δ(b − a)

2π= ε.

So I is continuous.

Problem 8.25: Let γ1 and γ2 be closed curves as in Problem 8.23. Suppose that

|γ1(t) − γ2(t)| < |γ1(t)|for t ∈ [a, b]. Prove that Ind(γ1) = Ind(γ2).

Hint: Put γ(t) = γ1(t)/γ2(t). Then |1 − γ(t)| < 1 for all t, so Problem 8.23implies that Ind(γ) = 0. Also,

γ′(t)γ(t)

=γ′2(t)

γ2(t)− γ′

1(t)γ1(t)

for all t ∈ [a, b].

Comment: In fact, Ind(γ) can be defined for arbitrary continuous closed curves inC \ {0}, and Ind(γ) is a homotopy invariant of the curve. (Problem 8.26 containsenough to prove this.) Since Ind(γ) ∈ Z, it follows that two homotopic curves inC\{0} have the same index. Therefore the map [γ] 7→ Ind(γ) defines a function fromπ1(C \ {0}) to Z. It is easy to check that this map is a surjective homomorphism.(Injectivity is harder, I think, unless of course you already know the fundamentalgroup.)

Solution (Sketch): The first part of the hint is proved by writing

|1 − γ(t)| =∣∣∣∣γ2(t)γ2(t)

− γ1(t)γ2(t)

∣∣∣∣ .

It follows that the range of γ does not intersect the negative real axis. So Ind(γ) = 0by Problem 8.23. The equation (gotten from the quotient rule)

γ′(t)γ(t)

=γ′2(t)

γ2(t)− γ′

1(t)γ1(t)

shows that Ind(γ) = Ind(γ2) − Ind(γ1).

Problem 8.12: Let δ ∈ (0, π), and let fδ : R → C be the 2π-periodic functiongiven on [−π, π] by

fδ(x) ={

1 |x| ≤ δ0 δ < |x| ≤ π

.

(a) Compute the Fourier coefficients of fδ.

Solution (Sketch): This is a computation, and gives cn = 1πn sin(nδ) for n 6= 0, and

c0 = 1π δ.

(b) Conclude from Part (a) that∞∑

n=1

sin(nδ)n

=π − δ

2.

Solution (Sketch): Apply Theorem 8.14 of Rudin to the function fδ at x = 0 to get

1 =δ

π+ 2

∞∑n=1

sin(nδ)πn

,

and rearrange.

(c) Deduce from Parseval’s Theorem that∞∑

n=1

[sin(nδ)]2

n2δ=

π − δ

2.

Solution (Sketch): Applying Parseval’s Theorem to the result of Part (a) gives(δ

π

)2

+ 2∞∑

n=1

(sin(nδ)

πn

)2

=(

12π

) ∫ π

−π

|fδ|2 =δ

π.

Now multiply by π2δ−1 and rearrange.

(d) Let δ → 0 and prove that∫ ∞

0

(sin(x)

x

)2

dx =π

2.

Date: 15 March 2002.

1

Solution (Sketch): Essentially, we want to interpret the sum in Part (c) as a Rie-mann sum for the improper integral. To to this, we must effectively interchangelimit operations.

Let ε > 0. Choose an integer M such that M > 4ε−1. Note that

0 ≤∫ ∞

M

(sin(x)

x

)2

dx ≤∫ ∞

M

(1x

)2

dx =1M

< 14ε

and, for any integer n > 0,

0 ≤ 1n

∞∑k=nM+1

(sin

(kn

)(

kn

))2

≤ 1n

∞∑k=nM+1

(1(kn

))2

≤∫ ∞

M

(1x

)2

dx =1M

< 14ε.

Using uniform continuity and the Riemann sum interpretation, choose an integerN so large that if n ≥ N then∣∣∣∣∣∣

1n

nM∑k=1

(sin

(kn

)(

kn

))2

−∫ M

0

(sin(x)

x

)2

dx

∣∣∣∣∣∣ < 14ε.

Also require that 12N < 1

4ε. Using the triangle inequality several times, this gives,when n ≥ N , ∣∣∣∣∣∣

1n

∞∑k=1

(sin

(kn

)(

kn

))2

−∫ ∞

0

(sin(x)

x

)2

dx

∣∣∣∣∣∣ < 34ε.

Part (c) (with δ = 1n ) therefore gives, when n ≥ N ,∣∣∣∣∣

π

2−

∫ ∞

0

(sin(x)

x

)2

dx

∣∣∣∣∣ < ε.

(e) Put δ = 12π in Part (c). What do you get?

Solution (Sketch): Only the terms with odd n appear. Therefore we get∞∑

n=0

1(2n + 1)2

=(π

2

) (π

4

)=

π2

8.

Problem 8.13: Let f : R → C be the 2π-periodic function given on [0, 2π) byf(x) = x. Apply Parseval’s Theorem to f to conclude that

∞∑n=1

1n2

=π2

6.

Solution (Sketch): A computation (integration by parts) shows that f has theFourier coefficients cn = i

n for n 6= 0. Also, c0 = π. So Parseval’s Theorem gives

π2 + 2∞∑

n=1

1n2

=12π

∫ 2π

0

x2 dx = 43π2.

Now solve for∑∞

n=11

n2 .

Problem 8.14: Define f : [−π, π] → R by f(x) = (π − |x|)2. Prove that

f(x) =π2

3+

∞∑n=1

4n2

cos(nx)

for all x, and deduce that∞∑

n=1

1n2

=π2

6and

∞∑n=1

1n4

=π4

90.

Solution (Sketch): We take f to be the 2π periodic function defined on all of R byf(x) = (π − |x − 2πn|)2 for n ∈ Z and x ∈ [(2n − 1)π, (2n + 1)π]. This formulagives two different definitions at each point (2n + 1)π with n ∈ Z, but both agree;it is then easy to check that f is continuous, and in particular Riemann integrableover any interval of length 2π.

Next, we find the Fourier coefficients cn. For this, observe that f(x) = (π − x)2

for all x ∈ [0, 2π]. For n 6= 0, set

gn(x) =i

ne−inx(π − x)2 − 2

n2e−inx(π − x) − 2i

n3e−inx

for x ∈ R. Then a calculation shows that g′n(x) = e−inx(π − x)2 for all n and x.(The formula can be found by integrating by parts twice. However, that isn’t partof the proof of the problem.) So the Fundamental Theorem of Calculus gives

cn =12π

∫ π

−π

e−inxf(x) dx =12π

∫ 2π

0

e−inx(π − x)2 dx =12π

[gn(2π) − gn(0)] =2n2

.

(When calculating gn(2π) − gn(0), most of the terms cancel out.) Similarly,

c0 =12π

∫ π

−π

f(x) dx =12π

∫ 2π

0

(π − x)2 dx =π2

3.

It follows that the partial sum sn(f ;x) (in the notation of Section 8.13 of Rudin’sbook) is given by

sn(f ;x) =n∑

k=−n

ckeikx =π2

3+

n∑k=1

2k2

[eikx + e−ikx] =π2

3+

n∑k=1

4k2

cos(kx).

We now verify the hypotheses of Theorem 8.14 of Rudin’s book, for every x ∈ R.For x 6∈ 2πZ, this is easy from the differentiability of f at x. (For x ∈ (2n + 1)πZ,use f(x) = [(2n+1)π−x]2 for x ∈ [2πn, 2π(n+2)].) For x = 0 we estimate directly.If |t| < 2π, then

f(t) − f(0) = (π − |t|)2 − π2 = |t|(|t| − 2π)

and −2π ≤ |t| − 2π ≤ 0, so |f(t)− f(0)| ≤ 2π|t|. This is the required estimate. Forother values of x ∈ 2πZ, the required condition follows from periodicity.

It now follows from Theorem 8.14 of Rudin’s book that limn→∞ sn(f ;x) = f(x)for all x ∈ R. That is,

f(x) =π2

3+

∞∑n=1

4n2

cos(nx)

for all x ∈ R.


Putting x = 0 gives

π2 =π2

3+

∞∑n=1

4n2

,

from which it follows that∞∑

n=1

1n2

=π2

6.

To get the formula for∑∞

n=11

n4 , compute ‖f‖22 two ways: directly and using

Parseval’s Theorem. Then compare the results.

Problem 8.15: With

DN (x) =N∑

n=−N

einx =sin((N + 1

2 )x)sin(1

2x)

as in Section 8.13 of Rudin’s book, define

KN (x) =1

N + 1

N∑n=0

Dn(x).

Prove that

KN (x) =(

1N + 1

)(1 − cos((N + 1)x)

1 − cos(x)

)

for x ∈ R \ 2πZ.

Solution (Sketch): This may not be the best way, but it will work. Write

KN (x) =(

1N + 1

)(1

sin(12x)

) N∑n=0

12i

(exp(i(n + 1

2 )x) − exp(−i(n + 12 )x)

)

and use the formula for the sum of a geometric series. Then do a little rearranging.

a. Prove that KN (x) ≥ 0.

Solution (Sketch): In the formula above, both the numerator and denominator arenonnegative.

b. Prove that12π

∫ π

−π

KN (x) dx = 1.

Solution (Sketch): This is immediate from the relations

KN (x) =1

N + 1

N∑n=0

Dn(x) and12π

∫ π

−π

Dn(x) dx = 1.

c. If δ > 0, then

KN (x) ≤(

1N + 1

)(2

1 − cos(δ)

)

for δ ≤ |x| ≤ π.

Solution (Sketch): Since KN is an even function, it suffices to prove this for δ ≤x ≤ π. Use

1 − cos(x) ≥ 1 − cos(δ) and 1 − cos((N + 1)x) ≤ 2.

Now let sn(f ;x) be as in Section 8.13 of Rudin’s book, and define

σN (f ;x) =1

N + 1

N∑n=0

sn(f ;x).

Prove that

σN (f ;x) =12π

∫ π

−π

f(x − t)KN (t) dt,

and use this to prove Fejer’s Theorem: If f : R → C is continuous and 2π periodic,then σN (f ;x) converges uniformly to f(x) on [−π, π].

Hint: Use (a), (b), and (c) to proceed as in the proof of Theorem 7.26 of Rudin’sbook.

Solution (Sketch): The formula

σN (f ;x) =12π

∫ π

−π

f(x − t)KN (t) dt

follows from the definitions of σN (f ;x) and KN (x) and the formula

sn(f ;x) =12π

∫ π

−π

f(x − t)Dn(t) dt.

This last formula was proved in Section 8.13 of Rudin’s book.Now assume f is not the zero function. (The result is trivial in this case, and

I want to divide by ‖f‖∞.) Let ε > 0. Since f is continuous and periodic, it isuniformly continuous. (Check this!) So there is δ0 > 0 such that whenever |t| < δ0

then |f(x − t) − f(x)| < 12ε. Set δ = 1

2δ0. Choose N so large that(2π‖f‖∞n + 1

) (2

1 − cos(δ)

)< ε.

For any n ≥ N and x ∈ R, write

|σn(f ;x) − f(x)| =∣∣∣∣ 12π

∫ π

−π

f(x − t)Kn(t) dt − 12π

∫ π

−π

f(x)Kn(t) dt

∣∣∣∣=

12π

∫ π

−π

|f(x − t) − f(t)| dt

=12π

(∫ −δ

−π

|f(x − t) − f(t)|Kn(t) dt +∫ δ

−δ

|f(x − t) − f(t)|Kn(t) dt

+∫ π

δ

|f(x − t) − f(t)|Kn(t) dt

)

≤ 12π

(2‖f‖∞

∫ −δ

−π

Kn(t) dt +ε

2

∫ δ

−δ

Kn(t) dt + 2‖f‖∞∫ π

δ

Kn(t) dt

).

The estimates at the last step are obtained as follows. For the first and third terms,

|f(x − t) − f(t)| ≤ |f(x − t)| + |f(t)| ≤ 2‖f‖∞.

For the middle term, |f(x − t) − f(x)| < 12ε because δ < δ0.

Now ∫ −δ

−π

Kn(t) dt ≤ (π − δ)(

1n + 1

) (2

1 − cos(δ)

)<

ε

2‖f‖∞ .

Similarly ∫ π

δ

Kn(t) dt <ε

2‖f‖∞ .

Also, ∫ δ

−δ

Kn(t) dt ≤ 2π.

Inserting these estimates, we get

|σn(f ;x) − f(x)| ≤ 12π

(2‖f‖∞

∫ −δ

−π

|Kn(t) dt +ε

2

∫ δ

−δ

Kn(t) dt + 2‖f‖∞∫ π

δ

Kn(t) dt

)

≤ 12π

(ε + 2π · 1

2ε + ε)

=(

2 + π

2π

)ε < ε.

This proves uniform convergence.

Problem B:(1) Let X be a complete metric space, let x0 ∈ X, let r > 0, and let C < 1.

Let f : Nr(x0) → X be a function such that d(f(x), f(y)) ≤ Cd(x, y) for all x, y ∈Nr(x0) and d(f(x0), x0) < (1−C)r. Prove that f has a unique fixed point z, thatis, there is a unique z ∈ Nr(x0) such that f(z) = z. Further prove that

d(z, x0) ≤ df(x0), x0)1 − C

.

Solution (sketch): This is essentially the same as Problem A(2). Uniqueness followseasily from the condition C < 1.

We prove by induction on n the combined statement:(1) fn(x0) is defined.(2) d(fn(x0), fn−1(x0)) ≤ Cn−1d(f(x0), x0)).

(3) d(fn(x0), x0) ≤ 1 − Cn−1

1 − Cd(f(x0), x0).

For n = 1, this is immediate. Suppose it is true for n. Condition (3) for n showsthat

d(fn(x0), x0) <

(1 − Cn

1 − C

)(1 − C)r = (1 − Cn−1)r < r,

so fn(x0) ∈ Nr(x0) and fn+1(x0) = f(fn(x0)) is defined. This is Condition (1) forn + 1. Then Condition (2) for n and the hypotheses imply

d(fn+1(x0), fn(x0)) ≤ Cd(fn(x0), fn−1(x0))

≤ C · Cn−1d(f(x0), x0)) = Cnd(f(x0), x0)),

which is Condition (2) for n + 1. Finally

d(fn+1(x0), x0) ≤ d(fn+1(x0), fn(x0)) + d(fn(x0), x0)

≤ Cnd(f(x0), x0)) +1 − Cn−1

1 − Cd(f(x0), x0))

=1 − Cn

1 − Cd(f(x0), x0)).

This is Condition (3) for n + 1.It follows as in previous similar problems that the sequence (fn(x0)) is Cauchy,

and hence has a limit z. Moreover,

d(z, x0) ≤ supn∈N

d(fn(x0), x0) ≤ supn∈N

1 − Cn

1 − Cd(f(x0), x0))

≤(

11 − C

)d(f(x0), x0)) < r,

so z ∈ Nr(x0) and f(z) is defined. It now follows from continuity, as in previoussimilar problems, that f(z) = z.

(2) Let I, J ⊂ R be open intervals, let x0 ∈ I, and let y0 ∈ J . Let ϕ : I×J → Rbe a continuous function and assume that there is a constant M such that.

|ϕ(x, y2) − ϕ(x, y1)| ≤ M |y2 − y1|for all x ∈ I and y1, y2 ∈ J . Prove that there is δ > 0 such that there is a uniquefunction f : (x0 − δ, x0 + δ) → J satisfying f(x0) = y0 and f ′(x) = ϕ(x, f(x)) forall x ∈ (x0 − δ, x0 + δ).

In this problem, you may assume the standard properties of∫ b

af when a ≤ b,

including the correct version of the Fundamental Theorem of Calculus.Hint: For a suitable δ > 0 and a suitable subset N ⊂ CR([x0 − δ, x0 + δ]), define

a function F : N → CR([x0 − δ, x0 + δ]) by

F (g)(x) = y0 +∫ x

x0

ϕ(t, g(t)) dt.

If δ and N are chosen correctly, the first part will imply that F has a unique fixedpoint f . Prove that this fixed point solves the differential equation.

Solution (Sketch): Choose r > 0 such that [y0 − r, y0 + r] ⊂ J . Choose δ0 > 0 suchthat [x0 − δ0, x0 + δ0] ⊂ I. Set

K = supt∈[x0−δ0, x0+δ0]

|ϕ(t, y0)|.

Let δ > 0 be any number satisfying

δ ≤ min(

12M

,r

3K

).

In the metric space CR([x0 − δ, x0 + δ]), let g0 be the constant function g0(x) = y0

for all x ∈ [x0 − δ, x0 + δ], and set N = Nr(g0). Following the hint, define F : N →CR([x0 − δ, x0 + δ]) by

F (g)(x) = y0 +∫ x

x0

ϕ(t, g(t)) dt.

One needs to check that F (g)(x) is defined, that is, that (t, g(t)) is in the domainof ϕ and that t 7→ ϕ(t, g(t)) is Riemann integrable. Both are easy; in fact, t 7→ϕ(t, g(t)) is continuous. One also needs to observe that F (g) is in fact a continuousfunction.

We now verify the hypotheses of Part (1) for this function. Let g1, g2 ∈ N . Forx ∈ [x0 − δ, x0 + δ], we then have

|F (g1)(x) − F (g2)(x)| =∣∣∣∣∫ x

x0

[ϕ(t, g1(t)) − ϕ(t, g2(t))] dt

∣∣∣∣≤ |x − x0| sup

t∈[x0−δ, x0+δ]

|ϕ(t, g1(t)) − ϕ(t, g2(t))|

≤ δM supt∈[x0−δ, x0+δ]

|g1(t) − g2(t)| = δM‖g1 − g2‖∞.

Therefore

‖F (g1) − F (g2)‖∞ ≤ δM‖g1 − g2‖∞ ≤ 12‖g1 − g2‖∞.

Thus, F is a contraction with constant C = 12 . Moreover, for x ∈ [x0 − δ, x0 + δ]

we have

|F (g0)(x) − g0(x)| =∣∣∣∣∫ x

x0

ϕ(t, y0) dt

∣∣∣∣≤ |x − x0|

(sup

t∈[x0−δ, x0+δ]

|ϕ(t, y0)|)

≤ δK ≤ 13r < (1 − C)r.

We can now apply Part (1) to conclude that there is a unique g ∈ CR([x0 −δ, x0 + δ]) such that F (g) = g, that is,

g(x) = y0 +∫ x

x0

ϕ(t, g(t)) dt

for all x ∈ [x0 − δ, x0 + δ]. Since t 7→ ϕ(t, g(t)) is continuous, the FundamentalTheorem of Calculus shows that the right hand side of this equation is differentiableas a function of x, with derivative x 7→ ϕ(x, g(x)). Thus g′(x) = ϕ(x, g(x)) for allx ∈ [x0 − δ, x0 + δ]. That g(x0) = y0 is immediate.

We have proved existence of a solution on (x0 − δ, x0 + δ) with

δ = min(

12M

,r

3K

).

Since g′(x) = ϕ(x, g(x)) and g(x0) = y0 imply

g(x) = y0 +∫ x

x0

ϕ(t, g(t)) dt,

we have also proved uniqueness among all continuous functions g : (x0−δ, x0+δ) →R satisfying in addition |g(x) − y0| < r for all x.

We now show that this implies uniqueness among all continuous functions g : (x0−δ, x0 + δ) → J . Suppose that, with δ as above, there is some solution h : (x0 −δ, x0 + δ) → J , with |h(x) − y0| ≥ r for some x. Set

ρ = inf({x ∈ (x0 − δ, x0 + δ) : |h(x) − y0| ≥ r}).

SOLUTIONS TO SELECTED PROBLEMS FROM

RUDIN

DAVID SEAL

Abstract. I became bored sometime this August 2006, and de-cided to review some of my analysis, by reading my old analysisbook[1] from Junior year. While I’m at it, I decided to type upsome solutions to a few problems that I scratched out solutionsto on notepaper in order to pick up some LATEXpractice along theway. This paper has solutions to some of the problems I was ableto solve, indeed many of the problems in this book were too chal-lenging to solve in a weekend. All of these problems were selectedfrom Principles of Mathematical Analysis[1] by Walter Rudin.

Contents

1. The Real and Complex Number System 12. Basic Topology 13. Numerical Sequences and Series 34. Continuity 85. Differentiation 10References 11

Ok, now the real work. But, first a definition. A metric space iscalled separable if it contains a countable dense subset. Problem #21on page 45 asks you to prove that Rk is separable. The second definitionon this page is given in the following problem.

1. The Real and Complex Number System

OK fine, I didn’t write up any of these. I just wanted the numbersto line up.

2. Basic Topology

Problem 1 (pg. 45, #23). A collection {Vα} of open subsets of X issaid to be a base for X if the following is true: For every x ∈ X andevery open set G ⊂ X such that x ∈ G, we have x ∈ Vα ⊂ G for some

1

2 DAVID SEAL

α. In other words, every open set in X is the union of a subcollectionof {Vα}.

Prove that every separable metric space has a countable base. Hint:Take all neighborhoods with rational radius and center in some count-able dense subset of X.

The hint tells us exactly what to do. The set

(1) V :={

Br(x) : r ∈ Q+, x ∈ C}

where C is the countable dense subset gives the desired result.

Solution. Set V as in (1). This set is countable since both Q and Care countable. Suppose G ⊂ X is open. Let x ∈ G and find an ǫ > 0such that B2ǫ(x) ⊂ G. Since C is dense, there exists a p ∈ C, whered(p, x) < ǫ. In addition, the density of the rationals on the real lineprovides an r ∈ Q where ǫ > r > d(p, x) > 0. Therefore, x ∈ Br(p).Furthermore, if y ∈ Br(p), we have that

d(y, x) ≤ d(y, p) + d(p, x) < 2r < 2ǫ,

so that Br(p) ⊂ G, the desired result. �

This problem set continues in the next one. Rudin[1] sure does liketo give important definitions and theorems in the problem sets!

Problem 2 (pg. 45, #24). Let X be a metric space in which everyinfinite subset has a limit point. Prove that X is separable. (A hint isgiven in the book. I won’t type it here, but I will follow it).

The best course of action is to follow the hint.

Solution. Fix 1/n > 0. Start with any x1 ∈ X. Having chosenx1, x2, . . . , xj ∈ X, choose xj+1 ∈ X, if possible, so that d(xi, xj+1) ≥1/n for i = 1, . . . , j. Assume this course of action is infinite, and we’llshow a contradiction from this assumption. The set defined as

X1/n := {xk}∞k=1

is infinite because each xk is distinct; it therefore has a limit point dueto the hypothesis. Let’s say xN ∈ {xk}∞k=1 is the limit point. However,xN cannot be a limit point, for

B 1

2n(x) ∩

(

{xk}∞k=1 \ {xN})

= ∅.because this set has the property that each xi ∈ X1/n is at least 1/naway from every other point in the set. Therefore, the process ofchoosing the xi’s is a finite one for any positive 1/n; we also havethe nice property that

⋃

k B 1

n(xk) ⊃ X for every positive 1/n, (oth-

erwise we would have been able to choose an additional xi). The set

SOLUTIONS TO SELECTED PROBLEMS FROM RUDIN 3

X∗ := ∪∞n=1X 1

nyields a countable dense subset. Therefore, X is sep-

arable (see pbm. #22) which shows that it has a countable base (seepbm #23). �

Gee, that was fun. I see it takes a lot longer to type these problemsup in LATEXrather than by hand, but I think it’ll be a useful tool forme to learn anyway. OK, on to the next problem.

3. Numerical Sequences and Series

Problem 3 (pg.78, #7). Prove that the convergence of∑

an impliesthe convergence of

∑

√an

n,

if an > 0.

I’ll define a sequence

bn := max{

1/n2, an

}

,

and claim that this sequence converges.

Proof of claim: Let ǫ > 0. From the Cauchy criterion of convergence,(both

∑

1n2 and

∑

an converge absolutely), there exists an N ∈ N suchthat

N+j∑

k=N

1

n2< ǫ/2 and

N+j∑

k=N

an < ǫ/2

for any j ∈ N. From the definition of bn, the following inequalities arefairly imminent. We will replace any missing terms with the remainingterms in the convergent sequences:

N+j∑

k=N

bk ≤N+j∑

k=N

1/n2 +

N+j∑

k=N

ak < ǫ/2 + ǫ/2 = ǫ.

By the cauchy criterion for convergence,∑

bn must converge. �

With that bit out of the way, we can provide the solution.

Solution. Starting with bn ≥ 1/n2 and bn ≥ an, we have that b2n ≥ an

n2

after multiplying these inequalities together. After taking square roots,

we have that bn ≥√

an

nand the comparison test implies the convergence

of∑

√an

n. �

4 DAVID SEAL

GWEN, HERE IS MY PROOF:

Theorem 1. If 0 < a ≤ b for some a, b ∈ R, then√

a ≤√

b.

Proof. Suppose not. That is, suppose there exist an 0 < a 

√b.

If we multiply the previous inequality together by itself, we obtain

√a√

a >√

b√

b,

In other words,

a > b.

This is a contradiction. �

Problem 4 (pg. 79, #12). Suppose an > 0 and∑

an converges. Put

rn :=∞

∑

m=n

am.

(1) Prove thatam

rm+ · · ·+ an

rn> 1 − rn

rm

if m < n, and deduce that∑

an

rndiverges.

(2) Prove thatan√rn

< 2 (√

rn −√rn+1)

and deduce that∑

an√rn

converges.

Solution (Part (1)). Rather than follow the books use of an m < n,I will choose to fix an m, j ∈ N, and prove this with a m < m + j.I believe this is much more explicit. Hence, the equation we seek toprove can be reformulated as

(2)am

rm+ · · ·+ am+j

rm+j> 1 − rm+j

rm.

Since an > 0 for every n, we must have that rn is a decreasing sequence:rm > rm+i for any i ∈ N. Hence, for any i ∈ N,

(3)rm

rm+i> 1.

Applying equation (3) a few times, in addition to the fact am+j > 0,we have that

rm − rm+j = am + · · ·+ am+j−1

< am + · · ·+ am+j

< am + am+1

(

rm

rm+1

)

+ · · ·+ am+j

(

rm

rm+j

)

.

Whence,

1 − rm+j

rm<

am

rm+

am+1

rm+1+ · · · + am+j

rm+j;

which is the desired result for half the problem.In order to show

∑

an

rndiverges, it is enough show that it fails the

cauchy criterion for convergence. Fix ǫ = 1/2 and let N ∈ N. Thereexists a j ∈ N such that rN+j ≤ rN

2, whence 1 − rN+j

rN≥ 1/2. If we

apply the inequality just shown, we have

N+j∑

k=N

ak

rk> 1 − rN+j

rN≥ 1/2.

This shows that∑

an

rnwill never pass the cauchy criterion for conver-

gence, which means it must be divergent. �

Solution (Part (2)). We will start by proving the inequality asked for.The following inequalities each proceed from the previous one:

0 < (√

rn −√rn+1)

2;

0 < rn − 2√

rn+1rn + rn+1 ;

−rn+1 < rn − 2√

rn+1rn ;

rn − rn+1 < 2 (rn −√rn+1rn) ;

an < 2 (rn −√rn+1rn) .

The final inequality is a restatement of the previous one since an =rn − rn+1. The desired result comes after dividing by

√rn;

(4)an√rn

< 2 (√

rn −√rn+1) .

6 DAVID SEAL

To prove the second part for this question; namely the convergence of∑

an√rn

, we might ask: what if we add up a bunch of these an√rn

’s?

an√rn

+an+1√rn+1

+ · · · + an+j√rn+j

< 2(

(√

rn −√rn+1) + · · ·+ (

√rn+j −

√rn+j+1)

)

= 2(√

rn −√rn+j+1

)

< 2√

rn.

The sequence sn :=∑n

k=1ak√rk

is bounded above by√

r1, and monoton-

ically increasing. Therefore the sum,∑∞

n=1an√rn

converges. �

The next problem is a multi-series problem.

Problem 5 (pg. 82, #24). Let X be a metric space.

(1) Call two Cauchy sequences {pn}, {qn} in X equivalent if

limn→∞

d(pn, qn) = 0.

Prove this is an equivalence relation.(2) Let X∗ be the set of all equivalence classes so obtained. If P ∈

X∗, Q ∈ X∗, {pn} ∈ P , {qn} ∈ Q, define

∆(P, Q) := limn→∞

d(pn, qn);

by excercise 23, this limit exists. Show that the number ∆(P, Q)is unchanged if {pn} and {qn} are replaced by equivalent se-quences, and hence that ∆ is a distance function in X∗.

(3) Prove that the resulting metric space X∗ is complete.**I FIND THIS QUESTION DIFFICULT, AND HAVEN’T

SOLVED IT YET; BUT I HAVE WRITTEN SOME IDEASDOWN**

(4) for each p ∈ X, there is a Cauchy sequence all of whose termsare p; let Pp be the element of X∗ which contains this sequence.Prove that

∆(Pp, Qq) = d(p, q)

for all p, q ∈ X. In other words, the mapping ϕ : X → X∗

defined by p 7→ Pp is an isometry.

Solution (Part (1)). There are three items we need for an equivalencerelation.

(1) {pn} ∼ {pn} since

limn→∞

d(pn, pn) = limn→∞

0 = 0.

(2) Suppose {pn} ∼ {qn}. Then {qn} ∼ {pn} because d(pn, qn) =d(qn, pn), so the limit still goes to 0.

(3) Suppose {pn} ∼ {qn} and {qn} ∼ {xn} for some {xn}. Thend(pn, xn) ≤ d(pn, qn) + d(qn, xn). After taking limits, the righthand side goes to zero, so we have that {pn} ∼ {xn}. �

Solution (Part (2)). Suppose both {pn}, {pn′} ∈ P and that {qn},

{qn′} ∈ Q. Set the following values:

a := limn→∞

d(pn, qn);

a′ := limn→∞

d(p′n, q′n).

These values exist (from problem #23), and we desire to show thata = a′ in order for ∆(P, Q) to be well defined.

Two applications of the triangle inequality give

0 ≤ d(pn, qn) ≤ d(pn, p′n) + d(p′n, q′n) + d(q′n, qn).

Sending n → ∞, we obtain

(5) 0 ≤ a ≤ a′.

Applying the triangle inequality again:

0 ≤ d(p′n, q′n) ≤ d(p′n, pn) + d(pn, qn) + d(qn, q

′n);

so sending n → ∞, we obtain

(6) 0 ≤ a′ ≤ a.

Together, (5) and (6) show a = a′. �

Solution (Part (3)). In order to show that X∗ is complete, we mustshow that given a cauchy sequence {Pn} ∈ X∗, (under the ∆ metric),there is a set P ∈ X∗ such that limn→∞ Pn = P .

Suppose {Pn} ∈ X∗ is cauchy. For every ǫ > 0 there exists a corre-sponding N ∈ N such that for every j ∈ N, and every n ≥ N

∆(Pn, Pn+j) < ǫ.

Refering to the definition for ∆, if {p(n)k } ∈ Pn and {p(n+j)

k } ∈ Pn+j,

limk→∞

d(

p(n)k , p

(n+j)k

)

= ∆(Pn, Pn+j) < ǫ.

Let’s start by defining P to be the set which would have the desiredproperty for which Pn 7→ P . Define:

P :={

{pk}∞k=1 : ∀ǫ > 0, ∃N ∈ N, ∀n ≥ N, ∃{q(n)k } ∈ Pn,(7)

limk→∞

d(

pk, q(n)k

)

< ǫ}

.

8 DAVID SEAL

The first task is to show this set is non-empty. To start, there existsan N1 < N2 such that every n1 ≥ N1, n2 ≥ N2,

∆(Pn1, Pn1+j) < 2−1;

∆(Pn2, Pn2+j) < 2−2.(8)

In order to fill in the whole sequence, take any p1, p2, . . . , pN1−1 ∈ X.

According to (8), for N1 ≤ j1 ≤ N2 there are sequences {p(j1)k } such

thatlimk→∞

d(p(j1)k , p

(N1)k ) < 2−1.

For these values, N1 ≤ k ≤ N2, set pk = p(N1)k . We proceed by in-

duction. Let i ∈ N, and suppose pk has been chosen for each k = 1,2, . . . , Ni. There exists an Ni+1, Ni < Ni+1 such that

(9) ∆(PNi, Pki

) < 2−(i+1) forNi ≤ ki;

THIS ISN’T QUITE RIGHT. I’M INDEXING ON THE WRONGVARIABLE... �

The final part (4) will not be proven. We’re taking the limit of aconstant, so ∆(Pp, Qq) = d(p, q) is obvious. This is pretty cool nonethe-less! OK - these problems are taking SUPER long to type up. MaybeI should start defining keyboard macros? Ok, I have a few defined, buta lot of the keys are already being used...

4. Continuity

Problem 6 (pg. 101 #20). If E is a nonempty subset of a metricspace X, define the distance from x ∈ X to E by

pE(x) := infd

(x, z).

(1) Prove that pE(x) = 0 iff x ∈ E.(2) Prove that pE is a uniformly continuous function on X, by

showing that

|pE(x) − pE(y)| ≤ d(x, y)

for all x, y ∈ X.

The book gives a hint. We’ll solve this problem by following it.

Solution (Part (1)). First, a proof of ⇐.

Proof (⇐). Suppose x ∈ E. For all r > 0, there exists a

y ∈ (Br(x) \ {x}) ∩ E 6= ∅.If pE(x) > 0, there exists a y, d(x, y) < pE(x)/2, so pE(x) is not alower bound of {d(x, z) : z ∈ E}. �


Now, a proof of (⇒).

Proof (⇒). If x /∈ E, then there exists an r > 0, Br(x) ⊂ Ec. (SinceEc is open). This implies that

pE(x) = infz∈E

≥ r > 0.

Therefore, by the contrapositive, pE(x) = 0 implies x ∈ E. �

Solution (Part (2)). Let x, y ∈ X, and take any ǫ > 0. There exists az ∈ E such that

(10) d(y, z) ≤ pE(y) + ǫ.

(Use the greatest lower bound property). Since z ∈ E, we have, afterone application of the triangle inequality, as well as an application ofequation (10):

pE(x) ≤ d(x, z) ≤ d(x, y) + d(y, z)

≤ d(x, y) + pE(y) + ǫ.

Sending ǫ ↓ 0, we obtain that

pE(x) − pE(y) ≤ d(x, y).

Since d(x, y) = d(y, x), we can exchange the roles of x, y in all theprevious lines in order to obtain the desired result. Namely, that

|pE(x) − pE(y)| ≤ d(x, y).

�

Problem 7 (pg. 101, #21). Suppose K and F are disjoint sets ina metric space X, K is compact, F is closed. Prove that there existsδ > 0 such that d(p, q) > δ if p ∈ K, q ∈ F .

Show that the conclusion may fail for two disjoint closed sets.

Solution. By problem #20, we know pF : K → R is uniformly contin-uous. Since K compact, we may apply the Max/Min theorem. Thereexists an x0 ∈ K, pF (x0) ≥ pF (x) for every x ∈ K. Whence, if we setδ = pF (x0)/2 we have for x ∈ K, y ∈ F ,

d(x, y) ≥ pF (x) ≥ pF (x0) > δ.

�

The conclusion fails if we consider subsets of R2:

K := {(x, 0) : x ∈ R} ;

F :={

(x, 1/(1 + x2) : x ∈ R}

;

both of which are closed sets, and fail to meet the criterion displayedin the problem.

10 DAVID SEAL

5. Differentiation

Problem 8 (pg. 114, #4). If

C0 +C1

2+ · · ·+ Cn−1

n+

Cn

n + 1,

where C0, C1, . . . , Cn are real constants, prove that the equation

C0 + C1x + · · · + Cn−1xn−1 + Cnx

n = 0


Solution. Consider the function, defined by

F (x) :=

∫ x

0

f(t) dt,

where f(x) := C0 + C1x + · · · + Cn−1xn−1 + Cnx

n. Apply the meanvalue theorem to F on the interval [0, 1].

Problem 9. Suppose f ′ is continuous on [a, b] and ǫ > 0. Prove thatthere exists a δ > 0 such that

∣

∣

∣

∣

f(t) − f(x)

t − x− f ′(x)

∣

∣

∣

∣

< ǫ

whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b.

Solution. Let ǫ > 0. As f ′ is a continuous function on a compact set,f ′ is uniformly continuous. There exists a δ > 0, for every x, y ∈ [a, b],|f ′(x) − f ′(y)| < ǫ if |x − y| < δ. Using this δ, apply the mean valuetheorem to the interval [min{t, x}, max{t, x}]. �


References

[1] Rudin, Walter “Principles of Mathematical Analysis,” McGraw Hill, 1990.

Plover’s Solution SeriesVolume 1:

Rudin, Principles of Mathematical Analysis

SYLeeMeng-Gen Tsai

i

Preface

This solution is written for one special book named Principles of Mathemat-ical Analysis. It is written by Rudin. The purpose I wrote this solution isjust for fun; surely, you will find a lot of errors containing in my solution :ppFinally, I dedicate this solution to SYLee, wwli, ljl, and many other friends.

Meng-Gen, Tsai.

Contents

Preface i

1 The Real and Complex Number Systems 3

2 Basic Topology 9

3 Numerical Sequences and Series 31

4 Continuity 35

5 Differentiation 41

6 Sequences and Series of Functions 67

1

2 CONTENTS

Chapter 1

The Real and ComplexNumber Systems

3

4 CHAPTER 1. THE REAL AND COMPLEX NUMBER SYSTEMS

Written by Men-Gen Tsaiemail: [email protected]

1.

2.

3.

4.

5.

6. Fix b > 1.(a) If m,n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, provethat

(bm)1/n = (bp)1/q.



(c) If x is real, define B(x) to be the set of all numbers bt, where t isrational and t ≤ x. Prove that

br = sup B(r)

where r is rational. Hence it makes sense to define

bx = sup B(x)

for every real x.

(d) Prove that bx+y = bxby for all real x and y.

Proof: For (a): mq = np since m/n = p/q. Thus bmq = bnp.By Theorem 1.21 we know that (bmq)1/(mn) = (bnp)1/(mn), that is,(bm)1/n = (bp)1/q, that is, br is well-defined.

For (b): Let r = m/n and s = p/q where m, n, p, q are integers, andn > 0, q > 0. Hence (br+s)nq = (bm/n+p/q)nq = (b(mq+np)/(nq))nq =bmq+np = bmqbnp = (bm/n)nq(bp/q)nq = (bm/nbp/q)nq. By Theorem 1.21

5

we know that ((br+s)nq)1/(nq) = ((bm/nbp/q)nq)1/(nq), that is br+s =bm/nbp/q = brbs.

For (c): Note that br ∈ B(r). For all bt ∈ B(r) where t is rational andt ≤ r. Hence, br = btbr−t ≥ bt1r−t since b > 1 and r − t ≥ 0. Hence br

is an upper bound of B(r). Hence br = sup B(r).

For (d): bxby = sup B(x) sup B(y) ≥ btxbty = btx+ty for all rationaltx ≤ x and ty ≤ y. Note that tx + ty ≤ x + y and tx + ty is rational.Therefore, sup B(x) sup B(y) is a upper bound of B(x + y), that is,bxby ≥ sup B(x + y) = b(x + y).

Conversely, we claim that bxbr = bx+r if x ∈ R1 and r ∈ Q. Thefollowing is my proof.

bx+r = sup B(x + r) = sup{bs : s ≤ x + r, s ∈ Q}= sup{bs−rbr : s− r ≤ x, s− r ∈ Q}= br sup{bs−r : s− r ≤ x, s− r ∈ Q}= br sup B(x)

= brbx.

And we also claim that bx+y ≥ bx if y ≥ 0. The following is my proof:(r ∈ Q)

B(x) = {br : r ≤ x} ⊂ {br : r ≤ x + y} = B(x + y),

Therefore, sup B(x + y) ≥ sup B(x), that is, bx+y ≥ bx.

Hence,

bx+y = sup B(x + y)

= sup{br : r ≤ x + y, r ∈ Q}= sup{bsbr−s : r ≤ x + y, s ≤ x, r ∈ Q, s ∈ Q}≥ sup{sup B(x)br−s : r ≤ x + y, s ≤ x, r ∈ Q, s ∈ Q}= sup B(x) sup{br−s : r ≤ x + y, s ≤ x, r ∈ Q, s ∈ Q}= sup B(x) sup{br−s : r − s ≤ x + y − s, s ≤ x, r − s ∈ Q}


= sup B(x) sup B(x + y − s)

≥ sup B(x) sup B(y)

= bxby

Therefore, bx+y = bxby.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

7

26.

27.

28.

29.

30.

Chapter 2

Basic Topology

9

10 CHAPTER 2. BASIC TOPOLOGY


1. Prove that the empty set is a subset of every set.

Proof: For any element x of the empty set, x is also an element ofevery set since x does not exist. Hence, the empty set is a subset ofevery set.

2. A complex number z is said to be algebraic if there are integers a0, ..., an,not all zero, such that

a0zn + a1z

n−1 + ... + an−1z + an = 0.

Prove that the set of all algebraic numbers is countable. Hint: Forevery positive integer N there are only finitely many equations with

n + |a0|+ |a1|+ ... + |an| = N.

Proof: For every positive integer N there are only finitely many equa-tions with

n + |a0|+ |a1|+ ... + |an| = N.

(since 1 ≤ n ≤ N and 0 ≤ |a0| ≤ N). We collect those equations asCN . Hence

⋃CN is countable. For each algebraic number, we can form

an equation and this equation lies in CM for some M and thus the setof all algebraic numbers is countable.

3. Prove that there exist real numbers which are not algebraic.

Proof: If not, R1 = { all algebraic numbers } is countable, a contra-diction.

4. Is the set of all irrational real numbers countable?

Solution: If R−Q is countable, then R1 = (R−Q)⋃

Q is countable,a contradiction. Thus R−Q is uncountable.

11

5. Construct a bounded set of real numbers with exactly three limit points.

Solution: Put

A = {1/n : n ∈ N}⋃{1 + 1/n : n ∈ N}

⋃{2 + 1/n : n ∈ N}.

A is bounded by 3, and A contains three limit points - 0, 1, 2.

6. Let E ′ be the set of all limit points of a set E. Prove that S ′ isclosed. Prove that E and E have the same limit points. (Recall thatE = E

⋃E ′.) Do E and E ′ always have the same limit points?

Proof: For any point p of X − E ′, that is, p is not a limit point E,there exists a neighborhood of p such that q is not in E with q 6= p forevery q in that neighborhood.

Hence, p is an interior point of X − E ′, that is, X − E ′ is open, thatis, E ′ is closed.

Next, if p is a limit point of E, then p is also a limit point of E sinceE = E

⋃E ′. If p is a limit point of E, then every neighborhood Nr(p)

of p contains a point q 6= p such that q ∈ E. If q ∈ E, we completedthe proof. So we suppose that q ∈ E − E = E ′ − E. Then q is a limitpoint of E. Hence,

Nr′(q)

where r′ = 12min(r−d(p, q), d(p, q)) is a neighborhood of q and contains

a point x 6= q such that x ∈ E. Note that Nr′(q) contains in Nr(p)−{p}.That is, x 6= p and x is in Nr(p). Hence, q also a limit point of E. Hence,E and E have the same limit points.

Last, the answer of the final sub-problem is no. Put

E = {1/n : n ∈ N},

and E ′ = {0} and (E ′)′ = φ.

7. Let A1, A2, A3, ... be subsets of a metric space. (a) If Bn =⋃n

i=1 Ai,prove that Bn =

⋃ni=1 Ai, for n = 1, 2, 3, ... (b) If B =

⋃∞i=1, prove that

B ⊃ ⋃∞i=1 Ai. Show, by an example, that this inclusion can be proper.

Proof of (a): (Method 1) Bn is the smallest closed subset of X thatcontains Bn. Note that

⋃Ai is a closed subset of X that contains Bn,

thus

Bn ⊃n⋃

i=1

Ai.


If p ∈ Bn − Bn, then every neighborhood of p contained a point q 6= psuch that q ∈ Bn. If p is not in Ai for all i, then there exists someneighborhood Nri

(p) of p such that (Nri(p)−p)

⋂Ai = φ for all i. Take

r = min{r1, r2, ..., rn}, and we have Nr(p)⋂

Bn = φ, a contradiction.Thus p ∈ Ai for some i. Hence

Bn ⊂n⋃

i=1

Ai.

that is,

Bn =n⋃

i=1

Ai.

(Method 2) Since⋃n

i=1 Ai is closed and Bn =⋃n

i=1 Ai ⊂⋃n

i=1 Ai,Bn ⊂

⋃ni=1 Ai.

Proof of (b): Since B is closed and B ⊃ B ⊃ Ai, B ⊃ Ai for all i.Hence B ⊃ ⋃

Ai.

Note: My example is Ai = (1/i,∞) for all i. Thus, Ai = [1/i,∞), andB = (0,∞), B = [0,∞). Note that 0 is not in Ai for all i. Thus thisinclusion can be proper.

8. Is every point of every open set E ⊂ R2 a limit point of E? Answerthe same question for closed sets in R2.

Solution: For the first part of this problem, the answer is yes.

(Reason): For every point p of E, p is an interior point of E. Thatis, there is a neighborhood Nr(p) of p such that Nr(p) is a subset ofE. Then for every real r′, we can choose a point q such that d(p, q) =1/2 min(r, r′). Note that q 6= p, q ∈ Nr′(p), and q ∈ Nr(p). Hence, everyneighborhood Nr′(p) contains a point q 6= p such that q ∈ Nr(p) ⊂ E,that is, p is a limit points of E.

For the last part of this problem, the answer is no. Consider A ={(0, 0)}. A′ = φ and thus (0, 0) is not a limit point of E.

9. Let Eo denote the set of all interior points of a set E.

(a) Prove that Eo is always open.(b) Prove that E is open if and only if Eo = E.

13

(c) If G is contained in E and G is open, prove that G is contained inEo.(d) Prove that the complement of Eo is the closure of the complementof E.(e) Do E and E always have the same interiors?(f) Do E and Eo always have the same closures?

Proof of (a): If E is non-empty, take p ∈ Eo. We need to show thatp ∈ (Eo)o. Since p ∈ Eo, there is a neighborhood Nr of p such thatNr is contained in E. For each q ∈ Nr, note that Ns(q) is contained inNr(p), where s = min{d(p, q), r − d(p, q)}. Hence q is also an interiorpoint of E, that is, Nr is contained in Eo. Hence Eo is always open.

Proof of (b): (⇒) It is clear that Eo is contained in E. Since E isopen, every point of E is an interior point of E, that is, E is containedin Eo. Therefore Eo = E.

(⇐) Since every point of E is an interior point of E (Eo(E) = E), Eis open.

Proof of (c): If p ∈ G, p is an interior point of G since G is open. Notethat E contains G, and thus p is also an interior point of E. Hencep ∈ Eo. Therefore G is contained in Eo. (Thus Eo is the biggest openset contained in E. Similarly, E is the smallest closed set containingE.)

Proof of (d): Suppose p ∈ X − Eo. If p ∈ X − E, then p ∈ X − Eclearly. If p ∈ E, then N is not contained in E for any neighborhoodN of p. Thus N contains an point q ∈ X − E. Note that q 6= p, thatis, p is a limit point of X − E. Hence X − Eo is contained in X − E.

Next, suppose p ∈ X − E. If p ∈ X − E, then p ∈ X − Eo clearly. Ifp ∈ E, then every neighborhood of p contains a point q 6= p such thatq ∈ X − E. Hence p is not an interior point of E. Hence X − E iscontained in X − Eo. Therefore X − Eo = X − E.

Solution of (e): No. Take X = R1 and E = Q. Thus Eo = φ andE

o= (R1)o = R1 6= φ.

Solution of (f): No. Take X = R1 and E = Q. Thus E = R1, andEo = φ = φ.


10. Let X be an infinite set. For p ∈ X and q ∈ X, define

d(p, q) =

{1 (if p 6= q)0 (if p = q)

Prove that this is a metric. Which subsets of the resulting metric spaceare open? Which are closed? Which are compact?

Proof: (a) d(p, q) = 1 > 0 if p 6= q; d(p, p) = 0. (b) d(p, q) = d(q, p)since p = q implies q = p and p 6= q implies q 6= p. (c) d(p, q) ≤d(p, r) + d(r, q) for any r ∈ X if p = q. If p 6= q, then either r = p orr = q, that is, r 6= p or r 6= q. Thus, d(p, q) = 1 ≤ d(p, r) + d(r, q). By(a)-(c) we know that d is a metric.

Every subset of X is open and closed. We claim that for any p ∈ X,p is not a limit point. Since d(p, q) = 1 > 1/2 if q 6= p, there exists anneighborhood N1/2(p) of p contains no points of q 6= p such that q ∈ X.Hence every subset of X contains no limit points and thus it is closed.Since X − S is closed for every subset S of X, S = X − (X − S) isopen. Hence every subset of X is open.

Every finite subset of X is compact. Let S = {p1, ..., pn} be finite.Consider an open cover {Gα} of S. Since S is covered by Gα, pi iscovered by Gαi

, thus {Gα1 , ..., Gαn} is finite subcover of S. Hence S iscompact. Next, suppose S is infinite. Consider an open cover {Gp} ofS, where

Gp = N 12(p)

for every p ∈ S. Note that q is not in Gp if q 6= p. If S is compact,then S can be covered by finite subcover, say

Gp1 , ..., Gpn .

Then there exists q such that q 6= pi for all i since S is infinite, a con-tradiction. Hence only every finite subset of X is compact.

11. For x ∈ R1 and y ∈ R1, define

d1(x, y) = (x, y)2,

d2(x, y) =√|x− y|,

15

d3(x, y) = |x2 − y2|,d4(x, y) = |x− 2y|,

d5(x, y) =|x− y|

1 + |x− y|.


Solution: (1) d1(x, y) is not a metric. Since d1(0, 2) = 4, d1(0, 1) = 1,and d1(1, 2) = 1, d1(0, 2) > d1(0, 1) + d1(1, 2). Thus d1(x, y) is not ametric.(2) d2(x, y) is a metric. (a) d(p, q) > 0 if p 6= q; d(p, p) = 0. (b)

d(p, q) =√|p− q| =

√|q − p| = d(q, p). (c) |p − q| ≤ |p − r| + |r − q|,√

|p− q| ≤√|p− r|+ |r − q| ≤

√|p− r|+

√|r − q|. That is, d(p, q) ≤

d(p, r) + d(r, q). (3) d3(x, y) is not a metric since d3(1,−1) = 0.(4) d4(x, y) is not a metric since d4(1, 1) = 1 6= 0.(5) d5(x, y) is a metric since |x− y| is a metric.

Claim: d(x, y) is a metric, then

d′(x, y) =d(x, y)

1 + d(x, y)

is also a metric.

Proof of Claim: (a) d′(p, q) > 0 if p 6= q; d(p, p) = 0. (b) d′(p, q) =d′(q, p). (c) Let x = d(p, q), y = d(p, r), and z = d(r, q). Then x ≤ y+z.

d′(p, q) ≤ d′(p, r) + d′(r, q)

⇔ x

1 + x≤ y

1 + y+

z

1 + z

⇔ x(1 + y)(1 + z) ≤ y(1 + z)(1 + x) + z(1 + x)(1 + y)

⇔ x + xy + xz + xyz ≤ (y + xy + yz + xyz) + (z + xz + yz + xyz)

⇔ x ≤ y + z + 2yz + xyz

⇐ x ≤ y + z

Thus, d′ is also a metric.

12. Let K ⊂ R1 consist of 0 and the numbers 1/n, for n = 1, 2, 3, ....Prove that K is compact directly from the definition (without usingthe Heine-Borel theorem).

Proof: Suppose that {Oα} is an arbitrary open covering of K. LetE ∈ {Oα} consists 0. Since E is open and 0 ∈ E, 0 is an interior pointof E. Thus there is a neighborhood N = Nr(0) of 0 such that N ⊂ E.Thus N contains

1

[1/r] + 1,

1

[1/r] + 2, ...

Next, we take finitely many open sets En ∈ {Oα} such that 1/n ∈ En

for n = 1, 2, ..., [1/r]. Hence {E, E1, ..., E[1/r] is a finite subcover of K.Therefore, K is compact.

Note: The unique limit point of K is 0. Suppose p 6= 0 is a limit pointof K. Clearly, 0 < p < 1. (p cannot be 1). Thus there exists n ∈ Z+

such that1

n + 1< p <

1

n.

Hence Nr(p) where r = min{ 1n− p, p − 1

n+1} contains no points of K,

a contradiction.

13. Construct a compact set of real numbers whose limit points form acountable set.

Solution: Let K be consist of 0 and the numbers 1/n for n = 1, 2, 3, ...Let xK = {xk : k ∈ K} and x + K = {x + k : k ∈ K} for x ∈ R1. Itake

Sn = (1− 1

2n) +

K

2n+1,

S =∞⋃

n=1

Sn

⋃{1}.

Claim: S is compact and the set of all limit points of S is K⋃{1}.

Clearly, S lies in [0, 1], that is, S is bounded in R1. Note that Sn ⊂[1 − 1

2n , 1 − 12n+1 ]. By Exercise 12 and its note, I have that all limit

points of S⋂

[0, 1) is

0,1

2, ...,

1

2n, ...

17

Clearly, 1 is also a limit point of S. Therefore, the set of all limit pointsof S is K

⋃{1}. Note that K⋃{1} ⊂ S, that is, K is compact. I com-

pleted the proof of my claim.

14. Give an example of an open cover of the segment (0, 1) which has nofinite subcover.

Solution: Take {On} = {(1/n, 1)} for n = 1, 2, 3, .... The following ismy proof. For every x ∈ (0, 1),

x ∈ (1

[1/x] + 1, 0) ∈ {On}

Hence {On} is an open covering of (0, 1). Suppose there exists a finitesubcovering

(1

n1

, 1), ..., (1

nk

, 1)

where n1 < n2 < ... < nk, respectively. Clearly 12np

∈ (0, 1) is not in

any elements of that subcover, a contradiction.

Note: By the above we know that (0, 1) is not compact.

15. Show that Theorem 2.36 and its Corollary become false (in R1, for ex-ample) if the word ”compact” is replaced by ”closed” or by ”bounded.”

Theorem 2.36: If {Kα} is a collection of compact subsets of a metricspace X such that the intersection of every finite subcollection of {Kα}is nonempty, then

⋂Kα is nonempty.

Corollary: If {Kn} is a sequence of nonempty compact sets such thatKn contains Kn+1 (n = 1, 2, 3, ...), then

⋂Kn is not empty.

Solution: For closed: [n,∞). For bounded: (−1/n, 1/n)− {0}.

16. Regard Q, the set of all rational numbers, as a metric space, withd(p, q) = |p − q|. Let E be the set of all p ∈ Q such that 2 < p2 < 3.Show that E is closed and bounded in Q, but that E is not compact.Is E open in Q?


Proof: Let S = (√

2,√

3)⋃

(−√

3,−√

2). Then E = {p ∈ Q : p ∈ S}.Clearly, E is bounded in Q. Since Q is dense in R, every limit point ofQ is in Q. (I regard Q as a metric space). Hence, E is closed in Q.

To prove that E is not compact, we form a open covering of E asfollows:

{Gα} = {Nr(p) : p ∈ E and (p− r, p + r) ⊂ S}

Surely, {Gα} is a open covering of E. If E is compact, then there arefinitely many indices α1, ..., αn such that

E ⊂ Gα1

⋃...

⋃Gαn .

For every Gαi= Nri

(pi), take p = max1≤i≤n pi. Thus, p is the nearestpoint to

√3. But Nr(p) lies in E, thus [p + r,

√3) cannot be covered

since Q is dense in R, a contradiction. Hence E is not compact.

Finally, the answer is yes. Take any p ∈ Q, then there exists aneighborhood N(p) of p contained in E. (Take r small enough whereNr(p) = N(p), and Q is dense in R.) Thus every point in N(p) is alsoin Q. Hence E is also open.

17. Let E be the set of all x ∈ [0, 1] whose decimal expansion contains onlythe digits 4 and 7. Is E countable? Is E dense in [0, 1]? Is E compact?Is E perfect?

Solution:

E ={ ∞∑

n=1

an

10n: an = 4 or an = 7

}Claim: E is uncountable.Proof of Claim: If not, we list E as follows:

x1 = 0.a11a12...a1n...

x2 = 0.a21a22...a2n...

... ...

xk = 0.ak1ak2...akn...

... ...

19

(Prevent ending with all digits 9) Let x = 0.x1x2...xn... where

xn =

{4 if ann = 77 if ann = 4

By my construction, x /∈ E, a contradiction. Thus E is uncountable.

Claim: E is not dense in [0, 1].Proof of Claim: Note that E

⋂(0.47, 0.74) = φ. Hence E is not dense

in [0, 1].

Claim: E is compact.Proof of Claim: Clearly, E is bounded. For every limit point p of E,I show that p ∈ E. If not, write the decimal expansion of p as follows

p = 0.p1p2...pn...

Since p /∈ E, there exists the smallest k such that pk 6= 4 and pk 6= 7.When pk = 0, 1, 2, 3, select the smallest l such that pl = 7 if possible.(If l does not exist, then p < 0.4. Thus there is a neighborhood of psuch that contains no points of E, a contradiction.) Thus

0.p1...pl−14pl+1...pk−17 < p < 0.p1...pk−14.

Thus there is a neighborhood of p such that contains no points of E, acontradiction.

When pk = 5, 6,

0.p1...pk−147 < p < 0.p1...pk−174.

Thus there is a neighborhood of p such that contains no points of E, acontradiction.

When pk = 8, 9, it is similar. Hence E is closed. Therefore E iscompact.

Claim: E is perfect.Proof of Claim: Take any p ∈ E, and I claim that p is a limit pointof E. Write p = 0.p1p2...pn... Let

xk = 0.y1y2...yn...

where

yn =

pk if k 6= n4 if pn = 77 if pn = 4

Thus, |xk−p| → 0 as k →∞. Also, xk 6= p for all k. Hence p is a limitpoint of E. Therefore E is perfect.

18. Is there a nonempty perfect set in R1 which contains no rational num-ber?

Solution: Yes. The following claim will show the reason.

Claim: Given a measure zero set S, we have a perfect set P containsno elements in S.

Proof of Claim: (due to SYLee). Since S has measure zero, thereexists a collection of open intervals {In} such that

S ⊂⋃

In and∑|In| < 1.

Consider E = R1 − ⋃In. E is nonempty since E has positive mea-

sure. Thus E is uncountable and E is closed. Therefore there existsa nonempty perfect set P contained in E by Exercise 28. P

⋂S = φ.

Thus P is our required perfect set.

19. (a) If A and B are disjoint closed sets in some metric space X, provethat they are separated.

(b) Prove the same for disjoint open sets.

(c) Fix p ∈ X, δ > 0, define A to be the set of all q ∈ X for whichd(p, q) < δ, define B similarly, with > in place of <. Prove that A andB are separated.

(d) Prove that every connected metric space with at least two pointsis uncountable. Hint: Use (c).

Proof of (a): Recall the definition of separated: A and B are sep-arated if A

⋂B and A

⋂B are empty. Since A and B are closed sets,

A = A and B = B. Hence A⋂

B = A⋂

B = A⋂

B = φ. Hence A andB are separated.

21

Proof of (b): Suppose A⋂

B is not empty. Thus there exists p suchthat p ∈ A and p ∈ B. For p ∈ A, there exists a neighborhood Nr(p) ofp contained in A since A is open. For p ∈ B = B

⋃B′, if p ∈ B, then

p ∈ A⋂

B. Note that A and B are disjoint, and it’s a contradiction.If p ∈ B′, then p is a limit point of B. Thus every neighborhood of pcontains a point q 6= p such that q ∈ B. Take an neighborhood Nr(p)ofp containing a point q 6= p such that q ∈ B. Note that Nr(p) ⊂ A,thus q ∈ A. With A and B are disjoint, we get a contradiction. HenceA

⋂(B) is empty.

Similarly, A⋂

B is also empty. Thus A and B are separated.

Proof of (c): Suppose A⋂

B is not empty. Thus there exists x suchthat x ∈ A and x ∈ B. Since x ∈ A, d(p, x) < δ. x ∈ B = B

⋃B′, thus

if x ∈ B, then d(p, x) > δ, a contradiction. The only possible is x is alimit point of B. Hence we take a neighborhood Nr(x) of x contains y

with y ∈ B where r = δ−d(x,p)2

. Clearly, d(y, p) > δ. But,

d(y, p) ≤ d(y, x) + d(x, p)

< r + d(x, p)

=δ − d(x, p)

2+ d(x, p)

=δ + d(x, p)

2

<δ + δ

2= δ.

A contradiction. Hence A⋂

B is empty. Similarly, A⋂

B is also empty.Thus A and B are separated.

Note: Take care of δ > 0. Think a while and you can prove the nextsub-exercise.

Proof of (d): Let X be a connected metric space. Take p ∈ X, q ∈ Xwith p 6= q, thus d(p, q) > 0 is fixed. Let

A = {x ∈ X : d(x, p) < δ}; B = {x ∈ X : d(x, p) > δ}.

Take δ = δt = td(p, q) where t ∈ (0, 1). Thus 0 < δ < d(p, q). p ∈ Asince d(p, p) = 0 < δ, and q ∈ B since d(p, q) > δ. Thus A and B arenon-empty.

By (c), A and B are separated. If X = A⋃

B, then X is not connected,a contradiction. Thus there exists yt ∈ X such that y /∈ A

⋃B. Let

E = Et = {x ∈ X : d(x, p) = δt} 3 yt.

For any real t ∈ (0, 1), Et is non-empty. Next, Et and Es are disjointif t 6= s (since a metric is well-defined). Thus X contains a uncount-able set {yt : t ∈ (0, 1)} since (0, 1) is uncountable. Therefore, X isuncountable.

Note: It is a good exercise. If that metric space contains only onepoint, then it must be separated.

Similar Exercise Given by SYLee: (a) Let A = {x : d(p, x) < r}and B = {x : d(p, x) > r} for some p in a metric space X. Show thatA, B are separated.(b) Show that a connected metric space with at least two points mustbe uncountable. [Hint: Use (a)]

Proof of (a): By definition of separated sets, we want to show A⋂

B =φ, and B

⋂A = φ. In order to do these, it is sufficient to show A

⋂B =

φ. Let x ∈ A⋂

B = φ, then we have:

(1) x ∈ A ⇒ d(x, p) ≤ r(2) x ∈ B ⇒ d(x, p) > r

It is impossible. So, A⋂

B = φ.

Proof of (b): Suppose that C is countable, say C = a, b, x3, .... Wewant to show C is disconnected. So, if C is a connected metric spacewith at least two points, it must be uncountable. Consider the setS = {d(a, xi) : xi ∈ C}, and thus let r ∈ R− S and inf S < r < sup S.And construct A and B as in (a), we have C = A

⋃B, where A and B

are separated. That is C is disconnected.

Another Proof of (b): Let a ∈ C, b ∈ C, consider the continuousfunction f from C into R defined by f(x) = d(x, a). So, f(C) is con-nected and f(a) = 0, f(b) > 0. That is, f(C) is an interval. Therefore,C is uncountable.

20. Are closures and interiors of connected sets always connected? (Lookat subsets of R2.)

23

Solution: Closures of connected sets is always connected, but interiorsof those is not. The counterexample is

S = N1(2, 0)⋃

N1(−2, 0)⋃{x− axis} ⊂ R2.

Since S is path-connected, S is connect. But So = N1(2)⋃

N1(−2) isdisconnected clearly.

Claim: If S is a connected subset of a metric space, then S is con-nected.

Pf of Claim: If not, then S is a union of two nonempty separated setA and B. Thus A

⋂B = A

⋂B = φ. Note that

S = S − T

= A⋃

B − T

= (A⋃

B)⋂

T c

= (A⋂

T c)⋃

(B⋂

T c)

where T = S − S. Thus

(A⋂

T c)⋂

B⋂

T c ⊂ (A⋂

T c)⋂

B⋂

T c

⊂ A⋂

B

= φ.

Hence (A⋂

T c)⋂

B⋂

T c = φ. Similarly, A⋂

T c⋂

(B⋂

T c) = φ.

Now we claim that both A⋂

T c and B⋂

T c are nonempty. Supposethat B

⋂T c = φ. Thus

A⋂

T c = S ⇔ A⋂

(S − S)c = S

⇔ A⋂

(A⋃

B − S)c = S

⇔ A⋂

((A⋃

B)⋂

Sc)c = S

⇔ A⋂

((Ac⋂

Bc)⋃

S) = S

⇔ (A⋂

S)⋃

(A⋂

Ac⋂

Bc) = S

⇔ A⋂

S = S.

Thus B is empty, a contradiction. Thus B⋂

T c is nonempty. Similarly,A

⋂T c nonempty. Therefore S is a union of two nonempty separated

sets, a contradiction. Hence S is connected.

21. Let A and B be separated subsets of some Rk, suppose a ∈ A, b ∈ B,and define

p(t) = (1− t)a + tb

for t ∈ R1. Put A0 = p−1(A), B0 = p−1(B). [Thus t ∈ A0 if and onlyif p(t) ∈ A.]

(a) Prove that A0 and B0 are separated subsets of R1.(b) Prove that there exists t0 ∈ (0, 1) such that p(t0) /∈ A

⋃B.

(c) Prove that every convex subset of Rk is connected.

Proof of (a): I claim that A0⋂

B0 is empty. (B0⋂

A0 is similar). Ifnot, take x ∈ A0

⋂B0. x ∈ A0 and x ∈ B0. x ∈ B0 or x is a limit

point of B0. x ∈ B0 will make x ∈ A0⋂

B0, that is, p(x) ∈ A⋂

B, acontradiction since A and B are separated.

Claim: x is a limit point of B0 ⇒ p(x) is a limit point of B. Take anyneighborhood Nr of p(x), and p(t) lies in B for small enough t. Moreprecisely,

x− r

|b− a|< t < x +

r

|b− a|.

Since x is a limit point of B0, and (x − r/|b − a|, x + r/|b − a|) is aneighborhood N of x, thus N contains a point y 6= x such that y ∈ B0,that is, p(y) ∈ B. Also, p(y) ∈ Nr. Therefore, p(x) is a limit point ofB. Hence p(x) ∈ A

⋂B, a contradiction since A and B are separated.

Hence A0⋂

B0 is empty, that is, A0 and B0 are separated subsets ofR1.

Proof of (b): Suppose not. For every t0 ∈ (0, 1), neither p(t0) ∈ Anor p(t0) ∈ B (since A and B are separated). Also, p(t0) ∈ A

⋃B for

all t0 ∈ (0, 1). Hence (0, 1) = A0⋃

B0, a contradiction since (0, 1) isconnected. I completed the proof.

Proof of (c): Let S be a convex subset of Rk. If S is not connected,then S is a union of two nonempty separated sets A and B. By (b),there exists t0 ∈ (0, 1) such that p(t0) /∈ A

⋃B. But S is convex, p(t0)

25

must lie in A⋃

B, a contradiction. Hence S is connected.

22. A metric space is called separable if it contains a countable dense sub-set. Show that Rk is separable. Hint: Consider the set of points whichhave only rational coordinates.

Proof: Consider S = the set of points which have only rational coor-dinates. For any point x = (x1, x2, ..., xk) ∈ Rk, we can find a rationalsequence {rij} → xj for j = 1, ..., k since Q is dense in R1. Thus,

ri = (ri1 , ri2 , ..., rik) → x

and ri ∈ S for all i. Hence S is dense in Rk. Also, S is countable, thatis, S is a countable dense subset in Rk, Rk is separable.

23. A collection {Vα} of open subsets of X is said to be a base for X if thefollowing is true: For every x ∈ X and every open set G ⊂ X such thatx ∈ G, we have x ∈ Vα ⊂ G for some α. In other words, every open setin X is the union of a subcollection of {Vα}.Prove that every separable metric space has a countable base. Hint:Take all neighborhoods with rational radius and center in some count-able dense subset of X.

Proof: Let X be a separable metric space, and S be a countable densesubset of X. Let a collection {Vα} = { all neighborhoods with rationalradius and center in S }. We claim that {Vα} is a base for X.

For every x ∈ X and every open set G ⊂ X such that x ∈ G, thereexists a neighborhood Nr(p) of p such that Nr(p) ⊂ G since x is aninterior point of G. Since S is dense in X, there exists {sn} → x. Takea rational number rn such that rn < r

2, and {Vα} 3 Nrn(sn) ⊂ Nr(p)

for enough large n. Hence we have x ∈ Vα ⊂ G for some α. Hence{Vα} is a base for X.

24. Let X be a metric space in which every infinite subset has a limitpoint. Prove that X is separable. Hint: Fix δ > 0, and pick x1 ∈X. Having chosen x1, ..., xj ∈ X, choose xj+1, if possible, so thatd(xi, xj+1) ≥ δ for i = 1, ..., j. Show that this process must stop after

finite number of steps, and that X can therefore be covered by finitemany neighborhoods of radius δ. Take δ = 1/n(n = 1, 2, 3, ...), andconsider the centers of the corresponding neighborhoods.

Proof: Fix δ > 0, and pick x1 ∈ X. Having chosen x1, ..., xj ∈ X,choose xj+1, if possible, so that d(xi, xj+1) ≥ δ for i = 1, ..., j. Ifthis process cannot stop, then consider the set A = {x1, x2, ..., xk}. Ifp is a limit point of A, then a neighborhood Nδ/3(p) of p contains apoint q 6= p such that q ∈ A. q = xk for only one k ∈ N . If not,d(xi, xj) ≤ d(xi, p) + d(xj, p) ≤ δ/3 + δ/3 < δ, and it contradicts thefact that d(xi, xj) ≥ δ for i 6= j. Hence, this process must stop afterfinite number of steps.

Suppose this process stop after k steps, and X is covered by Nδ(x1),Nδ(x2), ..., Nδ(xk), that is, X can therefore be covered by finite manyneighborhoods of radius δ.

Take δ = 1/n(n = 1, 2, 3, ...), and consider the set A of the centers ofthe corresponding neighborhoods.

Fix p ∈ X. Suppose that p is not in A, and every neighborhoodNr(p). Note that Nr/2(p) can be covered by finite many neighborhoodsNs(x1), ..., Ns(xk) of radius s = 1/n where n = [2/r] + 1 and xi ∈ Afor i = 1, ..., k. Hence, d(x1, p) ≤ d(x1, q) + d(q, p) ≤ r/2 + s < r whereq ∈ Nr/2(p)

⋂Ns(x1). Therefore, x1 ∈ Nr(p) and x1 6= p since p is not

in A. Hence, p is a limit point of A if p is not in A, that is, A is acountable dense subset, that is, X is separable.

25. Prove that every compact metric space K has a countable base, andthat K is therefore separable. Hint: For every positive integer n, thereare finitely many neighborhood of radius 1/n whose union covers K.

Proof: For every positive integer n, there are finitely many neighbor-hood of radius 1/n whose union covers K (since K is compact). Collectall of them, say {Vα}, and it forms a countable collection. We claim{Vα} is a base.

For every x ∈ X and every open set G ⊂ X, there exists Nr(x) such thatNr(x) ⊂ G since x is an interior point of G. Hence x ∈ Nm(p) ∈ {Vα}for some p where m = [2/r] + 1. For every y ∈ Nm(p), we have

d(y, x) ≤ d(y, p) + d(p, x) < m + m = 2m < r.

27

Hence Nm(p) ⊂ G, that is, Vα ⊂ G for some α, and therefore {Vα} is acountable base of K. Next, collect all of the center of Vα, say D, andwe claim D is dense in K (D is countable since Vα is countable). For allp ∈ K and any ε > 0 we can find Nn(xn) ∈ {Vα} where n = [1/ε] + 1.Note that xn ∈ D for all n and d(p, xn) → 0 as n → ∞. Hence D isdense in K.

26. Let X be a metric space in which every infinite subsets has a limitpoint. Prove that X is compact. Hint: By Exercises 23 and 24, X hasa countable base. It follows that every open cover of X has a countablesubcover {Gn}, n = 1, 2, 3, .... If no finite subcollection of {Gn} coversX, then the complement Fn of G1

⋃...

⋃Gn is nonempty for each n,

but⋂

Fn is empty. If E is a set contains a point from each Fn, considera limit point of E, and obtain a contradiction.

Proof: By Exercises 23 and 24, X has a countable base. It follows thatevery open cover of X has a countable subcover {Gn}, n = 1, 2, 3, ....If no finite subcollection of {Gn} covers X, then the complement Fn ofG1

⋃...

⋃Gn is nonempty for each n, but

⋂Fn is empty. If E is a set

contains a point from each Fn, consider a limit point of E.

Note that Fk ⊃ Fk+1 ⊃ ... and Fn is closed for all n, thus p lies in Fk

for all k. Hence p lies in⋂

Fn, but⋂

Fn is empty, a contradiction.

27. Define a point p in a metric space X to be a condensation point of a setE ⊂ X if every neighborhood of p contains uncountably many pointsof E.

Suppose E ⊂ Rk, E is uncountable, and let P be the set of all conden-sation points of E. Prove that P is perfect and that at most countablymany points of E are not in P . In other words, show that P c ⋂

E is atmost countable. Hint: Let {Vn} be a countable base of Rk, let W bethe union of those Vn for which E

⋂Vn is at most countable, and show

that P = W c.

Proof: Let {Vn} be a countable base of Rk, let W be the union ofthose Vn for which E

⋂Vn is at most countable, and we will show that

P = W c. Suppose x ∈ P . (x is a condensation point of E). Ifx ∈ Vn for some n, then E

⋂Vn is uncountable since Vn is open. Thus

x ∈ W c. (If x ∈ W , then there exists Vn such that x ∈ Vn and E⋂

Vn

is uncountable, a contradiction). Therefore P ⊂ W c.

Conversely, suppose x ∈ W c. x /∈ Vn for any n such that E⋂

Vn iscountable. Take any neighborhood N(x) of x. Take x ∈ Vn ⊂ N(x),and E

⋂Vn is uncountable. Thus E

⋂N(x) is also uncountable, x is a

condensation point of E. Thus W c ⊂ P . Therefore P = W c. Note thatW is countable, and thus W ⊂ W

⋂E = P c ⋂

E is at most countable.

To show that P is perfect, it is enough to show that P contains noisolated point. (since P is closed). If p is an isolated point of P , thenthere exists a neighborhood N of p such that N

⋂E = φ. p is not a

condensation point of E, a contradiction. Therefore P is perfect.

28. Prove that every closed set in a separable metric space is the unionof a (possible empty) perfect set and a set which is at most countable.(Corollary: Every countable closed set in Rk has isolated points.) Hint:Use Exercise 27.

Proof: Let X be a separable metric space, let E be a closed set onX. Suppose E is uncountable. (If E is countable, there is nothingto prove.) Let P be the set of all condensation points of E. Since Xhas a countable base, P is perfect, and P c ⋂

E is at most countable byExercise 27. Since E is closed, P ⊂ E. Also, P c ⋂

E = E − P . HenceE = P

⋃(E − P ).

For corollary: if there is no isolated point in E, then E is perfect. ThusE is uncountable, a contradiction.

Note: It’s also called Cauchy-Bendixon Theorem.

29. Prove that every open set in R1 is the union of an at most countablecollection of disjoint segments. Hint: Use Exercise 22.

Proof: (due to H.L.Royden, Real Analysis) Since O is open, for each xin O, there is a y > x such that (x, y) ⊂ O. Let b = sup{y : (x, y) ⊂ O}.Let a = inf{z : (z, x) ⊂ O}. Then a < x < b, and Ix = (a, b) is an openinterval containing x.

Now Ix ⊂ O, for if w ∈ Ix, say x < w < b, we have by the definition ofb a number y > w such that (x, y) ⊂ O, and so w ∈ O).

29

Moreover, b /∈ O, for if b ∈ O, then for some ε > 0 we have (b−ε, b+ε) ⊂O, whence (x, b + ε) ⊂ O, contradicting the definition of b. Similarly,a /∈ O.

Consider the collection of open intervals {Ix}, x ∈ O. Since each x ∈ Ois contained in Ix, and each Ix ⊂ O, we have O =

⋃Ix.

Let (a, b) and (c, d) be two intervals in this collection with a point incommon. Then we must have c < b and a < d. Since c /∈ O, it does notbelong to (a, b) and we have c ≤ a. Since a /∈ O and hence not to (c, d),we have a ≤ c. Thus a = c. Similarly, b = d, and (a, b) = (c, d). Thustwo different intervals in the collection {Ix} must be disjoint. ThusO is the union of the disjoint collection {Ix} of open intervals, and itremains only to show that this collection is countable. But each openinterval contains a rational number since Q is dense in R. Since wehave a collection of disjoint open intervals, each open interval containsa different rational number, and the collection can be put in one-to-onecorrespondence with a subset of the rationals. Thus it is a countablecollection.

30. Imitate the proof of Theorem 2.43 to obtain the following result:

If Rk =⋃∞

1 Fn, where each Fn is a closed subset of Rk, thenat least one Fn has a nonempty interior.

Equivalent statement: If Gn is a dense open subset of Rk, forn = 1, 2, 3, ..., then

⋂∞1 Gn is not empty (in fact, it is dense

in Rk).

(This is a special case of Baire’s theorem; see Exercise 22, Chap. 3, forthe general case.)

Proof: I prove Baire’s theorem directly. Let Gn be a dense opensubset of Rk for n = 1, 2, 3, .... I need to prove that

⋂∞1 Gn intersects

any nonempty open subset of Rk is not empty.

Let G0 is a nonempty open subset of Rk. Since G1 is dense and G0

is nonempty, G0⋂

G1 6= φ. Suppose x1 ∈ G0⋂

G1. Since G0 and G1

are open, G0⋂

G1 is also open, that is, there exist a neighborhood V1

such that V1 ⊂ G0⋂

G1. Next, since G2 is a dense open set and V1

is a nonempty open set, V1⋂

G2 6= φ. Thus, I can find a nonempty


open set V2 such that V2 ⊂ V1⋂

G2. Suppose I have get n nonemptyopen sets V1, V2, ..., Vn such that V1 ⊂ G0

⋂G1 and Vi+1 ⊂ Vi

⋂Gn+1

for all i = 1, 2, ..., n − 1. Since Gn+1 is a dense open set and Vn is anonempty open set, Vn

⋂Gn+1 is a nonempty open set. Thus I can find

a nonempty open set Vn+1 such that Vn+1 ⊂ Vn⋂

Gn+1. By induction, Ican form a sequence of open sets {Vn : n ∈ Z+} such that V1 ⊂ G0

⋂G1

and Vi+1 ⊂ Vi⋂

Gi+1 for all n ∈ Z+. Since V1 is bounded and V1 ⊃V2 ⊃ ... ⊃ Vn ⊃ ..., by Theorem 2.39 I know that

∞⋂n=1

Vn 6= φ.

Since V1 ⊂ G0⋂

G1 and Vn+1 ⊂ Gn+1, G0⋂

(⋂∞

n=1 Gn) 6= φ. Proved.

Note: By Baire’s theorem, I’ve proved the equivalent statement. Next,Fn has a empty interior if and only if Gn = Rk − Fn is dense in Rk.Hence we completed all proof.

Chapter 3

Numerical Sequences andSeries

31

32 CHAPTER 3. NUMERICAL SEQUENCES AND SERIES


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13. Prove that the Cauchy product of two absolutely convergent series con-verges absolutely.

Note: Given∑

an and∑

bn, we put cn =∑n

k=0 akbn−k and call∑

cn

the Cauchy product of the two given series.

Proof: Put An =∑n

k=0 |ak|, Bn =∑n

k=0 |bk|, Cn =∑n

k=0 |ck|. Then

Cn = |a0b0|+ |a0b1 + a1b0|+ ... + |a0bn + a1bn−1 + ... + anb0|≤ |a0||b0|+ (|a0||b1|+ |a1||b0|) + ...

+(|a0||bn|+ |a1||bn−1|+ ... + |an||b0|)= |a0|Bn + |a1|Bn−1 + ... + |an|B0

33

≤ |a0|Bn + |a1|Bn + ... + |an|Bn

= (|a0|+ |a1|+ ... + |an|)Bn

= AnBn

≤ AB

where A = lim An and B = lim Bn. Hence {Cn} is bounded. Notethat {Cn} is increasing, and thus Cn is a convergent sequence, thatis, the Cauchy product of two absolutely convergent series convergesabsolutely.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23. Suppose {pn} and {qn} are Cauchy sequences in a metric space X.Show that the sequence {d(pn, qn)} converges. Hint: For any m, n,

d(pn, qn) ≤ d(pn, pm) + d(pm, qm) + d(qm, qn);

it follows that|d(pn, qn)− d(pm, qm)|

is small if m and n are large.

Proof: For any ε > 0, there exists N such that d(pn, pm) < ε andd(qm, qn) < ε whenever m,n ≥ N . Note that

d(pn, qn) ≤ d(pn, pm) + d(pm, qm) + d(qm, qn).

34 CHAPTER 3. NUMERICAL SEQUENCES AND SERIES

It follows that

|d(pn, qn)− d(pm, qm)| ≤ d(pn, pm) + d(qm, qn) < 2ε.

Thus {d(pn, qn)} is a Cauchy sequence in X. Hence {d(pn, qn)} con-verges.

24.

25.

Chapter 4

Continuity

35

36 CHAPTER 4. CONTINUITY


1. Suppose f is a real function define on R1 which satisfies

limh→0

[f(x + h)− f(x− h)] = 0

for every x ∈ R1. Does this imply that f is continuous?

Solution: No. Take f(x) = 1, if x ∈ Z; f(x) = 0. otherwise.

2.

3.

4.

5. If f is a real continuous function defined on a closed set E ⊂ R1, provethat there exist continuous real function g on R1 such that g(x) = f(x)for all x ∈ E. (Such functions g are called continuous extensions of ffrom E to R1.) Show that the result becomes false if the word ”closed”is omitted. Extend the result to vector valued functions. Hint: Let thegraph of g be a straight line on each of the segments which constitutethe complement of E (compare Exercise 29, Chap. 2). The result re-mains true if R1 is replaced by any metric space, but the proof is notso simple.

Proof: Note that the following fact:

Every open set of real numbers is the union of a countablecollection of disjoint open intervals.

Thus, consider Ec =⋃

(ai, bi), where i ∈ Z, and ai < bi < ai+1 < bi+1.We extend g on (ai, bi) as following:

g(x) = f(ai) + (x− ai)f(bi)− f(ai)

bi − ai

(g(x) = f(x) for x ∈ E). Thus g is well-defined on R1, and g is contin-uous on R1 clearly.

37

Next, consider f(x) = 1/x on a open set E = R−0. f is continuous onE, but we cannot redefine f(0) = any real number to make new f(x)continue at x = 0.

Next, consider a vector valued function

f(x) = (f1(x), ..., fn(x)),

where fi(x) is a real valued function. Since f is continuous on E, eachcomponent of f , fi, is also continuous on E, thus we can extend fi, saygi, for each i = 1, ..., n. Thus,

g(x) = (g1(x), ..., gn(x))

is a extension of f(x) since each component of g, gi, is continuous onR1 implies g is continuous on Rn.

Note: The above fact only holds in R1. If we change R1 into anymetric spaces, we have no the previous fact.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19. Suppose f is a real function with domain R1 which has the interme-diate value property: If f(a) < c < f(b), then f(x) = c for some xbetween a and b.

Suppose also, for every rational r, that the set of all x with f(x) = ris closed. Prove that f is continuous.

Hint: If xn → x0, but f(xn) > r > f(x0) for some r and all n, thenf(tn) = r for some tn between x0 and xn; thus tn → x0. Find a contra-diction. (N. J. Fine, Amer. Math. Monthly, vol. 73, 1966, p. 782.)

Proof: Let S = {x : f(x) = r}. If xn → x0, but f(xn) > r > f(x0)for some r and all n since Q is dense in R1, then f(tn) = r for some tnbetween x0 and xn; thus tn → x0. Hence x0 is a limit point of S. SinceS is closed, f(x0) = r, a contradiction. Hence, lim sup f(xn) ≤ f(x0).Similarly, lim inf f(xn) ≥ f(x0). Hence, lim f(xn) = f(x0), and f iscontinuous at x0.

Note: Original problem is stated as follows:

Let f be a function from the reals to the reals, differentiableat every point. Suppose that, for every r, the set of pointsx, where f ′(x) = r, is closed. Prove that f ′ is continuous.

If we replace Q into any dense subsets of R1, the conclusion also holds.

20.

21.

22. Let A and B be disjoint nonempty closed sets in a metric space X, anddefine

f(p) =ρA(p)

ρA(p) + ρB(p), (p ∈ X).

39

Show that f is a continuous function on X whose range lies in [0, 1],that f(p) = 0 precisely on A and f(p) = 1 precisely on B. Thisestablishes a converse of Exercise 3: Every closed set A contained in Xis Z(f) for some continuous real f on X. Setting

V = f−1([0, 1/2)), W = f−1((1/2, 1]),

show that V and W are open and disjoint, and that A is contained inV , B is contained in W . (Thus pairs of disjoint closed set in a metricspace can be covered by pairs of disjoint open sets. This property ofmetric spaces is called normality.)

Proof: Note that ρA(p) and ρB(p) are (uniformly) continuous on X,and ρA(p) + ρB(p) > 0. (Clearly, ρA(p) + ρB(p) ≥ 0 by the definition.If ρA(p) + ρB(p) = 0, then p ∈ A

⋂B by Exercise 20, a contradiction).

Thus f(p) = ρA(p)/(ρA(p)+ρB(p)) is continuous on X. Next, f(p) ≥ 0,and f(p) ≤ 1 since ρA(p) ≤ ρA(p) + ρB(p). Thus f(X) lies in [0, 1].

Next, f(p) = 0 ⇔ ρA(p) = 0 ⇔ p ∈ A precisely, and f(p) = 1 ⇔ρB(p) = 0 ⇔ p ∈ B precisely by Exercise 20.

Now we prove a converse of Exercise 3: Every closed set A ⊂ X is Z(f)for some continuous real f on X. If Z(f) = φ, then f(x) = 1 for allx ∈ X satisfies our requirement. If Z(f) 6= φ, we consider two possiblecases: (1) Z(f) = X; (2) Z(f) 6= X. If Z(f) = X, then f(x) = 0 forall x ∈ X. If Z(f) 6= X, we can choose p ∈ X such that f(p) 6= 0.Note that Z(f) and {p} are one pair of disjoint closed sets. Hence welet

f(x) =ρZ(f)(x)

ρZ(f)(x) + ρ{p}(x).

By the previous result, we know that f(x) satisfies our requirement.Hence we complete the whole proof.

Note that [0, 1/2) and (1/2, 1] are two open sets of f(X). Since f iscontinuous, V = f−1([0, 1/2)) and W = f−1((1/2, 1]) are two open sets.f−1({0}) ⊂ f−1([0, 1/2)), and f−1({1}) ⊂ f−1((1/2, 1]). Thus, A ⊂ Vand B ⊂ W . Thus a metric space X is normal.


23.

24.

25.

26.

Chapter 5

Differentiation

41

42 CHAPTER 5. DIFFERENTIATION


1. Let f be defined for all real x, and suppose that

|f(x)− f(y)| ≤ (x− y)2

for all real x and y. Prove that f is constant.

Proof: |f(x)−f(y)| ≤ (x−y)2 for all real x and y. Fix y, |f(x)−f(y)x−y

| ≤|x− y|. Let x → y, therefore,

0 ≤ limx→y

f(x)− f(y)

x− y≤ lim

x→y|x− y| = 0

It implies that (f(x)− f(y))/(x− y) → 0 as x → y. Hence f ′(y) = 0,f = const.

2. Suppose f ′(x) > 0 in (a, b). Prove that f is strictly increasing in (a, b),and let g be its inverse function. Prove that g is differentible, and that

g′(f(x)) =1

f ′(x)(a < x < b).

Proof: For every pair x > y in (a, b), f(x)− f(y) = f ′(c)(x− y) wherey < c < x by Mean-Value Theorem. Note that c ∈ (a, b) and f ′(x) > 0in (a, b), hence f ′(c) > 0. f(x) − f(y) > 0, f(x) > f(y) if x > y, f isstrictly increasing in (a, b).

Let ∆g = g(x0 + h)− g(x0). Note that x0 = f(g(x0)), and thus,

(x0 + h)− x0 = f(g(x0 + h))− f(g(x0)),

h = f(g(x0) + ∆g)− f(g(x0)) = f(g + ∆g)− f(g).

Thus we apply the fundamental lemma of differentiation,

h = [f ′(g) + η(∆g)]∆g,

1

f ′(g) + η(∆g)=

∆g

h

43

Note that f ′(g(x)) > 0 for all x ∈ (a, b) and η(∆g) → 0 as h → 0, thus,

limh→0

∆g/h = limh→0

1

f ′(g) + η(∆g)=

1

f ′(g(x)).

Thus g′(x) = 1f ′(g(x))

, g′(f(x)) = 1f ′(x)

.

3. Suppose g is a real function on R1, with bounded derivative (say|g′| ≤ M). Fix ε > 0, and define f(x) = x + εg(x). Prove that fis one-to-one if ε is small enough. (A set of admissible values of ε canbe determined which depends only on M .)

Proof: For every x < y, and x, y ∈ R, we will show that f(x) 6= f(y).By using Mean-Value Theorem:

g(x)− g(y) = g′(c)(x− y) where x < c < y,

(x− y) + ε((x)− g(y)) = (εg′(c) + 1)(x− y),

that is,f(x)− f(y) = (εg′(c) + 1)(x− y). (∗)

Since |g′(x)| ≤ M , −M ≤ g′(x) ≤ M for all x ∈ R. Thus 1 − εM ≤εg′(c) + 1 ≤ 1 + εM , where x < c < y. Take c = 1

2M, and εg′(c) + 1 > 0

where x < c < y for all x, y. Take into equation (*), and f(x)−f(y) < 0since x− y < 0, that is, f(x) 6= f(y), that is, f is one-to-one (injective).

4. If

C0 +C1

2+ ... +

Cn−1

n+

Cn

n + 1= 0,

where C0, ..., Cn are real constants, prove that the equation

C0 + C1x + ... + Cn−1xn−1 + Cnx

n = 0


Proof: Let f(x) = C0x + ... + Cn

n+1xn+1. f is differentiable in R1

and f(0) = f(1) = 0. Thus, f(1) − f(0) = f ′(c) where c ∈ (0, 1) byMean-Value Theorem. Note that

f ′(x) = C0 + C1x + ... + Cn−1xn−1 + Cnx

n.

Thus, c ∈ (0, 1) is one real root between 0 and 1 of that equation.

5. Suppose f is defined and differentiable for every x > 0, and f ′(x) → 0as x → +∞. Put g(x) = f(x + 1) − f(x). Prove that g(x) → 0 asx → +∞.

Proof: f(x + 1) − f(x) = f ′(c)(x + 1 − x) where x < c < x + 1by Mean-Value Theorem. Thus, g(x) = f ′(c) where x < c < x + 1,that is,

limx→+∞

g(x) = limx→+∞

f ′(c) = limc→+∞

f ′(c) = 0.

6. Suppose(a) f is continuous for x ≥ 0,(b) f ′(x) exists for x > 0,(c) f(0) = 0,(d) f ′ is monotonically increasing.Put

g(x) =f(x)

x(x > 0)


Proof: Our goal is to show g′(x) > 0 for all x > 0

⇔ g′(x) = xf ′(x)−f(x)x2 > 0 ⇔ f ′(x) > f(x)

x.

Since f ′(x) exists, f(x) − f(0) = f ′(c)(x − 0) where 0 < c < x by

Mean-Value Theorem. ⇒ f ′(c) = f(x)x

where 0 < c < x. Since f ′ is

monotonically increasing, f ′(x) > f ′(c), that is, f ′(x) > f(x)x

for allx > 0.

7. Suppose f ′(x), g′(x) exist, g′(x) 6= 0, and f(x) = g(x) = 0. Prove that

limt→x

f(t)

g(t)=

f ′(x)

g′(x).

45

(This holds also for complex functions.)

Proof:

f ′(t)

g′(t)=

limt→xf(t)−f(x)

t−x

limt→xg(t)−g(x)

t−x

=limt→x f(t)

limt→x g(t)= lim

t→x

f(t)

g(t)

Surely, this holds also for complex functions.

8. Suppose f ′(x) is continuous on [a, b] and ε > 0. Prove that there existsδ > 0 such that

|f(t)− f(x)

t− x− f ′(x)| < ε

whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. (This could beexpressed by saying f is uniformly differentiable on [a, b] if f ′ is contin-uous on [a, b].) Does this hold for vector-valued functions too?

Proof: Since f ′(x) is continuous on a compact set [a, b], f ′(x) is uni-formly continuous on [a, b]. Hence, for any ε > 0 there exists δ > 0such that

|f ′(t)− f ′(x)| < ε

whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. Thus, f ′(c) =f(t)−f(x)

t−xwhere c between t and x by Mean-Value Theorem. Note that

0 < |c− x| < δ and thus |f ′(c)− f ′(x)| < ε, thus,

|f(t)− f(x)

t− x− f ′(x)| < ε

whenever 0 < |t− x| < δ, a ≤ x ≤ b, a ≤ t ≤ b.Note: It does not hold for vector-valued functions. If not, take

f(x) = (cos x, sin x),

[a, b] = [0, 2π], and x = 0. Hence f ′(x) = (− sin x, cos x). Take any1 > ε > 0, there exists δ > 0 such that

|f(t)− f(0)

t− 0− f ′(0)| < ε

whenever 0 < |t| < δ by our hypothesis. With calculating,

|(cos t− 1

t,sin t

t)− (0, 1)| < ε

|(cos t− 1

t,sin t

t− 1)| < ε

(cos t− 1

t)2 + (

sin t

t− 1)2 < ε2 < ε

2

t2+ 1− 2(cos t + sin t)

t< ε

since 1 > ε > 0. Note that

2

t2+ 1− 4

t<

2

t2+ 1− 2(cos t + sin t)

t

But 2t2

+ 1− 4t→ +∞ as t → 0. It contradicts.

9. Let f be a continuous real function on R1, of which it is known thatf ′(x) exists for all x 6= 0 and that f ′(x) → 0 as x → 0. Dose it followthat f ′(0) exists?

Note: We prove a more general exercise as following.Suppose that f is continuous on an open interval I containing x0, sup-pose that f ′ is defined on I except possibly at x0, and suppose thatf ′(x) → L as x → x0. Prove that f ′(x0) = L.

Proof of the Note: Using L’Hospital’s rule:

limh→0

f(x0 + h)− f(x0)

h= lim

h→0f ′(x0 + h)

By our hypothesis: f ′(x) → L as x → x0. Thus,

limh→0

f(x0 + h)− f(x0)

h= L,

Thus f ′(x0) exists andf ′(x0) = L.

47

10. Suppose f and g are complex differentiable functions on (0, 1), f(x) → 0,g(x) → 0, f ′(x) → A, g′(x) → B as x → 0, where A and B are complexnumbers, B 6= 0. Prove that

limx→0

f(x)

g(x)=

A

B.

Compare with Example 5.18. Hint:

f(x)

g(x)= (

f(x)

x− A)

x

g(x)+ A

x

g(x).

Apply Theorem 5.13 to the real and imaginary parts of f(x)x

and g(x)x

.

Proof: Write f(x) = f1(x)+ if2(x), where f1(x), f2(x) are real-valuedfunctions. Thus,

df(x)

dx=

df1(x)

dx+ i

df2(x)

dx,

Apply L’Hospital’s rule to f1(x)x

and f2(x)x

, we have

limx→0

f1(x)

x= lim

x→0f ′1(x)

limx→0

f2(x)

x= lim

x→0f ′2(x)

Combine f1(x) and f2(x), we have

limx→0

f1(x)

x+ i lim

x→0

f2(x)

x= lim

x→0

f1(x)

x+ i

f2(x)

x= lim

x→0

f(x)

x

or

limx→0

f1(x)

x+ i lim

x→0

f2(x)

x= lim

x→0f ′1(x) + i lim

x→0f ′2(x) = lim

x→0f ′(x)

Thus, limx→0f(x)

x= limx→0 f ′(x). Similarly, limx→0

g(x)x

= limx→0 g′(x).Note that B 6= 0. Thus,

limx→0

f(x)

g(x)= lim

x→0(f(x)

x− A)

x

g(x)+ A

x

g(x)

= (A− A)1

B+

A

B=

A

B.


Note: In Theorem 5.13, we know g(x) → +∞ as x → 0. (f(x) = x,

and g(x) = x + x2ei

x2 ).

11. Suppose f is defined in a neighborhood of x, and suppose f”(x) exists.Show that

limh→0

f(x + h) + f(x− h)− 2f(x)

h2= f ′′(x).

Show by an example that the limit may exist even if f”(x) dose not.Hint: Use Theorem 5.13.

Proof: By using L’Hospital’s rule: (respect to h.)

limh→0

f(x + h) + f(x− h)− 2f(x)

h2= lim

h→0

f ′(x + h)− f ′(x− h)

2h

Note that

f ′′(x) =1

2(f ′′(x) + f ′′(x))

=1

2(limh→0

f ′(x + h)− f ′(x)

h+ lim

h→0

f ′(x− h)− f ′(x)

−h)

=1

2limh→0

f ′(x + h)− f ′(x− h)

h

= limh→0

f ′(x + h)− f ′(x− h)

2h

Thus,f(x + h) + f(x− h)− 2f(x)

h2→ f ′′(x)

as h → 0. Counter-example: f(x) = x|x| for all real x.

12. If f(x) = |x|3, compute f ′(x), f ′′(x) for all real x, and show that f (3)(0)does not exist.

Proof: f ′(x) = 3|x|2 if x 6= 0. Consider

f(h)− f(0)

h=|h|3

h

49

Note that |h|/h is bounded and |h|2 → 0 as h → 0. Thus,

f ′(0) = limh→0

f(h)− f(0)

h= 0.

Hence, f ′(x) = 3|x|2 for all x. Similarly,

f ′′(x) = 6|x|.

Thus,f ′′(h)− f(0)

h= 6

|h|h

Since |h|h

= 1 if h > 0 and = −1 if h < 0, f ′′′(0) does not exist.

13. Suppose a and c are real numbers, c > 0, and f is defined on [−1, 1] by

f(x) =

{xa sin (x−c) (if x 6= 0),0 (if x = 0).

Prove the following statements:

(a) f is continuous if and only if a > 0.(b) f ′(0) exists if and only if a > 1.(c) f ′ is bounded if and only if a ≥ 1 + c.(d) f ′ is continuous if and only if a > 1 + c.(e) f ′′(0) exists if and only if a > 2 + c.(f) f ′′ is bounded if and only if a ≥ 2 + 2c.(g) f ′′ is continuous if and only if a > 2 + 2c.

Proof: For (a): (⇒) f is continuous iff for any sequence {xn} → 0with xn 6= 0, xa

n sin x−cn → 0 as n →∞. In particular, take

xn = (1

2nπ + π/2)

1c > 0

and thus xan → 0 as n →∞. Hence a > 0. (If not, then a = 0 or a < 0.

When a = 0, xan = 1. When a < 0, xa

n = 1/x−an → ∞ as n → ∞. It

contradicts.)

(⇐) f is continuous on [−1, 1]− {0} clearly. Note that

−|xa| ≤ xa sin (x−c) ≤ |xa|,

and |xa| → 0 as x → 0 since a > 0. Thus f is continuous at x = 0.Hence f is continuous.

For (b): f ′(0) exists iff xa−1 sin (x−c) → 0 as x → 0. In the previousproof we know that f ′(0) exists if and only if a−1 > 0. Also, f ′(0) = 0.

14. Let f be a differentiable real function defined in (a, b). Prove that f isconvex if and only if f ′ is monotonically increasing. Assume next thatf ′′(x) exist for every x ∈ (a, b), and prove that f is convex if and onlyif f ′′(x) ≥ 0 for all x ∈ (a, b).

15. Suppose a ∈ R1, f is a twice-differentiable real function on (a,∞),and M0, M1, M2 are the least upper bounds of |f(x)|, |f ′(x)|, |f ′′(x)|,respectively, on (a,∞). Prove that

M21 ≤ 4M0M2.

Hint: If h > 0, Taylor’s theorem shows that

f ′(x) =1

2h[f(x + 2h)− f(x)]− hf ′′(ξ)

for some ξ ∈ (x, x + 2h). Hence

|f ′(x)| ≤ hM2 +M0

h.

To show that M21 = 4M0M2 can actually happen, take a = −1, define

f(x) =

{2x2 − 1 (−1 < x < 0),x2−1x2+1

(0 ≤ x < ∞),

and show that M0 = 1, M1 = 4, M2 = 4. Does M21 ≤ 4M0M2 hold for

vector-valued functions too?

Proof: Suppose h > 0. By using Taylor’s theorem:

f(x + h) = f(x) + hf ′(x) +h2

2f ′′(ξ)

for some x < ξ < x + 2h. Thus

h|f ′(x)| ≤ |f(x + h)|+ |f(x)|+ h2

2|f ′′(ξ)|

51

h|f ′(x)| ≤ 2M0 +h2

2M2.

h2M2 − 2h|f ′(x)|+ 4M0 ≥ 0 (∗)

Since equation (*) holds for all h > 0, its determinant must be non-positive.

4|f ′(x)|2 − 4M2(4M0) ≤ 0

|f ′(x)|2 ≤ 4M0M2

(M1)2 ≤ 2M0M2

Note: There is a similar exercise:Suppose f(x)(−∞ < x < +∞) is a twice-differentiable real function,and

Mk = sup−∞<x<+∞

|f (k)(x)| < +∞ (k = 0, 1, 2).

Prove that M21 ≤ 2M0M2.

Proof of Note:

f(x + h) = f(x) + f ′(x)h +f ′′(ξ1)

2h2

(x < ξ1 < x + h or x > ξ1 > x + h) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (*)

f(x− h) = f(x)− f ′(x)h +f ′′(ξ2)

2h2

(x− h < ξ2 < x or x− h > ξ2 > x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (**)(*) minus (**):

f(x + h)− f(x− h) = 2f ′(x)h +h2

2(f ′′(ξ1)− f ′′(ξ2)).

2h|f ′(x)| ≤ |2hf ′(x)|

2h|f ′(x)| ≤ |f(x + h)|+ |f(x− h)|+ h2

2(|f ′′(ξ1)|+ |f ′′(ξ2)|)

2h|f ′(x)| ≤ 2M0 + h2M2

M2h2 − 2|f ′(x)|h + 2M0 ≥ 0


Since this equation holds for all h, its determinant must be non-positive:

4|f ′(x)|2 − 4M2(2M0) ≤ 0,

|f ′(x)|2 ≤ 2M0M2

ThusM2

1 ≤ 2M0M2

16. Suppose f is twice-differentiable on (0,∞), f ′′ is bounded on (0,∞),and f(x) → 0 as x → ∞. Prove that f ′(x) → 0 as x → ∞. Hint: Leta →∞ in Exercise 15.

Proof: Suppose a ∈ (0,∞), and M0, M1, M2 are the least upperbounds of |f(x)|, |f ′(x)|, |f ′′(x)| on (a,∞). Hence, M2

1 ≤ 4M0M2. Leta →∞, M0 = sup |f(x)| → 0. Since M2 is bounded, therefore M2

1 → 0as a →∞. It implies that sup |f ′(x)| → 0 as x →∞.

17. Suppose f is a real, three times differentiable function on [−1, 1], suchthat

f(−1) = 0, f(0) = 0, f(1) = 1, f ′(0) = 0.

Prove that f (3)(x) ≥ 3 for some x ∈ (−1, 1).Note that equality holds for 1

2(x3 + x2).

Hint: Use Theorem 5.15, with α = 0 and β = 1,−1, to show thatthere exist s ∈ (0, 1) and t ∈ (−1, 0) such that

f (3)(s) + f (3)(t) = 6.

Proof: By Theorem 5.15, we take α = 0, β = 1,

f(1) = f(0) + f ′(0) +f ′′(0)

2+

f (3)(s)

6

where s ∈ (0, 1). Take α = 0, andβ = −1,

f(−1) = f(0)− f ′(0) +f ′′(0)

2− f (3)(t)

6

53

where t ∈ (−1, 0). Thus

1 =f ′′(0)

2+

f (3)(s)

6, s ∈ (0, 1) (∗)

0 =f ′′(0)

2− f (3)(s)

6, s ∈ (−1, 0) (∗∗)

Equation (*) - equation (**):

f (3)(s)

6+

f (3)(t)

6, s ∈ (0, 1), t ∈ (−1, 0).

f (3)(s) + f (3)(t) = 6, s, t ∈ (−1, 1).

f (3)(x) ≥ 3 for some x ∈ (−1, 1).

Theorem 5.15: Suppose f is a real function on [a, b], n is a positiveinteger, f (n−1) is continuous on [a, b], f (n)(t) exists for every t ∈ (a, b).Let α, β be distinct points of [a, b], and define

P (t) =n−1∑k=0

f (k)(α)

k!(t− α)k.

Then there exists a point x between α and β such that

f(β) = P (β) +f (n)(x)

n!(β − α)n.

18. Suppose f is a real function on [a, b], n is a positive integer, and f (n−1)

exists for every t ∈ [a, b]. Let α, β, and P be as in Taylor’s theorem(5.15). Define

Q(t) =f(t)− f(β)

t− β

for t ∈ [a, b], t 6= β, differentiate

f(t)− f(β) = (t− β)Q(t)

n − 1 times at t = α, and derive the following version of Taylor’stheorem:

f(β) = P (β) +Q(n−1)(α)

(n− 1)!(β − α)n.

19. Suppose f is defined in (−1, 1) and f ′(0) exists. Suppose −1 < αn <βn < 1, αn → 0, and βn → 0 as n →∞. Define the difference quotients

Dn =f(βn)− f(αn)

βn − αn

Prove the following statements:

(a) If αn < 0 < βn, then lim Dn = f ′(0).(b) If 0 < αn < βn and {βn/(βn−αn)} is bounded, then lim Dn = f ′(0).(c) If f ′ is continuous in (−1, 1), then lim Dn = f ′(0).

Give an example in which f is differentiable in (−1, 1) (but f ′ is notcontinuous at 0) and in which αn, βn tend to 0 in such a way thatlim Dn exists but is different from f ′(0).

Proof: For (a):

Dn =f(βn)− f(0)

βn

βn

βn − αn

+f(αn)− f(0)

αn

−αn

βn − αn

Note that

f ′(0) = limn→∞

f(αn)− f(0)

αn

= limn→∞

f(βn)− f(0)

βn

Thus for any ε > 0, there exists N such that

L− ε <f(αn)− f(0)

αn

< L + ε,

L− ε <f(βn)− f(0)

βn

< L + ε,

55

whenever n > N where L = f ′(0) respectively. Note that βn/(βn−αn)and −αn/(βn − αn) are positive. Hence,

βn

βn − αn

(L− ε) <f(βn)− f(0)

βn

βn

βn − αn

<βn

βn − αn

(L + ε)

−αn

βn − αn

(L− ε) <f(αn)− f(0)

αn

−αn

βn − αn

<−αn

βn − αn

(L + ε)

Combine two inequations,

L− ε < Dn < L + ε

Hence, lim Dn = L = f ′(0).

For (b): We process as above prove, but note that −αn/(βn−αn) < 0.Thus we only have the following inequations:

βn

βn − αn

(L− ε) <f(βn)− f(0)

βn

βn

βn − αn

<βn

βn − αn

(L + ε)

−αn

βn − αn

(L + ε) <f(αn)− f(0)

αn

−αn

βn − αn

<−αn

βn − αn

(L− ε)

Combine them:

L− βn + αn

βn − αn

ε < Dn < L +βn + αn

βn − αn

ε

Note that {βn/(βn − αn)} is bounded, ie,

| βn

βn − αn

| ≤ M

for some constant M . Thus

|βn + αn

βn − αn

| = | 2βn

βn − αn

− 1| ≤ 2M + 1

Hence,

L− (2M + 1)ε < Dn < L + (2M + 1)ε

Hence, lim Dn = L = f ′(0).

For (c): By using Mean-Value Theorem,

Dn = f ′(tn)

where tn is between αn and βn. Note that

min{αn, βn} < tn < max{αn, βn}

and

max{αn, βn} =1

2(αn + βn + |αn − βn|)

min{αn, βn} =1

2(αn + βn − |αn − βn|)

Thus, max{αn, βn} → 0 and min{αn, βn} → 0 as αn → 0 and βn → 0.By squeezing principle for limits, tn → 0. With the continuity of f ′,we have

lim Dn = lim f ′(tn) = f ′(lim tn) = f ′(0).

Example: Let f be defined by

f(x) =

{x2 sin(1/x) (x 6= 0),0 (x = 0).

Thus f ′(x) is not continuous at x = 0, and f ′(0) = 0. Take αn =1

π/2+2nπand βn = 1

2nπ. It is clear that αn → 0, and βn → 0 as n →∞.

Also,

Dn =−4nπ

π(π/2 + 2nπ)→ − 2

π

as n →∞. Thus, lim Dn = −2/π exists and is different from 0 = f ′(0).

20.

21.

22. Suppose f is a real function on (−∞,∞). Call x a fixed point of f iff(x) = x.

(a) If f is differentiable and f ′(t) 6= 1 for every real t, prove that f hasat most one fixed point.

57

(b) Show that the function f defined by

f(t) = t + (1 + et)−1

has no fixed point, although 0 < f ′(t) < 1 for all real t.

(c) However, if there is a constant A < 1 such that |f ′(t)| ≤ A for allreal t, prove that a fixed point x of f exists, and that x = lim xn, wherex1 is an arbitrary real number and

xn+1 = f(xn)

for n = 1, 2, 3, ...

(d) Show that the process describe in (c) can be visualized by the zig-zag path

(x1, x2) → (x2, x2) → (x2, x3) → (x3, x3) → (x3, x4) → ....

Proof: For (a): If not, then there exists two distinct fixed points, sayx and y, of f . Thus f(x) = x and f(y) = y. Since f is differeniable,by applying Mean-Value Theorem we know that

f(x)− f(y) = f ′(t)(x− y)

where t is between x and y. Since x 6= y, f ′(t) = 1. It contradicts.

For (b): We show that 0 < f ′(t) < 1 for all real t first:

f ′(t) = 1 + (−1)(1 + et)−2et = 1− et

(1 + et)2

Sinceet > 0

(1 + et)2 = (1 + et)(1 + et) > 1(1 + et) = 1 + et > et > 0

for all real t, thus(1 + et)−2et > 0

(1 + et)−2et < 1

for all real t. Hence 0 < f ′(t) < 1 for all real t.

Next, since f(t) − t = (1 − et)−1 > 0 for all real t, f(t) has no fixedpoint.

For (c): Suppose xn+1 6= xn for all n. (If xn+1 = xn, then xn = xn+1 =... and xn is a fixed point of f).

By Mean-Value Theorem,

f(xn+1)− f(xn) = f ′(tn)(xn+1 − xn)

where tn is betweem xn and xn+1. Thus,

|f(xn+1)− f(xn)| = |f ′(tn)||(xn+1 − xn)|

Note that |f ′(tn)| is bounded by A < 1, f(xn) = xn+1, and f(xn+1) =xn+2. Thus

|xn+2 − xn+1| ≤ A|xn+1 − xn||xn+1 − xn| ≤ CAn−1

where C = |x2 − x1|. For two positive integers p > q,

|xp − xq| ≤ |xp − xp−1|+ ... + |xq+1 − xq|= C(Aq−1 + Aq + ... + Ap−2)

≤ CAq−1

1− A.

Hence

|xp − xq| ≤CAq−1

1− A.

Hence, for any ε > 0, there exists N = [logAε(1−A)

C] + 2 such that

|xp − xq| < ε whenever p > q ≥ N . By Cauchy criterion we know that{xn} converges to x. Thus,

limn→∞

xn+1 = f( limn→∞

xn)

since f is continuous. Thus,

x = f(x).

x is a fixed point of f .

For (d): Since xn+1 = f(xn), it is trivial.

59

23.

24.

25. Suppose f is twice differentiable on [a, b], f(a) < 0, f(b) > 0, f ′(x) ≥δ > 0, and 0 ≤ f ′′(x) ≤ M for all x ∈ [a, b]. Let ξ be the unique pointin (a, b) at which f(ξ) = 0.

Complete the details in the following outline of Newton’s method forcomputing ξ.

(a) Choose x1 ∈ (ξ, b), and define {xn} by

xn+1 = xn −f(xn)

f ′(xn).

Interpret this geometrically, in terms of a tangent to the graph of f .

(b) Prove that xn+1 < xn and that

limn−>oo

xn = ξ.

(c) Use Taylor’s theorem to show that

xn+1 − ξ =f ′′(tn)

2f ′(xn)(xn − ξ)2

for some tn ∈ (ξ, xn).

(d) If A = M2δ

, deduce that

0 ≤ xn+1 − ξ ≤ 1

A[A(x1 − ξ)]2

n

.

(Compare with Exercise 16 and 18, Chap. 3)

(e) Show that Newton’s method amounts to finding a fixed point of thefunction g defined by

g(x) = x− f(x)

f ′(x).

How does g′(x) behave for x near ξ?

(f) Put f(x) = x1/3 on (−∞,∞) and try Newton’s method. Whathappens?

Proof: For (a): You can see the picture in the following URL:http://archives.math.utk.edu/visual.calculus/3/newton.5/1.html.

For (b): We show that xn ≥ xn+1 ≥ ξ. (induction). By Mean-ValueTheorem, f(xn) − f(ξ) = f ′(cn)(xn − ξ) where cn ∈ (ξ, xn). Sincef ′′ ≥ 0, f ′ is increasing and thus

f(xn)

xn − ξ= f ′(cn) ≤ f ′(xn) =

f(xn)

xn − xn+1

f(xn)(xn − ξ) ≤ f(xn)(xn − xn+1)

Note that f(xn) > f(ξ) = 0 since f ′ ≥ δ > 0 and f is strictly increasing.Thus,

xn − ξ ≤ xn − xn+1

ξ ≤ xn+1

Note that f(xn) > 0 and f ′(xn) > 0. Thus xn+1 < xn. Hence,

xn > xn+1 ≥ ξ.

Thus, {xn} converges to a real number ζ. Suppose ζ 6= ξ, then

xn+1 = xn −f(xn)

f ′(xn)

Note that f(xn)f ′(xn)

> f(ζ)δ

. Let α = f(ζ)δ

> 0, be a constant. Thus,

xn+1 < xn − α

for all n. Thus, xn < x1 − (n − 1)α, that is, xn → −∞ as n → ∞. Itcontradicts. Thus, {xn} converges to ξ.

For (c): By using Taylor’s theorem,

f(ξ) = f(xn) + f ′(xn)(ξ − xn) +f ′′(tn)

2(xn − ξ)2

0 = f(xn) + f ′(xn)(ξ − xn) +f ′′(tn)

2(xn − ξ)2

0 =f(xn)

f ′(xn)− xn + ξ +

f ′′(tn)

2f ′(xn)(xn − ξ)2

61

xn+1 − ξ =f ′′(tn)

2f ′(xn)(xn − ξ)2

where tn ∈ (ξ, xn).

For (d): By (b) we know that 0 ≤ xn+1 − ξ for all n. Next by (c) weknow that

xn+1 − ξ =f ′′(tn)

2f ′(xn)(xn − ξ)2

Note that f ′′ ≤ M and f ′ ≥ δ > 0. Thus

xn+1 − ξ ≤ A(xn − ξ)2 ≤ 1

A(A(x1 − ξ))2n

by the induction. Thus,

0 ≤ xn+1 − ξ ≤ 1

A[A(x1 − ξ)]2

n

.

For (e): If x0 is a fixed point of g(x), then g(x0) = x0, that is,

x0 −f(x0)

f ′(x0)= x0

f(x0) = 0.

It implies that x0 = ξ and x0 is unique since f is strictly increasing.Thus, we choose x1 ∈ (ξ, b) and apply Newton’s method, we can findout ξ. Hence we can find out x0.

Next, by calculating

g′(x) =f(x)f ′′(x)

f ′(x)2

0 ≤ g′(x) ≤ f(x)M

δ2.

As x near ξ from right hand side, g′(x) near f(ξ) = 0.

For (f): xn+1 = xn − f(xn)f ′(xn)

= −2xn by calculating. Thus,

xn = (−2)n−1x1

for all n, thus {xn} does not converges for any choice of x1, and wecannot find ξ such that f(ξ) = 0 in this case.

26. Suppose f is differentiable on [a, b], f(a) = 0, and there is a real numberA such that |f ′(x)| ≤ A|f(x)| on [a, b]. Prove that f(x) = 0 for allx ∈ [a, b]. Hint: Fix x0 ∈ [a, b], let

M0 = sup |f(x)|, M1 = sup |f ′(x)|

for a ≤ x ≤ x0. For any such x,

|f(x)| ≤ M1(x0 − a) ≤ A(x0 − a)M0.

Hence M0 = 0 if A(x0 − a) < 1. That is, f = 0 on [a, x0]. Proceed.

Proof: Suppose A > 0. (If not, then f = 0 on [a, b] clearly.) Fixx0 ∈ [a, b], let

M0 = sup |f(x)|, M1 = sup |f ′(x)|for a ≤ x ≤ x0. For any such x,

f(x)− f(a) = f ′(c)(x− a)

where c is between x and a by using Mean-Value Theorem. Thus

|f(x)| ≤ M1(x− a) ≤ M1(x0 − a) ≤ A(x0 − a)M0

Hence M0 = 0 if A(x0 − a) < 1. That is, f = 0 on [a, x0] by takingx0 = a + 1

2A. Repeat the above argument by replacing a with x0, and

note that 12A

is a constant. Hence, f = 0 on [a, b].

27. Let φ be a real function defined on a rectangle R in the plane, givenby a ≤ x ≤ b, α ≤ y ≤ β. A solution of the initial-value problem

y′ = φ(x, y), y(a) = c (α ≤ c ≤ β)

is, by definition, a differentiable function f on [a, b] such that f(a) = c,α ≤ f(x) ≤ β, and

f ′(x) = φ(x, f(x)) (a ≤ x ≤ b)

Prove that such a problem has at most one solution if there is a constantA such that

|φ(x, y2)− φ(x, y1)| ≤ A|y2 − y1|

63

whenever (x, y1) ∈ R and (x, y2) ∈ R.

Hint: Apply Exercise 26 to the difference of two solutions. Note thatthis uniqueness theorem does not hold for the initial-value problem

y′ = y1/2, y(0) = 0,

which has two solutions: f(x) = 0 and f(x) = x2/4. Find all othersolutions.

Proof: Suppose y1 and y2 are solutions of that problem. Since

|φ(x, y2)− φ(x, y1)| ≤ A|y2 − y1|,

y(a) = c, y′1 = φ(x, y1), and y′2 = φ(x, y2), by Exercise 26 we know thaty1 − y2 = 0, y1 = y2. Hence, such a problem has at most one solution.

Note: Suppose there is initial-value problem

y′ = y1/2, y(0) = 0.

If y1/2 6= 0, then y1/2dy = dx. By integrating each side and noting thaty(0) = 0, we know that f(x) = x2/4. With y1/2 = 0, or y = 0. Allsolutions of that problem are

f(x) = 0 and f(x) = x2/4

Why the uniqueness theorem does not hold for this problem? Onereason is that there does not exist a constant A satisfying

|y′1 − y′2| ≤ A|y1 − y2|

if y1 and y2 are solutions of that problem. (since 2/x → ∞ as x → 0and thus A does not exist).

28. Formulate and prove an analogous uniqueness theorem for systems ofdifferential equations of the form

y′j = φj(x, y1, ..., yk), yj(a) = cj (j = 1, ..., k)

Note that this can be rewritten in the form

y′ = φ(x, y), y(a) = c


where y = (y1, ..., yk) ranges over a k-cell, φ is the mapping of a (k+1)-cell into the Euclidean k-space whose components are the functionφ1, ..., φk, and c is the vector (c1, ..., ck). Use Exercise 26, for vector-valued functions.

Theorem: Let φj(j = 1, ..., k) be real functions defined on a rectangleRj in the plane given by a ≤ x ≤ b, αj ≤ yj ≤ βj.

A solution of the initial-value problem

y′j = φ(x, yj), yj(a) = cj (αj ≤ cj ≤ βj)

is, by definition, a differentiable function fj on [a, b] such that fj(a) =cj, αj ≤ fj(x) ≤ βj, and

f ′j(x) = φj(x, fj(x)) (a ≤ x ≤ b)

Then this problem has at most one solution if there is a constant Asuch that

|φj(x, yj2)− φj(x, yj1)| ≤ A|yj2 − yj1|

whenever (x, yj1) ∈ Rj and (x, yj2) ∈ Rj.

Proof: Suppose y1 and y2 are solutions of that problem. For eachcomponents of y1 and y2, say y1j and y2j respectively, y1j = y2j byusing Exercise 26. Thus, y1 = y2

29. Specialize Exercise 28 by considering the system

y′j = yj+1 (j = 1, ..., k − 1),

y′k = f(x)−k∑

j=1

gj(x)yj

where f, g1, ..., gk are continuous real functions on [a, b], and derive auniqueness theorem for solutions of the equation

y(k) + gk(x)y(k−1) + ... + g2(x)y′ + g1(x)y = f(x),

65

subject to initial conditions

y(a) = c1, y′(a) = c1, ..., y

(k−1)(a) = ck.

Theorem: Let Rj be a rectangle in the plain, given by a ≤ x ≤ b,min yj ≤ yj ≤ max yj. (since yj is continuous on the compact set, say[a, b], we know that yj attains minimal and maximal.) If there is aconstant A such that{

|yj+1,1 − yj+1,2| ≤ A|yj,1 − yj,2| (j < k)|∑k

j=1 gj(x)(yj,1 − yj,2)| ≤ A|yk,1 − yk,2|

whenever (x, yj,1) ∈ Rj and (x, yj,2) ∈ Rj.

Proof: Since the system y′1, ..., y′k with initial conditions satisfies a fact

that there is a constant A such that |y′1−y′2| ≤ A|y1−y2|, that systemhas at most one solution. Hence,

y(k) + gk(x)y(k−1) + ... + g2(x)y′ + g1(x)y = f(x),

with initial conditions has at most one solution.

Chapter 6

Sequences and Series ofFunctions

67

68 CHAPTER 6. SEQUENCES AND SERIES OF FUNCTIONS

Rudin’s Principles of Mathematical Analysis:Solutions to Selected Exercises

Sam BlinsteinUCLA Department of Mathematics

March 29, 2008

Contents

Chapter 1: The Real and Complex Number Systems 2

Chapter 2: Basic Topology 10

Chapter 3: Numerical Sequences and Series 16

Chapter 4: Continuity 23

Chapter 5: Differentiation 32

Chapter 6: The Riemann-Stieltjes Integral 39

Chapter 7: Sequences and Series of Functions 47

1

The following are solutions to selected exercises from Walter Rudin’s Principles of MathematicalAnalysis, Third Edition, which I compiled during the Winter of 2008 while a graduate student atUCLA. Equations are numbered within each Chapter and their labels correspond to the questionnumber and equation number, so (2.3) refers to the third equation in question #2 of the currentChapter.

Chapter 1: The Real and Complex Number Systems

1. If r ∈ Q, r 6= 0, and x is irrational, prove that r + x and rx are irrational.

Solution: Suppose that r + x ∈ Q. Since Q is a field, −r ∈ Q and we have that −r + r + x =x ∈ Q, a contradiction. Therefore r+ x ∈ R\Q, i.e. r+ x is irrational. Similarly, suppose thatrx ∈ Q. Since Q is a field, 1

r ∈ Q and we have that 1r · rx = x ∈ Q, a contradiction. Therefore

rx ∈ R\Q, i.e. rx is irrational. �

2. Prove that there is no rational number whose square is 12.

Solution: Suppose there exist m,n ∈ Z such that neither are divisible by 3 (i.e. we assumetheir ratio is in simplified form) and

m

n=√

12 =⇒ m2 = 12n2 = 22 · 3n2.

Therefore, 3 | m2 =⇒ 3 | m because 3 is prime. Therefore, 32 | m2 =⇒ 32 | 12n2 =⇒3 | n2 =⇒ 3 | n, again because 3 is prime. This contradicts that m and n were chosen suchthat their ratio is in simplified form because both are divisible by 3.

We can investigate this further, discovering that Q lacks the greatest/least upper boundproperty, as is done analogously in Example 1.1 on page 2 of the text. Let A = {p ∈ Q | p >0, p2 < 12} and B = {p ∈ Q | p > 0, p2 > 12}. Define

q = p− p2 − 12p+ 12

=p(p+ 12)− p2 + 12

p+ 12=

12(p+ 1)p+ 12

. (2.1)

Then, we have that

q2 − 12 =144(p+ 1)2 − 12(p+ 12)2

(p+ 12)2=

132(p2 − 12)(p+ 12)2

. (2.2)

Now, if p ∈ A then p2 − 12 < 0 so by Equation (2.1) q > p and by Equation (2.2) q2 < 12 soq ∈ A. Therefore, A contains no largest number. Similarly, if p ∈ B then p2 − 12 > 0 so byEquation (2.1) 0 < q 12 so q ∈ B. Therefore, B contains nosmallest number. The elements of B are precisely the upper bounds of A, therefore, since Bhas no least element, A has no least upper bound. Similarly, the elements of A are the lowerbounds of B and since A has no greatest element, B has no greatest upper bound. �

3. Prove the following using the axioms of multiplication in a field.

(a) If x 6= 0 and xy = xz then y = z.

(b) If x 6= 0 and xy = x then y = 1.

(c) If x 6= 0 and xy = 1 then y = 1/x.

(d) If x 6= 0 the 1/(1/x) = x.

2

Solution: To show part (a) we simply use associativity of multiplication and the existenceof multiplicative inverses.

y = 1y =(

1x· x)y =

1x

(xy) =1x

(xz) =(

1x· x)z = 1z = z.

The statement of (b) and (c) follow directly from (a) if we let z = 1 and z = 1/x, respectively.Part (d) follows from (c) if we replace x with 1/x and y with x. �

4. Let E be a nonempty subset of an ordered set; suppose α is a lower bound of E and β is anupper bound of E. Prove that α ≤ β.

Solution: Since E is nonempty, let x ∈ E. Then, by definition of upper and lower bounds,we have that α ≤ x ≤ β. �

5. Let A be a nonempty set of real numbers which is bounded below. Let −A be the set of allnumbers −x, where x ∈ A. Prove that inf A = − sup(−A).

Solution: Let β = sup(−A), which exists because A is a nonempty subset of the real num-bers which is bounded below, hence −A is a nonempty subset of the real numbers which isbounded above. Thus,

∀ x ∈ A,−x ≤ β =⇒ x ≥ −β

so −β is a lower bound for A. Now, suppose there exists γ such that γ > −β and ∀ x ∈A, x ≥ γ. This implies that −γ < β and −x ≤ −γ, ∀ x ∈ A. But then, −γ is a smallerupper bound (than β) for −A, contradicting that β = sup(−A). Therefore no such γ existsand so − sup(−A) = −β = inf(A). The reversing of inequalities throughout follows fromProposition 1.18 on page 8 of the text. �

6. Fix b > 1.

(a) If m,n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, prove that

(bm)1/n = (bp)1/q.



(c) If x is real, define B(x) to be the set of all numbers bt, where t is rational and t ≤ x.Prove that

br = supB(r)

where r is rational. Hence, it makes sense to define

bx = supB(x)

for every real x.

(d) Prove that bx+y = bxby for every real x and y.

Solution:

(a) Frst observe that (bn1)n2 = bn1n2 for n1, n2 ∈ Z by direct expansion of either term. Letyn = bm so that y = (bm)1/n. Then, ynq = bmq = bnp =⇒ (yq)n = (bp)n, becausem/n = p/q =⇒ mq = np. By Theorem 1.12 on page 10, since nth roots are unique, wehave yq = bp. Then, (bm)1/n = y = (bp)1/q.

3

(b) Let r, s ∈ Q and write r = m/n and s = p/q form,n, p, q ∈ Z. Then, r+s = m/n+p/q =(mq + np)/nq. Note, for n1, n2 ∈ Z we have bn1bn2 = bn1+n2 by direct expansion.Then, by (a), br+s = (bmq+np)1/nq = (bmqbnp)1/nq = (bmq)1/nq(bnp)1/nq, where the lastequality follows from the Corollary to Theorem 1.12 on page 11 of the text. Again usingthe result of (a) we can simplify this last expression and obtain (bmq)1/nq(bnp)1/nq =(bm)1/n(bp)1/q = brbs, showing (b).

(c) Let t, r ∈ Q and write t = p/q, r = m/n, where p, q,m, n ∈ Z. Then, t < r =⇒ p/q <m/n =⇒ pn < mq. Thus, since b > 1, we have that bpn < bmq =⇒ bp < (bmq)1/n =⇒(bp)1/q < ((bmq)1/n)1/q. Using the result of part (a), we can simplify the last expressionto obtain bt = (bp)1/q < (bm)1/n = br, showing that

br > {bt | t < r} =⇒ br = sup {bt | t ≤ r} = B(r)

and proving part (c).

(d) We first observe that brbt = br+t by part (b) and so

bxby = supr≤x,t≤y

brbt = supr≤x,t≤y

br+t ≤ supr+t≤x+y

br+t = bx+y.

Here, supr≤x,t≤y br+t ≤ supr+t≤x+y br+t because we are taking the sup over a more re-stricted set on the right since r ≤ x, t ≤ y =⇒ r + t ≤ x + y but r + t ≤ x + y doesn’timply r ≤ x and t ≤ y. But, we in fact have equality because if supr≤x,t≤y br+t <supr+t≤x+y br+t then there exist r′, t′ ∈ Q such that r′ + t′ ≤ x+ y and

supρ≤x,τ≤y

bρ+τ < br′+t′ ≤ sup

r+t≤x+ybr+t. (6.1)

If r′ + t′ = x + y we get a contradiction to Equation (6.1) by choosing ρ = x, τ = y(this follows from Exercise 1 above). Otherwise, r′ + t′ < x+ y and then we can chooseρ < x, τ < y such that r′ + t′ < ρ + τ < x + y, because Q is dense in R. But then,bρ+τ > br

′+t′ , also contradicting Equation (6.1). Therefore we have equality and part (d)is shown. �

7. Omitted.

8. Prove that no order can be defined in the complex field that turns it into an ordered field.Hint: −1 is a square.

Solution: Suppose there is an ordered defined on the complex field that turns it into anordered field. Then, we get a contradiction to Proposition 1.18(d) on page 8 because i 6= 0yet i2 = −1 < 0, where −1 < 0 is forced by part (a) of the same Proposition. �

9. Suppose z = a+ bi, w = c+ di. Define z < w if a < c, and also if a = c but b < d. Prove thatthis turns the set of all complex numbers into an ordered set. (This type of order relationis called a dictionary order, or lexicographic order, for obvious reasons.) Does this ordered sethave the least-upper-bound property?

Solution: Let z = a+ bi, w = c+ di ∈ C. Then, since R is an ordered set, either a < b, a = b,or a > b. If a < b, then z < w. If a > b then z > w. If a = b we check the relationship betweenb and d and similarly will conclude that either z < w if b < d, z > w if b > d, or z = w if b = d.To check transitivity, let u = e + fi ∈ C and suppose that z < w and w < u. Thus a < c ora = c and b < d. Also, c < e or c = e and d < f . If a < c and c < e then a < e because R is an

4

ordered set and in this case z < u. If a < c and c = e with d < f then a < c = e so z < u. Ifa = c and b < d with c < e then a = c < e and z < u. If a = c and b < d with c = e and d < fthen a = e but b < d < f so b < f because R is an ordered set and therefore z < u. Thus, Cis an ordered set under this order.

This ordered set does have the least-upper-bound property. To see this, let E ⊆ C and defineα = sup {a | a+bi ∈ E} and β = sup {b | a+bi ∈ E}. Then, it is clear α+βi ∈ C becauseR hasthe least-upper bound property and furthermore ∀ a+bi ∈ E, a+bi ≤ α+βi, (because a ≤ αand b ≤ β) so α+βi is an upper bound for E. Now, suppose there exists γ+ δi ∈ C such thatγ + δi < α + βi and γ + δi is an upper bound for E. If γ < α, then we have a ≤ γ < α ∀ asuch that a + bi ∈ E, contradicting the definition of α. An identical contradiction with thedefinition of β occurs if γ = α and δ < β. Therefore, α + βi = sup(E) ∈ C and this orderedset does indeed have the least-upper-bound property. �

10. Suppose z = a+ bi, w = u+ vi, and

a =(|w|+ u

2

)1/2

, b =(|w| − u

2

)1/2

.

Prove that z2 = w if v ≥ 0 and that (z)2 = w if v ≤ 0. Conclude that every complex number(with one exception!) has two complex square roots.

Solution: First, suppose that v ≥ 0. Then

z2 = (a+ bi)(a+ bi) = (a2 − b2) + 2abi =(

(u2 + v2)1/2 + u

2

)−(

(u2 + v2)1/2 − u2

)+ 2abi

= u+ 2(

(u2 + v2)− u2

4

)1/2

i

= u+ vi

= w

where we needed to use that v ≥ 0 because (v2)1/2 = |v|. If v ≤ 0 then

(z)2 = (a− bi)(a− bi) = (a2 − b2)− 2abi = u− 2abi = u− 2(

(u2 + v2)− u2

4

)1/2

i

= u− |v|i= w.

Then it is clear each complex number (except 0 of course!) has two complex roots becausewe can take either both positive or both negative square roots defining a and b above since ifwe choose−a and−b, the minus signs vanish in w = (a2−b2)±2abi. It is clear we can’t havemore than 2 complex roots because if we mixed taking the positive square root defining aand the negative square root defining b, or vice versa, the minus signs would not cancel inthe ±2abi term. �

11. If z is a complex number, prove that there exists an r ≥ 0 and a complex number w with|w| = 1 such that z = rw. Are w and r always uniquely determined by z?

Solution: Simply let w = z/|z| and r = |z|. w is not always uniquely determined sinceif z = 0 we can choose any w ∈ C such that |w| = 1 and r = 0. r is always uniquely

5

determined since |z| = |rw| = |r||w| = |r| = r because r ≥ 0 by hypothesis. If z 6= 0 then wis uniquely determined because it must lie on the radius which z lies on, where it intersectsthe unit circle. To see this, let z = a+ bi and w = c+ di. Then, a+ bi = rc+ rdi and thereforec = a/r and d = b/r. Hence, since r is uniquely determined, so is w provided z 6= 0. �

12. If z1, . . . , zn are complex, prove that

|z1 + · · ·+ zn| ≤ |z1|+ · · ·+ |zn|.

Solution: We proceed by induction on n. For n = 1 there is nothing to show and the casen = 2 is the triangle inequality which is given by Theorem 1.33 on page 14 of the text.Suppose the result holds for n− 1. Then

|z1 + · · ·+ zn| = |z1 + (z2 + · · ·+ zn)| ≤ |z1|+ |z2 + · · ·+ zn| ≤ |z1|+ |z2|+ · · ·+ |zn|

where the last inequality follows by the inductive hypothesis applied to the term |z2 + · · ·+zn|. �

13. If x, y are complex, prove that ∣∣|x| − |y|∣∣ ≤ |x− y|.Solution: Let x, y ∈ C and let z = y − x. Then, by the triangle inequality we have that

|x+ z| ≤ |x|+ |z| =⇒ |x+ z| − |x| ≤ |z|,

hence substituting in the expression for z we obtain that

|y| − |x| ≤ |y − x|. (13.1)

Now, set z = x− y and proceeding analogously we find that

|y + z| ≤ |y|+ |z| =⇒ |y + z| − |y| ≤ |z|,

hence substituting in the expression for z we obtain that

|x| − |y| ≤ |x− y| = |y − x|,

thus combined with Equation (13.1) above we see that∣∣|x| − |y|∣∣ ≤ |x− y|,as desired.

Alternate Solution: Let x = a+ bi and y = c+diwith a, b, c, d ∈ R. By the triangle inequality(Theorem 1.33(e) on page 14 of the text) we have that

|x+ y| ≤ |x|+ |y| ⇐⇒ |x+ y|2 ≤ |x|2 + |y|2 + 2|x||y|.

Substituting our expressions for x and y and noting that by Definition 1.32 on page 14 of thetext we have that |x| = (a2 + b2)1/2 we find

a2 + c2 + 2ac+ b2 + d2 + 2bd = (a+ c)2 + (b+ d)2

= |(a+ c) + (b+ d)i|2

= |x+ y|2

≤ |x|2 + |y|2 + 2|x||y|= a2 + b2 + c2 + d2 + 2(a2 + b2)1/2(c2 + d2)1/2

6

which immediately implies that

ac+ bd ≤ (a2 + b2)1/2(c2 + d2)1/2. (13.2)

Therefore,∣∣|x| − |y|∣∣2 =∣∣(a2 + b2)1/2 − (c2 + d2)1/2

∣∣2 =[(a2 + b2)1/2 − (c2 + d2)1/2

]2= a2 + b2 + c2 + d2 − 2

((a2 + b2)1/2(c2 + d2)1/2

)≤ a2 + b2 + c2 + d2 − 2(ac+ bd)

= a2 + c2 − 2ac+ b2 + d2 − 2bd

= (a− c)2 + (b− d)2

= |(a− c) + (b− d)i|2

= |(a+ bi)− (c+ di)|2

= |x− y|2

hence we have ∣∣|x| − |y|∣∣ ≤ |x− y|.Since the triangle inequality was the fundamental trick in both solutions, we might suspectthat there is a geometric interpretation. There is and it is easy to visualize if we treat x andy as vectors in R2. Then, this merely states that the magnitude of the vector given by theirdifference is greater than or equal to the difference in their magnitudes. This makes sense ifwe consider x and y to point in opposite directions (i.e. y = −αx where α > 0). Then themagnitude of the vector given by their difference is the sum of their magnitudes which isgreater than or equal to the difference of their magnitudes because magnitudes are positiveor zero. When x and y point in the same direction, we get equality, otherwise we haveinequality. When they point in opposite directions we get “maximal inequality”. �

14. If z is a complex number such that |z| = 1, that is, such that zz = 1, compute

|1 + z|2 + |1− z|2.

Solution: Write z = a+ bi with a, b ∈ R. Then

|1 + z|2 + |1− z|2 = |(1 + a) + bi|2 + |(1− a) + bi|2 = (1 + a)2 + b2 + (1− a)2 + b2

= 1 + a2 + 2a+ b2 + 1 + a2 − 2a+ b2

= 2 + 2a2 + 2b2

= 2(a2 + b2) + 2

= 2|z|2 + 2= 4. �

15. Under what conditions does equality hold in the Schwarz inequality?

Solution: If a1, . . . , an and b1, . . . , bn are complex numbers then the Schwarz inequality statesthat ∣∣∣∣ n∑

i=1

aibi

∣∣∣∣2 ≤ n∑i=1

|ai|2n∑i=1

|bj |2.

7

We claim that equality holds if there exists λ ∈ C such that bi = λai ∀ i = 1, . . . , n. Supposingthis condition holds we have∣∣∣∣ n∑

i=1

aibi

∣∣∣∣2 =∣∣∣∣ n∑i=1

aiaiλ

∣∣∣∣2 = |λ|2∣∣∣∣ n∑i=1

|ai|2∣∣∣∣2 = |λ|2

(n∑i=1

|ai|2)(

n∑i=1

|ai|2)

=

(n∑i=1

|ai|2)(

n∑i=1

|λ|2|ai|2)

=

(n∑i=1

|ai|2)(

n∑i=1

|λai|2)

=n∑i=1

|ai|2n∑i=1

|bi|2. �

16. Suppose k ≥ 3,x,y ∈ Rk, |x− y| = d > 0, and r > 0. Prove:

(a) If 2r > d, there are infinitely many z ∈ Rk such that

|z− x| = |z− y| = r.

(b) If 2r = d, there is exactly one such z.

(c) If 2r < d, there are no such z.

How must these statements be modified if k is 2 or 1?

Solution:

(a) If z ∈ Rk such that |z− x| = |z− y| = r then z ∈ ∂B(x, r) ∩ ∂B(y, r), where ∂B(x, r) ={z ∈ Rk | |z−x| = r} is the boundary of the ball of radius r centered at x. If 2r > d =⇒r > d/2 then the two boundaries intersect in a surface (of dimension k − 2) and thusthere are infinitely many points in this intersection which satisfy the given equationbecause k ≥ 3.

(b) If r = d/2 then the two boundaries intersect in a single point, the midpoint of thesegment joining x and y.

(c) If r < d/2 then the boundaries never intersect and so no z ∈ Rk can be on both bound-aries simultaneously.

If k = 1 then there are no z ∈ R satisfying both equations simultaneously unless r = d/2.This is because there are only two points that are a distance r from x for a given x ∈ R.Without loss of generality we can assume x < y. If r = d/2, we see that z = x+ r is the onlypoint that is a distance r from both x and y. If r > d/2 or r < d/2 then the boundaries (set’sof two points) never intersect. This is because the points of distance r from x are x + r andx− r. But, y − (x+ r) = y − x− r = d− r < d− d/2 = d/2 < r so x+ r is not of distance rfrom y. Similarly y− (x− r) = y− x+ r = d+ r > r so x− r is also not of distance r from y.Analogous results hold when r < d/2 with inequalities reversed.

If k = 2 then when r > d/2 instead of infinitely many points in the intersection of theboundaries, there are just two since in this case the boundaries of the ball’s around x andy are circles and two different circles can maximally intersect in only 2 points. The casesr = d/2 and r < d/2 give the same results as they did for k ≥ 3. �

8

17. Prove that|x + y|2 + |x− y|2 = 2|x|2 + 2|y|2

if x,y ∈ Rk. Interpret this geometrically, as a statement about parallelograms.

Solution: If we write x = (x1, . . . , xk) and y = (y1, . . . , yk) then expanding the left hand sidegives

|x + y|2 + |x− y|2 =k∑i=1

(xi + yi)2 +k∑i=1

(xi − yi)2

=k∑i=1

(xi + yi)2 + (xi − yi)2

=k∑i=1

(x2i + 2xiyi + y2

i + x2i − 2xiyi + y2

i )

=k∑i=1

(2x2i + 2y2

i )

= 2k∑i=1

x2i + 2

k∑i=1

y2i

= 2|x|2 + 2|y|2.

If we consider the vectors x and y as representing two sides of a parallelogram then x− yand x + y represent the diagonals. The equation above is then seen to be nothing more thenthe generalized Pythagorean Theorem. Considering the case k = 2 the equation is merely thesum of the ordinary Pythagorean Theorem for the two diagonals. Since for both diagonalsthe sides of the triangle for which they are the hypotenuse are x and y we have that

|x + y|2 = |x|2 + |y|2

|x− y|2 = |x|2 + |y|2

hence our original equation is just the sum of these two. �

18. If k ≥ 2 and x ∈ Rk, prove that there exists y ∈ Rk such that y 6= 0 but x · y = 0. Is this alsotrue if k = 1?

Solution: We can see right away that the statement must fail for k = 1 because in this casethe dot product is nothing more than multiplication of real numbers and since the reals area field xy = 0 =⇒ x = 0 or y = 0, i.e. fields have no zero divisors. This means there is nonotion of perpendicularity in a 1-dimensional space as we’d expect since perpendicularityimplies linear independence and hence extra dimension.

For k ≥ 2 let x = (x1, . . . , xk) and without loss of generality suppose x 6= 0 because other-wise any non-zero y ∈ Rk will suffice. We consider two cases based on the parity of k. If k iseven then we let y = (x2,−x1, x4,−x3, . . . , xj+1,−xj , . . . , xk,−xk−1) where xj+1 is in the jth

slot. We then have

x · y = x1x2 − x2x1 + · · ·+ xjxj+1 − xj+1xj + · · ·+ xk−1xk − xkxk−1 = 0

and y 6= 0 because x 6= 0 by hypothesis. Now, suppose that k is odd. Since x 6= 0 by hypothe-sis we have that xj 6= 0 for some 1 ≤ j ≤ k. Supposing that xj is not the only non-zero entry,

9

we repeat the same argument and choose y = (x2,−x1, . . . , xj+1, 0,−xj−1, . . . , xk,−xk−1)where the 0 is in the jth slot and y 6= 0 because we have supposed that xj is not the onlynon-zero entry. Then we have

x · y = x1x2 − x2x1 + · · ·+ xj−1xj+1 + 0− xj+1xj−1 + · · ·+ xk−1xk − xkxk−1 = 0.

If xj is the only non-zero entry then we simply choose y = (1, . . . , 1, 0, 1, . . . , 1) where the 0is in the jth slot. �

Chapter 2: Basic Topology

1. Prove that the empty set is a subset of every set.

Solution: Let A be a set and let ∅ be the empty set. To show ∅ ⊂ A we need to show that∀ p ∈ ∅, p ∈ A. Since there are no p ∈ ∅, this statement is vacuously true and therefore ∅ ⊂ A.�

2. A complex number z is said to be algebraic if there are integers a0, . . . , an not all zero suchthat

a0zn + a1z

n−1 + · · ·+ an−1z + an = 0.

Prove that the set of all algebraic numbers is countable.

Solution: Let

Pn ={ n∑i=0

aiti

∣∣∣∣ ai ∈ Z, i = 0, . . . , n}

so Pn is the set of all degree n or smaller polynomials with integer coefficients. It is clear that

Pn ∼ Zn+1 given byn∑i=0

aiti 7→ (a0, . . . , an)

where ∼means bijective as sets, so that in particular, Pn is countable for each n. Now, let

Rn = {a ∈ C | p(a) = 0 for some p ∈ Pn}

so that Rn is the set of all complex roots of all degree n or smaller polynomials with integercoefficients. Since Pn is countable we have that Pn ∼ N so that in particular to each poly-nomial in Pn we can associate a unique natural number. Since every degree n or smallerpolynomial has at most n complex roots we see that

|Rn| ≤∣∣∣∣ ∞⋃m=0

{1m, . . . , nm}∣∣∣∣

where m ∈ N is the index for the elements of the countable set Pn and the set {1m, . . . , nm}is a set of n elements for each m representing the maximum number of roots of any degreen or smaller polynomial. Since the set above on the right is a countable union of countablesets it is also countable by Theorem 2.12 on page 20 of the text and so we have that Rn iscountable. But since the set of all complex roots of all polynomials with integer coefficientsis simply ∪∞n=0Rn, it is also countable by the same Theorem. �

10

3. Prove that there exist real numbers which are not algebraic.

Solution: We see that the set of all real algebraic numbers is at most countable because it isa subset of the set of all complex algebraic numbers which is countable by Exercise 3 above.But, since R is uncountable (see Theorem 2.14 on page 30 of the text) there must exist realnumbers which are not algebraic. �

4. Is the set of all irrational real numbers countable?

Solution: We have by the Corollary to Theorem 2.13 on pages 29-30 of the text that Q iscountable. Since R is uncountable (see Theorem 2.14 on page 30 of the text) we must havethat the irrationals are uncountable because the union of two countable sets is countable byTheorem 2.12 on page 29 of the text. �

5. Construct a bounded set of real numbers with exactly three limit points.

Solution: Consider the set S = {1/n | n ∈ N} ∪ {1 + 1/n | n ∈ N} ∪ {2 + 1/n | n ∈ N}. ThenS is bounded because d(p, 0) < 3 ∀ p ∈ S. It is also clear that 0, 1, and 2 are limit points ofS. No other points are limit points of S since around any other real number we can choosea small enough neighborhood such that it excludes all the points of S, except possibly itself,since we can always choose a radius smaller then 1

n −1

n+1 for any fixed n. �

6. Let E′ be the set of all limit points of a set E. Prove that E′ is closed. Prove that E and Ehave the same limit points. (Recall that E = E ∪ E′). Do E and E′ always have the samelimit points?

Solution: Let p be a limit point of E′. Thus, for each n ∈ N there exists pn ∈ E′ such thatpn 6= p and d(p, pn) < 1

2n . Since each pn ∈ E′ we have that for each pn there exists a pointsn ∈ E such that sn 6= pn and d(pn, sn) < min{ 1

2n , d(pn, p)}. Since d(pn, sn) < d(pn, p) wehave that sn 6= p. By the triangle inequality we then have that

d(p, sn) ≤ d(p, pn) + d(pn, sn) <1

2n+

12n

=1n.

Thus, every neighborhood of p contains a point in E not equal to p itself so p is a limit pointof E and hence p ∈ E′. Therefore, E′ is closed.

Now, let p be a limit point of E. Then any neighborhood of p contains a point q ∈ E suchthat q 6= p. Since E ⊂ E we have that q ∈ E so that in particular p is also a limit point of E.Now, suppose that p is a limit point of E. Then, for each n ∈ N there exists a point pn ∈ Esuch that pn 6= p and d(p, pn) < 1

2n . If pn /∈ E then pn ∈ E′ because pn ∈ E and we replacepn with an element qn ∈ E such that qn 6= pn and d(pn, qn) < min{ 1

2n , d(pn, p)}. Then qn 6= pbecause d(pn, qn) < d(pn, p) and d(p, qn) ≤ d(p, pn) + d(pn, qn) < 1

2n + 12n = 1

n . Therefore, p isalso a limit point of E and so E and E have the same limit points.

We can see that E and E′ do not always have the same limit points by the simple exampleE = (0, 1). Here, the limit points of E are 0 and 1 so that E′ = {0, 1}. But, E′ itself has nolimit points by the Corollary to Theorem 2.20 on pages 32-33 of the text because it is a finitepoint set. We can see this directly because if s 6= 0, 1 then any neighborhood of s of radiusr < min{d(s, 0), d(s, 1)} contains no elements of E′ and any neighborhood of 0 or 1 of radiusless than 1 contains no other points of E′ except 0 and 1, respectively, hence E′ has no limitpoints. �

7. Let A1, A2, . . . be subsets of a metric space.

11

(a) If Bn =⋃ni=1Ai, prove that Bn =

⋃ni=1Ai, for n = 1, 2, . . .

(b) If B =⋃∞i=1Ai, prove that B ⊃

⋃∞i=1Ai.

Show, by an example, that the inclusion in (b) can be proper.

Solution:

(a) We use the notation B′ for the set of limit points of B. For (a), suppose that p ∈ Bn.Then, either p ∈ Bn or p ∈ B′n (or both). If p ∈ Bn =

⋃ni=1Ai =⇒ p ∈ Ai for some i

hence p ∈ Ai =⇒ p ∈⋃ni=1Ai. So, suppose that p ∈ B′n so that p is a limit point of Bn.

Now, suppose that p /∈ Ai ∀ i = 1, . . . , n, i.e. that p is not a limit point for any of theAi’s. This means that for each i ∃ εi > 0 such that N(p, εi) ∩Ai ⊂ {p}. Here N(p, εi) isthe ball around p of radius εi. That is, there is some non-zero neighborhood of p whichintersects Ai in at most p itself. Letting 0 < ε < mini{εi} we see that N(p, ε) ∩ Ai ⊂{p} ∀ i = 1, . . . , n =⇒ N(p, ε)∩∪ni=1Ai = Bn ⊂ {p} contradicting that p is a limit pointof Bn. Therefore, we must have that p ∈ Ai for at least one i hence p ∈ ∪ni=1Ai and weobtain the inclusion Bn ⊆ ∪ni=1Ai. Now, let p ∈ ∪ni=1Ai. Then we have that p ∈ Ai forsome i. If p ∈ Ai ⊂ Bn then p ∈ Bn. If p ∈ A′i then every neighborhood of p contains apoint q ∈ Ai such that q 6= p. Since Ai ⊂ Bn, we have that q ∈ Bn also and hence p ∈ Bn

showing the other inclusion ∪ni=1Ai ⊆ Bn and completing part (a).

(b) The proof is identical to the second inclusion just shown in (a).

As an example, if we let Ai = {1/i} and B = ∪∞i=1Ai we see that 0 ∈ B′ yet 0 /∈ A′i for any ibecause for each i we can choose 0 < ε < 1/i and then N(0, ε)∩Ai = ∅. Hence the inclusionin part (b) is proper. �

8. Is every point of every open set E ⊂ R2 a limit point of E? Answer the same question forclosed sets in R2.

Solution: Let p ∈ E and let δ > 0 be such that B(p, δ) ⊂ E so that in particular there existsq ∈ B(p, δ) such that q 6= p and of course q ∈ E because B(p, δ) ⊂ E. For example, ifp = (p1, p2) we can choose q = (p1 + δ/2, p2). Now, let ε > 0 be given. If ε = δ then byconstruction B(p, ε) contains a point of E distinct from p. If ε < δ then B(p, ε) ⊂ B(p, δ) ⊂ Eand so p 6= q = (p1 + ε/2, p2) ∈ B(p, ε) ⊂ E so that B(p, ε) contains a point of E distinct fromp. If ε > δ thenB(p, δ) ⊂ B(p, ε) and so choosing q = (p1 +δ/2, p2) ∈ B(p, δ) ⊂ B(p, ε) showsthat B(p, ε) contains a point in E distinct from p. Thus every neighborhood of p contains apoint of E distinct from p hence p is a limit point of E and so every point of E is a limit pointof E. The same is not true of closed sets. Consider A = {(n, 0) | n ∈ Z} ⊂ R2. Then A isclosed because A contains no limit points at all. This is because any neighborhood of anypoint in A with radius less than 1 contains no other points of A. Yet, since A is not empty, itcontains points which are not limit points. �

9. Let E◦ denote the set of all interior points of a set E. [See Definition 2.18(e) on page 32 of thetext; E◦ is called the interior of E.]

(a) Prove that E◦ is always open.

(b) Prove that E is open if and only if E = E◦.

(c) If G ⊂ E and G is open, prove that G ⊂ E◦.(d) Prove that the complement of E◦ is the closure of the complement of E.

(e) Do E and E always have the same interiors?

12

(f) Do E and E◦ always have the same closures?

Solution:

(a) Let p ∈ E◦. Then by definition there exists δ > 0 such that N(p, δ) ⊂ E. By Theorem2.19 on page 32 of the text we know that every neighborhood is open and thus if q ∈N(p, δ) there exists ε > 0 such that N(q, ε) ⊂ N(p, δ) ⊂ E. Therefore q ∈ E◦ and soN(p, δ) ⊂ E◦ and hence E◦ is open.

(b) If E = E◦ then E is open by part (a). Now, suppose that E is open. If p ∈ E then thereexists δ > 0 such that N(p, δ) ⊂ E and therefore p ∈ E◦ by definition. Thus we havethe inclusion E ⊂ E◦. By definition, interior points of E are elements of E because anyneighborhood of a point contains that point. Therefore E◦ ⊂ E and we obtain E = E◦.

(c) Now, let p ∈ G ⊂ E. If G is open, there exists δ > 0 such that N(p, δ) ⊂ G ⊂ E. Thus,p ∈ E◦ by definition and we obtain G ⊂ E◦, showing part (c).

(d) The result follows from

p ∈ (E◦)c ⇐⇒ ∀ ε > 0, N(p, ε) ∩ Ec 6= ∅⇐⇒ p ∈ Ec or ∀ ε > 0, ∃ q ∈ N(p, ε) ∩ Ec such that q 6= p

⇐⇒ p ∈ Ec or p ∈ (Ec)′

⇐⇒ p ∈ Ec.

(e) Consider Q, which has no interior points because any neighborhood of a rational mustcontain irrationals because irrationals are dense in R, hence any such neighborhoodcannot be contained entirely in Q. Thus Q◦ = ∅. But, since Q = R (because the limitpoints of Q are precisely irrationals) we see that (Q)◦ = R◦ = R (because R is open initself) hence Q◦ 6= (Q)◦.

(f) The same example of Q shows that (f) is also false. Q = R 6= ∅ = ∅ = Q◦. �

10. Let X be an infinite set. For p, q ∈ X , define

d(p, q) ={

1 if p 6= q0 if p = q

.

Prove that this is a metric. Which subsets of the resulting metric space are open? Which areclosed? Which are compact?

Solution: To show it is a metric, only the triangle inequality is not obvious from the defini-tion. So, let p, q, r ∈ X . Then if p = q we have that d(p, q) = 0 ≤ d(p, r) + d(r, q) becaused(p, r), d(r, q) ≥ 0. If p 6= q we have that d(p, q) = 1 ≤ d(p, r) + d(r, q) because if r = q wehave that r 6= p and then d(p, r) = 1 and d(r, q) = 0 so the inequality holds. If r = p thenwe similarly get that r 6= q so d(p, r) = 0 and d(r, q) = 1 so the inequality still holds. If wefinally have that r 6= p, q then both d(p, r) = 1 = d(p, q) and the inequality holds. Thus, thisis a metric. First, note that N(p, r) = X if r ≥ 1 for any p ∈ X and N(p, r) = {p} if r < 1for any p ∈ X . Now, let ∅ 6= A ⊂ X and let p ∈ A. Then N(p, 0.5) = {p} ⊂ A thus everypoint in A has a neighborhood entirely contained in A, thus A is open and so every subsetof X is open. Since a set is closed if and only if its complement is open by Theorem 2.23 onpage 34 of the text, we see that every subset of X is also closed. It is clear from the definitionthat any finite subset of X will be compact. But, if A ⊂ X is infinite we see that it cannotbe compact since by Theorem 2.37 on page 38 of the text an infinite subset of a compact set

13

has a limit point in the compact set, yet from this metric we see that there are no limit pointsfor any sets. This is because any neighborhood of any point of radius less then 1 contains noother points. Hence, the only compact subsets of X are finite ones. �

11. For x ∈ R and y ∈ R, define

d1(x, y) = (x− y)2

d2(x, y) =√|x− y|

d3(x, y) = |x2 − y2|d4(x, y) = |x− 2y|

d5(x, y) =|x− y|

1 + |x− y|.


Solution: For each of these we must determine whether di(x, y) > 0 for x 6= y, di(x, x) =0 ∀ x ∈ R, di(x, y) = di(y, x), and di(x, y) ≤ d(x, z) + d(z, y).

d1(x, y) is not a metric because it does not satisfy the triangle inequality since d1(0, 2) =4, d1(0, 1) = 1, and d1(1, 2) = 1 thus 4 = d1(0, 2) > d1(0, 1) + d1(1, 2) = 1 + 1 = 2.

d2(x, y) is a metric from the following: d2(x, y) > 0 if x 6= y and d2(x, x) = 0 for all x ∈ R.Also, d2(x, y) =

√|x− y| =

√|y − x| = d2(y, x). Finally, since |x− y| ≤ |x− z|+ |z − y| we

have that√|x− y| ≤

√|x− z|+ |z − y| ≤

√|x− z| +

√|y − z|. The last inequality follows

since a+ b ≤ a+ b+ 2√ab = (

√a+√b)2 =⇒

√a+ b ≤

√a+√b.

d3(x, y) is not a metric since d(1,−1) = 0.

d4(x, y) is not a metric since d4(1, 1) = 1 6= 0.

To show that d5(x, y) is a metric we prove the more general result that whenever d(x, y) is ametric then

d′(x, y) =d(x, y)

1 + d(x, y)

is also be a metric. Since d(x, y) = |x − y| is the ordinary metric on R, it will follow thatd5(x, y) is a metric. Because d(x, y) is a metric we have that d′(x, y) > 0 if x 6= y, d′(x, x) = 0for all x ∈ R, and d′(x, y) = d′(y, x). Now, let p = d(x, y), q = d(x, z), r = d(z, y). Since d(x, y)is a metric we have that p, q, r ≥ 0 and

p ≤ q + r =⇒ p ≤ q + r + 2qr + pqr

=⇒ p+ pq + pr + pqr ≤ (q + pq + qr + pqr) + (r + pr + qr + pqr)=⇒ p(1 + q)(1 + r) ≤ q(1 + r)(1 + p) + r(1 + q)(1 + p)

=⇒ p

1 + p≤ q

1 + q+

r

1 + r

=⇒ d′(x, y) ≤ d′(x, z) + d′(z, y).

Thus, d′(x, y) is a metric. �

12. Let K ⊂ R consist of 0 and the numbers 1/n, for n = 1, 2, 3, . . . Prove that K is compactdirectly from the definition (without using the Heine-Borel theorem).

Solution: Let {Uα} be an open cover of K. Let 0 ∈ Uα0 . Since Uα0 is open there existsδ > 0 such that B(0, δ) ⊂ Uα0 hence we can choose any n such that 1/n < δ and then

14

1/n ∈ B(0, δ) ⊂ Uα0 . Thus, we also have that 1/m ∈ B(0, δ) ⊂ Uα0 for all m > n. Then,let 1/i ∈ Uαi for i = n − 1, n − 2, . . . , 1. Since there are only finitely many i’s we have that{Uαj}n−1

j=1 ∪ Uα0 is a finite subcover. �

13. Construct a compact set of real numbers whose limit points form a countable set.

Solution: Consider the set

Am = {m+ 1/n | n = 1, 2, . . .}.

Then Am is compact for all m ∈ N by Exercise 12 above (it is merely a translation of the setin that exercise) and the set of limit points of Am is {m}. Therefore, if we let

A =∞⋃i=1

Ai

A is a countable union of countable sets so it is countable by Theorem 2.12 on page 29 of thetext. The set of limit points of A is then precisely N so is countable also. �

14. Give an example of an open cover of the segment (0, 1) which has no finite subcover.

Solution: Consider the open cover {(1/n, 1)}∞n=2. This covers (0, 1), because if r ∈ (0, 1), letn ≥ 2 be such that 1/n < r. Then r ∈ (1/n, 1). Now, consider any finite collection in thisopen cover {(1/i, 1)}i∈I where |I| <∞. Let n ≥ 2 be such that n > maxi∈I{i}. Then we havethat 1/n ∈ (0, 1) yet 1/n < 1/i ∀ i ∈ I and thus 1/n /∈ ∪i∈I(1/i, 1) hence {(1/i, 1)}i∈I is notan open cover of (0, 1). Therefore, no finite subcover of this cover exists. �

15. Show that Theorem 2.36 and its Corollary become false (in R, for example) if the word“compact” is replaced by “closed” or by “bounded”.

Solution: Theorem 2.36, on page 38 of the text, states that If {Kα} is a collection of com-pact subsets of a metric space X such that the intersection of every finite subcollection of{Kα} is nonempty, then

⋂Kα is nonempty. Its Corollary states that if {Kn} is a sequence of

nonempty compact sets such that Kn ⊃ Kn+1(n = 1, 2, . . .), then⋂∞n=1Kn is nonempty.

Consider the sets Kn = (0, 1/n). Then each Kn is bounded and for any finite subcollection{Ki}i∈I where |I| < ∞ let n = mini∈I{i}. Then

⋂i∈I Ki = (0, 1/n) 6= ∅. Yet we have

that⋂∞n=1Kn = ∅ since if 0 < r ∈ R we can choose n ∈ N such that 1/n < r. But then

r /∈ (0, 1/n) =⇒ r /∈⋂∞n=1Kn.

Now, let Kn = {m ∈ N |m ≥ n}. Then each Kn is closed since it has no limit points (becauseit is discrete), so it vacuously contains them. Then, if {Ki}i∈I is a finite subcollection where|I| <∞, let m ∈ N such that m > maxi∈I{i}. Then, m ∈ Ki ∀ i ∈ I and so m ∈

⋂i∈I Ki. But,

we have that⋂∞n=1Kn = ∅ because if m ∈ N, let i ∈ N be such that i > m. Then m /∈ Ki and

so m /∈⋂∞n=1Kn. �

16. Omitted.

17. Omitted.

18. Omitted.

19. Omitted.

20. Omitted.

15

21. Omitted.

22. A metric space is called separable if it contains a countable dense subset. Show that Rk isseparable. Hint: Consider the set of points which have only rational coordinate.

Solution: Let Qk = {(x1, . . . , xk) ∈ Rk | xi ∈ Q ∀ i = 1, . . . , k}. If (x1, . . . , xk) ∈ Rk let ε > 0be given. Then, since Q is dense in R, choose p1, . . . , pn ∈ Q such that |xi − pi| < 1√

kε. Then,

writing p = (p1, . . . , pn) ∈ Qk and x = (x1, . . . , xn) ∈ Rk we see that

d(p,x) =( k∑i=1

|xi − pi|2)1/2

<

( k∑i=1

1kε2)1/2

= ε.

Therefore, Qk is dense in Rk and Qk is countable because it can be realized as the disjointunion of Q, k times. �

23. A collection {Vα} of open subsets of X is said to be a base if the following is true: For everyx ∈ X and every open set G ⊂ X such that x ∈ G, we have x ∈ Vα ⊂ G for some α. Inother words, every open set in X is the union of a subcollection of {Vα}. Prove that everyseparable metric space has a countable base. Hint: Take all neighborhoods with rationalradius and center in some countable dense subset of X .

Solution: Since X is separable, by definition it has a countable dense subset S ⊂ X . LetS = {p1, p2, . . .} and enumerate Q = {q1, q2, . . .}. Let Vij = N(pi, qj). Then, {Vij}∞i,j=1 is acountable collection of open sets in X by Theorem 2.12 on page 29 of the text. Let x ∈ Xand let G ⊂ X be open such that x ∈ G. Since G is open, there exists δ > 0 such thatN(x, δ) ⊂ G. Let pi ∈ S such that d(pi, x) = ε < δ/2 and then choose qj ∈ Q such thatε < qj < δ/2, both of which are possible because S is dense in X and Q is dense in R. Then,consider Vij = N(pi, qj). Since d(x, pi) = ε < qj we see that x ∈ Vij . Now, let a ∈ Vij . Then,d(a, x) ≤ d(a, pi) + d(pi, x) < qj + δ/2 < δ. Therefore a ∈ N(x, δ) ⊂ G and so we have thatx ∈ Vij ⊂ N(x, δ) ⊂ G showing that {Vij}∞i,j=1 is a countable base. �

Chapter 3: Numerical Sequences and Series

1. Prove that convergence of {sn} implies convergence of {|sn|}. Is the converse true?

Solution: Suppose that sn → s. Let ε > 0 be given and let N be such that

|sn − s| < ε ∀ n ≥ N.

Then, by Exercise #13 in Chapter 1 above, we have that∣∣|sn| − |s|∣∣ ≤ |sn − s| < ε ∀ n ≥ N

Thus |sn| → |s|. The converse is false because if we let sn = (−1)n then |sn| converges, yetsn does not. �

2. Omitted.

3. If s1 =√

2 andsn+1 =

√2 +√sn

prove that {sn} converges, and that sn < 2 for all n ≥ 1.

16

Solution: s1 =√

2 < 2. Suppose inductively that sn < 2. Then,

sn+1 =√

2 +√sn <

√2 +√

2 < 2

because 2 +√

2 < 2 + 2 = 4 =⇒√

2 +√

2 <√

4 = 2. Thus we see that sn < 2 for all n ≥ 1.We also have that

s2 =

√2 +

√√2 >√

2 = s1.

Suppose inductively that sn > sn−1. Then

sn+1 =√

2 +√sn >

√2 +√sn−1 = sn

and therefore we have that sn > sn−1 for all n ≥ 1 by induction. Thus the sequence isincreasing and since · · · > sn > sn−1 > · · · > s1 > 0 we have an increasing sequence ofnonnegative terms which is bounded above. Hence by Theorem 3.24 on page 60 of the text{sn} converges. �

4. Find the upper and lower limits of the sequence {sn} defined by

s1 = 0, s2m =s2m−1

2, s2m+1 =

12

+ s2m.

Solution: If we write out the first few terms we obtain

s1 = 0s2 = 0

s3 =12

s4 =14

s5 =12

+14

s6 =14

+18

s7 =12

+14

+18

s8 =14

+18

+116

thus we are lead to write

s2m+1 =m∑k=1

(12

)km ≥ 1

s2m =m∑k=2

(12

)km ≥ 2.

(4.1)

To prove this we proceed by induction onm. Form = 1 we have that s3 = 1/2 =∑1

k=1(1/2)k.For m = 2 we have that s5 = 1/2 + 1/4 =

∑2k=1(1/2)k and s4 = 1/4 =

∑2k=2(1/2)k. Now,

17

suppose inductively that Equation (4.1) holds for m ≥ 2. Then,

s2(m+1)+1 =12

+ s2(m+1) =12

+s2m+1

2=

12

+m∑k=1

(12

)k+1

=m+1∑k=1

(12

)k

s2(m+1) =s2m+1

2=

12

m∑k=1

(12

)k=

m∑k=1

(12

)k+1

=m+1∑k=2

(12

)kand so Equation (4.1) holds. We thus see that there are only two subsequential limits

limm→∞

s2m =∞∑k=2

(12

)k=

12

(4.2)

limm→∞

s2m+1 =∞∑k=1

(12

)k= 1 (4.3)

where the two limiting values are given by Theorem 3.26 on page 61 of the text (geometricseries). These are the only possible limits because if we have a subsequence {snk

} theneither {nk} has infinitely many evens, infinitely many odds, or infinitely many of both. Ifthere are infinitely many of both even and odd numbers, then {snk

} wouldn’t be Cauchysequence because for all N ≥ 1 we could find nk1 , nk2 ≥ N such that nk1 was even andnk2 was odd and then we would have that |snk1

− snk2| ≥ 1/2. If there are only finitely

many odds in {nk} then we would have that limk→∞ snk= limm→∞ s2m. Specifically, letting

N > maxk{nk odd } we see that |snk− snj | =

∑nkk=nj

(1/2)k for all nk ≥ nj ≥ N preciselybecause both nj and nk are even. Since

∑∞k=2(1/2)k converges, it is a Cauchy sequence by

Theorem 3.11 on page 53 of the text and so∑nk

k=nj(1/2)k can be made arbitrarily small for

large enough nj and nk. If there are only finitely many evens in {nk} then we would havethat limk→∞ snk

= limm→∞ s2m+1. Specifically, letting N > maxk{nk even } we see that|snk− snj | =

∑nkk=nj

(1/2)k for all nk ≥ nj ≥ N precisely because both nj and nk are odd.Since

∑∞k=1(1/2)k converges, it is a Cauchy sequence by Theorem 3.11 on page 53 of the

text and so∑nk

k=nj(1/2)k can be made arbitrarily small for large enough nj and nk. Since

Equations (4.2) and (4.3) are the only two subsequential limits we find that the upper andlower limits are

lim infm→∞

sm =12

lim supm→∞

sm = 1.

�

5. For any two real sequences {an}, {bn}, prove that

lim supn→∞


(an) + lim supn→∞

(bn),

provided the sum on the right is not of the form∞−∞.

Solution: If either lim sup on the right side above is infinite, we are done, so we can assumethat both are finite. Assume also that the left side is finite (hence our discussion here isincomplete). We make some general remarks about this statement, but we will prove it by

18

actually using an equivalent definition of lim sup which is nonetheless easier to work withthan Rudin’s. First, let a = lim supn→∞(an), b = lim supn→∞(bn), and c = lim supn→∞(an +bn). Then, let ank

+ bnk→ γ be a convergent subsequence of the sequence {an + bn}. If the

component subsequences {ank} and {bnk

} both converge, say to α and β, respectively, thenby Theorem 3.3 on page 49 of the text, we have that

γ = limk→∞

(ank+ bnk

) = limk→∞

ank+ limk→∞

bnk= α+ β ≤ a+ b.

where the last inequality follows by the definition of a and b as lim sup’s. Thus, we need onlyconsider subsequences {ank

+bnk}where one, or both, of the the component subsequences do

not converge. But, we can actually do better than this because we can show that if one of thecomponent subsequences converges, then both must. To see this, without loss of generality,let ank

+ bnk→ γ and suppose that bnk

→ β. Then, by the triangle inequality,

|ank− (γ − β)| ≤ |ank

− (γ − bnk)|+ |(γ − bnk

)− (γ − β)|= |(ank

+ bnk)− γ|+ |(γ − bnk

)− (γ − β)|.

Since bnk→ β we see that γ − bnk

→ γ − β because γ is a constant and so both terms onthe right above can be made arbitrarily small for large enough k. This shows that ank

→γ − β and therefore if one of the component subsequences of the convergent subsequence{ank

+ bnkconverges, both must. Hence, we are reduced to considering only those conver-

gent subsequences {ank+ bnk

} in which both component subsequences do not converge. Thatis, sequences such as ank

= (−1)k and bnk= (−1)k+1 so that neither converges individually,

but their sum does. Now, we would be done if we could show that γ = limk→∞(ank+ bnk

) =lim supk→∞(ank

+ bnk) ≤ lim supk→∞(ank

) + lim supk→∞(bnk) ≤ a+ b which is the statement

of the problem, but restricted to the subsequence {ank+ bnk

}. Thus, we could say repeatthe above argument and for every convergent sub-subsequence ankj

+ bnkjsuch that nei-

ther component sub-subsequence converges, repeat the argument again. Eventually, it willend because eventually some sub-· · · -subsequence will have only convergent componentsequences, since we are assuming that a, b, and c are finite. But, this is hard to make rigorousand so we will now prove the result by resorting to a different, but equivalent, definition ofthe lim sup.

Consider a sequence {an} and let Ak = sup{am |m ≥ k} and define A = limk→∞Ak (allow-ing it to be ±∞). Now, choose n1 such that |an1 − A1| < 1, which is possible by the defini-tion of A1 as the sup, so n1 ≥ 1. Now, choose ank

inductively such that |ank− Ak| < 1/k

and such that ank> ank−1

. That is, by definition of Ak, we have that there exists K suchthat |am − Ak| < 1/k for all m ≥ K. So, simply require that nk > max{K,nk−1}. Then,this defines a subsequence of {an} which by construction converges to A, thus we havethat lim supn→∞(an) ≥ A. Now, by Theorem 3.17(a) on page 56 of the text we see thatlim supn→∞(an) is actually a subsequential limit of the sequence {an} hence there exists somesubsequence ank

→ lim supn→∞(an). Since by construction we have that ank≤ Ank

takingthe limit as k → ∞ of both sides gives lim supn→∞(an) ≤ A thus A = lim supn→∞(an) andour A is an equivalent formulation of the lim sup to Rudin’s. The proof of the above inequal-ity is now nearly trivial since if ank

+ bn+k is any subsequence of {an+ bn}whose limit existsthen we of course have that ank

+ bnk≤ supm≥k{anm}+ supm≥k{bnm} = Ank

+Bnk. Taking

the limit of both sides gives limk→∞(ank+bnk

) ≤ A+B, whereB is defined analogously toAabove. Since {ank

+bnk}was any subsequence with a limit we see that lim supn→∞(an+bn) ≤

A+B = lim supn→∞(an) + lim supn→∞(bn). �

19

6. Omitted.

7. Prove that the convergence of∑∞

n=1 an implies the convergence of

∞∑n=1

√ann

,

if an ≥ 0.

Solution: By the Cauchy-Schwarz inequality (Theorem 1.35 on page 15 of the text), we havethat

n∑j=1

ajbj ≤( n∑j=1

|aj |2)( n∑

j=1

|bj |2).

Thus, for each n = 1, 2, . . . we have

n∑j=1

√an ·

1n≤( n∑j=1

an

)( n∑j=1

1n2

)≤( ∞∑j=1

an

)( ∞∑j=1

1n2

)<∞

where the second inequality follows because all terms are positive and the last inequalityfollows because both sequences on the right are convergent. Thus the partial sums, sn =∑n

i=1

√an · 1

n , form a bounded sequence. Since all the terms are nonnegative, by Theorem3.24 on page 60 of the text the series

∑∞i=1

√an · 1

n converges. �

8. If∑∞

n=1 an converges, and if {bn}∞n=1 is monotonic and bounded, prove that∑∞

n=1 anbn con-verges.

Solution: Since∑∞

n=1 an converges, its partial sums form a bounded sequence by defini-tion. By Theorem 3.14 on page 55 of the text, we see that {bn}∞n=1 converges because it ismonotonic and bounded, so let bn → b. Since {bn}∞n=1 is monotonic, it is either increasingor decreasing. If it is increasing, define cn = b − bn and in this case the sequence {cn}∞n=1 isdecreasing and we have that c1 ≥ c2 ≥ · · · and also cn → 0 because bn → b. On the otherhand, if {bn}∞n=1 is decreasing then let cn = bn−b and in this case {cn}∞n=1 is still a decreasingsequence so c1 ≥ c2 ≥ · · · and we still have that cn → 0. Thus, in either case, by Theorem3.42 on page 70 of the text, we have that the sequence

∑∞n=1 ancn converges. First, suppose

that {bn}∞n=1 was increasing and so cn = b− bn. Then, the partial sums

Cm =m∑n=1

ancn =m∑n=1

(anb− anbn) = bm∑n=1

an −m∑n=1

anbn

are a convergent sequence, say Cm → c. But then

m∑n=1

anbn → b

∞∑n=1

an − c,

hence∑∞

n=1 anbn converges. In the other case we would obtain

m∑n=1

anbn → c′ + b

∞∑n=1

an

where here c′ =∑∞

n=1 ancn and cn = bn − b and so∑∞

n=1 anbn still converges. �

20

9. Omitted.

10. Omitted.

11. Suppose that an > 0 and sn = a1 + · · ·+ an, and∑∞

n=1 an diverges.

(a) Prove that∑∞

n=an

1+andiverges.

(b) Prove thataN+1

sN+1+ · · ·+ aN+k

sN+k≥ 1− sN

sN+k

and deduce that∑∞

n=1ansn

diverges.

(c) Prove thatans2n≥ 1sn−1

− 1sn

and deduce that∑∞

n=1ans2n

converges.

(d) What can be said about

∞∑n=1

an1 + nan

and∞∑n=1

an1 + n2an

?

Solution:

(a) Suppose that∑∞

n=1an

1+anconverges. Observe that

an1 + an

→ 0 ⇐⇒ 11an

+ 1→ 0 ⇐⇒ 1

an→∞ ⇐⇒ an → 0.

Since we are supposing that∑∞

n=1an

1+anconverges we have by Theorem 3.23 on page

60 0f the text that an1+an

→ 0 =⇒ an → 0 by the above. Hence, there exists N1 such thatan < 1 for all n ≥ N1. Now, let ε > 0 be given. Then there exists N2 such that

am1 + am

+ · · ·+ an1 + an

<ε

2∀ n,m ≥ N2.

Letting N = max{N1, N2}we have that

ε

2>

am1 + am

+ · · ·+ an1 + an

>am

1 + 1+ · · ·+ an

1 + 1=am2

+ · · ·+ an2

for all n,m ≥ N thus

ε > am + · · ·+ an ∀ n,m ≥ N =⇒∞∑n=1

an converges,

a contradiction. Thus∑∞

n=1an

1+andiverges.

(b) Because an > 0 for all n we have that sn+1 > sn for all n. Thus,

aN+1

sN+1+ · · ·+ aN+k

sN+k>aN+1 + · · ·+ aN+k

sN+k=sN+k − sNsN+k

= 1− sNsN+k

.

21

Suppose now that∑∞

n=1ansn

converges and let ε > 0 be given. Then, there exists an Nsuch that for all n,m ≥ N we have that

ε >amsm

+ · · ·+ ansn.

In particular, for m = N + 1 and n = N + k we obtain that

ε >aN+1

sN+1+ · · ·+ aN+k

sN+k> 1− sN

sN+k.

Now, since∑∞

n=1 an diverges and an > 0 so the sn are increasing, we see that sN+k →∞as k → ∞. Thus we can choose k large enough such that sN+k > 2sN =⇒ sN

sN+k< 1

2

because N is fixed. But then, we would obtain

ε > 1− sNsN+k

> 1− 12

=12,

contradicting that ε can be chosen arbitrary, i.e. just choose ε = 12 at the beginning.

Thus,∑∞

n=1ansn

diverges.

(c) As in part (b), since an > 0 for all n we have that sn > sn−1 for all n. Thus

1sn−1

− 1sn

=sn − sn−1

snsn−1>sn − sn−1

s2n=ans2n

∀ n ≥ 2.

Then since sn > 0 for all n,

k∑n=2

ans2n

<k∑

n=2

(1

sn−1− 1sn

)=

1s1− 1sk

<1s1

=1a1, (11.1)

thusk∑

n=1

ans2n

<2a1

so the partial sums of∑∞

n=1ans2n

form a bounded sequence and since an > 0 for all n,we have that

∑∞n=1

ans2n

converges by Theorem 3.24 on page 60 of the text. We could alsotake the limit of both sides of Equation (11.1) as k →∞ and note that 1

sk→ 0 as k →∞

because∑∞

n=1 an diverges and an > 0 for all n so sn →∞.

(d)∑∞

n=1an

1+nanmay either diverge or converge. If we let an = 1

n then

∞∑n=1

an1 + nan

=∞∑n=1

1/n2

=12

∞∑n=1

1n

=∞.

On the other hand, if we let an = 1n log(n)p where p > 1 and n ≥ 2 then

an1 + nan

=1

n log(n)p· 1

1 + n(

1n log(n)p

) =1

n log(n)p(1 + log(n)−p)

=1

n log(n)p + n

<1

n log(n)p

22

thus∑∞

n=1an

1+nanconverges by Theorem 3.29 on page 62 of the text for this choice of an.

On the other hand,∞∑n=1

an1 + n2an

=∞∑n=1

11/an + n2

≤∞∑n=1

1n2

where the inequality follows because an > 0 for all n. Thus,∑∞

n=1an

1+n2anconverges. �

Chapter 4: Continuity

1. Suppose f is a real function defined on Rwhich satisfies

limh→0

[f(x+ h)− f(x− h)] = 0

for every x ∈ R. Does this imply f is continuous?

Solution: No, this does not imply f is continuous because this statement merely says thatlimt→x+ f(t) = limt→x− f(t), that is, the right and left handed limits of f are equal, but it saysnothing about whether these actually equal f(x) itself. That is, by Theorem 4.6 on page 86,f is continuous at x ⇐⇒ limt→x f(t) = f(x) and by the comment in Definition 4.25 on page94 we see that limt→x f(t) exists ⇐⇒ limt→x+ f(t) = limt→x− f(t) = limt→x f(t). So, if wedefine

f(x) ={

1 if x ∈ Z0 if x /∈ Z

then at any x ∈ Z we have that limh→0 f(x + h) = limt→x+ f(t) = 0 = limt→x− f(t) =limh→0 f(x − h) but of course limt→x+ f(t) = limt→x− f(t) = 0 6= 1 = f(x). That is,limt→x f(t) = 0 6= 1 = f(x) so f is not continuous at any x ∈ Z. �

2. If f is a continuous mapping of a metric space X into a metric space Y , prove that f(E) ⊂f(E). Show by an example that f(E) can be a proper subset of f(E).

Solution: If E is empty then the inclusion holds trivially, so suppose that it is not empty. Letx ∈ E so f(x) ∈ f(E) ⊂ f(E). Thus, x ∈ f−1(f(E)) and since x was arbitrary we see thatE ⊂ f−1(f(E)). But, f(E) is closed in Y (Theorem 2.27(a) on page 35 of the text) and since fis continuous we have that f−1(f(E)) is then closed inX by the Corollary to Theorem 4.8 onpage 87 of the text. But, this implies that E ⊂ f−1(f(E)) because E is the smallest closed setcontaining E so it is a subset of any closed set containing E, i.e. this follows from Theorem2.27(c) on page 35 of the text. But, this is precisely the statement that f(E) ⊂ f(E)).

To see that this inclusion can be proper, let E = Z and define f : Z → R by f(n) = 1/n.Then, f is continuous (in fact, it is uniformly continuous if we take δ < 1 because thend(n,m) < 1 =⇒ n = m when n,m ∈ Z and thus d(f(n), f(m)) = 0 < ε ∀ ε > 0). See thecomment at the end of the proof of Theorem 4.20, on page 92 of the text. But, sinceZ is closed,we have that f(Z) = f(Z) = {1/n | n ∈ Z} ( f(Z) = {1/n | n ∈ Z} = {1/n | n ∈ Z} ∪ {0},hence the inclusion is proper. �

Alternate Solution: If f(E) is empty then the conclusion holds trivially, so suppose that it isnot and let y ∈ f(E). Thus, there exists a point p ∈ E such that f(p) = y. If p ∈ E then wehave that y = f(p) ∈ f(E) ⊂ f(E). So, suppose that p ∈ E′, so p is a limit point of E. Since fis continuous at p, for all ε > 0 there exists a δ > 0 such that d(x, p) < δ =⇒ d(f(x), f(p)) <ε. But, since p is a limit point of E we see that there always exists x ∈ N(p, δ) for some x ∈ Eand for any δ > 0. Thus, for such an x, we have that d(f(x), f(p)) < ε which implies that

23

f(p) is a limit point of f(E) since we can always find points in f(E) arbitrarily close to it.Thus y = f(p) ∈ f(E) and so f(E) ⊂ f(E). �

3. Let f be a continuous real function on a metric space X . Let Z(f) (the zero set of f ) be the setof all p ∈ X at which f(p) = 0. Prove that Z(f) is closed.

Solution: Let pn → p in Z(f) so that pn ∈ Z(f) for all n = 1, 2, . . . Of course p ∈ X and sincef is continuous on X we see that limq→p f(q) = f(p) by Theorem 4.6 on page 86 of the text.Then, by Theorem 4.2 on page 84 of the text, we see that 0 = limn→∞ f(pn) = f(p) (wherepn 6= p ∀ n because otherwise we are done) hence p ∈ Z(f) and so Z(f) is closed.

Another approach is to show that Z(f)c = X\Z(f) is open. To this end, let x ∈ E = Z(f)c

so that f(x) 6= 0. Without loss of generality, suppose that f(x) > 0. Since f is continuouson X we see that for ε = f(x)/2 > 0 there exists a δ > 0 such that d(f(x), f(y)) = |f(x) −f(y)| < ε = f(x)/2 whenever d(x, y) < δ. But, this implies that f(y) 6= 0 because otherwise|f(x)− f(y)| = f(x) > ε, a contradiction. This means that N(x, δ) ⊂ E so that in particular,E is open. Thus, Z(f) = Ec is closed. If we had that f(x) < 0 we would simply let ε =−f(x)/2 > 0. �

4. Let f and g be continuous mappings of a metric space X into a metric space Y , and let E bea dense subset of X . Prove that f(E) is dense in f(X). If g(p) = f(p) for all p ∈ E, provethat g(p) = f(p) for all p ∈ X . (In other words, a continuous mapping is determined by itsvalues on a dense subset of its domain.)

Solution: First, let y ∈ f(X), let y = f(x) and let ε > 0. Then, because f is continuous, thereexists a δ > 0 such that d(f(x), f(p)) < ε whenever d(x, p) < δ. But, since E is dense in Xthere exists a p ∈ E such that d(x, p) < δ, which gives that d(y, f(p)) = d(f(x), f(p)) < εhence f(E) is dense in f(X) because we can find an element of f(E) arbitrarily close to anyelement of f(X).

Now, let ε > 0 be given and suppose that g(p) = f(p) for all p ∈ E. Let q ∈ X . Because f iscontinuous at q, there exists δ1 > 0 such that d(f(p), f(q)) < ε/2 whenever d(p, q) < δ1. Sinceg is also continuous at q let δ2 > 0 be such that d(g(p), g(q)) < ε/2 whenever d(p, q) < δ2 andlet δ = min{δ1, δ2}. Since E is dense in X we can find p ∈ E such that d(p, q) < δ. Then,

d(g(q), f(q)) ≤ d(g(q), g(p)) + d(g(p), f(p)) + d(f(p), f(q))= d(g(q), g(p)) + d(f(p), f(q))

<ε

2+ε

2= ε,

where p ∈ E so g(p) = f(p). Thus, since εwas arbitrary we must have that g(q) = f(q) hencethey agree on all of X .

Another way to see this is that if q ∈ X\E then we can find a sequence pn → q such thatpn ∈ E for all n and pn 6= q for all n because E is dense in X , i.e. every element of X is apoint or a limit point of E, so q must be a limit point of E. But then because g and f are bothcontinuous we can interchange limits (by Theorem 4.6 on page 86 and Theorem 4.2 on page84 of the text) so we obtain,

g(q) = g( limn→∞

pn) = limn→∞

g(pn) = limn→∞

f(pn) = f( limn→∞

pn) = f(q)

because g(pn) = f(pn) ∀ n since they agree on E. �

24

5. If f is a real continuous function defined on a closed set E ⊂ R, prove that there existcontinuous real functions g on R such that g(x) = f(x) for all x ∈ E. (Such functions g arecalled continuous extensions of f fromE to R.) Show that the result becomes false if the world“closed” is omitted. Extend the result to vector-valued functions. Hint: Let the graph of gbe a straight line on each of the segments which constitute the complement of E (compareExercise 29, Chapter 2). The result remains true if R is replaced by any metric space, but theproof is not so simple.

Solution: By Exercise 20 in Chapter we see that every open set of R is an at most countableunion of disjoint open sets. Hence, since E is closed, we can write

Ec =n⋃i=1

(ai, bi)

where ai < bi < ai+1 < bi+1 and we allow n to be∞. Then, on each (ai, bi) define

g(x) = f(ai) + (x− ai)f(bi)− f(ai)

bi − ai(5.1)

for x ∈ (ai, bi). That is, the graph of g is nothing more than the straight line connectingthe points f(ai) and f(bi), where ai, bi ∈ E because ai, bi /∈ Ec. We let g(x) = f(x) forx ∈ E. Then it is clear that g is continuous on the interior of E and since it is a linearfunction on Ec it is continuous there also. Hence, we need only check that it is continuousat the boundary of E, namely the points {ai, bi}ni=1. But, from Equation (5.1) we see thatlimt→ai g(t) = f(ai) = g(ai) because ai ∈ E, hence g is continuous at ai by Theorem 4.6on page 86 of the text. Similarly limt→bi g(t) = f(bi) = g(bi) because bi ∈ E hence g is acontinuous extension of f .

The fact that E is closed is essential because if we consider f(x) = 1/x on E = (0, 1), forexample, then f is continuous on E yet there is no continuous extension g of f to R such thatg = f on E yet g(0) ∈ R since as x → 0 we have that f(x) → ∞ hence we would also havethat g(x)→∞ as x→ 0.

The vector-valued case is simply a natural extension of the single variable case since if f is acontinuos vector valued function on R we can write f = (f1, . . . , fn) and so fi is continuousfor each i = 1, . . . , n because f is continuous. Hence, by the above, we can extend each fi toa continuous function gi on R and letting g = (g1, . . . , gn) we see that g is then a continuousextension of f to Rn. �

6. Omitted.

7. Omitted.

8. Let f be a real uniformly continuous function on the bounded set E in R. Prove that f isbounded on E. Show that the conclusion is false if boundedness of E is omitted from thehypothesis.

Solution: Since f is uniformly continuous, there exists a single δ > 0 such that whenever|x − y| < δ we have that |f(x) − f(y)| < 1 for all x, y ∈ E. Since E is bounded thereexists M > 0 such that −M < x < M for all x ∈ E. Now, note that if N ≥ 2M/δ isan integer, then N neighborhoods of size δ put side by side cover at least all but a finitenumber of points in (−M,M) (missing the points that lie between the boundaries of two

25

adjacent neighborhoods). Thus, we can cover (−M,M) with finitely many neighborhoodsof radius δ, and so we must be able to do the same with E but centering the neighborhoodsat elements of E. Hence, some finite subcollection of {B(x, δ) | x ∈ E} must cover E, sayE ⊆ {B(xi, δ)}ni=1. Now, let x ∈ E. Then, for some 1 ≤ i ≤ n we have that x ∈ B(xi, δ) hence|f(x)− f(xi)| < 1 and thus −1 < f(x)− f(xi) < 1. If we let M ′ = maxi{f(xi)} then we seethat −1 < f(x)−M ′ < 1 =⇒ |f(x)| < M ′ + 1. Hence, f is bounded.

To see that boundedness ofE is crucial, consider f(x) = x defined on R. Then f is uniformlycontinuous because if ε > 0 is given simply let δ = ε. Then if |x − y| < δ we have that|f(x)− f(y)| = |x− y| < δ = ε. But, f is clearly unbounded. �

9. Omitted.

10. Omitted.

11. Suppose f is a uniformly continuous mapping of a metric spaceX into a metric space Y andprove that {f(xn)}∞n=1 is a Cauchy sequence in Y for every Cauchy sequence {xn}∞n=1 in X .Use this to prove the following theorem (Exercise 13): If E ⊂ X is a dense subset of a metricspace X and f is a uniformly continuous function with values in a complete metric space Ythen f has a continuous extension from E to X . By Exercise 4 then, this extension will beunique because if we have two such extensions they agree on a dense subset ofX hence theyagree on all of X .

Solution: Let ε > 0 be given and let δ > 0 be such that d(f(x), f(y)) < ε whenever d(x, y) <δ. Let {xn}∞n=1 be a Cauchy sequence in X . Thus, there exist N such that d(xn, xm) < δ forall n,m ≥ N . But this implies that d(f(xn), f(xm)) < ε for all n,m ≥ N , hence {f(xn)}∞n=1 isa Cauchy sequence.

Now, let x ∈ X and let {pn}∞n=1 be a sequence in E such that pn → x, which exists because Eis dense inX so every element ofX is a limit point ofE or a point ofE, or both (see Definition2.18(j) on page 32 of the text). If x ∈ E then let pn = x for all n. Since pn → x we have that{pn}∞n=1 is a Cauchy sequence by Theorem 3.11(a) on page 53 of the text. By the first resultwe showed above we then have that {f(pn)}∞n=1 is Cauchy in Y , hence converges becauseY is complete by hypothesis. Therefore, for x ∈ X we can define g(x) = limn→∞ f(pn)where pn → x and pn ∈ E for all n. We have to verify that this definition is well-defined,i.e. it does not depend on the choice of the sequence {pn}∞n=1 converging to x. So, supposethat sn → x ← pn both converge to x. Let ε > 0 be given and choose δ > 0 such thatd(f(x), f(y)) < ε whenever d(x, y) < δ. Let N1 be such that d(pn, x) < δ/2 for all n ≥ N1 andsimilarly choose N2 such that d(sn, x) < δ/2 for all n ≥ N2. Letting N = max{N1, N2} wesee that

d(pn, sn) ≤ d(pn, x) + d(x, sn) <δ

2+δ

2= δ ∀ n ≥ N =⇒ d(f(pn), f(sn)) < ε ∀ n ≥ N.

Since ε was arbitrary this means that

limn→∞

d(f(pn), f(sn)) = 0. (11.1)

We now show the following general result: if xn → x and yn → y in a metric space X thend(xn, yn)→ d(x, y). Let ε > 0 be given and choose N such that d(xn, x) < ε/2 for all n ≥ N .Then, for y ∈ X , we have that

d(x, y) < d(x, xn) + d(xn, y) =⇒ d(x, y)− d(xn, y) < d(x, xn) < ε. (11.2)

26

Reversing the roles of xn and x we obtain

d(xn, y) < d(xn, x) + d(x, y) =⇒ d(xn, y)− d(x, y) < d(xn, x) < ε. (11.3)

Thus, combining Equations (11.2) and (11.3) we see that |d(x, y) − d(xn, y)| < ε and sinceε was arbitrary this shows that d(xn, y) → d(x, y) for all y ∈ X . In particular, it holdsfor yj ∈ {yn}∞n=1 hence we obtain that limn→∞ d(xn, yn) = limm→∞(limn→∞ d(xn, ym)) =limm→∞ d(x, ym) = d(x, y). Applying this to our result in Equation (11.1) we have that

d(

limn→∞

f(pn), limn→∞

f(sn))

= limn→∞

d(f(pn), f(sn)) = 0.

Thus, limn→∞ f(pn) = limn→∞ f(sn) and so defining g(x) = limn→∞ f(pn) is well-definedsince it does not depend on the particular choice of sequence converging to x. Observe thatif x ∈ E then letting pn = x for all n we see that g(x) = limn→∞ f(pn) = f(x) hence g = fon E. Now, we have only to verify that g is continuous. To this end, let x ∈ X and let ε > 0be given. Let δ > 0 be such that d(x, y) < δ =⇒ d(f(x), f(y)) < ε/3, for all x, y ∈ X . Lety ∈ X such that d(x, y) < δ/3 and let pn → x and sn → y where {pn}∞n=1 and {sn}∞n=1 aresequences in E. Choose N1 such that d(pn, x) < δ/3 and N2 such that d(sn, y) < δ/3. LetN = max{N1, N2}. Then,

d(pn, sn) ≤ d(pn, x) + d(x, y) + d(y, sn) <δ

3+δ

3+δ

3= δ ∀ n ≥ N. (11.4)

Now, since g(x) = limn→∞ f(pn) and g(y) = limn→∞ f(sn) let M1 and M2 be such thatd(g(x), f(pn)) < ε/3 for all n ≥ M1 and d(g(y), f(sn)) < ε/3 for all n ≥ M2. Let M =max{M1,M2} and set K = max{M,N}. Then, fixing n ≥ K we have that

d(g(x), g(y)) ≤ d(g(x), f(pn)) + d(f(pn), f(sn)) + d(f(sn), g(y)) <ε

3+ε

3+ε

3= ε,

where d(f(pn), f(sn)) < ε/3 because of Equation (11.4) which holds since n ≥ K ≥ N . Thus,g is continuous on X and is the desired extension of f . �

12. A uniformly continuous function of a uniformly continuous function is uniformly continu-ous. State this more precisely and prove it.

Solution: The statement can be rephrased as follows. Let f : X → Y and g : Y → Z be twouniformly continuous functions such that the image of f is in the domain of g. Then, thecomposition (g ◦ f) : X → Z is uniformly continuous. To see this, let ε > 0 be given. Then,there exists δ1 > 0 such that d(g(y1), g(y2)) < ε whenever d(y1, y2) < δ1. Similarly, thereexists δ2 > 0 such that d(f(x1), f(x2)) < δ1 whenever d(x1, x2) < δ2. But, this means that ifwe choose x1, x2 ∈ X such that d(x1, x2) < δ2 then we have that d(g(f(x1)), g(f(x2))) < εbecause d(f(x1), f(x2)) < δ1. Hence, (g ◦ f) is uniformly continuous. �

13. Omitted.

14. Let I = [0, 1] be the closed unit interval. Suppose f is a continuous mapping of I into I .Prove that f(x) = x for at least one x ∈ I .

Solution: Let g(x) = f(x) − x be defined on I . If g(1) = 0 or g(0) = 0 then we are finishedso suppose this doesn’t happen. Then, since f maps into I we see that f(1) ∈ [0, 1) sincef(1) 6= 1 by hypothesis, and therefore g(1) = f(1) − 1 < 0. Similarly, g(0) = f(0) − 0 > 0because f(0) ∈ (0, 1] since f maps into I and f(0) 6= 0 by hypothesis. But, g is continuous

27

on I by Theorem 4.4(a) on page 85 of the text, and by Theorem 4.6 on page 86 of the text.Hence, by the Intermediate Value Theorem (Theorem 4.23 on page 93 of the text), there existsa c ∈ (0, 1) such that g(c) = 0. But this says precisely that f(c) = c, hence f has a fixed point.�

15. Call a mapping of X into Y open if f(V ) is an open set in Y whenever V is an open set in X .Prove that every continuous open mapping of R into R is monotonic.

Solution: Suppose that we have a continuous open mapping f : R→ R which is not mono-tonic. This means that there exists x1 < x2 < x3 such that f(x2) > f(x1) and f(x2) > f(x3),or f(x2) < f(x1) and f(x2) < f(x3). More precisely, there exists two points x0 < x1 suchthat f(x0) < f(x1), and two other points x2 < x3 such that f(x2) > f(x3). But, since f is con-tinuous, by the Intermediate Value Theorem (Theorem 4.23 on page 93 of the text) f achievesall values in [f(x0), f(x3)], and so we simply choose x2 and then we have that f(x2) > f(x1)and f(x2) > f(x3).

First, consider the case f(x2) > f(x1) and f(x2) > f(x3). Since f is continuous on R, for

ε =f(x2)− f(x1)

2> 0

there exists δ1 > 0 such that |f(x)− f(x1)| < ε whenever |x− x1| < δ1. In particular,

f(x) <f(x1) + f(x2)

2< f(x2), x1 < x < x1 + δ1 (15.1)

Note that δ1 < x2−x1 since otherwise we could choose x2 itself, but then the above inequalitywould fail. That is, f(x2)− f(x1) ≮ (f(x2)− f(x1))/2. So, let y1 ∈ (x1, x1 + δ1). Similarly, for

ε =f(x2)− f(x3)

2> 0

there exists a δ2 > 0 such that |f(x)− f(x3)| < ε whenever |x− x3| < δ2. In particular,

f(x) <f(x2) + f(x3)

2< f(x2), x3 − δ2 < x < x3. (15.2)

Again, note that δ2 < x3− x2 as before since otherwise we could choose x2 itself and contra-dict the above inequality. That is, f(x2)−f(x3) ≮ (f(x2)−f(x3))/2. Choose y2 ∈ (x3−δ2, x3).Then, because δ1 < x2 − x1 and δ2 < x3 − x2 we have that y2 > y1. Since f is continuous onthe compact set [y1, y2] by Theorem 4.16 on page 89 of the text, f achieves a maximum valueat some p ∈ [y1, y2]. Now, observe that

supx∈(x1,x3)

f(x) ≤ supx∈[y1,y2]

f(x)

because of Equations (15.1) and (15.2). That is, f(x) < f(x2) for all x ∈ (x1, x1 + δ1) andfor all x ∈ (x3 − δ2, x3) hence the sup must occur on (x1 + δ1, x3 − δ2) ⊂ [y1, y2] becausex2 ∈ (x1, x3) . But, since [y1, y2] ⊂ (x1, x3) we have that other inclusion

supx∈(x1,x3)

f(x) ≥ supx∈[y1,y2]

f(x)

hencesup

x∈(x1,x3)f(x) = sup

x∈[y1,y2]f(x) = f(p), (15.3)

28

since f achieves its maximum on the compact set [y1, y2]. Now, f(p) ∈ f((x1, x3)) hence f(p)must be an interior point because (x1, x3) is open and f is an open mapping. Thus, thereexists some ε > 0 such that f(p) + ε ∈ f((x1, x3)). But, since f(p) + ε > f(p) this contradictsEquation (15.3) above which shows that f(p) is the maximum value of f on (x1, x3). Hencef must be monotonic. �

16. Omitted.

17. Omitted.

18. Every rational x can be written in the form x = m/n, where n > 0, and m and n are integerswithout any common divisors. When x = 0, we take n = 1. Consider the function f definedon R by

f(x) ={

0 if x is irrational1n if x = m

n

.

Prove that f is continuous at every irrational point, and that f has a simple discontinuity atevery rational point.

Solution: Let x be irrational. Then, we see that for any other irrational y we have that |f(x)−f(y)| = 0 hence to show continuity at xwe need only consider the difference |f(x)−f(r)| forr a rational. Let ε > 0 be given and choose n such that 1/n < ε. For each 1 ≤ j ≤ n, considera neighborhood of x of radius 1/j. Then, there are at most two rational numbers in lowestterms having j as a denominator in this neighborhood, because the size of the neighborhoodis 2/j and x is irrational, and the distance between any two adjacent rationals with j in thedenominator, i.e. between an m/j and (m+ 1)/j, is 1/j. That is, if there were three rationalshaving j as a denominator in this neighborhood, then that would force x to be the middleone, contradicting that x was irrational. Hence we can write

mj

j< x <

mj + 1j

,

for some mj so there are no rationals in lowest terms having a denominator j between mj/jand (mj + 1)/j because this gap is of size 1/j and rationals with denominators j are evenlyspaced, i.e. the distance between any such two is at least 1/j. Then, let

Ij =(mj

j,mj + 1j

),

and consider I = ∩nj=1Ij . Then, I contains no rationals with denominator j for all 1 ≤ j ≤ nby construction (because each Ij excludes all rationals with denominator j) yet still containsx because x ∈ Ij for all 1 ≤ j ≤ n and is open because it is a finite intersection of open sets.Now, let δ > 0 be such that B(x, δ) ⊂ I . Then, for all rational r ∈ B(x, δ) ⊂ I we see thatr has a denominator which is larger than n when written in lowest terms, say r = p/q withq > n, for otherwise if q ≤ n then q = j for some 1 ≤ j ≤ n, a contradiction that I containsno rationals with any such j in the denominator. Therefore,

|f(x)− f(r)| = 1q<

1n< ε,

and since r was arbitrary in B(x, δ), ε was arbitrary, and x was arbitrary, we see that f istherefore continuous at all irrational x.

29

Another way to say this is, pick mj such that

δj =∣∣∣∣x− mj

j

∣∣∣∣ is smallest.

We know it is greater than 0 because x is irrational, and there exists a smallest one becauserationals with denominator j do not get arbitrarily close but are always separated by a dis-tance of at least 1/j hence there exists a closest rational with denominator 1/j to x. Then, ifwe let 0 < δ < min{δ1, . . . , δn} we see that B(x, δ) has the same property as the analogousneighborhood constructed above.

Now, suppose that r = p/q ∈ R is rational. Then, since the irrationals are dense in R anyneighborhood around r will contain an irrational x. But then, |f(r) − f(x)| = 1/q for anysuch rational and we see that no matter how small a neighborhood we choose around rthere will be points in it such that |f(r) − f(x)| cannot be made arbitrarily small, hence fis discontinuous at every rational point. To see that this is a simple discontinuity, i.e. thatf(x+) and f(x−) exist, we observe that limx→y f(x) = 0 for any point y ∈ R, in particular arational point, because as discussed above, approaching any number arbitrarily close withrationals implies their denominators in lowest terms are getting arbitrarily large, and if weapproach with irrationals then f is 0 when evaluated at them. Since the value f(r) at arational point is never zero, this shows the discontinuity is simple because both limits exist,they just don’t equal the value of the function at the point. In particular, this also shows thatf is continuous at irrational points because then there, the value of the limits does equal thevalue of the function. �

19. Suppose f is a real function with domain R which has the intermediate value property: Iff(a) < c < f(b), then f(x) = c for some a < x < b. Suppose also for every rational r, thatthe set of all x with f(x) = r is closed. Prove that f is continuous.

Hint: If xn → x0 but f(xn) > r > f(x0) for some r and for all n ≥ 1, then f(tn) = r for sometn between x0 and xn; thus tn → x0. Find a contradiction.

Solution: If f is not continuous at some x0 ∈ R then that means that f(xn) 9 f(x0) for somesequence xn → x0, by Theorem 4.6 on page 86 with Theorem 4.2 on page 84 of the text. Sincef is not constant (otherwise it is continuous), we can assume without loss of generality thatf(xn) > r > f(x0) for some rational r for all n ≥ 1. That is, we rename our sequence to onlyinclude terms such that f(xn) > f(x0) and if there are only finitely many such terms thenwe choose r such that f(x0) > r > f(xn). This is possible because Q is dense in R, and ifwe only include terms where f(xn) 6= f(x0) (there are infinitely many such terms otherwisewe’d have convergence). Since f has the intermediate value property we see that there existstn between xn and x0 such that f(tn) = r for all n ≥ 1. But, since xn → x0 we must have thattn → x0 because d(tn, x0) ≤ d(xn, x0), which goes to 0. Since S = {x | f(x) = r} is closed byhypothesis and tn ∈ S for all n ≥ 1 by construction we see that x0 ∈ S so that f(x0) = r, acontradiction to the choice of r. Therefore, f is continuous at every x ∈ R. �

20. If E is a nonempty subset of a metric space X , define the distance from x ∈ X to E by

ρE(x) = infz∈E

d(x, z)

(a) Prove that ρE(x) = 0 if and only if x ∈ E.

30

(b) Prove that ρE(x) is a uniformly continuous function on X , by showing that

|ρE(x)− ρE(y)| ≤ d(x, y)

for all x, y ∈ X .Hint: ρE(x) ≤ d(x, z) ≤ d(x, y) + d(y, z), so that ρE(x) ≤ d(x, y) + ρE(y).

Solution:

(a) First, suppose that ρE(x) = 0. This means that for all n ≥ 1 there exists zn ∈ E suchthat d(x, zn) < 1/n by definition of inf hence zn → x and so x is a limit point of Ethus x ∈ E. On the other hand, suppose that x ∈ E so there exists a sequence zn → xwith {zn}∞n=1 ⊂ E. Then, for all ε > 0 there exists an N such that d(x, zn) < ε for alln ≥ N , hence the distance from x to some point of E can be made arbitrarily small, i.e.ρE(x) = infz∈E d(x, z) = 0. Otherwise, if ρE(x) = δ > 0 simply choose 0 < ε < δ. Then,we can find n such that d(x, zn) < ε < δ = infz∈E d(x, z), contradicting the definition ofthe inf .

(b) By the triangle inequality for x ∈ X we have that for all z ∈ E and for all y ∈ X

ρE(x) = infz′∈E

d(x, z′) ≤ d(x, z) ≤ d(x, y) + d(y, z).

Since z was arbitrary we must have that

ρE(x) ≤ d(x, y) + infz∈E

d(y, z) = d(x, y) + ρE(y).

HenceρE(x)− ρE(y) ≤ d(x, y).

Reversing the roles of x and y above we obtain ρE(y)− ρE(x) ≤ d(x, y) hence we have

|ρE(x)− ρE(y)| ≤ d(x, y).

This is precisely uniform continuity because if ε > 0 is given and x, y ∈ X let δ = ε.Then, d(x, y) < δ = ε =⇒ |ρE(x)− ρE(y)| ≤ d(x, y) < δ = ε, hence ρE(x) is uniformlycontinuous on X . �

21. Suppose K and F are disjoint sets in a metric space X such that K is compact and F isclosed. Prove that there exists δ > 0 such that d(p, q) > δ if p ∈ K and q ∈ F . Show that theconclusion may fail for two disjoint closed sets if neither is compact.

Solution: We show that inf{d(p, q) | p ∈ K, q ∈ F} > 0 and hence is greater than somepositive δ. Since K ∩ F = ∅ we have that d(x, y) > 0 for all x ∈ K, y ∈ F because d(x, y) =0 ⇐⇒ x = y. Therefore

0 ≤ inf{d(x, y) | x ∈ K, y ∈ F}.

So, suppose that0 = inf{d(x, y) | x ∈ K, y ∈ F}.

This means that for each n ∈ N, there exists xn ∈ K, yn ∈ F such that d(xn, yn) < 1/n bydefinition of the inf . Then, {xn} ⊂ K is a sequence and since K is compact, there exists a

31

convergent subsequence {xnk} → x0 ∈ K. Then, {ynk

} ⊂ F is a sequence. Let ε > 0 begiven. Then, there exists a K1 such that

d(xnk, ynk

) <ε

2, k ≥ K1

and since xnk→ x0, there exists an K2 such that

d(xnk, x0) <

ε

2, k ≥ K2.

If we let K = max{K1,K2} then

d(ynk, x0) ≤ d(ynk

, xnk) + d(xnk

, x0) <ε

2+ε

2= ε, ∀ k ≥ K.

Thus, ynk→ x0 also. Since F is closed it contains its limit points and therefore x0 ∈ F ,

contradicting that K ∩ F = ∅. Therefore, we must have that 0 < inf{d(x, y) | x ∈ K, y ∈ F}.Now, consider the sets A = Z+\{1, 2} = {n ∈ Z | n ≥ 3} and B = {n + 1/n | n ∈ Z+}.Both are closed and A ∩ B = ∅. But, for all ε > 0 if we choose n > 1/ε then we see that|n + 1/n − n| = 1/n < ε hence the conclusion fails because we can find points in A and Bthat are arbitrarily close. �

Alternate Solution: We can see this result in another way by using the results of Exercise 20above. First, by Exercise 20(a) we see that ρF (x) = 0 ⇐⇒ x ∈ F = F because F is closed,hence we have that ρF (k) > 0 for all k ∈ K because K ∩ F = ∅. By Exercise 20(b) ρF (k) iscontinuous on K and so by Theorem 4.16 on page 89 of the text, since K is compact, ρF (k)achieves its minimum on K, namely there exists k ∈ K such that ρF (k) = m > 0, where wehave the inequality because k ∈ K and we saw above that ρF was positive on K. Hence, forp ∈ K and q ∈ F we have that 0 < m/2 < ρF (p) = infq′∈F d(p, q′) ≤ d(p, q). �

Chapter 5: Differentiation

1. Let f be defined for all real x, and suppose that

|f(x)− f(y)| ≤ (x− y)2

for all real x and y. Prove that f is a constant.

Solution: We can rewrite this condition as

|f(x)− f(y)| ≤ |x− y||x− y|,

and therefore we can divide by |x−y| since it is positive and preserve the inequality to obtain∣∣∣∣f(x)− f(y)x− y

∣∣∣∣ =|f(x)− f(y)||x− y|

≤ |x− y|.

Thus the limit as x→ y of the left hand side exists (for a given ε > 0 choose δ = ε) and takingthis limit gives

|f ′(y)| ≤ 0 =⇒ f ′(y) = 0.

Since y was arbitrary we see that f is constant by Theorem 5.11 on page 108 of the text. �

32

2. Suppose that f ′(x) > 0 on (a, b). Prove that f is strictly increasing in (a, b), and let g be itsinverse function. Prove that g is differentiable, and that

g′(f(x)) =1

f ′(x)(a < x < b).

Solution: Let a < x < y < b. Then, f is continuous on [x, y] (because it is differentiable on(a, b) ) [x, y]) and differentiable on (x, y). Thus, by The Mean Value Theorem (Theorem 5.10on page 108 of the text), there exists c ∈ (x, y) such that

f(y)− f(x) = (y − x)f ′(c) > 0,

where the inequality follows because f ′(c) > 0 by hypothesis and y− x > 0 by construction.Since f is strictly increasing on (a, b), it is injective and therefore the inverse function g iswell-defined. Let a < x < y < b. Then,

g′(f(x)) = limf(y)→f(x)

g(f(y))− g(f(x))f(y)− f(x)

= limy→x

y − xf(y)− f(x)

=1

f ′(x),

where the second equality follows since f is injective (because it is strictly increasing) thusf(y)→ f(x) ⇐⇒ y → x. �

3. Suppose g is a real function on R, with bounded derivative (say |g′| ≤ M ). Fix ε > 0, anddefine f(x) = x+ εg(x). Prove that f is one-to-one if ε is small enough. (A set of admissiblevalues of ε can be determined which depends only on M .)

Solution: Let x < y. By the Mean Value Theorem (Theorem 5.10 on page 108 of the text),there exists a x < c < y such that

g(x)− g(y) = g′(c)(x− y).

Thus, we obtain

f(x)− f(y) = x− y + ε(g(x)− g(y))= x− y + εg′(c)(x− y)= (x− y)(1 + εg′(c)).

(3.1)

Since |g′| ≤ M we have that −M ≤ g′(x) ≤ M for all x ∈ R. Thus, 1 − εM ≤ 1 + εg′(x) ≤1 + εM . If we choose ε ≤ 1

2M then this inequality becomes

12

= 1− 12≤ 1− εM ≤ 1 + εg′(x) ≤ 1 + εM ≤ 1 +

12

=⇒ 1 + εg′(x) > 0, ∀ x ∈ R. (3.2)

In particular, this holds for x = c and thus from Equation (3.1) we see that f(x) − f(y) < 0since x − y < 0 by construction and we see in Equation (3.2) that 1 + εg′(c) > 0. That is,f(x) 6= f(y) so f is one-to-one. �

4. IfC0 +

C1

2+ · · ·+ Cn−1

n+

Cnn+ 1

= 0,

where C0, . . . , Cn are real constants, prove that the equation

C0 + C1x+ · · ·+ Cn−1xn−1 + Cnx

n = 0

33

Solution: Consider the polynomial defined by

p(x) = C0x+C1

2x2 + · · ·+ Cn

n+ 1xn+1, x ∈ [0, 1].

Then, p(1) = C0 + C12 + · · · + Cn−1

n + Cnn+1 = 0 = p(0) and by The Mean Value Theorem

(Theorem 5.10 on page 108 of the text) we see that

C0 + C1c+ · · ·+ Cn−1cn−1 + Cnc

n = p′(c) =p(1)− p(0)

1− 0= 0, c ∈ (0, 1),

hence c is a root of C0 + C1x+ · · ·+ Cn−1xn−1 + Cnx

n. �

5. Suppose f is defined and differentiable for every x > 0, and f ′(x) → 0 as x → ∞. Putg(x) = f(x+ 1)− f(x). Prove that g(x)→ 0 as x→∞.

Solution: Let ε > 0 be given and let N be such that |f ′(x)| < ε for all x ≥ N . Then, for suchx, by The Mean Value Theorem (Theorem 5.10 on page 108 of the text) we have that

|g(x)| = |f(x+ 1)− f(x)| =∣∣∣∣f(x+ 1)− f(x)

x+ 1− x

∣∣∣∣ = |f ′(c)|, c ∈ (x, x+ 1).

Thus |g(x)| = |f ′(c)| < ε since c > x ≥ N . Since ε was arbitrary we have that g(x) → 0 asx→∞. �

6. Suppose

(a) f is continuous for x ≥ 0,

(b) f ′(x) exists for x > 0,

(c) f(0) = 0,

(d) f ′ is monotonically increasing.

Put

g(x) =f(x)x

x > 0


Solution: Let 0 < x < y. By The Mean Value Theorem (Theorem 5.10 on page 108 of the text)we see that there exist c1 ∈ (0, x) and c2 ∈ (0, y) such that

f(x)x

=f(x)− f(0)

x− 0= f ′(c1),

f(y)y

=f(y)− f(0)

y − 0= f ′(c2).

Therefore,

g(y)− g(x) =f(y)y− f(x)

x= f ′(c2)− f ′(c1) ≥ 0,

where the inequality follows since c2 ≥ c1 (because y > x) and f ′ is monotonically increas-ing. Therefore, g is monotonically increasing. �

34

7. Suppose f ′(x), g′(x) exist, g′(x) 6= 0, and f(x) = g(x) = 0. Prove that

limt→x

f(t)g(t)

=f ′(x)g′(x)

.

(This holds also for complex functions.)

Solution: Simply consider

f ′(x)g′(x)

=limt→x

f(t)−f(x)t−x

limt→xg(t)−g(x)t−x

= limt→x

f(t)g(t)

.

Note, we cannot trivially use L’Hospital’s Rule because we don’t know that limt→xf ′(t)g′(t) =

f ′(x)g′(x) , that is, we don’t know that the derivatives are continuous at x, only that they exist atx. We can see that the same proof works for complex functions if we write f(x) = f1(x) =if2(x) where f1, f2 are real-valued functions. �

8. Suppose f ′ is continuous on [a, b] and ε > 0. Prove that there exists δ > 0 such that∣∣∣∣f(t)− f(x)t− x

− f ′(x)∣∣∣∣ < ε

whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. (This could be expressed by saying that f isuniformly differentiable on [a, b] if f ′ is continuous on [a, b].) Does this hold for vector-valuedfunctions too?

Solution: Since f ′ is continuous on the compact set [a, b] it is uniformly continuous on [a, b]and therefore there exists δ > 0 such that |f ′(t) − f ′(x)| < ε for all 0 < |t − x| < δ, a ≤t ≤ b, a ≤ x ≤ b. By The Mean Value Theorem (Theorem 5.10 on page 108 of the text) thereexists a c between t and x such that f ′(c) = f(t)−f(x)

t−x . Since c is between t and x we have that

0 < |c− x| < δ because |c− x| < |t− x|. Therefore∣∣f(t)−f(x)

t−x − f ′(x)∣∣ = |f ′(c)− f ′(x)| < ε.

This does not hold for vector-valued functions. We can use f(x) = (cos(x), sin(x)) as acounter example. �

9. Let f be a continuous real function on R, of which it is known that f ′(x) exists for all x 6= 0and that f ′(x)→ 3 as x→ 0. Does it follow that f ′(0) exists?

Solution: Yes, f ′(0) must exist. This follows from Corollary to Theorem 5.12 on page 109of the text. For suppose that f ′(0) did not exist. Since f ′(x) → 3 as x → 0 this means thatf ′ has a discontinuity of the first kind (i.e. a simple discontinuity) at x = 0 because the leftand right hand limits exist yet f ′ is not continuous at 0 because it is not defined there. Butthis contradicts the Corollary. We also give a direct proof of the following more generalstatement: Suppose that f is continuous on an open interval I containing x0, and let f ′ bedefined on I except possibly at x0 and f ′(x)→ L as x→ x0. Then f ′(x0) = L. To see this, letg(x) = x. Then we have that limh→0 g(h) = 0 and limh→0 f(x0 + h) − f(x0) = 0. Therefore,

35

we can apply L’Hospital’s Rule (Theorem 5.13 on page 109 of the text) and obtain that

f ′(x0) = limh→0

f(x0 + h)− f(x0)h

= limh→0

f(x0 + h)− f(x0)g(h)

= limh→0

(f(x0 + h)− f(x0))′

g′(h)= lim

h→0f ′(x0 + h)

= L,

where the last equality follows because by hypothesis f ′(x) → L as x → x0 so we have thatlimh→0 f

′(x0 + h) = L. �

10. Suppose f and g are complex differentiable functions on (0, 1), f(x) → 0, g(x) → 0, f ′(x) →A, g′(x)→ B as x→ 0, where A and B are complex numbers and B 6= 0. Prove that

limx→0

f(x)g(x)

=A

B.

Compare with Example 5.18 on page 112 of the text. Hint:

f(x)g(x)

=(f(x)x−A

)· x

g(x)+A · x

g(x).

Apply Theorem 5.13 on page 109 of the text to the real and imaginary parts of f(x)/x andg(x)/x.

Solution: Write f(x) = f1(x) + if2(x) where f1, f2 are real-valued functions. Then, we havethat

df

dx=

d

dx(f1(x) + if2(x)) =

df1(x)dx

+ idf2(x)dx

. (10.1)

Now, since f1 and f2 are real-valued, we can apply L’Hospital’s Rule (Theorem 5.13 on page109 of the text) to the functions f1(x)

x and f2(x)x , since f1(x), f2(x), x→ 0 as x→ 0. This shows

that

limx→0

f1(x)x

= limx→0

f ′1(x)

limx→0

f2(x)x

= limx→0

f ′2(x).

Therefore, we obtain that

limx→0

f(x)x

= limx→0

(f1(x)x

+ if2(x)x

)= lim

x→0(f ′1(x) + if ′2(x)) = lim

x→0f ′(x) = A,

where the last equality follows from Equation (10.1). An analogous construction shows thatlimx→0 g(x)/x = g′(x) = B thus limx→0 x/g(x) = 1/B. Then, we have that

limx→0

f(x)g(x)

= limx→0

(f(x)x−A

)· x

g(x)+A · x

g(x)= (A−A) · 1

B+A · 1

B=A

B.

In Example 5.18, we have that g′(x)→∞ as x→ 0, not a complex number. �

36

11. Suppose f is defined in a neighborhood of x, and suppose that f ′′(x) exists. Show that

limh→0

f(x+ h) + f(x− h)− 2f(x)h2

= f ′′(x).

Show by an example that the limit may exist even if f ′′(x) does not. Hint: Use Theorem 5.13on page 109 of the text.

Solution: Applying L’Hospital’s Rule (Theorem 5.13 on page 109 of the text), with respect toh, to the functions F (h) = f(x+ h) + f(x− h)− 2f(x) and G(h) = h2 we see that

limh→0

f(x+ h) + f(x− h)− 2f(x)h2

= limh→0

F (h)G(h)

= limh→0

F ′(h)G′(h)

= limh→0

f ′(x+ h)− f ′(x− h)2h

,

(11.1)where we have a minus sign in the numerator of the last expression from the chain rule:(f(x− h))′ = −f ′(x− h). Now, we have that

f ′′(x) =12

(f ′′(x) + f ′′(x))

=12

(limh→0

f ′(x+ h)− f ′(x)h

+ limh→0

f ′(x− h)− f ′(x)−h

)=

12

limh→0

f ′(x+ h)− f ′(x− h)h

= limh→0

f ′(x+ h)− f ′(x− h)2h

= limh→0

f(x+ h) + f(x− h)− 2f(x)h2

,

where the last equality follows by Equation (11.1). To see that the limit may exist even whenf ′′(x) does not, consider the function f(x) = x|x|. Then f ′′(0) does not exist yet

limh→0

f(0 + h) + f(0− h)− 2f(0)h2

= limh→0

h|h| − h|h|h2

= 0.

�

12. Omitted.

13. Omitted.

14. Let f be a differentiable real function defined on (a, b). Prove that f is convex if and onlyif f ′ is monotonically increasing. Assume next that f ′′(x) exists for all x ∈ (a, b), and provethat f is convex if and only if f ′′(x) ≥ 0 for all x ∈ (a, b).

Solution: First, suppose that f is convex so that f(λx+(1−λ)y) ≤ λf(x)+(1−λ)f(y) for all0 < λ < 1 and a < x < b, a < y < b. This means that f(λ(x− y) + y) ≤ λ(f(x)− f(y)) + f(y).If we choose a < y < x 0 then subtracting f(y) from both sides anddividing by λ(x− y) gives

f(λ(x− y) + y)− f(y)λ(x− y)

≤ f(x)− f(y)x− y

,

where the inequality is preserved because λ(x − y) > 0. Taking the limit as y → x of bothsides then shows that

f ′(y) ≤ f ′(x), a < y < x < b,

37

where we get f ′(y) on the left because y → x =⇒ λ(x− y)→ 0. Hence, f ′(x) is monotoni-cally increasing on (a, b).

Now, suppose that f ′ is monotonically increasing on (a, b). Let x, y ∈ (a, b) and 0 < λ < 1.Without loss of generality assume x < y. Set x0 = y − λ(y − x) which means that

λ =y − x0

y − x,

1− λ =y − xy − x

− y − x0

y − x=y − x− y + x0

y − x=x0 − xy − x

.

(14.1)

This shows that

λx+ (1− λ)y = x

(y − x0

y − x

)+ y

(x0 − xy − x

)=xy − xx0 + yx0 − yx

y − x

=yx0 − xx0

y − x= x0.

(14.2)

Write F (x, y, λ) = f(λx + (1 − λ)y) − (λf(x) + (1 − λ)f(y)) for notational purposes (soF (x, y, λ) is the convexity relation). Then, using Equations (14.1) and (14.2) we have that

F (x, y, λ) = f(x0)−(y − x0

y − x

)f(x)−

(x0 − xy − x

)f(y)

=(y − x0

y − x

)(f(x0)− f(x)) +

(x0 − xy − x

)(f(x0)− f(y)),

(14.3)

where the second equality follows because y−x0

y−x + x0−xy−x = 1. Now, two applications of the

Mean Value Theorem yields

f(x0)− f(x) = f ′(η)(x0 − x) η ∈ (x, x0)f(x0)− f(y) = f ′(ξ)(x0 − y) ξ ∈ (x0, y).

(14.4)

Substituting this into Equation (14.3) we obtain

F (x, y, λ) =(y − x0

y − x

)(x0 − x)f ′(η) +

(x0 − xy − x

)(x0 − y)f ′(ξ)

=(y − x0

y − x

)(x0 − x)f ′(η)−

(x0 − xy − x

)(y − x0)f ′(ξ)

=[(

y − x0

y − x

)(x0 − x)

](f ′(η)− f ′(ξ))

≤ 0,

(14.5)

where the inequality follows because y−x0, y−x, x0−x > 0 but from Equation (14.4) we seethat η < ξ =⇒ f ′(η) ≤ f ′(ξ) (because f ′ is monotonically increasing by hypothesis) =⇒f ′(η)− f ′(ξ) ≤ 0. But, Equation (14.5) is precisely

f(λx+ (1− λ)y)− (λf(x) + (1− λ)f(y)) ≤ 0,

38

by the definition of F (x, y, λ), hence f is convex. By Theorem 5.11(a) on page 108 we seethat f ′′(x) ≥ 0 ⇐⇒ f ′(x) is monotonically increasing because f ′′(x) = (f ′(x))′, hence f isconvex if and only if f ′′(x) ≥ 0 by the above result. �

Chapter 6: The Riemann-Stieltjes Integral

Before we present the solutions to selected exercises in this chapter, we prove one property of theintegral which was not shown in the text, namely Theorem 6.12(c) which states that if f ∈ R(α)on [a, b] and a < c < b then f ∈ R(α) on [a, c] and on [c, b] and∫ b

afdα =

∫ c

afdα+

∫ b

cfdα.

First, since f ∈ R(α) on [a, b] we can choose a partition P = {a = x0 ≤ x1 ≤ · · · ≤ xn = b} of [a, b]such that

U(P, f, α)− L(P, f, α) < ε.

Now, let P ∗ = P ∪ {c}, a (possible) refinement of P , and let xj = c for some 1 ≤ j ≤ n− 1. Then,by Theorem 6.4 on page 123 of the text we see that L(P, f, α) ≤ L(P ∗, f, α) and U(P ∗, f, α) ≤U(P, f, α). If we let P1 = {a = x0 ≤ x1 ≤ · · · ≤ xj = c} then,

ε > U(P, f, α)− L(P, f, α)

=n∑i=1

(Mi −mi)∆αi

≥j∑i=1

(Mi −mi)∆αi

= U(P1, f, α)− L(P1, f, α),

where the second inequality follows since Mi−mi ≥ 0 for all i. Therefore, f ∈ R(α) on [a, c] sinceP1 is a partition of [a, c]. Modifying the above procedure accordingly by letting P2 = {c = xj ≤xj+1 ≤ · · · ≤ xn = b} and keeping

∑ni=j+1(Mi −mi)∆αi instead, we similarly see that f ∈ R(α)

on [c, b] also. Now, let P be any partition of [a, b] and P ∗ = P ∪ {c}, with P1 and P2 analogousdefinitions as above. Then, we have that

U(P, f, α) ≥ U(P ∗, f, α)

=n∑i=1

Mi∆αi

=j∑i=1

Mi∆αi +n∑

i=j+1

Mi∆αi

= U(P1, f, α) + U(P2, f, α)

≥∫ c

afdα+

∫ b

cfdα.

Since P was arbitrary, we see that∫ b

afdα = inf

PU(P, f, α) ≥

∫ c

afdα+

∫ b

cfdα.

39

Now, if we use L(P, f, α) instead we find that

L(P, f, α) ≤ L(P ∗, f, α)

=n∑i=1

mi∆αi

=j∑i=1

mi∆αi +n∑

i=j+1

mi∆αi

= L(P1, f, α) + L(P2, f, α)

≤∫ c

afdα+

∫ b

cfdα.

Since P was arbitrary, we see that∫ b

afdα = sup

PL(P, f, α) ≤

∫ c

afdα+

∫ b

cfdα,

which together with the result above shows that∫ ba fdα =

∫ ca fdα+

∫ bc fdα as desired. �

1. Suppose α increases on [a, b], a ≤ x0 ≤ b, α is continuous at x0, f(x0) = 1, and f(x) = 0 ifx 6= x0. Prove that f ∈ R(α) and that

∫ ba fdα = 0.

Solution: By Theorem 6.10 on page 126 of the text we see that f ∈ R(α) since α is continuousat the only point where f is discontinuous. Now, we see that

U(P, f, α) =n∑i=1

sup[xi−1,xi]

f(x) ∆αi = ∆αi, x0 ∈ [xi−1, xi],

L(P, f, α) =n∑i=1

inf[xi−1,xi]

f(x) ∆αi = 0,

where L(P, f, α) = 0 because for any partition we have that x0 is contained in some intervalof positive length hence the inf of f on that interval will be 0 since it will contain points notequal to x0. Since P was arbitrary this implies that

∫ ba fdα = supP L(P, f, α) = 0, because f

was already shown to be Riemann integrable. We also have that that U(P, f, α) = αi + αi+1

if x0 = xi. So, we see that infP U(P, f, α) = 0 since taking finer and finer partitions meansthat ∆αi → 0. Since this is equal to supP L(P, f, α) we again see that

∫ ba fdα = 0. �

2. Suppose f ≥ 0, f is continuous on [a, b], and∫ ba f(x)dx = 0. Prove that f(x) = 0 for all

x ∈ [a, b]. (Compare this with Exercise 1).

Solution: Suppose there exists x0 ∈ [a, b] such that f(x0) = L > 0 (because f ≥ 0). Letting0 < ε < L we see that there exists δ > 0 such that f(x) ∈ (L − ε, L + ε) whenever x ∈(x0 − δ, x0 + δ) by the continuity of f . We then have that∫ b

af(x)dx =

∫ x0−δ

af(x)dx+

∫ x0+δ

x0−δf(x)dx+

∫ b

x0+δf(x)dx > 0,

where the inequality follows since on (x0 − δ, x0 + δ), f(x) > 0 since L− ε > 0 by the choiceof ε. This contradicts that

∫ ba f(x)dx = 0, hence no such x0 can exist. Exercise 1 shows that

the continuity of f is necessary for this result since in Exercise 1 we have a function whoseintegral is zero, but f(x) 6= 0 for some x ∈ [a, b]. �

40

3. Omitted.

4. If f(x) = 0 for all irrational x and f(x) = 1 for all rational x, prove that f /∈ R on [a, b] forany a < b.

Solution: Let a < b and let P be a partition of [a, b]. Because rational and irrational numbersare both dense in R we see that any interval of positive length will contain both types ofnumbers hence [xi−1, xi] contains both types of numbers for at least one i, because P is apartition and a < b so there is at least one interval of positive length with endpoints elementsof P . Hence,

U(P, f) =n∑i=1

sup[xi−1,xi]

f(x) ∆xi ≥ 1,

L(P, f) =n∑i=1

inf[xi−1,xi]

f(x) ∆xi = 0.

Therefore, supL(P, f) = 0 < 1 ≤ inf U(P, f) therefore f /∈ R by definition of R. �

5. Suppose f is a bounded real function on [a, b], and f2 ∈ R on [a, b]. Does it follow thatf ∈ R? Does the answer change if we assume f3 ∈ R?

Solution: Defining f(x) = −1 for irrational x and f(x) = 1 for rational x shows that f /∈ Ron [a, b] by Exercise 2 above, yet f2(x) = 1 for all x ∈ [a, b] hence f2 ∈ R on [a, b]. Therefore,we cannot conclude that f ∈ R if f2 ∈ R.

By Theorem 6.11 on page 127 of the text we see that if f ∈ R on [a, b] and m ≤ f ≤ M , ifφ is continuous on [m,M ] and h(x) = φ(f(x)) on [a, b] then h ∈ R on [a, b]. Thus, supposethat f3 ∈ R on [a, b]. Since f is bounded we have that m ≤ f ≤ M for some m andM . The function φ(x) = x1/3 is continuous on [m,M ] because it is continuous on all of R.Therefore, h(x) = φ(f3(x)) = f(x) ∈ R on [a, b] as desired. This fails if we only assumef2 ∈ R precisely because x1/2 is not defined for negative x. If f > 0 and bounded, thenf2 ∈ R =⇒ f ∈ R because x1/2 is continuous on [0,M ] for all M . �

6. Omitted.

7. Suppose that f is a real function on (0, 1] and f ∈ R on [c, 1] for every c > 0. Define∫ 1

0f(x)dx = lim

c→0

∫ 1

cf(x)dx

if this limit exists (and is finite).

(a) If f ∈ R on [0, 1], show that this definition of the integral agrees with the old one.

(b) Construct a function f such that the above limit exists, although it fails to exist with |f |in place of f .

Solution:

(a) If f ∈ R on [0, 1] then f is bounded on [0, 1] because the Riemann integral is definedonly for bounded functions. That is, if f were not bounded then the upper-Riemann

41

sums would be infinite, hence f would not be integrable. Thus, we can write |f(x)| ≤M for all x ∈ [0, 1]. Also, (the old)

∫ 10 f(x)dx exists hence we can form the difference∣∣∣∣ ∫ 1

0f(x)dx−

∫ 1

cf(x)dx

∣∣∣∣ =∣∣∣∣ ∫ c

0f(x)dx

∣∣∣∣ ≤ ∫ c

0|f(x)|dx ≤ cM → 0 as c→ 0.

So, the old integral is the limit of∫ 1c f(x)dx as c→ 0.

(b) Let

f(x) = n(−1)n,1

(n+ 1)< x ≤ 1

n.

Then this function works. �

8. Suppose that f ∈ R on [a, b] for every b > a where a is fixed. Define∫ ∞a

f(x)dx = limb→∞

∫ b

af(x)dx

if this limit exists (and is finite). In that case, we say that the integral on the left converges. Ifit also converges after f has been replaced by |f |, it is said to converge absolutely. Assumethat f(x) ≥ 0 and that f decreases monotonically on [1,∞). Prove that∫ ∞

1f(x)dx

converges if and only if∞∑n=1

f(n)

converges. (This is the so-called “integral test” for convergence of series).

Solution: First, suppose that∫∞1 f(x)dx converges. Thus, we can write∫ ∞

1f(x)dx = C.

Since f ≥ 0 on [1,∞) we have that∫∞1 f(x)dx =

∫ N1 f(x)dx +

∫∞N f(x)dx ≥

∫ N1 f(x). Now,

let N ≥ 1 and let P ′ = {x0 = 1, . . . , xN = N} be a partition of [1, N ]. Then,

N∑i=2

f(i) = L(P ′, f) ≤ supPL(P, f) =

∫ N

1f(x)dx ≤

∫ ∞1

f(x)dx = C, (8.1)

where the first equality follows because f is decreasing so inf [n,n+1] f(x) = f(n + 1). Thus,∑Ni=1 f(i) is bounded (because f(1) is finite). Since f ≥ 0 on [1,∞) we see that SN =∑Ni=1 f(i) is an increasing sequence, which is bounded above by Equation (8.1), hence it

converges.

Conversely, suppose that∑∞

i=1 f(i) converges. Then, because f is decreasing on [1,∞) wehave that

f(x) ≤ f(n) ∀ x ∈ [n, n+ 1].

42

Therefore, taking integrals preserves the inequalities, giving∫ n+1

nf(x)dx ≤

∫ n+1

nf(n)dx = f(n).

Since f ≥ 0 we arrive at∫ N

1f(x)dx =

N−1∑n=1

∫ n+1

nf(x)dx ≤

N−1∑n=1

f(n) ≤∞∑n=1

f(n) <∞.

Hence, the sequence aN =∫ N1 f(x)dx is bounded above. Since f ≥ 0 this sequence is also

increasing(aN+1 =

∫ N+11 f(x)dx =

∫ N1 f(x)dx+

∫ N+1N f(x)dx ≥

∫ N1 f(x)dx = aN

), hence it

converges. �

9. Omitted.

10. Let p and q be real numbers such that

1p

+1q

= 1. (10.1)

Prove the following statements.

(a) If u ≥ 0 and v ≥ 0 then

uv ≤ up

p+vq

q.

Equality holds if and only if up = vq.

(b) If f ∈ R(α), g ∈ R(α), f ≥ 0, g ≥ 0, and∫ b

afpdα = 1 =

∫ b

agqdα,

then ∫ b

afg dα ≤ 1.

(c) If f and g are complex functions in R(α) then∣∣∣∣ ∫ b

afg dα

∣∣∣∣ ≤ (∫ b

a|f |pdα

)1/p(∫ b

a|g|qdα

)1/q

.

This Holder’s inequality. When p = q = 2 it is usually called the Schwarz inequality.(Note that Theorem 1.35 is a very special case of this.)

We omit part (d).

Solution:

(a) If u = 0 or v = 0 then the inequality holds trivially, hence we can assume that both arepositive. From Equation (10.1) we can write

q + p

qp= 1 =⇒ q + p = pq =⇒ p =

q

q − 1, (10.2)

43

hence we have that 1 < p, q < ∞ because both are positive. Define a = up/vq. ByEquation (10.2) above we have that q/p = q − 1 hence

a1/p =(up

vq

)1/p

=u

vq/p=

u

vq−1= uv1−q.

Therefore, if we show that

a1/p ≤ 1pa+

1q, 0 < a <∞, (10.3)

then this implies that uv1−q ≤ (up/(vq)(1/p) + 1/q which after multiplying through byvq gives uv ≤ up/p+ vq/q, our desired inequality. Hence, we need only show Equation(10.3) holds. Consider the continuous function

f(a) =1pa+

1q− a1/p,

which is positive at a = 0 and tends to ∞ as a → ∞. It is continuous because x1/p iscontinuous for p > 1 and x > 0. Therefore, if it has a minimum on (0,∞) it must occurwhere its derivative vanishes. Calculating we have that

f ′(a) =1p− 1pa1/p−1.

Finding the zero of this, we wish to solve 1/p = (1/p)a1/p−1 which implies that a1/p−1 =1 hence a = 1. Since f(1) = 1/p + 1/q − 1 = 0 and f is positive at a = 0 and tends toinfinity as a→∞, we see that f(1) is in fact the minimum of f for 0 < a <∞ hence wefind that

0 = f(1) ≤ f(a) =1pa+

1q− a1/p, 0 < a <∞,

which is precisely Equation (10.3), hence the desired inequality also holds.

(b) If f ≥ 0 and g ≥ 0 then fp ∈ R(α) and gq ∈ R(α) by Theorem 6.11 on page 127 of thetext. We also have that fg ∈ R(α) by Theorem 6.13(a) on page 129 of the text. By part(a) we have that

f(x)g(x) ≤ f(x)p

p+g(x)q

q, a ≤ x ≤ b, (10.4)

hence we find that∫ b

afg dα ≤ 1

p

∫ b

afpdα+

1q

∫ b

agqdα =

1p

+1q

= 1.

(c) If f and g are complex-valued in R(α) then |f | and |g| are nonnegative elements ofR(α) and as in part (b), fg ∈ R(α) as well. Moreover, by Theorem 6.13(b) on page 129of the text we have that ∣∣∣∣ ∫ b

afg dα

∣∣∣∣ ≤ ∫ b

a|f ||g| dα.

Now, define

I =∫ b

a|f |pdα, J =

∫ b

a|g|qdα.

44

If I 6= 0 and J 6= 0 then let c = I1/p and d = J1/q. Then, we have that∫ b

a

(|f |c

)pdα =

1cp

∫ b

a|f |pdα =

1cpI = 1,

and similarly for g ∫ b

a

(|g|d

)qdα =

1dq

∫ b

a|g|qdα =

1dqJ = 1.

Hence, we can apply the results of part (b) to the two functions |f |/c and |g|/d to obtain∫ b

a

|f |c

|g|ddα ≤ 1,

which implies that∫ b

a|f ||g|dα ≤ cd = I1/pJ1/q =

(∫ b

a|f |pdα

)1/p(∫ b

a|g|qdα

)1/q

.

If one of I or J vanish, assume without loss of generality that I vanishes. Then, if c > 0is a constant, applying Equation (10.4) to the two functions |f | and c|g|we obtain that∫ b

a|f |(c|g|)dα ≤ 1

p

∫ b

a|f |pdα+

1q

∫ b

acq|g|qdα = cq

1q

∫ b

a|g|qdα,

since I = 0 by assumption. Thus, since q > 1 (this comes from part (a) where we sawthat 1 < p, q <∞),∣∣∣∣ ∫ b

afg dα

∣∣∣∣ ≤ ∫ b

a|f ||g|dα ≤ cq−1 1

q

∫ b

a|g|qdα→ 0 as c→ 0,

thus the inequality still holds. An analogous argument if J = 0 proves the result. Notethat if both I and J vanish the result holds trivially. �

11. Let α be a fixed increasing function on [a, b]. For u ∈ R(α), define

||u||2 =(∫ b

a|u|2dα

)1/2

.

Suppose that f, g, h ∈ R(α). Prove the triangle inequality,

||f − h||2 ≤ ||f − g||2 + ||g − h||2

as a consequence of the Schwarz inequality of Exercise 10 above, as in the proof of Theorem1.37.

Solution: Letting p = 2 = q in Exercise 10 above we obtain∫ b

a|uv|dα ≤

(∫ b

a|u|2dα

)1/2(∫ b

a|v|2dα

)1/2

, (11.1)

45

because |uv| ≥ 0 so we don’t need the absolute values on the outside of the term on the leftfrom Exercise 10. Then, we have that∫ b

a|u+ v|2dα ≤

∫ b

a|u|2dα+ 2

∫ b

a|uv| dα+

∫ b

a|v|2dα

≤∫ b

a|u|2dα+ 2

(∫ b

a|u|2dα

)1/2(∫ b

a|v|2dα

)1/2

+∫ b

a|v|2dα

=((∫ b

a|u|2dα

)1/2

+(∫ b

a|v|2dα

)1/2)2

,

appling Equation (11.1) to∫ ba |uv| dα. This means that

||u+ v||2 =(∫ b

a|u+ v|2dα

)1/2

≤(∫ b

a|u|2dα

)1/2

+(∫ b

a|v|2dα

)1/2

= ||u||2 + ||v||2.

Letting u = f − g and v = g − h then gives the desired inequality. �

12. Omitted.

13. Omitted.

14. Omitted.

15. Suppose f is a real, continuously differentiable function on [a, b], f(a) = f(b) = 0, and∫ b

af2(x)dx = 1.

Prove that ∫ b

axf(x)f ′(x)dx = −1

2

and that ∫ b

a[f ′(x)]2dx ·

∫ b

ax2f2(x)dx ≥ 1

4.

Solution: First, if we let F (x) = xf2(x) then F ′(x) = 2xf(x)f ′(x) + f2(x). Since f is con-tinuously differentiable on [a, b] so is F hence we can apply the Fundamental Theorem ofCalculus, Theorem 6.21 on page 134 of the text, and obtain∫ b

a[2xf(x)f ′(x) + f2(x)]dx =

∫ b

aF ′(x)dx = F (b)− F (a) = bf(b)2 − af(a)2 = 0.

Therefore,

2∫ b

axf(x)f ′(x)dx = −

∫ b

af2(x)dx = −1 =⇒

∫ b

axf(x)f ′(x)dx = −1

2.

46

We can see the same result by using Integration by Parts, Theorem 6.22 on page 134 of thetext, on the functions F (x) = xf(x) and G(x) = f(x) since we’d obtain∫ b

a(xf(x))f ′(x)dx = −

∫ b

a(xf ′(x) + f(x))f(x)dx = −

∫ b

axf(x)f ′(x)dx−

∫ b

af(x)2dx.

Finally, by Exercise 10(c) (with p = 2) we have that

14

=(∫ b

axf(x)f ′(x)dx

)2

≤∫ b

a(xf(x))2dx ·

∫ b

a[f ′(x)]2dx.

�

Chapter 7: Sequences and Series of Functions

1. Prove that every uniformly convergent sequence of bounded functions is uniformly bounded.

Solution: Suppose that fn → f uniformly on some set E and that for each n there exists anMn such that |fn(x)| ≤Mn for all x ∈ E. Let N be such that

|fn(x)− fm(x)| < 1, ∀ n,m ≥ N, ∀ x ∈ E.

Thus, for n > N we have that

|fn(x)| ≤ |fn(x)− fN (x)|+ |fN (x)| < 1 +MN , ∀ x ∈ E.

Letting M = max{M1, . . . ,MN} we also have that |fn(x)| ≤ M for all 1 ≤ n ≤ N and for allx ∈ E. Therefore, combining these two results we see that |fn(x)| < max{1 +MN ,M} for alln ≥ 1 and for all x ∈ E, hence {fn} is uniformly bounded. �

2. If {fn} and {gn} converge uniformly on a set E, prove that {fn + gn} converges uniformlyon E. If, in addition, {fn} and {gn} are sequences of bounded functions, prove that {fngn}converges uniformly on E.

Solution: Let fn → f uniformly and gn → g uniformly. Let ε > 0 be given. Let N1 besuch that |fn(x) − f(x)| < ε/2 for all n ≥ N1 and for all x ∈ E and let N2 be such that|gn(x) − g(x)| < ε/2 for all n ≥ N2 and for all x ∈ E. Let N = max{N1, N2}. Then, for allx ∈ E and for all n ≥ N we have that

|fn(x) + gn(x)− f(x)− g(x)| ≤ |fn(x)− f(x)|+ |gn(x)− g(x)| < ε,

hence fn + gn → f + g uniformly on E.

Now, suppose that for each n there exist Mn and Ln such that |fn(x)| ≤Mn for all x ∈ E and|gn(x)| ≤ Ln for all x ∈ E. By Exercise 2 above, we therefore have that {fn} and {gn} areboth uniformly bounded so that there exist M and L such that |fn(x)| ≤M for all n ≥ 1 andfor all x ∈ E and |gn(x)| ≤ L for all n ≥ 1 and for all x ∈ E. Let N be such that

|fn(x)− f(x)| < 1, ∀ n ≥ N, ∀ x ∈ E.

Then, we have that|f(x)| ≤ |f(x)− fN (x)|+ |fN (x)| < 1 +M,

47

hence f(x) is bounded. Now, let ε > 0 be given and choose K1 such that

|fn(x)− f(x)| < ε

2L, ∀ n ≥ K1, ∀ x ∈ E.

Similarly, choose K2 such that

|gn(x)− g(x)| < ε

2(1 +M), ∀ n ≥ K2, ∀x ∈ E.

Let K = max{K1,K2}. Then, for all n ≥ K and for all x ∈ E we have that

|fn(x)gn(x)− f(x)g(x)| ≤ |fn(x)gn(x)− f(x)gn(x)|+ |f(x)gn(x)− f(x)g(x)|= |gn(x)||fn(x)− f(x)|+ |f(x)||gn(x)− g(x)|

< L · ε2L

+ (1 +M) · ε

2(1 +M)

=ε

2+ε

2= ε,

hence fngn → fg uniformly on E. Notice, we needed that both {fn} and {gn}were boundedbecause we needed that f(x) was bounded and that gn(x) was bounded. �

3. Omitted.

4. Omitted.

5. Omitted.

6. Prove that the series∞∑n=1

(−1)nx2 + n

n2

converges uniformly in every bounded interval, but does not converge absolutely for anyvalue of x.

Solution: We can write the partial sums of this series as a sum of two series:

N∑n=1

(−1)nx2 + n

n2=

N∑n=1

(−1)nx2

n2+

N∑n=1

(−1)n1n.

Let x ∈ [a, b]. Then x2 ≤M for some M and so∣∣∣∣(−1)nx2

n2

∣∣∣∣ ≤ M

n2, ∀ x ∈ [a, b].

Since∑∞

n=1M/n2 converges, we see by Theorem 7.10 on page 148 of the text that∑∞

n=1(−1)nx2/n2

converges uniformly on [a, b]. That is,∣∣∣∣ m∑n=p

(−1)nx2

n2

∣∣∣∣ ≤ m∑n=p

M

n2,

48

and the series on the right can be made arbitrarily small for all x ∈ [a, b] because it converges.Hence, by the triangle inequality∣∣∣∣ m∑

n=p

(−1)nx2 + n

n2

∣∣∣∣ ≤ ∣∣∣∣ m∑n=p

(−1)nx2

n2

∣∣∣∣+∣∣∣∣ m∑n=p

(−1)n1n

∣∣∣∣,and so the term on the left can be made arbitrarily small for all x ∈ [a, b] because each of thetwo terms on the right can be, since

∑∞n=1(−1)n1/n converges. It is clear the series does not

converge absolutely for any value of x because if we write

m∑n=p

x2 + n

n2=

m∑n=p

x2

n2+

m∑n=p

1n,

we can see that the second term on the right cannot be made arbitrarily small no matter howlarge p and m are because the series

∑∞n=1 1/n diverges, and this of course holds for all x

because that particular series has no dependence on x. Hence also the sum on the left cannotbe made arbitrarily small since the first term on the right is always positive. �

7. For n = 1, 2, . . . and x real, putfn(x) =

x

1 + nx2.

Show that {fn} converges uniformly to a function f , and that the equation

f ′(x) = limn→∞

f ′n(x)

is correct if x 6= 0, but false if x = 0.

Solution: First we show that fn → 0 uniformly. For each n, we have that

f ′n(x) =1

1 + nx2− 2nx2

(1 + nx2)2=

1− nx2

(1 + nx2)2. (7.1)

To find the critical points we set f ′n(x) = 0 hence we wish to solve 1− nx2 = 0 which meansthat x = ±1/

√n. Observe that fn(−1/

√n) < fn(1/

√n) because of the numerator. Note also

that for every n, limx→±∞ fn(x) = 0 (because we have an x in the numerator but an x2 inthe denominator). Thus, we know that fn achieves its maximum at 1/

√n. That is, fn(x)

is bounded and goes to 0 at ±∞ thus its maximum is achieved and occurs at one of thecalculated critical points. In particular, we have that

||fn|| = supx∈R

fn(x) = fn(1/√n) =

1/√n

1 + n(1/n)=

12√n→ 0 as n→∞, ∀ x ∈ R,

thus fn → 0 uniformly. Using Equation (7.1) above we have that f ′n(0) = 1 for all n ≥ 1 andif x 6= 0 then limn→∞ f

′n(x) = 0 (because we have an n in the numerator but an n2 in the

denominator). So the equality limn→∞ f′n(x) = f ′(x) does indeed hold for x 6= 0 but fails at

x = 0 (f being the 0 function here). �

8. If

I(x) ={

0 if x ≤ 01 if x > 0

,

49

and if {xn}∞n=1 is a sequence of distinct points in (a, b), and if∑∞

n=1 |cn| converges, provethat the series

f(x) =∞∑n=1

cnI(x− xn), a ≤ x ≤ b

converges uniformly, and that f is continuous at every x 6= xn.

Solution: Write fn(x) = cnI(x − xn). Then, we wish to show that∑∞

n=1 fn(x) convergesuniformly. We have that for all x ∈ [a, b]

|fn(x)| = |cn||I(x− xn)| ≤ |cn|,

and by hypothesis∑∞

n=1 |cn| converges, therefore∑∞

n=1 fn(x) = f(x) converges uniformlyby the Weierstrass M-Test, Theorem 7.10 on page 148 of the text. Now, let gm(x) =

∑mn=1 cnI(x−

xn). Since x 6= xn for all n ≥ 1 let 0 < δ < min{|x − x1|, . . . , |x − xm|}. Then, we see thatB(x, δ) ∩ {xn}mn=1 = ∅ and therefore if |t− x| < δ we have that

I(x− xn) = I(t− xn) ∀ n = 1, . . . ,m. (8.1)

This is because if x− xn > 0 then

t− xn = t− x+ x− xn > −δ + x− xn > 0

by the definition of δ and because x− xn > 0. Similarly, if x− xn < 0 then

t− xn = t− x+ x− xn < δ + x− xn < 0

again by the definition of δ and because x − xn < 0. By the definition of I then, Equation(8.1) holds. Hence, for all ε > 0 we see that if |t− x| < δ then by Equation (8.1) we have that

|gm(x)− gm(t)| =∣∣∣∣ m∑n=1

cnI(x− xn)−m∑n=1

cnI(t− xn)∣∣∣∣ = 0 < ε,

therefore gm(x) is continuous at all x such that x 6= xn for all m ≥ 1. But then, as we sawabove, f is the uniform limit of these functions, hence is continuous at the same points. Thatis, choosem large enough such that |f(x)−gm(x)| < ε/2 for all x ∈ [a, b]. Then, for |x−t| < δwe have that

|f(x)− f(t)| = |f(x)− gm(x) + gm(x)− gm(t) + gm(t)− f(t)|≤ |f(x)− gm(x)|+ |gm(x)− gm(t)|+ |gm(t)− f(t)|

<ε

2+ 0 +

ε

2= ε.

�

9. Let {fn}∞n=1 be a sequence of continuous functions which converges uniformly to a functionf on a set E. Prove that

limn→∞

fn(xn) = f(x)

for every sequence of points xn ∈ E such that xn → x, and x ∈ E. Is the converse of thistrue?

50

Solution: Let ε > 0 be given. Since fn → f uniformly f is also continuous by Theorem 7.11on page 149 of the text. Specifically, for a large enough (fixed) n we have the inequality

|f(t)− f(x)| ≤ |f(t)− fn(t)|+ |fn(t)− fn(x)|+ |fn(x)− f(x)|,

and all three terms on the right can be made small for t close to x, since fn is continuous,and because fn → f uniformly. Now, let N1 be such that |f(x)− fn(x)| < ε/2 for all n ≥ N1

and for all x ∈ E, i.e. write ||f − fn||∞ = supx∈E |f(x) − fn(x)| < ε/2 for all n ≥ N1.Now, let N2 be such that |f(x) − f(xn)| < ε/2 for all n ≥ N2, which is possible because f iscontinuous hence f(xn) → f(x) as n → ∞ (see Theorem 4.2 on page 84 of the text). Now,let N = max{N1, N2}. Then, for each n ≥ N (i.e. fix an n ≥ N ) we have by the triangleinequality that

|f(x)− fn(xn)| ≤ |f(x)− f(xn)|+ |f(xn)− fn(xn)|≤ |f(x)− f(xn)|+ ||f − fn||∞<ε

2+ε

2= ε,

where we obtain the second inequality with the sup norm because we have fixed an n ≥ N ,hence xn isn’t changing. But, since n was arbitrary, this means that |f(x)−fn(xn)| < ε for alln ≥ N , i.e. fn(xn)→ f(x) as n→∞. Note, we needed that x ∈ E because it was continuityof f at x that allows us to write f(xn) → f(x) as n → ∞ and f must be defined at x to becontinuous there.

�

10. Omitted.

11. Omitted.

12. Omitted.

13. Omitted.

14. Omitted.

15. Suppose f is a real continuous function on R and fn(t) = f(nt) for n = 1, 2, . . . and supposealso that {fn}∞n=1 is equicontinuous on [0, 1]. What conclusion can you draw about f?

Solution: We must have that f is constant on [0,∞). This follows from the following. Letx, y ∈ [0,∞) such that x 6= y and suppose that f(x) 6= f(y). Let ε < |f(x) − f(y)| and letδ > 0 be such that |f(nt)− f(ns)| = |fn(t)− fn(s)| < ε for all t, s ∈ [0, 1] such that |t− s| < δand for all n ≥ 1. There exists an N1 such that

|x− y|N1

< δ.

Furthermore, since both x and y are greater than or equal to 0 we see that there exist N2 andN3 such that

x

N2∈ [0, 1] and

y

N3∈ [0, 1].

51

If we choose N > max{N1, N2, N3} then all three of these conditions hold simultaneously.But then we have that

|f(x)− f(y)| =∣∣∣∣f(N x

N

)− f

(Ny

N

)∣∣∣∣ =∣∣∣∣fN( x

N

)− fN

(y

N

)∣∣∣∣ < ε,

because |x/N − y/N | = |x − y|/N < δ and x/N, y/N ∈ [0, 1]. But, this contradicts thechoice of ε to be smaller than |f(x) − f(y)|, hence we must have that f(x) = f(y). Sincex, y ∈ [0,∞) were arbitrary this shows that f is constant on [0,∞). Note that we cannotuse the equicontinuity to conclude anything about the behavior of f(x) when x < 0 sincet ∈ [0, 1] and n ≥ 1 hence nt ≥ 0 for all n and t where the equicontinuity is known. If,though, we had that the fn’s were equicontinuous on [−1, 1] then we would have found thatf was constant on all of R by analogous arguments. �

16. Suppose {fn}∞n=1 is an equicontinuous sequence of functions on a compact setK and {fn}∞n=1

converges pointwise on K. Prove that {fn}∞n=1 converges uniformly on K.

Solution: Let ε > 0 be given and let δ > 0 be such that |fn(x) − fn(y)| < ε/3 whenever|x − y| < δ for all n ≥ 1 and for all x, y ∈ K, which is possible by the equicontinuity of thesequence {fn}∞n=1. Since K is compact and {B(x, δ)}x∈K is an open cover of K there exists afinite subcover, {B(pi, δ)}ki=1. Since fn(x) → f(x) for every x ∈ K we have that {fn(x)}∞n=1

is a Cauchy sequence for every x ∈ K so, specifically, if 1 ≤ i ≤ k then there exists an Ni

such that|fn(pi)− fm(pi)| <

ε

3, ∀ n,m ≥ Ni,

hence if we let N = maxi{Ni} then we see that

|fn(pi)− fm(pi)| <ε

3, ∀ n,m ≥ N, ∀ 1 ≤ i ≤ k. (16.1)

Now, let x ∈ K be arbitrary and let n,m ≥ N . Since {B(pi, δ)}ki=1 cover K we see thatx ∈ B(pj , δ) for some 1 ≤ j ≤ k. Thus, we have that

|fn(x)− fm(x)| ≤ |fn(x)− fn(pj)|+ |fn(pj)− fm(pj)|+ |fm(pj)− fm(x)|

<ε

3+ε

3+ε

3= ε.

Here, we have that |fn(x) − fn(pj)| < ε/3 because |x − pj | < δ by construction of the opencover {B(pi, δ)}ki=1 hence the inequality follows from equicontinuity, and identically for thelast term, |fm(pj) − fm(x)| < ε/3 also by equicontinuity. The inequality |fn(pj) − fm(pj)| <ε/3 follows from Equation (16.1) and hence our choice of N does not depend on x becauseEquation (16.1) holds for all 1 ≤ i ≤ k and every x ∈ K is in someB(pj , δ) because they coverK. Hence {fn(x)}∞n=1 is uniformly Cauchy and therefore converges uniformly by Theorem7.8 on page 147 of the text. Note, we could not have done the analog of applying the triangleinequality to |fn(x)− f(x)| instead (where f(x) is defined as the pointwise limit of fn(x) foreach x) because this would have produced

|fn(x)− f(x)| ≤ |fn(x)− fn(pj)|+ |fn(pj)− f(pj)|+ |f(pj)− f(x)|.

Here, we can make the first term on the right small by equicontinuity and we can make themiddle term small by choosing N larger than the finitely many Ni’s which make each of

52

|fn(pi)− f(pi)| small, which can be done for each 1 ≤ i ≤ k because fn(pi)→ f(pi) for eachsuch i. But, we have no control over the behavior of the last term |f(pj)− f(x)| since we donot know that f is continuous (because we don’t yet know that fn → f uniformly). �

17. Omitted.

18. Omitted.

19. Omitted.

20. If f is continuous on [0, 1] and if∫ 1

0f(x)xndx = 0, n = 0, 1, 2 . . . ,

prove that f(x) = 0 on [0, 1]. Hint: The integral of the product of f with any polynomial iszero. Use the Weierstrass theorem to show that

∫ 10 f

2(x)dx = 0.

Solution: Let P (x) be any polynomial. Then, we can write

P (x) = anxn + an−1x

n−1 + · · ·+ a1x+ a0.

Hence, we have, by linearity of the integral, that∫ 1

0f(x)P (x)dx =

∫ 1

0f(x)(anxn + an−1x

n−1 + · · ·+ a1x+ a0)dx

= an

∫ 1

0f(x)xndx+ · · ·+ a1

∫ 1

0f(x)x dx+ a0

∫ 1

0f(x)x0dx

= 0.

By The Weierstrass Theorem (Theorem 7.26 on page 159 of the text) we see that there existsa sequence of polynomials Pn such that Pn(x) → f(x) uniformly on [0, 1]. By the above, wehave that

∫ 10 f(x)Pn(x)dx = 0 for all n ≥ 1. By Theorem 7.16 on page 151 of the text, because

this convergence is uniform we can interchange limits and integration and obtain

0 = limn→∞

∫ 1

0f(x)Pn(x)dx =

∫ 1

0limn→∞

f(x)Pn(x)dx =∫ 1

0f2(x)dx.

But, f2(x) ≥ 0 for all x ∈ [0, 1] and since f2 is continuous (because f is and so is the functionx2) we must have that f2(x) = 0 for all x ∈ [0, 1], because otherwise it would be nonzero insome neighborhood within [0, 1] by continuity, hence the integral would be positive becauseit would be positive over that neighborhood and greater than or equal to zero over the restof [0, 1]. But, f2 = 0 on [0, 1] =⇒ f = 0 on [0, 1]. �

53

Rudin: Chapter 1 Solutions

♠ Problem 1. If r is rational and x is irrational, then r + x and rx are irrational.Proof: Suppose r is rational and x is irrational. Then r = m/n for two integers m and n. If r + x wererational, then there would exist integers p and q such that

p

q= r + x,

which would imply that

x =p

q− m

n

=np−mqnq

.

Thus, x is exhibited to be rational, a contradiction, and so r + x must be irrational. Similarly, if rx wererational, then rx = s/t for non-zero integers s and t, which would imply that x = (sn)/(mt) and therebycontradict the assumption that x is irrational. Thus rx is irrational.

♠ Problem 2. Prove that there is no rational number whose square is 12.

Proof: Suppose that there exists positive integers m and n, relatively prime, such that

m2

n2= 12.

Then m2 must be even, which implies that m is even; since m2 = 3(4n2), 3 divides m2 and thus 3 dividesm. It follows that m is divisible by 6, so that m = 6k for some integer k. But then, (6k)2 = 12n2 and so3k2 = n2, which implies that n2, and therefore n is divisible by 3. But this contradicts the premise that nand m are relatively prime.

♠ Problem 3. Prove Proposition 1.15: Let F be a field and let x, y, z ∈ F . Then the following are alltrue:

(a) If x 6= 0 and xy = xz, then y = z.(b) If x 6= 0 and xy = x, then y = 1.(c) If x 6= 0 and xy = 1, then y = 1/x.(d) If x 6= 0, then 1/(1/x)) = x.

Proof: For (a), suppose that xy = xz. Then

y = 1 · y = (1/x)(x)y = (1/x)(xy) = (1/x)(xz) = (1/x)(x)z = z,

as claimed. For (b), take z = 1 in (a). For (c), take z = 1/x in (a). For (d), take y = 1/x in (c).

♠ Problem 4. Let E be a nonempty subset of an ordered set; suppose α is a lower bound of E and β isan upper bound of E. Prove that α ≤ β.Proof: Let E be a nonempty subset of an ordered set, and let α and β be, respectively, a lower bound andan upper bound of E. Since E is nonempty, there exists some element e ∈ E. Since α is a lower bound ofE, α ≤ e; since β is a lower bound of E, e ≤ β. Thus, α ≤ β.

♠ Problem 5. Let A be a nonempty set of real numbers which is bounded below. Let −A be the set ofall numbers −x, where x ∈ A. Prove that inf A = − sup(−A).

1

Proof: Suppose that A is a set of real numbers that is bounded below. Let α = inf A. If y ∈ −A then thereis an x ∈ A such that y = −x ≤ −α. Consequently −α is an upper bound for −A. Now, if β is any otherupper bound for −A, then y ≤ β for all y ∈ −A and this implies that −x ≤ β for all x ∈ A. Hence, −β ≤ xfor every x ∈ A, which means that −β is a lower bound for A. Since α was the greatest lower bound for A,we must have that −β ≤ α, or −α ≤ β. Therefore −α is the least upper bound for −A. In other words,

α = inf A = − sup(−A).

♠ Problem 6. Fix b > 1.A. If m, n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, prove that

(bm)1/n = (b p)1/q.

Proof of A: Suppose that m, n, p, q are integers, n > 0, q > 0, and r = m/n = p/q. Let x = (bm)1/n andy = (bp)1/q. Then xnq = bmq = bnp = ynq, and hence x = y. It therefore makes sense to define br = (bm)1/n.

B. Prove that br+s = brbs if r and s are rational.Proof of B: Suppose that m, n, p, q are integers, n > 0, q > 0, and r = m/n = p/q. Then r + s =(mq + np)/(nq) and hence, using the result of (a),

br+s = (bmq+np)1/nq

= (bmqbnp)1/nq

= (bmq)1/nq(bnp)1/nq

= brbs.

C. If x is real, define B(x) to be the set of all numbers bt where t is rational and t ≤ x. Prove that

br = supB(r),

for rational r.Proof of C: Suppose r and s are rational with s < r. Then br = b(r−s)+s = br−sbs > bs since r − s > 0and b > 1. Therefore, if B(x) = {bt | t ∈ Q and t ≤ x}, we see that supB(r) = br. It therefore makes senseto define bx = supB(x) for all real numbers x.

D. Define bx = supB(x). Prove that bx+y = bxby for all real x and y.Proof of D: Suppose x and y are real and s and t are rational, with s ≤ x and t ≤ y. Then s+ t ≤ x+ yand hence bsbt = bs+t ≤ bx+y. Taking the supremum over rational s ≤ x and t ≤ y shows that bxby ≤ bx+y.To obtain the opposite inequality, let ε > 0 be given. We will show that bxby ≥ bx+y − ε; since ε is arbirary,this will demonstrate that bx+y ≤ bxby. To this end, choose a rational r < x + y so that br ≥ bx+y − ε. Ifδ = x+ y − r > 0, choose rationals s and t so that

x− δ

2< s < x and y − δ

2< t < y.

We then have thatx+ y − δ = r < s+ t < x+ y,

2

and hencebxby ≥ bsbt = bs+t > br > bx+y − ε.

We conclude that bxby ≥ bx+y, which combined with the earlier inequality establishes the result.

♠ Problem 7. Fix b > 1, y > 0, and prove that there is a unique real x such that bx = y by completingthe following outline:A. For any positive integer n, bn − 1 ≥ n(b− 1).Proof of A: Fix b > 1 and let n be a positive integer. By the binomial identity, bn− 1 = (b− 1)(bn−1 +· · ·+ b+ 1) ≥ n(b− 1), since bk > 1 when b > 1.

B. Conclude that b− 1 ≥ n(b1/n − 1).Proof of B: Replace b by b1/n > 1 in Part A.

C. If t > 1 and n > (b− 1)/(t− 1) then b1/n < t.Proof of C: Let t > 1 and suppose that n > (b− 1)/(t− 1). Thus, by Part B,

n(t− 1) > (b− 1) ≥ n(b1/n − 1)

which implies that t > b1/n.

Part D. If w is such that bw < y, then bw+(1/n) < y for sufficiently large n.Proof of D. Let w be such that bw < y. Then 1 < yb−w and we take t = yb−w in Part C. Thus, ifn > (b− 1)/(yb−w), then b1/n < t, which is the same as bw+(1/n) < y.

E. If bw > y, the bw−(1/n) > y for sufficiently large n.Proof of E: Let t = bw/y > 1. Part C implies that if n is sufficiently large, b1/n < t. But this is equivalentto y < bw−(1/n).

F. Let A be the set of all w such that bw < y. Then x = supA satisfies the bx = y.

Proof of F: If bx < y then by Part d, there exists n such that bx+(1/n) < y. But then x is not an upperbound of A, a contradiction. On the other hand, if y < bx, there exists by Part E an integer n such thaty < bx−(1/n), and so x is not the least upper bound of A, a contradiction. We conclude that bx = y.G. Prove that x is unique.Proof of G: Suppose bx = y and bz = y. Then bx = bz, in which case x = z, which establishes uniqueness.

♠ Problem 8. Prove that no order can be defined in the complex plane that turns it into an ordered field.Proof: Let < be any order relation on the complex numbers. Then either i < 0 or i > 0. If i > 0, then−1 = i2 = i · i > 0, which is impossible in an ordered field. On the other hand, if i < 0, then i · i > 0i = 0,i.e., −1 > 0, which is absurd in an ordered field. Thus, no ordering of the complex numbers turns it into anordered field.

♠ Problem 9. Suppose that z = a+ bi, w = c+ di. Define z < w if a < c, and also if a = c and b < d.Prove that this order relation turns the set of complex numbers into an ordered set.Proof: Let z = a+ bi and w = c+ di be two complex numbers. Then either a < c, a = c or c < a. In theformer case we have z < w and in the latter case w < z. If a = c, then either b < d, in which case z < w,or b = d, whence z = w or d < b giving thereby w < z. We have shown that eithe z < w, z = w or w < z.

3

Now suppose that z = a + bi, w = c + di, and u = e + fi are complex numbers with z < w and w < u. Itfollows that either a < c or a = c and b < d; furthermore, either c < e or c = e and d < f . Comparing thesepossible sets of inequalities we observe that either a < e or a = e and b < f ; thus z < u. The set of complexnumbers is thus an ordered set.

The complex numbers with this ordering do not satisfy the Least Upper Bound property. For example,the set

A = {z = a+ bi | a < 0}

is clearly bounded above by 0. But A has no least upper bound. To see this, note that no complex numberwith a nonzero real part can be the least upper bound of A. Thus, if it exists, supA = bi for some b. Sincethis set of complex numbers is not bounded below in the lexical ordering, there is no least upper bound ofA.

♠ Problem 11. If z is a complex number, prove that there exists a real number r ≥ 0 and a complexnumber w with |w| = 1 such that z = rw. Are r and w always uniquely determined by z?Proof: If z = 0, let r = 0 and let w be any complex number satisfying |w| = 1. Then clearly z = rw; notethat w is not uniquely determined in this case. On the other hand, if z 6= 0, then take r = |z| and w = z/|z|;r and w are uniquely determined in this instance.

♠ Problem 12. If z1, . . . , zn are complex, prove that

|z1 + · · ·+ zn| ≤ |z1|+ · · ·+ |zn|.

Proof: We use induction on n. For n = 2 the proposition reduces to the triangle inequality. Suppose thatthe given inequality is satisfied for n ≥ 2. If z1, · · · zn+1 are complex numbers, then

|z1 + · · · zn+1| = |(z1 + · · ·+ zn) + zn+1|≤ |z1 + · · ·+ zn|+ |zn+1| (Triangle inequality)

≤ |z1|+ · · ·+ |zn|+ |zn+1| (Induction hypothesis)

The inequality therefore holds for n+ 1, and by induction, for all n ≥ 2.

♠ Problem 13. If x and y are complex, prove that

| |x| − |y| | ≤ |x− y|.

Proof: Let x and y be complex. Write |x| = |(x − y) + y| and apply the triangle inequality to obtain|x| ≤ |x− y|+ |y|, which we rearrange to obtain

|x| − |y| ≤ |x− y|.

Interchanging the roles of x and y gives the inequality

|y| − |x| ≤ |y − x| = |x− y|.

Combining these two inequalites yields

−|x− y| ≤ |x| − |y| ≤ |x− y|,

4

which is equivalent to | |x| − |y| | ≤ |x− y|.

♠ Problem 14. If z is a complex number with |z| = 1, compute |1 + z|2 + |1− z|2.Solution: Let z be a complex number with |z| = 1. Since |w|2 = ww for any complex number w, we have

|1 + z|2 + |1− z|2 = (1 + z)(1 + z) + (1− z)(1− z)= (1 + z)(1 + z) + (1− z)(1− z)= (1 + z + z + zz) + (1− z − z + zz)= 1 + 1 + 2zz= 4.

♠ Problem 15. Under what conditions does equality hold in the Cauchy-Schwartz inequality?Solution: If either a1 = · · · = an = 0 or if b1 = · · · = bn = 0 both sides of the Cauchy-Schwartz inequalitywill be zero. Suppose that aj ≥ 0 for at least one j, and bk ≥ 0 for at least one k. From the proof of theSchwartz inequality in Rudin we have the identity∑

j

|Baj − Cbj |2 = B(AB − |C|2)

whereA =

∑j

|aj |2, B =∑j

|bj |2, C =∑j

aj bj .

Equality in the Schwarz inequality occurs if, and only if, AB = |C|2; since neither A nor B is 0, this equalitycan occur if, and only if, |Baj − Cbj | = 0 for all j. Equality will therefore hold in the Cauchy inequality if,and only if, aj = ubj for some complex number u.

♠ Problem 16. Suppose k ≥ 3, x, y ∈ Rk, |x− y| = d > 0. Prove:Part A. If 2r > d, there exists infinitely many z ∈ Rk such that

|z − x| = |z − y| = r.

Proof of A: We proceed by proving that |z − x| = |z − y| if, and only if(z − x+ y

2

)· (y − x) = 0.

What is our intuition here: z is equidistant from x and y if, and only if, it lies on the “hyperplane”perpendicular to the line segment between x and y at its midpoint. We now verify the claim: Note that(z − x+y

2

)· (y − x) = 0 if, and only if,

[(z − x) + (z − y)] · [(z − x)− (z − y)] = 0.

Distributing the dot product and simplifying shows that the last equality is equivalent to |z − x| = |z − y|,as claimed. Given this fact, we now see that any z that satisfies this equality must also satisfy

|z − x|2 =∣∣∣∣x− y2

∣∣∣∣2 +∣∣∣∣z − x− y2

∣∣∣∣2 .5

If we put |z − x| = r and |x− y| = d, then the last equality is equivalent to∣∣∣∣z − x− y2

∣∣∣∣2 = r2 − (d/2)2.

This equation has infinitely many solutions for z whenever d < 2r. For Part B, if d = 2r, there is exactlyone solution, namely z = 1

2 (x + y). There are no solutions if r < 2d, since the right hand side of the lastdisplayed equation is negative; this gives Part C. The results for k = 2 and k = 1 are straightforward.

♠ Problem 17. Prove that |x+ y|2 + |x− y|2 = 2|x|2 + 2|y|2 for x, y ∈ Rk. Interpret geometrically.Proof: Computing norms using the scalar product, we have

|x+ y|2 + |x− y|2 = (x+ y) · (x+ y) + (x− y) · (x− y)= (x · x+ 2x · y + y · y) + (x · x− 2x · y + y · y)= 2x · x+ 2y · y= 2|x|2 + 2|y|2.

A geometric interpretation: the sum of the squares of length of the sides of a parallelogram is equal to thesum of the squares of the lengths of the diagonals.

♠ Problem 18. If k ≥ 2 and x ∈ Rk prove that there exists y ∈ Rk such that y 6= 0 but x · y = 0. Isthis result true for k = 1?Proof: In R1, if x 6= 0 then xy = 0 iff y = 0, so the claim is false unless x = 0. Let k ≥ 2, and let x ∈ Rk.If x = 0, choose y ∈ Rk with y 6= 0; then x · y = 0. Now suppose that x = (x1, x2, . . . , xk) where xi 6= 0for some i with 1 ≤ i ≤ k. Let y = (y1, y2, . . . , yk) ∈ Rk, in which case x · y = x1y1 + · · ·+ xiyi + · · ·xkyk.Should xj = 0 for all j 6= i, then let yi = 0 and yj = 1 for j 6= i, whence x · y = 0. If xn 6= 0 for some n 6= i,choose yi = 1, yn = −xi/xn and yj = 0 for j 6= i, n. Then x · y = 0 and y 6= 0.

♠ Problem 19. Suppose a ∈ Rk and b ∈ Rk. Find c ∈ Rk and r > 0 such that

|x− a| = 2|x− b|

if, and only if, |x− c| = r.

Proof: We proceed directly, using properties of the dot product:

|x− a| = 2|x− b| ⇔ |x− a|2 = 4|x− b|2

⇔ (x− a) · (x− a) = 4(x− b) · (x− b)⇔ 3x · x− 8x · b+ 2x · a+ 4b · b− a · a = 0 (Expand and collect)

⇔ x · x− (83b− 2

3a) · x+

43b · b− 1

3a · a = 0

⇔ x · x− 2 (43b− 1

3a)︸︷︷︸

c

·x+43b · b− 1

3a · a = 0

⇔ x · x− 2c · x+ c · c− c · c+43b · b− 1

3a · a = 0

⇔ (x− c) · (x− c) = c · c− 43b · b+

13a · a

6

⇔ (x− c) · (x− c) =49

(b · b− 2b · a+ a · a) (Expand c and collect)

⇔ (x− c) · (x− c) =49

(b− a) · (b− a)

⇔ |x− c|2 =49|b− a|2

⇔ |x− c| = r ≡ 23|b− a|

7

Rudin: Chapter 2 Solutions

Problem 1. Prove that the empty set is a subset of every set.

Proof: Let A be a set. If φ is not a subset of A, then there would exista ∈ φ 3 a /∈ A. But φ is empty. Thus φ is a subset of A.

Problem 2. Prove that the set of algebraic numbers is countable.

Proof: Given an integer n ≥ 0, and a set of integers {a0, a1, . . . , an}, theequation

a0zn + a1z

n−1 + · · ·+ an−1z + an = 0

has at most n distinct roots in C, by the Fundamental Theory of Algebra,which we assume. Since Z × Z × · · · × Z is countable by Theorem 2.13, theset of roots is a countable union of finite sets and is therefore countable.

Problem 3. Prove there exist real numbers that are not algebraic.

Proof: Since R is uncountable and the set of algebraic numbers is countable,there are clearly are real numbers that are not algebraic. In fact, “most” ofthe them are not algebraic!

Problem 4. Is the set of irrational real numbers countable?

Proof: Clearly the set or irrational numbers is uncountable, since R isuncountable and Q ⊂ R is countable. Thus Qc is uncountable.

Problem 5. Constuct a bounded set of real numbers with exactly threelimit points.

Proof:

E = { 1

n| n ∈ Z+} ∪ {1 +

1

n| n ∈ Z+} ∪ {2 +

1

n| n ∈ Z+}.

Clearly 0,1, and 2 are the only limit points of E.

Problem 6. Let E ′ be the set of limit points of E. Prove (a) E ′ is closedand (b) E and E have the same limit points. (c) Do E and E ′ always havethe same limit points?

Proof: (a) To show E ′ is closed we show (E ′)c is open. Let x ∈ (E ′)c. Thenx is not a limit point of E. Thus there exists a neighborhood N ⊃ x 3 N ∩Ehas at most finitely many points of E. This implies that no limit point of Ecan be contained in N , i.e. N ⊂ (E ′)c ∴ (E ′)c is open and E ′ is closed.

(b) Since E ⊂ E, clearly a limit point of E is a limit point of E. Thus

E ′ ⊂ E′. Suppose x is a limit point of E. Since E is closed, x ∈ E = E ∪E ′.

1

If x ∈ E ′, we are done. If x ∈ E, then, since x is a limit point of E, everyneighborhood of x contains infinitely many points of E (and hence of E),

∴ x ∈ E ′ in any case. Thus E′ ⊂ E ′ ∴ E ′ = E

′.

(c) No. Consider E = { 1n| n ≥ 1}. Then E ′ = {0}. Thus E has 1 limit

point, while E ′ has none.

Problem 7. Suppose A1, A2, . . . are subsets of a metric space.(a) If Bn =

⋃ni=1 Ai, prove that Bn =

⋃ni=1 Ai.

(b) If B =⋃∞i=1 Ai, prove that B ⊃

⋃∞i=1 Ai.

Show that the inclusion can be proper.

Proof: (a) For each Ai, Ai ⊂ Ai ∴⋃ni=1 Ai ⊂

⋃ni=1 Ai. But

⋃ni=1 Ai is

closed ∴⋃ni=1 Ai = Bn ⊂

⋃ni=1 A1. To prove the opposite inclusion, for

each i, Ai ⊂⋃ni=1 Ai ∴ Ai ⊂

⋃ni=1 Ai ∴

⋃ni=1 Ai ⊂ Bn =

⋃ni=1 Ai. Thus

Bn =⋃ni=1 Ai. Notice that we used the fact that the collection of sets is

finite, since⋃ni=1 Ai is assumed to be closed.

(b) For every i, Ai ⊂⋃ni=1 Ai ∴ Ai ∈

⋃ni=1 Ai ∴

⋃ni=1 Ai ⊂ B.

Our counterexample: An =[

1n, 1− 1

n

], n ≥ 2. Ai = Ai.

⋃∞i=1 Ai =

(0, 1) 6=⋃ni=1 Ai = [0, 1] .

Problem 8. Is every point of every open set E ⊂ R2 a limit point of E?

Proof: Yes. Let x ∈ E. Given 1n, n ≥ 1, select xn 6= x, with xn ∈ N 1

n(n)∪E.

Thus every neighborhood of X will clearly contain infinitely many of the {xn}and so x is a limit point of E.

Is the same thing true for closed sets?Proof: No. Let E = {(0, 0)}. E is closed but has no limit points.

Problem 9. Let E◦ = {x ∈ E | x is an interior point}.a) Prove that E◦ is open.b) Prove that E is open iff E◦ = E.c) If G ⊂ E and G is open, prove that G ⊂ E◦.d) Prove that the complement of E◦ is the closure of the complement of E.e) Do E and E always have the same interiors?f) Do E and E◦ always have the same closures?

Proof: a) To show that E◦ is open, we need to show that every x ∈ E◦

is an interior point of E◦, i.e., that there is a neighborhood of x containedin E◦. Since x ∈ E◦, ∃Nr (x) ⊂ E. Let y ∈ Nr (x). Since Nr (x) is open,∃Ns (y) ⊂ Nr (x) ∴ Ns (y) ⊂ E ∴ y ∈ E◦ ∴ Nr (x) ∈ E◦.

2

b) For any E, we have E◦ ⊂ E. If E is open, then every point of E isan interior point of E ∴ E ⊂ E◦ ∴ E = E◦. If E = E◦, since E◦ is open wehave that E is open.

c) Suppose G ⊂ E and G is open. If x ∈ G, then ∃Nr (x) ⊂ G ⊂ E ∴x ∈ E◦ ∴ g ⊂ E◦.

d) Since E◦ ⊂ E, we have that Ec ⊂ (E◦)c. Since E◦ is open, (E◦)c

is closed ∴ Ec ⊂ (E◦)c. For the opposite inclusion, let x ∈ (E◦)c. Thenx ∈ Ec or x ∈ E − E◦. If x ∈ Ec, then x ∈ Ec. Otherwise, x ∈ E − E◦.Consequently, since x is not an interior point of E, every neighborhood of xmust have a non-empty intersection with Ec ∴ x is a limit point of Ec andhence x ⊂ Ec ∴ (E◦)c ⊂ Ec and we are done.

e) No. Let E = {r ∈ Q | 0 < r < 1}. Then E◦ = φ, but E◦

= (0, 1).f) No. Let E = Q. Then E◦ = φ and E◦ = φ but E = R.

Problem 10. Let X an infinite set. For p ∈ X, q ∈ X, define

d (p, q) =

{0 if p = q1 if p 6= q.

Prove (X, d) is a metric space.

Proof: Clearly (p, q) = 0 iff p = q, and d (p, q) = d (q, p). Let p, q, r ∈ X.Then d (p, q) ≤ d (p, r) + d (r, q), as we easily see by examining the variouscases: p 6= q, p = r; p 6= q, r = q; p 6= q, r 6= p, r 6= q.

Note that singleton sets in (X, d) are open, since ∀p ∈ X,N 12

(p) = {p} ⊂{p}. Since singleton subsets of X are open, and arbitrary unions of opensets are open, then all subsets of X are open. Since any subset of X is thecomplement of its complement, all subsets of X are also closed!

Compact sets in this metric space are finite.

Problem 11. a) d1 (x, y) = (x− y)2 is not a metric: d1 (0, 2) = 4.d1 (0, 1) + d1 (1, 2) = 1 + 1 = 2, so the triangle inequality fails.

b) This is a metric. We prove the triangle inequality. Because the absolutevalue satisfies the triangle inequality,

√|x− y| ≤

√|x− z|+ |z − y|. Then,

because√a+ b ≤

√a+√b for any positive numbers a and b, this is less than√

|x− z|+√|z − y|; therefore, d2 (x, y) ≤ d2 (x, z) + d2 (z, y).

c) d3 (x, y) = |x2 − y2| is not a metric since d3 (1,−1) = 0 and 1 6= −1.d) d4 (x, y) = |x− 2y| is not a metric as d4 (1, 1) 6= 0

e) d5 (x, y) = |x−y|1+|x−y| is a metric. It can be easily verified that d5 (x, y) = 0

iff x = y and that d5 (x, y) = d5 (y, x). Thus we only need prove that thetriangle inequality holds.

3

Note that if 0 < a < b, we have a1+a

< b1+b

. Thus,

|x− y|1 + |x− y|

≤ |x− z|+ |z − y|1 + |x− z|+ |z − y|

=|x− z|

1 + |x− z|+ |z − y|+

|z − y|1 + |x− z|+ |z − y|

≤ |x− z|1 + |x− z|

+|z − z|

1 + |z − y|.

Thus d5 (x, y) ≤ d5 (x, z) + d5 (z, y), and the triangle inequality holds.

Problem 12. Let K = {0} ∪ { 1n| n ≥ 1}. Prove directly that K is

compact.

Proof: Let {Gα} be any open cover of K. Then for some α0, α0 ∈ Gα0 .Since Gα0 is open, ∃r > 0 3 Nr (0) ⊂ Gα0 . Choose an integer n ≥ 1 suchthat 1

n< r.

Clearly, for all m ≥ n, 1m∈ Nr (0). For k = 1, 2, . . . , n1, select αk 3 1

k∈

Gαk . Thus K ⊂ Gα0 ∪Gα1 ∪ · · · ∪Gαn and K is thus compact.

Problem 13. Construct a compact set of real numbers whose limit pointsform a countable set.

Proof: Let E = {0}∪{

12n

m+2m+1

, n = 1, 2, 3, . . . ,m = 0, 1, 2, 3, . . .}

. Note thatm+2m+1≤ 2 for m ≥ 0. E is obviously bounded, and the limit points of E are

{0} ∪{

12n| n ≥ 1

}⊂ E ∴ E is closed.

Problem 14. Give an example of an open cover of the segment (0, 1)which has no finite subcover.

Proof: Let Gn =(

1n, 1). Then (0, 1) ⊂

⋃∞n=1 Gn and no finite subset of

(Gn)n≥1 covers (0, 1).

Problem 15. Show that Theorem 2.36 and its Corollary become false(in R1, for example) if the word “compact” is replaced by “closed” or by“bounded.”

Proof: Let Fn =[

1n,∞). Then Fn satisfies the finite intersection property,

but⋂∞n=1 Fn = φ.

Let Vn =(0, 1

n

). Then the Vn are bounded, and satisfy the finite inter-

section property, but⋂∞n=1 Vn = φ.

Problem 16. Regard Q, the set of all rational numbers as a metric space,with d (p, q) = |p − q|. Let E be the set of all p ∈ Q such that 2 < p2 < 3.

4

Show that E is closed and bounded in Q, but that E is not compact. Is Eopen in Q?

Proof: E = {p ∈ Q+ | 2 < p2 < 3} ⊂ (Q, d) where (Q, d) is the metricspace of rationals with the Euclidian metric. That E is closed follows fromthe fact that E =

[√2,√

3]∩ Q and

[√2,√

3]

is closed in R ∴ E is closedrelative to (Q, d). That E is not compact follows from the fact that E is notcompact in R. E is open because it is open in R.

Problem 17. Let E be the set of all x ∈ [0, 1] whose decimal expansioncontains only the digits 4 and 7. Is E countable? Is E dense in [0, 1]? Is Ecompact? Is E perfect?

Proof: E is not countable, by Cantor’s diagonal argument. E is clearly notdense in [0, 1]. E is closed and bounded, and therefore compact. Let x ∈ E.Then x = 0.α1α2α3 . . . , where αk = 4 or 7 for all k. For every n ≥ 1, letxn = 0.α1α2 . . . αn44. Then, given ε = 10−n, |xn − x| < 3.3 × 10−(n+1),andclearly x is a limit point of E. Thus E is perfect.

Problem 18. Is there a nonempty perfect set it R which contains norational numbers?

Proof: In problem 17, we constructed a perfect set E, whose elements areof the form o.α1α2α3 . . . , where αk = 4 or 7. However, E contains rationalnumbers, like x = 0.4.

Clearly, if we consider the set E∗ = {x+ c | x ∈ E}, for any c ∈ R, thenE∗ is perfect if E is perfect. Thus, we try to select c ∈ Qc 3 x+c ∈ Qc∀x ∈ E.

Let c = 0.101001000100001 . . . where the number of zeroes between eachsuccessive pair of ones increases by one. Let x ∈ E and examine x + c.Suppose x ∈ Qc. Then x has an infinite number of 4’s and 7’s. Thusx + c has an infinite number of 5’s or 8’s or both. Suppose without loss ofgenerality that x+ c has an infinite number of 5’s. Suppose x+ c ∈ Q. Thenx+ c = 0.d1d2d3 . . . dkr1r2 . . . rn where d1, . . . , dk are leading digits followedby a repeating block r1, . . . , rn of some length. Since x + c has an infinitenumber of 5’s, one of the digits r1, . . . , rn is a 5. This requires a 5 to occurevery n digits. But beyond some point, there are always more than n 0’sbetween each successive 1; a contradiction. Thus x + c ∈ Qc and so E∗ hasno rationals.Problem 22. Show that Rk is separable.

Proof: Let Qk = {(r1, . . . , rk) | ri ∈ Q, 1 ≤ i ≤ k}. Let x = (x1, . . . , xk) ∈Rk and let Nr be a neighborhood containing x. Given this r > 0, for each i

5

choose ri ∈ Q so that |xi − ri| < r/√n. Let r = (r1, . . . , rk). Thus,

|x− r| = [(x1 − r1)2 + · · ·+ (xk − rk)2]1/2

≤ [n(r/√n)2]1/2

= r,

which implies that Qk is dense in Rk.

Problem 23. Prove that every separable metric space has a countablebase.

Proof: Let X be a separable metric space, and let Y = {yk} be a countabledense subset of X. Given yk ∈ Y , define a family of neighborhoods {Nr(yk) |r ∈ Q, r > 0}. We prove that the collection {Nr(yk)|r ∈ Q+, yk ∈ Y } is acountable base forX. Let x ∈ X and suppose thatG is an open set containingx. Since G is open, there exists a rational δ > 0 such that Nδ(x) ∈ G. SinceY is dense in X, given δ/2 there exists yk such that yk ∈ Nδ/2(x). But thenx ∈ Nδ/2(yk) ∈ Nδ(x) ∈ G, which is what we sought to prove.

Problem 25. Every compact metric space K has a countable base.

Proof: Let n ≥ 1 be an integer. The collection {N 1n(x)}x∈X surely covers

X. Since X is compact, a finite subcollection {N 1n(xnk)} covers X. We will

show that {N 1n(xnk) | n ≥ 1} is a base for X. Let x ∈ X and let G be any

open set containing x. There then exists a neighborhood Nr(x) ∈ G. Choosen sufficiently large so that 1/n < r/2; then x ∈ {N 1

n(xnk)} for some k. Let

y ∈ {N 1n(xnk)}. Then d(y, x) ≤ d(y, xnk) + d(xnk, x) < r/2 + r/2 = r and so

{N 1n(xnk)} ∈ Nr(x) ∈ G.

Problem 26. Let X be a metric space in which every infinite subset hasa limit point. Then X is compact.

Proof: Using Problem 23 and Problem 24, X has a countable base{Vn}∞1 . Let {Gα}α∈A be an open cover of X; it follows that a countablesubcover {Gn}∞1 exists. Suppose that no finite subcollection of {Gn} coversX. Then for each n ≥ 1, select xn ∈ (G1 ∪ G2 ∪ · · · ∪ Gn)c. The set {xn}is infinite, and therefore has a limit point x ∈ X. Since {Gn} covers X,x ∈ Gk for some k. Since Gk is open, and x is a limit point of {xn}, thereexist infinitely many integers j such that xj ∈ Gk. In particular there existssome j > k for which xj ∈ Gk. But, by construction, xj ∈ Gc

k, which is acontradiction. Thus X is compact.

6

Problem 27. Suppose E ⊂ Rk is uncountable, and let P be the set ofcondensation points of E. Then P is perfect and that at most a countablenumber of points of E are not in P .

Proof: Let E ⊂ Rk be uncountable for some Rk, and let P be the set ofcondensation points of E. We first prove that P excludes at most countablymany points of E. Since Rk is separable, it has a countable base {Vn}. Thus∪nVn ⊃ E. Let W be the union of all Vn for which E∩Vn is at most countable.We will show that P = W c. Let p ∈ P . Because p is a condensation pointof E, given Vk ⊃ p there are uncountably many points of E in Vk. ThusVk ∩ W = ∅ and so p ∈ W c =⇒ P ⊂ W c. Let q ∈ W c and let Vk ⊃ q.Then Vk ∩E is uncountable, and so q ∈ P . Thus W c ⊂ P and together withthe previous inclusion we have P = W c. We now show that P is perfect.Since P is the complement of an open set it is closed. Let q be a limit pointof P . We will show that q is a condensation point of E and hence is in P .Let q be a limit point of P . Given a neighborhood Nr(q) ⊃ q, there existsp ∈ P , p 6= q such that p ∈ Nr(q). But p is a condensation point of E. Thusthe neighborhood Nr−d(p,q)(p) ⊃ p has uncountably many points of E, andis a subset of Nr(q). Thus q is a condensation point of E, and therefore amember of P .

Problem 28. Every closed set in a separable metric space is the unionof a (possibly empty) perfect set and a set which is at most countable.

Proof: Let F be a closed subset of a separable metric space. If F is count-able the proposition is evident. If F is uncountable, let P be the set ofcondensation points of F . Since F is closed, P ⊂ F . By Problem 27, Pis perfect and excludes at most countably many points of F . Thus we canwrite F = P ∪ C, where P is perfect and C is at most countable.

Problem 29. Prove that every open set in R1 is the union of an at mostcountable collection of disjoint segments.

Proof: Let E ⊂ R1 be open. Since R1 is separable, it has a countable base.In fact, if we let ψr(p) = {x : p− r < x < p+ r} then{

ψr(p) : p ∈ Q, r ∈ Q+}.

is a countable base of segments. We may therefore write E = ∪αψα for anat most countable sub-collection {ψα} of this particular base. We need toshow that we may rewrite ∪αψα as a disjoint union of segments. To thisend, we define a component of {ψα} as a maximal sub-collection {ψβ} of

7

{ψα} with the property that, if p, q ∈ ∪βψβ then tp + (1 − t)q ∈ ∪βψβfor all t ∈ [0, 1]. Clearly {ψα} has at most a countable number of distinctcomponents, denoted {ψβ}, {ψγ}, . . . . If we let S1 = ∪βψβ, S2 = ∪γψγ, . . . ,then ⋃

α

ψα =⋃n

Sn,

because every member of {ψα} is in at lest one component. We claim that{Sn} is a disjoint collection of segments. Sn is clearly open; given any twopoints p, q ∈ Sn, the segment (p, q) ⊂ Sn, so that Sn must itself be a seg-ment. It is equally clear that segments Sn and Sm corresponding to distinctcomponents must be disjoint, for otherwise the component corresponding toSn ∪ Sm would properly contain the components corresponding to Sn andSm, contradicting the requirement of maximality for a component.

Problem 30. If Gn is a dense, open subset of Rk for n = 1, 2, 3, . . . ,then ∩∞1 Gn is dense in Rk.

Proof: Let G1, G2, . . . be open dense subsets of some Rk. We will showthatB0∩(∩∞1 Gn) 6= ∅ for each openB0 ⊂ Rk. Start withB0 andG1. SinceG1

is dense, B0 ∩ G1 6= ∅, and so there exists a non-empty open neighborhoodB1 such that B1 ⊂ B1 ⊂ B0 ∩ G1. Now use B1 and G2. Because G2 isdense, B1 ∩ G2 6= ∅, and there exists an open neighborhood B2 such thatB2 ⊂ B2 ⊂ B1 ∩ G2. Proceeding by induction, there exists a collection{Bn}∞1 of open sets such that Bn ⊂ Bn−1∩Gn for each n ≥ 1. The collectionof closed (compact) sets {Bn} satisfy the finite intersection property, whichimplies that ∩∞1 Bn 6= ∅. But

∞⋂1

Bn ⊂ B0 ∩

(∞⋂1

Gn

)

since B1 ⊂ B0 ∩G1 and Bn ⊂ Gn for n ≥ 1, which establishes the result.

8

18.100B, FALL 2002SOLUTIONS TO RUDIN, CHAPTER 4, PROBLEMS 2,3,4,6

Problem 2

If f : X −→ Y is a continuous map then f−1(C) ⊂ X is closed for each closedsubset C ⊂ Y. For any map and any subset G ⊂ Y , f(f−1(G)) = G. Now, if E ⊂ X

then C = f(E) is closed and E ⊂ f−1(C) (since x ∈ E implies f(e) ∈ f(E) ⊂ Cimplies e ∈ f−1(C)). By the continuity condition f−1(C) is closed so E ⊂ f−1(C)which implies f(E) ⊂ f(E).

Consider X = [0, 1) ∪ [1, 2] as a subset of R with the usual metric. Then f :X −→ [0, 1] given by

f(x) =

{x x ∈ [0, 1)1 x ∈ [1, 2]

is continuous since a convergent sequence {xn} in X is eventually in [0, 1) or [1, 2]and so {f(xn)} converges by the continuity of x and constant maps. On the otherhand, E = [0, 1) is closed and f(E) is not, with f(E) = [0, 1] so f(E) is strictlycontained in f(E).

Problem 3

By definition Z(f) = f−1({0}). The set {0} is closed and f is continuous, soZ(f) is closed.

Problem 4

If y ∈ f(X) then there exists x ∈ X such that f(x) = y. By the density of Ein X there is a sequence {xn} in E with xn → x in X. By the continuity of f,f(xn) → f(x) = y so f(E) is dense in f(X).

Suppose g(p) = f(p) for all p ∈ E. Given x ∈ X, by the result above, there exists{xn} in E such that xn → x and f(xn) → f(x). The continuity of g means thatg(xn) = f(xn) → g(x) so f(x) = g(x) for all x ∈ X.

Problem 6

The distance on X ×Y is the sum of the distances on X and Y. I will do it withsequences.

Suppose E is compact and f : E −→ Y is continuous. Now suppose {pn} is asequence in graph(f). Thus, pn = (xn, f(xn)) for some sequence {xn} in E. By thecompactness of E, there is a convergent subsequence {xn(k)}. By the continuity off, f(xn(k)) is convergent, and hence pn = (xn(k), f(xn(k))) is convergent, so eachsequence in graph(f) has a convergent subsequence. It follows that it is compact.

Conversely, suppose that E and graph(f) are both compact. Let {xn} be aconvergent sequence in E, xn → x. Then {(xn, f(xn))} is a sequence in graph(f)so by its compactness has a convergent subsequence, (xn(k), f(xn(k))) → (x, q).Since the graph is closed, this must be a point in it, so q = f(x). This argument

1

2 18.100B, FALL 2002 SOLUTIONS TO RUDIN, CHAPTER 4, PROBLEMS 2,3,4,6

applies to any subsequence of {xn}, so we see that any subsequence of {f(xn)} hasa convergent subsequence with limit f(x). This however implies that f(xn) → f(x),since if not there would exist a sequence f(xn(k) with d(f(x), f(xn(k)) ≥ c > 0 andthis cannot have such a convergent subsequence. Thus in fact xn → x implies thatf(xn) → f(x), so f is continuous.

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