partial solutionsforquestions in appendixkof a companionto

Partial Solutions for Questions in

Appendix K of

A Companion to Analysis

T. W. Korner

1

2

Introduction

Here is a miscellaneous collection of hints, answers, partial answersand remarks on some of the exercises in the book. I expect that thereare many errors both large and small and would appreciate the oppor-tunity to correct them. Please tell me of any errors, unbridgable gaps,misnumberings etc. I welcome suggestions for additions. ALL COM-MENTS GRATEFULLY RECEIVED. (If you can, please use LATEX2εor its relatives for mathematics. If not, please use plain text. My e-mailis [email protected]. You may safely assume that I am bothlazy and stupid so that a message saying ‘Presumably you have alreadyrealised the mistake in question 33’ is less useful than one which says‘I think you have made a mistake in question 33 because not all leftobjects are right objects. One way round this problem is to quote X’stheorem.’)

To avoid disappointment note that a number like K15* means thatthere is no comment. A number marked K15? means that I still needto work on the remarks. Note also that what is given is at most asketch and often very much less.

Please treat the answers as a last resort. You will benefit more fromthinking about a problem than from reading a solution. I am inveteratepeeker at answers but I strongly advise you to do as I say and not asI do.

It may be easiest to navigate this document by using the table ofcontents which follow on the next few pages.

3

Contents

Introduction 2K1 11K2 12K3 13K4 14K5 15K6 16K7 17K8 18K9 19K10 20K11 21K12 22K13* 23K14* 24K15 25K16 26K17 27K18 28K19 29K20 30K21 31K22 32K23* 33K24 34K25 35K26 36K27 37K28 38K29 39K30 40K31 41K32 42K33 43K34* 44K35* 45K36 46K37* 47K38 48K39 49K40 50K41 51K42 52

4

K43 53K44 54K45 55K46* 56K47 57K48 58K49 59K50 60K51 61K52 62K53 63K54 64K55 65K56 66K57 67K58 68K59 69K60 70K61 71K62 72K63 73K64 74K65 75K66 76K67 77K68 78K69 79K70 81K71 82K72 83K73 84K74 85K75 86K76 87K77 88K78 89K79* 90K80 91K81 92K82 93K83 94K84 95K85 96K86 97K87 98

5

K88 99K89 100K90 101K91 102K92 103K93 104K94 105K95 106K96 107K97 108K98 110K99 111K100 112K101 113K102 114K103 115K104 116K105 117K106 118K107 119K108 120K109 121K110 122K111 123K112 124K113 125K114 126K115 127K116 128K117 129K118 130K119 131K120 132K121 133K122 134K123 135K124 136K125 137K126 138K127 139K128 140K129 141K130 142K131 143K132 144

6

K133 145K134 146K135* 147K136 148K137* 149K138 150K139 151K140 152K141 153K142 154K143 155K144 158K145 160K146* 161K147 162K148 163K149 164K150 165K151 166K152* 167K153 168K154 169K155 170K156 171K157 172K158 173K159 175K160 176K161 177K162 178K163 179K164 180K165 181K166 183K167 184K168 185K169 186K170 187K171 188K172 190K173* 191K174 192K175* 193K176 194K177 195

7

K178 196K179 198K180 199K181 200K182 202K183 203K184* 204K185 205K186* 206K187 207K188 208K189 209K190 210K191 211K192 213K193 214K194 215K195 216K196 218K197 219K198 220K199 221K200 222K201 223K202 224K203 226K204* 227K205 228K206 229K207 231K208 232K209 234K210 235K211 236K212 237K213 239K214 241K215 242K216 243K217 244K218 245K219 246K220 247K221 248K222 249

8

K223 250K224 251K225 252K226 253K227 254K228 255K229 256K230 258K231 260K232 262K233 263K234 264K235 266K236 267K237 269K238 271K239 272K240 273K241 274K242 276K243* 277K244 278K245 279K246 280K247 281K248* 283K249* 284K250 285K251* 286K252* 287K253 288K254 289K255 290K256 291K257 292K258 293K259 295K260 296K261 298K262 300K263 302K264 304K265 306K266 307K267 308

9

K268 309K269 310K270 311K271 312K272 313K273 315K274 317K275 318K276 319K277 320K278 321K279 322K280 323K281 324K282 325K283 326K284 327K285 328K286 330K287 331K288 333K289 335K290 336K291 337K292 338K293 339K294 340K295 341K296* 342K297 343K298 344K299 345K300 346K301 347K302 348K303 349K304 351K305 352K306 354K307 355K308 356K309 358K310 360K311 362K312 363

10

K313 364K314 365K315 367K316 368K317 369K318 370K319 372K320 373K321 374K322 376K323 377K324 378K325 380K326 381K327 383K328 384K329* 385K330 386K331 387K332 388K333 390K334 391K335 392K336 393K337 394K338 395K339 399K340 400K341 401K342 402K343 403K344 404K345 405

11

K1

(ii) Every non-empty set of positive integers has a least member.Since x is a rational number, qx is an integer for some strictly positiveinteger. We can certainly find a k such that k+ 1 > x ≥ k (ultimatelyby the Axiom of Archimedes). But x is not an integer so k+1 > x > k.

Since mx, m and k are integers, m′ = mx−mk is. Also

m′x = mx2 −mkx = mN − k(mx)

is an integer since mx is. Since 1 > x−k > 0 we have m > mx−mk =m′ > 0 and m > m′ ≥ 1 contradicting the definition of m as leaststrictly positive integer with mx an integer.

12

K2

Write cn = dn + d−1n . Then

⋆ d2n − cndn + 1 = 0

dn =cn ±

√

c2n − 4

2.

Since the product of the two roots of ⋆ is 1, we must take the biggerroot. Thus

dn =cn +

√

c2n − 4

2→ c+

√c2 − 4

2where c is the limit of cn.

For the second paragraph, just take

d2n = (1 + k)/2 and d2n+1 = 2/(1 + k).

In the third paragraph the condition |dn| ≥ 1 will not do. Take anyd with |d| = 1 and d 6= ±1. Then, if d2n = d and d2n+1 = d∗, the resultfails.

The condition |dn| > 1 will do but we must argue carefully. Lookfirst to see which root of d2 − cd+ 1 = 0 lies outside the unit circle.

13

K3

(Second part of problem.) Choose a1 = 2, b1 = 1, say.

14

K4

Take x = 0, f(t) = H(t), g(t) = 0. Take f(t) = 0, g(t) = H(t).

(H(t) = 1 for t > 0, H(t) = 0 for t ≤ 0.)

15

K5

(i), (ii) and (iii) are true and can be proved by, e.g. arguing by cases.

(iv) is false since f is continuous at 1/2. (f is, however, discontinuouseverywhere else.)

(v) is false. Take e.g. x = 16−1 + 17−1/2, y = 3−1 − 17−1/2.

16

K6

∑Nn=1 2

−nH(x − qn) is an increasing sequence in N bounded aboveby 1.

Observe that if x ≥ y then H(x− qn)−H(y− qn) ≥ 0 for all n ≥ 0.Thus

N∑

n=1

2−nH(x− qn)−N∑

n=1

2−nH(y − qn)

=

N∑

n=1

2−n(

H(x− qn)−H(y − qn))

≥ 0

for all N so f(x) ≥ f(y).

Observe that, if x > qm, then H(x − qn) − H(qm − qn) ≥ 0 for alln ≥ 0 and H(x− qm)−H(qm − qm) = 1 so

N∑

n=1


n=1

2−nH(qm − qn) ≥ 2−m

for all N ≥ m so f(x) ≥ f(qm) + 2−m for all x > qm. Thus f is notcontinuous at qm.

If x is irrational, we can find an ǫ > 0 such that |qm − x| > ǫ for allm ≤M and so

∣

∣

∣

∣

∣

N∑

n=1


n=1

2−nH(y − qn)

∣

∣

∣

∣

∣

≤ 2−M

whenever |x− y| < ǫ. Thus f is continuous at x.

17

K7

Observe that an+1 − an is decreasing. Thus either

(a) an+1 − an > 0 for all n and, if an is unbounded an → ∞, or

(b) There exists an N such that an+1 − an ≤ 0 for all n ≥ N and, ifan is unbounded, an → −∞.

For the reverse inequality, observe that −an satisfies the originalinequality.

18

K8

A decreasing sequence bounded below tends to a limit, so, for anyfixed x ∈ (0, 1),

xn = fn(x) → y, say.

Since x > xn > 0 we have 1 > y ≥ 0. If y > 0 then by continuity

xn+1 = f(xn) → f(y)

and so y = f(y). Thus y = 0.

If

f(x) =

14+ x

2for x > 1/2,

x4

for x ≤ 1/2,

then fn(x) → 1/2 for 1 > x > 1/2.

We suppose our calculator works in radians. If an+1 = sin an then,whatever a0 we start from, we have |a1| ≤ 1. If a1 = −b1 and bn+1 =sin bn then an = −bn for n ≥ 1. If 0 < a < 1, then 0 < sin a < a.

(If our calculator works in degrees, matters are simpler and less in-teresting.)

19

K9

(ii) we need only look at positive integers. If n ≥ 2(k + 1)

(1 + δ)n =n∑

j=0

(

n

j

)

δj

≥(

n

k + 1

)

δk+1

≥ (n/2)k+1

(k + 1)!δk+1

≥ nk+1

2k+1(k + 1)!δk+1

so

n−k(1 + δ)n ≥ nδk+1

2k+1(k + 1)!→ ∞

as n→ ∞.

20

K10

Observe that

xnn

=1

n

(

nx1 + x2 + · · ·+ xn

n− (n− 1)

x1 + x2 + · · ·+ xn−1

n− 1

)

=

(

x1 + x2 + · · ·+ xnn

−(

1− 1

n

)

x1 + x2 + · · ·+ xn−1

n− 1

)

→ 1− (1− 0)1 = 0

as n→ ∞.

Either there exists an N such that m(n) = m(N) for all n ≥ N andthe result is immediate or m(n) → ∞ and

xm(n)

n≤ xm(n)

m(n)→ 0

as n→ ∞.

If α > 1, then

xα1 + xα2 + · · ·+ xαnnα

≤ x1 + x2 + · · ·+ xnn

(xm(n)

n

)α−1

→ 1× 0 = 0.

If α < 1, then

xα1 + xα2 + · · ·+ xαnnα

≥ x1 + x2 + · · ·+ xnn

(xm(n)

n

)α−1

→ ∞.

21

K11

(i) True. If a+ b = c and a, c ∈ Q then b = c− a ∈ Q.

(ii) False. 0×√2 = 0. (However the product of an irrational number

with a non-zero rational number is irrational.)

(iii) True. By the axiom of Archimedes we can find a strictly positiveinteger N with N−1 < ǫ and an integer M such that mN−1 ≤ x <(m+ 1)N−1. Set y = mN−1 +

√2/(2N).

(iv) False. Take xn = 21/2/N .

(v) False. If a√2 is rational, then, since a is irrational, the statement

is false. If a√2 is irrational, then, since (a

√2)

√2 = 2 the statement is

false.

22

K12

(i) |xn − yn| = |(x− y)∑n−1

j=0 xjyn−1−j| ≤ |x− y|∑n−1

j=0 |x|j |y|n−1−j|.

(ii) If P (t) =∑N

j=0 ajtj, then

|P (x)− P (y)| ≤ |x− y|N∑

j=0

aj |j|Rj−1.

(vi) An increasing sequence bounded above converges.

N∑

k=J+1

10−k! ≤ 10−(J+1)!

N∑

k=J+1

10k−(J+1) ≤ 10

9× 10−(J+1)!.

23

K13*

No comments.

24

K14*

No comments.

25

K15

Apply the intermediate value theorem.

an converges (as in lion hunting) to c say.

Suppose f ′(c) > 0. We can find an ǫ > 0 such that∣

∣

∣

∣

f(t)− f(c)

t− c− f ′(c)

∣

∣

∣

∣

<f ′(c)

2

and sof(t)− f(c)

t− c>f ′(c)

2for all t with |t − c| < ǫ. Thus f(t) > f(c) for c < t < c + ǫ andf(t) < f(c) for c− ǫ < t < c

Choose n such that 2n < ǫ. We have f(an) < f(bn) which is absurd.

26

K16

Repeat the lion hunting argument for the intermediate value theoremwith f(an) ≥ g(an), f(bn) ≤ g(bn). If an → c, say, then, since f isincreasing this gives

g(bn) ≥ f(bn) ≥ f(c) ≥ f(an) ≥ g(an).

If an → c, say, then since g is continuous g(an) → g(c) and so f(c) =g(c).

First sentence of second paragraph. False. Consider [a, b] = [−1, 1],g(x) = 0, f(x) = x− 1 for x ≤ 0, f(x) = x+ 1 for x > 0.

Second sentence false. Consider [a, b] = [−1, 1], g(x) = −x + 3,f(x) = x− 1 for x ≤ 0, g(x) = −x+ 3, f(x) = x+ 1 for x > 0.

Third sentence true. Apply intermediate value theorem to f − g.

27

K17

Let ǫ > 0. Choose δk > 0 such that |fk(x) − f(c)| < ǫ for all|x− c| < δk. Set δ = min1≤k≤N δk. We claim that |g(x)− g(c)| < ǫ forall |x− c| < δ.

To see this, suppose, without loss of generality, that fN(c) = g(c).Then

g(x) ≥ fN (x) > fN(c)− ǫ = g(c)− ǫ

andfj(x) ≤ fj(c) + ǫ ≤ g(c) + ǫ

for all 1 ≤ j ≤ N and so

g(x) ≤ g(c) + ǫ

for all |x− c| < δ.

For the example we can take (a, b) = (−1, 1), c = 0 and

fn(x) =

0 if x ≤ 0,

nx if 0 ≤ x < 1/n,

1 if 1/n ≤ x.

28

K18

(i) Observe that

x ∈ R : E ∩ (−∞, x] is countableis a non-empty bounded set and so has a supremum α say. Observethat En = E ∩ (−∞, α− 1/n] is countable so E ∩ (−∞, α) =

⋃∞n=1En

is countable and so E ∩ (−∞, α] is countable. Thus E ∩ (−∞, γ) isuncountable if and only if γ > α. Similarly we can find a β such thatE∩ (γ,∞) is uncountable if and only if γ < β. Since E is uncountable,β > α.

(ii) We have one of four possibilities.

(a) There exist α and β with α > β and

e ∈ E : e < γ and e ∈ E : e > γare uncountable if and only if α < γ < β.

(b) There exists an α with

e ∈ E : e < γ and e ∈ E : e > γare uncountable if and only if α < γ.

(c) There exists a β with

e ∈ E : e < γ and e ∈ E : e > γare uncountable if and only if γ < β.

(d) We have

e ∈ E : e < γ and e ∈ E : e > γuncountable for all γ.

All these possibilities can occur. Look at Z∪ [−1, 1], (0,∞), (−∞, 0)and R.

(iii) Set E = 1/n : n ≥ 1, n ∈ Z

29

K19

(i) False. Consider xj = 1.

(ii) False. Consider xj = −j.(iii) True. Follows from proof of Lemma 3.19.

(iv) False. Consider xj = (−1)j .

30

K20

(i) Let ǫ > 0. There exists an N such that an ≤ lim supr→∞ ar + ǫfor n ≥ N . There exists an M such that n(p) ≥ N for p ≥ M . Nowlim supr→∞ ar + ǫ ≥ an(p) for p ≥M . Since ǫ was arbitrary,

lim supr→∞

ar ≥ a.

(ii) Take an = (1 + (−1)n)/2.

Take e.g. a2n+r = r/2n for 0 ≤ r < 2n, n ≥ 0.

(iii) Take an = −bn = (−1)n.

(True.) Let ǫ > 0. If n is large enough

lim supr→∞

ar + ǫ > an and lim supr→∞

br + ǫ > bn

solim supr→∞

ar + lim supr→∞

br + 2ǫ > an + bn.

Thuslim supr→∞

ar + lim supr→∞

br + 2ǫ > lim supn→∞

(an + bn).

But ǫ is arbitrary so

lim supr→∞

ar + lim supr→∞

br ≥ lim supn→∞

(an + bn).

(True) Similarly if ǫ > 0 and n is large enough

bn ≥ lim infr→∞

br − ǫ

and soan + bn ≥ an + lim inf

r→∞br − ǫ.

Thuslim supn→∞

(an + bn) ≥ lim supn→∞

an + lim infr→∞

br − ǫ

and, since ǫ was arbitrary,

lim supn→∞

(an + bn) ≥ lim supn→∞

an + lim infr→∞

br.

31

K21

∥

∥

∥

∥

x

‖x‖2 − y

‖y‖2∥

∥

∥

∥

2

=‖x‖2‖x‖2 − 2

x · y‖x‖‖y‖ +

‖y‖2‖y‖2

=‖y‖2 − 2x · y + ‖x‖2

‖x‖2‖y‖2

=

(‖x− y‖‖x‖‖y‖

)2

.

Thus∥

∥

∥

∥

x

‖x‖2 − y

‖y‖2∥

∥

∥

∥

=‖x− y‖‖x‖‖y‖ .

Ptolomey’s result now follows from the triangle inequality for points ofthe form ‖x‖−2x.

32

K22

(i), (ii) and (iii) are true. Proof by applying the definitions.

(iv) is false. Take, for example, n = 1, U = x : |x| < 1.

33

K23*

No comments.

34

K24

(i) If (xn, yn) ∈ E and ‖(xn, yn) − (x, y)‖ → 0, then xn → x andyn → y. Thus yn = 1/xn → x and x = y so (x, y) ∈ E.

π1(E) = x : x > 0 is not closed since 1/n ∈ π1(E) and 1/n→ 0 /∈π1(E) as n→ 0.

(ii) Let E = (x, 1/x) : x > 0 ∪ (1, 0). Then E is closed (uniontwo closed sets), π1(E) = x : x > 0 is not closed but π2(E) = y :y ≥ 0 is.

(iii) If xn ∈ π1(E) and xn → x we can find yn such that (xn, yn) ∈ E.Since yn is a bounded sequence there exists a convergent subsequenceyn(j) → y as j → ∞. Since xn(j) → x the argument of (i) shows that(x, y) ∈ E and x ∈ π1(E). (It is worth looking at why this argumentfails in (ii).)

35

K25

(i) If G ⊆ [−K,K]n+m, then E ⊆ [−K,K]n.

(ii) Take n = m = 1, E = (0,∞), f(x) = 1/x¿ See question K.24.

(iii) Suppose ‖(xj , f(xj))− (x,y)‖ → 0. Then ‖xj − x‖ → 0. SinceE is closed, x ∈ E. Since f is continuous, ‖f(xj) − f(x)‖ → 0. Thusy = f(x) and (x,y) ∈ G.

(iv) Take n = m = 1, E = R and f(x) = x−1 for x 6= 0, f(0) = 0.

(v) Suppose xj ∈ E and ‖xj−x‖ → 0. Since G is closed and boundedwe can find a subsequence j(k) → ∞ such that

‖(xj(k), f(xj(k)))− (x,y)‖ → 0

for some (x,y) ∈ G. Since (x,y) ∈ G, we have y = f(x).

Observe that x ∈ E, so E is closed. If f was not continuous atx, we could find a δ > 0 and xj ∈ E and ‖xj − x‖ → 0 such that‖f(xj)− f(x)‖ > δ and so

‖(xj(k), f(xj(k)))− (x,y)‖ > δ

for every j(k). Our result follows by reductio ad absurdum.

36

K26

If we write f(x) = 0 for x 6= 0 and f(0) = 1, then f is uppersemicontinuous but not continuous.

If f is not bounded we can find xn ∈ E such that f(xn) ≥ n. Extracta convergent subsequence and use upper semicontinuity to obtain acontradiction. To show that the least upper bound is attained, take asequence for which f approaches its supremum, extract a convergentsubsequence and use upper semicontinuity again.

Take n = 1, K = [0, 1], f(x) = −1/x for x 6= 0 , f(0) = 0 to obtainan upper semicontinuous function not bounded below.

Take n = 1, K = [0, 1], f(x) = x for x 6= 0 , f(0) = 1 to obtain anupper semicontinuous function bounded below which does not attainits infimum.

(As usual it is informative to see how the proof of the true resultbreaks down when we loosen the hypotheses.)

37

K27

(i) f is continuous on [0, a] so is bounded and attains its bounds on[0, a]. By periodicity f is everywhere bounded and attains its bounds.

(ii) Let f(x) = (x− [x])−1) if x is not an integer and f(x) = 0 if x isan integer. (Here [x] is the integer part of x.)

(iii) By considering g(x) − kx we see that we may suppose k = 0.Given ǫ > 0 there exists an X > 0 such that |g(x+1)− g(x)| < ǫ (andso |g(x + n) − g(x)| < nǫ for x > X . By hypothesis there exists a Ksuch that |g(y)| ≤ K for 0 ≤ y ≤ X + 1. Write n(x) for the integersuch that X < x− n(x) ≤ X + 1. If x > X + 1 we have

∣

∣

∣

∣

g(x)

x

∣

∣

∣

∣

≤∣

∣

∣

∣

g(x)− g(x− n(x))

x

∣

∣

∣

∣

+

∣

∣

∣

∣

g(x− n(x))

x

∣

∣

∣

∣

≤ n(x)ǫ

x+K

x→ ǫ

as x→ ∞. Since ǫ was arbitrary, g(x)/x→ 0 as x→ ∞.

(iv) Any example for (ii) will work here.

(v) False. Consider, for example, h(x) = kx+ x1/2 sin(πx/2).

38

K28

No comments.

39

K29

Observe that xn can lie in at most two of the three intervals

[an−1, an−1 + kn−1], [an−1 + kn−1, an−1 + 2kn−1]and [an−1 + 2kn−1, bn−1]

where kn−1 = (bn−1 − an−1)/3.

40

K30

(i) Lots of different ways. Observe that the map J given by (x,y) 7→x · y is continuous since

|(x+ h) · (y + k)− x · y| ≤ |x · k|+ |h · y|+ |h · k|≤ ‖h‖‖x‖+ ‖y‖‖k‖+ ‖h‖‖k‖ → 0

as ‖(h,k)‖ → 0. Since α is continuous, the mapA given by x 7→ (x, αx)is continuous so, composing the two maps A and J , we see that thegiven map is continuous.

Last part, continuous real valued function on a closed bounded setattains a maximum.

(ii) Observe that, if h · e = 0, then ‖e+ δh‖2 = (1 + δ2) so by (i)

(1 + δ2)e · (αe) ≥ (e+ δh) · (α(e+ δh))

and soδ(e · (αh) + h · (αe+ δ(e · (αe)− h · (αh)) ≥ 0

for all δ soe · (αh) + h · (αe) = 0.

(iv) If we assume the result true for Rn−1, then, since U has dimen-sion n − 1 and β|U is self adjoint, U has an orthonormal basis e2, e3,. . . , en of eigenvectors for β|U . Taking e1 = e, gives the desired result.

41

K31

(i) Observe that|g(u)− g(v)| ≤ ‖uv‖

since, as simple consequence of the triangle inequality,

|‖a‖ − ‖b‖| ≤ ‖a− b‖.

Choose any z ∈ E. Set R = ‖z − y‖. The continuous function g

attains a minimum on the closed bounded set B(y, R) ∩ E at x0, say.If x ∈ E then either x ∈ B(y, R) and g(x) ≥ g(x0) automatically, orx /∈ B(y, R) and

g(x) > R = g(z) ≥ g(x0).

(ii) Let u = x0 − y. Use the definition of x0 to show that

u · u ≥ (u+ δh) · (u+ δh)

whenever h ∈ E. By considering what happens when δ is small, deducethat u · h = 0.

If ‖x1 − y‖ = ‖x0 − y‖ and x1 ∈ E, then, setting h = x1 − x0, weget h · h = 0.

(iii) Since α 6= 0 we can find a y /∈ E. Set u = y − x0 and b =(‖u‖)−1u. Observe that

(x− (b · x)b) · b = 0

so, since E has dimension n− 1, we have

x− (b · x)b ∈ E

andα(x− (b · x)b) = 0

soαx = b · xαb.

Set a = (αb)b.

42

K32

(i) Argument as in K31.

(ii) Suppose x ∈ E and x · y > 0. Then, if δ is small and positive,we have

δx = (1− δ)0+ δx ∈ E

but the same kind of argument as in K30 and K31 shows that ‖y‖ >‖y − δx‖.

(iii) Translation.

43

K33

Write B for the closed unit ball. If ‖z‖ < 1 then

z = (1− ‖z‖)z+ ‖z‖0so z is not an extreme point.

On the other hand, if ‖x‖ = 1 and

z = λx+ (1− λ)y

with 0 < λ < 1 and x, y ∈ B, then, by the triangle inequality,

‖z‖ ≤ ‖λx‖+ ‖(1− λ)y‖with equality only if λx, (1− λ)y and z are collinear. But

‖λx‖+ ‖(1− λ)y‖ ≤ λ+ (1− λ) = 1

and ‖z‖ = 1. Thus ‖x‖ = ‖y‖ = 1 and λx, (1−λ)y and z are collinear.Inspection now shows that x = y = z.

A simpler pair of arguments gives the extreme points of the cube.

(ii) Since K is closed and bounded the continuous function x 7→ x ·aattains a maximum on K.

Suppose u0 is an extreme point of K ′. If v, w ∈ K, 1 > λ > 0 and

x0 + u0 = λv + (1− λ)w

then

x0 · a = (x0 + u0) · a= λv · a+ (1− λ)w) · a≥ λx0 · a+ (1− λ)x0 · a= x0 · a.

The inequality in the last set of equations must thus be replaced byequality and so v, w ∈ K ′. Since u0 is an extreme point of K ′, v =w = u0.

(iv) Use a similar argument to (ii) (or (ii) itself) to show that extremepoints of

x ∈ K : T (x) = T (x0),are extreme points of K.

44

K34*

No comments.

45

K35*

No comments

46

K36

(i) If neither K1 nor K2 have the finite intersection property, we canfind K1, K2, . . . KN , KN+1, . . . , KN+M ∈ K such that

N⋂

j=1

Kj ∩ [a, c] = ∅ andM⋂

j=N+1

Kj ∩ [c, b] = ∅

and soN+M⋂

j=1

Kj ∩ [a, b] = ∅.

(ii) We find [an, bn] such that

Ln = K ∩ [an, bn] : K ∈ Khas the finite intersection property

a = a0 ≤ a1 ≤ a2 ≤ · · · ≤ an ≤ bn ≤ · · · ≤ b2 ≤ b1 ≤ b0 = b

and bn − an = 2−n(b− a). We have an, bn → α for some α ∈ [a, b].

We claim that α ∈ K for eachK ∈ K. For, if not, we can findK0 ∈ Ksuch that α /∈ K0. Since K0 is closed, we can find a δ > 0 such that(α−δ, α+δ)∩K0 = ∅. If N is sufficiently large, [aN , bN ] ⊆ (α−δ, α+δ)so [aN , bN ] ∩ K0 = ∅ contradicting the finite intersection property ofLN .

Thus, by reductio ad absurdum, α ∈ ⋂K∈KK.

47

K37*

No comments

48

K38

Observe that, if [aj , bj] ∈ K, then

N⋂

j=1

[aj , bj] = [ max1≤j≤N

aj, min1≤j≤N

bj ] ∈ K

so, in particular, K has the finite intersection property.

If c ∈ ⋂K∈KK then c ∈ [a, b], that is to say a ≤ c ≤ b whenevera ∈ E and b ≥ e for all e ∈ E. Thus c is a (and thus the) greatestlower bound for E.

49

K39

(i) f ′1(t) = 1 − cos t ≥ 0 for all t. Thus f1 is everywhere increasing.

But f1(0) = 0 so f1(t) ≥ 0 for t ≥ 0 so t ≥ sin t for t ≥ 0.

(ii) f ′2(t) = f1(t) ≥ 0 for t ≥ 0 and f2(0) = 0.

(iv) We have

2N∑

j=0

(−1)jt2j+1

(2j + 1)!≥ sin t ≥

2N+1∑

j=0

(−1)jt2j+1

(2j + 1)!

for t ≥ 0 and, using the fact that sin(−t) = − sin t,

2N∑

j=0

(−1)jt2j+1

(2j + 1)!≤ sin t ≤

2N+1∑

j=0

(−1)jt2j+1

(2j + 1)!

for t ≤ 0.

(v) Thus∣

∣

∣

∣

∣

sin t−N∑

j=0

(−1)jt2j+1

(2j + 1)!

∣

∣

∣

∣

∣

≤ |t|N+1

(2N + 3)!→ 0

as N → ∞.

50

K40

(i) Observe that g′ is increasing since g′′ is positive. If g′(x1) > 0then g′(t) ≥ g′(x1) > 0 for all t ∈ [x1, x2] so 0 = g(x2) > g(x1) = 0which is absurd. Thus g′(x1) ≤ 0 and g′(x2) ≥ 0. By the intermediatevalue theorem there exist a c ∈ [x1, x2] such that g′(c) = 0. Since g′

is increasing g′(t) ≤ 0 and g is decreasing on [x1, c] whilst g′(t) ≥ 0

and g is increasing on [c, x2]. Thus g(t) ≤ g(x1) = 0 on [x1, c] andg(t) ≤ g(x2) = 0. We use a similar argument to get the result on[c, x2].

(ii) Look at g(t) = f(t)−A−Bt with A and B chosen to make thehypotheses of (i) hold.

(iii) We may assume that 1 > λn+1 > 0. The key algebraic manipu-lation in a proof by induction runs as follows.

f

(

n+1∑

j=1

λjxj

)

= f

(

λn+1xn+1 + (1− λn+1)

n∑

j=1

λj∑n

k=1 λkxj

)

≤ λn+1f(xn+1) + (1− λn+1)f

(

n∑

j=1

λj∑n

k=1 λkxj

)

≤ λn+1f(xn+1) + (1− λn+1)n∑

j=1

λj∑n

k=1 λkf(xj)

=

n+1∑

j=1

λjf(xj)

sincen∑

j=1

λj∑n

k=1 λk= 1 and

n∑

k=1

λk = (1− λn+1).

(iv) Apply Jensen as suggested and take exponentials.

51

K41

The area of the inscribed quadrilateral is

α =

4∑

j=1

a2 sin θj cos θj =12

4∑

j=1

a2 sin 2θj .

with a the radius of the circle. Also∑4

j=1 θj = π.

Now sin′′ t = − sin t ≤ 0 for t ∈ [0, π], so − sin is convex on [0, π] andJensen’s inequality gives

α = 2a24∑

j=1

14sin 2θj ≤ 2a2 sin

(

4∑

j=1

14(2θj)

)

= 2a2 sin(π/2) = 2a2.

The area is attained when the θj are all equal (and with a little thought,observing that sin′′ t < 0 on (0, π)) only then.

Consider a circumscribing n-gon A1A2 . . . An. If AjAj+1 touches thecircle at Xj let φ2j−1 be the angle ∠Aj−1OXj and φ2j be the angle∠XjOAj. Then

∑

r=1 2nφr = 2π and the area of the circumscribedpolygon is

β = 2−1a22n∑

r=1

tanφr ≥ 2−1a22n tan

(

2n∑

r=1

φr/2n

)

= na2 tan(π/n)

(since tan is convex on (0, π/2)). Equality is attained for a regularpolygon.

52

K42

Since g is continuous on [0, 1], it attains a maximum M say. LetE = x ∈ [0, 1] : f(x) =M. Since E is non-empty it has a supremumγ. Since f is continuous, g(γ) =M . If γ = 0 or γ = 1 then M = 0. Ifnot then we can find a k with γ − k, γ + k ∈ (0, 1) and

M = g(γ) = 12(g(γ − k) + g(γ + k)) ≥M

with equality if and only if g(γ − k) = g(γ + k) =M . Thus γ + k ∈ Econtradicting the definition of γ. We have shown thatM = 0 so g(t) ≤0 for all t. Similarly g(t) ≥ 0 for all t so g = 0.

Suppose that we drop the end conditions. By replacing g(t) byG(t) = g(t) − A − Bt with G(0) = G(1) = 0, we can show that gis linear.

[In higher dimensions the condition becomes ‘the average of g overthe surface of any sufficiently small sphere equals the value at thecentre’ and this turns out to be equivalent to saying that g satisfiesLaplace’s equation 2g = 0. Observe that in one dimension this re-duces to saying that g′′ = 0 so g is linear as we saw above. However inhigher dimensions the solutions of Laplace’s equation are much morediverse.]

53

K43

Observe that∣

∣

∣

∣

f(h)− f(0)

h

∣

∣

∣

∣

= |h sinh−4| ≤ |h| → 0

as h→ 0.

If x 6= 0, then f ′(x) = 2x sin x−4 − 4x−2 cos x−4. Thus we havef(

((2n+ 1)π)−1/4) → ∞ as n→ ∞.

54

K44

Since g is continuous on [a, b], it attains a maximum at some c ∈ [a, b].Since g′(c) = 0, we have c ∈ (a, b) and f ′(c) = k.

For the final paragraph, note that f ′′(a) ≤ 0 ≤ f ′′(b).

55

K45

(i) The statement ‘f must jump by at least 1 at z’ is not welldefined. If the reader feels that there should be some way of mak-ing this statement well defined in a useful way she should considerf(x) = 2−1 sin(1/x) for x 6= 0, f(0) = 0.

However f can not be continuous at z. Observe that

max(|f(z)− f(xn)|,|f(z)− f(yn|)≥ 1

2(|f(z)− f(xn)|+ |f(z)− f(yn|) ≥ 1

2.

Thus we can find zn → z with |f(z)− f(Zn)| ≥ 1/2.

(ii) The behaviour of f ′(zn(j)) does not control the value of f ′(z).Proposed result is false, see K43.

56

K46*

No comments.

57

K47

Since g′ is constant, the conclusions of intermediate value theoremhold trivially for g′.

58

K48

Observe that

cos nθ = ℜ(cos θ + i sin θ)n

=∑

0≤2r≤n

(

n

2r

)

(−1)r cosn−2r θ sin2r θ

=∑

0≤2r≤n

(

n

2r

)

(−1)r cosn−2r(1− cos2 θ)r

a real polynomial in cos θ of degree at most n. (Part (ii) shows thatthe degree is exactly n.)

T0(t) = 1, T1(t) = t, T2(t) = 2t2 − 1, T3(t) = 4t3 − 3t.

(a) Observe that

cosnθ cos θ = (cos(n + 1)θ + cos(n− 1)θ)/2.

Thus tTn(t) = (Tn+1(t) + Tn−1(t))/2 for all t ∈ [−1, 1].

(b) But a polynomial of degree at most n+1 which vanishes at n+2points is identically zero, so tTn(t) = (Tn+1(t) + Tn−1(t))/2 for all t.

(c) Induction using (i).

(d) |Tn(cos θ)| = | cosnθ| ≤ 1.

(e) Tn vanishes at cos((r + 1/2)π/n) with 0 ≤ r ≤ n− 1.

For the last part make the change of variables x = cos θ with range0 ≤ θ ≤ π.

If m = 0 replace π/2 by π.

59

K49

Two polynomials with the same roots and the same leading coeffi-cient are equal.

Use the fact that |Tn(t)| ≤ 1 for T ∈ [−1, 1].

Choose α and β so that −α + β = −1, α + β = 1. If P is theinterpolating polynomial of degree n for f on [a, b], then taking g andQ such that

g(t) = f(αt+ β), Q(t) = P (αt+ β),

we see that Q is the interpolating polynomial of degree n for g on[−1, 1]. If |fn(x)| ≤ A on [a, b] then |gn(x)| ≤ Aαn = A((b− a)/2)n on[−1, 1] so

|f(t)− P (t)| ≤ (b− a)nA

22n−1

on [a, b].

60

K50

(ii) P (t) =n∑

j=0

f j(a)

j!tj.

(vii) Observe thatf(t)

g(t)=f (n+1)(θt)

g(n+1)(φt)with θt, φt → a as t→ a and use continuity.

61

K51

If we set

P (t) =n∑

j=0

Ajtj(1− t)n+1 +

n∑

j=0

Bjtn+1(1− t)j,

then the resulting system of equations for Aj is triangular with non-vanishing diagonal entries and thus soluble. The same holds for Bj.

If y ∈ (0, 1), set

F (x) = f(x)− P (x)− E(y)xn+1(1− x)n+1

yn+1(1− y)n+1.

F has vanishing r-th derivative at 0 and 1 for 0 ≤ r ≤ n and vanishes aty. By Rolle’s theorem F ′ vanishes at least twice in (0, 1), F ′′ vanishesat least three times in (0, 1), F ′′′ vanishes at least four times in (0, 1),. . . , F (n+1) vanishes at least n + 2 times in (0, 1), F (n+2) vanishes atleast n+1 times in (0, 1), F (n+3) vanishes at least n times in (0, 1), . . . ,F 2n+2 vanishes at least once (at ξ say) and this gives the result.

62

K52

(i) g(b)− g(a) = g′(c)(b− a) 6= 0 for some c ∈ (a, b).

(ii) A = (f(b)− f(a))/(g(b)− g(a)).

(iv) Observe thatf(t)− f(a)

g(t)− g(a)=f ′(ct)

g′(ct)with ct → a as t→ a.

63

K53

(i) We have

f(b) = f(a) + f ′(a)h+ f ′′(c)h2

2!for some c ∈ (a, b).

(ii) Thus

f ′(a)h = f(b)− f(a)− f ′′(c)h2

2!and so

|f ′(a)||h| ≤ 2M0 + 2−1M2|h|2for all a and all h. Thus

|f ′(t)| ≤ 2M0

h+M2

2h

for all h > 0 and, choosing h = 2(M0M1)1/2, we have the required

result.

(iii) Take f(t) = t to see that (a) and (b) are false.

(c) is true by the method of (ii), provided that L ≥ 2(M0M1)1/2.

(d) True. g(t) = sin t.

(e) True. Scaling (d), G(t) =M0 sin(M−1/20 M

1/22 t) will do.

(f) Fails scaling. If G as in (e) then

G′(0)

(M0M2)1/4= (M0M1)

1/4

can be made as large as we wish.

65

K55

(i) True. By the intermediate value theorem, we see that f(x) → lfor some l as |x| → ∞. Given ǫ > 0, we can find a K such that|f(x) − l| < ǫ/2 for all |x| ≥ K. Since a continuous function ona closed bounded set is uniformly continuous, we can find a δ with1 > δ > 0 such that |f(x) − f(y)| < ǫ for all x, y ∈ [−K − 2, K + 2]with |x− y| < δ. By going through cases, we see that |f(x)− f(y)| < ǫfor all x, y ∈ R with |x− y| < δ.

(ii) False. Take f(t) = sin t.

(iii) False. Take f(t) = sin t2.

(iv) False. Take f(t) = t.

(v) False. Take f(z) = exp(i|z|).(vi) True. Observe that ||f(z)| − |f(w)|| ≤ |f(z)− f(w)|.(vii) True. Let ǫ > 0. Choose δ > 0 such that |x − y| < δ implies

||f(x)| − |f(y)|| < ǫ/2.

Now suppose y > x and |x− y| < δ. If f(x) and f(y) have the samesign, then automatically, |f(x)−f(y)| < ǫ/2. If they have opposite sign,then by the intermediate value theorem, we can find u with y > u > xand f(u) = 0. Since |x− u|, |y − u| < δ we have

|f(y)− f(x)| ≤ |f(y)− f(u)|+ |f(u)− f(x)| < ǫ/2 + ǫ/2 = ǫ.

(vii) False. Take f(x) = g(x) = x. If δ > 0 then, taking y = 8δ−1

and x = y+ δ/2, we have |x− y| < δ but |x2− y2| = (x+ y)(x− y) > 1so x 7→ x2 is not uniformly continuous.

(viii) True. Let ǫ > 0. We can find a δ > 0 such that |u − w| < δimplies |f(u)− f(w)| < ǫ and an η > 0 such that |x − y| < η implies|g(x)− g(y)| < δ. Thus |x− y| < η implies |g(f(x))− g(f(y))| < ǫ.

66

K56

Observe that

‖(x + u)− y‖ ≤ ‖x− y‖+ ‖u‖so that

f(x+ u) ≤ f(x) + ‖u‖.Thus

‖f(a)− f(b)‖ ≤ ‖a− b‖and f is uniformly continuous.

If E is closed, then we can find yn ∈ E such that ‖x − yn‖ ≤f(x)+1/n. The yn belong to the closed bounded set E∩ B(x, 2) so wecan find n(j) → ∞ and a y with ‖yn(j) − y‖ → 0. Since E is closed,y ∈ E. We observe that

f(x) + 1/n(j) ≥ ‖x− yn‖≥ ‖x− y‖ − ‖y − yn‖→ ‖x− y‖ ≥ f(x)

as j → ∞. Thus ‖x− y‖ = f(x).

Conversely, if the condition holds, suppose yn ∈ E and ‖yn−x‖ → 0.We have f(x) = 0 so there exists a y ∈ E with ‖x − y‖ = f(x) = 0.We have x = y ∈ E so E is closed.

68

K58

(i) If |z| < min(R, S) then∑N

n=0 anzn and

∑Nn=0 bnz

n tends to limitand so

N∑

n=0

(an + bn)zn =

N∑

n=0

anzn +

N∑

n=0

bnzn

tends to limit as N → ∞.

Suppose R < S If R < |z| < S then∑∞

n=0 anzn converges and

∑∞n=0 bnz

n diverges so∑∞

n=0(an + bn)zn diverges. Thus the sum has

radius of convergence R.

(ii) The argument of the first paragraph of (i) still applies.

(iii) We deal with the case ∞ > T ≥ R > 0. Let an = R−n + T−n

and bn = −R−n.

(iv) The radius of convergence is R unless λ = 0 in which case it isinfinite.

(v) If |z| > 1 then, provided n is sufficiently large, |z|n2 ≥ (2R)n so∑∞

n=0 anzn diverges. If |z| < 1 then, provided n is sufficiently large,

|z|n2 ≤ (R/2)n so∑∞

n=0 anzn converges. The radius of convergence is

1.

If R = 0, then any radius of convergence ρ with 0 ≤ ρ ≤ 1 is possible.For ρ = 0 take an = nnn

. For 0 < ρ < 1 take an = ρ−n2

. For ρ = 1take an = ρ−n2

n where ρn tends to 1 very slowly from below.

Similar ideas for R = ∞.

(vi) Let |cn| = max(|an|, |bn|). Since∑n

j=0 |cnzn| ≥ ∑nj=0 |anzn|

radius of convergence R′ at most R. Thus R′ ≤ min(R, S). But |cn| ≤|an|+ |bn| so R′ ≥ min(R, S). Thus R′ = min(R, S).

If |cn| = min(|an|, |bn|) then similar arguments show that the radiusof convergence R′′ ≥ max(R, S). But every value A ≥ max(R, S) ispossible for R′′. (If A < ∞ and R, S > 0 can take eg. a2n = R−2n,b2n = A−2n, a2n+1 = A−2n−1, b2n = S−2n−1.

69

K59

Let ǫ > 0. We can find an N such that |h(n)− γ| < ǫ for n ≥ N .

Observe that if n ≥ N then h(n) < γ + ǫ and so

an > 2

(

−(γ+ǫ)+(1−(γ+ǫ))

n = 21−2(γ+ǫ)n.

Thus if |z| > 22(γ+ǫ)−1 we have |anzn| → ∞. Thus the radius of con-vergence R satisfies R ≤ 22(γ+ǫ)−1. Since ǫ was arbitrary R ≤ 22γ−1.A similar argument shows that, if |z| < 22(γ−ǫ)−1 |anzn| → 0, and thusR ≥ 22γ−1 so R = 22γ−1.

Suppose 1 ≥ γ ≥ 0. Define E inductively by taking an+1 = an/2 ifan/2 ≥ 21−2γ , an+1 = 2an otherwise. Then h(n) → γ and |anzn| ≥ 1whenever |z| = 22γ−1 and n is large. Thus we have divergence every-where on the circle of convergence.

Suppose 1 ≥ γ ≥ 0. Define E inductively by taking an+1 = 2an if2an ≤ 2(1−2γ)nn−2, an+1 = an/2 otherwise. Then |anzn| ≤ n−2 when-ever |z| = 22γ−1 and n is large. Thus we have convergence everywhereon the circle of convergence. When n is large we have

an ≤ 4× 2(1−2γ)nn−2

so2n−2h(n)n ≤ 22+(1−2γ)nn−2

whence

(log 2)(n− 2h(n)n) ≤ (log 2)(2 + (1− 2γ)n− 2 logn

and dividing by n log 2 and allowing n→ ∞ we see that

1− 2 lim supn→∞

h(n) = 1− 2γ

so lim supn→∞ h(n) = γ. A similar argument (but there are alternativeways of proceeding) shows that lim infn→∞ h(n) = γ so h(n) → γ.

71

K61

Write α = lim supn→∞ an+1/an. If ǫ > 0, we can find an N withan+1/an ≤ α + ǫ for n ≥ N . Thus, by induction, an ≤ A(α + ǫ)n forn ≥ N where A = aN (α+ ǫ)−N . Thus, if n ≥ N ,

a1/nn ≤ A1/n(α + ǫ) → α + ǫ

as n→ ∞. Since ǫ was arbitrary

lim supn→∞

a1/nn ≤ lim supn→∞

an+1/an.

Remaining inequalities similar or simpler.

All 8 possibilities can arise. Here are d.

(1) Set an = 1.

lim infn→∞

an+1

an= lim inf

n→∞a1/nn = lim sup


n→∞

an+1

an= 1.

(2) Set a1 = 1, a2n+1/a2n = 2, ar+1/ar = 1 otherwise.

lim infn→∞

an+1

an= lim inf


n→∞a1/nn = 1 < 2 = lim sup

n→∞

an+1

an.

(3) Set a1 = 1, a2n+1/a2n = 2, a2n+2/a2n+1 = 2−1 for n ≥ 2 ar+1/ar =1 otherwise.

lim infn→∞

an+1

an= 1/2 < lim inf


n→∞a1/nn = 1 < 2 = lim sup

n→∞

an+1

an.

(4) Set a1 = 1, ar+1/ar = 2 if 222n ≤ r < 22

2n+1

, ar+1/ar = 1otherwise

lim infn→∞

an+1

an= lim inf

n→∞a1/nn = 1 < 2 = lim sup


n→∞

an+1

an.

(5) Set a1 = 1, ar+1/ar = 2 if 222n

+ 1 ≤ r < 222n+1

, ar+1/ar = 4 if

r = 222nar+1/ar = 1 otherwise

lim infn→∞

an+1

an= lim inf

n→∞a1/nn = 1 < 2 = lim sup

n→∞a1/nn < 4 = lim sup

n→∞

an+1

an.

(6) Set a1 = 1, ar+1/ar = 2 if 222n

+ 2 ≤ r < 222n+1

, ar+1/ar = 4 if

r = 222n, ar+1/ar = 1/2 if r = 22

2n+ 1 ar+1/ar = 1 otherwise

lim infn→∞

an+1

an= 2−1 < lim inf

n→∞a1/nn = 1 < 2 = lim sup

n→∞a1/nn < 4 = lim sup

n→∞

an+1

an.

Other two obtained similarly.

Note lim sup formula will always give radius of convergence. Ratiomay not.

72

K62

(iii) Could write bn = cn + dn with dn = 0 for n large and |bn| ≤ ǫ.Thus lim supn→∞ |Cn| ≤ ǫ.

(v) If bj = (−1)j limit is 0.

(viii) Pick N(j),Mj , sj and rj inductively so that N(0) = 0,M(0) =

1, s0 = 1/2. we pick rj+1 so that 0 < sj < rj+1 < 1 and 1 − rM(j)+1j+1 <

2−j. Pick anN(j+1) > M(j) such that rN(j+1)j+1 < 2−j. Pick sj+1 so that

0 < rj+1 < sj+1 < 1 and sN(j+1)j+1 > 1−2−j−1. PickM(j+1) > N(j+1)

so that sM(j+1)j+1 < 2−j−1 Set bn = 1 if N(j) ≤ n ≤ M(j), bn = 0

otherwise.

Ar(j+1) ≤M(j)∑

n=0

(1− rj+1)rnj+1 +

∞∑

n=N(j+1)

(1− rj+1)rnj+1

= (1− rM(j)+1j+1 ) + r

N(j)j+1 < 2−j + 2−j = 2−j+1

and

As(j+1) ≥M(j+1)∑

n=N(j+1)

(1− sj+1)snj+1

= sN(j+1)j+1 − s

M(j+1)+1j+1 > 1− 2−j−1 − 2−j−1 = 1− 2−j.

Thus Ar does not converge.

73

K63

(ii) C can be identified with R2.

(iii) (a) If z 6= 1,∑N

n=0 zn = (1 − zN+1)/(1 − z) tends to a limit if

and only if |z| < 1. The limit is 1/(1− z). If z = 1 we have divergence.

(b) Observe that, if z 6= 1,

Cn = (n+ 1)−1n∑

k=0

k∑

j=0

zj

= (n+ 1)−1n∑

k=0

1− zk+1

1− z

=1

n+ 1

(

(n+ 1) + z1 − zn+1

1− z

)

1

1− z

=1

1− z+

z

n + 1

1− zn+1

1− z.

Thus we have convergence if and only if |z| ≤ 1, z 6= 1 and the limit is1/(1− z).

(c) Ar = (1 − rz)−1. Abel sum exists if and only if |z| ≤ 1, z 6= 1and is then 1/(1− z).

74

K64

(v) If λk → ∞ then, using part (i), Bλk→ b. Since this is true for

every such sequence, Bλ → b as λ→ ∞.

(vi) We seek to Borel sum zj . Observe that, if z 6= 1∞∑

n=0

λn

n!e−λ1− zn+1

1− z=

1

1− z(1− ze−λ(1−z)).

We have convergence if and only if ℜ(1− z) < 1 and the Borel sum isthen 1/(1− z).

If z = 1,

Bλ =

∞∑

n=0

(n+ 1)λn

n!e−λ =

∞∑

n=0

λn

n!e−λ + λ

∞∑

n=0

λn

n!e−λ = 1− λ→ ∞.

75

K65

(i) True. Since the sequence is bounded, there exists a b and a strictlyincreasing sequence n(j) such that bn(j) → b. Set ujn(j) = 1, ujk = 0otherwise.

(ii) True. We can extract subsequences with different limits and use(i).

(iii) False. Let bj = 1 for 2n − n ≤ j ≤ 2n − 1, bj = −1 for2n + 1 ≤ j ≤ 2n + n [n ≥ 4], bj = 0 otherwise Observe that, if unk = 1for 2n −N ≤ k ≤ 2n, unk = −1 for 2n + 1 ≤ k ≤ 2n +N , then U ∈ Gbut

∑∞k=0 ujkbk = 2N for j sufficiently large. But N is arbitrary.

(iv) True. Define N(r) and j(r) inductively as follows. Set N(0) =j(0) = 1. Choose N(r + 1) > N(r) such that

∑∞k=N(r+1) |uj(r)k| < 2−r

and j(r + 1) > j(r) such that∑N(r+1)

k=0 |uj(r+1)k| < 2−r−1. Set bk =(−1)r+1 for N(r) ≤ k < N(r + 1). Then

(−1)r+1

∞∑

k=0

uj(r)kbk

= (−1)r+1

N(r)∑

k=0

uj(r)kbk +

N(r+1)∑

k=N(r)+1

uj(r)kbk +

∞∑

k=N(r+1)+1

uj(r)kbk

≤∞∑

k=0

uj(r)k − 2

N(r)∑

k=0

|uj(r)k| − 2

∞∑

k=N(r+1)+1

|uj(r)k|

≤∞∑

k=0

uj(r)k − 2−r+3 → 1.

as r → ∞.

(v) False. Try bj = 0.

76

K66

(i) Try bk = 1 for k = N , bk = 0 otherwise. Try bk = 1 for all k.

77

K67

(i) implies (ii) since absolute convergence implies convergence.

For the converse observe that, writing xn = (xn1, xn2, . . . , xnm), wehave

‖xj‖ ≤ |xj1|+ |xj2|+ · · ·+ |xjm|.Thus, if

∑∞j=1 ‖xj‖ diverges, we must be able to find a k with 1 ≤ k ≤ m

such that∑∞

j=1 |xjk| diverges. Choose ǫj so that ǫjxjk ≥ 0. Then∥

∥

∥

∥

∥

N∑

j=1

ǫjxj

∥

∥

∥

∥

∥

≥∣

∣

∣

∣

∣

N∑

j=1

ǫjxjk

∣

∣

∣

∣

∣

=

N∑

j=1

|xjk| → ∞.

as N → ∞.

78

K68

If∑∞

n=1 an converges, then

∑

n∈S1,n≤N

an ≤∑

n≤N

an ≤∞∑

n=1

an

and, since an increasing sequence bounded above converges,∑

n∈S1,n≤N antends to a limit as N → ∞.

If∑

n∈Sj ,n≤N an tends to a limit so does∑

n≤N

an =∑

n∈S2,n≤N

an +∑

n∈S2,n≤N

an.

Second paragraph does not depend on positivity aj so ‘if’ remainstrue. However, taking a2n−1 = −a2n = 1/n, S1 evens and S2 odds thenwe see ‘only if’ false.

Let S1 = n : a1/2n /n ≤ an and S2 = n : a

1/2n /n > an.

∑

n∈S1a1/2n /n converges by comparison with

∑∞n=1 an. If n ∈ S2 then

an = (a1/2n )2 < (n−1)2 = n−2

so, since∑∞

n=1 n−2 converges, so does

∑

n∈S2a1/2n /n. Hence the result.

By Cauchy Schwarz,(

N∑

n=1

a1/2n

n

)2

≤N∑

n=1

an

N∑

n=1

1

n2≤

∞∑

n=1

an

∞∑

n=1

1

n2

so, since an increasing sequence bounded above converges, we are done.

Observe that cn ≥ (an + bn)/2 and use comparison.

79

K69

(i) Diverges2N∑

n=1

1 + (−1)n

n=

N∑

n=1

1

n→ ∞

as N → ∞.

(ii) Converges by comparison since p−2n ≤ n−2.

(iii) Diverges by comparison. We have n1/n → 1 (e.g. by takinglogarithms) so n−1/n ≥ 1/2 for n large and so n−1−1/n > 2−1n−1 for nlarge.

(iv) If∑∞

n=1 an converges, then takingN(j) = j we have that∑N(j)

n=1 antends to a limit.

If∑N(j)

n=1 an tends to a limit A, then for any M we can find a j withN(j) > M so that

M∑

n=1

an ≤N(j)∑

n=1

an ≤ A

so, since an increasing sequence bounded above converges, we are done.

(v) If∫∞1f(t) dt converges then

N∑

n=1

an ≤N∑

n=1

∫ n+1

n

f(t) dt =

∫ N+1

1

f(t) dt ≤∫ ∞

0

f(t) dt

since an increasing sequence bounded above converges∑∞

n=1 an con-verges.

For the converse observe that if∑∞

n=1 an converges then, if X ≤ N ,

∫ X

1

f(t) dt ≤∫ N

1

f(t) dt =

N−1∑

n=1

∫ n+1

n

f(t) dt ≤ K

N−1∑

n=1

an ≤∞∑

n=1

an.

Now use something like Lemma 9.4.

(vi) an = (−1)n.

(vii) Let∑MN

n=1 an tend to b. Given ǫ > 0 we can find an N0 such

that ‖∑MNn=1 an − b‖ ≤ ǫ/2 for N ≥ N0. We can also find an N1 such

that ‖an‖ ≤ ǫ/2M for n ≥ N1. Let N2 = (M + 1)(N1 + N0). Ifm ≥ N2 we can write m = NM + u with N ≥ N0, 0 ≤ u ≤ M − 1 andMN + r ≥ N1 for all r ≥ 0. We then have

‖m∑

n=1

an − b‖ ≤ ‖MN∑

n=1

an − b‖+MN+u∑

n=MN+1

‖an‖ <ǫ

2+

uǫ

2M≤ ǫ

as desired.

80

(viii) Set N(j) = 22j, an = −j−1 if 22j ≤ n < 2 × 22j, an = j−1 if2× 22j ≤ n < 3× 22j, an = 0 otherwise.

81

K70

(i) Observe that (anbn)1/2 ≤ (an + bn)/2 and use comparison.

(ii) Take bn = an+1 in (i).

(iii) Observe that an+1 ≤ (anan+1)1/2 and use comparison.

(iv) a2n = 1, a2n+1 = n−4.

82

K71

(i) True. a4n → 0 so there exists an N with |an| ≤ 1 for n ≥ N . Thusa4n ≥ |a5N | for n ≥ N and the result follows by comparison and the factthat absolute converge implies convergence.

(ii) False. Take an = (−1)n−1/6 and use the alternating series test.

(iii) False. Take a2k = 2−k, ar = 0 otherwise.

(iv) False. Same counterexample as (iii).

(v) True. Given ǫ > 0 we can find an N such that∑M

j=N aj < ǫ/2for M ≥ N . Thus, if n ≥ 2N + 2

nan ≤ 2(n−N − 1)an ≤ 2

M∑

j=N

aj < ǫ.

(vi) False. Take an = (n logn)−1 for n ≥ 3 and use the integralcomparison test.

(vii) True. Abel’s test.

(viii) True. By Cauchy–Schwarz,(

N∑

n=1

|an|n−3/4

)2

≤N∑

n=1

|an|2N∑

n=1

n−3/2 ≤∞∑

n=1

|an|2∞∑

n=1

n−3/2.

(ix) False. Take an = (−1)n(log n)−1 for n ≥ 3. Use alternatingseries test and integral test.

(x) True. We must have an → 0 so there exists a K such that|an| ≤ K. Since |n−5/4an| ≤ Kn−5/4 the comparison test tells us that∑∞

n=1 n−5/4an is absolutely convergent and so convergent.

83

K72

akn → 0 so there exists an N with an ≤ 1 for n ≥ N . Thus akn ≥ |ak+1n |

for n ≥ N and the result follows by comparison.

Set f(n) = log(n+ 1)n and observe that, if k ≥ 2

3N∑

j=1

akj ≥N∑

j=1

(2k − 2)(log(n+ 1))−k) → ∞.

(Use comparison and the fact that n ≥ (logn)k for n sufficiently large.)

84

K73

(i) Write

Sn =n∑

j=1

µ−1j bj .

Then, by partial summation (or use Exercise 5.17),n∑

j=1

bj =

n∑

j=1

µj(Sj − Sj−1)

=

n−1∑

j=1

(µj − µj+1)Sj + µnSn

so that∥

∥

∥

∥

∥

n∑

j=1

bj

∥

∥

∥

∥

∥

≤n−1∑

j=1

(µj+1 − µj)‖Sj‖+ µn‖Sn‖

≤n−1∑

j=1

(µj+1 − µj)K + µnK = (2µn − µ1)K ≤ 2kµn

whence the result.

(ii) If ǫ > 0 then we can find an N such that ‖∑nj=N µ

−1j bj‖ ≤ ǫ for

n ≥ N . Thus, by (i),

µ−1n ‖

n∑

j=1

bj‖ ≤ µ−1n

N−1∑

j=1

‖bj‖+ µ−1n ‖

n∑

j=N

bj‖

≤ µ−1n

N−1∑

j=1

‖bj‖2ǫ→ 2ǫ

as n→ ∞. Kronecker’s lemma follows.

(iii) ((a) fails) Let bj = j4, µj = j6 for j 6= 2r, b2r = 1 and µ2r = 2−2r.

((b) fails) Let bj = j−2, µj = 1.

((c) fails) Let bj = j4, µj = j2.

85

K74

Consider the set An of integers between 10n and 10n+1 − 1 inclusivewho’s decimal expansion does not contain the integer 9. We observethat 0 ∪⋃n

j=0Aj contains (9/10)n+1 × 10n+1 = 9n+1 elements. Thus

An contains at most 9n+1 elements and∑

j∈An

j−1 ≤ 10−n × 9n+1.

The partial sums of the given series thus can not exceed∑∞

n=0 10−n ×

9n+1 = 90 and the series converges.

86

K75

(i) Observe that 1/x ≥ 1/(n+ 1) for n ≤ x ≤ n + 1 and integrate.

Tn − Tn+1 =1

n + 1−∫ n+1

n

1

xdx.

Observe that T1 = 1 and

Tn ≥n−1∑

r=1

(

1

r−∫ r+1

r

1

xdx

)

.

Decreasing sequence bounded below tends to a limit.

(v) Given l we can choose integers pn, qn ≥ 1 such that pn+1 > 2pn,qn+1 > 2qn, pn/qn is strictly decreasing and log 2+(1/2) log(pn/qn) → l.At the n-th stage, if the ratio of positive terms to negative exceedspn/qn use pn positive terms followed by qn + 1 terms if the ratio is lessthan pn/qn, use pn + 1 positive terms followed by qn negative terms,otherwise usepn positive terms followed by qn negative terms. Switchfrom the n-th stage to the n+1-th stage when the size of each unusedterms is less than (pn+1+qn+1+1)−12−n and the ratio of positive termsto negative differs from pn/qn by at most 2−n.

87

K76

(ii) Observe that, if∑∞

j=0 bj converges, we can find an N(ǫ) such

that |∑Nj=M) bj ≤ ǫ| for N ≤M ≤ N(ǫ) so

∣

∣

∣

∣

∣

∞∑

j=M

bjxj

∣

∣

∣

∣

∣

≤ ǫxM ≤ ǫ

for M ≥ N(ǫ).

(iii) If M ≥ N(ǫ), then∣

∣

∣

∣

∣

∞∑

j=0

bjxj −

∞∑

j=0

bj

∣

∣

∣

∣

∣

≤M−1∑

j=0

|bjxj − bj |+∣

∣

∣

∣

∣

∞∑

j=M

bj

∣

∣

∣

∣

∣

+

∣

∣

∣

∣

∣

∞∑

j=M

bjxj

∣

∣

∣

∣

∣

≤M−1∑

j=0

|bjxj − bj |+ 2ǫ→ 2ǫ

as x→ 1−. Since ǫ is arbitrary the required result follows.

(iv) Observe that, if 0 < x < 1, then∑∞

j=0 ajxj ,∑∞

j=0 bjxj and

∑∞j=0 cjx

j are absolutely convergent so (using Exercise 5.38) we maymultiply them to get

∞∑

j=0

ajxj

∞∑

j=0

bjxj =

∞∑

j=0

cjxj.

Now let x→ 1−.

88

K77

(a) Since∑∞

n=0

∑∞m=0 cn,mx

nym converges for |x| = |y| = δ, we have|cn,m|δn+m = |cn,mxnym| → 0 for n, m → ∞ so there exists a K with|cn,m|δn+m < K for all n, m ≥ 0. Set ρ = δ−1.

(b) (i)∑

n≤N, m≤N |xnym| =∑n≤N |x|n∑m≤N |y|m.E is the square (−1, 1)× (−1, 1).

(ii)∑∑

n+m≤N

(

n+mn

)

|xnym| =∑Nr=0(|x|+ |y|)r

E is the square (x, y) : |x|+ |y| < 1.(iii)

∑∑

n+m≤N

(

n+mn

)

|x2ny2m| =∑Nr=0(|x|2 + |y|2)r.

E is the circle (x, y) : x2 + y2 < 1.(iv)

∑

n,m≤N |xnym/(n!m!)| =∑n≤N |x|n/n!∑m≤N |y|m/m!.

E = R2.

(v)∑

nm≤N |xnymn!m!| =∑n≤N |x|nn!∑m≤N |y|mm!.

E = 0.(vi)

∑

n≤N |xnyn| =∑n≤N |xy|n.E = (x, y) : |xy| < 1 a set bounded by branches of hyperbola.

(c) Observe that |cnmxnym| ≤ |cnmxn0ym0 | when |x| ≤ |x0|, |y| ≤ |y0|.

89

K78

(i) Choose Nj(ǫ) so that |aj − aj,n| < ǫ/M for n ≥ Nj(ǫ) and setN(ǫ,M) = max1≤j≤M Nj(ǫ).

Observe that∞∑

j=1

aj ≥M∑

j=1

aj ≥M∑

j=1

aj,n

for all M .

(ii) If∑∞

j=1 aj converges with value A, then given ǫ > 0 we can find

anM such that∑M

j=1 aj > A−ǫ. By (i) we can find N(ǫ,M) such that

∞∑

j=1

aj,n ≥M∑

j=1

aj − ǫ

and so ∞∑

j=1

aj ≥∞∑

j=1

aj,n ≥ A− 2ǫ

for all n ≥ N(ǫ,M). Thus∑∞

j=1 aj,n → A.

If∑∞

j=1 aj fails to converge, then given any K > 0, we can find

an M such that∑M

j=1 aj > 2K. But we can find an N such that∑M

j=1 aj,n ≥∑Mj=1 aj −K for n ≥ N and so

∑∞j=1 aj,n ≥ K for n ≥ N .

Thus∑∞

j=1 aj,n can not converge.

(iii) Yes we can drop the condition. Consider bj,n = aj,n−aj,1 instead.

90

K79*

No comments

91

K80

It is possible to do this by fiddling with logarithms but I prefer tocopy the proof of the radius of convergence. Show that if

∑∞n=0 bne

nz

converges then∑

n=0 bnenw converges absolutely when ℜw < ℜz.

Next observe that the sum∑∞

n=0 bnenz converges everywhere (set

X = −∞) or nowhere (set X = ∞) or we can find z1 and z2 suchthat

∑∞n=0 bne

nz1 converges and∑∞

n=0 bnenz2 diverges. Explain why

this means that

X = supℜz :

∞∑

n=0

bnenz converges

exists and has the desired properties.

For X = −∞ can take bn = 0, for X = ∞ can take bn = en2

, forX = R with R finite can take bn = e−Rn.∑∞

n=0 2nenz/(n + 1)2 converges when z = − log 2. Also 2nenx/(n +

1)2 → ∞ when x is real and x > − log 2 so X = − log 2.

By the first paragraph there exists a Y ≥ 0 such that∑∞

n=0 cneinz

converges for ℑz < Y and diverges forℑz > Y . By taking complex con-jugates we see that

∑∞n=0 cne

−inz converges for ℑz > −Y and divergesfor ℑz < −Y .

Since∑∞

n=0 sn + tn diverges if exactly one of∑∞

n=0 sn and∑∞

n=0 tnconverges and converges if both converge,

∑∞n=0 cn(e

inz + e−inz) andthus

∑∞n=0 cn cosnz converges if ℑz < Y and diverges if ℑz > Y .

92

K81

(i) By integration by parts,

In+1 = [− sinn x cosx]π/20 + n

∫ π/2

0

sinn−1 x cos2 x dx

= n

∫ π/2

0

sinn−1 x(1− sin2 x) dx

= nIn−1 − nIn+1.

(ii) Observe that sinn−1 x ≤ sinn x ≤ sinn+1 x for x ∈ [0, π/2]. Inte-grating gives In+1 ≤ In ≤ In−1 so

n

n + 1=In+1

In−1≤ InIn−1

≤ 1.

so In/In−1 → 1 as n→ ∞.

(iii) I0 = π/2, I1 = 1

I2n =(2n− 1)(2n− 3) . . . 1

2n(2n− 2) . . . 2

π

2, andI2n+1 =

2n(2n− 2) . . . 2

(2n+ 1)(2n− 1) . . . 1.

Thusn∏

k=1

4k2

4k2 − 1

2

π=I2n+1

I2n→ 1

and the Wallis formula follows.

93

K82

(i) Power series or set f(x) = exp x − x and observe f(0) = 0,f ′(x) ≥ 0 for all x ≥ 0.

(iii) Observe that∣

∣

∣

∣

∣

n∏

j=1

(1 + aj)−m∏

j=1

(1 + aj)

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

n∏

j=1

(1 + aj)

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

m∏

j=n+1

(1 + aj)− 1

∣

∣

∣

∣

∣

≤n∏

j=1

(1 + |aj|)(

m∏

j=n+1

(1 + |aj |)− 1

)

.

Now use the inequality (1 + |aj|) ≤ exp |aj |.(iv) Use the general principle of convergence and (iv).

(v) We must have an → 0. If |an| < 1/2 then

|bn| =∣

∣

∣

∣

an1 + an

∣

∣

∣

∣

≤ 2|an|.

Thus∑∞

n=1 bn converges absolutely.

1 =

n∏

j=1

(1 + aj)

n∏

j=1

(1 + bj) →∞∏

j=1

(1 + aj)

∞∏

j=1

(1 + bj)

so∏∞

j=1(1 + aj)∏∞

j=1(1 + bj) = 1.

(vi) Observe that

N∏

j=1

(1 + aj/Rj) =

N∏

j=1

Rj−1

Rj=

R0

RN→ 0

as N → ∞ and use (v).

94

K83

(i) Left to reader.

(ii) Let an = n−α, an,N = n−α if all the prime factors of N lie amongthe first N primes, an,N = 0 otherwise. Observe that

0 ≤ an,N ≤ an,N+1 ≤ an,

that an,N → an as N → ∞ and that∑∞

n=1 an converges.

(iii) If∏

p∈P, p≤N

1

1− p−1≤ K

for all N then∏

p∈P, p≤N

1

1− p−α≤ K

for all N and all α > 1 so∏

p∈P

1

1− p−α≤ K

and ∞∑

n=1

1

nα≤ K

for all α > 1. ThusN∑

n=1

1

nα≤ K

for all α > 1 and all N soN∑

n=1

1

n≤ K

for all∑∞

n=11nwhich is absurd.

(iv) Let E be a set of positive integers. Write

En = r ∈ E : 2n ≤ r < 2n+1.If En has M(n) elements and M(n) ≤ A2βn then

∑

r∈E,r≤2N

1

r=

N−1∑

j=0

∑

r∈Ej

1

r≤

N−1∑

j=0

2−jM(j) ≤ A

N−1∑

j=0

2n(β−1) =A

1− 2β−1

for all N .

95

K84

Suppose that Pn > 1 for all n ≥ N . Then Pm = PN

∏mj=N(1 − an).

If∑∞

j=1 an diverges then (see K82) Pm → 0 which is impossible by the

definition of N . Thus∑∞

j=1 an converges and Pm tends to a limit. Asimilar argument applies if Pn ≤ 1 for all n ≥ N .

Thus we need only consider the case Pn − 1 changes sign (or is zero)infinitely often. But an → 0 and if (A) PN−1 − 1 ≤ 0 ≤ PN − 1 orPN−1 − 1 ≥ 0 ≥ PN − 1 and (B) 0 ≤ an ≤ ǫ for n ≥ N − 1 then wehave 1 + ǫ ≥ Pn ≥ 1− ǫ for n ≥ N . Thus Pn → 1.

If∑∞

j=1 an diverges we can say nothing about l. (If l ≥ 1 consider

a1 = l − 1, an = 0 for n ≥ 2. If l ≥ 1 consider a1 = 1, a2 = 1 − l/2,an = 0 for n ≥ 3.)

96

K85

Let an = zn, an,N = zn if 0 ≤ n ≤ 2N+1 − 1 and an,N = 0, otherwise.Observe that 0 ≤ |an,N ≤ |an|, and that an,N → an as N → ∞. Bydominated convergence (Lemma 5.25)

N∏

j=1

(1 + z2j

) =∞∑

n=0

an,N →∞∑

n=0

an = (1− z)−1.

97

K86

Observe that

2 cot 2φ =cos2 φ− sin2 φ

cosφ sinφ= cotφ+ tanφ

and use induction.

Observe that

1

2ncot

θ

2n=

1

θ

2−nθ

sin 2−nθcos 2−nθ → 1

θas n→ ∞.

98

K87

Very similar to K86. To obtain the formula for un, observe that2 cos2 φ = 1 + cos 2φ and cos(π/4) = 2−1/2.

99

K88

The square of a rational is rational.

(ii) Observe, by induction or Leibniz’s rule, that f (k) is the sum ofpowers of x with integral coefficients plus terms of the form

Mxr(1− x)s

(min(r, s))!

with M an integer and r, s ≥ 1. Thus f (k)(0) and f (k)(1) are alwaysintegers. Since bnπ2n−r is an integer, G(0) +G(1) is always an integer.

0 < πanf(x) sin πx ≤ an

n!→ 0

as n→ ∞.

100

K89

(a)(i) Given c ∈ C we can find b ∈ B with g(b) = c and a ∈ A withf(a) = b.

(ii) g f(a) = g f(a′) implies f(a) = f(a′) and so a = a′.

(iii) f(a) = f(a′) implies g f(a) = g f(a′) and so a = a′.

(iv) Given c ∈ C, we can find a ∈ A with g f(a) = b. Observe thatf(a) ∈ B and g

(

f(a))

= c

(b) Let A = a, B = a, b, C = a with a 6= b f(a) = a,g(a) = g(b) = a.

101

K90

If f satisfies the conditions of the first sentence of the second para-graph, we can write f = g J where J(x) = −x and g satisfies theconditions of the first paragraph. Thus

f ′(x) = J ′(x)g′(J(x)) = −g′(J(x)) < 0

for all x ∈ E.

If ad− bc = 0 then f is constant and f ′ = 0.

102

K91

(i) Suppose that f is continuous. If f is not strictly monotonic wecan find x1 < x2 < x2 so that f(x1), f(x2) and f(x3) are not a strictlyincreasing or a strictly decreasing sequence. Without loss of generality(or by considering f(1−x) and 1− f(x) if necessary), we may supposef(x3) ≥ f(x1) ≥ f(x2). If either of the inequalities are equalities thenf is not injective. Thus we may suppose f(x3) > f(x1) > f(x2). Butby the intermediate value theorem we can find c ∈ (x2, x3) so thatf(c) = f(x1) and f is not injective.

Suppose f is strictly monotonic. Without loss of generality we sup-pose f increasing. We shall show that f is continuous at t ∈ (0, 1)(the cases t = 0 and t = 1 are handled similarly). Observe thatf(1) > f(t) > f(0). Let ǫ > 0. Set ǫ′ = min(1 − f(t), f(t), ǫ)/2.We can find s1 and s2 with f(s1) = f(t) − ǫ′ f(s2) = f(t) + ǫ′. If/ta = min(t − s1, t + s2), then |s − t| > δ implies s2 > s > s1 sof(s2) > f(s) > f(s1) and |f(s)− f(t)| ≤ ǫ′ < ǫ.

(ii) The same proof as in (i) shows that, if g is continuous and in-jective, it must be strictly monotonic. The example of g(t) = t/2 fort ≤ 1/2, g(t) = t/2 + 1/2 shows that the converse is false.

(iii) The same proof as in (i) shows that, if g is strictly monotonicand surjective, it must be continuous. The example g(t) = sin πt showsthat the converse is false.

103

K92

Write f(x) = (log x)/x. By looking at f ′ we see that f is strictlyincreasing from −∞ to e−1 on the interval (0, e], has a maximum valueof e−1 at e and is strictly decreasing from e−1 to 0 on the interval(0,∞). The function f has a unique zero at 1,

We have xy = yx if and only if f(x) = f(y). The set

(x, y) : xy = yx, x, y > 0is thus the union of the straight line (x, x) : x > 0 and a curve withreflection symmetry in that line, having asymptotes x = 1 and y = 1and passing through (e, e).

If f(x) = f(y) and x < y, then x ≤ e so we need only examine thecases m ≤ 2 to see that the integer solutions of nm = mn are n = m,(m,n) = (2, 4) and (m,n) = (4, 2).

Since f(π) < f(e) we have eπ > πe.

104

K93

Observe thatxy = 1

4

(

(x+ y)2 − (x− y)2.

105

K94

(i) Choose γ with β > γ > 1. If we set f(x) = xαγx, then (recallingthat γx = exp(x log γ))

f ′(x) = αxα−1γx + xα(log γ)γx = αxα−1γx(α+ γx) > 0

for x sufficiently large (x ≥ x0, say). Thus

xαβx = f(x)(β/γ)x ≥ f(x0)(β/γ)x → ∞.

(ii) Set

f4(x) = exp

(

1

x

)

, f3(x) =

(

1

x

)1/2

, f2(x) = exp

(

log1

x

)1/2

and

f1(x) =

(

log1

x

)3

.

Set y = 1/x so y → ∞ as x→ 0+.

Observe that log f4(x)/ log f3(x) = 2y/(log y) → ∞ as x→ ∞.

Observe that log f3(x)/ log f2(x) = (log y)1/2/2 → ∞ as x→ ∞.

Observe that log f2(x)/ log f1(x) = (log y)1/2/(3 log log y) → ∞ asx→ ∞.

106

K95

(ii) If Q has a root α 6= 0 of order n ≥ 1 then differentiating the givenequation n − 1 times and setting z = α we get −P (α)Q(n)(α) = 0 soP (α) = 0 and P and Q have a common factor. Thus Q(z) = AzN forsome N ≥ 1, A 6= 0 and it is easy to check that this too is impossible.

(iv) If the degree of P is no smaller than that of Q then

P (x)

Q(x)log x→ ∞.

If the degree of P is smaller than that of Q then

P (x)

Q(x)log x→ 0.

If the degree of P is no smaller than that of Q then

P (x)

Q(x)log x→ ∞.

107

K96

Set f(x) = x−log(1+x). Then f(0) = 0 and f ′(x) = 1−(1+x)−1 ≥ 0for 0 < x and f ′(x) ≤ 0 for x > 0. so f(x) ≥ 0 and −x ≥ log(1 − x)for 0 ≤ x ≤ 1.

Set g(x) = log(1 + x)− x+ x2. Then g(0) = 0 and

g′(x) = (1 + x)−1 − 1 + 2x = (x+ 2x2)(1 + x)−1 ≥ 0

for x > 0 and g′(x) ≤ 0 for −1/2 ≤ x ≤ 0 so g(x) ≥ 0 and log(1+x) ≥x− x2 for x ≥ −1/2.

Since

logkn∏

r=1

(

1− r

n2

)

=kn∑

r=1

log(

1− r

n2

)

we have

−kn∑

r=1

r

n2≥ log

kn∏

r=1

(

1− r

n2

)

≥ −kn∑

r=1

r

n2−

kn∑

r=1

r2

n4.

Butkn∑

r=1

r

n2=kn(kn+ 1)

2n2=k

2

(

k +1

n

)

→ k2

2

andkn∑

r=1

r2

n4≤ (kn)3

n4=k

n→ 0

as n→ ∞ so

logkn∏

r=1

(

1− r

n2

)

→ −k2

2

and the result follows on applying exp.

108

K97

Suppose that x ≥ 1. Then xn ≥ 1 and

2n(xn − 1)− 2n+1(xn+1 − 1) = 2n(x2n+1 − 2xn+1 + 1)

= 2n(xn+1 − 1)2 ≥ 0.

Thus the sequence 2n(xn− 1) is decreasing bounded below by 0 and sotends to a limit. A similar argument applies if 1 > x > 0.

Since2n(xn − 1)

2n(1− 1/xn)= xn → 1

2n(xn − 1) and 2n(1− 1/xn) tend to the same limit.

(i) If x = 1, then xn = 1 so log 1 = 0.

(ii) If xn = 1 + δn, y = 1 + ηn then(

(1 + δn)(1 + ηn)

1 + δn + ηn

)2n

=

(

1 +δnηn

1 + δn + ηn

)2n

.

Since 2nδn tends to a limit we can find an A such that |δn ≤ A2−n.Similarly we can find a B such that |ηn ≤ B2−n. Thus we can find aC such that

∣

∣

∣

∣

δnηn1 + δn + ηn

∣

∣

∣

∣

≤ C2−2n.

It follows that∣

∣

∣

∣

∣

(

(1 + δn)(1 + ηn)

1 + δn + ηn

)2n

− 1

∣

∣

∣

∣

∣

≤2n∑

r=1

(

2n

r

)

Cr2−2nr

≤2n∑

r=1

2nrCr2−2nr

=≤2n∑

r=1

2nrCr2−nr

≤ C2−n

1− C2−n→ 0

as n→ ∞.

By the mean value theorem

(b2−n − a2

−n

) = 2−nc2−n−1(b− a)

for some c ∈ (a, b) so

2n((1 + δn)(1 + ηn)− (1 + δn + ηn) → 0

that is to say

2n((xy)2−n − (xn + yn − 1)) → 0

109

so writing z = xy

2n(zn − 1)− 2n(xn − 1)− 2n(yn − 1) → 0

andlog xy = log x+ log y.

Result (ii) shows that we need only prove (iii), (iv), (v) and (vi) atthe point x = 1. To do this observe that a simple version of Taylor’stheorem

|f(1 + δ)− f(1)− f ′(1)δ| ≤ sup|η|≤|δ|

|f ′′(1 + η)||δ2|

applied to f(x) = x−2n yields

|(1 + δ)−2n − 1− 2−nδ| ≤ 2−n × 4× |δ|2

for all |δ| ≤ 10−1 and n ≥ 3. (We can do better but we do not needto.) Thus if |δ| ≤ 10−1 and we set x = 1 + δ we obtain

|2n(xn − 1)− δ| ≤ 4|δ|2

for n ≥ 3 so that| log(1 + δ)− δ| ≤ 4|δ|2.

Thus log is continuous and differentiable at 1 with derivative 1.

Since log(x+t) = log x+log(1+t/x) the chain rule gives log′ x = 1/xand since log′ x > 0 we see that log is strictly increasing.

(vii) Since log 2 > log 1 = 0 we have log 2n = n log 2 → ∞ solog x → ∞ as x → ∞ and log y = − log y−1 → −∞ as y → 0+. Sincelog is continuous and strictly monotonic, part (vii) follows.

110

K98

If(

f(x+ η)

f(x)

)1/η

→ l

then, taking logarithms,

log f(x+ η)− log f(x)

η→ log l

so log f is differentiable and, by the chain rule, f = exp log f is differ-entiable at x. Since

log l =d

dxlog f(x) =

f ′(x)

f(x)

we have

l = expf ′(x)

f(x).

111

K99

If

f(x) =γxβ−1

xβ − aβ− 1

x− atends to a limit as x→ a then (x− a)f(x) → 0. Since

xβ − aβ

x− a→ βaβ−1

as x→ a this givesγ

β− 1 = 0

so β = γ.

If β = γ then f(x) = F (x)/G(x) with

F (x) = βxβ − βaxβ−1 − xβ + aβ

G(x) = xβ+1 − aβx− axβ + aβ+1

We find F (a) = F ′(a) = 0, G(a) = G′(a) = 0 and use L’Hopital’srule to give

f(x) → F ′′(a)

G′′(a)=β − 1

2a.

112

K100

We only consider xα when x is real and we define it as exp(α log x)where log is the real logarithm.

113

K101

(iii) Set f(x) = log g(x) and apply (i) to get g(x) = xb for some realb. Easy to see that this is a solution.

(iv) Observe that, if x > 0, g(x) = g(x1/2)2 > 0. Thus, by (iii)g(x) = bx for all x > 0 and some b > 0.

Now g(−1)2 = g(1) = 1 so g(−1) = ±1 and either g(x) = b|x| for allx or g(x) = sgn(x)b|x| for all x. (Recall sgn x = 1 if x > 0, sgn x = −1if x < 0.) It is easy to see that these are solutions.

114

K102

(i) We know thatχ(t/2)2 = χt

but the equation (exp is)2 = exp iu, where u ∈ R, has two real solutionsin s with −π < s ≤ π (the solutions being s/2 and one of s/2 ± π).As a specific example, if χ(t) = exp 2it, then θ(3π/4) = −π/2 butθ(3π/8) = 3π/4.

However since χ is continuous we can find a δ > 0 such that |χ(t)−1| ≤ 100−1 for |t| ≤ δ. If |t| ≤ δ then |θ(t)| ≤ π/4 and |θ(t/2)| ≤ π/4so θ(t/2) = θ(t)/π (since |θ(t/2)± π| > π/4).

(iii) The same ideas show that the continuous homomorphisms areprecisely the maps λ 7→ λn for some integer n.

115

K103

(i) f(0) = 2f(0) so f(0) = 0.

(ii) Set F (2kp−1j ) = 2kj whenever j and k are integers with j ≥ 1

and F (t) = 0 otherwise. (By the uniqueness of factorisation F , is welldefined.)

Since F (p−1j ) → ∞ and p−1

j → 0 as j → ∞, F is not continuous at0.

(iii)

f(t) = 2nf(t2−n) = tf(t2−n)− f(0)

t2−n→ f ′(0)t

as n→ ∞ so f(t) = f ′(0)t for all t 6= 0 and so for all t.

(iv) Set F (2kp−1j ) = 2kp−1

j whenever j and k are integers with j ≥ 1and F (t) = 0 otherwise. Since |F (t)− F (0)| = |f(t)| ≤ |t|, we have Fcontinuous at 0. Since

F (p−1j )− F (0)

p−1j

= 1 → 1 butF (21/2j−1)− F (0)

21/2j−1→ 0

as j → ∞, F is not differentiable at 0.

(v) Observe that u(t) = u(t/2)2 ≥ 0 for all t. If u(x) ≤ 0 then

u(2−nx) = u(x)2−n → 1 so by continuity u(0) = 1. Thus by continuity

there exists a δ > 0 such that u(t) > 0 for |t| < δ. Thus u(t) =u(2−nt)2

n> 0 for |t| < 2nδ and so u is everywhere strictly positive.

Now set f(t) = log u(t) and apply part (iii).

116

K104

Consider the case x 6= 0. Observe that

1

(1 + δ)1/2= 1− δ

2+ ǫ(δ)|δ|

with ǫ(δ) → 0 as δ → 0. (Use differentiation or Taylor series.) Observealso that |x · h| ≤ ‖x‖‖h‖. Thus

f(x + h) =x+ h

(‖x‖2 + 2x · h+ ‖h‖2)1/2

=x+ h

‖x‖(

1 + (2x · h+ ‖h‖2)/‖x‖2)1/2

=x+ h

‖x‖

(

1− 2x · h+ ‖h‖22‖x‖2

)

+ ǫ1(h)‖h‖

= f(x)− h

‖x‖ − (x · h)x‖x‖3 + ǫ2(h)‖h‖

where ‖ǫj(h)‖ → 0 as ‖h‖ → 0. Thus f is differentiable at x and

Df(x)h =h

‖x‖ − (x · h)x‖x‖3 .

(Df(x)h) · x =h · x‖x‖ − x · h‖x‖

2

‖x‖3 = 0.

Since‖f(x)− f(0)‖ = 1 9 0

as ‖x‖ → 0, f is not even continuous at 0

Observe that f is constant along radii.

117

K105

Without loss of generality take z0 = 0.

(i)⇒(ii) If f is complex differentiable at 0, then writing f ′(0) =A+Bi with A and B real, and taking h and k real

f(h+ ik) = f(0) + (A+Bi)(h + ik) + ǫ(h + ik)|h+ ik|with ǫ(h + ik) → 0 as |h+ ik| → 0. Taking real and imaginary parts,(

u(h, k)v(h, k)

)

=

(

u(0.0)v(0, 0)

)

+

(

Ah− BkAk +Bh

)

+

(

ǫ1(h, k)ǫ2(h, k)

)

(h2 + k2)1/2

with∥

∥

∥

∥

(

ǫ1(h, k)ǫ2(h, k)

)∥

∥

∥

∥

→ 0

as (h2+k2)1/2 → 0. Thus F is differentiable at 0 with Jacobian matrix(

A −BB A

)

If A = B = 0 take λ = 0 and θ = 0. Otherwise take λ = (A2 +B2)1/2

and θ such that cos θ = A, sin θ = B.

(ii)⇒(iii)⇒(iv) Just a matter of correct interpretation.

(iv)⇒(i) Carefully reverse the first proof.

118

K106

(i) Observe that

g(x+ h) = f(x+ h, c− x− h)

= f(x, c− x)f,1(x, c− x)h + f,2(x, c− x)(−h) + ǫ(h, h)(h2 + h2)1/2

= g(x) +(

f,1(x, c− x)− f,2(x, c− x))

+ 21/2ǫ(h, h)|h|where ǫ(h, k) → 0 as (h2 + k2)1/2 → 0. Thus g is differentiable withderivative

g′(x) = f,1(x, c− x)− f,2(x, c− x).

(ii) Observe that g = f u where

u(x) =

(

xc− x

)

.

Thus u has Jacobian matrix(

1−1

)

.

By chain the rule g is differentiable with 1× 1 Jacobian matrix

(

f,1(x, c− x) f,2(x, x− c))

(

1−1

)

= (f,1(x, c− x)− f,2(x, c− x)).

Define g as stated. If f,1 = f,2, then g′ = 0 so, by the constant value

theorem, g is constant. Thus f(x, x−c) = f(y, y−c) for all x, y, c ∈ R.Writing h(x+ y) = f(0, x+ y), we have the required result.

119

K107

(i) Differentiating with respect to λ and applying the chain rule, wehave

αλα−1f(x) =

m∑

j=1

xjf,j(λx).

Taking λ = 1 gives the required result.

(ii) We observe that v is differentiable with

v′(λ) =m∑

j=1

xjf,j(λx) = λ−1m∑

j=1

λxjf,j(λx) = αλ−1v(λ).

Thusd

dλ(λ−αv(λ)) = 0

and by the constant value theorem

λ−αv(λ) = v(1)

and f(λx) = λαf(x).

120

K108

This is really just a question for thinking about.

(i) We seek to maximise the (square root of)

f(θ) =

∥

∥

∥

∥

(

a bc d

)(

cos θsin θ

)∥

∥

∥

∥

2

= (a cos θ + b sin θ)2(c sin θ + b cos θ)2

and this we can do, in principle, by examining the points with f ′(θ) = 0.

(ii) We seek to maximise the (square root of)

p∑

r=1

(

m∑

s=1

arsxs

)2

subject to the constraintm∑

s=1

x2s = 1

which can, in principle, be handled using Lagrange multipliers.

However, this involves solving an m ×m set of linear equations in-volving a parameter λ leading to polynomial of degree m in λ. The mroots must then be found and each one inspected. This neither soundsnor is a very practical idea even when m is quite small.

121

K109

If e1, e2, . . . , em is the standard basis (or any orthonormal basis)and (aij) is the associated matrix of α, then

aij =

m∑

r=1

arj〈er, ei〉 =⟨

n∑

r=1

arjer, ei

⟩

= 〈αej, ei〉 = 〈ej, αei〉 = 〈αei, ej〉 = aji

Conversely if aij = aji for all i, j, then essentially the same calculationshows that

〈αej, ei〉 = 〈αei, ej〉and so by linearity

⟨

α

(

n∑

r=1

xrer

)

,m∑

s=1

yses

⟩

=m∑

r=1

m∑

s=1

xrys〈αer, es〉

=m∑

r=1

m∑

s=1

xrys〈er, αes〉

=

⟨

n∑

r=1

xrer, α

(

m∑

s=1

yses

)⟩

.

Let u1, u2, . . . , em be an orthonormal basis of eigenvectors witheigenvalues λ1, λ2, . . .λm with |λ1| ≥ |λ2| ≥ · · · ≥ |λm|. Then

∥

∥

∥

∥

∥

α

(

m∑

j=1

xjuj

)∥

∥

∥

∥

∥

2

=

∥

∥

∥

∥

∥

m∑

j=1

xjλjuj

∥

∥

∥

∥

∥

2

=m∑

j=1

x2jλ2j ≥ λ21

m∑

j=1

x2j

with equality when x2 = x3 = · · · = xm = 0 (and possibly for othervalues). This proves †.

The matrix A given has eigenvalue 0 only. However ‖A‖ = 1.

122

K110

(ii) Let x =∑n

j=1 xjej. Then

yk =n∑

j=1

λkjxj

(∑n

r=1(λkrxr)

2)1/2

→ e1

unless x1 = 0.

If the largest eigenvalue is negative, then (in general) ‖xk‖ → |λ1|and ‖(−1)kyk − u1‖ → 0 where u1 = ±e1.

(iii) (c) The number of operations for each operation is bounded byAn2 for some A (A = 6 will certainly do),

Suppose |λ1| > µ ≥ |λj | for j ≥ 2. Then the e1 component of αkx

will grow at a geometric rate λj/µ relative to the other components.

(v) Look at (iii). If the two largest eigenvalues are very close thesame kind of thing will occur.

(iv) Suppose e1 the eigenvector with largest eigenvalue known. Usethe iteration

x 7→ (α(x− x · e1)or something similar. However, the accuracy to which we know e1 willlimit the accuracy to which we can find e2 and matters will get rapidlyworse if try to find the third largest eigenvalue and so on.

(vii) No problem in real symmetric case because eigenvectors orthog-onal.

123

K111

(ii) 〈x, α∗αx〉 = 〈αx, αx〉 = ‖αx‖2.Thus

‖α∗α‖‖x‖2 = ‖x‖‖α∗αx‖ ≥ |〈x, α∗αx〉| = ‖αx‖2 ≥ ‖α‖2‖‖x‖2

for all x and so |α∗α‖ ≥ ‖α‖2.(iii) ‖α∗‖‖α‖ ≥ ‖α∗α‖ ≥ ‖α‖2 and so either α = 0 (in which case

the result is trivial) or ‖α‖ 6= 0 and ‖α∗‖ ≥ ‖α‖. But α∗∗ = α so ‖α‖ =‖α∗∗‖ ≥ ‖α‖. Thus ‖α‖ = ‖α∗‖ and ‖α‖2 = ‖α∗‖‖α‖ ≥ ‖α∗α‖ ≥ ‖α‖2so ‖α∗α‖ = ‖α‖2.

Since (α∗α)∗ = α∗α∗∗ = α∗α we have α∗α symmetric. If αe = λewith e 6= 0 then

λ‖e‖2 = 〈e, λe〉 = 〈e, α∗αe〉 == 〈αe, αe〉 = ‖αe‖2.Thus λ ≥ 0.

(vi) Multiplication of A with A∗ takes of the order of m3 operations(without tricks) and dominates.

(vii) We have A∗A =

(

1 11 5

)

.

A has eigenvalues 1 and 2. A∗A has eigenvalues 4 ± 51/2. ‖A‖ =(4 + 51/2)1/2.

124

K112

Follow the proof of Rolle’s theorem. If ‖c|| < 1 and f has a maximumat c then f,1(c) = f,2(c) = 0 so Df(c) is the zero map.

g(x, y) = x2 will do. Observe that cos2 θ − a cos θ − b sin θ is notidentically zero (consider θ = π/2 and other values of θ).

We can not hope to have a ‘Rolle’s theorem proof’ in higher dimen-sions.

125

K113

Since (x, y) 7→ x − y, (x, y) 7→ xy, x 7→ x−1 and x 7→ f(x) are allcontinuous we see that F which can be obtained by composition of suchfunctions is continuous except perhaps at points (x, x).

If F is continuous at (x, x) then F (x+ h, x) must tend to the limitF (x, x) as h → 0 so f is differentiable and F (x, x) = f ′(x). SinceF (x+ h, x+ h) → F (x, x) as h→ 0, f ′ must be continuous.

Conversely if f ′ exists and is continuous then, setting F (x, x) =f ′(x), we saw above that F is continuous at all points (x, y) with x 6= y.If h 6= k the mean value theorem gives

F (x+ h, x+ k)− F (x, x) =f(x+ h)− f(x+ k)

x− y− f ′(x)

= f ′(x+ (θh,kh+ (1− θh,k)k))

− f ′(x)

for some θh,k with 0 < θh,k < 1. But it is also true that, if we setθh,h = 1/2,

F (x+ h, x+ h)− F (x, x) = f ′(x+ h)− f ′(x)

= f ′(x+ (θh,hh+ (1− θh,h)h))

− f ′(x).

Thus

F (x+ h, x+ k)− F (x, x) = f ′(x+ (θh,kh+ (1− θh,k)k))

− f ′(x) → 0

as (h2 + k2)1/2 → 0.

Suppose now that f is twice continuously differentiable. Much asbefore, the chain rule shows that F is differentiable at all points (x, y)with x 6= y. If f is twice continuously differentiable then the local formof Taylor’s theorem shows us that

f(x+ h) = f(x) + f ′(x)h+f ′′(x)

2h2 + ǫ(h)h2

with ǫ(h) → 0 as h→ 0. Thus, if h 6= k,

F (x+ h, x+ k)

=(f(x) + f ′(x)h+ 1

2f′′(x)h2 + ǫ(h)h2)− (f(x) + f ′(x)k + 1

2f′′(x)k2 + ǫ(k)k2

h− k

= f ′(x) +f ′′(x)2

(h+ k) +ǫ(h)h2 + ǫ(k)k2

h− k

= F (x, x) +f ′′(x)2

(h+ k) + ǫ′(h, k)(h2 + k2)1/2

where ǫ′(h, k) → 0 as (h2 + k2)1/2 → 0. The formula can be extendedto the case h = k by applying the appropriate Taylor theorem to f ′.Thus F is differentiable everywhere.

126

K114

If k 6= 0, then we can find an X > 0 such that |f ′(t)−k| < |k|/4 andso |f ′(t)| > 3|k|/4 for t ≥ X . Thus, using the mean value inequality,

∣

∣

∣

∣

f(x)

x

∣

∣

∣

∣

=

∣

∣

∣

∣

f(x)− F (X)

x−X

x−X

x+F (X)

x

∣

∣

∣

∣

≥∣

∣

∣

∣

f(x)− f(X)

x−X

∣

∣

∣

∣

∣

∣

∣

∣

x−X

x

∣

∣

∣

∣

−∣

∣

∣

∣

f(X)

x

∣

∣

∣

∣

≥ 3|k|4

∣

∣

∣

∣

x−X

x

∣

∣

∣

∣

−∣

∣

∣

∣

f(X)

x

∣

∣

∣

∣

≥ |k|2

− |k|4

=|k|4

whenever x ≥ max(3X, (4|f(X)|)−1).

g(x) = x−2 sin x4 will do.

127

K115

The local form of Taylor’s theorem

f(x+ h) = f(x) + f ′(x)h+f ′′(x)

2h2 + ǫ(h)h2

applied to log (or other arguments) show us that

log(1− h) = −h− h2

2+ ǫ(h)h2

with ǫ(h) → 0 as h→ 0.

Thus since qn → ∞ we can find an N such that

−2−1q−1n ≥ log(1− q−1

n ) and 0 ≥ log(1− q−1n ) + q−1

n ≥ −2q−2n ≥ −2n−2

for all n ≥ N , so the results follow from the comparison test.

128

K116

If a = b = 0 maximum c.

If a = 0 maximum c if b ≤ 0, 10b+ c if b ≥ 0

If a > 0 then maximum c if −b/2a ≥ 5, 100a+10b+ c if −b/2a ≤ 5.

If a < 0 then maximum c if−b/2a ≤ 0, c−b2/(4a) if 0 ≤ −b/2a ≤ 10,100a+ 10b+ c if 10 ≤ −b/2a.

129

K117

(i) Take x1 = 1, xj = 0 otherwise.

(ii) To see that B is positive definite if A is, take

x1 = −(a−111 a12x2 + a−1

11 a13x3 + · · ·+ a−111 a1nxn).

130

K118

(i) One way of seeing that the λj are continuous is to solve thecharacteristic equation. Now use the intermediate value theorem.

(iii) B(t) =

(

cos t − sin tsin t cos t

)

will do.

131

K119

The reduction to Figure K2 can not be rigorous unless we stateclearly what paths are allowed. However if we have a path which is notsymmetric under reflection in the two axes of symmetry joining midpoints of opposite sides then by considering the reflected path we canreconstruct a shorter symmetric path.

[A more convincing route involves the observation that the shortestpath system for three points M , N , P is three paths MZ, NZ, PZmaking angle 2π/3 to each other provided the largest angle of thetriangle MNP is less than 2π/3 and consists of the two shortest sidesotherwise. (See Courant and Robbins What is Mathematics?.)]

One we have the diagram we see that the total path length is

f(θ) = 2a sec θ + (a− a tan θ)

where a is the length of the side of the square and θ is a base angle ofthe isosceles triangles.

f ′(θ) =1− 2 sin θ

cos2 θso f increases on [0, π/6] and decreases on [π/6, π/4] so the best angle isθ/6 (check that this gives an arrangement consistent with the statementin square brackets). IN PARTICULAR θ = π/4 is not best.

There are two such road patterns the other just being a rotationthrough π/4

132

K120

The time taken is

f(x) =h− x

u+

(1 + x2)1/2

vfor 0 ≤ x ≤ h. We observe that

f ′(x) = −1

u+

x

v((1 + x2)1/2

so f ′ is negative and f is decreasing for 0 ≤ x ≤ α with 0 ≤ x ≤ (1−v2/u2)−1/2 and f ′ is positive and f is increasing for x ≥ (1−v2/u2)−1/2.Thus if (1 − v2/u2)−1/2 < h (i.e. (1 − v2/u2)h2 > 1) he should takex = (1 − v2/u2)−1/2. However, if (1 − v2/u2)h2 < 1 his best course if0 ≤ x ≤ h is to take x = h. Since choosing x outside this range isclearly worse he should take x = h.

If h is large his initial run is almost directly towards the tree and hisgreater land speed makes this desirable. If h is small his run is almostperpendicular to the direction of the tree and does him very little good.

133

K121

We wish to find the maxima and minima of f(x, y) = y2 − 13x3 + ax

in the region x2 + y2 ≤ 1. We first examine the region x2 + y2 < 1.

The stationary points are given by

0 = f,1(x, y) = −x2 + a, 0 = f,2(x, y) = 2y.

Thus, if a < 0 or a ≥ 1, there are no interior stationary points andboth the global maximum and minimum must occur on the boundary.

If 0 < a < 1, then there are two interior stationary points at x =±a1/2, y = 0. The Hessian is

(

−2x 00 2

)

so (a1/2, 0) is a saddle and (−a1/2, 0) a minimum. Thus the globalmaximum must occur on the boundary and the global minimum mayoccur at (−a1/2, 0) with value −2a3/2/3 or may occur on the boundary.

If a = 0 then there is one stationary point at (0, 0), the Hessian issingular there but inspection of the equation f(x, y) = y2 − 1

3x3 shows

that we have a saddle and both the global maximum and minimummust occur on the boundary.

Now we look at the boundary x2 + y2 = 1. Here

f(x, y) = g(x) = 1− x2 − 13x3 + ax

with x ∈ [−1, 1]. Notice that g is a cubic so we can draw sketches tohelp ourselves.

Now g′(x) = −2x−x2+a and g′ has roots at −1±(1+a2). If |a| ≥ 1then g is decreasing on (−1, 1) so has maximum at x = −1 (which isthus the global maximum for f at (−1, 0)) and a minimum at x = 19which is thus the global minimum for f at (0, 1)).

If |a| < 1, g increases for x running from −1 to −1 ± (1 + a2)(which is thus a global maximum and f has its global maximum at

x = −1+ (1+ a2) y =(

1− (1+ (1+ a2))2)1/2

) and decreases as x runsfrom from −1± (1+a2) to 1. Thus g has minima at x = −1 and 1 and(by evaluating g at these points) a global minimum at 1 if a ≤ −1/3.Since a ≤ −1/3 we know that this gives a global minimum for f at(0, 1). If a ≥ −1/3, g has a global minimum at −1 and if 0 ≥ a ≥ −1/3we know that it gives a global minimum for f at (0, 1).

If 1 > a > 0 we observe that we know that the global minimumoccurs when y = 0 and by looking at h(x) = f(x, 0) = −1

3x3 + ax we

see that it occurs at (−a1/2, 0).

134

K122

The composition of differentiable functions is differentiable and theproduct of differentiable functions (with values in R) are differentiableso since the maps (x, y) 7→ r and (x, y) 7→ θ are differentiable exceptat 0, f is. (To see that (x, y) 7→ r is differentiable observe that r =(x2 + y2)1/2 and use basic theorems again. To see that (x, y) 7→ θ isdifferentiable on

S = (x, y) : |y| > 9xobserve that on S, θ = tan−1 x/y. To extend to R2 \ 0 either userotational symmetry or cover R2\0 with rotated copies of S on whichsimilar formulae hold.

Sincef(r cos θ, r sin θ)− f(0, 0)

r→ g(θ)

as r → 0+, f has a directional derivative at 0 if and only if g(θ) =−g(−θ).

If f is differentiable at 0 then

f(r cos θ, r sin θ)− f(0, 0)

r→ cos θf,1(0, 0) + sin θf,2(0, 0)

sog(θ) = g(0) cos θ + g(π/2) sin θ

ieg(θ) = A sin θ +B cos θ

for some constants A and B so

f(x, y) = Ax+By.

The necessary condition is sufficient by inspection.

135

K123

(i) Observe that.

f(αt, βt) = t2(β − aα2t)(β − bα2t)

which has a strict local minimum at t = 0.

However, if a > c > b, then

f(t, ct2) = (c− a)(c− b)t4

which has a strict local maximum at t = 0.

By looking at the behaviour of f along curves (x, y) = (αt, βt2) thefunction f has no minima.

(ii) In part (i), f has Hessian(

f,11(0) f,12(0)f,21(0) f,22(0)

)

=

(

0 00 2

)

which is singular.

136

K124

(i) The second sentence holds by Exercise 8.22 which says that if fis Riemann integrable so is |f |.

(ii) First sentence false. Take f = −g = F .

Second sentence false. Take

f(x) = F (x), g(x) = 0 for x < 1/2

f(x) = 0, g(x) = F (x) for x ≥ 1/2.

(iii) By considering functions of the form

f(x) = F (x), g(x) = 0 for x < 1/2

f(x) = 0, g(x) = F (x) for x ≥ 1/2.

and of the form

f(x) = F (x), g(x) = 0 for x < 1/2

f(x) = 0, g(x) = 1 for x ≥ 1/2.

and functions like f(1 − x) and g(x) we see that the product of twofunctions that are not Riemann integrable may or may not be Riemannintegrable and that the product of one that is Riemann integrable withone that is not may or may not be Riemann integrable.

137

K125

(i) Consider the dissection DN = r/N : 0 ≤ r ≤ N where N ≥M2.At most 2M2 intervals [(r−1)/N,R/N ] contain points of the form p/qwith p and q coprime and 1 ≤ q ≤M . Thus

0 = s(f,DN) ≤ S(f,DN) ≤ 2M2N−1 +M−1 →M−1

as N → ∞. Since M is arbitrary f is Riemann integrable with∫ 1

0f(t) dt = 0.

(ii) f is continuous at irrational points x. Since there are only finitelymany points of the form p/q with p and q coprime and 1 ≤ q ≤ M andx is not one of them we can find a δ > 0 such that (x− δ, x+ δ)∩ [0, 1]contains no such points. Thus |f(x) − f(y)| = f(y) ≤ M−1 for all|x− y| < δ.

f is discontinuous at rational points x. Choose xn irrational suchthat xn → x. Since 0 = f(xn) 9 f(x), we are done.

(iii) Nowhere differentiable. By (ii) need only look at irrationalpoints x. If q is a prime we know that there exists a p such that(p− 1)/q ≤ x < p/q so

(f(p/q)− f(x)

p/q − x≥ 1/p

1/p= 1

but choosing xn irrational such that xn → x we have

(f(xn)− f(x)

xn − x= 0 → 0.

138

K126

Observe that, if 0 ≤ u < v ≤ 1, then un−vn ≤ n(y−x) (by the meanvalue theorem or algebra). Write fn(x) = f(xn). If D is a dissectionwrite

Dn = y1/n : y ∈ D.By the observation of the first sentence

S(fn,Dn)− s(fn,Dn0 ≤ n(

S(f,D)− s(f,D))

so, if f is Riemann integrable so is fn.

(ii) and so (i) is true

Suppose that |f(x)| ≤ K. Given ǫ > 0, we can find a 1 > δ > 0 suchthat |f(x)− f(0)| ≤ ǫ/2 for 0 ≤ x ≤ δ. Since δ1/n → 1 as n → ∞ wecan find an N such that 1− δ1/n ≤ (K + |k|)−1ǫ/2 for n ≥ N .

If N ≥ n then |fn(x)− k| ≤ ǫ/2 for x ∈ [0, δ1/n] and |fn(x)| ≤ K forx ∈ [δ1/n, 1] so∣

∣

∣

∣

∣

∫ δ1/n

0

f(x) dx− k(1− δ1/n

∣

∣

∣

∣

∣

≤ ǫ/2 and

∣

∣

∣

∣

∫ 1

δ1/nf(x) dx− kδ1/n

∣

∣

∣

∣

≤ ǫ/2

so that∣

∣

∣

∣

∫ 1

0

f(x) dx− k

∣

∣

∣

∣

≤ ǫ.

(iii) is false. Take f(x) = 0 for x 6= 0 and f(0) = 1.

139

K127

Without loss of generality, suppose that f(

(a + b))

≥ 0. Observethat, since f(a) = 0 and |f ′(t)| ≤ K it follows that, by the mean valueinequality, |f(s)| ≤ K(s− a) for a ≤ s ≤ (a+ b)/2 with equality whens = (b+ a)/2 if and only if f(s) = K(s− a) for all a ≤ s ≤ (a + b)/2.Similarly |f(s)| ≤ K(b − s) for b ≥ s ≥ (a + b)/2 with equality whens = (b+ a)/2 if and only if f(s) = K(b− s) for all a ≤ s ≤ (a + b)/2.Unless K = 0 (in which case the result is trivial) the conditions forequality give f non-differentiable at (b+ a)/2. Thus if we write

K(b− a)/2− |f(

(b+ a)/2)

= 4η(b− a)

we have 4η > 0 and, by continuity, we can find a δ > 0 such that

|f(s)| ≤ (K − η)(s− a) for (b+ a)/2 ≥ s ≥ (b+ a)/2− η

|f(s)| ≤ (K − η)(b− s) for (b+ a)/2 ≤ s ≤ (b+ a)/2 + η

Thus∫ b

a

|f(s)| ≤∫ (a+b)/2−η

a

+

∫ (b+a)/2

(a+b)/2−η

+

∫ (a+b)/2+η

(a+b)/2

+

∫ b

(a+b)/2+η

|f(s)| ds

≤ K(

(b− a)/2− η)2

+ (K − η)((

b− a)/2)2 −

(

(b− a)/2− η)2

< K(b− a)2/4

as required.

To see that this is best possible define

f(x) = K(x− a) for a ≤ x ≤ (a+ b)/2 − ǫ,

f(x) = C − τ(x− (a+ b)/2)2 for (a + b)/2− ǫ < x ≤ (a + b)/2,

and f((a + b)/2 − t) = f((a + b)/2 + t) where C is chosen to make fcontinuous and τ is chosen to make f differentiable at (a + b)/2 − ǫ(thus we take τ = K/(2ǫ)).

140

K128

(v) Observe that once we know that f is Riemann integrable we

know that b−an

∑n−1j=0 f(a + j(b − a)/n) converges but the fact that

b−an

∑n−1j=0 f(a + j(b − a)/n) converges does not imply that f is Rie-

mann integrable.

141

K129

For (A) note that, if Fj and Gj are positive bounded Riemann in-tegrable functions with F1 − G1 = F2 − G2 then F1 + G2 = F2 + G1

and∫ b

a

F1(t) dt+

∫ b

a

G2(t) dt =

∫ b

a

F1(t) +G2(t) dt

=

∫ b

a

F2(t) +G1(t) dt

=

∫ b

a

F2(t) dt+

∫ b

a

G1(t) dt

and so∫ b

a

F1(t) dt−∫ b

a

G1(t) dt =

∫ b

a

F2(t) dt−∫ b

a

G2(t) dt.

The problems with (C) begin (I think) when we try to prove that if fand g are Riemann integrable so is f+g. We also get problems (which,I think have the same cause) when we try to show that anything whichis Riemann integrable under the old definition are Riemann integrableunder the new. One way forward is to show that anything which is(A) integrable is (C) integrable. To do this we can first prove that abounded positive function f is Riemann integrable if and only if givenany ǫ > 0 we can find a dissection

D = x0, x1, . . . , xNwith a = x0 < x1 < · · · < xN = b and a subset Θ of r :< 1 ≤ r ≤ Nsuch that

|f(x)− f(y)| ≤ ǫ for x, y ∈ [xr−1, xr] when r ∈ Θ.

and∑

r /∈Θ(xr − xr−1) ≤ ǫ.

142

K130

(iii) Everything works well until we try to define the upper and lowerintegrals but the set of s(D, f) has no supremum in Q and S(D, f) hasno infimum.

143

K131

(i) Since F is continuous on a closed bounded interval the infimumand supremum are attained at α and β say. Now use the intermediatevalue theorem.

(ii) Observe that

sups∈[a,b]

f(s)w(t) ≥ f(t)w(t)

for all t ∈ [a, b] so

sups∈[a,b]

f(s)

∫ b

a

w(t) dt ≥∫ b

a

f(t)w(t) dt

Using a similar result for the infimum we have

sups∈[a,b]

f(s)

∫ b

a

w(t) dt ≥∫ b

a

f(t)w(t) dt ≥ infs∈[a,b]

f(s)

∫ b

a

w(t) dt.

Now use part (i).

(iii) Take a = −1, b = 1, F (t) = w(t) = t.

If w everywhere negative, consider −w.

144

K132

The graph of f splits the rectangle [a, b]× [f(a), f(b)] into two partswhose areas correspond to the given integrals.

Consider a dissection of [a, b]

D = x0, x1, . . . , xNwith a = x0 < x1 < · · · < xN = b. This gives rise to a dissection of[f(a), f(b)]

D′ = f(x0), f(x1), . . . , f(xN )with f(a) = F (x0) < f(x1) < · · · < f(xN) = f(b).

We haveS(D, f) + s(D′, g) = bf(b)− af(a)

so, taking an infimum over all D, we get

I∗(f) + I∗(g) ≥ bf(b)− af(a)

Similarlys(D, f) + S(D′, g) = bf(b)− af(a)

soI∗(f) + I∗(g) ≤ bf(b)− af(a).

But I∗(f) = I∗(f) =∫ b

af(t) dt and I∗(g) = I∗(g) =

∫ f(b)

f(a)g(s) ds so

∫ b

a

f(t) dt+

∫ f(b)

f(a)

g(s) ds = bf(b)− af(a).

(ii) Without loss of generality, assume Y ≥ F (X). Then∫ X

0

f(x) dx+

∫ Y

0

g(s) ds =

∫ X

0

f(x) dx+

∫ f(X)

0

g(s) ds+

∫ Y

f(X)

g(s) ds

≥ XF (X) + (Y − F (X))X = Y X

with equality if and only if Y = f(X).

(iii) Take f(x) = xp−1−1, g(y) = yq

−1−1. Equality if and only ifX1/q = Y 1/p.

145

K133

Since f is continuous at c, we can find a δ > 0 such that |f(c) −f(x)| < κ/4 when |x− c| < δ. Thus if [α, β] ⊆ (c− δ, c+ δ)

I∗[α,β](f)− I∗[α,β](f) ≤ κ(β − α)/2.

147

K135*

No comments.

148

K136

The formula

supx∈[xj−1,xj ]

f ′(t)(xj −xj−1) ≥ f(xj)− f(xj−1) ≥ infx∈[xj−1,xj ]

f ′(t)(xj −xj−1)

is a version of the mean value inequality.

Consider a dissection of [a, b]

D = x0, x1, . . . , xNwith a = x0 < x1 < · · · < xN = b. By summing the formula of the firstparagraph we get

S(D, f ′) ≥ f(b)− f(a) ≥ s(D, f ′)

soI∗(f ′) ≥ f(b)− f(a) ≥ I∗(f

′)

and the result follows.

149

K137*

No comments.

150

K138

Let N ≥ 4. Take a dissection DN consisting of the points −1, 0, 1,1−N−3 and rN−1 ±N−3 with 1 ≤ r ≤ N − 1. Then

0 = s(DN , f) ≤ S(DN , f) ≤ N−1 +N × 2N−3 = N−1 + 2N−2 → 0

as N → ∞. Thus f is Riemann integrable and∫ 1

−1f(t) dt = 0.

Similarly F (t) = 0 for all t. Thus F ′(0) exists with value 0 = f(0).Since f(1/n) = 1 9 0, f is not continuous at 0.

151

K139

(i)n∑

r=1

r(r−1) =n∑

r=1

(r + 1)r(r − 1)− r(r − 1)(r − 2)

3=

(n+ 1)n(n− 1)

3.

n∑

r=1

r =

n∑

r=1

(r + 1)r − r(r − 1)

2=

(n+ 1)n

2n∑

r=1

r2 =n∑

r=1

r(r − 1) +n∑

r=1

r =n(n+ 1)(2n+ 1)

6.

(ii)n∑

r=1

n3 =n2(n+ 1)2

4.

(iii) If f(x) = xm

S(D, f) =n∑

r=1

n−1(r

n)m =

1

m+ 1+Pm(n)

nm+1→ 1

m+ 1

and

s(D, f) =n−1∑

r=0

n−1(r

n)m =

1

m+ 1(1− n−1)m+1 +

Pm(n− 1)

nm+1→ 1

m+ 1

so I∗f = I∗f = (m+ 1)−1.

(iv) If f(x) = xm and F (x) = xm+1/(m+ 1) then F ′ = f and∫ b

a

f(x) dx = F (b)− F (a).

152

K140

(v) We have |g(t)− g(−1)(t+ 1)| ≤ K(t− 1) so∣

∣

∣

∣

∫ 1

−1

g(t) dt− g(−1)

∣

∣

∣

∣

≤ K

∫ 1

−1

(t− 1) dt = 2K.

Scaling, we get∣

∣

∣

∣

∫ a−rh

a−(r−1)h

f(t) dt− f(a− (r − 1)h)

∣

∣

∣

∣

≤ Kh2

if |f ′(t)| ≤ K for t ∈ [a, b] and Nh = b− a. Thus, summing,∣

∣

∣

∣

∫ b

a

f(t) dt− Sh(f)

∣

∣

∣

∣

≤ K(b− a)h.

Taking a = 0, b = π, Nh = π and F (t) = sin2(nt), we see that theresult cannot be substantially improved.

153

K141

(ii) Let s = yt.

(viii) We have

1

1− t− (1 + t+ · · ·+ tn) =

tn+1

1− t.

So, integrating both sides from 0 to x, we have

− log(1− x)−(

x+x2

2+ · · ·+ xn+1

n + 1

)

=

∫ x

0

tn+1

1− tdt→ 0

as n→ ∞ for all |x| < 1. Thus

log(1− x) = −∞∑

r=1

xr+1

r + 1

for |x| < 1. [The result also holds if x = −1 but the argument abovedoes not prove this without extra work.]

(ix) By (viii)

log

(

1 + x

1− x

)

= log(1 + x)− log(1− x)

=

∞∑

r=1

(−1)rxr+1

r + 1+

∞∑

r=1

xr+1

r + 1

= 2

∞∑

r=1

x2r+1

2r + 1

for |x| < 1.

If y > 0 and we seek an x with

y =1 + x

1− x

we obtain

x =y − 1

y + 1so |x| < 1 and substitution in the formula of (ix) gives the desiredresult.

154

K142

Power series can be multiplied term by term within their radius ofconvergence. We observe that

n−1∑

r=1

(−1)r

r× (−1)n−r

n− r=

(−1)n

n

n−1∑

r=1

(

1

r+

1

n− r

)

=(−1)n2Sn−1

n.

The result is valid for |x| < 1.

155

K143

(i) Write

λ =x3 − x2x3 − x1

and observe that 1 > λ > 0, so

λf(x1) + (1− λ)f(x3) ≥ f(λx3 + (1− λ)x1)

so that

(x3 − x2)f(x1) + (x2 − x1)f(x3) ≥ (x3 − x1)f(x2)

and the result follows on rearrangement.

Observe, either similarly, or as a consequence that, if y1 < y2 < y3,thenf(y3)− f(y1)

y3 − y1≥ f(y2)− f(y1)

y2 − y1and

f(y3)− f(y2)

y3 − y2≥ f(y3)− f(y1)

y3 − y1.

(ii) By (i), if h > 0, then

f(c+ h)− f(c)

h≥ f(c)− f(d)

c− d

where d is fixed with c > d > a. Since a non-empty set bounded belowhas an infimum, σf(c+) exists.

By definition, given ǫ > 0 we can find a δ > 0 such that

σf (c+) + ǫ ≥ f(c+ δ)− f(c)

δ.

If h > δ > 0 then (looking at the last paragraph of our discussion of (i))

σf (c+) + ǫ ≥ f(c+ δ)− f(c)

δ≥ f(c+ h)− f(c)

h≥ σf (c+)

sof(c+ h)− f(c)

h→ σf (c+)

as h→ 0+.

(iii) We can find a δ > 0 such that, if δ > h > 0, then

σf (c+) + 1 ≥ f(c+ h)− f(c)

h≥ σf (c+)

and so|f(c+ h)− f(c)| ≤ (|σf(c+)|+ 1)|h| → 0

as h→ 0+. Similarly f(c+ h)− f(c) → 0 as h→ 0− so we are done.

(iv) We have f(x) − f(c) ≥ B(x − c) for all x whenever σf (c+) ≥B ≥ σf (c−).

(v) Observe thatαg(t) ≥ tg(t) ≥ βg(t)

156

and integrate to obtain c ∈ [α, β].

We know that there exists a B such that f(x)−f(c) ≥ B(x− c) andso

(f(t)− f(c))g(t) ≥ B(t− c)g(t).

Now integrate.

(vi) We use the notation of Exercise K40 (iii). Let c = λjxj . Weknow that there exists a B such that f(x)− f(c) ≥ B(x− c) and so

f(xj)− f(c) ≥ B(xj − c)

whencen∑

j=1

λjf(xj)−f(

n∑

j=1

λjxj

)

=n∑

j=1

λj(f(xj)−f(c)) ≥n∑

j=1

λj(xj−c) = 0.

Similarly we know that there exists a B such that f(x)− f(EX) ≥B(x− EX) and so

f(X)− f(EX) ≥ B(X − EX).

Taking expectations gives the result.

(vii) Pick a′, b′ so that a < a′ < c < b′ < b Observe that,

E = (x, y) ∈ R2 : y ≥ f(x), x ∈ [a′, b′]is a closed convex set. Applying Exercise 32 (iii) with y = (c, f(c) −n−10 we see that there exist an, bn and cn such that

bnx+ any ≤ cn whenever (x, y) ∈ E

and bnc+an(f(c)−n−1) > cn. By considering what happens when y isvery large, we see that an ≤ 0 and the conditions given are incompatibleif an = 0. Thus an < 0 and, setting Bn = bn/an,Cn = cn/an, we obtain

Bnx+ y ≥ Cn whenever (x, y) ∈ E

and Bnc+ (f(c)− n−1) < Cn.

In particular we have

Bnx+ f(x) > Bnc+ (f(c)− n−1)

for all x ∈ [a′, b′]. Thus

f(x)− f(c) > Bn(x− c)− n−1

for all x ∈ [a′, b′]. If δ > 0 is such that [c− δ, c+ δ] ⊆ [a′, b′] then

f(c+ δ)− f(c) > Bnδ − n−1 and f(c)− f(c+ δ)− f(c) < Bnδ + n−1.

Thus Bn is bounded and we can find a convergent subsequence Bn(j) →B. We have

f(x)− f(c) ≥ B(x− c)

for all x ∈ [a′, b′].

157

To extend the result to (a, b), observe that the result just provedshows that that if a < a + n−1 < c < b − n−1 < b we can find a Bn

such thatf(x)− f(c) ≥ Bn(x− c)

for all x ∈ [a + n−1, b − n−1]. A subsequence argument now gives theexistence of a B with

f(x)− f(c) ≥ B(x− c)

for all x ∈ (a, b).

158

K144

(ii) Suppose that γ < β. If 0 < h, k < (β − γ)/2 then

f(β)− f(β − h)

h≥ f(β − h)− f(γ + h

β − γ − h− k≥ f(γ + k)− f(γ)

k

so

σf(β−) ≥ σf (γ+).

Thus

N∑

j=1

(

σf (cj+)− σf(cj−))

= σf (cN+)−N∑

j=1

(

σf (σf(cj−)− σf (cj−1+)))

− σf(c1−)

≤ σf(cN+)− σf (c1−)

≤ σf(α+)− σf(β−).

In particular the set

En = c ∈ (a, b) : σf (c−)− σf (c+)

contains at most [(σf (α+)− σf (β−))/N ] points (with [x] meaning theinteger part of x). Thus

E = a, b ∪∞⋃

j=1

Ej

is countable. If c /∈ E, then σf (c−) = σf (c+) = σf (c), say, and

f(c+ h)− f(c)

h→ σf (c)

as h→ 0.

(iv) Enumerate the rational points in (−1, 1) as q1, q2, . . . . If we set

fn(x) = n−2|x− qn|

then fn is convex and

0 ≤ fn(x) ≤ 2n−2

Thus, using the comparison test,∑∞

n=1 fn(x) converges everywhere tof(x), say.

159

Noting that, by convexity, fm(x+ h) + fm(x − h)− 2fm(x) ≥ 0 forall m when x, x− h, x+ h ∈ (−1, 1), we have

f(qn + h)− f(qn)

h− f(qn)− f(qn − h)

h=f(qn + h) + f(qn − h)− 2f(qn)

h

≥ fn(qn + h) + f(qn − h)− 2f(qn)

h= 2n−2 9 0

as h→ 0+, so f is not differentiable at qn

160

K145

(i) Given ǫ > 0, we can find a δ > 0 such that |f(x)− f(c)| < ǫ for|x− c| < δ. If [an, bn] ⊆ (c− δ, c+ δ), then

∣

∣

∣

∣

1

|In|

∫

In

f(t) dt− f(c)

∣

∣

∣

∣

=

∣

∣

∣

∣

1

|In|

∫

In

f(t)− f(c) dt

∣

∣

∣

∣

≤ 1

|In|

∫

In

|f(t)− f(c)| dt

≤ 1

|In|

∫

In

ǫ dt = ǫ

(ii) The following is one generalisation (interpreting the notationappropriately). Let B(c, r) be the ball centre c then (provided theintegrals exist)

1

VolB(c, r)

∫

B(c,r)

f(t) dV (t) → f(c)

as r → 0+ whenever f is continuous at c.

161

K146*

No comments.

162

K147

(i) The range of integration is not fixed.

(ii) If x is fixed the fundamental theorem of the calculus shows thatF,2 exists and

F,2(x, y) = g(x, y)

so F,2 is continuous.

Theorem 8.57 tells us that, if x is fixed we may differentiate underthe integral to obtain

F,1(x, y) =

∫ y

0

g,1(x, y) dx.

If (x0, y0) is fixed Take R ≥ 2+ |y0|+ |x0|. Since g,1 is continuous onK = [−R,R]× [−R,R] it is uniformly continuous and bounded. Thus

|F,1(x, y)− F,1(x0, y0)|≤ |F,1(x, y)− F,1(x0, y)|+ |F,1(x0, y)− F,1(x0, y0)|≤ |y| sup

|t|≤|y||g,1(x, t)− g,1(x0, t)|+ |y − y0| sup

|t|≤max(|y|,|y0|)|g,1(x0, t)|

≤ R sup(s,t),(s′,t)∈K,|s−s′|≤|x−x0|

|g,1(s, t)− g,1(s′, t)|

+ |y − y0| sup(s,t),(s′,t)∈K

|g,1(x0, t)| → 0

as ‖(x, y)− (x0, y0)‖ → 0. Thus F,2 is continuous and F is once differ-entiable.

(iii) The chain rule shows that G is differentiable and

G′(x) = F1(x, x) + F2(x, x) =

∫ x

0

g,1(x, t) dt+ g(x, x).

(iv) We have H(x) = F (x, h(x)) so the chain rule gives

H ′(x) = F1(x, h(x))+F2(x, h(x))h′(x) =

∫ h(x)

0

g,1(x, t) dt+h′(x)g(x, h(x)).

163

K148

Observe that differentiating under the integral, using the symmetryof partial derivatives and the fundamental theorem of the calculus

E ′(t) =

∫ 1

0

∂

∂t

(

(

∂u

∂t(x, t)

)2

+

(

∂u

∂x(x, t)

)2)

dx

= 2

∫ 1

0

u,2(x, t)u,22(x, t) + u,1(x, t)u,12(x, t) dx

= 2

∫ 1

0

−u,2(x, t)u,11(x, t) + u,1(x, t)u,21(x, t) dx

=

∫ 1

0

d

dx

(

(

∂u

∂t(x, t)

)2

+

(

∂u

∂x(x, t)

)2)

dx

=

[

(

∂u

∂t(x, t)

)2

+

(

∂u

∂x(x, t)

)2]1

0

= 0

so by the constant value theorem (and thus the mean value theorem)E is constant.

164

K149

Hδ is a smooth function with

Hδ(t) = 0 for t ≤ 0

Hδ(t) = cδ for t ≥ 2δ

and 0 ≤ Hδ(t) ≤ cδ for all t, where cδ is a strictly positive number.

Let Lδ(t) = c−1δ (Hδ(t)−Hδ(t + 1− 2δ). If 0 < δ < 1/4 then

Lδ(t) = 0 for t /∈ [0, 1]

Lδ(t) = 1 for t ∈ [2δ, 1− 2δ]

and 0 ≤ Lδ(t) ≤ 1 for all t.

Now let kn(x) = (−1)[2nx]Lǫ/20(2nx−2[nx]) and follow Exercise 8.66.

165

K150

We seek to minimise∫ b

a

F (x, y(x), y′(x)) dx with F (u, v, w) = g(u, v)(1 + w2)1/2.

The Euler-Lagrange equation gives

0 = F,2(x, y(x), y′(x))− d

dxF,3(x, y(x), y

′(x))

= g,2(x, y)(1 + y′2)1/2 − d

dx

g(x, y)y′

(1 + y′2)1/2

= g,2(x, y)(1 + y′2)1/2 − g,1(x, y)y

′ + g,2(x, y)y′2 + g(x, y)y′′

(1 + y′2)1/2+g(x, y)y′2y′′

(1 + y′2)3/2.

Multiplying through by 1 + y′2 gives the required result.

We seek to minimise

2π

∫ b

a

x ds.

We take g(x, y) = x and obtain

−y′ − xy′′

1 + y′2= 0

ord

dx

(

xy′

(1 + y′2)1/2

)

= 0

(observe that g(x, y) has no direct dependence on y and recall Exer-cise 8.63). Thus

xy′

(1 + y′2)1/2= c

whencey′ =

c

(x2 − c2)1/2

andy + a = c cosh−1 x

cfor appropriate constants a and c.

166

K151

Precisely as in the standard case, we consider

G(η) =

∫ b

a

f(x, y(x) + ηg(x), y′(x) + ηg′(x), y′′(x) + ηg′′(x)) dx

and require

0 = G′(0) =

∫ b

a

f,2(x, y, y′, y′′)+g′(x)f,3(x, y, y

′, y′′)+g′′(x)f,4(x, y, y′, y′′) dx

for all well behaved g with g(a) = g′(a) = 0, g(b) = g′(b) = 0.

Integrating by parts once and twice gives

0 =

∫ b

a

g(x)

(

f,2(x, y(x), y′(x), y′′(x))− d

dxf,3(x, y(x), y

′(x), y′′(x))

+d2

dx2f,4(x, y(x), y

′(x), y′′(x))

)

dx

for all g as above, so we get an Euler-Lagrange type equation

0 = f,2(x, y(x), y′(x), y′′(x))− d

dxf,3(x, y(x), y

′(x), y′′(x))

+d2

dx2f,4(x, y(x), y

′(x), y′′(x)).

In the case given this becomes

0 = −24 +d2

dx2y′′(x)

that is to sayy′′′′(x) = 24

so y = A + Bx + Cx2 + Dx3 + x4. The boundary conditions give usA = B = 0, 0 = C +D + 1, 4 = 2C + 3D + 4 so C = −3, D = 2 and

y(x) = −3x2 + 2x3 + x4.

If we consider y(x) + ǫx2(1− x)2 sinNx, we can make small changesin y increasing I so we can not have a maximum.

167

K152*

No comments.

168

K153

(i) g(1/2) = 1, g(x) = 0 otherwise.

(ii) Could take

g(x) =

1−m2n+3|x− r2−n| if |x− r2−n| < m−12−n−3, 0 ≤ r ≤ n,

0 otherwise.

169

K154

Second sentence done exactly as the case h = 1 (Lemma 9.6).

If∫∞1f(t) dt converges then, since

h

N−1∑

n=0

f(nh) ≥∫ Nh

0

f(t) dt ≥ h

N−1∑

n=1

f(nh),

we have (allowing N → ∞)

h∞∑

n=0

f(nh) ≥∫ ∞

0

f(t) dt ≥ h∞∑

n=1

f(nh),

and so∣

∣

∣

∣

∣

h

∞∑

n=1

f(nh)−∫ ∞

0

f(t) dt

∣

∣

∣

∣

∣

≤ hf(0) → 0

as h→ 0+.

Without loss of generality, suppose 1 > ǫ > 0. Choose K > 8Nǫ−1

and set

g(x) =

1−K|x− rN−1| if |x− rN−1| < K, 0 ≤ r ≤ n,

0 otherwise.

We could set

G(x) =

2−k/2(1− 24k)|x− r2−k| if |x− r2−k| < 2−4k, 2k ≤ r ≤ 2k+1 − 1, k ≥ 2

0 otherwise.

170

K155

(i) Observe that

f(x) ≤ f(n)+

∫ x

n

f ′(t) dt ≤ f(n)+

∫ x

n

|f ′(t)| dt ≤ f(n)+

∫ n+1

n

|f ′(t)| dt

for n ≤ x ≤ n + 1 and so∫ n+1

n

f(x) dx ≤ f(n) +

∫ n+1

n

|f ′(t)| dt.

Similarly∫ n+1

n

f(x) dx ≥ f(n)−∫ n+1

n

|f ′(t)| dt.

It follows that,∣

∣

∣

∣

∣

m∑

r=n

f(r)

∣

∣

∣

∣

∣

≤∣

∣

∣

∣

∫ m+1

n

f(x) dx

∣

∣

∣

∣

+

∫ m+1

n

|f ′(x)| dx

and, if∫∞1f(x) dx converges,

∑nr=0 f(r) is a Cauchy sequence and con-

verges by the general principle of convergence.

Conversely, if∑∞

r=0 f(r) converges, a similar argument shows that∫ n

1f(x) dx tends to a limit. Further, if

∑∞r=0 f(r) converges we have

f(n) → 0 and the arguments above show that∫ n+1

n

|f(x)| dx ≤ |f(n)|+∫ n+1

n

|f ′(x)| dx→ 0

as n→ ∞ and so∫∞1f(x) dx converges.

(ii) If f is decreasing, positive and has a continuous derivative∫ X

1

|f ′(x)| dx = −∫ X

1

f ′(x) dx = f(1)− f(X) ≤ f(1).

(iii) False. We consider a continuously differentiable function g suchthat

g(x) =

(2n)−12−2n if 22n + 2−1 ≤ x ≤ 22n+1 − 2−1, n ≥ 0,

−(2n + 1)−12−2n−1 if 22n+1 + 2−1 ≤ x ≤ 22n+2 − 2−1, n ≥ 0,

and, in addition, g is decreasing on [22n+1−2−1, 22n+1+2−1], increasingon [22n − 2−1, 22n + 2−1] and g(2m) = 0 for all m.

171

K156

∫ n+1/2

n−1/2

log x dx− log n =

∫ n+1/2

n

(log x− logn) dx+

∫ n

n−1/2

(log x− log n) dx

=

∫ n+1/2

n

log(x/n) dx+

∫ n

n−1/2

log(x/n) dx

=

∫ 1/2

0

(

log

(

1 +t

n

)

+ log

(

1− t

n

))

dt

making substitutions like x = t+ n and x = n− t.

But by the mean value inequality,∣

∣

∣

∣

log

(

1 +t

n

)

+ log

(

1− t

n

)∣

∣

∣

∣

≤ 2−1 supt∈(0,1/2)

∣

∣

∣

∣

1

t + n− 1

n− t

∣

∣

∣

∣

= 2−1 supt∈(0,1/2)

2t

n2 − t2≤ 4

3n2.

Thus by the comparison test∞∑

n=1

(

log

(

1 +t

n

)

+ log

(

1− t

n

))

is absolutely convergent and so convergent to c, say. Thus∫ N+1/2

1/2

log x dx− logN ! → c.

But, integrating by parts,∫ N+1/2

1/2

log x dx = [x log x]N+1/21/2 −

∫ N+1/2

1/2

dx

= (N + 12) log(N + 1

2)− 1

2) log 1

2− (N + 1).

Thus(N + 1

2) log(N + 1

2)−N − logN ! → c′

for some constant C and the result follows on taking exponentials (andnoting that exp is continuous).

172

K157

With the notation of the previous question

2n!

(2n+ 1/2)(2n+1/2)e−2n

/(

n!

(n + 1/2)(n+1/2)e−n

)2

→ C−1

where C is the constant of K156. Thus(

2n

n

)

(1 + 14n)2n

(1 + 12n)2n

22nn1/2 → C ′

and

4nn1/2

(

2n

n

)

→ C ′′

as n → ∞ for appropriate constants C ′ and C ′′. In particular, we canfind constants K1 and K2 with K1 > K2 > 0 such that

K14−nn−1/2 ≥

(

2n

n

)

≥ 4−nn−1/2.

Thus∣

∣

∣

∣

nα

(

2n

n

)

zn∣

∣

∣

∣

→ 0 if |z| < 1/4 and

∣

∣

∣

∣

nα

(

2n

n

)

zn∣

∣

∣

∣

→ 0 if |z| > 1/4

so∑∞

n=0 nα(

2nn

)

zn has radius of convergence 1/4.

If α < −1/2 we have |nα(

2nn

)

zn| ≤ K1nα−1/2 for all |z| = 1/4 so,

by the comparison test,∑∞

n=0 nα(

2nn

)

zn converges absolutely and soconverges everywhere on the circle of convergence.

If α ≥ 1/2 we have |nα(

2nn

)

zn| ≥ K2 for all |z| = 1/4 so∑∞

n=0 nα(

2nn

)

zn

diverges everywhere on the circle of convergence.

If α ≥ −1/2 then nα(

2nn

)

≥ K2nα−1/2 and the comparison test tells

us that∑∞

n=0 nα(

2nn

)

zn diverges when z = 1/4.

The only remaining case is |z| = 1/4, z 6= 1/4 and 1/2 > α ≥ −1/2.Let us write w = 4z and un = 4−nnα

(

2nn

)

. We know that un → 0 so, ifwe can show that un > un+1 for all sufficiently large n, Abel’s test willtell us that

∑∞n=0 unw

n. But

un+1

un=

(1 + 12n)

(1 + 1n)1−α

< 1

so the conditions of Abel’s test are satisfied. (Observe that (1 + x)β ≥1 + βx for β ≥ 0 and x ≥ 0.) If 1/2 > α ≥ −1/2 then

∑∞n=0 n

α(

2nn

)

zn

converges everywhere on the circle of convergence except the pointz = 1/4.

173

K158

(i) Observe that∫ g(X)

0

f(s) ds =

∫ X

0

f(x)g′(x) dx

and that g(X) → ∞ as X → ∞ and, if g(h(Y )) → ∞ as Y → ∞, thenh(Y ) → ∞ as Y → ∞.

Only one limit so no problem of interchange.

(ii) (sin x)/x → 1 as x → 0 (by considering differentiation, if youwish).

Observe that

0 ≤ (−1)n+2 sin(x+ (n + 1)π)

x+ (n + 1)π≤ (−1)n+1 sin(x+ nπ)

x+ nπ

so

0 ≤ (−1)n+2

∫ π

0

sin(x+ (n+ 1)π)

x+ (n+ 1)πdx ≤ (−1)n+1

∫ π

0

sin(x+ nπ)

x+ nπdx

and, if we write

vn = (−1)n+1

∫ (n+1)π

nπ

sin x

xdx,

the vN are a decreasing sequence with

0 ≤ vn ≤∫ (n+1)π

nπ

1

xdx ≤ (nπ)−1 → 0

as n→ ∞. Thus, by the alternating series test,

∫ nπ

0

sin x

xdx =

n−1∑

r=0

(−1)rvr

converges to a limit L with v0 ≥ L ≥ v0 − v1 > 0.

Since∫ (n+1)π

nπ

∣

∣

∣

∣

sin x

x

∣

∣

∣

∣

dx ≤ (nπ)−1 → 0,

we have∫ X

0

sin x

xdx→ L

as X → ∞.

(iii) Change of variable s = xt with t > 0 gives I(t) = I(1) = L.Since (sin−x)/x = −(sin x)/x, I(−t) = −I(t).

174

(vi) By convexity sin x ≤ 2x/π for 0 ≤ x ≤ π/2, so g(x) = g(π−x) =2x/π2 for 0 ≤ x ≤ π/2 will do. Observe that

∫ nπ

π

∣

∣

∣

∣

sin x

x

∣

∣

∣

∣

dx =n−1∑

r=1

∫ (r+1)π

rπ

∣

∣

∣

∣

sin x

x

∣

∣

∣

∣

dx

≥n−1∑

r=1

∫ (r+1)π

rπ

x− rπ

rdx

=

∫ π

0

g(x) dx

n−1∑

r=1

1

r→ ∞

as n→ ∞.

(v) If α > 0, then the arguments above show that∫∞1

sinxxα dx con-

verges. If α ≤ 0, then∣

∣

∣

∣

∣

∫ (n+1)π

nπ

sin x

xαdx

∣

∣

∣

∣

∣

≥∫ (n+1)π

nπ

| sin x| dx9 0

so∫∞1

sinxxα dx does not converge.

However, if 0 ≤ x ≤ 1 we have x ≥ sin x ≥ 2x/π so

x1−α ≥ sin x

xα≥ 2x1−α/π

and comparison shows that∫ 1

ǫ

sin x

xαdx

converges as ǫ → 0+ if and only if α < 2.

Thus∫∞0

sinxxα dx converges if and only if 2 > α > 0.

175

K159

(i) Observe that

1 + 2n∑

r=1

cos rx =n∑

r=−n

eirx = e−inx1− ei(2n+1)x

1− eix=

sin((n + 12)x)

x

for |x| ≤ π.

(ii) Change of variable t = λx.

(iii) Observe that the integrals∫ (r+1)π/(n+1/2)

rπ/(n+1/2)

sin(

(n+ 12)x)

sin 12x

dx

alternate in sign and decease in absolute value as r increases from1 to n and as r decreases from −! to −n. Observe further that ifǫ ≤ α ≤ β ≤ π and α− β ≤ π/(n+ 1/2) then∣

∣

∣

∣

∣

∫ β

α

sin(

(n+ 12)x)

sin 12x

dx

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∫ −alpha

−β

sin(

(n+ 12)x)

sin 12x

dx

∣

∣

∣

∣

∣

≤ π

(n+ frac12)| sin ǫ| → 0

as n→ ∞.

(iv) Write2

x− 1

sin 12x=

2(sin 12x− 1

2x)

x sin 12x

and use L’Hopital or Taylor expansions.

(v) Given any η > 0 we can find an ǫ > 0 with η > ǫ such that∣

∣

∣

∣

2

x− 1

sin 12x

∣

∣

∣

∣

< η

for |x| < ǫ and so∣

∣

∣

∣

∫ ǫ

−ǫ

sin((n+ 12)x)

sin x2

dx− 2 sin((n+ 12)x)

xdx

∣

∣

∣

∣

≤ 2ǫη ≤ 2η2.

Thus, allowing n→ ∞ and using (ii) and (iii) we have∣

∣

∣

∣

2π − 2

∫ ∞

−∞

sin x

x, dx

∣

∣

∣

∣

≤ 2η2.

Since η was arbitrary∫ ∞

−∞

sin x

x, dx = π

and we are done.

176

K160

(i) and (ii) (Here ‘permits’ means ‘permits but does not imply’.)

f(n) = O(g(n)), permits f(n) = o(g(n)), permits f(n) = Ω(g(n))and permits f(n) ∼ g(n).

f(n) = o(g(n)), implies f(n) = O(g(n)), forbids f(n) = Ω(g(n)) andforbids f(n) ∼ g(n).

f(n) = Ω(g(n)) permits f(n) = O(g(n)), permits f(n) ∼ g(n) andforbids f(n) = o(g(n)).

f(n) ∼ g(n) implies f(n) = O(g(n)), implies f(n) = Ω(g(n)) andforbids f(n) = o(g(n)).

Suitable examples will be found amongst the pairs f(n) = g(n) = 1;f(n) = n, g(n) = 1; f(n) = 1, g(n) = n.

(iii) f(n) = o(1) means f(n) → 0 as n→ 0. f(n) = O(1) means f isbounded.

(iv) Could take f(n) = 1+n(1+ (−1)n), g(n) = 1+n(1+ (−1)n+1).

177

K161

(i) True. If 0 ≤ fj(n) ≤ Ajgj(n) with Aj ≥ 0 then

0 ≤ f1(n) + f2 ≤ (A1 + A2)(g1(n) + g2(n)).

(ii) False. Take f1(n) = f2(n) = g1(n) = −g2(n).(iii) True. If 0 ≤ fj(n) ≤ Ajgj(n) with Aj ≥ 0 then

0 ≤ f1(n) + f2 ≤ (A1 + A2)max(g1(n), g2(n)).

(iv) True. n! ≤ nn ≤ (2n)n = 2n2

.

(v) True. Use L’Hopital or (I think better) observe that

cos x− 1 + x2/2 = O(x4), sin x = x+ α(x)x3,

with |α(x)| < 1 when |x| is small so

(sin x)2 = x2 + 2α(x)x4 + α(x)2x5

and(sin x)2 − x2 = O(x4).

(vi) False. Use L’Hopital or (I think better) observe that

cosx− 1 + x2/2− x4/4! = O(x6), sin x = x− x3/6 + β(x)x5,

with |γ(x)| < 1 when |x| is small so

sin2 x = x2 − x4/3 + γ1(x)x6, and sin4 x = x4 + γ2(x)x

6

where |γ1(x)|, |γ2(x)| ≤ 10 for |x| small (we can do better but we donot need to) so

cosx− 1 + sin2 x/2− sin4 x/4! + x4/6 = O(x6).

178

K162

(i) If −1 < α ≤ 0, then xα is decreasing so

n∑

r=1

rα ≥n∑

r=1

∫ r+1

r

xα dx =

∫ n+1

1

xα dx ≥n+1∑

r=2

rα.

Since∫ n

1

xα dx =nα+1 − 1

α + 1,

this gives∣

∣

∣

∣

(α + 1)∑n

r=1 rα

nα+1

∣

∣

∣

∣

≤ 1 + 2nα

nα+1→ 0

as n→ ∞.

If α ≥ 0, then xα is increasing son∑

r=1

rα ≤n∑

r=1

∫ r+1

r

xα dx =

∫ n+1

1

xα dx ≤n+1∑

r=2

rα

and much the same argument works.

(ii) If α = −1, then much the same argument givesn∑

r=1

1

r∼ log n.

(iv) No such modification, since∑∞

r=n r−1 diverges.

179

K163

Argument similar to that in K156. We have |g′′(x)| ≤ Ax−λ forx ≥ R say. If |t| ≤ 1

2and n ≥ R + 1, then applying the mean value

theorem twice gives

|(g(t+ n+ 12)− g(n+ 1

2)) + (g(−t+ n+ 1

2)− g(n+ 1

2))|

= |g′(s+ n+ 12)− g′(−s + n+ 1

2)|

= |2sg′(u+ n+ 12)| ≤ A(n− 1

2)−λ

for some s and u with 12> s > 0 and |u| < s. Thus we can find a B

such that

|(g(t+ n + 12)− g(n+ 1

2)) + (g(−t+ n + 1

2)− g(n+ 1

2))| ≤ Bn−λ

for n ≥ R and |t| ≤ 12.

We have∣

∣

∣

∣

∫ n+1

n

(g(x)− g(n+ 12) dx

∣

∣

=

∣

∣

∣

∣

∣

∣

∫

12

0

(g(t+ n+ 12)− g(n+ 1

2)) + (g(−t+ n+ 1

2)− g(n+ 1

2)) dt

∣

∣

∣

∣

∣

∣

≤∫

12

0

|(g(t+ n + 12)− g(n+ 1

2)) + (g(−t+ n+ 1

2)− g(n+ 1

2))| dt

≤ Bn−λ

so∫ n+1

n

(g(x)− g(n+ 12)dx = O(n−λ).

We have∣

∣

∣

∣

∫ n+1

n

g(x) dx− g(n+ 12)

∣

∣

∣

∣

≤ Bn−λ

for n ≥ R so adding and using K162 or direct argument∣

∣

∣

∣

∣

∫ n+1

1

g(x) dx−n∑

r=1

g(r)

∣

∣

∣

∣

∣

≤ C +Bn−λ+1

−λ + 1

for some constant C. Dividing through by∫ n+1

1g(x) dx gives the re-

quired result.

180

K164

Since sin x > 2x/π for 0 ≤ x ≤ π/2, we have

0 ≤ − log sin x ≤ log(π/2)− log x ≤ log(π/2) + x−1/2

when x is small and positive, so, by comparison,∫ π/2

0log(sin x) dx con-

verges. (Similar arguments apply to all the integrals of this question.)

Using the change of variables formula with t = 2x and symmetry wehave

∫ π/2

0

log(sin 2x) dx =1

2

∫ π

0

log(sin t) dt =

∫ π/2

0

log(sin x) dx.

But, using symmetry again,∫ π/2

0

log(sin 2x) dx =

∫ π/2

0

log(2 sin x cosx) dx

=

∫ π/2

0

log 2 dx+

∫ π/2

0

log(sin x) dx+

∫ π/2

0

log(cosx) dx

=π

2log 2 + 2

∫ π/2

0

log(sin x) dx.

Combining our formulae gives the result.

181

K165

Suppose that∫ a

0f(x) dx converges. Since f is decreasing

∫ a

0

f(x) dx ≥ a

N

N∑

r=1

f(ar

N

)

≥∫ a

a/N

f(x) dx→∫ a

0

f(x) dx

soN∑

r=1

f(ar

N

)

→∫ a

0

f(x) dx.

Suppose that∫ a

0f(x) dx diverges. Then given any K we can find an

ǫ > 0 such that∫ a

ǫ

f(x) dx ≥ K + 1.

However, since f is continuous

∑

ǫN≤r≤N

f(ar

N

)

→∫ a

ǫ

f(x) dx

so we can find an N0 such that∑

ǫN≤r≤N

f(ar

N

)

→∫ a

ǫ

f(x) dx− 1

and soN∑

r=1

f(ar

N

)

≥ K

for all N ≥ N0. ThusN∑

r=1

f(ar

N

)

→ ∞

as N → ∞.

(ii) Consider a = 1,

f(x)

2n(1− 24n|x− 2−n|) if |x− 2−n| ≤ 2−4n,

0 otherwise.

(iii) Observe that

(1− z)

N−1∑

r=0

zn = 1− zN =

N−1∏

r=0

(z − ωr)

so, dividing by z − 1,

N−1∑

r=0

zn = 1− zN =N−1∏

r=1

(z − ωr)

182

for z 6= 1. Since a polynomial of degree N − 1 which vanishes at Npoints vanishes identically we have

N−1∑

r=0

zn = 1− zN =N−1∏

r=1

(z − ωr)

for all z.

In particular, setting z = 1, we have

N =N−1∏

r=1

(1− ωr) = τ 1+2+···+(N−1)N−1∏

r=1

(τ−r − τ r)

where τ = π/N . Thus

N = τN(N−1)/2(−2i)N−1N−1∏

r=1

sinrπ

N= 2N−1

N−1∏

r=1

sinrπ

N.

(iv) Observe that

π

2N

N∑

r=1

log(

sin(πr

N

))

=π

2Nlog

(

N−1∑

r=1

sin(πr

N

)

)

=π

2Nlog

N

2N−1

= −πN − 1

2Nlog 2 + π

logN

2N→ −π

2log 2.

183

K166

Observe thatEk = x : f(x) = k

is the union of a finite set of disjoint intervals (of the form [a, b)) oftotal length (9/10)k−1/10. Thus (writing IA for the indicator functionof A)

fX =∑

1≤k≤X

kIEk+X

(

1−∑

1≤k≤X

IEk

)

is Riemann integrable with∫ 1

0

fX(x) dx = 10−1∑

1≤k≤X

k(9/10)k−1 +X

(

1− 9−1∑

1≤k≤X

(9/10)k−1

)

.

Now writing [X ] for the integer part of X ,

0 ≤ X

(

1− 10−1∑

1≤k≤X

(9/10)k−1

)

= X(9/10)[X] ≤ ([X ]+1)(9/10)[X] → 0

and using the Taylor expansion

(1− t)−2 =∞∑

j=0

(j + 1)tj

(Newton’s binomial expansion) we have∫ 1

0

fX(x) dx→ 10−1(1− 9/10)−2 = 10.

It makes no difference if we replace f by g with g(x) = f(x) exceptat points where f(x) = ∞. However we have to work a little harder toshow that gX is Riemann integrable. One way is the observe that

gX(x)− fX(x) = 0

for all x ∈ ⋃Mk=1Ek so, given any ǫ > 0, we know that gX(x)−fX(x) = 0

on the union of a finite set of disjoint intervals of total length 1 − ǫ/2and so

IfX −Xǫ = I∗fX −Xǫ ≤ I∗gX ≤ I∗gX ≤≤ I∗fX +Xǫ = IfX +Xǫ.

Since ǫ is arbitrary gX is integrable and IgX = IfX .

184

K167

(ii) Observe that if η > 0 is fixed, we can find an R0(η) such that∫ b

a

fR,S(x) dx ≥∫

ab−ηf(x) dx

for all R, S ≥ R0(η), and if R and S are fixed, we can find an η0(R, S) >0 such that

∫

ab−ηf(x) dx ≥∫ b

a

fR,S(x) dx

for all 0 < η < η0(R, S).

(iii) If (A) is true, the argument of (i) shows that, if we write f+(x) =max(f(x), 0) and f−(x) = min(f(x), 0), then (A) remains true for bothf+ and f−. There is this no loss of generality in assuming that f ispositive. The argument of (ii) now applies.

(iv) The substitution t = x−1 suggests

f(t) =sin t−1

t.

185

K168

(Note that, since (x, t) 7→ e−tx is continuous on [0, X ] × [a, b], theapparent singularity at x = 0 must be a trick of the notation.)

∫ X

0

e−ax − e−bx

xdx =

∫ X

0

∫ b

a

e−tx dt dx

=

∫ b

a

∫ X

0

e−tx dx dt

=

∫ b

a

1− e−tX

tdt.

Now

0 ≤ e−tX

t≤ e−aX

tfor t ∈ [a, b] so

0 ≤∫ b

a

e−tX

tdt ≤ e−aX log

(

b

a

)

→ 0

as X → 0 so∫ X

0

e−ax − e−bx

xdx =

∫ b

a

1

tdt−

∫ b

a

e−tX

tdt→ log

b

aas x→ ∞.

186

K169

By the fundamental theorem of the calculus

d

dt

∫ t

a

(∫ d

c

f(u, v) dv

)

du =

∫ d

c

f(t, v) dv.

By differentiating under the integral sign and using the fundamentaltheorem of the calculus

d

dt

∫ d

c

(∫ t

a

f(u, v) du

)

dv =

∫ d

c

∂

∂t

(∫ t

a

f(u, v) du

)

dv =

∫ d

c

f(t, v) dv.

Thus

d

dt

(∫ t

a

(∫ d

c

f(u, v) dv

)

du−∫ d

c

(∫ t

a

f(u, v) du

)

dv

)

= 0,

and by the constant value theorem∫ t

a

(∫ d

c

f(u, v) dv) du−∫ d

c

(∫ t

a

f(u, v) du

)

dv

=

∫ a

a

(∫ d

c

f(u, v) dv

)

du−∫ d

c

(∫ a

a

f(u, v) du

)

dv = 0.

It is weaker because we say nothing about the existence or value of∫

R

f(x) dA.

187

K170

(i) We have∫ 1

0

f(x, y) dx = 0

for all x and∫ 1

0

f(x, y) dy =

2 if 2−1 < y < 1,

0 otherwise,

so∫ 1

0

∫ 1

0

f(x, y) dx dy = 0 6= 1 =

∫ 1

0

∫ 1

0

f(x, y) dy dx.

(ii) We have∫ 1

0

g(x, y) dx = 0

for all x and∫ 1

0

g(x, y) dy =

2u(2y − 1) if 2−1 < y < 1,

0 otherwise,

so∫ 1

0

∫ 1

0

g(x, y) dx dy = 0 6= 1 =

∫ 1

0

∫ 1

0

g(x, y) dy dx.

Theorem 9.20 demands continuity everywhere. (The underlying prob-lem is that g is not bounded.)

(iii) We have∫ 1

0

h(x, y) dx =1

2

∫ 1+y2

y2

y(w − 2y2)

w3dw

=1

2

[

− y

w+y3

w2

]1+y2

y2

=−y

2(1 + y2)2

so that∫ 1

0

∫ 1

0

h(x, y) dx dy =

[

1

4(1 + y2)

]1

0

= −1/8.

Observe that h(y, x) = −h(x, y).

188

K171

(ii) Observe that, if R ≥ 1, then∣

∣

∣

∣

∫ ∞

0

f(x, y) dx−∫ R

0

f(x, y) dx

∣

∣

∣

∣

≤∫ ∞

R

A

(1 + x2)(1 + y2)dx⋆

= Aπ2− tan−1R

1 + y2

≤ A(π2− tan−1R)

Thus given ǫ > 0 we can find an R such that∣

∣

∣

∣

∫ ∞

0

f(x, y) dx−∫ R

0

f(x, y) dx

∣

∣

∣

∣

≤ ǫ/4

for all y. Since∫ R

0f(x, y) dx is a continuous function of y it follows

that given any y0 we can find a δ > 0 such that |y0 − y| < δ implies∣

∣

∣

∣

∫ R

0

f(x, y) dx−∫ R

0

f(x, y0) dx

∣

∣

∣

∣

≤ ǫ/2

and so∣

∣

∣

∣

∫ ∞

0

f(x, y) dx−∫ ∞

0

f(x, y0) dx

∣

∣

∣

∣

< ǫ.

If follows that∫∞0f(x, y) dx is a continuous function of y. (This can

be done better using uniform convergence.)

Since∫∞0f(x, y) dx is continuous in y and

∣

∣

∣

∣

∫ ∞

0

f(x, y) dx

∣

∣

∣

∣

≤ Aπ/2

1 + y2,

comparison shows that∫∞0

∫∞0f(x, y) dx dy exists. A similar but sim-

pler argument shows that∫∞0

∫ R

0f(x, y) dx exists Equation ⋆ shows

that∣

∣

∣

∣

∫ ∞

0

∫ ∞

0

f(x, y) dx dy −∫ ∞

0

∫ R

0

f(x, y) dx dy

∣

∣

∣

∣

≤ (π2− tan−1R)

πA

2.

We also know that∣

∣

∣

∣

∫ ∞

0

∫ R

0

f(x, y) dx dy −∫ R

0

∫ R

0

f(x, y) dx dy

∣

∣

∣

∣

≤ (π2− tan−1R)

πA

2

so that∣

∣

∣

∣

∫ ∞

0

∫ ∞

0

f(x, y) dx dy −∫ R

0

∫ R

0

f(x, y) dx dy

∣

∣

∣

∣

≤ (π2− tan−1R)πA.

Similarly∣

∣

∣

∣

∫ ∞

0

∫ ∞

0

f(x, y) dy dx−∫ R

0

∫ R

0

f(x, y) dy dx

∣

∣

∣

∣

≤ (π2− tan−1R)πA,

189

so, using Theorem 9.20 (Fubini on products of intervals),∣

∣

∣

∣

∫ ∞

0

∫ ∞

0

f(x, y) dy dx−∫ ∞

0

∫ ∞

0

f(x, y) dx dy

∣

∣

∣

∣

≤ 2(π2− tan−1R)πA.

Allowing R → ∞ gives the result.

190

K172

Just use the rectangles [0, 1]× [0, N−1] and [0, 1]× [N−1, 1].

F2(0) is not defined because x 7→ f(x, 0) is not Riemann integrable.

191

K173*

No comments

192

K174

(iv) Observe that U is open, [−2, 2] ⊇ U and U does not have Rie-mann length so [−3, 3] \ U is closed, bounded and does not have Rie-mann length.

193

K175*

No comments.

194

K176

(i) Observe that f(tj) ≤ f(b). Recall that an increasing sequencebounded above tends to limit.

(ii) Chose n(j) and xj so that n(1) = 1, x1 = tn(1), n(2j) > n(2j−1),x2j = sn(2j) > x2j−1, n(2j + 1) > n(2j), x2j+1 = tn(2j)+1 > x2j [j ≥ 1].Consider the limit of f(xj).

Let a = 0, b = 2, f(t) = 0 for t < 1, f(t) = 1 for t ≥ 1 tj = 1− j−1,sj = 1.

(vii) If x /∈ E ∪ a, then f(x−) = f(x+). But f(x−) ≤ f(x+) so

f(x) = f(x+). Thus f = f .

If x < y then, choosing x < xn < y < yn < b with xn, yn strictlydecreasing with xn → x, yn → y, we have f(xn) ≤ f(yn) so f(x) ≤f(y). A similar but simpler argument applies when y = b. Thus

f is increasing. If x ∈ [a, b) we can find xn → x with xn strictly

decreasing and xn /∈ E ∪ a. Thus f(xn) = f(xn) → f(x) and f isright continuous.

(ix) Observe that g(

(nπ)−1)

= 0 → 0 and g(

((2n+ 12)π)−1

)

= 1 → 1.

195

K177

(iii) The function f is Riemann integrable with respect to H if andonly if f(t) → f(0) as t→ 0−.

Observe that, if D is a dissection, then

SH(f,D)− sH(f,D) ≥ SH(f,D ∪ 0)− sH(f,D ∪ 0)= lim sup

t→0, t<0( supx∈(t,0]

f(x)− infx∈(t,0]

f(x))

so if I∗(f,H) = I∗(f,H) then f(t) → f(0) as t→ 0−.

Conversely if f(t) → f(0) as t→ 0−, then, taking DN = −N−1, 0we have

SH(f,DN), sH(f,DN) → f(0).

196

K178

(i) By adding a constant, we may suppose that u(t) → 0 as t→ −∞We know that u is continuous except at a countable set of points xj andand that, writing λj = u(xj)− supt<xj

u(t),∑∞

j=1 λj converges (indeed

∞∑

j=1

λj ≤ limt→∞

f(t).

Without loss of generality we may suppose λ1 ≥ λ2 ≥ . . . . Defineun(t) so that

u(t) = un(t) +

n∑

j=1

H(x− xj).

We observe that un(t) ≥ un+1(t) ≥ 0 so un(t) → U(t) for some U :R → R. Since un is increasing for each n, U is increasing. We observethat

0 ≤ infs<x<t

un(t)− un(s) ≤ λn

and

0 ≤ un+1(t)− un+1(s) ≤ un(t)− un(s)

for any fixed t and s with t > s, so

infs<x<t

U(t)− U(s) = 0

and U is continuous.

A very much simpler argument shows that∑n

j=1H(x−xj) convergesfor all x.

(ii) Suppose that f and g are bounded increasing left continuousfunctions with 0 ≤ f(t) − g(t) ≤ ǫ for all t and g is Riemann-Stieljesintegrable with respect to G. Then

I∗(f,G)− I∗(f,G) ≤ ǫ( limt→∞

G(t)− limt→−∞

G(t)).

By (i), we can write

f(t) = v(t) +

∞∑

j=1

λjH(t− xj)

where H(t) = 0 for t ≤ 0, H(t) = 1 for t > 0, λj > 0 and∑∞

j=1 λjconverges. By the first paragraph we need only show that the functionfN given by

fN(t) = v(t) +

N∑

j=1

λjH(t− xj)

197

is Riemann-Stieljes integrable. We know that v is Riemann-Stieljesintegrable so we need only show that t 7→ H(t − xj) is a Riemann-Stieljes integrable function and this follows much the same pattern asin K177.

(iii) f will be Riemann-Stieljes integrable if and only if no disconti-nuity of f coincides with that of G.

If f and G share a discontinuity x0 then

S(f,D)− s(f,D) ≥ S(f,D ∪ x0)− s(f,D ∪ x0)≥ inf

s<x0<t(G(t)−G(s)) inf

s<x0<t(f(t)− f(s)) > 0

so f is not Riemann-Stieljes integrable with respect to G.

If f and G have no points of discontinuity in common, the argumentof (ii) will work if appropriately modified.

198

K179

(ii) The condition is ‘G strictly increasing’.

If G is not strictly increasing, choose a < b such that G(a) = G(b).Take f(x) = max(0, 1−4(b−a)−1|x−(a+b)/2| to see that the theoremfails.

Suppose G is strictly increasing. If f is bounded continuous positivefunction with f(x0) 6= 0, observe that there exists an ǫ > 0 such that|f(t)− f(x0)| ≤ f(x0)/2 so f(t) ≥ f(x0/2 for |x0 − t| ≤ 2ǫ. Now

∫

R

f(x) dG(x) ≥ (G(x0 + ǫ)−G(x0 − ǫ))f(x0)

2> 0.

(iii) Same condition as (ii).

Suppose G is strictly increasing. If f(x0) 6= 0 there exists an ǫ > 0such that |f(t)−f(x0)| ≤ |f(x0)|/2 for |x0−t| ≤ ǫ. Choose g continuousso that g vanishes outside (x0 − ǫ, x0 + ǫ), g(t)f(x0) ≥ 0 for all t andg(X0)f(x0) = 1. By part (ii), g(t)f(t) = 0 for all t which is impossible.Thus f is identically zero.

200

K181

(ii) Given ǫ > 0, we can find a = x0 ≤ y0 ≤ x1 ≤ . . . xn ≤ yn = tsuch that

n∑

j=1

F (yj)− F (xj) ≥ F+(t)− ǫ

and a = x′0 ≤ y′0 ≤ x′1 ≤ . . . xm ≤ y′m = t such that

n∑

j=1

F (x′j)− F (y′j) ≥ F−(t)− ǫ.

Let a = z0 ≤ z1 ≤ . . . zN = t with

z0, z1, z2 . . . zN =x0, x1, x2 . . . xn ∪ y0, y1, y2 . . . yn∪ x′0, x′1, x′2 . . . x′m ∪ y′0, y′1, y′2 . . . y′m

Then

F (t)− F (a) =N∑

j=1

F (zj)− F (zj−1)

=∑

F (zj)−F (zj−1)>≥0

F (zj)− F (zj−1) +∑

F (zj)−F (zj−1)<0

F (zj)− F (zj−1),

F+(t) ≥∑

F (zj)−F (zj−1)>≥0

F (zj)− F (zj−1) ≥ F+(t)− ǫ

and

F−(t) ≥ −∑

F (zj)−F (zj−1)>≤0

F (zj)− F (zj−1) ≥ F−(t)− ǫ

so |F (t)−F+(t)+F+(t)| ≤ 2ǫ. Since ǫ is arbitrary, F (t) = F+(t)−F−(t).

The proof that VF = F+(t) + F−(t) is similar.

(iii) Observe that, if a = x0 ≤ y0 ≤ x1 ≤ . . . xn ≤ yn = t then

G+(t) =

n∑

j=1

G+(yj)−G+(xj) +

n∑

j=1

G+(xj)−G+(yj−1)

≥n∑

j=1

G+(yj)−G+(xj)

=

n∑

j=1

(f(yj)− f(xj)) + (G−(yj)−G−(xj))

≥n∑

j=1

f(yj)− f(xj)

201

so G+(t) ≥ F+(t). Exactly the same calculation gives G+(t)−G−(s) ≥F+(t)− F−(s) for t ≥ s so G+ − F+ is increasing.

202

K182

(ii) If F is right continuous at t then, since F = F (a) + F+ − F−,either both F+ and F− are right continuous or neither is (which isimpossible by (i)).

(iii) Write

G+(t) =

∫ t

a

max(F ′(x), 0) dx, G−(t) = −∫ t

a

min(F ′(x), 0) dx.

Then, by the fundamental theorem of the calculus, G+ is differentiablewith G′

+(t) = max(F ′(t), 0) so G+ is increasing. Similarly G− is de-creasing. Further G′

+(t) − G′−(t) = F ′(t), so using the fundamental

theorem of the calculus, F = F (a) +G+ −G−.

Now let ǫ > 0. Since F ′ is continuous on [a, t], it is uniformly con-tinuous so we can find an integer N such that

|F ′(x)− F ′(y)| < ǫ whenever |x− y| ≤ (t− a)/N.

Let us an integer r with 1 ≤ r ≤ n ‘good’ if there exists an x ∈[(r − 1)/N, rN ] with |F ′(x)| > ǫ. Let us call the other integers r with1 ≤ r ≤ n ‘bad’. Observe that, if r is good, then F ′ is single signed onx ∈ [(r − 1)/N, rN ]. Thus, using the mean value theorem

N∑

j=1

|F (xj)− F (xj−1)| ≥∑

j good

∫ xj

xj−1

|F ′(x)| dx

≥∫ t

a

|F ′(x)| dx−∑

j bad

ǫ(t− a)/N

≥∫ t

a

|F ′(x)| dx− ǫ(t− a)

so that

VF (t) ≥∫ t

a

|F ′(x)| dx = G+(t) +G−(t)

and F+(t) = G+(t), F−(t) = G−(t) as stated.

203

K183

Observe that fα,β is well behaved away from 0, so this exercise isabout behaviour near 0. The behaviour is different according as β > 0,β = 0 or β < 0.

(A) If β > 0, we have

|x|α sin(|x|β) = |x|α+β + |x|α+2βǫ(|x|)with ǫ(|x|) → 0 as x→ 0.

If α + β > 1, fα,β is differentiable at 0 with derivative 0. We checkthat f ′

α,β is continuous at 0 so fα,β is everywhere differentiable withcontinuous derivative (so of bounded variation).

If 1 ≥ α + β > 0, fα,β is not differentiable at 0 but is continuousthere (so everywhere continuous). By inspection of f ′

α,β, or otherwise,fα,β is of bounded variation.

If α+ β = 0, fα,β is not continuous at 0 but is of bounded variation.

If 0 > α + β then fα,β(x) → ∞ so fα,β is not continuous and not ofbounded variation.

(B) If β = 0. fα,0 = |x|α sin 1. The function is differentiable every-where with continuous derivative if and only if α > 1, and continuouseverywhere if and only if α > 0, of bounded variation if and only ifα ≥ 0.

(C) If β < 0 then the function differentiable at 0 so everywhere ifand only if α > 1. Derivative is then 0. Now

f ′α,β(x) = αxα−1 sin xβ + βxα+β−1 cosxβ

so derivative is continuous if and only if (consider x = (nπ/2)−1) α +β − 1 > 0.

The function is continuous at 0 (and so everywhere) if and only ifα > 0.

If α ≤ 0, inspection shows that fα,β is not of bounded variation. Ifα > 0 then by differentiating (or rescaling) we see that in each intervalIn = [((n+1)π)1/β, (nπ)1/β] the function fα,β either increases and thendecreases or decreases and then increases. The total variation Vn offα,β in the interval In is thus 2 supt∈In |fα,β(t)|. It follows that

2(nπ)α/β ≥ Vn ≥ 2((n+ 1)π)α/β

so fα,β is of bounded variation if and only if∑∞

n=1 nα/β converges, i.e.

α/β < −1, i.e. α > −β, i.e. α+ β > 0.

204

K184*

No comments.

205

K185

(i) Observe that, in a self explanatory notation,

S(f,D, G+ F ) = S(f,D, F ) + S(f,D, G)and

S(f,D1 ∪ D2, G+ F ) = S(f,D1 ∪ D2, F ) + S(f,D1 ∪ D2, G)

≥ S(f,D1, F ) + S(f,D2, G)

soI∗(f, F +G) = I∗(f, F ) + I∗(f,G)

and similarly I∗(f, F +G) = I∗(f, F ) + I∗(f,G).

(ii) Rewrite as∫

R

f(x) dF1(x) +

∫

R

f(x) dG2(x) =

∫

R

f(x) dF2(x) +

∫

R

f(x) dG1(x).

and observe that by part (i)∫

R

f(x) dF1(x) +

∫

R

f(x) dG2(x) =

∫

R

f(x) d(F1 +G2)(x)

and∫

R

f(x) dF2(x) +

∫

R

f(x) dG1(x) =

∫

R

f(x) d(F2 +G1)(x)

206

K186*

No comments.

207

K187

The main problem is the behaviour of θ(y) for y close to 1.

Observe that writing t = 1 − s we have, using appropriate Taylortheorems,

(1− (1− s)2)−1/2 = (2s− s)−1/2

= 2−1/2s−1/2(1− s/2)−1/2

2−1/2s−1/2(1− ǫ(s)s)

with ǫ(s) → 0 as ǫ→ 0. Thus∫ 1

y1

(1−t2)1/2dt converges and

∫ 1

y

1

(1− t2)1/2dt = −

∫ 1−y

0

1

(1− (1− s)2)1/2ds

= 21/2(1− y)1/2 + (1− y)3/2δ(1− y)

for 1 > y > 0, where δ(s) → 0 as s→ 0.

This proves the existence of ω and shows that

sin(ω − u) = 1− u2/2 + η(u)u2

for ω > u > 0 where η(u) → 0 as u → 0. Thus in (iii) we know thatthe extended function sin is differentiable at ω with sin′(ω) = 0. Ifω > u > 0

sin′(ω − u) = −(1− (sinω − u)2)

= −(1− (1− u2/2 + η(u)u2)2)1/2 = u+ β(u)u

andsin′(ω + u) = − sin′(ω + u) = −u − β(u)u

with β(u) → 0 as u→ 0+. Thus sin is twice differentiable at ω with

sin′′(ω) = −1 = − sinω.

208

K188

The points at issue are very similar to those of K187.

209

K189

(iv) Observe that

π

∫ R

1

f(x)2 dx = π

∫ R

1

x−2 dx = π(1−R−1) → π

but

2π

∫ R

1

f(x)(1 + f ′(x)2)1/2 dx ≥ 2π

∫ R

1

x−1 dx = 2π logR→ ∞

as R→ ∞.

(v) We have

VolK = π

∫ ∞

0

x−1 dx = ∞but

Vol(K ′ \K) = π

∫ ∞

0

(x−1/2 + x−1)2 − (x−1/2)2 dx

= π

∫ ∞

0

(x−2 + 2x−3/2) dx

= π limR→∞

(

(1−R−1) + 4(1−R−1/2))

= 5π.

210

K190

We have∫

γ1

f(x) ds =

∫

γ2

f(x) ds

=

∫

γ3

f(x) ds

=

∫

γ4

f(x) ds

=

∫ 1

0

cos(2πt)(−2π sin(2πt), 2π cos(2πt)) dt

= 2π

∫ 1

0

(− cos(2πt) sin(2πt), cos2(2πt)) dt

= π

∫ 1

0

(sin(4πt), 1 + cos(4πt)) dt

= (0, π)

Also∫

γ5

f(x) ds = −∫

γ1

f(x) ds = (0,−π)

and∫

γ6

f(x) ds = 2

∫

γ1

f(x) ds = (0, 2π).

211

K191

(ii) We haveun(k + 1)

unk=p(n− k)

(1− p)k

so un(k+1) > unk if p(n−k) > (1−p)k, i.e. if np > k and un(k+1) <unk if np < k (and, if np = k, then un(k + 1) = unk). We take k0 sothat np ≥ k0 − 1 and np ≤ k0.

(ii) If k ≥ n(p+ ǫ/2) then

un(k + 1)

unk=p(n− k)

(1− p)k≤ p(1− p− ǫ/2)

(1− p)(p+ ǫ/2)

Now 1 =∑n

j=0 un(j) ≥ un(k) so, writing k1 = k1(n) for the least

integer greater than n(p+ ǫ/2), we see that

un(k) ≤(

p(1− p− ǫ/2)

(1− p)(p+ ǫ/2)

)k−k1

for all k ≥ k1

If k2 = k2(n) is the greatest integer less than n(p + ǫ/2) we have

∑

k≥np+nǫ

un(k) ≤∑

k≥k2

(

p(1− p− ǫ/2)

(1− p)(p+ ǫ/2)

)k−k1

≤(

p(1− p− ǫ/2)

(1− p)(p+ ǫ/2)

)k2(n)−k1(n)(

1− p(1− p− ǫ/2)

(1− p)(p+ ǫ/2)

)−1

→ 0

as n→ ∞.

Similarly∑

k≤np−nǫ

un(k) → 0

so

Pr|Nn − np| ≥ ǫn =∑

|k−np|≥nǫ

un(k) → 0

as n→ ∞.

(iii) We haveEXj = p1 + (1− p)0 = p

and

EX2j = p1 + (1− p)0 = p

so varXj = p− p2 = p(1− p).

The sum of the expectations is the expectation of the sum so

ENn = En∑

j=1

Xj =

n∑

j=1

EXj.

212

For independent random variables, the sum of variances is the varianceof the sum so

varNn = np(1− p)

and Thchebychev gives

Pr|Nn − np| ≥ ǫn ≤ varNn

(ǫn)2=p(1− p)

ǫ2n→ 0

as n→ ∞.

213

K192

Recall that (A), (B) and (C) give

d(x, y) = (d(x, y) + d(y, x))/2 ≥ d(x, x)/2 = 0.

(i) Set x = y in (C)′ to obtain (B). (C) now follows from (B) and(C)′.

(ii) Setting d′(x, y) = −d(x, y), we see that d′ is a metric space. Ifd(x, y) ≥ 0 for x, y ∈ X then X has exactly one point.

214

K193

By (A)′, x ∼ x.

By (B), x ∼ y implies y ∼ x.

By (C) if x ∼ y and y ∼ z then

0 = d(x, y) + d(y, z) ≥ d(x, z) ≥ 0,

so d(x, z) = 0 and x ∼ z.

Observe x ∼ x′, y ∼ y′ gives

d(x, y) = d(x′, x) + d(x, y) + d(y, y′) ≥ d(x′, y′)

and d(x′, y′) ≥ d(x, y) so d(x, y) = d(x′, y′). Thus d is well defined.

215

K194

(i) The result is true with A1 = 1. We use induction on n to proveit. The result is trivially true for n = 1. Suppose it is true for n = m.If τ ∈ Sm+1 then writing σ = (m+ 1, τ(m+ 1)) and τ ′ = σ−1τ we seethat τ ′ fixes m+ 1 so is generated by at most m elements of the form(ij) (transpositions). Thus τ = στ ′ is is generated by at most m + 1transpositions. Any shuffle of m cards needs at most m swaps.

(ii) Observe that (1i)(1j) = (ij) and use (i).

(iii) Observe that

(123 . . . n)k(12)(123 . . . n)−k = (k k + 1)

so any transposition of the form (k k + 1) is a distance at most 2n+ 1from e. Next observe that

(k k + 1)(1 k)(k k + 1) = (1 k + 1)

so any transposition of the form (1 k) is a distance at most n(2n + 1)from e. Thus, using (ii) and (i), the diameter of Sn is at most 2n2(2n+1)with respect to X3.

(iv) Observe that arb = ba−r. Let N be the integer part of (n−1)/2.If N ≥ r then, by induction, the product of r elements of the form a, b,a−1, must have one of the forms au with |u| ≤ r or bav with |v| ≤ r−1.Looking at aN we see that the diameter of Dn is at least N .

(v) Sn has n! elements and Dn has 2n. There appears to be littlerelationship between size and diameter.

216

K195

(i) We wish to prove the statement P (N). If |x| ≥ 4−k for someinteger k. and

x =

N∑

j=1

xj with |xj| ≤ 4−n(j) for some integer n(j) ≥ 0 [1 ≤ j ≤ N ],

then∑N

j=1 2−n(j) ≥ 2−k.

If x =∑1

j=1 xj then x = x1 so P (1) is true.

Suppose now that P (r) is true for all 1 ≤ r ≤ N − 1. If |x| ≥ 4−k

and

x =N∑

j=1

xj with |xj | ≤ 4−n(j)

then if n(j) ≤ k for any j we are done. Thus we need only considerthe case n(j) ≥ k+1 for all 1 ≤ j ≤ N . Let M be the smallest m suchthat

m∑

j=1

|xj| ≥ 4−k−1.

We know thatM∑

j=1

|xj| ≤M−1∑

j=1

|xj|+ 4−k−1 ≤ 2× 4−k−1

soN∑

M+1

|xj | ≥ |x| −M∑

j=1

|xj | ≥ 2× 4−k−1

. Applying the inductive hypothesis to

x′ =M∑

j=1

|xj| and x′′ =N∑

M+1

|xj |

we see thatN∑

j=1

2−n(j) =

M∑

j=1

2−n(j) +

N∑

j=M+1

2−n(j) ≥ 2−k−1 + 2−k−1 = 2−k.

Thus P (N) is true and the induction is completed.

(ii) The key point is that d(x, 0) = 0 implies x = 0. But if x 6= 0then |x| ≥ 4−k for some k and so, by (i), d(x, 0) ≥ 2−k.

(iii) If 4−k+1 > |x| ≥ 4−k then taking x =∑1

j=1 xj with x1 = x, we

see that 2−k+1 ≥ d(x, 0). We also know, by (i), that d(x, 0) ≥ 2−k.

217

Thus4|x|2 ≥ d(x, 0) ≥ 4−1|x|2.

(iv) I think the graph is quite complex with f constant on intervalsof the form [4−k(1 − η), 4−k] (for some η > 0) and on other intervalswhich form a dense set (cf the ‘devil’s staircase’ in K251).

218

K196

(i) and (iv) are false. Let X = R with usual metric, X1 = x : x ≤0, X2 = x : x > 0, f(x) = 0 for x ∈ X1 and f(x) = 1 for x ∈ X2.

Parts (ii) and (iii) have one line solutions for the sufficiently sophis-ticated. Here are longer solutions.

(ii) True. If x ∈ X1 ∪ X2 then, without loss of generality, we maysuppose x ∈ X1. Thus there exists a δ1 > 0 such that, if z ∈ X andd(x, z) < δ1, we have z ∈ X1. Since f |X1

is continuous, we know that,given ǫ > 0 there exists a δ2(ǫ) > 0 such that d(x, z) < δ2(ǫ) impliesρ(f(x), f(z)) < ǫ for all z ∈ X1. Now d(x, z) < min(δ1, δ2(ǫ)) impliesρ(f(x), f(z)) < ǫ for all z ∈ X .

(iii) True. If x ∈ X \ Xj, the argument of (ii) shows that f iscontinuous at x. If x ∈ X1 ∩ X2, then given ǫ > 0 there exists aηj(ǫ) > 0 such that d(x, z) < ηj(ǫ) implies ρ(f(x), f(z)) < ǫ for allz ∈ Xj and d(x, z) < minj=1,2 ηj(ǫ) implies ρ(f(x), f(z)) < ǫ for allz ∈ X .

Set f(x) = g(x) when ‖g(x)‖ < 1, f(x) = ‖g(x)‖−1g(x) otherwise.Let

X1 = x ∈ X : ‖g(x)‖ ≤ 1 and X2 = x ∈ X : ‖g(x)‖ ≥ 1and apply part (iii).

219

K197

(i) Consider X = 1, 2 with the discrete metric and A = 1.(iii) By (ii), f is continuous. If A, B 6= ∅ choose a ∈ A, b ∈ B and

apply the intermediate value theorem to show that there exists a c withf(c) = 1/2.

(iv) If A is closed, then t ∈ [0, 1) : (1− t)a+ tb ∈ A is closed. IfA is open, then t ∈ [0, 1) : (1− t)a+ tb /∈ A is closed.

220

K198

Suppose A satisfies (i). Then, if ym ∈ A and ym → y, we can findxm ∈ A with d(xm, ym) < 1/m so xm → y and y ∈ A. Thus A ∈ F .But, if F ⊇ A and F is closed, we have F ⊇ A Thus A satisfies (ii).

Suppose A satisfies (ii). Then, since A is the intersection of closedsets, A ∈ F and automatically F ⊇ A whenever F ∈ F so A satis-fies (iii).

Suppose A satisfies (iii). If xn ∈ A and xn → x then xn ∈ A so sinceA is closed x ∈ A. Suppose x ∈ A, then (writing B(x, 1/n) for theopen ball centre x radius 1/n) if B(x, 1/n) ∩ A = ∅ A \ B(x, 1/n) is astrictly smaller closed set containing A contradicting our assumption.Thus B(x, 1/n)∩A 6= ∅ and we can find xn ∈ B(x, 1/n)∩A. We havexn ∈ A and xn → x.

(iii′ B is an open set with B ⊇ B such that, if U is open andB ⊇ U , we have B ⊇ U .

IntQ = ∅, IntZ = ∅, Int1/n : n ≥ 1 = ∅, Int[a, b] = Int(a, b) =Int[a, b) = (a, b).

ClQ = R, ClZ = Z, Cl1/n : n ≥ 1 = 0 ∪ 1/n : n ≥ 1,Cl[a, b] = Cl(a, b) = Cl[a, b) = [a, b].

Suppose f(A) ⊆ f(A) for every A. If xn → x but f(xn) 9 f(x),then we can find a δ > 0 and n(j) → ∞ such that ρ(f(xn(j)), f(x)) > δ.Now take

A = xn(j) : j ≥ 1to obtain a contradiction (x ∈ A but f(x) /∈ f(A)). Thus f is contin-uous.

Conversely, if f is continuous and x ∈ A, then we can find xn ∈ Asuch that xn → x and so, by continuity, f(xn) → f(x). Thus f(A) ⊆f(A).

Let X = 1, 2 with d(1, 2) = d(2, 1) = 1 and d(1, 1) = d(2, 2) = 0.Take y = 1.

221

K199

Observe that ClU ⊇ Int(ClU)) and ClU is closed so that

ClU ⊇ Cl(Int(ClU)).

However, since U is open, and ClU ⊇ U , we have Int(ClU)) ⊇ U so

Cl(Int(ClU)) ⊇ ClU

and Cl(Int(ClU)) = ClU .

It follows that, If B is any set,

Cl(Int(Cl(IntB))) = Cl(IntB)

and by taking complements, if C is any set

Int(Cl(Int(ClC))) = Int(ClC)

Since Cl ClA = ClB and Int IntA = IntB, the only possible dis-tinct sets that we can produce are A, IntA, ClA, Cl IntA, Int ClA,Int Cl IntA and Cl Int ClA.

The set

A = −3 ∪ (−2,−1] ∪ (Q ∩ [0, 1]) ∪ ((2, 4) \ 3)shows that these 7 can all be distinct.

222

K200

(i) (C) Take λ = (max(2, ‖x‖))−1, for example.

(iii) Since Γ is absorbing we can define

n(x) = inft ≥ 0 : t−1x ∈ Γ.We note that n(x) ≥ 0. Our object is to show that ‖x‖ = n(x) definesthe required norm.

If λ ≥ 0 the definition gives n(λx) = λn(x). The fact that Γ issymmetric now gives

n(λx) = |λ|n(x)for all λ ∈ R.

Condition (B) tells us that 0 ∈ Γ, and then (D)′ together withconvexity tells us that, if x 6= 0, there exists a µ > 0 such that tx /∈ Γfor all t ≥ µ. Thus n(x) = 0 if and only if x = 0.

Suppose that x, y 6= 0. Then, if ǫ > 0,(

(1 + ǫ)−1n(x)−1x, (1 + ǫ)−1n(y)−1y)

∈ Γ

and so, by convexity,

(1 + ǫ)−1 1

n(x) + n(y)(x+ y) = (1 + ǫ)−1 n(x)−1n(y)−1

n(x)−1 + n(y)−1(x+ y)

=n(y)−1

n(x)−1 + n(y)−1(1 + ǫ)−1n(x)−1x+

n(x)−1

n(x)−1 + n(y)−1(1 + ǫ)−1n(y)−1y ∈ Γ.

Since ǫ is arbitrary, n(x + y) ≤ n(x) + n(y).

(iv) A necessary and sufficient condition is

ΓB ⊆ KΓA

where KΓA = Kx : x ∈ ΓA and ΓC is the closed unit ball of thenorm ‖ ‖C .

223

K201

Observe that

xn = (1, 2−1, 3−1, . . . , n−1, 0, 0, . . . ) ∈ E

but that

xn → x = (1, 2−1, 3−1, . . . , m−1, (m+ 1)−1, (m+ 2)−1, . . . ) /∈ E.

If a(j) ∈ F and a(j) → a, then

|a2m| = |a2m − a(j)2m| ≤ ‖a− a(j)‖ → 0

as j → 0 so a2m = 0 for all m. Thus a ∈ F .

Suppose that (V, ‖ ‖) is a normed space and E is a finite dimensionalsubspace. Let e1, e2, . . . , eN be a basis for E. Then

∥

∥

∥

∥

∥

N∑

j=1

xjej

∥

∥

∥

∥

∥

∗

=

N∑

j=1

|xj|

is a norm on E. But all norms on a finite dimensional space are Lips-chitz equivalent so there exists K ≥ 1 with K‖x‖ ≥ ‖x‖∗ ≥ K−1‖x‖.

If yn ∈ E, y ∈ V and ‖yn−y‖ → 0, then yn is a Cauchy sequence in(V, ‖ ‖) so a Cauchy sequence in (E, ‖ ‖) so, by Lipschitz equivalence, aCauchy sequence in (E, ‖ ‖∗) so, since (E, ‖ ‖∗) is complete, convergesin (E, ‖ ‖∗) to some z ∈ E. Since ‖yn − z‖∗ → 0, it follows, byLipschitz equivalence, that ‖yn − z‖ → 0. By the uniqueness of limits,y = z ∈ E. Thus E is closed.

224

K202

(i) If

λ′y + e′ = λy + e

with λ, λ′ ∈ R and e, e′ ∈ E then

(λ′ − λ)y = e− e′

so, applying T to both sides,

(λ′ − λ)Ty = 0

so λ = λ′ and e = e′.

Also

x− Tx

Tyy ∈ E.

If T = 0 then N = E.

(ii) Observe that the statement that

B(y, δ) ∩N 6= ∅is equivalent to the statement that there exists an f ∈ N such that

‖y − f‖ < δ

and (setting e = −λf) this is equivalent to saying that

‖λy + e‖ < δλ

for some e ∈ N and λ 6= 0.

(iii) If T is continuous then since the inverse image of a closed setunder a continuous function is closed N is closed.

If N is closed pick a y /∈ N and observe that there exists a δ > 0such that

‖λy + e‖ ≥ δ|λfor all λ and e ∈ N .

By part (i), if x ∈ E we have

x = λy + e

with e ∈ N , so, by (ii),

‖x‖ ≥ δ|λ| whilst Tx = λTy.

Thus

‖Tx‖ ≤ δ−1|Tx|‖x‖for all x and T is continuous.

(iv) We have N = a ∈ s00 :∑∞

j=1 aj = 0.

225

(a) If we use the norm ‖ ‖∞, then N not closed. Take

ej(n) =

1 if j = 1

−n−1 if 2 ≤ j ≤ n + 1

0 otherwise.

Then e(n) ∈ N , but e → (1, 0, 0, . . . ) /∈ N . Thus N is not closed.

(b) If we use the norm ‖ ‖w, N not closed. Observe that∑∞

j=2 j−1

diverges so (since j−1 ≤ 1) we can find N(n) such that

n ≤N(n)∑

j=2

j−1 ≤ n+ 1.

Set A(n) =∑N(n)

j=2 j−1. Take

ej(n) =

1 if j = 1

−A(n)−1j−1 if 2 ≤ j ≤ N(n)

0 otherwise.

Then e(n) ∈ N but e → (1, 0, 0, . . . ) /∈ N . Thus N is not closed.

(c) If we use the norm ‖ ‖1, N is closed. If e(n) ∈ N and ‖e(n) −e‖1 → 0 then (remember all sums over a finite number of non-zeroterms)

∣

∣

∣

∣

∣

∞∑

j=1

ej

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∞∑

j=1

ej −∞∑

j=1

ej(n)

∣

∣

∣

∣

∣

= ‖e(n)− e‖1 → 0

so∑∞

j=1 ej = 0.

(d) If we use the norm ‖2, N not closed. Consider the same exampleas (a).

(e) If we use the norm ‖ ‖u then N is closed. We can use much thesame proof as (c).

[To see T not continuous in (a), (b) and (d) we can consider thenorms of e − e(n) and T (e − e(n)). In (c) and (e) simple inequalitiesgive ‖T‖ ≤ 1, indeed ‖T‖ = 1.]

226

K203

The key observation is that, since 21/2 is irrational, x + y21/2 =x′ + y′21/2 for x, y, x′, y′ ∈ Q if and only if x = x′ and y = y′.

If pn, qn are positive integers with pn/qn → 21/2 (in the usual Eu-clidean metric) then N

(

(−pn, qn))

→ 0 but ‖(−pn, qn)‖ → ∞.

On the other hand

N(

(x, y))

= |x+ y21/2| ≤ |x|+ 2|y| ≤ 2‖(x, y)‖for (x, y) ∈ Q2.

227

K204*

No comments.

228

K205

Pick xn ∈ Fn. If m ≥ n then

d(xn, xm) ≤ diam(Fn) → 0

as n→ ∞. Thus xn is Cauchy and we can find x ∈ X with d(xn, x) → 0as n→ ∞.

Since xn ∈ Fm for all n ≥ m and Fm is closed, x ∈ Fm for each m sox ∈ ⋂∞

j=1 Fj . On the other hand, if y ∈ ⋂∞j=1 Fj , then since x, y ∈ Fn

for each n,d(x, y) = diam(Fn) → 0

so d(x, y) = 0 and x = y. Thus⋂∞

j=1 Fj contains exactly one point.

(a) Take X = (0, 1] with the usual Euclidean metric d and Fn =(0, 1/n].

(b) Take X = Z with the discrete metric d (so d(m,n) = 1 form 6= n). Let Fn = m : m ≥ n.

(c) Take Fn = x : x ≥ n.

229

K206

(i) If ρ is a metric, then ρ(x, y) > 0 for x 6= y so f is strictlyincreasing.

If f is strictly increasing, then it is easy to check that ρ is a metric.

(ii) If f is not continuous at x, then without loss of generality, we maysuppose that f is not left continuous and there exist x1 < x2 < . . . withxj < x and xj → x together with a δ > 0 such that f(x) − δ ≥ f(xj)for all j. Since

N∑

j=1

ρ(xj+1, xj) = f(xN+1)− f(x1) < f(x)− f(x1),

we have∑∞

j=1 ρ(xj+1, xj) convergent and xj Cauchy with respect to ρ.

If t < x then there exists an N such that t > xN so ρ(t, xj) ≥ρ(t, xN ) > 0 for j ≥ N so ρ(t, xj) 9 0. If t > x, then ρ(t, xj) ≥ρ(t, x) > 0 for all j so ρ(t, xj) 9 0. If t = x, then ρ(t, xj) ≥ δ > 0for all j so ρ(t, xj) 9 0. Thus the xj form a non-convergent Cauchysequence for ρ.

If f(t) 9 ∞ then a similar argument shows that, if xj = j, thexj form a non-convergent Cauchy sequence for ρ. A similar argumentcovers the case f(t) 9 −∞.

(iii) We must have

g(t) + g(s) = d(t, 0) + d(0,−s) ≥ d(t,−s) = g(t+ s)

for all t, s ≥ 0,

g(t) + g(s) = d(t, 0) + d(0, s) ≥ d(t, s) = g(t− s)

for all t ≥ s ≥ 0 and

g(0) = d(0, 0) = 0.

We check that these conditions are sufficient.

(iv) If 2−n−1 ≤ x ≤ 2−n, then

2n+1g(x) ≥ 2n=1g(2−n−1) =2n+1∑

j=1

g(2−n−1) ≥ g

(

2n+1∑

j=1

2−n−1

)

= g(1)

so g(x) ≥ 2−n−1g(1) ≥ x2−1g(1). Take K = 2−1g(1).

If we take g(x) = x1/2 we see that no relation d(x, y) ≤ L|x− y| willhold in this case.

(v) Since g is increasing and g(t) ≥ 0, g(t) → l as t → 0+ for somel ≥ 0. If l > 0 then d(x, y) > l for all x 6= y so any Cauchy sequencefor d is eventually constant and so converges.

230

If l = 0 then, since g is decreasing, given any δ > 0 we can find andǫ > 0 such that ǫ > g(s) ≥ 0 implies |s| < δ. Thus if xn is Cauchy ford, it follows that xn is Cauchy for the Euclidean metric so there existsan x ∈ R with |xn − x| → 0 and so d(xn, x) = g(|xn − x|) → 0.

(vi) Sufficiency by direct verification. Not necessary. Let θ be thediscrete metric with θ(x, y) = 1 for x 6= y. Then g(t) = 0 for t 6= 1,g(1) = 1 gives a metric but g(1/2) + g(1/2) = 0 and g(1) = 1.

231

K207

(i) Write ej for the vector with 1 in the jth place and 0 elsewhere.Then

‖ajej‖ ≤ ‖a‖and so

|aj| ≤ ‖a‖‖ej‖−1.

232

K208

We have

N∑

j=1

(aj + bj)2 ≤

(

N∑

j=1

a2j

)1/2

+

(

N∑

j=1

a2j

)1/2

2

≤

( ∞∑

j=1

a2j

)1/2

+

( ∞∑

j=1

a2j

)1/2

2

,

so, since an increasing sequence bounded above converges we have∑∞

j=1(aj + bj)2 convergent. Moreover

∞∑

j=1

(aj + bj)2 ≤

( ∞∑

j=1

a2j

)1/2

+

( ∞∑

j=1

a2j

)1/2

2

.

(ii) To show completeness, suppose a(n) a Cauchy sequence in l2.We have

|aj(n)− aj(m)| ≤ ‖a(n)− a(m)‖2so aj(n) is Cauchy with respect to the usual Euclidean distance andthere exists an aj ∈ R such that |aj(n)− aj| → 0.

Now observe that,

(

M∑

j=1

a2j(n)

)1/2

≤(

M∑

j=1

(aj − aj(n))2

)1/2

+

(

M∑

j=1

aj(n)2

)1/2

≤(

M∑

j=1

(aj − aj(n))2

)1/2

+ ‖a(n)‖2

≤(

M∑

j=1

(aj − aj(n))2

)1/2

+ supr≥1

‖a(r)‖2

→ supr≥1

‖a(r)‖2

as n→ ∞. (Recall that a Cauchy sequence is bounded.) Thus

(

M∑

j=1

a2j

)1/2

≤ supr≥1

‖a(r)‖2

for all M and so a ∈ l2.

233

Finally, if n, m ≥ N(

M∑

j=1

(a2j − aj(n))2

)1/2

≤(

M∑

j=1

(a2j − aj(m))2

)1/2

+

(

M∑

j=1

(a2j (n)− aj(m))2

)1/2

≤(

M∑

j=1

(a2j − aj(m))2

)1/2

+ ‖a(n)− a(m)‖2

≤(

M∑

j=1

(a2j − aj(m))2

)1/2

+ supr,,s≥N

‖a(s)− a(r)‖2

so, allowing m→ ∞,(

M∑

j=1

(a2j − aj(n))2

)1/2

≤ supr,s≥N

‖a(s)− a(r)‖2

and, allowing M → ∞,

‖a− a(n)‖2 ≤ supr,s≥N

‖a(s)− a(r)‖2

for all n ≥ N . Thus‖a− a(n)‖2 → 0

as n→ ∞.

(iii) We have (by Cauchy-Schwarz in N dimensions)

N∑

j=1

|ajbj | ≤(

N∑

j=1

a2j

)1/2( N∑

j=1

a2j

)1/2

≤ ‖a‖2‖b‖

so, since an increasing sequence bounded above converges, we have∑∞

j=1 ajbj absolutely convergent and so convergent.

234

K209

(ii) we have

− log

(

x

p+y

q

)

≤ −1

plog x− 1

qlog y

for all x y > 0. Setting x = Xp, y = Y Q, multiplying by −1 and takingexponentials gives the inequality.

(iii) By (ii),

|f(t)g(t)| ≤ |f(t)|pp

+|g(t)|qq

.

Now integrate.

236

K211

It may be helpful to look at K209.

(iv) If ‖ ‖∗ is derived from an inner product, then

‖a+ b‖2∗ + ‖a− b‖2∗ = 2(‖a‖2 + (‖a‖2).Taking

a = (1, 0, 0, 0 . . . ), b = (0, 1, 0, 0, . . . ),

we see that this fails for ‖ ‖p unless p = 2.

In the case C([0, 1], ‖ ‖p) we look at (for example)

f(x) = g(1− x) = max(1− 4x, 0).

Observe that∫ 1

0

f(x)p dx =

∫

0

1/4(4t)p dt.

(v) We consider C([a, b]). It is easy to check that

‖fg‖1 ≤ ‖f‖1‖g‖∞and (by setting g = 1) that if

‖fg‖1 ≤ A‖g‖∞for all continuous g then ‖f‖1 ≤ 1. If f ∈ C([a, b]) and there existsan x ∈ [a, b] with |f(x)| ≥ A + ǫ. Without loss of generality, we maysuppose f(x) ≥ A + ǫ. By continuity we can find a δ > 0 such thatf(t) ≥ A = ǫ/2 for t ∈ [a, b] ∩ (x − δ, x + δ). Choose a non-zerog ∈ C([a, b]) with g(t) = 0 for all t /∈ [a, b] ∩ (x− δ, x+ δ). Then

|fg‖1 > (A+ ǫ/2)‖g‖1.Thus Holder and reverse Holder hold for p = 1, q = ∞ and p = ∞,q = 1.

237

K212

(i) p = 1. A square with sides at π/4 to the axes.

p = 2. A disc.

p = ∞ A square with sides parallel to the axes.

Other values of p give smooth boundary (as does p = 2).

(ii) If 1 < p < ∞, then the chain rule shows that fp differentiableexcept at (0, 0). Since

fp(x, 0)

x= sgn x

does not tend to a limit as x→ 0, fp is not differentiable at (0, 0).

If x, y > 0 then f1(x, y) = x + y so f1 is differentiable on (x, y) :x, y > 0. On the other hand

f1(h, y)− f1(0, y)

h= sgn h

does not tend to a limit as h → 0 so f1 is not differentiable at (0, y).Similar arguments show that f1 is differentiable at the points (x, y)with both x 6= 0 and y 6= 0 and only there.

If x > y ≥ 0 then f∞(x, y) = x so f1 is differentiable on (x, y) :x > y ≥ 0. However

f∞(x+ h, x)− f∞(x, x)

h=

1 if h > 0,

−1 if h < 0,

so f∞ is not differentiable at (x, x) if x > 0 and a similar argumentshows that f∞ is not differentiable at (0, 0). Similar arguments showthat f∞ is differentiable at the points (x, y) with x 6= y and only there.

(iii) If ‖x‖s = 1 then |xj | ≤ 1 and |xj |s ≤ |xj |r so1 = ‖x‖ss ≤ ‖x‖rr

so ‖x‖r ≥ 1.

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

(iv) We have

n∑

j=1

|xj|r ≤(

n∑

j=1

|xj |rp)1/p

n1/q.

Now set rp = s (so q(s− r) = s).

(vi) If s > r we can find an α with r−1 > α > s−1. If bj(N) = j−α forj ≤ N , bj(N) = 0 otherwise, then ‖b(N)‖s/‖b(N)‖r → ∞ as N → ∞.

238

If bj = j−α, then b ∈ lr but b /∈ ls.

(vii) Argument similar to (iii).

Consider gn = max(1− nx, 0).

Essentially same but there is scale change and we have the formula

(b− a)1/r‖f‖s ≤ (b− a)1/s‖f‖r.

239

K213

(ii) First look at maps from (Rn, ‖ ‖∞) to (Rm, ‖ ‖∞) Observe that

‖Tx‖∞ = max1≤i≤m

∣

∣

∣

∣

∣

n∑

j=1

aijxj

∣

∣

∣

∣

∣

≤ max1≤i≤m

n∑

j=1

|aij|‖x‖∞

and that, if we set yj(i) = sgn aij we have

‖Ty(i)‖∞ =

n∑

j=1

|aij| =n∑

j=1

|aij|‖y(i)‖∞.

Thus ‖T‖ = max1≤i≤m

∑nj=1 |aij|.

Now look at maps from (Rn, ‖ ‖1) to (Rm, ‖ ‖∞). Observe that

‖Tx‖∞ = max1≤i≤m

∣

∣

∣

∣

∣

n∑

j=1

aijxj

∣

∣

∣

∣

∣

≤ max1≤i≤m

n∑

j=1

|aij ||xj| ≤ max1≤i≤m

max1≤j≤n

|aij|‖x‖1

and that, if we set yj(k) = 1 if j = k, yj(k) = 0, otherwise, then

‖Ty(k)‖∞ = max1≤i≤m

|aik| = max1≤i≤m

|aik|‖y(k)‖1

Thus ‖T‖ = max1≤i≤mmax1≤j≤n |aij|.Now look at maps from (Rn, ‖ ‖1) to (Rm, ‖ ‖1). Observe that

‖Tx‖1 =m∑

i=1

∣

∣

∣

∣

∣

n∑

j=1

aijxj

∣

∣

∣

∣

∣

≤m∑

i=1

n∑

j=1

|aij||xj|

=n∑

j=1

m∑

i=1

|aij ||xj|

=n∑

j=1

|xj|m∑

i=1

|aij|

≤ max1≤j≤n

|m∑

i=1

|aij |‖x‖1

and that, if we set yk(j) = 1 if k = j, yk(j) = 1, otherwise then

‖Ty(j)‖1 =m∑

i=1

|aij| =m∑

i=1

|aij‖y(j)‖1.

Thus ‖T‖ = max1≤j≤n |∑m

i=1 |aij|.

240

If we look at maps from (Rn, ‖ ‖∞) to (Rm, ‖ ‖1) we have the obviousbound

‖T‖ ≤m∑

i=1

n∑

j=1

|aij|

but looking at the maps T1 and T2 with matrices(

1 11 1

)

and

(

1 1−1 1

)

we see that, in the first case the bound is attained but in the secondcase it is not since

(

1 11 −1

)(

xy

)

=

(

x+ yx− y

)

and |x + y| + |x − y| ≤ 2max(|x|, |y|). However, if we really need ato find ‖T‖ we can obtain it as a solution of a linear programmingproblem.

241

K214

Suppose a, b ∈ X . If ǫ > 0, we can find a y ∈ E such that

d(a, E) ≥ d(a, y)− ǫ.

Thus

d(b, E) ≤ d(b, y) ≤ d(a, b) + d(a, y) ≤ d(a, b) + d(a, E) + ǫ.

Since ǫ is arbitrary

d(b, E) ≤ d(a, b) + d(a, E)

and, similarly, d(a, E) ≤ d(a, b) + d(b, E). Thus |f(a)− f(b)| ≤ d(a, b)and f is continuous.

Suppose k ∈ K, l ∈ L. In the previous paragraph we saw that

d(k,M) ≤ d(k, l) + d(l,M)

so, by definition,

d(k,M) ≤ d(k, l) + ρ′(L,M)

so, allowing l to range freely,

d(k,M) ≤ d(k, L) + ρ′(L,M)

and so, by definition again,

d(k,M) ≤ ρ′(K,L) + ρ′(L,M)

soρ′(K,M) ≤ ρ′(K,L) + ρ′(L,M).

It follows at once that ρ obeys the triangle inequality. Also ρ issymmetric and positive. If ρ(E, F ) = 0 then ρ′(E, F ) = 0 and, ife ∈ E, we can find fn ∈ F such that d(fn, e) → 0. Since F is closede ∈ F so F ⊇ E. Similarly E ⊇ F so E = F . Thus ρ is a metric.

The continuous image of a closed bounded set in Rn is closed andbounded. If |fn(x)− f(x)| < ǫ then

ρ′(f(Kn), F (K)), ρ′(f(K), F (Kn)) < ǫ.

No. Let K = [−π, π] and fn(x) = sinnx.

242

K215

The intersection of closed bounded sets is closed and bounded. Choosexn ∈ Kn. Since K1 is closed and bounded, the theorem of Bolzano–Weierstrass says that we can pick n(j) → ∞ such that xn(j) → x forsome x ∈ K1. Since xn(j) ∈ Km for j sufficiently large and Km is closedx ∈ Km for all m so x ∈ K.

If ρ(K,Kn) 9 0 then we can find a δ > 0 such that there exists xn ∈Kn with d(xn, K) > δ (recall KN ⊇ KN+1). By the argument of thefirst paragraph we can find n(j) → ∞ and x ∈ K with d(xn(j),x) → 0which is absurd.

243

K216

(i) and (iii). It is helpful to observe that d(e, f) = ρ(e, f).(ii) Take y ∈ Y if and only if d(y, E) ≤ 2/N . Observe that the set

of possible Y is finite.

244

K217

Since E is closed, r = inf‖z− y‖ : y ∈ E > 0. If 1 > ǫ′ > 0 wecan find y0 ∈ E such that

‖z− y0‖ > r(1− ǫ′)

Set x0 = z− y0. Then ‖x0‖ < r(1 + ǫ′) and

‖x0 − e‖ = ‖z− (y0 + e)‖ ≥ r

for all e ∈ E.

If we set x = ‖x0‖−1x0‖ then ‖x‖ = 1 and

‖x− e‖ ≥ ‖x0‖−1r = (1 + ǫ′)−1 > 1− ǫ

for all e ∈ E if we choose ǫ′ appropriately.

Let x1 be any vector of norm 1. Let E1 be the subspace generated byx1. Now proceed inductively choosing xn+1 of norm 1 with ‖xn+1−e‖ ≥1/2 for all e ∈ En and taking En+1 be the subspace generated by xn+1

and En.

245

K218

(i) If κn 9 0 we can find a δ > 0 and n(j) → ∞ with κn(j) ≥ δ. Ifun(j)(j) = δ, uk(j) = 0 if k 6= n(j), then u(j) ∈ E but the sequencehas no convergent subsequence.

Suppose κn → 0 and xn ∈ E. Set n(0, j) = n(j). Then since[−κr, κr] is closed and bounded we can use Bolzano–Weierstrass tofind sequences n(r, j) and yr ∈ [−κr, κr] such that

(a) n(r, j) is a subsequence of n(r, j − 1) and n(r, j) > n(r − 1, j).

(b) xr(

n(r, j))

→ xr as j → 0.

(c) |xj(

n(r, j))

− xj | ≤ 2−r for all 1 ≤ j ≤ r.

Observe that x ∈ E. Now consider xn(r,r). We have

‖x(n(r, r))− x‖ ≤ max(

max1≤j≤r

|xj(

n(r, j))

− xj |, supj≥r+1

|κj|)

≤ max(2−r, supj≥r+1

|κj|) → 0

so x(n(r, r)) → x as r → ∞.

(ii) If∑∞

n=1 κn diverges, we can find N(j) such that∑N(j)

n=j κn > 1.

If uk(r) = κk/(∑N(j)

n=j κn) for j ≤ k ≤ N(j), uk(r) = 0 otherwise, then

u(k) ∈ E but the sequence has no convergent subsequence.

The proof of the positive result resembles that in (i).

246

K219

(i) Take complements.

(ii) If not we can find xn such that B(xn, 1/n) does not lie in any U .By the Bolzano–Weierstrass property we can find n(j) → 0 and x ∈ Xsuch that xn(j) → x. Now there must exist a U ∈ U with x ∈ U . SinceU is open there exists a δ > 0 such that B(x, δ) ⊆ U . But we can finda J such that n(J) > 2δ−1 and ‖xn(J)− x‖ < δ/2 so

B(xn(J), 1/n(J)) ⊆ B(x, δ) ⊆ U

which contradicts our initial assumption.

(iii) By Lemma 11.22 we can find y1, y2, . . . yM such that⋃M

m=1B(ym, δ) =X and by (ii) we can find Um ∈ U with B(ym, δ) ⊆ Um. Thus⋃M

m=1 Um = X .

247

K220

(i) Take contrapositives.

(ii) If the space has property (B) then, taking U to be the collection of

B(y, δy), we can find y1, y2, . . . , yM such that X =⋃M

m=1B(ym, δym).Now set N = max1≤m≤M Nm. We have xN /∈ B(ym, δym) for each1 ≤ m ≤M and this gives a contradiction.

(iii) Combine the results of this question with that of K219.

248

K221

‖ ‖B is not a norm since ‖1‖A = 0 but 1 6= 0. The rest can bechecked to be norms.

Recall that Lipschitz equivalent metrics are either both complete orneither complete.

‖f‖∞ ≤ ‖f‖A ≤ 2‖f‖∞so ‖ ‖∞ and ‖ ‖A are equivalent. Looking at fn(x) = (n−2+(x− 1

2)2)1/2

we see that fn is Cauchy. If it has a limit f in the uniform norm thenit is a pointwise limit of fn so f(x) = |x− 1

2|. But f /∈ C1([0, 1]).

‖f‖D ≤ ‖f‖C ≤ 2‖f‖Dso ‖ ‖C and ‖ ‖D are equivalent. If fn is Cauchy in ‖ ‖C then bothfn and f ′

n are Cauchy in ‖ ‖∞ converge uniformly to continuous f andF and standard theorems tell us that f is differentiable with f ′ = F .Thus f ∈ C1([0, 1]) and ‖fn − f‖C → 0.

By considering fN (x) = sinNx we see that ‖ ‖D is not Lipschitzequivalent to ‖ ‖∞ or ‖ ‖1. By considering fN(x) = xN we see that‖ ‖∞ is not equivalent to ‖ ‖1 (or observe that one norm is completeand the other is not).

249

K222

(i) By the inverse function rule

d

dxsin−1 x =

1

(1− x2)1/2

for |x| < 1. By the binomial expansion (or by some rigorous Taylorexpansion)

1

(1− t)1/2= 1 +

∞∑

n=1

n∏

r=1

(2r − 1)

2rtn

for |t| < 1. Thus

d

dxsin−1 x = 1 +

∞∑

n=1

n∏

r=1

(2r − 1)

2rx2n

for |x| < 1. But we may integrate term by term within the radius ofconvergence so (since sin−1 0 = 0)

sin−1 x = x+∞∑

n=1

n∏

r=1

(2r − 1)

2r

x2n+1

2n+ 1=

∞∑

n=0

2−2n

2n+ 1

(

2n

n

)

x2n+1

for |x| < 1.

(ii) The power series has radius of convergence 1 so the only otherpoints to consider are x = ±1. Taking logarithms (cf K96) shows that

n∏

r=1

(2r − 1)

2r=

n∏

r=1

(

1− 1

2r

)

→ n−1/2

so ∞∑

n=1

1

2n+ 1

n∏

r=1

(2r − 1)2r

converges by comparison with n−3/2 so we actually have (by the Weier-strass M-test) uniform convergence on [−1, 1] so the power series iscontinuous on [−1, 1] so, since sin−1 is also continuous on [−1, 1], wehave equality at x = 1 and x = −1 as well.

250

K223

(i) fn(x) = (n−2 + x−2)1/2 (with positive square root) will do.

(ii) Use (i).

(iii) If fn is Cauchy in ‖ ‖∗ then both fn and f ′n are Cauchy in ‖ ‖∞

converge uniformly to continuous f and F but then we know that f isdifferentiable with f ′ = F . Thus f ∈ C1([0, 1]) and ‖fn − f‖C → 0.

251

K224

Let hn : [−1, 1] → R be given by

hn(x) =

2n+3(x+ 1) for x ∈ [−1,−1 + 2−n−3],

1 for x ∈ [−1 + 2−n−3,−2−n−3]

−2n+3x for x ∈ [−2−n−3, 0]

−h(−x) for x ∈ [0, 1].

Let gn : [−1, 1] → R be given by

gn(x) =

∫ 1

−1

hn(t) dt

By the fundamental theorem of analysis, gn is differentiable with con-tinuous derivative hn so |g′n(x)| ≤ 1 for all x and |gn(x)| = 1 outsidethree intervals of total length 2−n−1. We observe that gn(x) → 1− |x|uniformly on [−1, 1]

We define kn : R → R to be the 1-periodic function with kn(x) =2−1gn(2(x − 1)) for x ∈ [−1, 1] and define fn : [0, 1] → R by fn =n−1kn(nx) for x ∈ [0, 1].

Observe that fn ∈ A, ‖fn‖∞ → 0 as n → 0 but I(fn) → 0 andI(0) = 1. Thus I is not continuous with respect to the uniform norm.

However if gn, g ∈ A and ‖gn−g‖∗ → 0 then g′n(x) → g(x) uniformly,so (1− (g′n(x))

4)2 → (1− (g′(x))4)2 uniformly, so

(1− (g′n(x))4)2 + gn(x)

2 → (1− (g′(x))4)2 + (x)2

uniformly as n → 0 so I(gn) → I(g) and I is continuous with respectto ‖ ‖∗.

(iii) Done in (i).

(iv) Since I is continuous it follows that, if fN → f with respect to‖ ‖∗ then Ifn → If , so If = 0 which is impossible. (If If = 0 then∫ 1

−1f(x)2 dx = 0 so f = 0 (since f is continuous) but I(0) = 1.)

252

K225

(i) If fN,ǫ(x) = ǫ sin 2πNx then

I(fN,ǫ) =

∫ 1

0

(

1− 2ǫ(2πN)4(cos 2πNx)4

+ ǫ2(2πN)8(cos 2πNx)8 + ǫ2(sin 2πNx)2)

dx

= 1− AǫN4 +Bǫ8N8 + Cǫ2

with

A = 2(2π)4∫ 1

0

(cos 2πx)4 dx > 0

B = (2π)8∫ 1

0

(cos 2πx)8 dx > 0

C = (2π)2∫ 1

0

(sin 2πx)2 dx > 0.

Also‖fN,ǫ‖∗ = (1 + 2πN)ǫ.

If we take gN = fN,N−4/3 then ‖gN‖∗ → 0, and (when N is sufficientlylarge) I(gN) < 1 = I(0). If we take hN = f1,N−1 , then ‖hN‖∗ → 0, and(when N is sufficiently large) I(hN ) > 1 = I(0).

(ii) Observe that g(t) = 6(t − 1)(t − 2) = 6t2 − 18t + 12 has zerosat 1 and 2. Thus f(t) = 2t3 − 9t2 + 12t has stationary points at 1 and2 so (from our knowledge of cubics or by looking more closely) has aunique zero at 0. Setting P (t) = f(t)2 gives a function of the requiredtype.

(iii) There exists a δ > 0 such that P (t) > 0 for all 0 < |t| < δ.Thus, if |f ′(t)| < δ1/2, the function h : [0, 1] → R defined by

h(x) = P (1− f ′(x)2) + f(x)2 − P (1)

is continuous and positive so

J(f)− I(0) =

∫ 1

0

h(x) dx ≥ 0

with equality only if h(x) = 0 for all x ∈ [0, 1] i.e. only if f = 0. Thusif ‖f‖∗ < δ1/2 we have J(f) ≥ 0 with equality if and only if f = 0.

Use the fn of K224.

(iv) This is really no more complicated than finding a smooth func-tion u : R → R with 0 < u(x) < 1 for all t and infx∈R u(x) = 0,supx∈R u(x) = 1.

Set v(x) = 12+ π−1 tan−1 x and G(s, t) = v(s).

253

K226

(iv) It remains true that d∞ is a metric and that the continuousfunctions form a closed subspace of B(E) and that the uniform limitof continuous functions is continuous since the proofs do not use thecompleteness of (Y, ρ).

Suppose that Y is not complete, so we can find a Cauchy sequencey(n) in Y which does not converge. Let E = e and fy(e) = y. (NoteB(E) = C = C(E) and the members of C(E) are precisely the fz.)Then d∞(fy, fz) = ρ(y, z) so the fy(n) do not converge. B(E), C(E)and C(E) are not complete and the general principle of convergencefails.

254

K227

Observe that, if 0 < v < u, then evxe−uxxγ−1 → 0 as x → 0 (expo-nentials beat powers) so we can find an A such that

0 ≤ e−uxxγ−1 ≤ Ae−vx

for x ≥ 1 so, by comparison,∫∞1e−uxxγ−1 dx converges. Also

0 ≤ e−uxxγ−1 ≤ xγ−1

so, by comparison,∫ 1

0e−uxxγ−1 dx converges. Thus

∫∞0e−uxxγ−1 dx

converges.

Our theorem on differentiating under a finite integral show that

d

du

∫ n

1/n

e−uxxγ−1 dx =

∫ n

1/n

e−uxxγ dx.

The argument of our proof on differentiating under an infinite integralnow shows that

d

du

∫ ∞

0

e−uxxγ−1 dx =

∫ ∞

0

e−uxxγ dx.

Now integrate by parts, to get∫ X

ǫ

e−uxxγ dx =[

−u−1xγe−ux]X

ǫ+ γu−1

∫ X

ǫ

e−uxxγ−1 dx

and, taking limits,φ′γ(u) = γu−1φγ(u).

Thusd

du

(

u−γφγ(u))

= 0

and, by the constant value theorem,

φγ(u) = Aγuγ.

(ii) We have

φn(u) =1

(n− 1)!uγ.

(iii) The arguments including the proofs of convergence for the ap-propriate integrals are similar to but simpler than those of (i).

ψ′(u) = −∫ ∞

0

xe−uxe−x2/2 dx =[

−e−uxe−x2/2]∞

−∞− uψ(u) = −uψ(u)

sod

du

(

ψ(u)eu2/2)

= 0

and ψ(u) = Ae−u2/2 for some constant A.

(iv) To obtain (i) make the substitution s = ux. To obtain (iii) makethe substitution y = x− u.

255

K228

Since f (n+1) → g uniformly on the closed interval with end points aand x, we have

∫ x

a

g(t) dt = g(x)− g(a).

Since g is continuous, the fundamental theorem of analysis gives

g(x) = g′(x)

sod

dx

(

e−xg(x))

= 0

so, by the constant value theorem, g(x) = Cex for some C.

No, f(x) = 1 + ex has the property.

256

K229

(i) Since exponentials beat polynomials, fn(x) → 0 for x 6= 0. Butfn(0) = 0 → 0, so fn → 0 pointwise for all α, β > 0.

(ii) We observe that (if x ∈ (0, 1))

f ′n(x) = nαxβ−1(1− x)n−1

(

β − (β + n)x)

so, examining the sign of f ′n, we see that there is a unique maximum

at x = β/(β + n). Thus

0 ≤ f(x) ≤ f

(

β

β + n

)

= nα−β

(

β

1 + β/n

)β (

1− β

β + nr

)n

so, remembering that (1 + γn−1)n → eγ, we see that fn → 0 uniformlyif and only if β > α.

(iii) We have∫ 1

0

fn(x) dx ≥∫ 2/n

1/n

fn(x) dx

≥∫ 2/n

1/n

nα(n−1)β(

1− 2

n

)n

dx

= nα−β−1

(

1− 2

n

)n

so∫ 1

0fn(x) dx9 0 if α ≥ β + 1.

Now suppose α < β + 1. Choose γ with

1 > γ >α

β + 1.

We have∫ 1

0

fn(x) dx =

∫ n−γ

0

fn(x) dx+

∫ 1

n−γ

fn(x) dx

≤∫ n−γ

0

nαxβ dx+

∫ 1

n−γ

nα(1− x)n dx

=nα−(β+1)γ

β + 1+

nα

n+ 1(1− n−γ)n+1

=nα−(β+1)γ

β + 1+

nα

n+ 1(1− n−γ)

(

(1− n−γ)nγ)n(1−γ) → 0,

as n→ ∞.

[An alternative approach is to show that, when α = β + 1,

∞ > lim supN→∞

∫ 1

0

fn(x) dx ≥ lim infN→∞

∫ 1

0

fn(x) dx > 0

257

and deduce the other cases.]

258

K230

(i) Suppose that gn → g uniformly and xn → x. Let ǫ > 0. By thecontinuity of g at x, we can find a δ > 0 such that |g(t)−gn(t)| < ǫ/2 for|x− t| < δ and t ∈ [a, b]. Now there exists an N such that |xn − x| < δfor n ≥ N and an M such that ‖gn − g‖∞ < ǫ/2 for n ≥ M . Thus ifn ≥ max(N,M)

|gn(xn)− g(x)| ≤ |gn(xn)− g(xn)|+ |g(xn)− g(x)| < ǫ.

If gn 9 g uniformly then we can find a δ > 0 and n(j) → ∞ suchthat

|gn(j)(xn(j))− g(xn(j))| > δ.

By Bolzano–Weierstrass we can find j(k) → ∞ such that xn(j(k)) → ysay. Set yn(j(k)) = xn(j(k)), yr = y otherwise. Then yr → y but, since

|gn(j)(yn(j))− g(y)| ≥ |gn(j)(xn(j))− g(xn(j))| − |g(xn(j))− g(y)|≥ δ − |g(xn(j))− g(y)| → δ

we have gr(yr) 9 g(y).

(ii) ‘Only if’ fails if g is not assumed continuous. Let [a, b] = [−1, 1]and gn(x) = g(x) = H(x) with H(x) = 0 for x ≤ 0 and H(x) = 1 forx > 0. Then gn → g uniformly but g(1/n) 9 g(0).

However, the ‘if’ part still works. Suppose gn 9 g uniformly asbefore and define δ > 0 as before. Now define n(j) andm(j) inductivelyas follows. Set m(0) = 0. For each j ≥ 1 choose n(j) > m(j − 1) andxn(j) such that

|gn(j)(xn(j))− g(xn(j))| > δ.

Now we know that gn(xn(j)) → g(xn(j)) so we can find m(j) > n(j)such that

|gm(j)(xn(j))− g(xn(j))| < δ/2,

and so, in particular

|gn(j)(xn(j))− gm(j)(xn(j))| > δ/2.

By Bolzano–Weierstrass we can find j(k) → ∞ such that xn(j(k)) → ysay. Set

yn(j(k)) = ym(j(k)) = xn(j(k)),

and yr = y otherwise. Since

|gn(j)(yn(j))− gm(j)(ym(j))| = |gn(j)(xn(j))− gm(j)(xn(j))| > δ/2 > 0

we know that gr(yr) does not converge.

(ii) The proof ‘only if’ in (i) still works. However ‘if’ part fails (evenif gn is continuous). Let (a, b) = (0, 1) and gn(x) = max(1 − nx, 0). Ifx ∈ (0, 1), then we can find a δ > 0 such that [x − δ, x + δ] ⊆ (0, 1).Since gn → 0 uniformly on [x − δ, x + δ] we apply part (i) to show

259

that xn → x implies gn(xn) → g(x). However gn does not convergeuniformly on [0, 1].

260

K231

(i) Since f is continuous on [0, 1 − δ], it attains its bounds so thereexists a y ∈ [0, 1 − δ] with f(y) ≥ f(x) ≥ 0. Since 0 ≤ f(y) < 1, wehave

n

∫ 1−δ

0

f(x)n dx ≤ nf(y)n → 0.

Observe that (if f ′(1) 6= 0) f ′(1) > 0 since f(1) > f(1 − x) forx ∈ (0, 1). Given any ǫ with f ′(1) > ǫ > 0 we can find a 1 > δ > 0such that

|f(x)− 1− f ′(0)(1− x)| = |f(x)− f(1)− f ′(1)(1− x)| ≤ ǫ(1− x)

and so

1− (f ′(1)− ǫ)(1− x) ≤ f(x) ≤ 1− (f ′(1)− ǫ)(1− x)

for x ∈ [1− δ, 1]. Thus∫ 1

1−δ

(1− (f ′(1)− ǫ)(1− x))n dx ≤∫ 1

1−δ

f(x)n

≤∫ 1

1−δ

(1− (f ′(1)− ǫ)(1 − x))n dx.

Now

n

∫ 1

1−δ

(1−A(1− x))n dx = n

∫ δ

0

(1− At)n dt

=n

A(n + 1)(1− (1−Aδ)n+1) → A−1

so, using the first paragraph,

n

∫ 1

0

f(x)n dx→ 1

f ′(1).

(ii) Suppose g(y) = 1 for some y ∈ (0, 1). By the argument ofRolle’s theorem, g′(y) = 0, so given any ǫ > 0, we can find a δ > 0with [y − δ, y + δ] ⊆ (0, 1) such that

g(t) > 1− ǫ|y − t|for all t with |y − t| ≤ δ. Thus

n

∫ 1

0

g(x)n dx ≥ n

∫ y+δ

y−δ

(1− ǫ|y − t|)ndt

=2n

ǫ(n+ 1)(1− (1− Aδ)n+1) → ǫ−1.

Since ǫ was arbitrary,

n

∫ 1

0

g(x)n dx→ ∞

261

as n→ ∞.

262

K232

Observe that, if η is fixed with 1/3 > η > 0,∫ 1

−1

Sm,η(x) dx ≥∫ η1/2/2

−η1/2/2

Sm,η(x) dx

≥∫ η1/2/2

−η1/2/2

(

1 +3η

4

)

dx→ ∞

but, if |x| ≥ 2η1/2, then

|Sm,η(x)| ≤ (1− 3η)n → 0

as m→ ∞. Thus Tm,η(t) → 0 uniformly on [−1,−2η1/2] ∪ [2η1/2, 1] asm→ ∞.

(vi) If F : [−1, 1] → R is continuous, then it is uniformly continuousand, given any ǫ > 0, we can find an integer N such whenever |t− s| ≤N−1 and t, s ∈ [−1, 1] we have |F (t) − F (s)| < ǫ/4. Thus if h is thesimplest piecewise linear function on [−1, 1] with h(r/N) = F (r/N)for N ≤ r ≤ N we have |h(t)− F (t)| < ǫ for all t ∈ [−1, 1].

263

K233

Define F : [−1, 1] → R by

F (x) =

f(x) if x ∈ [0, 1]

f(−x) if x ∈ [−1, 0].

Since F is continuous, we can find polynomials Qn with Qn(t) → F (t)uniformly on [−1, 1]. Set Rn(t) = (Qn(t) +Qn(−t))/2. Then

|Rn(t)− F (t)| = |(Qn(t) +Qn(−t))/2− (F (t) + F (−t))/2|≤ |Qn(t)− F (t)|/2 + |Qn(−t)− F (−t)|/2 → 0

uniformly on [−1, 1]. Also, since Rn(t) = Rn(−t), we can find a poly-nomial Pn with Pn(t

2) = Qn(t).

Let g : [−1, 1] → R be given by g(x) = x. Then

|g(1)−Q(12)|+ |g(−1)−Q((−12))| = |g(1)−Q(1)|+ |Q(1))− g(−1)|≥ ||g(1)− g(−1)| = 2.

264

K234

Write Xn(t) = Y1 + Y2 + · · · + Yn where Yj = 1 if the jth trial isa success and Yj = 0 otherwise. Observe that (since the expectationof the sum of random variable is the sum of the expectations, and thevariance of the sum of independent random variables is the sum of thevariances)

EXn(t) =n∑

j=1

EYj = nt

varXn(t) =n∑

j=1

var Yj = nt(1− t).

We know that there exists an M such that |f(t)| ≤ M for all t ∈[0, 1] and we know that given ǫ > 0 there exists a δ > 0 such that|f(t)−f(s)| ≤ ǫ whenever t, s ∈ [0, 1] and |t−s| ≤ δ. By Tchebychev’sinequality

Pr

(∣

∣

∣

∣

Xn(t)

n− t

∣

∣

∣

∣

≥ δ

)

= Pr

(∣

∣

∣

∣

Xn(t)

n− E

(

Xn(t)

n

)∣

∣

∣

∣

≥ δ

)

=var(Xn(t)/n)

δ2

=t(1− t)

nδ2

≤ 1

nδ2

so, writing IA for the indicator function of A,

Ef

(

Xn(t)

n

)

− f(t)E

((

f

(

Xn(t)

n

)

− f(t)

)

I[t−δ,t+δ]∪[0,1]

(

Xn(t)

n

))

+ E

((

f

(

Xn(t)

n

)

− f(t)

)

I[0,1]\[t−δ,t+δ]

(

Xn(t)

n

))

≤ ǫPr

(∣

∣

∣

∣

Xn(t)

n− t

∣

∣

∣

∣

≤ δ

)

+ 2M Pr

(∣

∣

∣

∣

Xn(t)

n− t

∣

∣

∣

∣

> δ

)

≤ ǫ+2M

nδ2.

Allowing n → ∞ and observing that ǫ is arbitrary gives the requiredresult.

We have

pn(t) =

n∑

r=0

f( r

n

)

Pr(Xn(t) = r) =

n∑

r=0

f( r

n

)

(

n

r

)

tr(1− t)n−r,

so pn is polynomial of degree n.

265

Since(

n

r

)

tr(1− t)n−r ≥ 0,

n∑

r=0

(

n

r

)

tr(1− t)n−r = 1

andn∑

r=0

(

n

r

)

r

ntr(1− t)n−r = t

we can use Jensen’s inequality (or we could just quote K144 (iv)).

266

K235

(i) Suppose that fn does not converge uniformly to f . Since the fn(x)are decreasing this means that there exists a δ > 0 and xn ∈ I suchthat fn(xn) > f(x)+δ. By the theorem of Bolzano–Weierstrass we canfind a subsequence n(j) → ∞ and an x ∈ I such that xn(j) → x. Sincefn(x) → f(x) we can find an N such that f(x) + δ/2 > fN(x) Sincef − fN is continuous we can find an η > 0 such that δ > fN (y)− f(y)for all y ∈ I with |y − x| < η. Choosing a J such that |xn(J) − x| < ηwe obtain a contradiction.

(ii) False without (a). Let I = [0, 1], fn(1/r) = 1 for all r ≥ n,fn(x) = 0 otherwise.

False without (b). Let I = [0, 1], fn(x) = xn.

False without (c). Witch’s hat.

False without (d). Let I = (0, 1), fn = 1/(nx).

(iii) Easy to check n = 0.

If t ∈ [−1, 1] and 0 ≤ pn−1(t) ≤ pn(t) ≤ |t| then

pn+1(t)− pn(t) = pn(t)− pn−1(t) +1

2(pn(t)

2 − pn−1(t)2)

=

(

1− pn(t) + pn(t)

2

)

(pn(t)− pn−1(t))

≤ (1− |t|)(pn(t)− pn−1(t) ≥ 0

and

pn+1(t) ≤t2 + |t|

2=

|t|(|t|+ 1)

2≤ |t|.

The result follows by induction.

Since pn(t) is increasing and bounded above, pn(t) → g(t) say. Wehave

0 = pn(t)− pn−1(t)−1

2(t2 − pn−1(t))

→ g(t)− g(t)− 1

2(t2 − g(t)) = −1

2(t2 − g(t))

so (since g(t) ≥ 0) g(t) = |t|.Setting f = −g, fn = −pn we see that part (i) applies and pn(t) → |t|

uniformly as n→ ∞.

(v) By induction pn is a polynomial.

(vi) Can extend to any bounded closed set I in Rn or any metricspace satisfying the Bolzano–Weierstrass condition.

267

K236

(ii) Proceed inductively. Write pn(x) = xn. Set s0 = 1. When s0, s1,. . . , sn−1 have been defined set

sn = pn −n−1∑

j=0

〈pn, sj〉〈sj, sj〉

.

(This is the Gramm-Schmidt technique.)

(iii) To prove uniqueness observe that

n∑

j=0

bjxj =

n∑

j=0

ajxj

implies bn = an and use induction. A similar induction proves exis-tence.

Observe that Q =∑n−1

j=0 cjxj .

(iv) Write Pn for the set of polynomials of degree or less. Since thepolynomials are uniformly dense, we have, using (iii),

infa∈Rn+1

∥

∥

∥

∥

∥

n∑

j=0

ajsj − f

∥

∥

∥

∥

∥

2

= infP∈Pn

‖P − f‖2

≤ (b− a)1/2 infP∈Pn

‖P − f‖∞ → 0

as n→ ∞.

(v) We have

‖n∑

j=0

bjsj − f‖22 = ‖f‖22 +n∑

j=0

(b2j‖sj‖22 − 2bj〈f, sj〉)

= ‖f‖22 +n∑

j=0

(bj‖sj‖2 − 〈f, sj〉‖sj‖−12 )2 −

n∑

j=0

(〈f, sj〉‖sj‖−12 )

=

∥

∥

∥

∥

∥

f −n∑

j=0

〈f, sj〉‖sj‖22

sj

∥

∥

∥

∥

∥

2

2

+

n∑

j=0

‖sj‖22(

bj −〈f, sj〉‖sj‖22

)2

=

∥

∥

∥

∥

∥

f −n∑

j=0

f(j)sj

∥

∥

∥

∥

∥

2

2

+n∑

j=0

‖sj‖22(bj − f(j))2

The result of the first sentence can now be read off.

The result of the second sentence follows from part (iv).

268

(vi) Observe that sn(t)q(t)w(t) is single signed (that is always posi-tive or always negative) and only zero at finitely many points. Thus

∫ b

a

sn(t)q(t)w(t) dt 6= 0

and, by part (iii), q must be a polynomial of degree at least n. Thusk ≥ n and we have exhibited n distinct zeros t1, t2, . . . , tn in (a, b).Since sn can have at most n distinct zeros we are done.

269

K237

Observe (by Leibniz rule or induction or otherwise) that, if n > r ≥0, then

dr

dxr(x2 − 1)n has (x2 − 1) as a factor and so vanishes when

x = ±1. Thus, if n ≥ m ≥ 0, integrating by parts n times gives∫ 1

−1

dn

dxn(x2 − 1)n

dm

dxm(x2 − 1)m dx

=

[

dn−1

dxn−1(x2 − 1)n

dn

dxn(x2 − 1)n

]1

−1

−∫ 1

−1

dn−1

dxn−1(x2 − 1)n

dm+1

dxm+1(x2 − 1)m dx

= −∫ 1

−1

dn−1

dxn−1(x2 − 1)n

dm+1

dxm+1(x2 − 1)m dx

= . . .

= (−1)n∫ 1

−1

(x2 − 1)ndm+n

dxm+n(x2 − 1)m dx.

Thus∫ 1

−1

dn

dxn(x2−1)n

dm

dxm(x2−1)m dx =

0 if n 6= m

(2n)!∫ 1

−1(1− x2)n dx if n = m.

Making the substitution x = sin θ, we have∫ 1

−1

(1− x2)n dx =

∫ π/2

−π/2

cos2n+1 θ dθ.

If In =∫ π/2

−π/2cosn θ dθ, and n ≥ 2 then integration by parts gives

In =

∫ π/2

−π/2

cosn−1 θ cos θ dθ

=[

cosn−1 θ sin θ]π/2

−π/2+ (n− 1)

∫ π/2

−π/2

cosn−2 θ sin2 θ dθ

= (n− 1)

∫ π/2

−π/2

cosn−2 θ(1− cos2 θ) dθ

= (n− 1)In−2 − (n− 1)In

so nIn = (n− 1)In−2. Since I1 = 2 we have

γn = (2n)!I2n+1 = (2n)!× 2×n−1∏

j=0

2n− 2j

2n+ 1− 2j=

2n+1(n!)2n

(2n+ 1)!.

By expanding (x2 − 1)n and differentiating n times we see that pnis a polynomial of degree n with leading coefficient (2n)!/(n!). Takecn = (n!)/(2n)!.

270

There are several ways of doing the last sentence including integrat-ing by parts along the lines of the first part of the question.

271

K238

(i) Observe that P −∑nj=1 P (xj)ej is a polynomial of degree at most

n− 1 which vanishes at the n points xk and so must be the zero poly-nomial.

aj =

∫ 1

−1

ej(t) dt.

(ii) Long division (more exactly, induction on the degree k of p)shows that, if q is a polynomial of degree r, any polynomial p of degreek can be written as p = sq+r where s has degree at most k−s if k ≥ sand is zero, otherwise, and r has degree at most k − 1).

Observe that∑n

j=1 ajQ(xj)pn(xj) =∑n

j=1 0 = 0.

(iv) Let p(t) =∏n

j=1(t− yj). Then p is polynomial of degree exactlyn with leading coefficient 1 such that

∫ 1

1

p(t)q(t) dt =n∑

j=1

bjp(yj)q(yj) = 0

whenever q has degree at most n−1 (and so, in particular, when q = prwith 0 ≤ r ≤ n− 1). Thus p = pn.

272

K239

(i) The argument of K236 shows that sn + λsn−1 must change signat least n− 1 times in (a, b). It changes sign only at zeros of odd orderand has exactly n zeros if multiple roots are counted multiply so, bycounting, it must have n simple zeros in (a, b).

(ii) Let the common root be y. We have P (x) = (x − y)p(x) andQ(x) = (x − y)q(x). If p(y) = 0, then P has a multiple root at 0 andwe set µ = 1, λ = 0. If not then (q(y)/p(y))p− q has a root at y and(q(y)/p(y))P −Q has a multiple root at y.

If sn and sn−1 have a common root at y then there exist λ and µboth non-zero (since sn and sn−1 do not have multiple roots) such thatµsn+λsn−1 and so sn+µ

−1λsn−1 has multiple roots which is impossible.

(iii) Without loss of generality we may suppose that P has no realroots in the open interval (x, y). Thus both P and Q are single signedand, without loss of generality, we may suppose that P and Q arepositive in the open interval (x, y). Since a continuous function on aclosed bounded interval is bounded and attains its bounds we can finds1, s2, s3 ∈ [x, y] such that

Q(s1) ≥ Q(s) ≥ Q(s2) > 0 and P (s3) ≥ P (s) ≥ 0

for s ∈ [x, y].

It follows that the set of real numbers

E = t ≥ P (s3)/Q(s1) : P (s)− tQ(s) ≥ 0 for some s ∈ [x, y]is a non-empty and bounded above by P (s3)/Q(s2). Thus E has asupremum λ′ say.We observe that, for each n ≥ 1 there exists an wn ∈[x, y] such that

P (wn)− (λ′ − n−1)Q(wn) ≥ 0.

By the Bolzano–Weierstrass theorem there exists a sequence n(j) → ∞and an w ∈ [x, y] such that wn(j) → w. By continuity

P (w)− λ′Q(w) = 0.

We note that this implies w ∈ (x, y). Since

P (t)− λ′Q(t) ≤ 0

for all t ∈ [x, y] we see that w is a multiple root. Set λ = −λ′.Last sentence much as for (ii).

273

K240

Fix M ≥ 1 temporarily. Observe that (if |x| < M/2) we have∣

∣

∣

∣

1

(x− n)−2

∣

∣

∣

∣

≤ 4

n2

so by Weierstrass M-test∑N

n=M(x−n)2 converges uniformly. We write

FM,N(x) =

N∑

n=M

1

(x− n)2+

1

(x+ n)2.

Since the uniform limit of continuous functions is continuous, fM iscontinuous.

∣

∣

∣

∣

−2

(x− n)3

∣

∣

∣

∣

+

∣

∣

∣

∣

−2

(x+ n)3

∣

∣

∣

∣

≤ 8

n3

so by theWeierstrassM test f ′M,N converges uniformly on [−M/2,M/2]

to limit gM say. Thus fM is differentiable with derivative gM on(−M,M). Since M was arbitrary we are done.

If y /∈ Z we can find a δ > 0 such that B(y, 2δ)∩Z = ∅ and an integerM such that |y|+ 2δ < M/4. Thus since the sum of two differentiablefunctions is differentiable

F (x) = FM (x) +M−1∑

n=0

1

(x− n)2+

1

(x+ n)2

is well defined, continuous and differentiable on B(y, 2δ). Since y waschosen arbitrarily, we are done.

(iii) Fix x /∈ Z.

0 =

N∑

n=−N

1

(x− n)2−

N+1∑

n=−N+1

1(

(x+ 1)− n)2 → F (x)− F (x+ 1)

as N → 0, so F (x) = F (x+ 1).

274

K241

(i) Since g is continuous on [0, 1] it is bounded, with |g(x)| ≤ K sayfor all x ∈ [0, 1]. Now use periodicity.

If we set f(x) = x−1 for x ∈ (0, 1] and define f(x) = f(x − n)whenever x− n ∈ (0, 1] then f is periodic but not bounded. If we seth(x) = x for all x then h is continuous but not bounded.

(ii) Observe that

|g(x)| = 1

4

(

∣

∣

∣g(x

2

)∣

∣

∣+

∣

∣

∣

∣

g

(

x+ 1

2

)∣

∣

∣

∣

)

≤ K

2

and repeat.

(iii) Use L’Hopital or power series manipulation

π2

(sin πx)2=

π2

(πx− (πx)3/3! + ǫ1(x)x3)2

=1

x2(1− (πx)2/6 + ǫ2(x)x3)2

=1

x2

(

1 +2(πx)2

6+ ǫ3(x)x

3

)

=1

x2+π2

3+ ǫ3(x)x

where ǫj(x) → 0 as x→ 0.

Observe that g(x) → g(0) as x→ 0 so g is continuous at 0. Now useperiodicity.

(iv) Observe that, if 2x /∈ Z, then

g(x

2

)

+ g

(

x+ 1

2

)

=

∞∑

n=−∞

(

1

(x2− n)2

+1

(x2− n+ 1

2)2

)

− π2(cosec2(πx/2)− cosec2(π(x+ 1)/2)

=∞∑

n=−∞

(

4

(x− 2n)2+ (x− 2n+ 1)2

)

− π2 cos2(πx/2)− sin2(πx/2)

cos2(πx/2) sin2(πx/2)

= 4g(x).

The result holds for all x by continuity.

By (i) and (ii) g(x) = 0 for all x. Thus if x /∈ Z we have

∞∑

n=−∞

1

(x− n)2= π2 cosec2(πx)

275

and looking at x = 0,

π2

3= f1(0) = 2

∞∑

n=1

1

n2.

276

K242

(i) f ′n(x) = x(x2 + n2)−1/2. Thus

f ′n(x) →

1 if x ∈ (0, 1],

0 if x = 0,

−1 if x ∈ [−1, 0).

We also have∣

∣fn − |x|∣

∣ =n−2

|x|+ (x2 + n2)1/2≤ 1

n→ 0

uniformly on [−1, 1].

(ii) We could take gn(x) = nmax(1 − |x − (2n)−1|, 0) and fn(x) =∫ x

0gn(t) dt. fn tends pointwise to a (continuous from below at 0) Heav-

iside function.

(iii) Demand uniform convergence of the derivatives.

277

K243*

No comments.

278

K244

Observe that

|t|−1/2v(t) = |t|1/2 v(t)− v(0)

t→ 0× v′(0) = 0.

(ii) Standard results (chain rule, product rule etc) show that g iscontinuous except possibly at points (x′, 0). Observe that since u isnon-zero only a bounded closed set we can find aK such that |u(t)| ≤ Kfor all t. Thus, if y 6= 0

|g(x, y)− g(x′, 0)| = |g(x, y)| ≤ K|y−1/2v(y)| → 0

as y → 0. Thus g is everywhere continuous.

(iii) Standard results (chain rule, product rule etc) show that g hascontinuous partial derivatives on the open set

A = (x, y) ∈ R2 : x, y 6= 0.Also we know that g is identically zero on the open set

B = (x, y) ∈ R2 : x2|y| < 1 ∪ (x, y) ∈ R2 : x2|y| > 3and so, trivially, has continuous partial derivatives there. The onlypoint not in A ∪ B is (0, 0). But g(x, 0) = 0 for all x and g(0, y) = 0for all y so partial derivatives exist at (0, 0)

(iv) If y 6= 0, and |y| ≤ 1/9

G(y) =

∫ 1

0

g(x, y) dx

= |y|−1/2v(y)

∫ 1

0

u(x|y|−1/2) dx

= v(y)

∫ y1/2

0

u(s) ds = v(y),

and G(0) = 0 = v(0) so G′(0) = v′(0).

Since (x, 0) ∈ B for x 6= 0, we have g,2(x, 0) = 0 for x 6= 0 (andfor the reasons given in the last sentence of (iii), g,2(0, 0) = 0 as well).

Thus∫ 1

0g,2(x, 0) dx = 0.

We do not have g,2 continuous at the origin. (More vividly, look atthe behaviour of g(x, y) when x is fixed and small and y runs from x2/9to x2.)

279

K245

By comparison,∫∞0

|fn(t)| dt converges and so, since absolute con-

vergence implies convergence,∫∞0fn(t) dt converges. Given ǫ > 0, we

can find an X such that∫∞Xg(t) dt ≤ ǫ/2 and then

∣

∣

∣

∣

∫ ∞

X

fn(t) dt

∣

∣

∣

∣

≤∫ ∞

X

|fn(t)| dt ≤∫ ∞

X

g(t) dt ≤ ǫ/2.

Similarly∫∞0f(t) dt converges and

∣

∣

∣

∣

∫ ∞

X

f(t) dt

∣

∣

∣

∣

≤ ǫ/2.

Thus∣

∣

∣

∣

∫ ∞

0

fn(t) dt−∫ ∞

0

fn(t) dt

∣

∣

∣

∣

≤∣

∣

∣

∣

∫ ∞

X

fn(t) dt−∫ ∞

X

fn(t) dt

∣

∣

∣

∣

+

∣

∣

∣

∣

∫ ∞

X

fn(t) dt

∣

∣

∣

∣

+

∣

∣

∣

∣

∫ ∞

X

f(t) dt

∣

∣

∣

∣

≤∣

∣

∣

∣

∫ ∞

X

fn(t)− f(t) dt

∣

∣

∣

∣

+ ǫ→ ǫ

as n→ 0 and so, since ǫ was arbitrary, the result follows.

The proof above works under the conditions of the last sentence.

280

K246

(ii) If follows by comparison and (using Dini’s theorem) dominatedconvergence (K245).

To prove only if, observe that if any of the∫∞0fn(t) dt fails to con-

verge then so does∫∞0f(t) dt. If

∫∞0fn(t) dt converges to an then we

can find Xn such that∫ Xn

0

fn(t) dt ≥ an − 2−n

and so∫ Xn

0

f(t) dt ≥ an − 2−n

. Thus if∫∞0f(t) dt converges so do all the

∫∞0fn(t) dt and the val-

ues an of the integrals form an increasing sequence bounded above soconverging to a limit A. We have

A ≥∫ ∞

0

fn(t) dt ≥∫ X

0

fn(t) dt→∫ X

0

f(t) dt

by part (i). Since X is arbitrary∫∞0f(t) dt converges. The fact that

∫∞0fn(t) dt→

∫∞0f(t) dt now follows from the ‘if’ part.

281

K247

(i) Observe that

0 ≤ 2−jdj(x, y)

1 + dj(x, y)≤ 2−j

so the sum converges and d is well defined. To prove the triangleinequality, observe that, if d is metric,

d(x, y)

1 + d(x, y)+

d(y, z)

1 + d(y, z)− d(x, z)

1 + d(x, z)

= A(

d(x, y)(1 + d(y, z))(1 + d(x, z)) + d(y, z)(1 + d(x, y))(1 + d(x, z))

− d(x, z)(1 + d(x, y))(1 + d(y, z)))

= A(

d(x, y) + d(y, z)− d(x, z) + d(x, y)d(x, z) + d(y, z)d(x, z)

+ d(x, z)d(x, y)d(y, z))

≥ 0

for some A = A(x, y, z) > 0.

(ii) Suppose that xn is Cauchy for d. Then, since 2jd(a, b) ≥ dj(a, b),the sequence xn is Cauchy for dj so there exists a zj ∈ X with dj(xj , zj) →0 for each j. We observe that dj+1(xn, zj+1) ≥ dj(xn, zj+1) so dj(xj , zj+1) →0 and by the uniqueness of limits zj+1 = zj . Thus

z1 = z2 = · · · = zj = · · · = z, say.

Since

0 ≤ d(xn, z) ≤N∑

j=1

2−jdj(xn, z) +

∞∑

j=N+1

2−j

=

N∑

j=1

2−jdj(xn, z) + 2−N → 2−N

and N is arbitrary, d(xn, z) → 0 as n→ 0.

(iii) Observe that any Cauchy sequence for (X, d∗) is either even-tually constant (and then converges to that constant value) or is asubsequence of 1/n and so converges to 0∗. Thus (X, d∗) is complete.

However 1/n is a Cauchy sequence in (X, d) which does not converge.(Observe that d(1/n, 1/m) ≥ 1/m when n ≥ 2m and d(1/n, 0∗) =d(1/n, 0∗∗) = 2.)

Take d1 = d∗, dj = d∗∗ otherwise.

(iv) Use the general principle of uniform convergence. If fj is Cauchythen there is an fj such that f (j) → fj . Now use the theorem oninterchanging derivative and limit.

282

(v) Could take

d(f, g) =

∞∑

j=1

2−j sup|z|≤1−j−1 |f(z)− g(z)|1 + 2−j sup|z|≤1−j−1 |f(z)− g(z)| .

283

K248*

No comments.

284

K249*

No comments

285

K250

(i) Suppose that the result is false for some j. Then we can find

gk ∈ D such that ‖gk‖ ≤ 1/k but ‖g(j)k ‖∞ ≥ 1. Since ‖gk‖ → 0 but g(j)k

does not converge uniformly to 0 as k → ∞ we have a contradiction.

(ii) Use the fact that we have a norm.

286

K251*

No comments.

287

K252*

No comments.

288

K253

(v) Observe that, given any ǫ > 0, we can find

0 = x0 < y0 < x1 < y1 < · · · < xn−1 < yn−1 < xn < yn = 1

such that f(yj−1) = f(xj) for 1 ≤ j ≤ n and∑n

j=0(yj − xj) ≤ ǫ. Thus,

if |f(x)−f(y)| ≤ K|x−y| for all x, y ∈ [0, 1], we have 1 = f(1)−f(0) ≤Kǫ which is impossible if ǫ is chosen sufficiently small.

289

K254

(i) Observe that rn(x) = xn is a strictly increasing, differentiablefunction. It thus has a differentiable inverse r1/n whose derivative isgiven by the rule for differentiation of inverses

(r−1n )′(x) =

1

r′n(r−1n (x))

=r−1n (x)

rn(nr−1n (x))

=r−1n (x)

nx.

(ii) Take nn′th powers of both sides.

(iii) (f) is trivial but must be checked. rn(x) = (r1(x))n = xn.

(iv) r−α(x) = (rα(x))−1.

290

K255

The key here is bounding |rα(x)− rβ(x)| with α, β ∈ Q. Since

|rα(x)− rβ(x)| = rα(x)|1− rβ−α(x)|.we have to bound rα(x) and |1− rγ(x)| (again for γ ∈ Q).

If N ≥ α we have N − α > 0 so rN−α(x) ≥ 1 so rN(x) ≥ rα(x) ≥ 0and so

0 ≤ rα(x) ≤ rN(x) = xN ≤ RN

for all 0 < x ≤ R.

If γ ≥ 0 then rγ(1) = 1 and rγ is increasing so

|1− rγ(x)| ≤ max(

rγ(R)− 1, 1− rγ(R−1))

for all R−1 ≤ x ≤ R. We obtain a similar formula for γ ≤ 0 andcombining this with the previous paragraph we obtain

⋆ |rα(x)− rβ(x)| ≤ RN max(

r|α−β|(R)− 1, 1− r|α−β|(R−1))

whenever N ≥ α and R ≥ x ≥ R−1.

Since r1/n(x) → 1 as n→ 0 (proof by, e.g. observing that (1+ δ)n ≥1+nδ so 1 ≤ r1/n(1+ δ) ≤ 1+ δ/n for δ > 0) it follows that rγ(x) → 1as γ → 0 through values of γ ∈ Q (proof by e.g. observing that if1/n ≥ γ ≥ 0 and x ≥ 1 then r1/n(x) ≥ rγ(x) ≥ 1).

(ii) Observe that r′αn→ r′α uniformly on [a, b].

291

K256

(v) Sincef ′(y) = y1−1/n − y1−1/(n+1) ≥ 0

for y ≥ 1, f is increasing on [1,∞). But f(1) = 0 so f(y) ≥ 0 fory ≥ 1. Thus

n(y1/n − 1) ≥ (n+ 1)(y1/(n+1) − 1) ≥ 0

and so, since a decreasing sequence bounded below tends to a limit,n(y1/n − 1) → L.

(iii) If 1/n ≥ x ≥ 1/(n+ 1) we also have

P (x)− P (0)

x≥ P (1/(n+ 1))− P (0)

x

≥ n

n+ 1× P (1/(n+ 1))− P (0)

1/(n+ 1)→ 1× L = L

as n→ ∞.

(viii) To deal with 1 > a > 0 repeat argument or use chain rule onthe maps t 7→ −t and t 7→ at.

If La = 0 then Pa is constant, if La > 0 then Pa is strictly increasingand if La < 0 then Pa is strictly decreasing. It follows that L1 = 0,La > 0 for a > 1 and La < 0 for a > 0.

(ix) By the chain rule,

d

dt(aλ)t =

d

dtaλt = λLaa

λt = λLa(aλ)t

so Laλ = λLa.

Choose λ = L−12 and set e = 2L

−12 .

(x) We step outside the context of the question and observe thatet = exp t so

La = Lelog a = log aLe = log a.

292

K257

(i) Observe that

cosh x cosx =ex + e−x

2

eix + e−ix

2

=1

4

(

e21/2ωx + e2

1/2ω3x + e21/2ω5x + e2

1/2ω7x)

with ω = eiπ/4. Now use the Taylor expansion of each of the four terms.

(ii) Since cosh x cosx is even the Taylor series has only even powers.The coefficient of x2n isn∑

r=0

(−1)r

(2r)!(2n− 2r)!=

1

(2n)!

n∑

r=0

(

2n

2r

)

(−1)r

=1

(2n)!

(1 + i)2n + (1− i)2n

2=

0 if n is odd(−1)n/22n/2

(2n)!if n is even.

(iii) Set f(x) = cosh x cosx. Our main problem is to calculate f (n)(x)and then to bound it. Using Leibniz’s rule gives a calculation for f (r)(0)which resembles the calculation for (ii). Also

|f (n)(x)| ≤n∑

r=0

(

n

r

)

cosh x ≤ 2n coshR

for |x| ≤ R. This gives an estimate for the remainder

|Rn(f, x)| ≤(2R)n coshR

(n− 1)!

for |x| ≤ R (or something similar) so Rn(f, x) → 0.

(iv) Solve y(4) − y = 0 with y(0) = 1, y′(0) = y′′(0) = y(3)(0) = 0.Remember to check the radius of convergence of the resulting powerseries.

293

K258

Since∑∞

n=0 anzn has radius of convergence at least R, we know that

∑∞n=0 anz

n converges absolutely. Thus if |h| < R− |z0| we know that∞∑

n=0

n∑

r=0

|an|(

n

r

)

|z0|n−r|h|r =∞∑

n=0

|an|(|z0|+ |h|)n.

Write cn,r = |an|(

nr

)

|z0|n−r|h|r if r ≤ n, cn,r = 0 otherwise. By Fubini’stheorem for sums,

∞∑

n=0

∞∑

r=0

cn,r =

∞∑

r=0

∞∑

n=0

cn,r

so∞∑

n=0

an(z0 + h)n =

∞∑

n=0

n∑

r=0

an

(

n

r

)

zn−r0 hr

=

∞∑

r=0

∞∑

n=r

an

(

n

r

)

zn−r0 hr

=

∞∑

r=0

hr∞∑

n=r

(

n

r

)

zn−r0

=

∞∑

r=0

brhr

with

br =

∞∑

m=0

(

n+m

m

)

zm0 .

Thus∑∞

r=0 brzr converges for |z| < R − |z0| and

∑∞n=0 an(z0 + h)n =

∑∞r=0 brh

r.

Let 2δ = R− |z0|. Since∑∞

r=0 brδr converges we can find an M such

that |br| ≤Mδ−r. It follows that, if |w| ≤ δ/2,∣

∣

∣

∣

g(z0 + w)− g(z0)

w− b1

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∞∑

r=2

brwr−1

∣

∣

∣

∣

∣

≤ |w|∞∑

k=0

bk−2|w|k

≤ |w|Mδ−2

∞∑

k=0

(w|δ−1)k

≤ |w|Mδ−2

∞∑

k=0

2−k

≤ 2Mδ−2|w| → 0

294

as |w| → 0.

295

K259

(i) We could set A0 = 4, B0 = 1, C0 = π/2, Bn = 2n+1(

(n!)2 + 1 +∑n−1

r=0 |f(n)r (x)|, An = Bn−1

n and take Cn so that sin(n)Cn = 1 for n ≥ 1.

Observe that|f (r)

n (x)| ≤ 2−n−1

for all n > r and all x so, by the Weierstrass M-test,∑∞

n=r+1 f(r)n (x)

converges uniformly and∣

∣

∣

∣

∣

∞∑

n=r+1

f (r)n (x)

∣

∣

∣

∣

∣

≤ 2−r.

Thus∑∞

n=1 f(r)n (x) converges uniformly for each r and f =

∑∞n=1 fn is

infinitely differentiable with

f (r)(x) =

∞∑

n=1

f (r)n (x).

Also

f (r)(0) ≥ f (r)(0)−r−1∑

n=1

|f (r)n (0)| −

∞∑

n=r+1

|f (r)n (0)| ≥ (n!)2.

296

K260

(i) Just observe that

∞∑

n=0

|anzn| ≤∞∑

n=0

|an|(R− ǫ)n.

if |z| ≤ R− ǫ.

(ii) Since |nanzn| ≥ |anzn|, the radius of convergence does not in-crease. Now observe that, if |z| < R and we choose r = (2R+|z0|)/3, wehave

∑∞n=0 anr

n convergent so anrn → 0 so we can find an M ≥ 0 with

|anrn| ≤ M for all n and so |an| ≤ Mr−n. Now set ρ = (R + 2|z0|)/3.Since r > ρ, n(ρ/r)n → 0 as n → 0 so there exists a K with |nan| ≤Kρ−n for all n ≥ 0. Thus |nanzn| ≤ K(|z|/ρ)n and, by comparison,∑∞

n=0 nanzn converges absolutely and so converges.

Thus∑∞

n=0 nanzn has radius of convergence R and so

∑∞n=1 nanz

n−1

has radius of convergence R. The result for∑∞

n=2 n(n − 1)anzn−2 fol-

lows.

(iii) and (iv). Observe that

n(n− 1)

(

n− 2

j

)

=n(n− 1)(n− 2)!

j!(n− 2− j)!≥ n!

j!(n− j)!=

(

n

j

)

.

so

|(z + h)n − zn − nzn−1h| =∣

∣

∣

∣

∣

n−2∑

j=0

(

n

j

)

zjhn−j

∣

∣

∣

∣

∣

≤n−2∑

j=0

(

n

j

)

|z|j |h|n−j

≤ n(n− 1)|h|2n−2∑

j=0

(

n− 2

j

)

|z|j |h|n−j−2

≤ n(n− 1)(|z|+ |h|)n−2.

Suppose δ > 0 and |z|+ |h| < r − δ. By parts (ii) and (i), there existsan A(δ) independent of h and z such that

∞∑

n=2

n(n− 1)|an|(|z|+ |h|)n ≤ A(δ).

297

Thus∣

∣

∣

∣

∣

(

N∑

n=0

an(z + h)n −N∑

n=0

anzn

)

− h

N∑

n=1

nanzn−1

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

N∑

n=0

an(

(z + h)n − zn − nzn−1h)

∣

∣

∣

∣

∣

≤ |h|2N∑

n=2

n(n− 1)|an|(|z|+ |h|)n

≤ A(δ)|h|2

and, allowing N → ∞,∣

∣

∣

∣

∣

( ∞∑

n=0

an(z + h)n −∞∑

n=0

anzn

)

− h

∞∑

n=1

nanzn−1

∣

∣

∣

∣

∣

≤ A(δ)|h|2

(v) Divide by h and allow |h| → 0.

298

K261

We try, formally, y =∑∞

j=0 ajxj obtaining, on substituting in the

Legendre equation,

(1− x2)∞∑

j=2

j(j − 1)ajxj−2 − 2x

∞∑

j=1

jajxj−1 + l(l + 1)

∞∑

j=0

ajxj = 0.

Rearranging gives∞∑

j=0

(

(

− j(j − 1)− 2j + l(l + 1))

aj + (j + 2)(j + 1)aj+2

)

xj = 0

so we must have(

(j + 1)(j + 2)− l(l + 1))

aj = (j + 2)(j + 1)aj+2

that is

aj+2 =(j + l + 2)(j − l)

(j + 2)(j + 1)aj .

This gives

y(x) = a0

∞∑

r=0

∏rk=0(l − 2k)(l + 2k + 1)

(2r)!x2r

+ a1

∞∑

r=0

∏rk=0(l − 2k − 1)(l + 2k + 2)

(2r + 1)!x2r+1

where a0 and a1 can be chosen freely. When l is not a positive integerwe observe that

(j + l + 2)(j − l)

(j + 2)(j + 1)=

(1 + (l + 2)j−1)(1− j−1)

(1 + 2j−1)(1 + j−1)→ 1

so careful application of the ratio test to the sums of even and oddpowers shows that the power series has radius of convergence 1 andjustifies our formal manipulation for |x| < 1.

If l is positive integer we see that one of the two infinite sums inthe first paragraph actually terminates to give a polynomial of degreel. The other does not terminate and the argument above shows it hasradius of convergence 1,

If v(x) = (x2 − 1)n then v′(x) = 2nx(x2 − 1)n−1 so

(x2 − 1)v′(x) = 2nxv(x)

as required. Using Leibniz’s rule we have

dn+1

dx(1− x2)v′(x)

= (1− x2)v(n+2)(x)− 2(n+ 1)xv(n+1)(x) + n(n+ 1)v(n)(x)

299

anddn+1

dx2nxv′(x) = 2nxv(n+1)(x) + 2n(n+ 1)v(n)(x).

Thus differentiating the final equation of the first sentence of this para-graph n + 1 times gives

(1− x2)v′′n(x)− 2xv′n(x) + n(n+ 1)v(x) = 0

as required. Observe that vn is the pn of K237.

300

K262

(this is a special case of Bessel’s equation.)

We try, formally, w =∑∞

j=0 ajzj obtaining, on substituting in the

given equation,

z2∞∑

j=2

j(j − 1)ajzj−2 + z

∞∑

j=1

jajzj−1 + (z2 − 1)

∞∑

j=0

ajzj = 0

that is to say

a0 +∞∑

j=2

(

− (j − 1)(j + 1)aj − aj−2

)

zj = 0

Thus a0 = 0 and

aj = − aj−2

(j − 1)(j + 1)

for j ≥ 2. It follows that a2j = 0 for all j ≥ 0 and a2j+1 = a1(−1)j

22j(j!)2(j+1).

Thus

w(z) = a1

∞∑

j=0

(−1)j

22j(j!)2(j + 1)z2j+1

Since1

(j − 1)(j + 1)→ 0

the ratio test shows that our power series has infinite radius of conver-gence so our solution is valid everywhere.

We note that we can only choose one constant a1 freely. Observethat setting z = 0 in the original equation shows that automaticallyw(0) = 0.

Let us try and find a solution w =∑∞

j=0 bjzj for the equation

z2d2w

dz2+ z

dw

dz+ (z2 − 1)w = z2.

We obtain

b0 + (3b2 − b0 − 1)z3∞∑

j=3

(

(j − 1)(j + 1)bj − bj−2

)

zj = 0

so, equating coefficients, b0 = 0, b2 = 1/3 and

bj = − bj−2

(j − 1)(j + 1)

for j ≥ 3. Thus

w(z) = b1W (z) +∑

j=1

(−1)j

(2j + 1)∏j−1

k=1(2k + 1)2z2j

301

with

W (z) =∑

j=0

(−1)j

22j(j!)2(j + 1)z2j+1.

(We can talk about a particular integral plus a complementary func-tion.) As before the solution is valid everywhere.

If we try and find a solution w =∑∞

j=0 cjzj for

z2d2w

dz2+ z

dw

dz+ (z2 − 1)w = z,

looking at the coefficient of z gives 0 = 1 which is impossible.

302

K263

By Lemma 11.81 we have (provided |x| < 1)

(1 + x)1/2 =

∞∑

n=0

anxn

with

an = (−1)n1

n!

n−1∏

j=0

(12− j).

We observe that (if z 6= 0)

|an+1zn+1|

|anzn|=

|12− n||z|n

→ |z|

so, by the ratio test,∑∞

n=0 anzn has radius of convergence 1. Since

power series can be multiplied within their circles of convergence

( ∞∑

n=0

anzn

)2

=

∞∑

n=0

cnzn

for |z| > 1 with

1 + x =∞∑

n=0

cnxn

for all |x| < 1 and x real. By the uniqueness of power series (or wecould use K268 below) we have c0 = c1 = 1 and cj = 0 otherwise so

( ∞∑

n=0

anzn

)2

= 1 + z

has radius of convergence infinity.

Take K > 1. Observe that (1 + z)−1/2 has power series radius ofconvergence 1 (proof much as above). Since (1 +K−1z)1/2 has powerseries has radius of convergence K the function (1+z)−1/2(1+K−1z)1/2

has power series radius of convergence at least 1. Since power seriesare continuous within their circle of convergence and

(1− z)−1/2(1 +K−1z)1/2

303

is not continuous at −1, the power series∑∞

n=0 cnzn of (1− z)−1/2(1 +

K−1z)1/2 has radius of convergence exactly 1. Now

|∞∑

n=0

anzn

∞∑

n=0

cnzn = (1 + z)1/2(1 + z)−1/2(1 +K−1z)1/2

= (1 +K−1z)1/2

=∞∑

n=0

anK−nzn

for |z| < 1. So we have an example of two power series of radiusof convergence 1 whose product has radius of convergence K for anyK > 1. An example with radius of convergence infinity was given in thefirst paragraph, An example with radius of convergence 1 is given byconsidering (1+ z)−1/2(1+ z)−1/2. General result obtained by rescaling(i.e. looking at

∑∞j=0 bjR

−jzj for appropriate R.

[For enthusiasts. Observe that (if we ignore convergence) to obtain∞∑

j=0

ujzj

∞∑

j=0

vjzj =

∞∑

j=0

wjzj

we merely have to solven∑

j=0

un−jvj = wn

Thus we can ensure that u2n = (2n)! and v2n+1 = (2n + 1)! (so∑∞

j=0 ujzj and

∑∞j=0 vjz

j certainly have radius of convergence 0 and∑∞

j=0wjzj is any series we please (so with any radius of convergence

we please).]

Repeating much the same arguments, if L > K > 1, then (1 +z)−1(1+K−1z)1/2 has power series with radius of convergence 1 and (1+z)(1+L−1z)1/2 has power series with radius of convergence L and theirproduct has power series with radius of convergence K. On the otherhand, ifK > L > 1 then (1+z)−1(1+L−1z)−1/2(1+K−1z)1/2 has powerseries with radius of convergence 1 and (1+ z)(1 +L−1z)1/2 has powerseries with radius of convergence L and their product has power serieswith radius of convergence K. If K > 1 then (1 + z)−1(1 +K−1z)−1/2

has power series with radius of convergence 1 and (1 +K−1z)−1/2 hashas power series with radius of convergence K and their product haspower series with radius of convergence K. Remaining cases (infiniteradius of convergence) dealt with similarly.

304

K264

If we write

Sn(t) =

n∑

j=1

aj(t),

thenq∑

j=p

λjaj(t) =

q∑

j=p

λj(Sj(t)− Sj−1(t))

= λq+1Sq(t) +

q∑

j=p

(λj − λj+1)Sj(t)− λpSp−1(t)

and so∥

∥

∥

∥

∥

q∑

j=p

λjaj(t)

∥

∥

∥

∥

∥

≤ λq+1K +

q∑

j=p

(λj − λj+1)K + λpK = 2λpK

for all q ≥ p ≥ 1 and so, by the general principle of uniform conver-gence,

∑∞j=1 λjaj(t) converges uniformly.

Observe that, since xn is a decreasing sequence,∣

∣

∣

∣

∣

q∑

j=p

bjxj

∣

∣

∣

∣

∣

≤ 2xp supq≥p

∣

∣

∣

∣

∣

q∑

j=p

bjxj

∣

∣

∣

∣

∣

≤ 2 supq≥p

∣

∣

∣

∣

∣

q∑

j=p

bjxj

∣

∣

∣

∣

∣

→ 0

as p→ ∞, so by the general principle of uniform convergence,∑∞

j=1 bjxj

converges uniformly.

We may thus integrate term by term to get

∫ 1

0

( ∞∑

n=0

bnxn

)

dx =∞∑

n=0

∫ 1

0

bnxn dx =

∞∑

n=0

bnn+ 1

.

If∑∞

n=0 bn/(n+1) converges, then the power series∑∞

n=0 bnxn/(n+1)

has radius of convergence at least 1 so the power series∑∞

n=0 bnxn has

radius of convergence at least 1 and so∑∞

n=0 bnxn converges uniformly

on [0, 1− ǫ] and

∫ 1−ǫ

0

( ∞∑

n=0

bnxn

)

dx =1−ǫ∑

n=0

∫ 1

0

bnxn dx =

∞∑

n=0

bn(1− ǫ)n+1

n+ 1

for all 1 > ǫ > 0. By the result of the last paragraph but one,∑∞

j=1 bjxj+1/(j + 1) converges uniformly on [0, 1]. The function

∞∑

j=1

bjxj+1

j + 1

305

is thus defined and continuous on [0, 1] so∞∑

j=1

bjxj

j + 1→

∞∑

j=1

bjj + 1

as x→ 1−. Thus∫ 1−ǫ

0

( ∞∑

n=0

bnxn

)

dx→∞∑

j=1

bjj + 1

as ǫ→ 0+.

Taking bj = (−1)j we know, by the alternating series test, that∑∞

j=0 bj/(j + 1) converges and∑∞

j=0 bjxj = (1 + x)−1 for 0 ≤ x < 1.

Also∫ 1

0

1

1 + xdx = [log(1 + x)]10 = log 2.

Taking b2j = (−1)j , b2j+1 = 0 we know, by the alternating seriestest, that

∑∞j=0 bj/(j + 1) converges and

∑∞j=0 bjx

j = (1 + x2)−1 for0 ≤ x < 1. Also

∫ 1

0

1

1 + x2dx =

[

tan−1 x]1

0=π

4.

Not good methods. Rate of convergence painfully slow.

Observe that log a loga b = log b∞∑

j=0

(−1)j+1 log2j+1 e =

∞∑

j=0

(−1)j+1

(j + 1) log 2=

log 2

log 2= 1.

306

K265

Can be done in several ways. Here is one. Write pn = Pr(X = n).Observe that, since pn ≥ 0 and

0 ≥ 1− tn

1− t= 1 + t+ · · ·+ tn−1 ≤ n

for 1 > t > 0, we haveN∑

j=0

pj1− tj

1− t≤∑M

j=0 pj −∑N

j=0 pjtj

1− t≤

M∑

j=0

jpj ≤ EX

for M ≥ N and soN∑

j=0

pj1− tj

1− t≤ φX(1)− φX(t)

1− t≤ EX

for all N ≥ 0 and all 0 < t < 1. Allowing t→ 1− we haveN∑

j=0

jpj ≤ lim inft→1−

φX(1)− φX(t)

1− t≤ lim sup

t→1−

φX(1)− φX(t)

1− t≤ EX.

Allowing N → ∞ gives the result.

307

K266

Just as in K82, if m ≥ n,∣

∣

∣

∣

∣

n∏

j=1

(1 + aj(z)) −m∏

j=1

(1 + aj(z))

∣

∣

∣

∣

∣

≤ exp

(

n∑

j=1

|aj(z)|)(

exp

(

m∑

j=n+1

|aj(z)|)

− 1

)

≤ exp

( ∞∑

j=1

Mj

)(

exp

( ∞∑

j=n+1

Mj

)

− 1

)

→ 0

as n→ ∞ so, by the general principle of uniform convergence, it followsthat

∏Nj=1(1 + aj(z)) converges uniformly.

308

K267

(i) We have|z2n−2| ≤ R2n−2

for all |z| ≤ R so, by the previous question, SN(z) converges uniformlyon D(R) = z : |z| < R. Since the uniform limit of continuousfunctions is continuous, S is continuous on D(R) for each R > 0 andso on all of C.

(ii) Observe that (if z is not an integer)

SN (z + 1)

SN(z)=

(N !)2∏N

r=−N(r − (1 + z))

(N !)2∏N

r=−N(r − z)

=N + 1− z

−N − z=

1− (z − 1)/N

−1− z/N→ −1.

Thus S(z + 1) = −S(z) for z not an integer. Extend to all z bycontinuity.

Recall from K82 that, if∑∞

n=1 |an| converges, and an 6= −1 for alln then

∏∞n=1(1 + an) 6= 0. Thus, if N > |z| we know that

∏∞n=N(1 −

z2n−2) 6= 0 and so S(Z) = 0 if and only if z∏N−1

n=1 (1 − z2n−2) = 0 i.e.if and only if z is an integer.

[We might guess that S(z) = A sin πz. Near 0, S(z) behaves like zso, if the guess is correct, A = π−1.]

309

K268

(i) Take R > r > 0 As usual we note that∑∞

n=0 anrn converges,

so anrn → 0 and there exists an M such that |an| ≤ Mr−n. Thus, if

|z| < r/2,∣

∣

∣

∣

∣

q∑

n=0

anzn

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

q∑

n=N

anzn

∣

∣

∣

∣

∣

≥ |aNzN | −q∑

n=N+1

|anzn|

= |z|N(

|aN | − |z|q∑

n=N+1

|anzn−N−1|)

≥ |z|N(

|aN | −M |z|r−N−1

q∑

n=N+1

(|z|r)n−N−1

)

≥ |z|N(|aN | − 2M |z|r−N−1).

Thus if we set δ = min(r/2, |aN |rN+1/4M) we have∣

∣

∣

∣

∣

q∑

n=0

anzn

∣

∣

∣

∣

∣

≥ |aN ||z|N2

for all q ≥ N and so∣

∣

∣

∣

∣

∞∑

n=0

anzn

∣

∣

∣

∣

∣

≥ |aN ||z|N2

.

310

K269

(i) Proof of K268 works.

(ii) Exercise 5.91.

(iii) sin x =

∞∑

n=0

(−1)nx2n+1

(2n+ 1)!.

311

K270

(i) Observe that(

n

r − 1

)

+

(

n

r

)

=n!

(r − 1)!(n− r)!

(

1

n− r + 1+

1

r

)

=n!

(r!(n+ 1− r)!(n+ 1) =

(

n + 1

r

)

.

(ii) Result true for n = 0. If the result is true for n then

(x+ y)n+1 = x(x+ y)n + (x+ y)ny

= xn+1 +

n∑

r=1

((

n

r − 1

)

+

(

n

r

))

xryn+1−r + yn+1

=

n+1∑

r=0

(

n+ 1

x

)r

yn+1−r

so the result is true by induction.

(iii) Result true for n = 0. If true for n then

n∏

q=0

(x+ y − q) =n∑

r=0

(x+ y − n)

(

n

r

) r−1∏

j=0

(x− j)n−r−1∏

k=0

(y − k)

=n∑

r=0

(

n

r

) r∏

j=0

(x− j)n−r−1∏

k=0

(y − k) +n∑

r=0

(

n

r

) r∏

j=0

(x− j)n−r∏

k=0

(y − k)

=

n∏

j=0

(x− j) +

n−1∑

r=0

((

n

r

)

+

(

n

r − 1

)) r∏

j=0

(x− j)

n−r∏

k=0

(y − k) +

n∏

k=0

(y − k)

=

n+1∑

r=0

(

n+ 1

r

) r−1∏

j=0

(x− j)

n−r∏

k=0

(y − k)

so the result is true by induction.

312

K271

(i) Since |w| < 1, we can find a ρ > 1 such that ρ|w| < 1. We cannow find an N ≥ 1 such that

∣

∣

∣

∣

x− j

j

∣

∣

∣

∣

≤ M + j

j= 1 +Mj−1 < ρ

for all |x| ≤M and all j ≥ N − 1. We observe that∣

∣

∣

∣

∣

∞∑

n=0

1

n!

n−1∏

j=0

(x− j)

∣

∣

∣

∣

∣

≤ (N +M)N = K, say

for all |x| ≤M and all n ≤ N . Thus∣

∣

∣

∣

∣

∞∑

n=0

1

n!

n−1∏

j=0

(x− j)wn

∣

∣

∣

∣

∣

≤ K(ρw)n

for all |x| ≤M and all n ≥ 1 so by the Weierstrass M-test,

∞∑

n=0

1

n!

n−1∏

j=0

(x− j)wn

converges uniformly. Since the uniform limit of continuous functionsis continuous, f is continuous on [−M,M ]. Since N is arbitrary, f isdefined and continuous everywhere. Note that we have shown that thesum is absolutely convergent.

(ii) By Cauchy’s lemma (Exercise 5.38) or considering the powerseries in w we have

f(x)f(y) =∞∑

j=0

cn

with

cn =

n∑

r=0

(

n

r

) r−1∏

j=0

(x− j)

n−r−1∏

k=0

(y − k).

Exercise K270 now gives f(x)f(y) = f(x+ y).

(iii) If x = 0 then x−0 = 0 and so∏n−1

j=0 (x− j) = 0. Thus f(0) = 1.

It follows that f(x)f(−x) = f(0) = 1 so f(x) 6= 0. By the intermediatevalue theorem, f(x) > 0 for all x.

(iv) Since f(x) > 0, we can define g(x) = log f(x) and, since f iscontinuous, g is. Thus by K101, g(x) = ax and f(x) = eax = bx forsome b.

(v) Since b = f(1) = 1 + w, f(x) = (1 + w)x and we are done.

313

K272

(i) We prove that AN ≥ |aN | inductively. The result is trivial forN = 0. If the result is true for N = k − 1, we observe that

kak =

k−1∑

n=0

k−1∑

m=0

cn,mPn,m(a1, a2, . . . , ak−1)

and

kAk =k−1∑

n=0

k−1∑

m=0

Kρn+mPn,m(A1, A2, . . . , Ak−1)

where the Pn.m are multinomials with all coefficients positive. Thus

k|ak| ≤k−1∑

n=0

k−1∑

m=0

|cn,m|Pn,m(|a1|, |a2|, . . . , |ak−1|)

≤k−1∑

n=0

k−1∑

m=0

Kρn+mPn,m(A1, A2, . . . , Ak−1) = kAk

and the induction can proceed.

(ii) Formally

dw

dt= K

∞∑

n=0

∞∑

m=0

(ρt)n(ρw)m

= K

( ∞∑

n=0

(ρt)n

)( ∞∑

m=0

(ρw)m

)

=K

(1− ρt)(1 − ρw).

To solve the differential equation we set, informally,

(1− ρw)dw =K

1− ρtdt

obtaining

w(1− ρw/2) = −Kρlog(1− ρt) + C.

Since w(0) = 0 this gives

w(1− ρw/2) = −Kρlog(1− ρt)

and

w(t) =−1 + (1− k2(log(1− ρt))2)1/2

ρ.

314

(iii) If we now fix w as in the previous paragraph then Exercise K79tells us that w has a power series expansion

w(t) =∞∑

j=0

Ajtj

with radius of convergence η > 0. (Note A0 = 0 since w(0) = 0.)Since w satisfies ⋆⋆ our standard methods show that the An satisfy⋆. By the arguments of (i), |an| ≤ An, so, by comparison,

∑∞n=0 ant

n

has radius of convergence at least η. If we set u(t) =∑∞

n=0 antn, then,

since a0 = 0, we have u(0) = 0. By continuity, we can find δ with0 < δ ≤ η and |u(t)| < ρ for |t| < δ. The required results can now beread off.

315

K273

(i) Observe that, since f is continuous on [0, 4π], it is uniformly con-tinuous on [0, 4π]. Since f is 2π periodic it is thus uniformly continuouson R. Similarly g is bounded. Thus

|f ∗ g(t+ u)− f ∗ g(t)| =∣

∣

∣

∣

1

2π

∫ π

−π

(f(t+ u− s)− f(t− s))g(s) ds

∣

∣

∣

∣

≤ ‖g‖∞ sup|x−y|≤|u|

|f(x)− f(y)| → 0

as u→ 0.

(iii) Observe, for example, that making the substitution x = t − swe have

f ∗ g(t) = 1

2π

∫ π

−π

f(t− s)g(s) ds =1

2π

∫ t+π

t−π

f(x)g(t−x) dx = g ∗ f(t).

Fubini gives

f ∗ (g ∗ h)(t) = 1

2π

∫ π

−π

f(t− s)

(∫ π

−π

g(s− u)h(u) du

)

ds

=1

(2π)2

∫ π

−π

∫ π

−π

f(t− s)g(s− u)h(u) du ds

=1

(2π)2

∫ π

−π

∫ π

−π

f(t− s)g(s− u)h(u) ds du

=1

(2π)2

∫ π

−π

∫ π−u

−π−u

f(t− u− v)g(v)h(u) dv du

=1

2π

∫ π

−π

f ∗ g(t− u)h(u) du = (f ∗ g) ∗ h(t).

(iv) If we set G(s, t) = f(t − s)g(s), then G has continuous partialderivative G,2(s, t) = f ′(t − s)g(s). We may thus apply our theoremon differentiating under the integral to show that f ∗ g is differentiableand (f ∗ g)′ = f ′ ∗ g.

317

K274

(i) Fubini gives

f ∗ g(n) = 1

2π

∫ π

−π

e−int

(

1

2π

∫ π

−π

f(t− s)g(s) ds

)

dt

1

(2π)2

∫ π

−π

∫ π

−π

e−intf(t− s)g(s) ds dt

=1

(2π)2

∫ π

−π

∫ π

−π

e−intf(t− s)g(s) dt ds

=1

2π

∫ π

−π

e−insg(s)

(

1

2π

∫ π

−π

f(t− s)e−in(t−s) dt

)

ds

g(n)f(n).

(iii) We have

|f(n)− g(n)| = |f − g(n)| ≤ ‖f − g‖∞.

(v) If e ∗ f = f for all f then

f(n) = e ∗ f(n) = e(n)f(n)

for all f and all n. Take f(t) = eimt to see that e(m) = 1 for all mcontradicting the Riemann-Lebesgue lemma.

318

K275

(i) We have

f1(n) =1

2π

∫ π

−π

g1(t)(eit − e−it)

2ie−int dt =

g1(n+ 1)− g1(n− 1)

2i

Thus, using Riemann-Lebesgue,

Sn(f1, 0) =n∑

j=−n

g1(j + 1)− g1(j − 1)

2i

=g1(n+ 1) + g1(n− 1)− g1(−n+ 1)− g1(−n− 1)

2i→ 0.

(ii) If we set

g2(t) =

f2(t)sin t

if t/π is not an integer,

f ′2(0) if t = 2nπ,

−f ′2(π) if t = (2n+ 1)π,

then g2 is automatically continuous except possibly at points nπ. But,

g2(t) =f2(t)− f2(π)

t− π

t− π(

− sin(t− π)) → −f ′

2(π) = g2(π)

as t→ π so g2 is continuous at π. Similarly g2 is continuous at all nπ.

(iii) Note that f4 ∈ CP (R). We have

f4(n) =1

2π

∫ π

−π

f3(t/2)e−int dt

=1

π

∫ 2π

−2π

f3(s)e−i2ns ds

1

2π

∫ π

−π

f3(s)e−i2ns dt = f4(2n).

Observe that f4 obeys the conditions of (iii).

(iv) Translate.

319

K276

(i) Let S(z) = z − f(z). We have S ′(z) = 1 − f ′(z) so |S ′(z)| ≤ 1/2for |z| ≤ δ and

|Tz| ≤ |w|+ |S(z)− S(0)| ≤ |w|+ |z| sup|u|≤δ

|S ′(u)| ≤ δ/2 + δ/2 = δ

for all |z| ≤ δ.

(ii) Observe that, if z, u ∈ X ,

|Tz − Tu| = |Sz − Su| ≤ |z − u|/2.Thus T is a contraction mapping and there is a unique z0 ∈ X withTz0 = z0 i.e. with f(z0) = w.

(iii) Observe that the continuity of F ′ at 0 gives us a δ > 0 such that|F ′(z)− 1| ≤ 1/2 for |z| ≤ δ.

(iv) Let F (z) = (g(z)− g(0))/g′(0) and apply (iii).

(v) No. g = 0 is a counterexample. [However, in more advancedwork, it is shown that this is the only counterexample!]

(vi) No. If g(z) = z2, the equation g(z) = reiθ has two roots r1/2eiθ/2

and −r1/2eiθ/2.

320

K277

(i) True. w∗ is the solution.

(ii) False. f2(z) is real so f2(z) = δi has no solution for δ > 0.

(iii) and (v) False. For (iii), observe ℜf3(z) ≥ 0 so f3(z) = −δ hasno solution for δ > 0.

(iv) and (vi) True. For (vi), observe that we can find a δ with 1/6 >δ′ > 0 such that |F ′(z)− 1| ≤ 1/3 for |z| ≤ δ′. Let X = z : |z| ≤ δ.Set δ = δ′/3. If |w| ≤ δ write Sz = F (z)− 1 and Tz = Sz+ |z|1/2 −w.Using the mean value inequality, we have

|Tz| ≤ |w|+ |Sz|+ |z|2 ≤ δ′/3 + sup|u|<δ′

|S ′(u)||z|+ |z|/3 ≤ δ′

whenever |z| ≤ δ′ and

|Tz − Tu| ≤ |Sz − Su|+ ||z|2 − |u|2|≤ |z − u|/3 + |z − u|(|z|+ |u|) = 2|z − u|/3,

so T restricted to X is a contraction mapping on X . Since X is closed,we are dealing with a complete metric space and we can find z0 ∈ Xwith Tz0 = z0 i.e. w = F (z0) + |z0|2.

321

K278

(i) Observe that

T ′(x) = 1− f ′(x)

f ′(x)+f(x)f ′′(x)

f ′(x)2=f(x)f ′′(x)

f ′(x)2

so, by the mean value inequality,

|Ta− Tb| ≤ |λ||a− b|.Since R is complete under the usual metric, the contraction mappingtheorem applies.

(ii) Much as in (i),

|Tx| = |Tx− T0| ≤ |x| sup|w|≤a

|T ′(w)| ≤ |λ||x|

so, if |x| ≤ a, we see, by induction that |T nx| ≤ |a| and |T nx| ≤|λ|n|x| → 0.

Observe that

T ′(x) =F (x)F ′′(x)

F ′(x)2

and

T ′′(x) =F ′′(x)

F ′(x)+F (x)F ′′′(x)

F ′(x)2− 2F (x)(F ′′(x))2

F ′(x)3,

so T (0) = T ′(0) = 0 and T ′′(0) = F ′′(0)/F ′(0). Thus∣

∣

∣

∣

Tx− F ′′(0)

2F ′(0)x2∣

∣

∣

∣

= |Tx− T (0)− T ′(0)x− 12T ′′(0)| ≤ ǫ(x)x2

with ǫ(x) → 0 as as x→ 0. Choose η so that ǫ(x) ≤ 1 for |x| ≤ η.

(iii) We have quadratic convergence (roughly doubling of correct dec-imal places) with Newton rather than linear convergence (increasing thenumber of correct decimal places by the same amount) as is implied bythe contraction mapping theorem. There may be problems if F ′(0) issmall.

322

K279

Suppose f has a unique fixed point x. Then f(g(x)) = g(f(x)) =g(x) and, since x is the unique fixed point of f , g(x) = x.

Let X = −1, 0, 1, f(t) = −t, g(t) = t. Then fg = gf = f , f hasonly one fixed point but g has three.

Let X = −1, 1, f(t) = t, g(t) = −t. Then fg = gf = f , f has twofixed points but g has none.

323

K280

We first prove uniqueness. If f(x) = x and f(y) = y then d(x, y) =d(f(x), f(y)) ≥ Kd(x, y) so d(x, y) = 0 and x = y.

For existence choose x0 ∈ X and then choose inductively xn+1 ∈ Xsuch that that Txn+1 = xn (this uses surjectivity). We observe that

d(xn+1, xn) = d(Txn+2, Txn+1) ≥ Kd(xn+2, xn+1)

soK−1d(xn+1, xn) ≥ d(xn+2, xn+1)

and precisely the same argument as in the contraction mapping the-orem shows that xn → z for some z ∈ X . Now z = Tw for some wand

d(z, w) ≤ d(xn, z) + d(xn, w) ≤ d(xn, z) +Kd(Txn, Tw)

= d(xn, z) +Kd(xn−1, z) → 0

as n→ ∞. Thus d(z, w) = 0 so z = w and Tz = z.

324

K281

(i) If T (a) = a and T (b) = b, then ‖T (a) − T (b)‖ = ‖a − b‖ soa = b.

(ii) Uniqueness much as in (i). Existence not implied see e.g. secondparagraph of K281.

(iii) Let E = −1, 1 be a subset of R. The mapping T : E → Egiven by Tx = −x preserves length but has no fixed point. The mapT : E → E given by Tx = x preserve length and has two fixed points.

(The reader may prefer the example of a circle E in R2. Non-trivialrotations have no fixed points, reflection in a diameter has two.)

(iv) Observe that, since T decreases distance, it is continuous. Themap

x 7→ ‖x− Tx‖is thus continuous and, since E is closed and bounded in Rn, thusattains a minimum at x0, say. By definition

‖x0 − Tx0‖ ≤ ‖Tx0 − T 2x0‖so Tx0 = x0.

(v) Let E∗ be the closure of the set T nc : n ≥ 0. Then E∗ is isclosed bounded set and T |E∗ is a distance decreasing map from E∗ toitself. By (iv), T |E∗ and so T has a fixed point.

325

K282

Observe that, if X = [0,∞) with the usual metric, the map f givenby x 7→ x+1 is an isometry but we cannot find an isometry g : X → 0such that f g(x) = x.

We now restrict ourselves to bijective isometries. Since the bijectivemaps on X form a group under composition we need only show wehave a subgroup.

If f and g are isometries then

d(f g(x), f g(y)) = d(g(x), g(y)) = d(x, y)

so the composition of bijective isometries is a bijective isometry. Theidentity map is a bijective isometry.

If f is a bijective isometry, f−1 is well defined and

d(x, y) = d(f f−1(x), f f−1(y)) = d(f−1(x), f−1(y))

so f−1 is a bijective isometry.

(i) Take X to be the vertices of an isosceles triangle. G(X) is thegroup of permutations of the vertices (isomorphic to S3) so non Abelian.

(ii) Take X to be the vertices of a triangle all of whose sides havedifferent length.

(iii) Take X to be a circle.

(iv) Take X to be a regular n-gon and T rotation about the centreof symmetry through 2π/n.

326

K283

(i) Observe that, since (X, d) has the Bolzano–Weierstrass property,we can find a sequence m(j) → ∞ of strictly positive integers and anα ∈ X such that d(am(j), α) → 0. Take a sequence j(r) → ∞ of strictlypositive integers such that m(j(r + 1)) > 3m(j(r)) + 1. Then settingn′(r) = m(j(r + 1))−m(j(r)) we have

d(an′(r), a) ≤ d(fm(j(r)an(r), fm(j(r)a) = d(am(j(r+1)), am(j(r)))

≤ d(am(j(r+1)), α) + d(am(j(r)), α) → 0

as r → ∞.

Taking n(j) to be subsequence of the n′(j) with

d(bn(r), b) → 0,

we have d(an(r), a), d(bn(r), b) → 0. Thus

d(a, b) ≤ d(f(a), f(b)) ≤ d(fn(r)(a), fn(r)(b)))

≤ d(a, b) + d(an(r), a) + d(bn(r), b) → d(a, b)

as r → ∞ so d(a, b) = d(f(a), f(b)).

Examples Y = R, ρ usual metric, g(x) = 2x, h(x) = 1+2x for x ≥ 0,h(x) = −1 + 2x if x < 0.

(ii) Since f(x) = f(y) implies d(x, y) = d(f(x), f(y)) = 0 impliesx = y, f is injective.

To show surjectivity take any a ∈ X . As before we can find asubsequence n(r) with n(r) ≥ 2 such that d(an(r), a) → 0. Let br =an(r)−1. By the Bolzano–Weierstrass property we can find b ∈ X andr(j) → ∞ such that d(br(j), b) → 0. Thus

d(a, f(b)) ≤ d(a, an(r(j)))+d(f(br(j), f(b)) = d(a, an(r(j)))+d(br(j), b) → 0

as j → ∞.

Example, Y = [0,∞) with usual metric and f(x) = x+ 1.

327

K284

Observe that g f : X → X is expansive so, by K283, g f is bijec-tive and an isometry. Thus g is surjective and (since it is expansive)injective. Thus f is bijective. Since

d(x, y) ≤ ρ(f(x), f(y)) ≤ d(g f(x), g f(y)) = d(x, y),

f is an isometric bijection so (Y, ρ) has the Bolzano–Weierstrass prop-erty and g is an isometric bijection.

If X = Y = x ∈ R2 : ‖x‖ = 1 with the usual metric, then, takingf = g to be rotation through π/4, we see that f and g need not beinverses.

If X = Y = [0,∞) with the usual metric and we set f(x) = 2x,g(x) = x+ 1 then f is a distance increasing bijection but not distancepreserving and g is distance preserving but not a bijection.

328

K285

We note that x 7→ f(x) and x 7→ ‖f(x)‖ are continuous maps on aclosed bounded subset of Rn. Thus there exists anM with ‖f(x)‖ ≤Mfor all x ∈ A.

By convexity, g maps A into A. We have

‖g(a)− g(b)‖ = (1− ǫ)‖f(a)− f(b)‖ ≤ (1− ǫ)‖a− b‖

so g is a contraction mapping and so, since A is closed and we aredealing with a complete metric space, g has a fixed point z, say. Weobserve that

‖f(z)− z‖ = ‖f(z)− gz‖≤ ǫ‖f(x)‖+ ǫ‖a0‖≤ (M + ‖a0‖)ǫ.

Thus

inf‖f(a)− a‖ : a ∈ A = 0.

But x 7→ ‖f(x)−x‖ is a continuous map on a closed bounded subset ofRn so attains its infimum. Thus ‖f(y)− y‖ = 0 for some y ∈ A whichis thus a fixed point.

Let V = R, A = [−2,−1] ∪ [1, 2] and Tx = −x. Then A is closedand bounded and T is distance non-increasing with no fixed point.

Let V = R, A = (0, 1) and Tx = x/2. Then A is convex and boundedand T is distance decreasing with no fixed point.

Let V = A = R and Tx = x + 1. Then A is closed and convex andT is distance non-increasing with no fixed point.

In final paragraph take

A = q ∈ Rn :n∑

i=1

qi = 1 and qi ≥ 0 for 1 ≤ i ≤ n

and f(q) = q with

qj =

n∑

i=1

qipij .

330

K286

(i) Note that

|d(x, Tx)− d(y, Ty) ≤ d(x, y) + d(Tx, Ty) ≤ 2d(x, y)

so S is continuous. Thus since our space has the Bolzano–Weierstrassproperty S must attain a minimum at z say. If Tz 6= z then

S(TN(z,Tz)z) = d(

TN(z,Tz)z, T (TN(z,Tz)z))

= d(

TN(z,Tz)z, TN(z,Tz)(Tz))

< d(z, T z) = S(z)

which is absurd. Thus Tz = z.

(ii) Write fn(x) = d(T nx, z). Then fn is continuous (by direct proofor by composition of continuous functions) so Un = f−1

n ((−ǫ, ǫ)) isopen. Since d(T n+1x, z) = d(T (T nx), T z) ≤ d(T nx, z) we have Un ⊆Un+1.

Observe that F is a closed subset of (X, d) and so inherits theBolzano–Weierstrass property. If Tx ∈ U then Tx ∈ Un for somen so x ∈ Un+1 ⊆ U . Thus, if F 6= ∅ we can apply (i) to T |F : F → Fto obtain w ∈ F with Tw = w contradicting uniqueness.

(iii) Thus U = X and, given x ∈ X , we can find an n such thatx ∈ Un so d(Tmx, z) < ǫ for m ≥ n. Since ǫ was arbitrary, the requiredresult follows.

331

K287

Observe that if q ∈ X then

m∑

i=1

qipij ≥ 0 and

m∑

j=1

m∑

i=1

qipij =

m∑

i=1

m∑

j=1

qipij = 1.

(i) Note all norms on Rm Lipschitz equivalent. X is a closed (e.g.because the intersection of closed sets of form Ei = x : xi ≥ 0 andE = x :

∑mi=1 xi = 1) and bounded (if x ∈ E then ‖x‖ ≤ 1).

(ii) We have

‖Tu− Tv‖1 =m∑

j=1

∣

∣

∣

∣

∣

m∑

i=1

(ui − vi)pij

∣

∣

∣

∣

∣

≤m∑

j=1

m∑

i=1

|ui − vi|pij

=m∑

i=1

m∑

j=1

|ui − vi|pij = ‖u− v‖1.

(iii) Show that

p(n+1)ij =

m∑

k=1

pnikpkj

and use induction.

Probabilistically, p(n)ij is the probability that, starting at i we are at

j after n steps.

(iv) Repeat the calculation of (ii) but observe that if u 6= v theremust exist I and I ′ such that uI − vI and uI′ − vI′ are non-zero andhave opposite signs. Thus

∣

∣

∣

∣

∣

m∑

i=1

(ui − vi)pNij

∣

∣

∣

∣

∣

<m∑

i=1

|ui − vi|pNij

for each j.

(vi) Tπ = π gives π1 = π2 so π1 = π2 = 1/2. T nq → π if and only

if q = π. Observe p(2n+1)11 = 0 and p

(2n)12 = 0 for all n so there is no n

with pnij 6= 0 for all i, j.

(vii) Tπ = π for all π ∈ X . We have T nq = q for all q ∈ X .

Observe that p(n)12 = 0 for all n.

(viii) Tπ = π has as solutions all π ∈ X with π1 = π2 and π3 = π4.

332

Given q ∈ X set π1 = π2 = (q1 + q2)/2 and π3 = π4 = (q3 + q4)/2.Then, Tq = π and (if n ≥ 1) T nq = π → π.

333

K288

We have∞∑

i=1

∞∑

j=1

|xipij | =∞∑

i=1

∞∑

j=1

|xi|pij =∞∑

i=1

|xi| = ‖x‖1

so Fubini’s theorem tells us that∑∞

i=1 xipij converges for each j. Fur-ther

∑∞i=1 |xi|pij converges and

∞∑

j=1

∣

∣

∣

∣

∣

∞∑

i=1

xipij

∣

∣

∣

∣

∣

≤∞∑

j=1

∞∑

i=1

|xi|pij

=∞∑

i=1

∞∑

j=1

|xi|pij = ‖x‖1.

Thus Tx is a well defined point of l1 and ‖Tx‖1 ≤ ‖x‖1. (Note thatthis only shows that ‖T‖ ≤ 1.) Since αn → α and βn → β implyλαn + µβn → λα+ µβ, it is easy to check that T is linear.

Further, using Fubini again, if q ∈ X∞∑

j=1

∞∑

i=1

qipij =

∞∑

i=1

∞∑

j=1

qipij = 1

and, trivially,∑∞

j=1 qipij ≥ 0 so Tq ∈ X .

(ii) If u, v ∈ X and u 6= v then we can find I and I ′ such thathat uI − vI and uI′ − vI′ are non-zero and have opposite signs. Ifk = maxI ′, I then as in (iv) of K287, d(TN(k)u, TN(k)v) < d(u,v).

(iii) If Tu = u and Tv = v then (ii) shows that u = v.

If Th = h and hi ≥ 0 for all i then either h = 0 or setting k =‖h‖−1h we have k ∈ X and Tk = k. If Tπ = π has a solution in Xthen the required solutions are λπ with λ ≥ 0. If not, the only solutionis 0.

(iv) We must have πI > 0 for some I. Take k = maxI, i

πi =

∞∑

r=1

πrpN(k)pri ≥ πIp

N(k)pIi > 0.

(v)(a) We have for i ≥ 2

πi =12πi+1 +

14πi +

14πi−1.

so2πi+1 − 3πi + πi−1 = 0

and πn = Aαn + Bβn with α and β roots of the auxiliary equation2θ2 − 3θ + 1 = 0. Thus πn = A+B(1/2)n.

334

Treating the previous paragraph as purely exploratory we observethat πn = 2−n is a solution.

(b) Proceeding as in (a) we get

πi+1 − 3πi + 2πi−1 = 0

with general solution πi = A + B2n. There are no constants A and Bgiving π ∈ X .

(c) Proceeding as in (a) we get

πi+1 − 2πi + πi−1 = 0

with general solution πi = A + Bn. There are no constants A and Bgiving π ∈ X .

335

K289

(i) Observe that πJ > 0 and set h = π−1J π.

(ii) If e(j)k = 1 if k = j and e(j)k = 0 otherwise, then the e(k)form a sequence in X with no convergent subsequence. (Observe that‖e(k) − e(l)‖1 = 2 for k 6= l.) Thus X does not have the Bolzano–Weierstrass property. However (see Exercise K218)

Y = y : hi ≥ |yi| for i ≥ 1does so, since X is closed, so does XJ = X ∩ Y . Observe that

π = ‖h‖−11 h ∈ XJ .

Also if q ∈ XJ then Tq ∈ X and

(Tq)j =∞∑

i=1

qipij

≤∞∑

i=1

hipij = hj

so Tq ∈ XJ . Thus by Exercise K186 T nq → π for all q ∈ XJ .

(iii) In particular T ne(J) → π. But J is arbitrary.

Now if q ∈ X and ǫ > 0 we can find M such that∑M

r=1 qr > 1 − ǫ.

Write u = q−∑Mr=1 qre(r). Then

‖T nq− π‖1 ≤∥

∥

∥

∥

∥

T nM∑

r=1

qre(r)−M∑

r=1

qrπ

∥

∥

∥

∥

∥

1

+

∥

∥

∥

∥

∥

(

1−M∑

r=1

qr

)

π

∥

∥

∥

∥

∥

1

+ ‖T nu‖1

≤M∑

r=1

qr‖T ne(r)− π‖1 + ǫ+ ‖u‖1

≤M∑

r=1

qr‖T ne(r)− π‖1 + 2ǫ

→ 2ǫ

as n→ ∞. Since ǫ was arbitrary, ‖T nq− π‖1 → 0 as required.

336

K290

For uniqueness. Suppose Tx = Ty. Then T nx = T n−1Tx =T n−1Ty = T ny so, by the contraction mapping theorem, x = y.

For existence. By the contraction mapping theorem we can find anx0 with T

nx0 = x0. Now T n(Tx0) = T (T nx0) = Tx0 so, by uniqueness,Tx0 = x0.

In question K282, T n has a fixed point but is not a contractionmapping.

Desired result trivially true for n = 0. If true for n then

|(T n+1h)(t)− (T n+1k)(t)| ≤∣

∣

∣

∣

∫ t

a

φ(h(s), s)− φ(k(s), s) ds

∣

∣

∣

∣

≤∫ t

a

|φ(h(s), s)− φ(k(s), s)| ds

≤∫ t

a

|M(h(s)− k(s))| ds

≤ Mn+1

n!‖h− k‖∞

∫ t

a

(s− a)n ds

≤ Mn+1

(n+ 1)!‖h− k‖∞(t− a)n+1.

The induction can proceed.

We thus have

‖T nh− T nk‖ ≤ 1

n!Mn(b− a)n‖h− k‖∞.

Since

un =1

n!Mn(b− a)n → 0

(observe that un+1/un → 0) we can find an N such that

‖TNh− TNk‖ ≤ 1

2‖h− k‖∞

for all h, k ∈ C([a, b]). Thus there exists a unique f ∈ C([a, b]) with

f(t) = g(t) +

∫ t

a

φ(f(s), s) ds

and so (if g(t) = c) a unique solution of

f ′(t) = φ(f(t), t)

with initial condition f(a) = c.

337

K291

Consider Rn with its usual Euclidean norm and recall that it iscomplete. Set

Tx = y, with yi =

n∑

j=1

aijf(xj) + bi

then, using Cauchy-Schwarz inequality and the mean value inequality,we have

‖Tu− Tv‖ =

n∑

i=1

(

n∑

j=1

aij(f(uj)− f(vj)

)2

1/2

≤(

n∑

i=1

(

n∑

j=1

a2ij

)(

n∑

j=1

(f(uj)− f(vj))2

))1/2

≤(

n∑

i=1

(

n∑

j=1

a2ij

)(

n∑

j=1

(M |uj − vj |2)))1/2

=M‖u− v‖(

n∑

j=1

a2ij

)1/2

= K‖u− v‖with K ≥ 0 and

K2 =M2n∑

j=1

a2ij .

If K < 1 the contraction mapping theorem applies.

338

K292

A little fiddling around (we could write y−αdy = dx obtaining, in apurely exploratory, non-rigorous manner (1 − α)y1−α = x + A) showsthat, if 0 < α < 1, we have y(t) = 0 for all t, and y(t) = 0 for t < s,y(t) = (1 − α)1/(1−α)(t − s)1/(1−α) for t ≥ s are solutions. (We needs ≥ 0 to satisfy y(0) = 0).

If α ≥ 1 then using the mean value theorem

||y1|α − |y2|α| = α|z|α−1||y1| − |y2|| ≤ α(max(|y1|, |y2|))α−1

and this Lipschitz condition shows that the obvious solution y(t) = 0for all t is unique.

339

K293

(i) If u > 0, then f(t, u) ≥ 0 for t > 0 so, by continuity, f(u, 0) ≥ 0.Similarly f(t, u) ≤ 0 so f(u, 0) ≤ 0 Thus f(u, 0) = 0 for u > 0.Similarly f(0, u) = 0 for u < 0 so f(0, 0) = 0 by continuity.

Observe, just as in the proof of Rolle’s theorem, that, if y is constanton [0, c], then y(t) = 0 for t ∈ [0, c]. Otherwise, since a continuous func-tion on a closed bounded interval is bounded and attains its bounds, ymust have a maximum or minimum at some α ∈ (0, c) with y(α) 6= 0.Without loss of generality, let α be a maximum. Then y(α) > 0 so,using the fact that at an interior maximum the derivative is zero,

0 = y′(α) = f(α, y(α)) > 0

which is impossible. Thus y(t) = 0 for t ∈ [0, c] for all c > 0 so y(t) = 0for t ≥ 0 and, similarly, for all t. Thus ⋆ has at most one solution.Since y = 0 is a solution it is the unique solution.

(ii) If E = (x, t) : ‖(x, t)‖ ≥ δ with δ > 0 then E is closed and therestriction of f is continuous on each of the three closed sets,

E1 = (x, t) ∈ E : x ≥ t2, E2 = (x, t) ∈ E : |x| ≤ t2and

E3 = (x, t) ∈ E : x ≤ −t2and so the restriction f |E is continuous on E =

⋃3i=1Ei. Thus f is

continuous at every (x, t) with ‖(x, t)‖ > δ with δ > 0 and so at every(x, t) 6= (0, 0) Since |f(t, u)| ≤ 2max(|x|, |t|), f is continuous at (0, 0)(Lots of other ways to do this, many probably better.)

Now

y1(t) =

∫ t

0

f(s, s2) ds =

∫ t

0

(−2s) ds = −y0(t)for t ≥ 0 and a similar calculation applies when t < 0. Thus yn =(−1)ny0.

340

K294

Suppose, if possible, that y(τ) ≤ 0 for some τ > 0. Let

σ = infs ≥ 0 : y(s) ≤ 0.By continuity, y(σ) = 0. By the mean value theorem, we can find an swith 0 < s < σ such that

|y(s)|β ≤ y′(s) = s−1(y(s)− y(0)) < 0

which is absurd. Thus y(t) > 0 for all t, so y′(t) > 0 for all t, so y isstrictly increasing, so y′(t) ≥ y(0)β = 1 so y(t) ≥ 1 + t for all t ≥ 0.

On (tj , tj+1), we have

y′(t) ≥ y(t)β ≥ y((tj)β = 2βj

so, by the mean value inequality,

(tj+1 − tj)2βj ≤ y(tj+1)− y(tj)) = 2j

and sotj+1 − tj ≤ 2j(1−β).

Thus

tr =

r−1∑

j=0

tj+1 − tj ≤r−1∑

j=0

2j(1−β) ≤ 1

1− 21−β

for all r and so

a ≤ 1

1− 21−β.

(ii) Observe that, if y′ = 1 + y2, then y′ ≥ y2.

(iii) Recall that∫

(w logw)−1 dw = log logw + a. We have y(t) =exp(exp t) as a solution.

341

K295

Observe thaty((r + 1)h)− y(rh)

h≈ y′(rh).

Observe that

|f(t, u)− f(t, v)| ≤ sups

|f,2(t, s)||u− v| ≤ K|u− v|

so we have a (global) Lipschitz condition.

By the chain rule y′′ exists and

y′′(t) = f,1(t, y(t)) + y′(t)f,2(t, y(t)) = f,1(t, y(t)) + f(t, y(t))f,2(t, y(t)).

Thus |y′′(t)| ≤ L for all t and so

|y(nh+ s)− y(nh)− y′(nh)s| ≤ Ls2

2whence

|y((n+ 1)h)− y(nh)− f(nh, y(nh))h| ≤ Lh2

2and

|y((n+ 1)h)− yn+1| ≤ |y(nh)− yn|+ |y((n+ 1)h)− yn+1 − y(nh) + yn|≤ |y(nh)− yn|+ |y((n+ 1)h)− y(nh)− f(nh, y(nh))h|+ |f(nh, y(nh))h− yn+1 + yn|≤ |y(nh)− yn|+ |y((n+ 1)h)− y(nh)− f(nh, y(nh))h|+ |f(nh, y(nh))h− f(nh, yn)h|

≤ |y(nh)− yn|+Lh2

2+Kh|y(nh)− yn|

= (1 +Kh)|y(nh)− yn|+Lh2

2.

Thus if

|y(nh)− yn| ≤Lh

2K

(

(1 +Kh)n+1 − 1)

,

it follows that

|y((n+ 1)h)− yn+1| ≤ (1 +Kh)LhN2K

(

(1 +Kh)n+1 − 1)

+Lh2

2

≤ Lh

2K

(

(1 +Kh)n+2 − 1)

.

Since the result is true for n = 0 it is true inductively.

In particular

|y(Nh)− yN | ≤Lh

2K

(

(1 +Kh)N+1 − 1)

.

342

K296*

No comments.

343

K297

As in K295,

|y((n+ 1)h)− y(nh)− y′(nh)h| ≤ Lh2

2,

and then, much as in K295,

|y((n+ 1)h)− yn+1| ≤ (1 +Kh)|y(nh)− yn|+Lh2

2+ ǫ

so, by induction,

|y(nh)− yn| ≤(

Lh

2K+

ǫ

Kh

)

(

(1 +Kh)n+1 − 1)

.

Thus

|y(a)− yN | ≤(

Lh

2K+

ǫ

Kh

)

(eaK − 1).

The error due to machine inaccuracy increases directly with the num-ber of computations.

(ii) and (iii) Recall that(

A

h+Bh

)

=

(

A1/2

h1/2− B1/2h1/2

)2

+ 2A1/2B1/2

takes its minimum value 2A1/2B1/2 when h = (A/B)1/2. We wanth = (2ǫ/L)1/2 giving minimum error (2L)1/2K−1ǫ1/2.

344

K298

(i) Set f(t, y(t)) = g(t),M = L. Either redo calculations with K = 0or observe that

∣

∣

∣

∣

∣

∫ a

0

g(t) dt− hN−1∑

n=0

g(rh)

∣

∣

∣

∣

∣

= |y(a)− y[hN ]N | ≤ LhN

2

eaK − 1

K

→ aLhN2

=aMh

2as K → 0+. Take κa =Ma/2.

(ii) We have∫ a

0

G(t) dt− h

N−1∑

n=0

G(rh) =

∫ π

0

1 + sin 2Nt

2dt =

π

2=π

2‖G‖∞.

(iii) No. Consider f(t, u) = G(t) in (ii).

345

K299

(i) The result is trivially true for n = 0. If it is true for n then

(uv)(n+1)(x) = ((uv)′)(n)(x) = (u′v + uv′)(n)(x)

=n∑

r=0

(

n

r

)

u(n+1−r)(x)v(r)(x) +n∑

r=0

(

n

r

)

u(n−r)(x)v(r+1)(x)

= u(n+1)(x)v(x) +n∑

r=1

((

n

r

)

+

(

n

r − 1

))

u(n+1−r)(x)v(r)(x) + u(x)v(n+1)(x)

=n+1∑

r=0

(

n+ 1

r

)

u(n+1−r)(x)v(r)(x)

so the induction can proceed.

(ii) Differentiating we get

x(1 + x2)−1/2y(x) + (1 + x2)1/2y′(x) =1 + x(1 + x2)−1/2

x+ (1 + x2)1/2

so, multiplying through by (1 + x2)1/2 we have

(1 + x2)y′(x) + xy(x) = 1.

Differentiating n times, using Leibniz’s formula, we have

(1+x2)y(n+1)(x)+2nxy(n)(x)+n(n−1)y(n−1)(x)+xy(n)(x)+ny(n−1)(x) = 0

for n ≥ 1. Setting x = 0, we have

y(n+1)(0) + n2y(n−1)(0) = 0

Since y(0) = 0 and y′(0) = 1 (by setting x = 0 in the first two displayedequations in the statement of (ii)) we have y(2n)(0) = 0 and

y(2n+1)(0) = 2nn!.

Set un = y(n)(0)/n!. Then

|u2n+1x2|

|u2n−1|x2

2n + 1

which tends to zero if |x| ≤ 1 but diverges otherwise. Thus the Taylorseries for y has radius 1.

Now∑∞

j=0 unxn may be differentiated term by term with in its circle

of convergence so u(x) =∑∞

j=0 unxn is a solution for the given dif-

ferential equation with the same value as y at 0. Since the Lipschitzconditions apply, y(t) = u(t) for |t| < 1 so

y(x) =∞∑

j=0

y(n)(0)

n!xn

for |x| < 1.

346

K300

Since K is continuous on the closed bounded subset [0, 1]2 of R2, itis bounded so there exists an M with M ≥ K(s, t) for (s, t) ∈ [0, 1]2.

If f ∈ C([0, 1]), set

(Tf)(t) = g(t) + λ

∫ 1

0

K(s, t)f(s) ds.

We observe that Tf ∈ C([0, 1]) since K is uniformly continuous (be-cause K is continuous on the closed bounded subset [0, 1]2) and so

|(Tf)(u)− (Tf)(v)| =∣

∣

∣

∣

∫ 1

0

(K(u, s)−K(v, s))f(s) ds

∣

∣

∣

∣

≤ ‖f‖∞ sup‖x−y‖≤|u−v|

|K(x)−K(y)| → 0

as u→ v.

Further, if |λ| < M−1, T is a contraction mapping on the completenormed space (C([0, 1]), ‖ ‖∞) since

|(Tf)(t)− (Th)(t)| = |λ|∣

∣

∣

∣

∫ 1

0

K(s, t)(f(s)− h(s) ds

∣

∣

∣

∣

≤ |λ|∫ 1

0

|K(s, t)(f(s)− h(s)| ds

≤ |λ|M‖f − h‖∞so ‖Tf−Th‖∞ ≤M‖f−h‖∞. The contraction mapping theorem nowgives the required result.

347

K301

(ii) Observe that

d

dt

∣

∣

∣

∣

∣

∣

u1(t) u2(t) u3(t)u′

1(t) u′

2(t) u′

3(t)u′′

1 (t) u′′

2 (t) u′′

3 (t)

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

u′

1(t) u′

2(t) u′

3(t)u′

1(t) u′

2(t) u′

3(t)u′′

1 (t) u′′

2 (t) u′′

3 (t)

∣

∣

∣

∣

∣

∣

+

∣

∣

∣

∣

∣

∣

u1(t) u2(t) u3(t)u′′

1 (t) u′′

2 (t) u′′

3 (t)u′′

1 (t) u′′

2 (t) u′′

3 (t)

∣

∣

∣

∣

∣

∣

+

∣

∣

∣

∣

∣

∣

u1(t) u2(t) u3(t)u′

1(t) u′

2(t) u′

3(t)u′′′

1 (t) u′′′

2 (t) u′′′

3 (t)

∣

∣

∣

∣

∣

∣

= 0 + 0 +

∣

∣

∣

∣

∣

∣

u1(t) u2(t) u3(t)u′

1(t) u′

2(t) u′

3(t)u′′′

1 (t) u′′′

2 (t) u′′′

3 (t)

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

u1(t) u2(t) u3(t)u′

1(t) u′

2(t) u′

3(t)−a(t)u′′

1 (t) − b(t)u′

1(t) − c(t)u1(t) −a(t)u′′

2 (t) − b(t)u′

2(t) − c(t)u2(t) −a(t)u′′

3 (t) − b(t)u′

3(t) − c(t)u3(t)

∣

∣

∣

∣

∣

∣

= −a(t)

∣

∣

∣

∣

∣

∣

u′

1(t) u′

2(t) u′

3(t)u′

1(t) u′

2(t) u′

3(t)u′′

1 (t) u′′

2 (t) u′′

3 (t)

∣

∣

∣

∣

∣

∣

so W ′(t) = −a(t)W (t).

348

K302

Observe that, since W (a) 6= 0, we have g(a) 6= 0. Similarly g(b) 6= 0.Thus, if g has no zeros in (a, b), g has no zeros in [a, b]. Thus f/gis well defined continuous function on [a, b] which is differentiable on(a, b). By Rolle’s theorem there exists a c ∈ (a, b) with

0 =

(

f

g

)′(c) =

f ′(c)g(c)− g′(c)f(c)

(g′(c))2=

W (c)

(g′(c))2.

Thus W (c) = 0 contrary to the hypothesis. Thus there must be a zeroof g between any two zeros of f and, similarly, a zero of f between anytwo zeros of g. [Moreover the zeros must be simple, if f(c) = f ′(c) = 0,then W (c) = 0.]

349

K303

(i) Without loss of generality suppose that λ1 6= 0. Then withoutloss of generality we may suppose λ1 = 1 and set λ2 = λ. Then

W (f1, f2)(x) = det

(

f1 f2f1 f ′

2

)

= det

(

f1 + λf2 f2f ′1 + λf ′

2 f ′2

)

= det

(

0 f20 f ′

2

)

= 0.

However, if g1(x) = (x − 1)2 for x ≥ 1 and g2(x) = g1(−x) (we couldplay the same trick with infinitely differentiable functions) then g1 andg2 are continuously differentiable and their Wronskian vanishes every-where but they are not linearly dependent.

(ii) By Exercise 12.24 applied to the vectorial equation

d

dt

(

y(t)v(t)

)

=

(

v1(t)−a1(t)v(t)− a2(t)y1(t)

)

,

it follows that if a1 and a2 are continuous (so bounded on [a, b]), theequation

y′′(x) + a1(x)y′(x) + a2y(x) = 0

has exactly one solution with y(a) = A and y′(a) = B.

If W (f1, f2)(x) never vanishes then setting

a1(x) =(

W (f1, f2)(x))−1

det

(

f1(x) f2(x)f ′′1 (x) f ′′

2 (x)

)

and

a2(x) =(

W (f1, f2)(x))−1

det

(

f ′1(x) f ′

2(x)f ′′1 (x) f ′′

2 (x)

)

we have a1 and a2 continuous. Thus the equation

⋆ y′′(x) + a1(x)y′(x) + a2y(x) = 0

has exactly one solution with y(a) = A and y′(a) = B.

But ⋆ is equivalent to

W (y, f1, f2) = 0

and so has f1 and f2 as solutions. Since W (f1, f2)(a) 6= 0 the equations

λ1f1(a) + λ2f2(a) = y(a)

λ1f′1(a) + λ2f

′2(a) = y′(a)

have a solution, so, if y is a solution, the uniqueness result of theprevious paragraph shows that

y(x) = λ1f1(x) + λ2f2(x)

350

for all x so f1 and f2 are indeed a basis.

Recall from e.g. Lemma 12.8 that if f1 and f2 are solutions of

y′′(x) + b1(x)y′(x) + b2y(x) = 0

then if W (f1, f2) vanishes at one point it vanishes everywhere.

351

K304

Observe that

0 = (u′′v + 2v′u′ + v′′u) + p(u′v + v′u) + qvu

= v(u′′ + pu′ + qu) + v′′u+ v′(2u′ + pu) = v′′u+ v′(2u′ + pu)

so, writing w = v′ we have

† w′(x)u(x) + w(x)(2u′(x) + p(x)u(x)) = 0

Observe that † is linear so, if w is a solution, so is Bw. Thus

y(x) = u(x)v(x) = u(x)

(

A +B

∫ x

0

w(t) dt

)

is a solution. Since it contains two arbitrary constants we would expectit to be the general solution.

(ii) We try y(x) = v(x)(exp x) obtaining

v′′(x) = (2x− 1)v′(x)

so v′(x) = B(exp(x2 − x)) and

y(x) = A exp(x) +B exp(x)

∫ t

0

exp(x2 − x) dx.

(iii) y(x) = (A+Bx) exp x.

(iv) If u is a solution, try y = uv, obtaining

0 = (u′′′v + 3v′u′′ + 3v′′u′ + v′′′u) + p1(u′′v + 2u′v′ + v′′u) + p2(u

′v + v′u) + p3vu

= w′′u+ w′(3u′ + p1u) + w(3u′′ + 2u′p1 + p2)

a second order equation for w = v′.

352

K305

(i) We have

f = (u′′1y1 + 2u′1y′1 + u1y

′′1 + p(u′1y1 + u1y

′1) + qu1y1)

+ (u′′2y2 + 2u′2y′2 + u2y

′′2) + p(u′2y2 + u2y

′2) + qu2y2

= 2u′1y′1 + u′′1y1 + pu′1y1 + 2u′2y

′2 + u′′2y2 + pu′2y2.

(ii) If we differentiate ††, then we get

0 = u′′1y1 + u′1y1 + u′′2y2 + u′2y2

so † becomesf = u′1y

′1 + u′2y

′2.

In other words, we have the pair of equations

u′1y′1 + u′2y

′2 = f

u′1y1 + u′2y2 = 0

so, since W never vanishes,

u′1 = −W−1fy2, and u′2 = W−1fy1.

(iii) Performing the integration we get

y(x) = A1y1(x) + A2y2(x)

+ y1(x)

∫ a

x

W−1(s)f(s)y2(s) dx+ y2(x)

∫ x

b

W−1(s)f(s)y1(s) dx

and, if y1 and y2 are chosen as in our discussion of Green’s function, weobtain our standard Green’s function solution plus the general solutionof the equation with f = 0.

(iv) In the case given, we can take y1(x) = ex and y2(x) = e−x,obtaining W (x) = −2 and

u′1(x) = − e−x

1 + ex=

ex

1 + ex+ e−x − 1 and u′2(x) =

ex

1 + ex.

Thus

u1(x) = A1 + log(1 + ex)− e−x + x and u2(x) = A2 + log(1 + ex)

and we have the general solution

y(x) = A1ex + A2e

−x + (e−x − ex) log(1 + ex)− ex + 1.

(v) Set y = y1u1 + y2u2 + y3u3 subject to

0 = y1u′1 + y′2u

′2 + y3u

′3

0 = y′1u′1 + y′2u

′2 + y′3u

′3.

We getf = y′′1u1 + y′′2u2 + y′′3u3.

353

Since the Wronskian never vanishes, we can solve the last three equa-tions to find u′1, u

′2 and u′3. Integrating now gives u1, u2, u3 and thus

the solution.

354

K306

Observe that since K is continuous on the closed bounded set S itis uniformly continuous. Thus

|(TK(f))(s)− (TK(f))(t)| =∣

∣

∣

∣

∫ 1

0

K(s, y)−K(t, y)f(y) dy

∣

∣

∣

∣

≤ ‖f‖∞ sup‖z−w‖≤|s−t|

|K(z)−K(w)| → 0

as s→ t. Thus Tk maps C(I) to C(I). We easily check linearity and

|(TK(f))(s)| ≤ ‖f‖∞‖K‖∞so TK is continuous and ‖T‖ ≤ ‖K‖∞.

(i) True, since ‖TKn − TK‖ = ‖TKn−K‖ ≤ ‖Kn −K‖∞.

(ii) False. If Kn(x, y) = max(0, 1− n2(x2 + y2)) then ‖Kn‖∞ = 1 forall n but

|(TKn(f))(s)| ≤∫ 1/n

0

|f(y)| dy ≤ ‖f‖∞/n

so ‖TKn‖ ≤ n−1 → 0.

(iii) True. Since TK − TL = TK−L, it suffices to show that TK = 0implies K = 0. Suppose K 6= 0. Then we can find (x, y) ∈ S such thatK(x, y) 6= 0. Without loss of generality suppose K(x, y) = m > 0. Bycontinuity we can find a δ > 0 such that (s, t) ∈ S and |s−x|, |t−y| ≤ δimplies K(s, t) ≥ m/2. Set f(t) = max(0, 1 − 2δ−1|t − y|). ThenTKf(x) > 0 so TK 6= 0.

(iv) False. Let Tf(x) = f(1/2) for all [x ∈ [0, 1]. Then T is linearand continuous with ‖T‖ ≤ 1. We claim that T 6= TK for all K. Forconsider fn(x) = max(0, 1− n|x− 1

2|). Tfn = 1 for all n but, if n ≥ 2,

|TKfn| ≤∫

12+1/n

12−1/n

‖K‖∞ ds = 2‖K‖∞/n→ 0

as n→ ∞.

355

K307

(ii) We apply Exercise 12.24 to the vectorial equation

d

dt

(

y1(t)v1(t)

)

=

(

v1(t)−a(t)v1(t)− b(t)y1(t)

)

to obtain the existence and uniqueness of y1.

(iii) We have (noting that∫ 1

tu(x) dx = −

∫ t

1u(x) dx)

y′(t) = H(t, t)f(t) +

∫ t

0

H,2(s, t)f(s) ds− H(t, t)f(t) +

∫ 1

t

H,2(s, t)f(s) ds

=

∫ t

0

H,2(s, t)f(s) ds+

∫ 1

t

H,2(s, t)f(s) ds

and

y′′(t) = H,2(t, t)f(t) +

∫ t

0

H,22(s, t)f(s) ds− H,2(t, t)f(t) +

∫ t

0

H,22(s, t)f(s) ds

= f(t) +

∫ t

0

H,22(s, t)f(s) ds+

∫ t

0

H,22(s, t)f(s) ds.

Summing gives

y′′(t) + a(t)y′(t) + b(t)y(t)

= f(t) +

∫ t

0

(b(t)H,2 + a(t)H,2(s, t) +H,22(s, t))f(s) ds

+

∫ t

0

(b(t)H,2 + a(t)H,2(s, t) + H,22(s, t))f(s) ds

= f(t).

Further

y(0) =

∫ 1

0

H(s, 1)f(s) ds =

∫ 1

0

0 ds = 0

and similarly y(1) = 0.

356

K308

We seek a twice continuously differentiable function y(x) = G(x, t)which is four times differentiable (using left and right derivatives atend points) on [0, t) and (t, 1] and satisfies y(0) = y′(0) = 0,

y(4)(x)− k4y(x) = 0 for x ∈ [0, t)

and y(1) = y′(1) = 0,

y(4)(x)− k4y(x) = 0 for x ∈ (t, 1]

whilsty′′′(t+)− y′′′(t−) = 1.

This gives

y(x) = a sinh kx+ b sin kx+ c cosh x+ d cosx for x ∈ [0, t]

(remember we have continuity at t) and

0 = c+ d, 0 = b+ d

so

y(x) = A(sinh kx− sin kx) +B(cosh kx− cos kx) for x ∈ [t, 1]

and, using appropriate symmetries

y(x) = C(sinh k(1−x)−sin k(1−x))+D(cosh k(1−x)−cos k(1−x)) for x ∈ (t, 1].

The continuity of the zeroth, first and second derivatives and thecondition on the discontinuity for the third gives (subject to the checkin the next paragraph)

M(

A B C D)

=(

0 0 0 1)

so(

A B C D)T

=M−1(

0 0 0 1)T

where

M =

(

sinh kt− sinkt) (

cosh kt− cos kt)

−(

sinh k(1− t) + sin k(1− t)) −(

cosh k(1− t)− cos k(1− t))

k(

cosh kt− cos kt)

k(

sinhkt+ sinkt)

k(

cosh k(1− t) + cos k(1− t)) k(

sinhk(1− t) + sink(1− t))

k2(

sinhkt− sinkt)

k2(

cosh kt− cos kt)

−k2(

sinhk(1− t)− sink(1− t)) −k2(

cosh k(1− t) + cos k(1− t))

k3(

cosh kt+ cos kt)

k3(

sinh kt+ sinkt)

k3(

cosh k(1− t) − cos k(1− t)) k3(

sinhk(1− t) − sin k(1− t))

.

We must check that the four functions(

sinh kx− sin kx)

,(

cosh kx+ cos kx)

and(

sinh k(1− x)− sin k(1− x))

,(

cosh k(1− x) + cos k(1− x))

are indeed linearly independent. If

A(

sinh kx− sin kx)

+B(

cosh kx− cos kx)

+ C(

sinh k(1− x)− sin k(1− x))

+D(

cosh k(1− x)− cos k(1− x))

= 0,

then writing

y(x) = A(

sinh kx− sin kx)

+B(

cosh kx− cos kx)

357

we must have and y(1) = y′(1) = 0 that is to say

A(

sinh k − sin k)

+B(

cosh k − cos k)

= 0

A(

cosh k − cos k)

+B(

sinh k + sin k)

= 0.

But

det

((

sinh k − sin k) (

cosh k − cos k)

(

cosh k − cos k) (

sinh k + sin k)

)

=(

sinh2 k − sin2 k)

−(

cosh2 k − cos2 k)

= −1 + cos2 k − sin2 k = cos 2k − 1.

Thus if k 6= nπ for some integer n, we have A = B = 0 (and soC = D = 0).

If k = nπ then, by direct calculation, the Wronskian of our fourfunctions (which is a multiple of detM) vanishes at 1 so detM = 0everywhere and our method fails. (The method will also fail for simplerreasons in the case k = 0.) [In fact we are in a situation like the lastexample of Chapter 12 and there is no solution.]

358

K309

We seek a continuous function y(x) = G(x, t) which is twice differ-entiable (using left and right derivatives at end points) on [0, t) and(t, 1]) and satisfies y(0) = y′(0) = 0,

y′′(x) + 2βy′(x) + ω2y(x) = 0 for x ∈ [0, t)

andy′′(x) + 2βy′(x) + ω2y(x) = 0 for x ∈ (t,∞)

whilsty′(t+)− y′(t−) = 1.

This givesy(x) = 0 for x ∈ [0, t]

and [∗]y(x) = Ae−β(x−t) sinh(α(x− t)) +Be−β(x−t) cosh(α(x− t))

for x ∈ [t,∞).

Continuity gives B = 0 and the condition on the jump of the firstderivative gives

1 = αA.

Thus

G(x, t) =

0 if x ≤ t

α−1e−β(x−t) sinh(α(x− t))

so, if β > ω, the general solution on [0,∞) of

y′′(x) + 2βy′(x) + ω2y(x) = f(x)

is

y(t) =

∫ ∞

0

G(s, t)f(s) ds = α−1

∫ t

0

f(s)e−β(t−s) sinh(α(t− s)) ds

with α the positive square root of β2 − ω2.

If ω > β the argument proceeds as far as [∗] in the paragraph above.We then get

y(x) = Ae−β(x−t) sin(λ(x−t))+Be−β(x−t) cos(λ(x−t))+ for x ∈ [t,∞),

where λ is the positive square root of ω2− β2. Arguing much as abovewe get B = 0, A = λ−1 and the general solution on [0,∞) of

y′′(x) + 2βy′(x) + ω2y(x) = f(x)

as

y(t) = λ−1

∫ t

0

f(s)e−β(t−s) sin(λ(t− s)) ds.

If ω = β the argument again proceeds as far as [∗]. We then get

y(x) = A(x− t)e−β(x−t) +Be−β(x−t).

359

Arguing much as before we get B = 0, A = 1 and the general solutionon [0,∞) of

y′′(x) + 2βy′(x) + ω2y(x) = f(x)

as

y(t) =

∫ t

0

(t− s)f(s)e−β(t−s) ds.

If β is large the pointer takes a long time to return to close to equi-librium. If β is small the pointer overshoots and we have violent oscil-lation so it also takes a long time before the pointer remains close toequilibrium. As a rule of thumb β close to ω is preferred.

360

K310

(i) Observe that, if m ≥ n

‖Sn − Sm‖ ≤m∑

r=n+1

‖T r‖ ≤m∑

r=n+1

‖T‖r ≤ ‖T‖n+1

1− ‖T‖ → 0

as n → ∞, so the sequence Sn is Cauchy. Thus we can find an S ∈L(U, U) with ‖Sn − S‖ → 0.

Now

‖S‖ ≤ ‖Sn − S‖+ ‖Sn‖ ≤ ‖Sn − S‖+n∑

r=0

‖T‖r

≤ ‖Sn − S‖+ (1− ‖T‖)−1 → (1− ‖T‖)−1

so ‖S‖ ≤ (1− ‖T‖)−1. A similar calculation gives

‖I − S‖ ≤ ‖T‖(1− ‖T‖)−1.

(ii) Observe that

‖S(I − T )− I‖ ≤ ‖Sn(I − T )− I‖+ ‖(S − Sn)(I − T )‖= ‖T n+1‖+ ‖(S − Sn)(I − T )‖≤ ‖T‖n+1 + ‖S − Sn‖‖I − T‖ → 0

as n → 0, so ‖S(I − T ) − I‖ = 0 and S(I − T ) = I. Similarly(I−T )S = I. Thus I−T is invertible with inverse S. Setting T = I−Agives the result.

(iii) Observe that

‖A− I‖ = ‖B−1B − B−1C‖ ≤ ‖B − C‖‖B−1‖ < 1

so A is invertible. Thus

A−1B−1C = I and CA−1B−1 = BB−1CA−1B−1 = BB−1 = I

so C is invertible with inverse A−1B−1.

We observe that

‖C−1‖ ≤ ‖B−1‖‖A−1‖ ≤ ‖B−1‖(1−‖I−A‖)−1 = ‖B−1‖(1−‖I−A‖)−1

but

1 > ‖B − C‖‖B−1‖ ≥ ‖A− I‖ ≥ 0

so

‖C−1‖ ≤ ‖B−1‖(1− ‖B − C‖‖B−1‖)−1.

Similar arguments give the second inequality.

(iv) If B ∈ E, then part (ii) shows that

Γ = C ∈ L(U, U) : ‖B − C‖ < ‖B−1‖−1 ⊆ E

361

so E is open. If C ∈ Γ then

‖Θ(C)−Θ(B)‖ ≤ ‖B‖−1‖‖B − C‖(1− ‖B‖−1‖B − C‖)−1 → 0

as ‖C −B‖ → 0. Thus Θ is continuous.

(iv) Calculations as in (ii).

(v) Use the chain rule or (essentially the same) observe that

Θ(B +H) = (B +H)−1 = (B(I +B−1H))−1

= (I +B−1H)−1B−1

= (I − B−1H + ǫ(B−1H)‖H‖)B−1

= B−1 −B−1HB−1 + η(H)‖H‖= Θ(B) + ΦB(H) + η(H)‖H‖.

where

‖η(H)‖ = ‖B−1ǫ(B−1H)B−1‖ ≤ ‖B−1‖2‖ǫ(B−1H)‖ → 0

as ‖H‖ → 0.

If n = 1, this reduces to the statement that

d

dxcx−1 = −cx−2.

362

K311

(ii) If ‖A‖ = 0, then A = 0 and there is nothing to prove so weassume ‖A‖ 6= 0. ∆ exists since any non-empty set of real numbersbounded below has an infimum. Now choose ǫ > 0. We can find a jsuch that ‖Aj‖1/j ≤ ∆+ ǫ. Thus

‖Ajk+r‖1/(jk+r) ≤(

‖Aj‖k‖A‖r)1/(jk+r)

≤ (∆ + ǫ)jk/(jk+r)‖A‖r/(jk+r) → ∆+ ǫ

for all 0 ≤ r ≤ j − 1. Thus

∆ ≤ lim infn→∞

‖An‖1/n ≤ lim supn→∞

‖An‖1/n ≤ ∆+ ǫ

and since ǫ > 0 ‖An‖1/n → ∆.

(iii) and (iv) If ρ(A) < 1, set λ = (ρ(A) + 1)/2. There exists anN such that ‖An‖1/n ≤ λ and so ‖A‖n ≤ λn for n ≥ N . The kindof arguments used in K310 show that

∑∞j=0A

j converges and that thelimit is the inverse of I − A.

If ρ(A) > 1 then there exists an N such that ‖An‖1/n ≥ 1 for alln ≥ N so, by the easy part of the GPC,

∑∞j=0A

j fails to converge.

363

K312

(i) ‖(λα)n‖1/n = |λ|‖αn‖1/n

(ii) Take a basis ej and let β be the unique linear map with βej =ej+1 for 1 ≤ j ≤ m− 1, βem = 0. We have ρ(β) = 0.

(iii) The m eigenvectors ej with eigenvalues λj are linearly indepen-dent. (If

∑mj=1 µjej = 0 apply

∏

j 6=k(α−λjι) to both sides.) They thusform a basis. If the standard basis is uj , let θ be the unique linear mapwith θej = uj .

Observe that

‖(θαθ−1)n‖1/n = ‖θαnθ−1‖1/n ≤ ‖θ−1‖1/n‖αn‖1/n‖θ‖1/n → ρ(α)

so ρ(θαθ−1) ≤ ρ(α). Similarly

ρ(α) = ρ(θ−1(θαθ−1)θ) ≤ ρ(θαθ−1)

so ρ(θαθ−1) = ρ(α). But if φ is a diagonal matrix with respect to thestandard basis, φ has norm equal to the absolute value of the largestdiagonal entry so ρ(θαθ−1) = max |λj|.

(iv) If u1 and u2 form the standard basis let α, β be the linear mapsgiven by

αu1 = u2, αu2 = 0, βu2 = u1, βu1 = 0.

364

K313

(i) This is just the formula for change of basis.

(ii) Here and elsewhere we use the fact that

m2 sup1≤i,j≤m

|aij| ≥ ‖α‖ ≥ sup1≤i,j≤m

|aij|

when (aij) is the matrix of α with respect to the matrix. (In fact wemerely use

K sup1≤i,j≤m

|aij| ≥ ‖α‖ ≥ K−1 sup1≤i,j≤m

|aij |

for some K and this follows from the fact that all norms on a finitedimensional space are Lipschitz equivalent.)

Given any η > 0, we can certainly find cij with cij = bij for i 6= j,|cii − bii| < η and the cii all distinct.

(iii) Use the notation of (i). Use (ii) to find βn with m distincteigenvalues and ‖βn − β‖ → 0. Let αn = θ−1βnθ. We know that αn

has the same eigenvalues as βn and

‖α− αn‖ = ‖θ−1(β − βn)θ‖ ≤ ‖θ−1‖‖β − βn‖‖θ‖ → 0

as n→ 0.

(iv) If α has m distinct eigenvalues we know that we can find a basisof eigenvectors. With respect to this basis, α is a diagonal matrix Dwith entries the eigenvalues so

χα(t) = det(tI −D) =

m∏

j=1

(t− λj)

and χα(D) = 0 the zero m×m matrix. Thus χα(α) = 0.

(v) We know that the map from L(Cm,Cm) to Cm2

given by α 7→ (aij)is continuous so the map α 7→ χα(A) is continuous. But the inversemap (aij) 7→ α is also continuous so the map α 7→ χα(α) is continuous.

By the previous parts of the question, given any α we can find αn

with ‖α− αn‖ → 0 with χαn(αn) = 0, so, by continuity, χα(α) = 0

(vi) We need αβ − βα non singular. One (among many ways) ofdoing this is as follows. Take e1, e2, . . . , em to be a basis of Cm. Letα be the linear map with αej = αej+1 for 1 ≤ j ≤ m− 1, αem = αe1and β be the linear map with βej = jαej for 1 ≤ j ≤ m.

365

K314

(i) We know that, given any ǫ > 0, we can find a(ǫ) ≥ 1 and b(ǫ) ≥ 1such that

‖αr‖ ≤ a(ǫ)(ρ(α) + ǫ)r and ‖βr‖ ≤ b(ǫ)(ρ(β) + ǫ)r

for all r ≥ 0. Thus

‖(α+ β)n‖ =

∥

∥

∥

∥

∥

n∑

j=0

(

n

j

)

αjβn−j

∥

∥

∥

∥

∥

≤n∑

j=0

(

n

j

)

‖αj‖‖βn−j‖

≤ a(ǫ)b(ǫ)

n∑

j=0

(

n

j

)

(ρ(α) + ǫ)j(ρ(β) + ǫ)n−j

= a(ǫ)b(ǫ)(ρ(α) + ρ(β) + 2ǫ)n

so, taking n-th roots and allowing n→ ∞,

ρ(α + β) ≤ (ρ(α) + ρ(β) + 2ǫ).

Since ǫ > 0 this gives the result.

(ii) Observe that, given an upper triangular matrix A, we can finda diagonal matrix B with entries in absolute value less than any giveη > 0 such that the diagonal entries of A+B are all distinct. Observethat A and B commute. Now argue as in K313 parts (i) to (iii).

(iii) If α has m distinct eigenvalues, then we may find a basis e1,e2, . . . of eigenvalues vectors with corresponding eigenvalues λj with|λ1| ≥ |λ2| ≥ . . . . Since ‖αne1‖/‖e1‖ = |λ1|n, we have ρ(α) ≥ |λ1|.However, if λ > |λ1| we have

λn

∥

∥

∥

∥

∥

αnm∑

j=1

xjej

∥

∥

∥

∥

∥

≤m∑

j=1

∣

∣

∣

∣

λjλ

∣

∣

∣

∣

‖ej‖ → 0

so ρ(α) ≤ λ. Thus ρ(α) = |λ1|.If α does not have n distinct eigenvalues, then, by (ii), we can find

βr such that the eigenvalues of α and α + βr differ by less than 1/r,‖βr‖ ≤ 1/r and α and βr commute.

Observe that, by (i),

ρ(α + βr) ≤ ρ(α) + ρ(βr) ≤ ρ(α) + ‖βr‖ = ρ(α) + 1/r

and

ρ(α) = ρ(α + βr − βr) ≤ ρ(α + βr) + 1/r

so ρ(α + βr) → ρ(α) and the desired result follows.

366

(iv) If α is represented by A with respect to one basis and by Bwith respect to another, we can find an invertible θ such that θαθ−1 isrepresented by B with respect to the first basis. Now

‖(θαθ−1)n‖ = ‖θαnθ−1‖ ≤ ‖θ‖‖αn‖‖θ−1‖,so, taking n-th roots and letting n→ ∞, we have ρ(θαθ−1) ≤ ρ(α) andsimilarly

ρ(α) = ρ(θ−1θαθ−1θ) ≤ ρ(θαθ−1)

so ρ(θαθ−1) = ρ(α) and there is no ambiguity.

367

K315

(i) Observe that

xn = αnx0 +

n−1∑

j=0

αjb

so xn → (ι− α)−1b.

(ii) Observe that the map τ : Cm → Cm given by τx = b + αxsatisfies

‖τnx− τny‖ = ‖αn(x− y)‖ ≤ ‖αn‖‖x− y‖.For large enough n we have τn a contraction mapping so xn → c wherec is the unique fixed point given by

τc = c

i.e. by αc = b.

Suppose ρ(α) > 1. If∑n−1

j=0 αjb diverges, then take x0 = 0. If not,

take x0 to be an eigenvector corresponding to a largest (in absolutevalue eigenvalue).

Suppose ρ(α) = 1. If∑n−1

j=0 αjb diverges, then take x0 = 0. If

not, then either all of the largest in absolute value eigenvalues are 1and we consider x0 = ±e with e an associated eigenvector to get twofixed points (with necessarily different limits) or one of the largest inabsolute value eigenvalues, is not 1. Take x0 = e with e an associatedeigenvector to get a non-convergent sequence.

368

K316

We require ρ(I − A) < 1. If ‖I − A‖ is small, convergence will berapid with the error roughly multiplied by ‖I − A‖ at each step.

Jacobi needs ρ(D−1(U+L)) < 1 and Gauss needs ρ((D−L)−1U) < 1.

Variation needs ρ(H) < 1 with

H = (D − ωL)−1((1− ω)D + ωU)).

Noting that the determinant of a triangular matrix is the product ofits diagonal elements we have

detH = (det(D − ωL))−1 det((1− ω)D + ωU) = (detD)−1 det((1− ω)D)

= (detD)−1(1− ω)n detD = (1− ω)n.

Now the determinant of a matrix is the product of its eigenvalues (mul-tiple roots counted multiply) (consider the coefficient of t0 in det(tI −A)) so, if | detH| > 1, we have ρ(H) > 1. Thus the scheme fails forω < 0 or ω > 2.

369

K317

From K310 we know that the map α 7→ α−1 is continuous on the opensubset of L(U, U) where the inverse is defined. Thus thus compositionof the maps

x 7→ Df(x), α 7→ α−1

is continuous.

370

K318

(i) Observe that, if m ≥ n,∥

∥

∥

∥

∥

m∑

r=0

αr

r!−

n∑

r=0

αr

r!

∥

∥

∥

∥

∥

≤m∑

r=n+1

‖α‖rr!

→ 0

(since∑m

r=0(xr/r!) converges for all x) so, by completeness,

∑nr=0 α

rr!converges.

(ii) Observe that∥

∥

∥

∥

∥

n∑

r=0

αr

r!

n∑

r=0

βr

r!−

n∑

r=0

(α + β)r

r!

∥

∥

∥

∥

∥

=

∥

∥

∥

∥

∥

∑

0≤r,s≤n

αr

r!

βs

s!−

∑

0≤r+s≤n r,s≥0

αr

r!

βs

s!

∥

∥

∥

∥

∥

=

∥

∥

∥

∥

∥

∑

r+s>n, 0≤r,s≤n

αr

r!

βs

s!

∥

∥

∥

∥

∥

≤∑

r+s>n, 0≤r,s≤n

‖α‖rr!

‖β‖ss!

≤∑

2n≥r+s>n, r,s≥0

‖α‖rr!

‖β‖ss!

=

2n∑

k=n+1

1

k!(‖α‖+ |β‖)k → 0

[Or use results from Chapter 3.]

(iii) Observe that, if 0 < ‖h‖ < ‖α‖/2 we have∥

∥

∥

∥

∥

n∑

r=0

hαr

r!− ι− hα

∣

∣

∣

∣

∣

=

∥

∥

∥

∥

∥

n∑

r=2

hαr

r!

∣

∣

∣

∣

∣

≤n∑

r=2

|h|‖α‖rr!

≤ |h|2‖α‖2∞∑

r=0

2−r−1 = |h|2‖α‖2.

Thus

exp(hα) = ι+ hα + h2θα(h)h2

where ‖θα(h)‖ ≤ ‖α‖2 for 0 < ‖h‖ < ‖α‖/2. Similarly

exp(hβ) = ι+ hβ + h2θβ(h)h2

where ‖θβ(h)‖ ≤ ‖β‖2 for 0 < ‖h‖ < ‖β‖/2.

371

We thus have

‖h−2(exp(hα) exp(hβ)− exp(hβ) exp(hα))− (αβ − βα)‖= ‖h−2

(

(ι+ hα + h2θα(h)h2)(ι+ hβ + h2θβ(h)h

2)

− (ι+ hβ + h2θβ(h)h2)(ι+ hα + h2θα(h)h

2))

− (αβ − βα)‖≤ 2|h|‖α‖‖θβ(h)‖+ ‖β‖‖θα(h)‖+ 2h2‖θα(h)‖‖θβ(h)‖ → 0

as h→ 0. Thus, if αβ 6= βα, we have exp(hα) exp(hβ) 6= exp(hβ) exp(hα)for h small.

If expα exp β = exp(α + β) and exp β expα = exp(β + α) thenexpα exp β = exp β expα.

(iv) Observe by considering the various terms in the expansion or byinduction that

‖(α + κ)n − αn‖ ≤n∑

j=1

(

n

j

)

‖κ‖j‖α‖n−j

≤n∑

j=1

n

(

n− 1

j − 1

)


= ‖κ‖n−1∑

r=0

n

(

n− 1

r

)

‖κ‖r‖α‖n−1−r

= n‖κ‖(‖α‖+ ‖κ‖)n−1

(We shall do something similar in K319.)

Thus∥

∥

∥

∥

∥

N∑

n=0

(α+ κ)n

n!−

N∑

n=0

(α)n

n!

∥

∥

∥

∥

∥

≤N∑

n=1

1

n!‖(α + κ)n − αn‖

≤ ‖κ‖N∑

n=1

n

n!(‖α‖+ ‖κ‖)n−1

= ‖κ‖N−1∑

n=0

1

n!(‖α‖+ ‖κ‖)n

= ‖κ‖ exp(‖α‖+ ‖κ‖)so that

‖ exp(α + κ)− exp(α)‖ ≤ ‖κ‖ exp(‖α‖+ ‖κ‖) → 0

as ‖κ‖ → 0. Thus exp is continuous.

372

K319

(ii) If α ∈ L(U, U) and β ∈ L(U, U) writeΦn(α)(β) = βαn−1 + αβαn−2 + α2βαn−3 + · · ·+ αn−1β.

It is easy to check that Φn(α) : L(U, U) → L(U, U) is a continuouslinear function with

‖Φn(α)‖ ≤ n‖α‖n−1,

The next paragraph (not surprisingly) uses K260 (iii). Observe byconsidering the various terms in the expansion or by induction that

‖(α+ κ)n − αn − Φn(α)(κ)‖ ≤n∑

j=2

(

n

j

)


≤n∑

j=2

n(n− 1)

(

n− 2

j − 2

)


= n(n− 1)‖κ‖2n−2∑

r=0

(

n− 2

r

)

‖κ‖r‖α‖n−2−r

= n(n− 1)‖κ‖2(‖α‖+ ‖κ‖)n−2

Thus Θn is differentiable at α with derivative Φn(α).

(iii) Since ‖Φn(α)‖ ≤ n‖α‖n−1 we have

N∑

j=1

‖Φn(α)‖n!

≤N∑

j=1

n‖α‖n−1

n!≤ e‖α‖

so∑N

j=1Φn(α)n!

converges in the space of linear maps from L(U, U) to

L(U, U) in the appropriate operator norm. Call the limit Φ(α).

We have∥

∥

∥

∥

∥

N∑

n=0

(α+ κ)n

n!−

N∑

n=0

(α)n

n!−

N∑

n=1

Φn(α)

n!κ

∥

∥

∥

∥

∥

≤N∑

n=1

1

n!‖(α+ κ)n − αn − Φn(α)(κ)‖

≤N∑

n=2

1

n!n(n− 1)‖κ‖2(‖α‖+ ‖κ‖)n−2

= ‖κ‖2N−2∑

n=0

1

n!(‖α‖+ ‖κ‖)n

≤ ‖κ‖2 exp(‖α‖+ ‖κ‖)so

‖ exp(α+ κ)− exp(α)− Φ(α)κ‖ ≤ ‖κ‖2 exp(‖α‖+ ‖κ‖)and exp is differentiable at α with derivative Φ(α).

373

K320

Use the fact that

m2 sup1≤i,j≤m

|aij| ≥ ‖α‖ ≥ sup1≤i,j≤m

|aij|

when (aij) is the matrix of α with respect to the matrix. (Or merelyuse

K sup1≤i,j≤m

|aij| ≥ ‖α‖ ≥ K−1 sup1≤i,j≤m

|aij |

for some K and this follows from the fact that all norms on a finitedimensional space are Lipschitz equivalent.)

If A is upper triangular then Ar is upper triangular with (j, j) thentry the rth power of ajj the (j, j) th entry of A. Thus expA is uppertriangular with (j, j) th entry exp ajj. Thus

det(expA) =∏

exp ajj = exp∑

ajj = expTraceA

and, since any α has an upper triangular representation and det andTrace depend on α and not the representation chosen,

det(expα) = expTraceα.

374

K321

(ii) Note, for example, that, writing r(B) for the rank of B, we have

2− r(A) ≥ r(A)− r(A2) ≥ r(A2)− r(A3) ≥ r(A3)− r(A4).

Since A4 = 0 and A2 6= 0 we have r(A2)− r(A3) ≥ 1 so 2 ≥ 3 which isabsurd.

(iii) We have

a2 + bc = 1

b(a + d) = 0

c(a + d) = 0

d2 + bc = 1

so either a+d 6= 0, in which case b = c = 0 and a = d = ±1 so A = ±Ior a+ d = 0 and, either a2 = 1 in which case bc = 0 and

A = ±(

1 β0 −1

)

or A = ±(

1 0β −1

)

for some β or a2 6= 1 and

A =

(

a bb−1(1− a2) −a

)

for some a and b.

Observe that

1 = det I = detA2 = (detA)2

so detA = ±1.

By inspection the only such matrices with detA = 1 are ±I and theonly such matrices with diagonal entries are

A =

(

±1 00 ±1

)

.

(iv) Observe that S(I) = I and that S has a continuous derivativeand thatDS(I)(C) = 2C soDS(I) is invertible. The result now followsfrom the inverse function theorem.

(v) The result fails for Y = 0 since(

1 0β −1

)

is a solution for all β.

375

We show that DS(Y ) is not invertible so the the hypotheses of theinverse function theorem fail. Observe that if

H =

(

u vw x

)

then

DS(Y )H = Y H + Y H

=

(

u v−w −x

)

+

(

u −vw −x

)

= 2

(

u 00 −x

)

so DS(Y ) is not bijective. (Consider, for example, u = x = 0, v = w =1. Then H 6= 0 but DS(Y )H = 0.)

(vi) Inspection of cases (along the lines of (v)) shows that, unlessB = ±I then, if B2 = I, there exist Cn with Cn 6= B, ‖Cn − B‖ → 0and C2

n = I. If B = −I then the argument of (iv) (or argument fromthe result of (iv)) shows that G and H exist.

376

K322

We have

〈αx,x〉 = 〈x, αTx〉 = −〈x, αx〉 = −〈αx,x〉.So 〈αx,x〉 = 0.

If α2x = λx, then

λ〈x,x〉 = 〈λx,x〉 = 〈α2x,x〉 = −〈αx, αx〉 ≤ 0.

Since a real cubic has at least one real root, the characteristic equa-tion det(tι − α) = 0 must have at least one real root. Thus α has aneigenvector e3 of norm 1. By the first paragraph αe3 = 0. We observethat e3 is an eigenvector of α2 and so since α2 is symmetric we can finde1 and e2 such that the ej form an orthonormal basis. We know thatαe1 is perpendicular to e1. Also

〈αe1, e3〉 = 〈e1,−αe3〉 = 0

Thus αe1 = µe2 for some µ ∈ R.

Similarly αe2 is perpendicular to e2 and e3 whilst

〈αe2, e1〉 = 〈e2,−αe1〉 = −µso αe2 = µe1.

With respect to the basis ej the linear map expα has the matrixrepresentation

∞∑

n=0

1

n!

0 µ 0−µ 0 00 0 0

n

=

∑∞r=0

(−1)rµ2r

(2r)!

∑∞r=0

(−1)rµ2r+1

(2r+1)!0

−∑∞r=0

(−1)rµ2r+1

(2r+1)!

∑∞r=0

(−1)rµ2r

(2r)!0

0 0 1

=

cosµ − sinµ 0sin µ cosµ 00 0 1

so expα is a rotation through µ about e3.

377

K323

We haveα = (α + αT )/2 + (α− αT )/2.

The first term is symmetric the second antisymmetric. Note that thedecomposition is unique (if α = θ + φ with θ symmetric and φ anti-symmetric then (α + αT )/2 = θ).

If α is as stated

‖(α− αT )/2‖ ≤ ‖(α− ι)/2‖+ ‖(α− ι)T/2‖ = ‖α− ι‖ < ǫ

so α = ι+ ǫβ + φ with β antisymmetric, ‖φ‖ symmetric and ‖β‖ ≤ 1.

Now

ι = ααT = (ι+ ǫβ + φ)(ι− ǫβ + φ) = ι+ ǫ(βφ− φβ)− ǫ2β2 + 2φ+ φ2

so2φ = −φ2 + ǫ(φβ − βφ) + ǫ2β2.

But

‖(α + αT )/2‖ ≤ ‖(α− ι)/2‖+ ‖(α− ι)T /2‖ = ‖α− ι‖ < ǫ

so2‖φ‖ ≤ ‖φ‖2 + 2ǫ‖φ‖β‖+ ǫ2β2 ≤ ǫ2 + 2ǫ2 + ǫ2 = 4ǫ2

so ‖φ‖ ≤ 2ǫ2. [We can reuse this estimate to show that ‖φ‖ = ǫ2/2 +O(ǫ3).]

378

K324

We call matrices U with UUT orthogonal. If, in addition, detU = 1we say it is special orthogonal.

(i) Observe that

1 = detUUT = detU detUT = (detU)2.

(ii) Observe that(

N∑

j=0

Bj

j!

)T

=N∑

j=0

(BT )j

j!=

N∑

j=0

(−B)j

j!

so, since the map A 7→ AT is continuous, (expB)T = exp(−B) =(expB)−1. (Recall that, if C andD commute, exp(C+D) = expC expD.)

(iii) The maps h 7→ hB, A 7→ expA and C 7→ detC are continuousso their composition θ is. Note if BT = −B we have (hB)T = −hB soexp hB is orthogonal so det(exp hB) = ±1. Since h 7→ det(exp hB) iscontinuous, the intermediate value theorem tells us that that the mapis constant so θ(1) = θ(0) and det(expB) = 1.

(iv) By (ii) and (iii) exp hB is special orthogonal so A(exp hB) =(exp hB)A. But, as in K318, we have

‖h−1(I − hB − exp hB)‖ → 0

so

‖A(h−1(I − hB − exp hB))− (h−1(I − hB − exp hB)A‖ → 0

so‖AB −BA‖ → 0

i.e. AB = BA.

(v) If n ≥ 2, we can have 1 ≤ I, J ≤ n with I 6= J . Considering Bwith matrix given by bIJ = −bJI = 1, bij = 0 otherwise, we see thataII = aJJ and aIJ = −aJI for all I 6= J .

If n ≥ 3, we can have 1 ≤ I, J ≤ n with I, J and K distinctConsidering B with matrix given by bIJ = −bJI = 1, bJK = −bKJ = 1,bij = 0 otherwise, we see that

aJK =n∑

r=1

bIrarK =n∑

s=1

aIsbsK = aIK .

On the other hand, if we consider B with matrix given by bIJ = −bJI =1, −bJK = bKJ = 1, bij = 0 otherwise, we see that

aJK =n∑

r=1

bIrarK =n∑

s=1

aIsbsK = −aIK .

379

The two last results show that aJK = 0 for all J 6= K.

Thus A = λI. Conversely, if A = λI, it is easy to see that A isisotropic.

(vi) When n = 2 we have shown that

A =

(

a b−b a

)

Let k be the positive square root of a2 + b2. Then

A = k

(

cos θ sin θ− sin θ cos θ

)

,

that is A is product of a dilation and a rotation. But the specialorthogonal matrices are precisely the rotations so the isotropic matricesare precisely those which are a product of a dilation and a rotation.The isotropic matrices of determinant 1 are the orthogonal matrices.

(vii) If n = 1 the special orthogonal matrices consist of one example(1) so all matrices are isotropic.

(viii) If we omit the restriction detU = 1, we get a stronger conditionso the new isotropic matrices are a subset of the old. If n ≥ 3 or n = 1this makes no difference, by inspection.

If n = 2, we observe that(

1 00 −1

)(

cos θ sin θ− sin θ cos θ

)(

1 00 −1

)

=

(

cos θ − sin θsin θ cos θ

)

so under the new definition only matrices of the form λI can be isotropic(and it is easily checked that they are).

380

K325

The area of the inscribed n-gon is given by

A =

n∑

k=1

r2 sinθk2cos

θk2

=r2

2

n∑

k=1

sin θk

subject ton∑

k=1

θk = 2π.

We form the Lagrangian

L =r2

2

n∑

k=1

sin θk − λ

n∑

k=1

θk = 2π

and observe that, at a stationary point,

0 =∂L

∂θk= cos θk − λ.

Thus θ1 = θ2 = · · · = θn. Inspection (this is methods question) showsthat we have the maximum.

The area of the inscribed n-gon is given by

A =n∑

k=1

r2 tanθk2

subject ton∑

k=1

θk = 2π.

We form the Lagrangian

L =

n∑

k=1

r2 tanθk2

− λ

n∑

k=1

θk = 2π

and observe that, at a stationary point,

0 =∂L

∂θk=

1

2 cos2 θk2

− λ

Thus θ1 = θ2 = · · · = θn. Inspection (this is methods question) showsthat we have the maximum.

381

K326

Observe that the set

E = x ∈ Rn : xj ≥ 0 for all j and∑

j=1

xpj = c

is closed (observe that the map x 7→ ∑

j=1 xpj is continuous) and

bounded (if x ∈ E then c1/p ≥ xj ≥ 0.) Thus the continuous func-tion f : E → R given by f(x) =

∑

j=1 xjyj attains a maximum. This

maximum does not occur if any of the xj = 0 (for, if xJ = 0, thenif delta is very small and positive and we take x′J = δ and take ǫ sothat taking x′r = xr − ǫ if xr 6= 0, x′r = 0 if xr = 0 and r 6= J wehave x′ ∈ E, we see that ǫ = O(δp) and f(x′) > f(x)). Thus, if theLagrange method gives us a unique stationary point with xj 6= 0 forall j, it will indeed be the maximum.

Form the Lagrangian

L =n∑

j=1

xjyj − λn∑

j=1

xpj .

At a stationary point,

0 =∂L

∂xj= yj − λpxp−1

j ,

so we have indeed got the maximum. Rewriting the last equations weobtain

yj = λpxp−1j

soxpj = kyqj

for some constant k. Since∑n

j=1 xpj = c, we have k = c

(

∑nj=1 y

qj

)−1

and

n∑

j=1

xjyj =

(

c1/pn∑

j=1

y1+q/pj

)(

n∑

j=1

yqj

)−1/p

=

(

c1/pn∑

j=1

yqj

)(

n∑

j=1

yqj

)−1/p

= c1/p

(

n∑

j=1

yqj

)1/q

.

Thus

c1/p

(

n∑

j=1

yqj

)1/q

≥n∑

j=1

tjyj

382

whenever tj ≥ 0 and∑n

j=1 tpj = c with equality if and only if tpj = kyqj .

Thus(

n∑

j=1

tpj

)(

n∑

j=1

yqj

)1/q

≥n∑

j=1

tjyj

with equality if and only if tpj = kyqj .

We assumed that all the yj where non-zero but a little reflectionshows that if tj ≥ 0 and yj ≥ 0 (and t, y 6= 0) then

(

n∑

j=1

tpj

)(

n∑

j=1

yqj

)1/q

≥n∑

j=1

tjyj

with equality if and only if tpj = kyqj .

Thus if t, y 6= 0 then(

n∑

j=1

|tj |p)(

n∑

j=1

|yj|q)1/q

≥n∑

j=1

|tjyj|

with equality if and only if |tj|p = k|yj|q.

383

K327

(iii) The sum of the squares of the four sides of a parallelogram equalsthe sum of the squares of the two diagonals.

(iv) Observe that‖x‖ ≤ ‖x− y‖+ ‖y‖

so‖x‖ − ‖y‖ ≤ ‖x− y‖.

Now interchange x and y.

Observe that

4〈x,y〉 = ‖x+ y‖2 + ‖x− y‖2

≤ (‖x‖+ ‖y‖)2 + (‖x‖ − ‖y‖)2= 4‖x‖‖y‖

(v) Observe, for example, that the parallelogram law fails for theuniform norm when we consider f(x) = max(1 − 4x, 0) and g(x) =f(1− x).

384

K328

(ii) We have

‖u+ v +w‖2 + ‖u+ v−w‖2 = ‖(u+ v) +w‖2 + ‖(u+ v)−w‖2

= 2‖u+ v‖2 + 2‖w‖2.Thus

4(〈(u+w,v〉+ 〈(u−w,v〉)= ‖u+w + v‖2 − ‖u+w− v‖2 + ‖u−w + v‖2 − ‖u−w − v‖2

= 2‖u+ v‖2 + 2‖w‖2 − 2‖u− v‖2 − 2‖w‖2= 8〈u,v〉

so (1) holds.

Setting w = u, we obtain (2). Now set u = (x + y)/2 and w =(x− y)/2 to obtain

〈x,v〉+ 〈y,v〉 = 〈u+w,v〉+ 〈u−w,v〉= 2〈u,v〉= 〈2u,v〉= 〈x+ y,v〉.

(iii) A simple induction now gives 〈nx,y〉 = n〈x,y〉 for n a positiveinteger. The remark that

〈−x,y〉 = 4−1(‖x− y‖2 − ‖x+ y‖2 = −〈x,y〉now gives the result for all integer n. Now we have

n〈mn−1x,y〉 = 〈mx,y〉 = m〈x,y〉so

〈mn−1x,y〉 = mn−1〈x,y〉for all integer m and n with n 6= 0.

Now observe that the Cauchy-Schwarz inequality holds (see K327 (iv))so

|〈λx,y〉 − 〈λkx,y〉| = |〈(λ− λk)x,y〉| ≤ ‖(λ− λk)x‖‖y‖ → 0

as λk → 0. Thus allowing λk → λ through rational values of λk givesthe full result.

(vi) Observe that if a norm obeys the parallelogram law on a densesubset it must obey it on the whole space.

(vii) In parts (ii) and (iii) start by proving the identities for ℜ〈x,y〉.

385

K329*

No comments.

387

K331

(ii) We have xn−1 = ME(n−1, x) = M(n−1, x) → M(0, x) as n → 0

by the continuity of M so M(0, x) = 0 for all x 6= 0. In particularM(0, 1) = M(0, 2), which is incompatible with group laws (ab = cbimplies a = b).

(iii) As for (ii).

(iv) If xn is Cauchy for d, then d(xn, 1) is bounded so there existA > 1 such that A−1 ≤ xn ≤ A. Observe that (by the mean valuetheorem) there exists aK > 1 such thatK−1|x−y| ≤ d(x, y) ≤ K|x−y|for all |x|, |y| ∈ [A−1, A]. Take M(x, y) = xy.

388

K332

(i) If E = 0 then we are done. If not, then since x ∈ E implies−x ∈ E, we know that x ∈ E : x > 0 is non-empty.

Let α = infx ∈ E : x > 0. If α = 0 then we can find xn ∈ E with0 < xn < 1/n. We have

E ⊇∞⋃

n=1

mxn : m ∈ Z

so, since E is closed, E = R.

If α > 0, then since E is closed α ∈ E. Given e ∈ E we can findk ∈ Z such that

α > e− kα ≥ 0.

Since e− kα ∈ E we have e− kα = 0 so E ⊆ αZ so E = αZ.

(ii) A similar argument shows that either E = S1 orE = exp 2πin/N :0 ≤ n ≤ N − 1 for some positive integer N .

(iii) If E is a subset of a one dimensional subspace of R2 then (i)tells that either

E = xu : x ∈ R or E = nu : n ∈ Z.

If not either

(A) xn 6= 0 with ‖xn‖ → 0, or (B) not.

In case (A) either

(A)′ There exists an N such that all the xn lie in a one dimensionalsubspace or (A)′′ not.

In case (A)′′, E will be dense in R2 and so E = R2.

In case (A)′ the argument of (i) tells us that there is a subspace

U = xu : x ∈ R ⊆ E.

We are considering the case U 6= E. Consider

α = inf‖x‖ : x ∈ E \ U.If α = 0, E will be dense in R2 and so E = R2. If α 6= 0 then sinceU is a subgroup of E, ‖y − z‖ ≥ α whenever y ∈ E \ U and z ∈ U .Thus, since E is closed, we can find v ∈ E \U with ‖v‖ = α. Supposee ∈ E. By simple geometry we can find x ∈ R and k ∈ Z such that‖e− xu− kv‖ ≤ ‖v‖/2 so e− xu− kv ∈ U so e− kv ∈ U . Thus

E = xu+ kv : x ∈ R, k ∈ Z.

389

In case (B) let u be an element of smallest norm E \ 0, writeU = lu : l ∈ Z

and let v an element of smallest norm in E \ U . If e ∈ E then bysimple geometry we can find integers l and k such that e− lu− kv liesin the parallelogram with vertices (±u± v)/2 and so is 0. Thus

E = lu+ kv : l ∈ Z, k ∈ Z.

390

K333

Second paragraph. Let X = [−1, 1] with the usual metric d. LetE = X \ 0 and let Y = E with ρ the restriction of d to Y . Let

f(x) = x so f is certainly uniformly continuous. Suppose f : (X, d) →(Y, ρ) is continuous and f(x) = f(x) for all x ∈ E. If f(x) = y then

|x− y| = ρ(f(x), f(0)) → 0

as |x| = d(x, 0) → 0 (x 6= 0) which is absurd.

We have the following theorem. If (x, d) is a metric space and E adense subset ofX then if (Y, ρ) is a complete metric space any uniformlycontinuous function f : E → Y can be extended to a continuous (indeed

uniformly continuous) function f : X → Y .

The proof follows a standard pattern. Suppose x ∈ X . Then we canfind xn ∈ E with d(xn, x) → 0. Since f is uniformly continuous on E,given any ǫ > 0 we can find a δ(ǫ) > 0 such that d(u, v) < δ impliesρ(f(u), f(v)) < ǫ. Choose N such that d(xn, x) < δ/2 for n ≥ N .Then d(xn, xm) < δ and ρ(f(xn), f(xm)) < ǫ for n, m ≥ N . Thusf(xn) is Cauchy and so converges in (Y, ρ) to limit lx, say. A similar,but simpler, argument shows that if yn ∈ E with d(yn, x) → 0 then

ρ(f(yn), lx) → 0. Thus we may define f(x) = lx without ambiguity.

Taking xn = x, we see that f(x) = f(x) for all x ∈ E.

Now suppose ǫ > 0, u, v ∈ X and d(u, v) < δ(ǫ)/3. We can findun, vn ∈ E with d(un, u), d(vn, v) < δ(ǫ)/3 and d(un, u), d(vn, v) → 0.By the triangle inequality d(un, vn) < δ(ǫ) so ρ(f(un), f(vn)) < ǫ and

ρ(f(u), f(v)) ≤ ǫ. Thus f is uniformly continuous.

391

K334

Observe thatf(x)− f(y)

x− y→ 0

as y → x. Thus f is everywhere differentiable with derivative zero andso is constant.

If we replace R by Q, then f is uniformly continuous on Q and socan be extended to a continuous function f on R. If x, y ∈ R andx 6= y choose xn, yn ∈ Q with xn → x and yn → y. Since

|f(xn)− f(yn)| ≤ (xn − yn)2

we have|f(xn)− f(yn)| ≤ (xn − yn)

2

so, allowing n→ ∞,

|f(x)− f(y)| ≤ (x− y)2

whence f is constant and so f is.

The function in Example 1.3 satisfies the condition:-

Given x ∈ Q there exists a δ(x) > 0 with

|f(x)− f(y)| ≤ (x− y)2

for all x, ∈ Q with |x− y| < δ(x) but the δ(x) is not uniform.

392

K335

This is the statement that a complete metric space is totally bounded(see e.g. Section 11.2).

If Fn is finite set such that⋃

x∈FnB(x, 1/n) = X , then E =

⋃∞n=1 Fn

is a countable dense subset.

If we give Z the usual metric, then, since the sequence xn = n has noconvergent subsequence, (|xn − xm| ≥ 1 for m 6= n), Z does not havethe Bolzano–Weierstrass property. However Z is a countable densesubset of itself and since any Cauchy sequence is eventually constantit is complete.

The space Q with the usual metric is not complete but is a countabledense subset of itself.

If we give R the discrete metric (d(x, y) = 1 if x 6= y) thenB(x, 1/2) =x so the only dense subset of R is R itself which is uncountable. Sinceany Cauchy sequence is eventually constant R with the discrete metricis complete.

393

K336

(i) and (ii). Let xn be a sequence of points forming a dense subset.If U is open, consider xn. If xn /∈ U , we do nothing, If xn ∈ U , thenput n ∈ Γ. Since U is open we can find a δ′n > 0 such that d(xn, y) > δ′nfor all y /∈ U . If U = X set δn = 1. Otherwise infy/∈U d(xn, y) is a welldefined strictly positive number. Set δn = infy/∈U d(xn, y)/2.

We claim that⋃

n∈ΓB(xn, δn) = U . Observe that if n ∈ Γ thenB(xn, δn) ⊆ U so

⋃

n∈ΓB(xn, δn) ⊆ U . Now suppose x ∈ U . Since Uis open we can find a δ > 0 with 1 > δ such that B(x, δ) ⊆ U . Sincethe sequence xn is dense we can find an N such that d(xN , x) < δ/4.Automatically infy/∈U d(xN , y) > 3δ/4 so δN ≥ 3δ/8 and x ∈ B(xN , δN).Thus

⋃

n∈ΓB(xn, δn) = U .

(iii) The space R with the discrete metric d(x, y) = 1 if x 6= y iscomplete. Any open ball of the form B(x, r) with r > 1 is R itself. Anyopen ball of the form B(x, r) with r < 1 is x. Thus the countableunions of open balls are countable subsets of R and R itself. But everyset is open so any uncountable set (e.g. x : x > 0) which is not thewhole space is a counterexample.

(iv) Observe that B(x, r) =⋃∞

n=1 B(x, (1− 2−n)r). Since the count-able union of countable sets is countable, part (ii) shows that everyopen set U is the countable union of closed balls L1, L2, . . . say. SetKj =

⋃jr=1Lr.

394

K337

(i) If x ∈ U , we can find a δ > 0 such that (x− δ, x+ δ) ⊆ U . Thusx ∼ x.

If x ∼ y, then x, y ∈ (a, b) ⊆ U so y ∼ x

If x ∼ y and y ∼ z, then x, y ∈ (a, b) ⊆ U and y, z ∈ (c, d) ⊆ U .We then have (a, b)∪ (c, d) an open interval (since both (a, b) and (c, d)contain y) lying within U so x ∼ z.

(ii) Since x ∈ [x], [x] is non-empty. If [x] is bounded above look atα = sup[x]. If α ∈ U then we can find δ > 0 such that (α−2δ, α+2δ) ∈U . We can find γ ∈ [x] such that γ ≥ α − δ so x ∼ γ, γ ∼ α − δ andα − δ ∼ α + δ. Thus x ∼ α + δ contradicting the definition of α. Byreductio ad absurdum,

y ∈ [x] : y ≥ x ⊆ [x, α).

But we can find αn ≥ α−1/n with αn ∼ x, so we can find (an, cn) ⊂ Uwith a, αn ∈ (an, cn). Thus

y ∈ [x] : y ≥ x ⊇∞⋃

n=1

[x, cn) ⊇∞⋃

n=1

[x, αn) = [x, α).

Thusy ∈ [x] : y ≥ x = [x, α).

If [x] is unbounded above we can find αn ≥ n with αn ∼ x so we canfind (an, cn) ⊂ U with x, αn ∈ (an, cn). Thus

y ∈ [x] : y ≥ x ⊇∞⋃

n=1

[x, cn) ⊇∞⋃

n=1

[x,∞)

soy ∈ [x] : y ≥ x = [x,∞).

(iii) Take C to be the set of equivalence classes.

(v) Every open interval contains a rational. Thus each element of Ccontains a rational which belongs to no other element so C is countable.(Or say that the map fQ → C given by f : (q) = [q] is surjective.)

If D is an open disc x : ‖x− x0‖ = r and ‖x− x0‖ = r then anyopen disc containing y intersects D but y /∈ D.

395

K338

(i) Observe that

n∑

j=1

|(f+g)(xj)−(f+g)(xj−1)| ≤n∑

j=1

|f(xj)−f(xj−1)|+n∑

j=1

|g(xj)−g(xj−1)|

so f, g ∈ BV implies f + g ∈ BV and V (f + g) ≤ V (f) + V (g).

Observe also that if V (f) = 0 then

0 ≤ |f(t)− f(0)| ≤ |f(1)− f(t)|+ |f(t)− f(0)| ≤ V (f) = 0

so f(t) = f(0) for all t.

The proof that BV is complete follows a standard form of argument.Suppose fn is Cauchy in BV . Arguing as in the previous paragraph

|fp(t)− fq(t)| ≤ |fp(0)− fq(0)|+ V (fp − V (fq) = ‖fp − fq‖BV

so fn(t) is Cauchy in R for each t. Thus fn → f(t) for some f(t). Wenow show that f ∈ BV . Since any Cauchy sequence is bounded wecan find a K with K ≥ ‖fn‖BV for all n. Now suppose 0 = x0 ≤ x1 ≤· · · ≤ xN = 1, Then

N∑

j=1

|f(xj)− f(xj−1)|

≤N∑

j=1

|(f − fn)(xj)− (f − fn)(xj−1)|+N∑

j=1

|fn(xj)− fn(xj−1)|

≤N∑

j=1

|(f − fn)(xj)− (f − fn)(xj−1)|+K

→ K

396

as n → ∞. Thus f ∈ BV . Finally we use an ‘irrelevant m argument’to observe that

|(f − fn)(0)|+N∑

j=1

|(f − fn)(xj)− (f − fn)(xj−1)|

≤ |(f − fm)(0)|+N∑

j=1

|(f − fm)(xj)− (f − fm)(xj−1)|+ |(fm − fn)(0)|

+

N∑

j=1

|(fm − fn)(xj)− (fm − fn)(xj−1)|

≤ |(f − fm)(0)|+N∑

j=1

|(f − fm)(xj)− (f − fm)(xj−1)|+ ‖fn − fm‖BV

≤ |(f − fm)(0)|+N∑

j=1

|(f − fm)(xj)− (f − fm)(xj−1)|+ supp,q≥n

‖fp − fq‖BV

→ supp,q≥n

‖fp − fq‖BV

as m→ ∞. Thus ‖fn − f‖BV → 0 as n→ ∞.

(ii) If f(t) = 0 for 0 ≤ t ≤ 1/2, f(t) = 1 for 1/2 < t ≤ 1 thenf ∈ BV but f /∈ C. If g(t) = t cosπt then g ∈ C but, examining∑n

j=1 |g(1/j)− g(1/(j + 1))|, we see that g /∈ BV .

C ∩ BV is the intersection of two vector spaces and so a vectorsubspace of BV . If f ∈ BV ∩ C then arguments like those in (i)show that ‖f‖∞ ≤ ‖f‖BV . Thus if fn ∈ BV ∩ C and f ∈ BV and‖fn − f‖BV → 0 we know that fn is Cauchy in (BV, ‖ ‖BV ) and so fnis Cauchy in (C, ‖ ‖∞). Thus fn converges uniformly to some g ∈ C.We now observe that

|fn(t)− f(t)| ≤ ‖f − fn‖BV → 0

and

|fn(t)− g(t)| ≤ ‖g − fn‖∞ → 0

and so f(t) = g(t) for each t ∈ [0, 1]. Thus f = g and f ∈ BV ∩ C.Thus BV ∩ C is closed in (BV, ‖ ‖BV ).

The mean value inequality shows that C1 is a subspace of BV . Weobserve that if g is continuous and G(t) =

∫ t

1/2g(x) dx then

|G(xj)−G(xj−1)| ≤ supt∈[xj−1,xj ]

|g(t)||xj − xj−1|.

397

Thus if we take

gn(x) =

−1 if 0 ≤ x ≤ 2−1 − 2−1n−1,

4n(x− 1/2) if 2−1 − 2−1n−1 < x < 2−1 + 2−1n−1,

1 if 2−1 + 2−1n−1 ≤ x ≤ 1,

and set Gn =∫ t

1/2gn(x) dx, G(t) = |t−2−1| we have Gn ∈ C1, G ∈ BV

and ‖G−Gn‖BV → 0 as n→ ∞ but G /∈ C1.

(iii) Since g ∈ C1 there exists an M such that |g′(t)| ≤ M for all t.Let ǫ > 0. By the definition of V (g). 0 = x0 ≤ x1 ≤ · · · ≤ xn = 1 suchthat

n∑

j=1

|g(xj)− g(xj−1)| ≥ V (g)− ǫ.

Let m be a strictly positive integer. By the staircase property of thefunction f we can find a finite collection I of disjoint closed sets oftotal length 1− (2/3)m such that f is constant on each I ∈ I. We canchoose 0 = y0 ≤ y1 ≤ · · · ≤ yN = 1 so that

xj ∈ y0, y1, . . . , yN

for each 0 ≤ j ≤ n and

either [yr−1, yr] ⊂ I for some I ∈ I or (yr−1, yr) ∩ I = ∅ for all I ∈ I

whenever 1 ≤ r ≤ N . We observe that

N∑

r=1

|g(yr)− g(yr−1)| ≥n∑

j=1

|g(xj)− g(xj−1)| ≥ V (g)− ǫ.

Let us say that r ∈ A if (yr−1, yr) ∩ I = ∅ for all I ∈ I. We have

∑

r∈A|(f + g)(yr)− (f + g)(yr−1)| ≥

∑

r∈A|f(yr)− f(yr−1)| −

∑

r∈A|g(yr)− g(yr−1)|

=∑

r∈A(f(yr)− f(yr−1))−

∑


= 1−∑

r /∈A(f(yr)− f(yr−1))−

∑


≥ 1−∑


≥ 1−∑

r∈AM |yr − yr−1|

≥ 1−M(2/3)m = V (f)−M(2/3)m

399

K339

(iii) f : R → (−1, 1) given by f(x) = (2/π) tan−1 x will do.

(iv) f : I → J given by f(x) = exp(iπx) will do. Observe that(−1, 1) is a dense subset of [−1, 1] which is complete. Observe that Jis a dense subset of z ∈ C : |z| = 1 which is complete.

The Cauchy sequences xn = 1 − 1/n and yn = −1 + 1/n have nolimits in I. But |xn − yn| 9 0 so they can not have the same limit inthe completion.

400

K340

(i) Suppose, if possible, that f−1 is not uniformly continuous. Thenwe can find an ǫ > 0 and un, vn ∈ Y such that ρ(un, vn) → 0 butd(f−1(un), f

−1(vn)) > ǫ. By the Bolzano–Weierstrass property of X ,applied twice to obtain a subsequence and then a subsequence of thatsubsequence, we can find n(j) → 0 and α, β ∈ X such that

d(f−1(un(j)), α) → 0 and d(f−1(vn(j)), β) → 0.

By the continuity of f , we have ρ(un(j), f(α)) → 0 and ρ(vn(j), f(β)) →0 so (since ρ(un, vn) → 0) we have f(α) = f(β). Since f is bijective,α = β so

ǫ < d(f−1(un(j)), f−1(vn(j))) ≤ d(f−1(un(j)), α) + d(f−1(vn(j)), α) → 0

which is absurd.

Suppose yn ∈ Y . Then f−1(yn) ∈ X and by the Bolzano–Weierstrassproperty ofX we can find x ∈ X and n(j) → ∞ such that d(f−1(yn(j)), x) →0 and so, by the continuity of f ,

ρ(yn(j), f(x)) = ρ(

f(f−1(yn(j))), f(x))

→ 0

as j → ∞. Thus Y has the Bolzano–Weierstrass property.

(ii) The fact that f−1 is continuous.

(iii) If f(x) = x1/3, the mean value theorem shows that |f(x) −f(y)| ≤ |x− y|/3 if |x|, |y| ≥ 1. Since f is continuous, it is uniformlycontinuous on [−2, 2] so f is uniformly continuous on R. Howeverf−1(x) = x3 so f(x+ δ) − f(x) ≥ 3xδ2 → ∞ as x → ∞ for all δ > 0,so f−1 is not uniformly continuous.

(iv) Observe that ρ(x, y) < 1/2 implies x = y implies d(f(x), f(y)) =0 but that d(1/n, 0) → 0 and ρ(f−1(1/n), f−1(0)) = ρ(1/n, 0) = 1 9 0.

401

K341

Since (X, d) has the Bolzano–Weierstrass property, there exists a Ksuch that d(x, y) ≤ K for all x, y ∈ X (direct proof or use total bound-edness), thus, by the Weierstrass M-test,

∑∞j=1(2

−jd(x, xj))2 converges

and f(x) ∈ l2.

Observe that

‖f(x)− f(y)‖22 =∞∑

j=1

(2−j(d(x, xj)− d(y, xj))2

≤∞∑

j=1

(2−jd(x, y))2 ≤ d(x, y)2

so f is continuous.

If x 6= y then d(x, y) > 0. Set 4k = d(x, y). We can find an N withd(x, xN) < k and so with d(y, xN) > 3k whence d(x, xN) 6= d(y, xN)and f(x) 6= f(y). Thus f is continuous. It follows that f : X → f(X) isa bijective continuous function and, by K339 (i) is a homeomorphism.

402

K342

(i) We can have (X, d) complete but (Y, ρ) not. Let X = R with theusual metric d, Y = (−π/2, π/2) with the usual metric ρ, and f(x) =tan−1(x). Using the mean value inequality |f(x)− f(y)| ≤ |x− y|.

We can have (Y, ρ) complete but (X, d) not. Let X = (−4, 4) withthe usual metric d and y = [−1, 1] with the usual metric ρ. Takef(x) = sin x and observe that by the mean value inequality |f(u) −f(v)| ≤ |u− v|.

(ii) If (X, d) is complete (Y, ρ) must be. Let yn be Cauchy in Y .Since f is surjective we can find xn ∈ X with f(xn) = yn. Since

d(xn, xm) ≤ K−1ρ(f(xn), f(xm) = ρ(yn, ym),

the xn are Cauchy so we can find x ∈ X with d(xn, x) → 0. Bycontinuity, ρ(yn, f(x)) = ρ(f(xn0, f(x)) → 0 as n→ ∞.

We can have (Y, ρ) complete but (X, d)) not. Let Y = R withthe usual metric ρ, and X = (−π/2, π/2) with the usual metric df(x) = tan(x). Using the mean value theorem, |f(x)− f(y)| ≥ |x− y|.

403

K343

Observe thatαβ ∪ βγ ⊇ αγ

so d(α, β) + d(β, γ) ≥ d(α, γ).

Suppose αn = (an, bn) forms a Cauchy sequence with respect to d.If there exists a c > 0 such that bn − an > c for all n, then we knowthat there exists an N such that d(αn, αm) < c/2 for n, m ≥ N . Thus,provided n, m ≥ N , we know that αn∩αm 6= ∅ so |an−am| ≤ d(αn, αm).It follows that an is Cauchy in the standard Euclidean metric, so an → afor some a. Similarly bn → b for some b. Since bn − an > c we haveb− a ≥ c > 0 and, writing α = (a, b), we have d(αn, α) → 0.

If there does not exist a c > 0 such that bn − an > c for all n, thenwe can find n(j) → ∞ such that bn(j) − an(j) → 0. If θ is any openinterval of length |θ|, d(θ, αn(j)) → |θ|. Thus (X, d) is not complete.

However if we takeX∗ = X∪∞ and define d∗ by d∗(α, β) = d(α, β)for α, β ∈ X , d∗(θ,∞) = d∗(∞, θ) = |θ| if θ ∈ X and d∗(∞,∞) = 0then d∗ is a metric and a Cauchy sequence which consists from somepoint on of intervals of length greater than some fixed c converges bythe arguments above and (since a Cauchy sequence with a convergentsubsequence converges) one which does not converges to ∞.

[The reader may prefer to replace ∞ by ∅.]

404

K344

(ii) Observe that the open unit ball E centre 1 consists of the oddnumbers. If r ∈ E and s /∈ E then d(r, s) = 1 so E is closed.

(iii) d(n + 2k, n) ≤ 2−k → 0 so the complement of n is not closedand n is not open.

(iv) d(r, s) ≤ 2−k implies r ≡ s mod 2k so r2 ≡ s2 mod 2k andd(r2, s2) ≤ 2−k. Thus f is continuous.

(v) d(n+ 2k, n) → 0 as k → ∞ but d(2n+2k , 2n) = 2−n 9 0. Thus gis nowhere continuous.

(vi) Let xn =∑n

r=0 22r. If 0 ≤ k ≤ 2m then (think of the binary

expansion of x− k) d(xn, y) ≥ 2−m−2 for all n ≥ k + 2 so the sequencexn can not converge. However d(xn, xm) ≤ 2−2r for all n, m ≥ r so thesequence is Cauchy.

405

K345

Recall that we can find E : R → R infinitely differentiable withE(x) = 0 for x < 0 and E strictly increasing on [0,∞). Thus, if weconsider F (x, y) = E(x + 1)E(−x+ 1)E(y + 1)E(−y + 1), we have Finfinitely differentiable, F (0, 0) 6= 0 and F (x, y) = 0 if |x| ≥ 1 and/or|y| ≥ 1.

Set (xn, yn) = (n−2, n−4) and δn = (10n)−8. Note that, if we write

An = [xn − 2δn, xn + 2δn]× [yn − 2δn, yn + 2δn, ]

no line through the origin can intersect both An and Am for n 6= m.

Set

G(x, y) =

∞∑

m=1

n−1F (4δ−1m (x− xm), 4δ

−1m (y − ym)).

For each (u, v) 6= 0 we can find a δ > 0 and an n such that

F (4δ−1m (x− xm), 4δ

−1m (y − ym)) = 0

for all m 6= n and all |u − x|, |v − y| < δ. Thus G is well definedeverywhere (observe that F (4δ−1

m (−xm), 4δ−1m (−ym)) = 0) and infinitely

differentiable except at (0, 0).

Since δn ≤ (10n)−8 we have G(x, y) ≤ n−1 for |x|, |y| ≤ (10n)−1 soG is continuous at (0, 0). Since no line through the origin can intersectboth An and Am for n 6= m, G(λt, µt) = 0 for sufficiently small |t|, soG has directional derivative zero at the origin.

However,

|G(xn, yn)−G(0, 0)|‖(xn, yn)‖

=n−1E(1)4

‖(xn, yn)‖≥ nE(1)4/2 → ∞,

so G is not differentiable at the origin.

partial solutionsforquestions in appendixkof a companionto

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