analysis solutions

Analysis Qualifying Exam Solutions

Thomas Goller

September 4, 2011

Contents

1 Spring 2011 21.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Fall 2010 62.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 8


4 Fall 2009 154.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 17


6 Fall 2008 246.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

7 Spring 2008 277.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1

Chapter 1

Spring 2011

1.1 Real Analysis

A1. (a) `1(Z) is separable. A countable set whose finite linear combinationsare dense is {en}nZ, where en has a 1 in the nth position and is 0 everywhereelse. If x `1(Z), then the sums Nk=N xkek approximate x arbitrarily wellin the norm as N since k |xk| 0, we have

|f |1((, a)) = f1((a, a)).Now, f measurable implies f1

((, a)) is measurable for each a R, so each

f1([a,)) is measurable by taking complements. Therefore each f1((a,))

is measurable since

f1((a,)) =

n=1

f1([a+

1

n,)),

so f1((a, a)) = f1((, a)) f1((a,)) is measurable, as desired.

2

(b) |f | measurable does not imply that f is measurable, as long as Xcontains non-measurable sets. For if N is a non-measurable set, define

f(x) =

{1 x N ;1 otherwise.

Then |f | 1 is measurable, but f1((, 0)) = N is not measurable.A3. Havent reviewed Fourier series.

A4. f L1 implies |f | < , and f uniformly continuous means thatfor any > 0, there is a > 0 such that |f(x) f(y)| < whenever |x y| 0 such that for each N > 0, there exists n Nsuch that |f(xn)| . Thus x has a subsequence {xnk} such that |f(xnk)| for each k 1.

We may assume the xnk are distinct, deleting repeats if necessary. Thenby uniform continuity, choose > 0 such that |f(x) f(y)| < 2 whenever|x y| < . Also, make small enough so that |xnk xnj | whenever k 6= j.Then taking a ball of radius 2 around each xk and letting E be the union ofthese balls, we see that

|f | E

|f | k=1

2

=,

contradicting the fact that f L1.

A5. Without (1), consider the sequence

{fn} ={[0,1],

1

2[0,2],

1

4[0,4], . . .

},

which converges pointwise to the function f 0 for all x. Then (1) fails and

limn

|fn f | = lim

n

|fn| = 1 6= 0.

With (1), we can prove the result. Set gn = inf{fn, fn+1, . . . }, which ismeasurable since the fn are measurable. Then 0 gn fn and gn f , so |gnf | 0 by the dominated convergence theorem, whence also gn f .Now we use (1) and the fact that |fn gn| = fn gn to deduce that

|fn f | |fn gn|+

|gn f | =

fn

gn +

|gn f | 0.

3

1.2 Complex Analysis

B6. g is holomorphic on U . Let z0 U . Since f is holomorphic at z0 U , wecan write f(z) =

n=0 an(z z0)n for z near z0. Thus

g(z) := f(z) =

n=0

an(z z0)n

is a power series expansion for g near z0.

B7. Since |f(z)| C|z|k for z outside DR, all the poles of f are inside DR,and there can be only finitely many such poles since DR is compact. Let f(z)be the sum of the principal parts of f at each of these poles. Then g := f fis an entire function that still satisfies a bound of the form |g(z)| C |z|k for|z| R since the principal parts are bounded away from their poles.

We claim that g is a polynomial of degree k, which will imply that f is arational function. Since g is entire, it suffices to prove that g(j)(0) = 0 for allj > k (look at the power series expansion). For each R R > 0, the Cauchyinequalities imply

|g(k+1)(0)| (k + 1)!Rk+1

C Rk 0,

as R , giving the desired result.

B8. (i) = (ii): Suppose a is a zero of f of order l k. Let f(z) =(za)ln0 an(za)n be the power series expansion of f near a, so that a0 6= 0.Let C = |a0| and choose 0 < < 1 sufficiently small such that B(a) U and|n1 ann| |a0|. Then for all z B(a),

|f(z)| |z a|l(|a0|+ |a0|) 2|a0||z a|k.

(ii) = (i): Suppose (ii) holds and let f(z) = n0 an(z a)n be thepower series expansion for f near a. Then for all z B(a) with z 6= a, f(z)(z a)k

Cimplies a0 = a1 = = ak1 = 0.

B9. The entire function sin z := 12i (eiz eiz) has zeroes at pik, k Z,

an essential singularity at, and no other zeroes or singularities. The functionsin 1z therefore has zeroes at

1pik and an essential singularity at 0, with no other

zeroes or singularities.

4

Since sin z = z z33! + z5

5! we also have sin 1z = z1 z33! +

z55! .

Thus the residue at 0 of the product of these two series, namely the coefficientof z1 of the Laurent series, is 0.

Is this the right answer?

B10. Set

f(z) =zeiz

z2 + 2z + 10.

The denominator factors into (z + 3i + 1)(z 3i + 1), so there are poles at3i 1 and 3i 1. We will integrate f over the upper-half-plane semicircleconsisting of , the real-axis interval [R,R], and the semicircle R of radiusR. The imaginary part of the integral over gives the real integral we want toevaluate. Since we are taking R large, we pick up the residue at 3i 1, which is

limz3i1

(z 3i+ 1) zeiz

(z + 3i+ 1)(z 3i+ 1) =(3i 1)e3i

6i

=1

6e3(3 cos 1 + sin 1 + i(cos 1 3 sin 1)).

What a mess! The integral over R vanishes thanks to the 1/z decay of fnear the real axis, together with the better decay of f due to eiz away from thereal axis.

5

Chapter 2

Fall 2010

2.1 Real Analysis

A1. Yes, the fact that f1((r,)) is measurable for each r Q implies that

f is measurable. Taking complements, we see that f1((, r]) is measurable

for each r Q. Now for any s R, let rk be an increasing sequence of rationalnumbers < s such that rk s. Then

f1((, s)) =

k=1

f1 ((, rk]) ,

so f is measurable.

A2. Suppose 1 p q such that 1p + 1q = 1, where we allowp = 1, q = . Since f, fn Lp and g, gn Lq, we also have f fn Lp andg gn Lq. Holders inequality implies fngn and fg are in L1. Note that|fg| |fngn| = |fg| |fng|+ |fng| |fngn| |g||f fn|+ |fn||g gn|.Thus using Holders inequality, |fg| |fngn| |g||f fn|+ |fn||g gn|

= g(f fn)1 + fn(g gn)1 gqf fnp + fnpg gnq.

Since gq and fnp are bounded but f fnp 0 and g gnq 0, wesee that fg fngn1 0, which gives the desired result since |fg| |fngn| |fg| |fngn| 0.

6

(Maybe Im missing something, but I dont think the case p = 1, q = isany different than p > 1.)

A3. Let HTn // C denote the bounded linear functional defined by

Tn(y) = (y, xn). Then the countable collection {Tn} is pointwise bounded byassumption, hence the norms are bounded by the uniform boundedness property.So there exists M 0 such that

|(y, xn)| My for all y H.In particular, choosing y = xn gives xn M , establishing the desired result.

A4. Suppose f is measurable such that |f |p


B6. We write

f(z) = 1 cos z = 1 eiz + eiz

2=(eiz2 e iz2

)2,

so the zeroes of f are at z = 2pi k for k Z and are each of order 2 since

limz0

eiz 1z

= limz0

ieiz

1= 1

by LHospitals rule (or apply LHospital to limz2pik(1 cos z)/(z 2pik)2 andnote that you get a constant).

B7. (i) We have

f(z) =1

2i

(eizz+1 e izz+1

).

Clearly the only singularity of f is at z = 1, and we claim that the singularityis essential. Consider the sequence {xn} := {1 + in}. Then xn 1 and

|f(xn)| = 12i (en+i eni)

en4 .On the other hand, let {yn} := {1 + 1n}, which also satisfies yn 1, but

|f(yn)| = 12i (ei(n+1) ei(n+1))

1.So z = 1 is an essential singularity of f .

(ii) Recall that

sin z = z z3

3!+z5

5!

and

cos z = 1 z2

2!+z4

4! ,

as well as the identity

sin(a+ b) = sin a cos b+ cos a sin b,

which holds for all a, b C (since the proof relies on the properties of exponen-tials). So

sinz

z + 1= sin

(1 1

z + 1

)= sin 1 cos

1

z + 1 cos 1 sin 1

z + 1

= sin 1

(1 1

2!(z + 1)2+

) cos 1

(1

z + 1 1

3!(z + 1)3+

)

8

is the Laurent series for f(z).

(iii) The residue of f at z = 1 is the coefficient of 1z+1 in the Laurentseries, namely cos 1.

B8. We integrate

f(z) =zeiz

(z + 3i 1)(z 3i 1)on a semicircle of radius R in the upper half plane, picking up the residue atz = 3i+ 1. The real part of the integral along the real axis goes to the desiredintegral as R, while the integral along the curved portion goes to 0 by thedecay of eiz away from the real axis. The residue is

res3i+1f = limz3i+1

(z 3i 1)f(z) = 16e3

(3 cos 1 + sin 1 + i( cos 1 + 3 sin 1)).

The real part of 2pii res3i+1f , which is the value of the desired integral, istherefore pi3e3 (cos 1 3 sin 1).

B9. (i) If f is entire and |f(z)| C|z|n for all z C, then f is a polynomialof degree n. To see this, it suffices to prove that f (k)(0) = 0 for all k > n(look at the power series). This in turn can be proved by the Cauchy inequality:for any R > 0 we have

|f (k)(0)| (k)!Rk

CRn 0as R when k > n.

(ii) The dimension is n+ 1 since a basis is given by {1, z, z2, . . . , zn}.

B10. The polynomials f(z) := 2z5 z3 + 3z2 z and g(z) = 8 are bothentire and have no zeroes on the circle |z| = 1. Indeed, |f(z)| 2+1+3+1 = 7on this circle, while |g(z)| = 8 on the circle. Thus by Rouches theorem, thenumber of zeroes of f + g inside the circle is the same as the number of zeroesof g inside the circle, namely zero. Thus f + g has all 5 of its zeroes (countedwith multiplicities) outside the circle.

9

Chapter 3

Spring 2010

3.1 Real Analysis

1. (a) Since the sequence is Cauchy, we can pick a subsequence {fnk} suchthat

fnk fnk+1p 0. But this means that fnq , contradicting theboundedness from (a).

3. Note that piV is the map that sends x H to the unique closest elementin V in the norm of H. One shows that x piV (x) V .

(a) For any x H, x piV (x) V , so by the Pythagorean theorem

x2 = (x piV (x)) + piV (x)2 = x piV (x)2 + piV (x)2.

Thus piV (x) x, whence piV 1. It follows that piV = 1 sincepiV (x) = x for nonzero x V .

piV is the identity on V since the unique closest element in V of an elementin V is itself. Thus piV is idempotent.

We need to show that (piV x, y) = (x, piV y) for all x, y H. This followseasily from

0 = (x piV x, piV y) = (x, piV y) (piV x, piV y)and likewise

0 = (piV x, y piV y) = (piV x, y) (piV x, piV y).

(b) Set V := im(P ), which is a subspace since it is the image of a linearoperator, and which we claim is closed. For if {xn} is a Cauchy sequence in V ,with limit y H, then P idempotent and continuous implies {xn} = {Pxn} Py, whence Py = y proves that y V . So V is closed.

Using the fact that P and piV are self-adjoint,

(Px, y) = (x, Py) = (x, piV Py) = (piV x, Py) = (PpiV x, y) = (piV x, y)

for all x, y H. Thus ((P piV )x, y) = 0 for all x, y, so (P piV )x is perpen-dicular to all of H, hence 0, and this holds for each x, whence P = piV .

11

(Did we not need the assumption that the operator norm is 1?)

4. (a) The map `1 // (`) defined by

()(1, 2, . . . ) =i

ii

is well defined since

|()()| i

|i||i| i

|i| = 1

must then also be almost disjoint. But has empty intersection with eachf j() by the definition of , whence

, f(), f2(), . . . , fk+1()

are almost disjoint.


6. (a) The singularities of f inside the circle of radius 2 are simple poles atz = 1. The corresponding residues are

res1(f) = limz1

(z 1) f = 1

andres1(f) = lim

z1(z + 1) f = 1.

Thusf = 2pii(1 + 1) = 4pii.

(b) In the annulus 1 < |z| < 3, we have1

z + 1=

1/z

1 + 1/z=

1

z

(1 1

z+

1

z2

);

1

z 1 =1/z

1 1/z =1

z

(1 +

1

z+

1

z2+

);

1

z + 3=

1/3

1 + z/3=

1

3

(1 z

3+z2

9

).

Expanding f by partial fractions, we get

f(z) =6z + 2

(z2 1)(z + 3) =1

z + 1+

1

z 1 2

z + 3,

so the coefficient of 1z of the Laurent series of f is 1 + 1 = 2, which agrees withthe sum of the residues in (a). To check the integral, note that every term ofthe Laurent series has a primitive (and hence its integral vanishes) except theone corresponding to 1z , and integrating

1z over any circle centered at the origin

gives 2pii. Multiplying by the coefficient, namely the residue, yields 4pii.

7. Skip.

13

8. (a) If f is analytic and injective on U , then f is nonzero on U . Assumefor contradiction that f (z0) = 0. Let the power series for f about z0 be

f(z) = a0 + a1(z z0) + a2(z z0)2 + .

Then a1 = 0 since the derivative vanishes, so

f(z) a0 = ak(z z0)k +G(z),

where ak 6= 0, k 2, and G vanishes to order at least k+1 at z0. Now for smallw write

f(z) a0 w = F (z) +G(z), where F (z) := ak(z z0)k w.

The idea is that we can choose a small enough circle C around z0 and w smallenough so that |F (z)| > |G(z)| on C and F has k roots inside C. Then byRouches theorem, F (z)+G(z) has k roots inside C, contradicting the injectivityof f .

One way to pick C and w is the following. By the vanishing of G, choose

> 0 so small that |G(z)| < |ak|2 k on the circle C(z0). Then choose 0 < < 12kand set w = ak, so that

|F (z)| = |ak||(z z0)k | |ak|12k > |G(z)|

on C(z0). Then F has its zeroes inside C(z0) since 1/k < , so we are done.

(b) f never 0 on U does not ensure that f is injective on U . Let U = C andf(z) = ez. Then f (z) = ez has no roots, but e0 = e2pii, so f is not injective.

Another example is f(z) = z2, on the punctured disc centered at the origin.

9. There is a mistake: the upper half disc should be defined as all z with|z| < 2 and Im z 0. First scale this map by 12 , then map to the upper halfplane by z 7 (z + 1z ), and finally map to the unit disc with z 7 izi+z . Allautomorphisms of the unit disc are rotations composed with Blaschke factors.Chasing i, we see that i 7 i2 7 32 i 7 15 . So to map 15 to 0, we use theBlaschke factor (z) =

z1z , with =

15 . Thereafter we can compose with

arbitrary rotations since they fix the origin.

10. Skip.

14

Chapter 4

Fall 2009

4.1 Real Analysis

1. All the fn, f are integrable since they are bounded by f1, which is integrable.By assumption

|fn f | =

(fn f) =fn

f 0.

SetE = {x X : fn(x) f(x) for all n},

which is measurable since it is the intersectionn(fn f)1([,]). Let E be

the set of all x X for which fn(x) 6 f(x), which is measurable since it is theunion

k E1/k. Then

|fn f | E1/k

|fn f | m(E1/k)

k

for all n, so m(E1/k) = 0. Thus m(E) = 0 as well.If we dont assume f1 is integrable, then all bets are off. For instance, if

X is a set with infinite measure (e.g. R), fn 2 for all n, and f 1, thenfn

f since both sides are infinite, but fn 6 f for all x.

2. For boundedness, simply note that

|f(x)| |f(y)| dy = f1

3. The function Hf // C sending y 7 limn (xn, y) = limn(y, xn)

is well-defined and linear. We will show that f is bounded/continuous. Thenby the Riesz representation theorem, there exists x H such that f(y) = (y, x)for all y H. Thus

limn(xn, y) = f(y) = (y, x) = (x, y)

gives the desired result.To prove boundedness, define the collection of bounded linear functionals

HTn // C by

Tny = (y, xn).

By assumption, supn |Tny| < for each y, so by the uniform boundednessprinciple, supn Tn < . Thus there is some M 0 such that |Tny| Myfor all y H, whence

|f(y)| = limn(y, xn)

limn |(y, xn)| My

proves that f is bounded.

4. Since f L(X), |f | f a.e. Thus fg |fg| f |g|by monotonicity and the fact that we can ignore a set of measure 0 when inte-grating. This proves that T is bounded and that T f. Next we willprove that T f. We need to show that given > 0, there is a g suchthat fg > (f ) |g|.By definition of f, the set

E := {x X : |f(x)| > f }, > 0

has positive measure, and it is measurable since f is measurable. Set g(x) :=E(x)f(x)/|f(x)|, which is in L1(X) since m(E) m(X) m(E)(f ) = (f )|g|,

so taking = gives the desired result. Note that the strict inequality dependson the fact that m(E) > 0.

16

5. The claim is false. Consider the sequence of intervals

{In} :={

[0, 1],

[0,

1

2

],

[1

2, 1

],

[0,

1

3

],

[1

3,

2

3

],

[2

3, 1

],

[0,

1

4

], . . .

}and let {In} be the corresponding sequence of characteristic functions. ThenIn2 =

(In) 0 but the sequence {In(x)} has infinitely many 0s and

1s for each x, and therefore does not converge.


6. We prove Schwarzs lemma.

Lemma 1 (Schwarzs lemma). Let f // be holomorphic such that f(0) =

0, where is the open unit disc. Then

(i) |f(z)| |z| for all z ;(ii) if equality holds in (i) for some z0 6= 0, then f is a rotation;(iii) |f (0)| 1, and if equality holds then f is a rotation.Proof. (i): Since f(0) = 0, f(z)/z is holomorphic in . If |z| = r < 1, we havef(z)z

1r ,and by maximum modulus this 1r bound holds for all |z| r. Letting r 1shows that 1 is an upper bound for all z .

(ii): By assumption |f(z)| = |z|, so by maximum modulus f is constant.Thus f(z) = cz, where |c| = 1, namely f is a rotation.

(iii): Write g(z) := f(z)/z. Then

g(0) = limz0

f(z) f(0)z

= f (0),

and we already showed in (i) that g is bounded by 1 in absolute value. If|g(0)| = 1, then by (ii) we see that f is a rotation.

7. We will compute the Laurent series of f(z) = (1 + z2)e1/z to find theresidue of f at the essential singularity z = 0. Since

ez = 1 + z +z2

2!+z3

3!+ ,

17

we have

e1/z = 1 +1

z+

1

2!z2+

1

3!z3+ .

Thus

f(z) = + ( 12!

+1

4!)

1

z2+ (1 +

1

3!)1

z+ (1 +

1

2!) + z + z2,

so by the residue theorem the integral of f on the ccw. unit circle is

2pii (1 + 13!

) =7pii

3.

8. We begin by proving that f is also holomorphic on the real axis. ByMoreras theorem, it is enough to check that the integral of g over every triangleis 0. If the triangle T has a side on the real axis, then slightly shifting it ensuresthat T lies entirely in one of the open half planes, hence the integral over anarbitrarily close contour is 0. Continuity then ensures the integral over T is 0.Similarly, if the triangle crosses the real axis, then splitting the triangle intosmaller triangles and shifting them slightly gives the result. So f is entire.

Now define a new function

g(z) :=

{f(z) Im(z) 0;f(z) otherwise.

Then g is holomorphic on the open set Im(z) < 0 since for z near z0 in the lowerhalf plane we can use the power series expansion for f in the upper half planeto write

f(z) =

an(z z0)n,whence

g(z) = f(z) =

an(z z0)n

is a power series expansion for g. So g is holomorphic except possibly on thereal axis, and continuous on the real axis since f is continuous and real-valuedthere. Replacing f by g in the above argument thus shows that g is entire.Since f and g agree on the entire upper half plane, they must agree everywhereby the uniqueness of analytic continuation.

9. Skip.

10. Let F (z) = z4 and G(z) = 6z 3. Then for all |z| = 2, we have

|F (z)| = 16 > 15 = 6 2 + 3 |G(z)|,

18

whence by Rouches theorem the functions F and F +G have the same numberof solutions within the unit circle, namely all 4.

19

Chapter 5

Spring 2009

5.1 Real Analysis

1. There are typos in the problem, which I assume is meant to be the follow-ing. Let {xn} be a sequence of measurable functions with 0 fn 1 for eachn such that

10fn 0. Then is it necessarily true that fn 0 a.e.?

No, this is false. Consider the sequence of intervals

{In} :={

[0, 1],

[0,

1

2

],

[1

2, 1

],

[0,

1

3

],

[1

3,

2

3

],

[2

3, 1

],

[0,

1

4

], . . .

}and let In be the corresponding characteristic functions. Then the assumptionshold but In 0 nowhere.

2. If f L, then for each p 1,

fpp =|f |p

fp fp

since (X) = 1. Thus f Lp for each p.For any > 0, we wish to show that there is an N such that for all p N ,

ffp < . Choose such that > > 0 and set E := {x X : |f(x)| >f }, which has positive measure. Then

fpp E

|f |p (f )p(E),

which impliesfp (f )(E)1/p.

Since (E) has positive measure 1, (E)1/p 1 as p, whence we getthe desired result since < .

20

3. First, we may assume T is injective. (Assuming the axiom of choice,we can extend a basis of ker(T ) to a basis for X, obtaining a decompositionX = ker(T ) X . T |X is a compact bijective linear map from X to Y .)Suppose for contradiction that Y is infinite dimensional. Let y1, y2, . . . be anormal basis of Y . Since T is bijective, there exist unique x1, x2, . . . such thatTxn = yn. Then {xn} cannot be bounded since if it were, compactness ofT would ensure the existence of a convergent subsequence {Txnk}, which isimpossible. So we may assume |xn|

This basis method does not seem so good. I cant even prove that the yksdont converge!

Help!

4. Define bounded linear functionals HTn // C by Tny = y, vn for

all y H. By assumption, this collection of linear functionals is bounded foreach y H, hence bounded in the norm by the uniform boundedness principle.Thus there exists M 0 such that |Tny| My for all n, y. In particular,vn2 = |Tnvn| Mvn implies vn M for all n.

5. Skip.


6. Since f(z) sin z 1 as z 0 and sin z = z z3/3! + has a zero of order1 at 0, the limit implies that f has a pole of order 1 at 0, with residue 1. Thusthe residue theorem shows that the integral of f on the ccw. unit circle is 2pii.

7. Since f is bounded by B outside DR, all its poles must be in DR,which is compact, so there can be only finitely many poles, at a1, . . . , an. Theng(z) = f(z)(z a1) (z an) is entire and satisfies

|g(z)| C|z|n

for some constant C 0. We claim that this implies that g is a polynomial ofdegree n, for which it suffices to show that g(k)(0) = 0 for all k > n (look atthe power series). This is proved by using the Cauchy inequalities:

|g(k)(0)| k!Rk

CRn 0

21

as R when k > n. Thus

f(z) =g(z)

(z a1) (z an)expresses f as a rational function.

8. Fix a point z0 . Define F (z) =f(z) dz, where is any curve con-

sisting of line segments parallel to the axes, starting at z0 and ending at z. Thisis well-defined since our assumption ensures that the definition is independentof the curve chosen. We now prove that F is holomorphic, with derivative f .

For h so small that z + h , note that

F (z + h) F (z) =

f(w) dw,

where is any curve from z to z + h composed of line segments parallel to theaxes. Actually, we insist that is of minimal length (as direct as possible fromz to z + h, to ensure so that we can use the continuity of f to bound part ofthe integral, as follows. Since f is continuous, write f(w) = f(z) +(w), where(w) 0 as w z. Thus

f(w) dw = f(z)

dw +

(w) dw.

Sincedw = h and

(w)

|h| supw|(w)|,

with the supremum tending to 0 as h 0, we see that

limh0

F (z + h) F (z)h

= f(z),

as desired.

9. As stated, the problem is false since, for instance, f(z) = z + z2 hasf(0) = 0, f (0) = 1, but f is not the identity map. Clearly we are meant toassume that f maps Q to Q. Let z0 Q denote the point at which f(z0) = z0and f (z0) = 1. I want to use an argument similar to the Schwarz lemma but Idont know how. Riemann mapping theorem to map Q to the unit disc? Butthen the derivative may change...

10. The idea is to rewrite the integral as a complex integral that reducesto the desired integral when we use the parametrization z = ei. Let be the

22

ccw. unit circle and consider the integral

1

i

1

z(a+

z 1z2i

) dz = 1i

2i

2aiz + z2 1 dz

= 2

1

(z + i(aa2 1))(z + i(a+a2 1)) dz.

From the left side, we see that taking z = ei and dz = iei d gives the desiredintegral. On the other hand, the right side shows that the integrand has polesat i(a a2 1). The one corresponding to (+) clearly does not lie in theunit circle, but we claim that i(a a2 1) always does for a > 1, namelythat f(a) := a a2 1 < 1. For this it suffices to argue that f(1) = 1, andthat f (a) < 0 for a > 1, hence f is strictly decreasing for a > 1. This is easy:

f (a) = 1 2a2a2 1 = 1

aa2 1 < 1 1 = 0.

Now we proceed to use the residue theorem. The residue at the pole i(aa2 1) is

limzi(aa21)

2

z i(a+a2 1) =2

2ai =i

a,

so the residue theorem shows that the integral is

2pii(i

a) = 2pi

a.

23

Chapter 6

Fall 2008

6.1 Real Analysis

1. (a) Ta is clearly linear since multiplication in R distributes over addition.Tax ` since

Tax = sup |(Tax)n| = supni=1

aixi

i=1

|aixi| xa1

Since |f | is measurable and non-negative, there is a sequence of simple func-tions {n} such that n |f |. Thus the monotone convergence theorem guar-antees that

n

|f |, and likewise 2n |f |2. If we can prove that1

(X)2

holds for all non-negative simple functions , then we get the inequality for allf L2 since

n1 (X)n2

(X)f2

by monotonicity, and the inequality is preserved upon taking the limit of theleft side.

So let be a simple function, with canonical form n =Nk=1 akEk , where

the Ek are disjoint finite measurable sets and the ak are distinct positive realnumbers. Then(

)2=

(Nk=1

ak(Ek)

)2=

Nk=1

a2k(Ek)2 +

1j

for each N , whence f / L2, contradicting the assumption.

3. For only if, suppose f = zg for some z C. If h H such thatg, h = 0, then

f, h = zg, h = zg, h = 0,so there is no h with the stated properties.

For if, suppose that for every h H orthogonal to g, f, h 6= 1. Then infact f, h = 0 is necessary since otherwise we could scale h appropriately. LetS be the subspace spanned by g, which is closed, and consider H = S S.Choose such that f = g+h, where h S. Then h g and therefore h fimply the equalities

g, g = f, g = 1f, f,

whence f = ||g. But the Pythagorean theorem givesf2 = ||2g2 + h2,

whence h = 0, so f = g as desired.

4. Define g(x, y) = f(x)[y,1][0,1](x, y), which is non-negative and mea-surable since f(x) is measurable on [0, 1] [0, 1] (see Stein 85). Note that g isreally just

g(x, y) =

{f(x) if x y;0 otherwise.

By Fubinis theorem, we obtain[0,1]

([0,1]

f(x)[y,1][0,1](x, y) dy

)dx =

[0,1]

([0,1]

f(x)[y,1][0,1](x, y) dx

)dy,

in the extended sense, where the left side yields[0,1]

xf(x) dx

and the right side yields [0,1]

([0,y]

f(x) dx

)dy.

This proves the claim.

5. Skip.

26

Chapter 7

Spring 2008

7.1 Real Analysis

1. Set fn := fEn , where

En := {x X : f(x) n}.

Then fn f , sofn

f . Thus given > 0, we can choose N > 0 such that

f fn < /2 whenever n N . Now if we choose > 0 such that < /2N ,then for any measurable E X with (E) < , we compute

E

f =

E

(f fn) +E

fn

(f fn) + n (E) < 2

+n

2N ,

which proves the result.

2. Since K is continuous on a compact subset of R2, K is bounded, namelyK L([0, 1] [0, 1]), with |K| having supremum K. Then

|(Tf)(x)| =

[0,1]

K(x, y)f(y) dy

[0,1]

|K(x, y)||f(y)| dy Kf1,

so Tf is in fact uniformly bounded, hence square integrable on [0, 1], a set withfinite measure. Since in fact [0, 1] has measure 1, the above inequality impliesthat

Tf1 :=

[0,1]

|Tf |

[0,1]

Kf1 Kf1,

whence T K.

27

3. We can write T = kerT (kerT ), where T |(kerT ) gives a bijectionwith im(T ). Then T |(kerT ) is bounded since linear transformations betweenfinite-dimensional Hilbert spaces are bounded, whence T has the same bound.

Bounded linear transformations map closed sets to closed sets since if xk x, then Txk Tx. This holds because

Tx Txk = T (x xk) Tx xk.

Now, every bounded closed set of a finite-dimensional normed space is com-pact. For this it suffices to show that every bounded sequence in a finite-dimensional vector space has a convergent subsequence. Choosing a normal ba-sis {e1, . . . , en}, we have a1e1 + + anen |a1|+ + |an|, so convergenceis analogous to convergence in Cn, which is guaranteed by Bolzano-Weierstrass.

Then if T is bounded and {xk} is bounded, {Txk} is bounded, hence has aconvergent subsequence, so T is compact.

28

analysis solutions

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