Partial Solutions for Questions in Appendix K of A Companion to Analysis

Partial Solutions for Questions in Appendix K of A Companion to Analysis T. W. K¨ orner 1 2 Introduction Here is a m...

Author: T. W. Korner

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Partial Solutions for Questions in Appendix K of A Companion to Analysis T. W. K¨ orner

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Introduction Here is a miscellaneous collection of hints, answers, partial answers and remarks on some of the exercises in the book. I expect that there are many errors both large and small and would appreciate the opportunity to correct them. Please tell me of any errors, unbridgable gaps, misnumberings etc. I welcome suggestions for additions. ALL COMMENTS GRATEFULLY RECEIVED. (If you can, please use LATEX 2ε or its relatives for mathematics. If not, please use plain text. My e-mail is [email protected]. You may safely assume that I am both lazy and stupid so that a message saying ‘Presumably you have already realised 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 left objects are right objects. One way round this problem is to quote X’s theorem.’) To avoid disappointment note that a number like K15* means that there is no comment. A number marked K15? means that I still need to work on the remarks. Note also that what is given is at most a sketch and often very much less. Please treat the answers as a last resort. You will benefit more from thinking about a problem than from reading a solution. I am inveterate peeker at answers but I strongly advise you to do as I say and not as I do. It may be easiest to navigate this document by using the table of contents which follow on the next few pages.

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Contents Introduction K1 K2 K3 K4 K5 K6 K7 K8 K9 K10 K11 K12 K13* K14* K15 K16 K17 K18 K19 K20 K21 K22 K23* K24 K25 K26 K27 K28 K29 K30 K31 K32 K33 K34* K35* K36 K37* K38 K39 K40 K41 K42 K43

2 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

4

K44 K45 K46* K47 K48 K49 K50 K51 K52 K53 K54 K55 K56 K57 K58 K59 K60 K61 K62 K63 K64 K65 K66 K67 K68 K69 K70 K71 K72 K73 K74 K75 K76 K77 K78 K79* K80 K81 K82 K83 K84 K85 K86 K87 K88

54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

5

K89 K90 K91 K92 K93 K94 K95 K96 K97 K98 K99 K100 K101 K102 K103 K104 K105 K106 K107 K108 K109 K110 K111 K112 K113 K114 K115 K116 K117 K118 K119 K120 K121 K122 K123 K124 K125 K126 K127 K128 K129 K130 K131 K132 K133

100 101 102 103 104 105 106 107 108 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145

6

K134 K135* K136 K137* K138 K139 K140 K141 K142 K143 K144 K145 K146* K147 K148 K149 K150 K151 K152* K153 K154 K155 K156 K157 K158 K159 K160 K161 K162 K163 K164 K165 K166 K167 K168 K169 K170 K171 K172 K173* K174 K175* K176 K177 K178

146 147 148 149 150 151 152 153 154 155 158 160 161 162 163 164 165 166 167 168 169 170 171 172 173 175 176 177 178 179 180 181 183 184 185 186 187 188 190 191 192 193 194 195 196

7

K179 K180 K181 K182 K183 K184* K185 K186* K187 K188 K189 K190 K191 K192 K193 K194 K195 K196 K197 K198 K199 K200 K201 K202 K203 K204* K205 K206 K207 K208 K209 K210 K211 K212 K213 K214 K215 K216 K217 K218 K219 K220 K221 K222 K223

198 199 200 202 203 204 205 206 207 208 209 210 211 213 214 215 216 218 219 220 221 222 223 224 226 227 228 229 231 232 234 235 236 237 239 241 242 243 244 245 246 247 248 249 250

8

K224 K225 K226 K227 K228 K229 K230 K231 K232 K233 K234 K235 K236 K237 K238 K239 K240 K241 K242 K243* K244 K245 K246 K247 K248* K249* K250 K251* K252* K253 K254 K255 K256 K257 K258 K259 K260 K261 K262 K263 K264 K265 K266 K267 K268

251 252 253 254 256 257 259 261 263 264 265 267 268 270 272 273 274 275 277 278 279 280 281 282 284 285 286 287 288 289 290 291 292 293 294 296 297 299 301 303 305 307 308 309 310

9

K269 K270 K271 K272 K273 K274 K275 K276 K277 K278 K279 K280 K281 K282 K283 K284 K285 K286 K287 K288 K289 K290 K291 K292 K293 K294 K295 K296* K297 K298 K299 K300 K301 K302 K303 K304 K305 K306 K307 K308 K309 K310 K311 K312 K313

311 312 313 314 316 318 319 320 321 322 323 324 325 326 327 328 329 331 332 334 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 352 353 355 356 357 359 361 363 364 365

10

K314 K315 K316 K317 K318 K319 K320 K321 K322 K323 K324 K325 K326 K327 K328 K329* K330 K331 K332 K333 K334 K335 K336 K337 K338 K339 K340 K341 K342 K343 K344 K345

366 368 369 370 371 373 374 375 377 378 379 381 382 384 385 386 387 388 389 391 392 393 394 395 396 400 401 402 403 404 405 406

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 positive integer. We can certainly find a k such that k + 1 > x ≥ k (ultimately by the Axiom of Archimedes). But x is not an integer so k +1 > x > k. Since mx, m and k are integers, m0 = mx − mk is. Also m0 x = mx2 − mkx = mN − k(mx)

is an integer since mx is. Since 1 > x − k > 0 we have m > mx − mk = m0 > 0 and m > m0 ≥ 1 contradicting the definition of m as least strictly positive integer with mx an integer.

12

K2 Write cn = dn + d−1 n . Then d2n − cn dn + 1 = 0 p cn ± c2n − 4 . dn = 2 Since the product of the two roots of F is 1, we must take the bigger root. Thus p √ cn + c2n − 4 c + c2 − 4 → dn = 2 2 where c is the limit of cn .

F

For the second paragraph, just take d2n = (1 + k)/2 and d2n+1 = 2/(1 + k). For the third paragraph, the condition |dn | ≥ 1 will do but we must argue carefully. Look first 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, discontinuous everywhere else.) (v) is false. Take e.g. x = 16−1 + 17−1/2 , y = 3−1 − 17−1/2 .

16

K6 PN

n=1

by 1.

2−n H(x − qn ) is an increasing sequence in N bounded above

Observe that if x ≥ y then H(x − qn ) − H(y − qn ) ≥ 0 for all n ≥ 0. Thus N N X X −n 2 H(x − qn ) − 2−n H(y − qn ) n=1

=

N X n=1

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 all n ≥ 0 and H(x − qm ) − H(qm − qm ) = 1 so N X

2

n=1

−n

H(x − qn ) −

N X n=1

2−n H(qm − qn ) ≥ 2−m

for all N ≥ m so f (x) ≥ f (qm ) + 2−m for all x > qm . Thus f is not continuous at qm . If x is irrational, we can find an ² > 0 such that |qm − x| > ² for all m ≤ M and so ¯ ¯ N N ¯X ¯ X ¯ ¯ 2−n H(x − qn ) − 2−n H(y − qn )¯ ≤ 2−M ¯ ¯ ¯ n=1

n=1

whenever |x − y| < ². 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, if an is unbounded, an → −∞. For the reverse inequality, observe that −an satisfies the original inequality.

18

K8 A decreasing sequence bounded below tends to a limit, so, for any fixed 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) =

(

1 4 x 4

+

x 2

for x > 1/2, 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 interesting.)

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K9 (ii) we need only look at positive integers. If n ≥ 2(k + 1) n µ ¶ X n j n (1 + δ) = δ j j=0 µ ¶ n ≥ δ k+1 k+1 (n/2)k+1 k+1 δ ≥ (k + 1)! nk+1 δ k+1 ≥ k+1 2 (k + 1)! so n−k (1 + δ)n ≥ n as n → ∞.

δ k+1 →∞ 2k+1 (k + 1)!

20

K10 Observe that µ ¶ x1 + x 2 + · · · + x n x1 + x2 + · · · + xn−1 xn 1 n = − (n − 1) n n n n−1 µ ¶ ¶ µ x1 + x 2 + · · · + x n 1 x1 + x2 + · · · + xn−1 = − 1− n n n−1 → 1 − (1 − 0)1 = 0

as n → ∞.

Either there exists an N such that m(n) = m(N ) for all n ≥ N and the result is immediate or m(n) → ∞ and xm(n) xm(n) ≤ →0 n m(n) as n → ∞. If α > 1, then x1 + x2 + · · · + xn ³ xm(n) ´α−1 xα1 + xα2 + · · · + xαn → 1 × 0 = 0. ≤ nα n n If α < 1, then x1 + x2 + · · · + xn ³ xm(n) ´α−1 xα1 + xα2 + · · · + xαn ≥ → ∞. nα n n

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 positive integer N with N −1 < ² 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.

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K12 P P j n−1−j j n−1−j |. | ≤ |x − y| n−1 (i) |xn − y n | = |(x − y) n−1 j=0 |x| |y| j=0 x y P j (ii) If P (t) = N j=0 aj t , then |P (x) − P (y)| ≤ |x − y|

N X j=0

aj |j|Rj−1 .

(vi) An increasing sequence bounded above converges. N X

k=J+1

10−k! ≤ 10−(J+1)!

N X

k=J+1

10k−(J+1) ≤

10 × 10−(J+1)! . 9

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 0 (c) > 0. We can find an ² > 0 such that ¯ ¯ ¯ f 0 (c) ¯ f (t) − f (c) 0 ¯< ¯ − f (c) ¯ ¯ t−c 2

and so

f (t) − f (c) f 0 (c) > t−c 2 for all t with |t − c| < ². Thus f (t) > f (c) for c < t < c + ² and f (t) < f (c) for c − ² < t < c

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

26

K16 Repeat the lion hunting argument for the intermediate value theorem with f (an ) ≥ g(an ), f (bn ) ≤ g(bn ). If an → c, say, then, since f is increasing 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 ² > 0. Choose δk > 0 such that |fk (x) − f (c)| < ² for all |x − c| < δk . Set δ = min1≤k≤N δk . We claim that |g(x) − g(c)| < ² for all |x − c| < δ. To see this, suppose, without loss of generality, that fN (c) = g(c). Then and

g(x) ≥ fN (x) > fN (c) − ² = g(c) − ² fj (x) ≤ fj (c) + ² ≤ g(c) + ²

for all 1 ≤ j ≤ N and so for all |x − c| < δ.

g(x) ≤ g(c) + ²

For the example we can take (a, b) = (−1, 1), c = 0 and   if x ≤ 0, 0 fn (x) = nx if 0 ≤ x < 1/n,  1 if 1/n ≤ x.

28

K18 (i) Observe that {x ∈ R : E ∩ (−∞, x] is countable}

is a non-empty bounded set and so has a supremum α say. S Observe that En = E ∩ (−∞, α − 1/n] is countable so E ∩ (−∞, α) = ∞ n=1 En is countable and so E ∩ (−∞, α] is countable. Thus E ∩ (−∞, γ) is uncountable if and only if γ > α. Similarly we can find a β such that E ∩ (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 > γ}

are uncountable if and only if α < γ < β. (b) There exists an α with

{e ∈ E : e < γ} and {e ∈ E : e > γ}

are uncountable if and only if α < γ. (c) There exists a β with

{e ∈ E : e < γ} and {e ∈ E : e > γ}

are uncountable if and only if γ < β. (d) We have

{e ∈ E : e < γ} and {e ∈ E : e > γ}

uncountable 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) False. Consider xj = (−1)j . (iv) False. Consider xj = (−1)j .

30

K20 (i) Let ² > 0. There exists an N such that an ≤ lim supr→∞ ar + ² for n ≥ N . There exists an M such that n(p) ≥ N for p ≥ M . Now lim supr→∞ ar + ² ≥ an(p) for p ≥ M . Since ² was arbitrary, lim sup ar ≥ a. r→∞

(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 ² > 0. If n is large enough lim sup ar + ² > an and lim sup br + ² > bn r→∞

r→∞

so lim sup ar + lim sup br + 2² > an + bn . r→∞

r→∞

Thus lim sup ar + lim sup br + 2² > lim sup(an + bn ). r→∞

r→∞

n→∞

But ² is arbitrary so lim sup ar + lim sup br ≥ lim sup(an + bn ). r→∞

r→∞

n→∞

(True) Similarly if ² > 0 and n is large enough bn ≥ lim inf br − ² r→∞

and so an + bn ≥ an + lim inf br − ². r→∞

Thus lim sup(an + bn ) ≥ lim sup an + lim inf br − ² n→∞

n→∞

r→∞

and, since ² was arbitrary, lim sup(an + bn ) ≥ lim sup an + lim inf br . n→∞

n→∞

r→∞

31

K21

Thus

°2 ° 2 2 ° ° x y ° = kxk − 2 x · y + kyk ° − ° kxk2 kyk2 ° kxk2 kxkkyk kyk2 kyk2 − 2x · y + kxk2 = kxk2 kyk2 ¶2 µ kx − yk . = kxkkyk

° ° ° x ° kx − yk y ° ° − ° kxk2 kyk2 ° = kxkkyk . Ptolomey’s result now follows from the triangle inequality for points of the form kxk−2 x.

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 k(xn , yn ) − (x, y)k → 0, then xn → x and yn → 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 (union two 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 subsequence yn(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 argument fails 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 k(xj , f (xj )) − (x, y)k → 0. Then kxj − xk → 0. Since E is closed, x ∈ E. Since f is continuous, kf (xj ) − f (x)k → 0. Thus y = 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 kxj −xk → 0. Since G is closed and bounded we can find a subsequence j(k) → ∞ such that k(xj(k) , f (xj(k) )) − (x, y)k → 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 at x, we could find a δ > 0 and xj ∈ E and kxj − xk → 0 such that kf (xj ) − f (x)k > δ and so k(xj(k) , f (xj(k) )) − (x, y)k > 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 upper semicontinuous but not continuous. If f is not bounded we can find xn ∈ E such that f (xn ) ≥ n. Extract a convergent subsequence and use upper semicontinuity to obtain a contradiction. To show that the least upper bound is attained, take a sequence for which f approaches its supremum, extract a convergent subsequence and use upper semicontinuity again. Take n = 1, K = [0, 1], f (x) = −1/x for x 6= 0 , f (0) = 0 to obtain an upper semicontinuous function not bounded below. Take n = 1, K = [0, 1], f (x) = x for x 6= 0 , f (0) = 1 to obtain an upper semicontinuous function bounded below which does not attain its infimum. (As usual it is informative to see how the proof of the true result breaks 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 is an integer. (Here [x] is the integer part of x.) (iii) By considering g(x) − kx we see that we may suppose k = 0. Given ² > 0 there exists an X > 0 such that |g(x + 1) − g(x)| < ² (and so |g(x + n) − g(x)| < n² for x > X. By hypothesis there exists a K such that |g(y)| ≤ K for 0 ≤ y ≤ X + 1. Write n(x) for the integer such that X < x − n(x) ≤ X + 1. If x > X + 1 we have ¯ ¯ ¯ ¯ ¯ ¯ ¯ g(x) ¯ ¯ g(x) − g(x − n(x)) ¯ ¯ g(x − n(x)) ¯ ¯ ¯≤¯ ¯+¯ ¯ ¯ x ¯ ¯ ¯ ¯ ¯ x x

n(x)² K + →² x x as x → ∞. Since ² 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|

≤ khkkxk + kykkkk + khkkkk → 0

as k(h, k)k → 0. Since α is continuous, the map A given by x 7→ (x, αx) is continuous so, composing the two maps A and J, we see that the given map is continuous. Last part, continuous real valued function on a closed bounded set attains a maximum. (ii) Observe that, if h · e = 0, then ke + δhk2 = (1 + δ 2 ) so by (i) and so for all δ so

(1 + δ 2 )e · (αe) ≥ (e + δh) · (α(e + δh))

δ(e · (αh) + h · (αe + δ(e · (αe) − h · (αh)) ≥ 0 e · (αh) + h · (αe) = 0.

(iv) If we assume the result true for Rn−1 , then, since U has dimension 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)| ≤ kuvk

since, as simple consequence of the triangle inequality, |kak − kbk| ≤ ka − bk. Choose any z ∈ E. Set R = kz − yk. The continuous function g ¯ R) ∩ E at x0 , say. attains a minimum on the closed bounded set B(y, ¯ R) and g(x) ≥ g(x0 ) automatically, or If x ∈ E then either x ∈ B(y, ¯ x∈ / 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, deduce that u · h = 0. If kx1 − yk = kx0 − yk and x1 ∈ E, then, setting h = x1 − x0 , we get h · h = 0. (iii) Since α 6= 0 we can find a y ∈ / E. Set u = y − x0 and b = −1 (kuk) u. Observe that (x − (b · x)b) · b = 0

so, since E has dimension n − 1, we have and so Set a = (αb)b.

x − (b · x)b ∈ E

α(x − (b · x)b) = 0 αx = b · xαb.

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 kyk > ky − δxk. (iii) Translation.

43

K33 Write B for the closed unit ball. If kzk < 1 then z = (1 − kzk)z + kzk0

so z is not an extreme point.

On the other hand, if kxk = 1 and

z = λx + (1 − λ)y

with 0 < λ < 1 and x, y ∈ B, then, by the triangle inequality, kzk ≤ kλxk + k(1 − λ)yk

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

and kzk = 1. Thus kxk = kyk = 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 · a attains a maximum on K. Suppose u0 is an extreme point of K 0 . If v, w ∈ K, 1 > λ > 0 and then

x0 + u0 = λv + (1 − λ)w

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 by equality and so v, w ∈ K 0 . Since u0 is an extreme point of K 0 , v = w = u0 . (iv) Use a similar argument to (ii) (or (ii) itself) to show that extreme points 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 can find K1 , K2 , . . . KN , KN +1 , . . . , KN +M ∈ K such that N \

j=1

and so

Kj ∩ [a, c] = ∅ and N\ +M j=1

M \

j=N +1

Kj ∩ [c, b] = ∅

Kj ∩ [a, b] = ∅.

(ii) We find [an , bn ] such that Ln = {K ∩ [an , bn ] : K ∈ K}

has the finite intersection property

a = a 0 ≤ a1 ≤ a2 ≤ · · · ≤ a n ≤ bn ≤ · · · ≤ b 2 ≤ 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 each K ∈ K. For, if not, we can find K0 ∈ K such 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 of LN . T Thus, by reductio ad absurdum, α ∈ K∈K K.

47

K37* No comments

48

K38 Observe that, if [aj , bj ] ∈ K, then N \

j=1

[aj , bj ] = [ max aj , min bj ] ∈ K 1≤j≤N

1≤j≤N

so, in particular, K has the finite intersection property. T If c ∈ K∈K K then c ∈ [a, b], that is to say a ≤ c ≤ b whenever a ∈ E and b ≥ e for all e ∈ E. Thus c is a (and thus the) greatest lower bound for E.

49

K39 (i) f10 (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) f20 (t) = f1 (t) ≥ 0 for t ≥ 0 and f2 (0) = 0.

(iv) We have 2N X (−1)j t2j+1 j=0

(2j + 1)!

≥ sin t ≥

2N +1 X

(−1)j t2j+1 (2j + 1)!

2N +1 X

(−1)j t2j+1 (2j + 1)!

j=0

for t ≥ 0 and, using the fact that sin(−t) = − sin t, 2N X (−1)j t2j+1 j=0

for t ≤ 0.

(2j + 1)!

≤ sin t ≤

j=0

(v) Thus

as N → ∞.

¯ ¯ N ¯ j 2j+1 ¯ X (−1) t |t|N +1 ¯ ¯ →0 ¯sin t − ¯≤ ¯ (2j + 1)! ¯ (2N + 3)! j=0

50

K40 (i) Observe that g 0 is increasing since g 00 is positive. If g 0 (x1 ) > 0 then g 0 (t) ≥ g 0 (x1 ) > 0 for all t ∈ [x1 , x2 ] so 0 = g(x2 ) > g(x1 ) = 0 which is absurd. Thus g 0 (x1 ) ≤ 0 and g 0 (x2 ) ≥ 0. By the intermediate value theorem there exist a c ∈ [x1 , x2 ] such that g 0 (c) = 0. Since g 0 is increasing g 0 (t) ≤ 0 and g is decreasing on [x1 , c] whilst g 0 (t) ≥ 0 and g is increasing on [c, x2 ]. Thus g(t) ≤ g(x1 ) = 0 on [x1 , c] and g(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 the hypotheses of (i) hold. (iii) We may assume that 1 > λn+1 > 0. The key algebraic manipulation in a proof by induction runs as follows. Ã n+1 ! Ã ! n X X λj Pn λj xj = f λn+1 xn+1 + (1 − λn+1 ) f xj λ k k=1 j=1 j=1 ! Ã n X λj Pn xj ≤ λn+1 f (xn+1 ) + (1 − λn+1 )f k=1 λk j=1 ≤ λn+1 f (xn+1 ) + (1 − λn+1 )

=

n+1 X

λj f (xj )

n X j=1

λ Pn j

k=1

λk

j=1

since n X j=1

λ Pn j

k=1

λk

= 1 and

n X k=1

λk = (1 − λn+1 ).

(iv) Apply Jensen as suggested and take exponentials.

f (xj )

51

K41 The area of the inscribed quadrilateral is α=

4 X

a2 sin θj cos θj =

1 2

4 X

a2 sin 2θj .

j=1

j=1

with a the radius of the circle. Also

P4

j=1 θj

= π.

Now sin00 t = − sin t ≤ 0 for t ∈ [0, π], so − sin is convex on [0, π] and Jensen’s inequality gives Ã 4 ! 4 X X 1 1 α = 2a2 sin 2θj ≤ 2a2 sin (2θj ) = 2a2 sin(π/2) = 2a2 . 4 4 j=1

j=1

The area is attained when the θj are all equal (and with a little thought, observing that sin00 t < 0 on (0, π)) only then. Consider a circumscribing n-gon A1 A2 . . . An . If Aj Aj+1 touches the circle at Xj let φP 2j−1 be the angle ∠Aj−1 OXj and φ2j be the angle ∠Xj OAj . Then r=1 2nφr = 2π and the area of the circumscribed polygon is Ã 2n ! 2n X X β = 2−1 a2 tan φr ≥ 2−1 a2 2n tan φr /2n = na2 tan(π/n) r=1

r=1

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

52

K42 Since g is continuous on [0, 1], it attains a maximum M say. Let E = {x ∈ [0, 1] : f (x) = M }. Since E is non-empty it has a supremum γ. Since f is continuous, g(γ) = M . If γ = 0 or γ = 1 then M = 0. If not 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 ∈ E contradicting the definition of γ. We have shown that M = 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) by G(t) = g(t) − A − Bt with G(0) = G(1) = 0, we can show that g is linear. [In higher dimensions the condition becomes ‘the average of g over the surface of any sufficiently small sphere equals the value at the centre’ and this turns out to be equivalent to saying that g satisfies Laplace’s equation O2 g = 0. Observe that in one dimension this reduces to saying that g 00 = 0 so g is linear as we saw above. However in higher dimensions the solutions of Laplace’s equation are much more diverse.]

53

K43 Observe that ¯ ¯ ¯ f (h) − f (0) ¯ ¯ = |h sin h−4 | ≤ |h| → 0 ¯ ¯ ¯ h as h → 0.

f 0 (x) = 2x sin x−4 − 4x−2 cos x−4 . Thus we have ¡If x 6= 0, then f ((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 0 (c) = 0, we have c ∈ (a, b) and f 0 (c) = k. For the final paragraph, note that f 00 (a) ≤ 0 ≤ f 00 (b).

55

K45 (i) The statement ‘f must jump by at least 1 at z’ is not well defined. If the reader feels that there should be some way of making this statement well defined in a useful way she should consider f (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 |)

≥ 12 (|f (z) − f (xn )| + |f (z) − f (yn |) ≥ 12 .

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

(ii) The behaviour of f 0 (zn(j) ) does not control the value of f 0 (z). Proposed result is false, see K43.

56

K46* No comments.

57

K47 Since g 0 is constant, the conclusions of intermediate value theorem hold trivially for g 0 .

58

K48 Observe that cos nθ = <(cos θ + i sin θ)n X µn¶ = (−1)r cosn−2r θ sin2r θ 2r 0≤2r≤n X µn¶ (−1)r cosn−2r (1 − cos2 θ)r = 2r 0≤2r≤n

a real polynomial in cos θ of degree at most n. (Part (ii) shows that the degree is exactly n.) T0 (t) = 1, T1 (t) = t, T2 (t) = 2t2 − 1, T3 (t) = 4t3 − 3t. (a) Observe that cos nθ 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 + 2 points is identically zero, so tTn (t) = (Tn+1 (t) + Tn−1 (t))/2 for all t. (c) Induction using (i). (d) |Tn (cos θ)| = | cos nθ| ≤ 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 range 0 ≤ θ ≤ π. If m = 0 replace π/2 by π.

59

K49 Two polynomials with the same roots and the same leading coefficient 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 the interpolating polynomial of degree n for f on [a, b], then taking g and Q 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 |f n (x)| ≤ A on [a, b] then |g n (x)| ≤ Aαn = A((b − a)/2)n on [−1, 1] so (b − a)n A |f (t) − P (t)| ≤ 22n−1 on [a, b].

60

K50 (ii) P (t) =

n X f j (a) j=0

j!

tj .

(vii) Observe that f (t) f (n+1) (θt ) = (n+1) g(t) g (φt ) with θt , φt → a as t → a and use continuity.

61

K51 If we set P (t) =

n X j=0

Aj tj (1 − t)n+1 +

n X j=0

Bj tn+1 (1 − t)j ,

then the resulting system of equations for Aj is triangular with nonvanishing diagonal entries and thus soluble. The same holds for Bj . If y ∈ (0, 1), set

xn+1 (1 − x)n+1 . y n+1 (1 − y)n+1 F has vanishing r-th derivative at 0 and 1 for 0 ≤ r ≤ n and vanishes at y. By Rolle’s theorem F 0 vanishes at least twice in (0, 1), F 00 vanishes at least three times in (0, 1), F 000 vanishes at least four times in (0, 1), . . . , F (n+1) vanishes at least n + 2 times in (0, 1), F (n+2) vanishes at least 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. F (x) = f (x) − P (x) − E(y)

62

K52 (i) g(b) − g(a) = g 0 (c)(b − a) 6= 0 for some c ∈ (a, b). (ii) A = (f (b) − f (a))/(g(b) − g(a)). (iv) Observe that

with ct → a as t → a.

f 0 (ct ) f (t) − f (a) = 0 g(t) − g(a) g (ct )

63

K53 (i) We have f (b) = f (a) + f 0 (a)h + f 00 (c)

h2 2!

f 0 (a)h = f (b) − f (a) − f 00 (c)

h2 2!

for some c ∈ (a, b). (ii) Thus

and so |f 0 (a)||h| ≤ 2M0 + 2−1 M2 |h|2

for all a and all h. Thus

2M0 M2 + h 2h for all h > 0 and, choosing h = 2(M0 M1 )1/2 , we have the required result. |f 0 (t)| ≤

(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(M0 M1 )1/2 . (d) True. g(t) = sin t. −1/2

(e) True. Scaling (d), G(t) = M0 sin(M0

1/2

M2 t) will do.

(f) Fails scaling. If G as in (e) then G0 (0) = (M0 M1 )1/4 1/4 (M0 M2 ) can be made as large as we wish.

64

K54 (i) True. Choose a δ > 0 such that |f (t) − f (s)| < 1 whenever |t − s| < 2δ and an integer N > 2 + δ −1 . If x ∈ (0, 1) we can find M ≤ N and x1 , x2 , . . . , xM ∈ (0, 1) such that x1 = x, xM = 1/2, |xj+1 − xj | ≤ δ for 1 ≤ j ≤ M − 1. Since |f (xj+1 ) − f (xj )| ≤ 1 we have |f (x) − f (1/2)| ≤ N and |f (x)| ≤ N + |f (1/2)|. (ii) False. If f (t) = t then f does not attain its bounds. (iii) False. Set f (t) = sin(1/t).

65

K55 (i) True. By the intermediate value theorem, we see that f (x) → l for some l as |x| → ∞. Given ² > 0, we can find a K such that |f (x) − l| < ²/2 for all |x| ≥ K. Since a continuous function on a closed bounded set is uniformly continuous, we can find a δ with 1 > δ > 0 such that |f (x) − f (y)| < ² for all x, y ∈ [−K − 2, K + 2] with |x − y| < δ. By going through cases, we see that |f (x) − f (y)| < ² for 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 ² > 0. Choose δ > 0 such that |x − y| < δ implies ||f (x)| − |f (y)|| < ²/2. Now suppose y > x and |x − y| < δ. If f (x) and f (y) have the same sign, then automatically, |f (x)−f (y)| < ²/2. If they have opposite sign, then by the intermediate value theorem, we can find u with y > u > x and f (u) = 0. Since |x − u|, |y − u| < δ we have |f (y) − f (x)| ≤ |f (y) − f (u)| + |f (u) − f (x)| < ²/2 + ²/2 = ².

(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 − y 2 | = (x + y)(x − y) > 1 so x 7→ x2 is not uniformly continuous. (viii) True. Let ² > 0. We can find a δ > 0 such that |u − w| < δ implies |f (u) − f (w)| < ² 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))| < ².

66

K56 Observe that so that Thus

k(x + u) − yk ≤ kx − yk + kuk f (x + u) ≤ f (x) + kuk. kf (a) − f (b)k ≤ ka − bk

and f is uniformly continuous.

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

≥ kx − yk − ky − yn k

→ kx − yk ≥ f (x)

as j → ∞. Thus kx − yk = f (x).

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

67

K57 (i) Given ² > 0 we can find a δ > 0 such that, if u, v ∈ Q, then |u − v| < δ implies |f (u) − f (v)| < ². We can now find an N such that |xn − x| < δ/2 for n ≥ N . If n, m ≥ N then |xn − xm | < δ so |f (xn )−f (xm )| < ². Thus the sequence f (xn ) is Cauchy and converges. (ii) Using the notation of (i), we can find an M such that |xm − x| < δ/2 and |ym − x| < δ/2 for m ≥ M . Thus |xm − ym | < δ and |f (xm ) − f (ym )| < ² for m ≥ M . Since ² was arbitrary, f (xm ) and f (ym ) tend to the same limit. (iii) Observe that (ultimately by the Axiom of Archimedes) any x ∈ R is the limit of xn ∈ Q. (iv) Take xn = x. (v) We use the notation of (i). If |x−y| < δ/3 we can find xn , yn ∈ Q with xn → x, yn → y and |xn − x| < δ/3, |yn − y| < δ/3. Then |xn − yn | < δ for all n so |f (xn ) − f (yn )| < ² and |F (x) − F (y)| ≤ ².

68

K58 PN P n n (i) If |z| < min(R, S) then N n=0 bn z tends to limit n=0 an z and and so N N N X X X bn z n an z n + (an + bn )z n = n=0

n=0

n=0

tends to limit as N → ∞.

P∞ n R < S If R < |z| < S then n=0 an z converges and PSuppose P ∞ ∞ n n n=0 bn z diverges so n=0 (an + bn )z 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 is infinite. 2

If |z| > 1 then, provided n is sufficiently large, |z|n ≥ (2R)n so P(v) ∞ n If |z| < 1 then, provided n is sufficiently large, n=0 an z diverges. P∞ 2 n n |z| ≤ (R/2) so n=0 an z n converges. The radius of convergence is 1. If R = 0, then any radius of convergence ρ with 0 ≤ ρ ≤ 1 is possible. 2 n For ρ = 0 take an = nn . For 0 < ρ < 1 take an = ρ−n . For ρ = 1 2 take an = ρ−n where ρn tends to 1 very slowly from below. n Similar ideas for R = ∞.

Pn Pn n n (vi) Let |cn | = max(|an |, |bn |). Since j=0 |an z | j=0 |cn z | ≥ radius of convergence R0 at most R. Thus R0 ≤ min(R, S). But |cn | ≤ |an | + |bn | so R0 ≥ min(R, S). Thus R0 = min(R, S). If |cn | = min(|an |, |bn |) then similar arguments show that the radius of convergence R00 ≥ max(R, S). But every value A ≥ max(R, S) is possible for R00 . (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 ² > 0. We can find an N such that |h(n) − γ| < ² for n ≥ N . Observe that if n ≥ N then h(n) < γ + ² and so ¡ ¢ an > 2 −(γ+²)+(1−(γ+²) n = 21−2(γ+²)n .

Thus if |z| > 22(γ+²)−1 we have |an z n | → ∞. Thus the radius of convergence R satisfies R ≤ 22(γ+²)−1 . Since ² was arbitrary R ≤ 22γ−1 . A similar argument shows that, if |z| < 22(γ−²)−1 |an z n | → 0, and thus R ≥ 22γ−1 so R = 22γ−1 . Suppose 1 ≥ γ ≥ 0. Define E inductively by taking an+1 = an /2 if an /2 ≥ 21−2γ , an+1 = 2an otherwise. Then h(n) → γ and |an z n | ≥ 1 whenever |z| = 22γ−1 and n is large. Thus we have divergence everywhere on the circle of convergence. Suppose 1 ≥ γ ≥ 0. Define E inductively by taking an+1 = 2an if 2an ≤ 2(1−2γ)n n−2 , an+1 = an /2 otherwise. Then |an z n | ≤ n−2 whenever |z| = 22γ−1 and n is large. Thus we have convergence everywhere on the circle of convergence. When n is large we have so whence

an ≤ 4 × 2(1−2γ)n n−2 2n−2h(n)n ≤ 22+(1−2γ)n n−2 (log 2)(n − 2h(n)n) ≤ (log 2)(2 + (1 − 2γ)n − 2 log n

and dividing by n log 2 and allowing n → ∞ we see that 1 − 2 lim sup h(n) = 1 − 2γ n→∞

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

70

K60 If |an |1/n is unbounded, the radius of convergence is zero.

We deal with the case (lim supn→∞ |an |1/n )−1 = R with 0 < R < ∞. The other cases are similar. Suppose R > ² > 0. Then we can find an N such that |an |1/n ≤ (R − ²/2)−1 for all n ≥ N . Thus if |z| < R − ² we have ¶n µ R−² n |an z | < R − ²/2 P∞ for n ≥ N and n=0 |an z n | converges by comparison with a convergent geometric sum. We can also find n(j) → ∞ such that |an |1/n(j) ≥ (R + ²/2)−1 . Thus, if |z| > R + ² we have µ ¶n(j) R+² n(j) |an z | > → ∞. R + ²/2 Since ² was arbitrary, R is the radius of convergence as required.

71

K61 Write α = lim supn→∞ an+1 /an . If ² > 0, we can find an N with an+1 /an ≤ α + ² for n ≥ N . Thus, by induction, an ≤ A(α + ²)n for n ≥ N where A = aN (α + ²)−N . Thus, if n ≥ N , a1/n ≤ A1/n (α + ²) → α + ² n

as n → ∞. Since ² was arbitrary

lim sup a1/n ≤ lim sup an+1 /an . n n→∞

n→∞

Remaining inequalities similar or simpler. All 8 possibilities can arise. Here are d. (1) Set an = 1. an+1 an+1 = lim inf a1/n = lim sup a1/n = lim sup = 1. lim inf n n n→∞ n→∞ an an n→∞ n→∞ (2) Set a1 = 1, a2n +1 /a2n = 2, ar+1 /ar = 1 otherwise. an+1 an+1 = lim inf a1/n = lim sup a1/n = 1 < 2 = lim sup . lim inf n n n→∞ n→∞ an an n→∞ n→∞ (3) Set a1 = 1, a2n +1 /a2n = 2, a2n +2 /a2n +1 = 2−1 for n ≥ 2 ar+1 /ar = 1 otherwise. an+1 an+1 lim inf = 1/2 < lim inf a1/n = lim sup a1/n = 1 < 2 = lim sup . n n n→∞ n→∞ an an n→∞ n→∞ 2n

2n+1

(4) Set a1 = 1, ar+1 /ar = 2 if 22 ≤ r < 22 , ar+1 /ar = 1 otherwise an+1 an+1 = lim inf a1/n = 1 < 2 = lim sup a1/n = lim sup . lim inf n n n→∞ n→∞ an an n→∞ n→∞ 2n

2n+1

2n

2n+1

(5) Set a1 = 1, ar+1 /ar = 2 if 22 + 1 ≤ r < 22 , ar+1 /ar = 4 if 22n r = 2 ar+1 /ar = 1 otherwise an+1 an+1 = lim inf a1/n = 1 < 2 = lim sup a1/n < 4 = lim sup . lim inf n n n→∞ n→∞ an an n→∞ n→∞ , ar+1 /ar = 4 if (6) Set a1 = 1, ar+1 /ar = 2 if 22 + 2 ≤ r < 22 22n 22n r = 2 , ar+1 /ar = 1/2 if r = 2 + 1 ar+1 /ar = 1 otherwise an+1 an+1 < 4 = lim sup lim inf = 1 < 2 = lim sup a1/n = 2−1 < lim inf a1/n . n n n→∞ n→∞ an an n→∞ n→∞ Other two obtained similarly. Note lim sup formula will always give radius of convergence. Ratio may not.

72

K62 (iii) Could write bn = cn + dn with dn = 0 for n large and |bn | ≤ ². Thus lim supn→∞ |Cn | ≤ ². (v) If bj = (−1)j limit is 0.

(viii) Pick N (j), Mj , sj and rj inductively so that N (0) = 0, M (0) = M (j)+1 1, s0 = 1/2. we pick rj+1 so that 0 < sj < rj+1 < 1 and 1 − rj+1 < N (j+1) −j −j < 2 . Pick sj+1 so that 2 . Pick an N (j+1) > M (j) such that rj+1 N (j+1) 0 < rj+1 < sj+1 < 1 and sj+1 > 1 − 2−j−1 . Pick M (j + 1) > N (j + 1) M (j+1) so that sj+1 < 2−j−1 Set bn = 1 if N (j) ≤ n ≤ M (j), bn = 0 otherwise. M (j)

Ar(j+1) ≤

X n=0

n (1 − rj+1 )rj+1 +

= (1 −

and

M (j)+1 rj+1 )

+

N (j) rj+1

∞ X

n=N (j+1)

n (1 − rj+1 )rj+1

< 2−j + 2−j = 2−j+1

M (j+1)

As(j+1) ≥

X

n=N (j+1) N (j+1)

= sj+1

(1 − sj+1 )snj+1 M (j+1)+1

− sj+1

Thus Ar does not converge.

> 1 − 2−j−1 − 2−j−1 = 1 − 2−j .

73

K63 (ii) C can be identified with R2 . P n N +1 (iii) (a) If z 6= 1, N )/(1 − z) tends to a limit if n=0 z = (1 − z 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)

−1

= (n + 1)−1

k n X X

zj

k=0 j=0 n X k=0

1 − z k+1 1−z

µ ¶ 1 − z n+1 1 1 (n + 1) + z = n+1 1−z 1−z z 1 − z n+1 1 + . = 1−z n+1 1−z Thus we have convergence if and only if |z| ≤ 1, z 6= 1 and the limit is 1/(1 − z).

(c) Ar = (1 − rz)−1 . Abel sum exists if and only if |z| ≤ 1, z 6= 1 and 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 z j . Observe that, if z 6= 1 ∞ X 1 λn −λ 1 − z n+1 e = (1 − ze−λ(1−z) ). n! 1 − z 1 − z n=0

We have convergence if and only if <(1 − z) < 1 and the Borel sum is then 1/(1 − z). If z = 1, ∞ ∞ ∞ X X λn −λ X λn −λ λn −λ Bλ = (n + 1) e = e +λ e = 1 − λ → ∞. n! n! n! n=0 n=0 n=0

75

K65 (i) True. Since the sequence is bounded, there exists a b and a strictly increasing sequence n(j) such that bn(j) → b. Set ujn(j) = 1, ujk = 0 otherwise. (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 for 2 + 1 ≤ j ≤ 2n + n [n ≥ 4], bj = 0 otherwise Observe that, if unk = 1 n for 2P − N ≤ k ≤ 2n , unk = −1 for 2n + 1 ≤ k ≤ 2n + N , then U ∈ G but ∞ k=0 ujk bk = 2N for j sufficiently large. But N is arbitrary. n

(iv) True. Define N (r) and j(r) inductivelyP as follows. Set N (0) = −r j(0) = 1. Choose N (r + 1) > N (r) such that ∞ k=N (r+1) |uj(r)k | < 2 PN (r+1) and j(r + 1) > j(r) such that k=0 |uj(r+1)k | < 2−r−1 . Set bk = (−1)r+1 for N (r) ≤ k < N (r + 1). Then ∞ X (−1)r+1 uj(r)k bk k=0



= (−1)r+1  ≤ ≤ as r → ∞.

∞ X k=0

∞ X k=0

N (r)

N (r+1)

X

X

uj(r)k bk +

k=0

uj(r)k − 2

k=N (r)+1

N (r)

X k=0

|uj(r)k | − 2

uj(r)k − 2−r+3 → 1.

(v) False. Try bj = 0.

uj(r)k bk + ∞ X

k=N (r+1)+1

∞ X

k=N (r+1)+1

|uj(r)k |



uj(r)k bk 

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 ), we have kxj k ≤ |xj1 | + |xj2 | + · · · + |xjm |. P∞ Thus, if j=1 kxj k diverges, we must be able to find a k with 1 ≤ k ≤ m P such that ∞ j=1 |xjk | diverges. Choose ²j so that ²j xjk ≥ 0. Then ° N ° ¯ N ¯ N °X ° ¯X ¯ X ° ° ¯ ¯ ²j x j ° ≥ ¯ ²j xjk ¯ = |xjk | → ∞. ° ° ° ¯ ¯ j=1

as N → ∞.

j=1

j=1

78

K68 If

P∞

n=1

an converges, then ∞ X X X an ≤ an ≤ an n∈S1 ,n≤N

n≤N

n=1

and, since an increasing sequence bounded above converges, tends to a limit as N → ∞. P If n∈Sj ,n≤N an tends to a limit so does X X X an = an + an . n≤N

n∈S2 ,n≤N

P

n∈S1 ,n≤N

an

n∈S2 ,n≤N

Second paragraph does not depend on positivity aj so ‘if’ remains true. However, taking a2n−1 = −a2n = 1/n, S1 evens and S2 odds then we see ‘only if’ false. 1/2

1/2

Let S1 = {n : an /n ≤ an } and S2 = {n : an /n > an }. P P∞ 1/2 n∈S1 an /n converges by comparison with n=1 an . If n ∈ S2 then so, since

P∞

n=1

n−2

2 −1 2 −2 an = (a1/2 n ) < (n ) = n P 1/2 converges, so does n∈S2 an /n. Hence the result.

By Cauchy Schwarz, !2 Ã N ∞ N ∞ N X X X X X a1/2 1 1 n ≤ ≤ an an 2 n n n2 n=1 n=1 n=1 n=1 n=1

so, since an increasing sequence bounded above converges, we are done. Observe that cn ≥ (an + bn )/2 and use comparison.

79

K69 (i) Diverges 2N X 1 + (−1)n

n

n=1

as N → ∞.

N X 1 = →∞ n n=1

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

(iii) Diverges by comparison. We have n1/n → 1 (e.g. by taking logarithms) so n−1/n ≥ 1/2 for n large and so n−1−1/n > 2−1 n−1 for n large. PN (j) P (iv) If ∞ n=1 an n=1 an converges, then taking N (j) = j we have that tends to a limit. PN (j) If n=1 an tends to a limit A, then for any M we can find a j with N (j) > M so that M X n=1

N (j)

an ≤

X n=1

an ≤ A

so, since an increasing sequence bounded above converges, we are done. R∞ (v) If 1 f (t) dt converges then Z N +1 Z ∞ N Z n+1 N X X f (t) dt = f (t) dt ≤ f (t) dt an ≤ n=1

n=1

n

1

0

P since an increasing sequence bounded above converges ∞ n=1 an converges. P For the converse observe that if ∞ n=1 an converges then, if X ≤ N , Z X Z N ∞ N −1 N −1 Z n+1 X X X an . an ≤ f (t) dt ≤ K f (t) dt ≤ f (t) dt = 1

1

n=1

n

n=1

n=1

Now use something like Lemma 9.4.

(vi) an = (−1)n . P N (vii) Let M a tend to b. Given ² > 0 we can find an N0 such PM N n=1 n that k n=1 an − bk ≤ ²/2 for N ≥ N0 . We can also find an N1 such that kan k ≤ ²/2M for n ≥ N1 . Let N2 = (M + 1)(N1 + N0 ). If m ≥ N2 we can write m = N M + u with N ≥ N0 , 0 ≤ u ≤ M − 1 and M N + r ≥ N1 for all r ≥ 0. We then have k

m X n=1

an − bk ≤ k

MN X n=1

an − bk +

MX N +u

n=M N +1

kan k <

² u² + ≤² 2 2M

80

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

81

K70 (i) Observe that (an bn )1/2 ≤ (an + bn )/2 and use comparison. (ii) Take bn = an+1 in (i). (iii) Observe that an+1 ≤ (an an+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 . Thus ≥ |a5N | for n ≥ N and the result follows by comparison and the fact that absolute converge implies convergence. a4n

(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 ² > 0 we can find an N such that for M ≥ N . Thus, if n ≥ 2N + 2 nan ≤ 2(n − N − 1)an ≤ 2

M X

PM

j=N

aj < ²/2

aj < ².

j=N

(vi) False. Take an = (n log n)−1 for n ≥ 3 and use the integral comparison test. (vii) True. Abel’s test. (viii) True. By Cauchy–Schwarz, Ã N !2 N N ∞ ∞ X X X X X −3/4 2 −3/2 2 |an |n ≤ |an | n ≤ |an | n−3/2 . n=1

n=1

n=1

n=1

n=1

(ix) False. Take an = (−1)n (log n)−1 for n ≥ 3. Use alternating series test and integral test. (x) True. We must have an → 0 so there exists a K such that |an | ≤ K. Since |n−5/4 an | ≤ Kn−5/4 the comparison test tells us that P ∞ −5/4 an is absolutely convergent and so convergent. n=1 n

83

K72 akn → 0 so there exists an N with an ≤ 1 for n ≥ N . Thus akn ≥ |ak+1 n | for n ≥ N and the result follows by comparison. Set f (n) = log(n + 1)n and observe that, if k ≥ 2 3N X j=1

akj ≥

N X j=1

(2k − 2)(log(n + 1))−k ) → ∞.

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

84

K73 (i) Write Sn =

n X

µ−1 j bj .

j=1

Then, by partial summation (or use Exercise 5.17), n n X X µj (Sj − Sj−1 ) bj = j=1

j=1

=

n−1 X j=1

(µj − µj+1 )Sj + µn Sn

so that ° n ° n−1 °X ° X ° ° bj ° ≤ (µj+1 − µj )kSj k + µn kSn k ° ° ° j=1

j=1

≤

n−1 X j=1

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

whence the result.

(ii) If ² > 0 then we can find an N such that k n ≥ N . Thus, by (i), µ−1 n k

n X j=1

bj k ≤

µ−1 n

≤ µ−1 n

N −1 X j=1

N −1 X j=1

kbj k +

Pn

µ−1 n k

j=N

n X

j=N

µ−1 j bj k ≤ ² for

bj k

kbj k2² → 2²

as n → ∞. Kronecker’s lemma follows.

(iii) ((a) fails) Let bj = j 4 , µj = j 6 for j 6= 2r , b2r = 1 and µ2r = 2−2r .

((b) fails) Let bj = j −2 , µj = 1. ((c) fails) Let bj = j 4 , µj = j 2 .

85

K74 Consider the set An of integers between 10n and 10n+1 − 1 inclusive who’s decimal contain the integer 9. We observe Sn expansion does not n+1 that {0} ∪ j=0 Aj contains (9/10) × 10n+1 = 9n+1 elements. Thus An contains at most 9n+1 elements and X j −1 ≤ 10−n × 9n+1 . j∈An

The partial sums of the given series thus can not exceed 9n+1 = 90 and the series converges.

P∞

n=0

10−n ×

86

K75 (i) Observe that 1/x ≥ 1/(n + 1) for n ≤ x ≤ n + 1 and integrate. Tn − Tn+1

1 − = n+1

Z

n+1 n

1 dx. x

Observe that T1 = 1 and ¶ Z r+1 n−1 µ X 1 1 Tn ≥ − dx . r x r r=1 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 exceeds pn /qn use pn positive terms followed by qn + 1 terms if the ratio is less than pn /qn , use pn + 1 positive terms followed by qn negative terms, otherwise usepn positive terms followed by qn negative terms. Switch from the n-th stage to the n + 1-th stage when the size of each unused terms is less than (pn+1 + qn+1 + 1)−1 2−n and the ratio of positive terms to negative differs from pn /qn by at most 2−n .

87

K76 P∞ (ii) Observe that, if j=0 bj converges, we can find an N (²) such PN that | j=M ) bj ≤ ²| for N ≤ M ≤ N (²) so ¯ ∞ ¯ ¯X ¯ ¯ j¯ b x ¯ ¯ ≤ ²xM ≤ ² j ¯ ¯ j=M

for M ≥ N (²).

(iii) If M ≥ N (²), then ¯∞ ¯ M −1 ¯ ∞ ¯ ¯ ∞ ¯ ∞ ¯X ¯ X ¯X ¯ ¯X ¯ X ¯ ¯ ¯ ¯ ¯ ¯ bj xj − bj ¯ ≤ |bj xj − bj | + ¯ bj ¯ + ¯ bj xj ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ j=0

j=0

j=0

≤

M −1 X j=0

j=M

j=M

|bj xj − bj | + 2² → 2²

as x → 1−. Since ² is arbitrary the required result follows. P∞ P∞ j j (iv) Observe that, if 0 < x < 1, then j=0 bj x and j=0 aj x , P∞ j j=0 cj x are absolutely convergent so (using Exercise 5.38) we may multiply them to get ∞ ∞ ∞ X X X j j aj x c j xj . bj x = j=0

Now let x → 1−.

j=0

j=0

88

K77 P∞ P n m converges for |x| = |y| = δ, we have (a) Since ∞ m=0 cn,m x y n=0 n+m n m |cn,m |δ = |cn,m x y | → 0 for n, m → ∞ so there exists a K with |cn,m |δ n+m < K for all n, m ≥ 0. Set ρ = δ −1 . P P P (b) (i) n≤N, m≤N |xn y m | = n≤N |x|n m≤N |y|m . E is the square (−1, 1) × (−1, 1). ¡n+m¢ n m P PP r |x y | = N (ii) r=0 (|x| + |y|) n+m≤N n

E is the square {(x, y) : |x| + |y| < 1}. ¡n+m¢ 2n 2m P PP 2 2 r |x y | = N (iii) r=0 (|x| + |y| ) . n+m≤N n

E is the circle {(x, y) : x2 + y 2 < 1}. P P P (iv) n, m≤N |xn y m /(n!m!)| = n≤N |x|n /n! m≤N |y|m /m!. E = R2 . P P P (v) n m≤N |xn y m n!m!| = n≤N |x|n n! m≤N |y|m m!.

E = {0}. P P (vi) n≤N |xn y n | = n≤N |xy|n .

E = {(x, y) : |xy| < 1} a set bounded by branches of hyperbola.

(c) Observe that |cnm xn y m | ≤ |cnm xn0 y0m | when |x| ≤ |x0 |, |y| ≤ |y0 |.

89

K78 (i) Choose Nj (²) so that |aj − aj,n | < ²/M for n ≥ Nj (²) and set N (², M ) = max1≤j≤M Nj (²). Observe that ∞ X j=1

aj ≥

M X j=1

aj ≥

M X

aj,n

j=1

for all M . P (ii) If ∞ j=1 aj converges with value A, then given ² > 0 we can find P an M such that M j=1 aj > A − ². By (i) we can find N (², M ) such that ∞ X j=1

and so

∞ X j=1

aj,n ≥

aj ≥

∞ X j=1

M X j=1

aj − ²

aj,n ≥ A − 2²

P for all n ≥ N (², M ). Thus ∞ j=1 aj,n → A. P∞ If j=1 aj fails to converge, then given any K > 0, we can find PM an M such that j=1 aj > 2K. But we can find an N such that PM PM P aj,n ≥ j=1 aj − K for n ≥ N and so ∞ j=1 P 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 P I prefer to nz copy the proof P of the radius of convergence. Show that if ∞ n=0 bn e converges then n=0 bn enw converges absolutely when <w <
exists and has the desired properties.

2

For X = −∞ can take bn = 0, for X = ∞ can take bn = en , for X = R with R finite can take bn = e−Rn . P∞ n nz 2 n nx n=0 2 e /(n + 1) converges when z = − log 2. Also 2 e /(n + 1)2 → ∞ when x is real and x > − log 2 so X = − log 2. P inz By the first paragraph there exists a Y ≥ 0 such that ∞ n=0 cn e converges for =z < YPand diverges for=z > Y . By taking complex con−inz converges for =z > −Y and diverges jugates we see that ∞ n=0 cn e for =z < −Y . P P∞ P∞ Since ∞ n=0 sn + tn diverges if exactly one n=0 sn and n=0 tn Pof ∞ inz −inz converges and converges if both converge, c (e + e ) and n=0 n P∞ thus n=0 cn cos nz converges if =z < Y and diverges if =z > Y .

92

K81 (i) By integration by parts, In+1 = [− sin =n

Z

n

x cos x]π/2 0

π/2 0

+n

Z

π/2

sinn−1 x cos2 x dx 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]. Integrating gives In+1 ≤ In ≤ In−1 so In+1 In n = ≤ ≤ 1. n+1 In−1 In−1 so In /In−1 → 1 as n → ∞. (iii) I0 = π/2, I1 = 1 (2n − 1)(2n − 3) . . . 1 π 2n(2n − 2) . . . 2 I2n = , andI2n+1 = . 2n(2n − 2) . . . 2 2 (2n + 1)(2n − 1) . . . 1 Thus n Y

4k 2 2 I2n+1 = →1 2 4k − 1 π I2n k=1 and the Wallis formula follows.

93

K82 (i) Power series or set f (x) = exp x − x and observe f (0) = 0, f 0 (x) ≥ 0 for all x ≥ 0. (iii) Observe that ¯ ¯¯ m ¯ ¯ n ¯ n m ¯ ¯¯ Y ¯ ¯Y ¯Y Y ¯ ¯¯ ¯ ¯ ¯ (1 + aj ) − 1¯ (1 + aj )¯ = ¯ (1 + aj )¯ ¯ ¯ (1 + aj ) − ¯ ¯¯ ¯ ¯ ¯ j=n+1 j=1 j=1 j=1 ! Ã m n Y Y (1 + |aj |) − 1 . (1 + |aj |) ≤ j=n+1

j=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 ¯ ¯ ¯ an ¯ ¯ ≤ 2|an |. |bn | = ¯¯ 1 + an ¯ P Thus ∞ n=1 bn converges absolutely. n n ∞ ∞ Y Y Y Y 1= (1 + aj ) (1 + bj ) → (1 + aj ) (1 + bj ) j=1

so

Q∞

j=1 (1

+ aj )

j=1

Q∞

j=1 (1

j=1

j=1

+ bj ) = 1.

(vi) Observe that N Y

(1 + aj /Rj ) =

j=1

as N → ∞ and use (v).

N Y Rj−1 j=1

Rj

=

R0 →0 RN

94

K83 (i) Left to reader. (ii) Let an = n−α , an,N = n−α if all the prime factors of N lie among the first N primes, an,N = 0 otherwise. Observe that 0 ≤ an,N ≤ an,N +1 ≤ an , P → an as N → ∞ and that ∞ n=1 an converges.

that an,N (iii) If

Y

1 ≤K 1 − p−1

Y

1 ≤K 1 − p−α

p∈P, p≤N

for all N then

p∈P, p≤N

for all N and all α < 1 so

Y

p∈P

and

∞ X 1 ≤K nα n=1

for all α < 1. Thus

for all α < 1 and all N so

for all

P∞

1 n=1 n

1 ≤K 1 − p−α

N X 1 ≤K α n n=1 N X 1 ≤K n n=1

which 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 X

r∈E,r≤2N

for all N .

N −1 N −1 N −1 X 1 X X 1 X −j A = ≤ 2 M (j) ≤ A 2n(β−1) = r r 1 − 2β−1 j=0 r∈E j=0 j=0 j

95

K84 Q Suppose that Pn > 1 for all n ≥ N . Then Pm = PN m j=N (1 − an ). P∞ If j=1 an diverges then (see K82) Pm → 0 which is impossible by the P definition of N . Thus ∞ j=1 an converges and Pm tends to a limit. A similar 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 or PN −1 − 1 ≥ 0 ≥ PN − 1 and (B) 0 ≤ an ≤ ² for n ≥ N − 1 then we have 1 + ² ≥ Pn ≥ 1 − ² for n ≥ N . Thus Pn → 1. P 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 = z n , an,N = z n if 0 ≤ n ≤ 2N +1 − 1 and an,N = 0, otherwise. Observe that 0 ≤ |an,N ≤ |an |, and that an,N → an as N → ∞. By dominated convergence (Lemma 5.25) N Y j=1

2j

(1 + z ) =

∞ X n=0

an,N →

∞ X n=0

an = (1 − z)−1 .

97

K86 Observe that 2 cot 2φ =

cos2 φ − sin2 φ = cot φ + tan φ cos φ sin φ

and use induction. Observe that

as n → ∞.

θ 1 2−n θ 1 1 cot = cos 2−n θ → n n −n 2 2 θ sin 2 θ θ

98

K87 Very similar to K86. To obtain the formula for un , observe that 2 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 of powers of x with integral coefficients plus terms of the form xr (1 − x)s M (min(r, s))! with M an integer and r, s ≥ 1. Thus f (k) (0) and f (k) (1) are always integers. Since bn π 2n−r is an integer, G(0) + G(1) is always an integer.

as n → ∞.

0 < πan f (x) sin πx ≤

an →0 n!

100

K89 (a)(i) Given c ∈ C we can find b ∈ B with g(b) = c and a ∈ A with f (a) = b. (ii) g ◦ f (a) = g ◦ f (a0 ) implies f (a) = f (a0 ) and so a = a0 .

(iii) f (a) = f (a0 ) implies g ◦ f (a) = g ◦ f (a0 ) and so a = a0 . (iv) Given c ∈¡ C, we ¢ can find a ∈ A with g ◦ f (a) = b. Observe that f (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 paragraph, we can write f = g ◦ J where J(x) = −x and g satisfies the conditions of the first paragraph. Thus for all x ∈ E.

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

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

102

K91 (i) Suppose that f is continuous. If f is not strictly monotonic we can find x1 < x2 < x2 so that f (x1 ), f (x2 ) and f (x3 ) are not a strictly increasing or a strictly decreasing sequence. Without loss of generality (or by considering f (1 − x) and 1 − f (x) if necessary), we may suppose f (x3 ) ≥ f (x1 ) ≥ f (x2 ). If either of the inequalities are equalities then f is not injective. Thus we may suppose f (x3 ) > f (x1 ) > f (x2 ). But by the intermediate value theorem we can find c ∈ (x2 , x3 ) so that f (c) = f (x1 ) and f is not injective. Suppose f is strictly monotonic. Without loss of generality we suppose f increasing. We shall show that f is continuous at t ∈ (0, 1) (the cases t = 0 and t = 1 are handled similarly). Observe that f (1) > f (t) > f (0). Let ² > 0. Set ²0 = min(1 − f (t), f (t), ²)/2. We can find s1 and s2 with f (s1 ) = f (t) − ²0 f (s2 ) = f (t) + ²0 . If O/ta = min(t − s1 , t + s2 ), then |s − t| > δ implies s2 > s > s1 so f (s2 ) > f (s) > f (s1 ) and |f (s) − f (t)| ≤ ²0 < ². (ii) The same proof as in (i) shows that, if g is continuous and injective, it must be strictly monotonic. The example of g(t) = t/2 for t ≤ 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 monotonic and surjective, it must be continuous. The example g(t) = sin πt shows that the converse is false.

103

K92 Write f (x) = (log x)/x. By looking at f 0 we see that f is strictly increasing from −∞ to e−1 on the interval (0, e], has a maximum value of 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 = y x if and only if f (x) = f (y). The set {(x, y) : xy = y x , x, y > 0}

is thus the union of the straight line {(x, x) : x > 0} and a curve with reflection symmetry in that line, having asymptotes x = 1 and y = 1 and passing through (e, e). If f (x) = f (y) and x < y, then x ≤ e so we need only examine the cases 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 > π e .

104

K93 Observe that xy =

1 4

¡

(x + y)2 − (x − y)2 .

105

K94 (i) Choose γ with β > γ > 1. If we set f (x) = xα γ x , then (recalling that γ x = exp(x log γ)) f 0 (x) = αxα−1 γ x + xα (log γ)γ x = αxα−1 γ x (α + γx) > 0 for x sufficiently large (x ≥ x0 , say). Thus

xα β x = f (x)(β/γ)x ≥ f (x0 )(β/γ)x → ∞.

(ii) Set (µ µ ¶1/2 ¶1/2 ) µ ¶ 1 1 1 , f3 (x) = , f2 (x) = exp f4 (x) = exp log x x x and f1 (x) =

µ

1 log 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) → ∞ as x → ∞.

106

K95 (ii) If Q has a root α 6= 0 of order n ≥ 1 then differentiating the given equation n − 1 times and setting z = α we get −P (α)Q(n) (α) = 0 so P (α) = 0 and P and Q have a common factor. Thus Q(z) = Az N for some 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) log x → ∞. Q(x) If the degree of P is smaller than that of Q then P (x) log x → 0. Q(x) If the degree of P is no smaller than that of Q then P (x) log x → ∞. Q(x)

107

K96 Set f (x) = x−log(1+x). Then f (0) = 0 and f 0 (x) = 1−(1+x)−1 ≥ 0 for 0 < x and f 0 (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 0 (x) = (1 + x)−1 − 1 + 2x = (x + 2x2 )(1 + x)−1 ≥ 0

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

kn kn ³ ³ Y r´ r´ X log 1 − 2 1− 2 = n n r=1 r=1

kn kn kn ³ kn X X Y X r r r´ r2 ≥ − ≥ log − . 1 − 2 2 2 4 n n n n r=1 r=1 r=1 r=1

But

and

as n → ∞ so

µ ¶ kn X 1 k2 kn(kn + 1) k r k + → = = n2 2n2 2 n 2 r=1 kn X r2 (kn)3 k ≤ = →0 4 4 n n n r=1 kn ³ Y k2 r´ log 1− 2 →− n 2 r=1

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 so tends to a limit. A similar argument applies if 1 > x > 0. Since 2n (xn − 1) = xn → 1 2n (1 − 1/xn )

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 ¶2 n µ ¶2 n µ δ n ηn (1 + δn )(1 + ηn ) = 1+ . 1 + δ n + ηn 1 + δ n + ηn Since 2n δn 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 a C such that ¯ ¯ ¯ ¯ δ n ηn −2n ¯ ¯ ¯ 1 + δn + ηn ¯ ≤ C2 . It follows that ¯µ ¯ 2n µ n ¶ ¯ (1 + δ )(1 + η ) ¶2n ¯ X 2 ¯ ¯ n n C r 2−2nr − 1¯ ≤ ¯ ¯ ¯ r=1 r 1 + δ n + ηn n

≤

2 X

2nr C r 2−2nr

r=1

=≤

n

2 X

2nr C r 2−nr

r=1

≤ as n → ∞.

C2−n →0 1 − C2−n

By the mean value theorem (b2 for some c ∈ (a, b) so

−n

−n

− a2 ) = 2−n c2

−n −1

(b − a)

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

109

that is to say 2n ((xy)2

−n

so writing z = xy and

− (xn + yn − 1)) → 0

2n (zn − 1) − 2n (xn − 1) − 2n (yn − 1) → 0 log xy = log x + log y.

Result (ii) shows that we need only prove (iii), (iv), (v) and (vi) at the point x = 1. To do this observe that a simple version of Taylor’s theorem |f (1 + δ) − f (1) − f 0 (1)δ| ≤ sup |f 00 (1 + η)||δ 2 | applied to f (x) = x

−2n

|η|≤|δ|

yields n

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

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

|2n (xn − 1) − δ| ≤ 4|δ|2

| 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 0 x = 1/x and since log0 x > 0 we see that log is strictly increasing. (vii) Since log 2 > log 1 = 0 we have log 2n = n log 2 → ∞ so log x → ∞ as x → ∞ and log y = − log y −1 → −∞ as y → 0+. Since log is continuous and strictly monotonic, part (vii) follows.

110

K98 If µ

f (x + η) f (x)

¶1/η

→l

then, taking logarithms, log f (x + η) − log f (x) → log l η so log f is differentiable and, by the chain rule, f = exp log f is differentiable at x. Since d f 0 (x) log l = log f (x) = dx f (x) we have f 0 (x) l = exp . f (x)

111

K99 If 1 γxβ−1 − β β x −a x−a tends to a limit as x → a then (x − a)f (x) → 0. Since f (x) =

as x → a this gives

xβ − a β → βaβ−1 x−a γ −1=0 β

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 − aβ x − axβ + aβ+1 We find F (a) = F 0 (a) = 0, G(a) = G0 (a) = 0 and use L’Hˆopital’s rule to give F 00 (a) β−1 f (x) → 00 = . G (a) 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 real b. 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 all x or g(x) = sgn(x)b|x| for all x. (Recall sgn x = 1 if x > 0, sgn x = −1 if x < 0.) It is easy to see that these are solutions.

114

K102 (i) We know that χ(t/2)2 = χt but the equation (exp is)2 = exp iu, where u ∈ R, has two real solutions in 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 but θ(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/4 so θ(t/2) = θ(t)/π (since |θ(t/2) ± π| > π/4). (iii) The same ideas show that the continuous homomorphisms are precisely the maps λ 7→ λn for some integer n.

115

K103 (i) f (0) = 2f (0) so f (0) = 0. k (ii) Set F (2k p−1 j ) = 2 j whenever j and k are integers with j ≥ 1 and F (t) = 0 otherwise. (By the uniqueness of factorisation F , is well defined.)

0.

−1 Since F (p−1 j ) → ∞ and pj → 0 as j → ∞, F is not continuous at

(iii) f (t2−n ) − f (0) → f 0 (0)t t2−n as n → ∞ so f (t) = f 0 (0)t for all t 6= 0 and so for all t. f (t) = 2n f (t2−n ) = t

k −1 (iv) Set F (2k p−1 j ) = 2 pj whenever j and k are integers with j ≥ 1 and F (t) = 0 otherwise. Since |F (t) − F (0)| = |f (t)| ≤ |t|, we have F continuous at 0. Since F (p−1 F (21/2 j −1 ) − F (0) j ) − F (0) →0 = 1 → 1 but 21/2 j −1 p−1 j

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 −n u(2−n x) = u(x)2 → 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) = n u(2−n t)2 > 0 for |t| < 2n δ 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 δ 1 = 1 − + ²(δ)|δ| 1/2 (1 + δ) 2 with ²(δ) → 0 as δ → 0. (Use differentiation or Taylor series.) Observe also that |x · h| ≤ kxkkhk. Thus x+h f (x + h) = (kxk2 + 2x · h + khk2 )1/2 x+h = ¢1/2 ¡ kxk 1 + (2x · h + khk2 )/kxk2 ¶ µ 2x · h + khk2 x+h + ²1 (h)khk 1− = kxk 2kxk2 h (x · h)x = f (x) − − + ²2 (h)khk kxk kxk3 where k²j (h)k → 0 as khk → 0. Thus f is differentiable at x and Df (x)h =

(x · h)x h . − kxk kxk3

kxk2 h·x −x·h = 0. (Df (x)h) · x = kxk kxk3 Since kf (x) − f (0)k = 1 9 0

as kxk → 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 (0) = A + Bi with A and B real, and taking h and k real f (h + ik) = f (0) + (A + Bi)(h + ik) + ²(h + ik)|h + ik| with ²(h + ik) → 0 as |h + ik| → 0. Taking real and imaginary parts, ¶ ¶ µ ¶ µ ¶ µ µ ²1 (h, k) Ah − Bk u(0.0) u(h, k) (h[ 2 + k 2 )1/2) + + = ²2 (h, k) Ak + Bh v(0, 0) v(h, k)

with

°µ ¶° ° ²1 (h, k) ° ° ° ° ²2 (h, k) ° → 0

as (h2 + k 2 )1/2 → 0. Thus F is differentiable at 0 with Jacobian matrix ¶ µ A −B B A If A = B = 0 take λ = 0 and θ = 0. Otherwise take λ = (A2 + B 2 )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) + ²(h, h)(h2 + h2 )1/2 ¡ ¢ = g(x) + f,1 (x, c − x) − f,2 (x, c − x) + 21/2 ²(h, h)|h|

where ²(h, k) → 0 as (h2 + k 2 )1/2 → 0. Thus g is differentiable with derivative g 0 (x) = f,1 (x, c − x) − f,2 (x, c − x). (ii) Observe that g = f ◦ u where µ ¶ x u(x) = . c−x

Thus u has Jacobian matrix

µ

¶ 1 . −1

By chain the rule g is differentiable with 1 × 1 Jacobian matrix µ ¶ ¡ ¢ 1 f,1 (x, c − x) f,2 (x, x − c) = (f,1 (x, c − x) − f,2 (x, c − x)). −1

Define g as stated. If f,1 = f,2 , then g 0 = 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, we have m X α−1 αλ f (x) = xj f,j (λx). j=1

Taking λ = 1 gives the required result.

(ii) We observe that v is differentiable with m m X X 0 −1 v (λ) = xj f,j (λx) = λ λxj f,j (λx) = αλ−1 v(λ). j=1

j=1

Thus

d (λ−α v(λ)) = 0 dλ and by the constant value theorem λ−α v(λ) = v(1) and f (λx) = λα f (x).

120

K108 This is really just a question for thinking about. (i) We seek to maximise the (square root of) °µ ¶µ ¶°2 ° a b ° cos θ ° = (a cos θ + b sin θ)2 (c sin θ + b cos θ)2 f (θ) = ° ° c d sin θ °

and this we can do, in principle, by examining the points with f 0 (θ) = 0. (ii) We seek to maximise the (square root of) Ã m !2 p X X ars xs r=1

s=1

subject to the constraint

m X

x2s = 1

s=1

which can, in principle, be handled using Lagrange multipliers. However, this involves solving an m × m set of linear equations involving a parameter λ leading to polynomial of degree m in λ. The m roots must then be found and each one inspected. This neither sounds nor 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 * n + m X X aij = arj er , ei arj her , ei i = r=1

r=1

= hαej , ei i = hej , αei i = hαei , ej i = aji

Conversely if aij = aji for all i, j, then essentially the same calculation shows that hαej , ei i = hαei , ej i

and so by linearity ! m * Ã n + X X xr er , α ys e s r=1

s=1

= = =

m X m X

r=1 s=1 m X m X

r=1 s=1 * n X

xr ys hαer , es i xr ys her , αes i

xr er , α

r=1

Ã

m X s=1

ys e s

!+

.

Let u1 , u2 , . . . , em be an orthonormal basis of eigenvectors with eigenvalues λ1 , λ2 , . . . λm with |λ1 | ≥ |λ2 | ≥ · · · ≥ |λm |. Then °2 ° Ã m !°2 ° m m m ° ° °X ° X X X ° ° ° ° 2 2 2 x2j xj λj ≥ λ 1 xj λj u j ° = xj u j ° = ° °α ° ° ° ° j=1

j=1

j=1

j=1

with equality when x2 = x3 = · · · = xm = 0 (and possibly for other values). This proves †. The matrix A given has eigenvalue 0 only. However kAk = 1.

122

K110 (ii) Let x =

Pn

j=1

xj ej . Then

yk =

n X j=1

unless x1 = 0.

λkj xj (

k 2 1/2 r=1 (λr xr ) )

Pn

→ e1

If the largest eigenvalue is negative, then (in general) kxk k → |λ1 | and k(−1)k yk − u1 k → 0 where u1 = ±e1 . (iii) (c) The number of operations for each operation is bounded by An2 for some A (A = 6 will certainly do), Suppose |λ1 | > µ ≥ |λj | for j ≥ 2. Then the e1 component of αk x 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 the same kind of thing will occur. (iv) Suppose e1 the eigenvector with largest eigenvalue known. Use the iteration x 7→ (α(x − x · e1 )

or something similar. However, the accuracy to which we know e1 will limit the accuracy to which we can find e2 and matters will get rapidly worse if try to find the third largest eigenvalue and so on. (vii) No problem in real symmetric case because eigenvectors orthogonal.

123

K111 (ii) hx, α∗ αxi = hαx, αxi = kαxk2 . Thus kα∗ αkkxk2 = kxkkα∗ αxk ≥ |hx, α∗ αxi| = kαxk2 ≥ kαk2 kkxk2

for all x and so |α∗ αk ≥ kαk2 .

(iii) kα∗ kkαk ≥ kα∗ αk ≥ kαk2 and so either α = 0 (in which case the result is trivial) or kαk 6= 0 and kα∗ k ≥ kαk. But α∗∗ = α so kαk = kα∗∗ k ≥ kαk. Thus kαk = kα∗ k and kαk2 = kα∗ kkαk ≥ kα∗ αk ≥ kαk2 so kα∗ αk = kαk2 .

Since (α∗ α)∗ = α∗ α∗∗ = α∗ α we have α∗ α symmetric. If αe = λe with e 6= 0 then λkek2 = he, λei = he, α∗ αei == hαe, αei = kαek2 .

Thus λ ≥ 0.

(vi) Multiplication of A with A∗ takes of the order of m3 operations (without tricks) and dominates. ¶ µ 1 1 ∗ . (vii) We have A A = 1 5 A has eigenvalues 1 and 2. A∗ A has eigenvalues 4 ± 51/2 . kAk = (4 + 51/2 )1/2 .

124

K112 Follow the proof of Rolle’s theorem. If kc|| < 1 and f has a maximum at 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 not identically zero (consider θ = π/2 and other values of θ). We can not hope to have a ‘Rolle’s theorem proof’ in higher dimensions.

125

K113 Since (x, y) 7→ x − y, (x, y) 7→ xy, x 7→ x−1 and x 7→ f (x) are all continuous we see that F which can be obtained by composition of such functions is continuous except perhaps at points (x, x). If F is continuous at (x, x) then F (x + h, x) must tend to the limit F (x, x) as h → 0 so f is differentiable and F (x, x) = f 0 (x). Since F (x + h, x + h) → F (x, x) as h → 0, f 0 must be continuous. Conversely if f 0 exists and is continuous then, setting F (x, x) = f 0 (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) − f (x + k) F (x + h, x + k) − F (x, x) = − f 0 (x) x−y ¡ ¢ 0 = f x + (θh,k h + (1 − θh,k )k) − f 0 (x)

for some θh,k with 0 < θh,k < 1. But it is also true that, if we set θh,h = 1/2, F (x + h, x + h) − F (x, x) = f 0 (x + h) − f 0 (x) ¡ ¢ = f 0 x + (θh,h h + (1 − θh,h )h) − f 0 (x).

Thus

¡ ¢ F (x + h, x + k) − F (x, x) = f 0 x + (θh,k h + (1 − θh,k )k) − f 0 (x) → 0

as (h2 + k 2 )1/2 → 0.

Suppose now that f is twice continuously differentiable. Much as before, 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 form of Taylor’s theorem shows us that f 00 (x) 2 h + ²(h)h2 f (x + h) = f (x) + f 0 (x)h + 2 with ²(h) → 0 as h → 0. Thus, if h 6= k, F (x + h, x + k)

(f (x) + f 0 (x)h + 12 f 00 (x)h2 + ²(h)h2 ) − (f (x) + f 0 (x)k + 12 f 00 (x)k 2 + ²(k)k 2 h−k 00 (x) 2 + ²(k)k 2 f ²(h)h = f 0 (x) + (h + k) + 2 h−k f 00 (x) = F (x, x) + (h + k) + ²0 (h, k)(h2 + k 2 )1/2 2

=

where ²0 (h, k) → 0 as (h2 + k 2 )1/2 → 0. The formula can be extended to the case h = k by applying the appropriate Taylor theorem to f 0 . Thus F is differentiable everywhere.

126

K114 If k 6= 0, then we can find an X > 0 such that |f 0 (t) − k| < |k|/4 and so |f 0 (t)| > 3|k|/4 for t ≥ X. Thus, using the mean value inequality, ¯ ¯ ¯ ¯ ¯ f (x) ¯ ¯ f (x) − F (X) x − X F (X) ¯ ¯=¯ ¯ ¯ + ¯ x ¯ ¯ x−X x x ¯ ¯ ¯¯ ¯ ¯ ¯ ¯ f (x) − f (X) ¯ ¯ x − X ¯ ¯ f (X) ¯ ¯¯ ¯ ¯ ¯ ≥ ¯¯ ¯¯ x ¯ − ¯ x ¯ x−X ¯ ¯ ¯ ¯ 3|k| ¯¯ x − X ¯¯ ¯¯ f (X) ¯¯ ≥ − 4 ¯ x ¯ ¯ x ¯ |k| |k| |k| − = ≥ 2 4 4 −1 whenever x ≥ max(3X, (4|f (X)|) ). g(x) = x−2 sin x4 will do.

127

K115 The local form of Taylor’s theorem f 00 (x) 2 h + ²(h)h2 2 applied to log (or other arguments) show us that f (x + h) = f (x) + f 0 (x)h +

log(1 − h) = −h −

h2 + ²(h)h2 2

with ²(h) → 0 as h → 0. Thus since qn → ∞ we can find an N such that

−2−1 qn−1 ≥ log(1 − qn−1 ) and 0 ≥ log(1 − qn−1 ) + qn−1 ≥ −2qn−2 ≥ −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 −1 −1 x1 = −(a−1 11 a12 x2 + a11 a13 x3 + · · · + a11 a1n xn ).

130

K118 (i) One way of seeing that the λj are continuous is to solve the characteristic equation. Now use the intermediate value theorem. ¶ µ cos t − sin t will do. (iii) B(t) = sin t cos t

131

K119 The reduction to Figure K2 can not be rigorous unless we state clearly what paths are allowed. However if we have a path which is not symmetric under reflection in the two axes of symmetry joining mid points of opposite sides then by considering the reflected path we can reconstruct a shorter symmetric path. [A more convincing route involves the observation that the shortest path system for three points M , N , P is three paths M Z, N Z, P Z making angle 2π/3 to each other provided the largest angle of the triangle M N P is less than 2π/3 and consists of the two shortest sides otherwise. (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 of the isosceles triangles. 1 − 2 sin θ f 0 (θ) = cos2 θ so f increases on [0, π/6] and decreases on [π/6, π/4] so the best angle is θ/6 (check that this gives an arrangement consistent with the statement in square brackets). IN PARTICULAR θ = π/4 is not best. There are two such road patterns the other just being a rotation through π/4

132

K120 The time taken is h − x (1 + x2 )1/2 + u v for 0 ≤ x ≤ h. We observe that x 1 f 0 (x) = − + u v((1 + x2 )1/2 so f 0 is negative and f is decreasing for 0 ≤ x ≤ α with 0 ≤ x ≤ (1 − v 2 /u2 )−1/2 and f 0 is positive and f is increasing for x ≥ (1−v 2 /u2 )−1/2 . Thus if (1 − v 2 /u2 )−1/2 < h (i.e. (1 − v 2 /u2 )h2 > 1) he should take x = (1 − v 2 /u2 )−1/2 . However, if (1 − v 2 /u2 )h2 < 1 his best course if 0 ≤ x ≤ h is to take x = h. Since choosing x outside this range is clearly worse he should take x = h. f (x) =

If h is large his initial run is almost directly towards the tree and his greater land speed makes this desirable. If h is small his run is almost perpendicular 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) = y 2 − 13 x3 + ax in the region x2 + y 2 ≤ 1. We first examine the region x2 + y 2 < 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 and both 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 µ ¶ −2x 0 0 2

so (a1/2 , 0) is a saddle and (−a1/2 , 0) a minimum. Thus the global maximum must occur on the boundary and the global minimum may occur 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 is singular there but inspection of the equation f (x, y) = y 2 − 13 x3 shows that we have a saddle and both the global maximum and minimum must occur on the boundary. Now we look at the boundary x2 + y 2 = 1. Here f (x, y) = g(x) = 1 − x2 − 13 x3 + ax

with x ∈ [−1, 1]. Notice that g is a cubic so we can draw sketches to help ourselves. Now g 0 (x) = −2x − x2 + a and g 0 has roots at −1 ± (1 + a2 ). If |a| ≥ 1 then g is decreasing on (−1, 1) so has maximum at x = −1 (which is thus the global maximum for f at (−1, 0)) and a minimum at x = 1 9which 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 ¡ ¢1/2 x = −1 + (1 + a2 ) y = 1 − (1 + (1 + a2 ))2 ) and decreases as x runs 2 from from −1 ± (1 + a ) 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 ≥ −1/3 we know that it gives a global minimum for f at (0, 1). If 1 > a > 0 we observe that we know that the global minimum occurs when y = 0 and by looking at h(x) = f (x, 0) = − 13 x3 + ax we see that it occurs at (−a1/2 , 0).

134

K122 The composition of differentiable functions is differentiable and the product of differentiable functions (with values in R) are differentiable so since the maps (x, y) 7→ r and (x, y) 7→ θ are differentiable except at 0, f is. (To see that (x, y) 7→ r is differentiable observe that r = (x2 + y 2 )1/2 and use basic theorems again. To see that (x, y) 7→ θ is differentiable on S = {(x, y) : |y| > 9x}

observe that on S, θ = tan−1 x/y. To extend to R2 \ {0} either use rotational symmetry or cover R2 \{0} with rotated copies of S on which similar formulae hold. Since f (r cos θ, r sin θ) − f (0, 0) → g(θ) r as r → 0+, f has a directional derivative at 0 if and only if g(θ) = −g(−θ).

so

If f is differentiable at 0 then f (r cos θ, r sin θ) − f (0, 0) → cos θf,1 (0, 0) + sin θf,2 (0, 0) r g(θ) = g(0) cos θ + g(π/2) sin θ

ie g(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 − aα2 t)(β − bα2 t)

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 ) the function f has no minima. (ii) In part (i), f has Hessian ¶ ¶ µ µ 0 0 f,11 (0) f,12 (0) = 0 2 f,21 (0) f,22 (0)

which is singular.

136

K124 (i) The second sentence holds by Exercise 8.22 which says that if f is 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,

for x ≥ 1/2.

g(x) = F (x)

(iii) By considering functions of the form f (x) = F (x), g(x) = 0

for x < 1/2

f (x) = 0,

for x ≥ 1/2.

g(x) = F (x)

and of the form f (x) = F (x), g(x) = 0

for x < 1/2

f (x) = 0,

for x ≥ 1/2.

g(x) = 1

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

137

K125 (i) Consider the dissection DN = {r/N : 0 ≤ r ≤ N } where N ≥ M 2 . At most 2M 2 intervals [(r − 1)/N, R/N ] contain points of the form p/q with p and q coprime and 1 ≤ q ≤ M . Thus 0 = s(f, DN ) ≤ S(f, DN ) ≤ 2M 2 N −1 + M −1 → M −1

as R 1 N → ∞. Since M is arbitrary f is Riemann integrable with f (t) dt = 0. 0

(ii) f is continuous at irrational points x. Since there are only finitely many points of the form p/q with p and q coprime and 1 ≤ q ≤ M and x 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 such that xn → x. Since 0 = f (xn ) 9 f (x), we are done. (iii) Nowhere differentiable. By (ii) need only look at irrational points 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) 1/p ≥ =1 p/q − x 1/p but choosing xn irrational such that xn → x we have (f (xn ) − f (x) = 0 → 0. xn − x

138

K126 Observe that, if 0 ≤ u < v ≤ 1, then un −v n ≤ n(y −x) (by the mean value theorem or algebra). Write fn (x) = f (xn ). If D is a dissection write Dn = {y 1/n : y ∈ D}.

By the observation of the first sentence ¡ ¢ S(fn , Dn ) − s(fn , Dn 0 ≤ 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 ² > 0, we can find a 1 > δ > 0 such that |f (x) − f (0)| ≤ ²/2 for 0 ≤ x ≤ δ. Since δ 1/n → 1 as n → ∞ we can find an N such that 1 − δ 1/n ≤ (K + |k|)−1 ²/2 for n ≥ N .

If N ≥ n then |fn (x) − k| ≤ ²/2 for x ∈ [0, δ 1/n ] and |fn (x)| ≤ K for x ∈ [δ 1/n , 1] so ¯ ¯Z 1/n ¯Z 1 ¯ ¯ ¯ δ ¯ ¯ ¯ ¯ f (x) dx − k(1 − δ 1/n ¯ ≤ ²/2 and ¯¯ f (x) dx − kδ 1/n ¯¯ ≤ ²/2 ¯ ¯ ¯ 0 δ 1/n so that

¯Z ¯ ¯ ¯

1

0

¯ ¯ f (x) dx − k ¯¯ ≤ ².

(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. Observe that, since f (a) = 0 and |f 0 (t)| ≤ K it follows that, by the mean value inequality, |f (s)| ≤ K(s − a) for a ≤ s ≤ (a + b)/2 with equality when s = (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 when s = (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 for equality 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 Z b Z |f (s)| ≤ a

(a+b)/2−η

+

a

Z

(b+a)/2

+ (a+b)/2−η ¢2

¡ ≤ K (b − a)/2 − η

< K(b − a)2 /4

Z

(a+b)/2+η

+ (a+b)/2

Z

b (a+b)/2+η ¢2 ¡

¡ + (K − η)( b − a)/2

|f (s)| ds

− (b − a)/2 − η

as required.

To see that this is best possible define f (x) = K(x − a)

f (x) = C − τ (x − (a + b)/2)2

for a ≤ x ≤ (a + b)/2 − ²,

for (a + b)/2 − ² < x ≤ (a + b)/2,

and f ((a + b)/2 − t) = f ((a + b)/2 + t) where C is chosen to make f continuous and τ is chosen to make f differentiable at (a + b)/2 − ² (thus we take τ = K/(2²)).

¢2

140

K128 (v) Observe P that once we know that f is Riemann integrable we n−1 b−a know that n j=0 f (a + j(b − a)/n) converges but the fact that Pn−1 b−a j=0 f (a + j(b − a)/n) converges does not imply that f is Rien mann integrable.

141

K129 For (A) note that, if Fj and Gj are positive bounded Riemann integrable functions with F1 − G1 = F2 − G2 then F1 + G2 = F2 + G1 and Z b Z b Z b F1 (t) dt + G2 (t) dt = F1 (t) + G2 (t) dt a a a Z b = F2 (t) + G1 (t) dt a Z b Z b G1 (t) dt F2 (t) dt + = a

a

and so

Z

b a

F1 (t) dt −

Z

b

G1 (t) dt = a

Z

b a

F2 (t) dt −

Z

b

G2 (t) dt. a

The problems with (C) begin (I think) when we try to prove that if f and 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 which is Riemann integrable under the old definition are Riemann integrable under 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 a bounded positive function f is Riemann integrable if and only if given any ² > 0 we can find a dissection D = {x0 , x1 , . . . , xN }

with a = x0 < x1 < · · · < xN = b and a subset Θ of {r : < 1 ≤ r ≤ N } such that and

|f (x) − f (y)| ≤ ² for x, y ∈ [xr−1 , xr ] when r ∈ Θ. X (xr − xr−1 ) ≤ ². r∈Θ /

142

K130 (iii) Everything works well until we try to define the upper and lower integrals but the set of s(D, f ) has no supremum in Q and S(D, f ) has no infimum.

143

K131 (i) Since F is continuous on a closed bounded interval the infimum and supremum are attained at α and β say. Now use the intermediate value theorem. (ii) Observe that sup f (s)w(t) ≥ f (t)w(t)

s∈[a,b]

for all t ∈ [a, b] so sup f (s) s∈[a,b]

Z

a

b

w(t) dt ≥

Z

b

f (t)w(t) dt a

Using a similar result for the infimum we have Z b Z b Z b sup f (s) w(t) dt ≥ f (t)w(t) dt ≥ inf f (s) w(t) dt. s∈[a,b]

a

a

Now use part (i). (iii) Take a = −1, b = 1, F (t) = w(t) = t. If w everywhere negative, consider −w.

s∈[a,b]

a

144

K132 The graph of f splits the rectangle [a, b] × [f (a), f (b)] into two parts whose areas correspond to the given integrals. Consider a dissection of [a, b] D = {x0 , x1 , . . . , xN }

with a = x0 < x1 < · · · < xN = b. This gives rise to a dissection of [f (a), f (b)] D0 = {f (x0 ), f (x1 ), . . . , f (xN )}

with f (a) = F (x0 ) < f (x1 ) < · · · < f (xN ) = f (b). We have

S(D, f ) + s(D 0 , g) = bf (b) − af (a)

so, taking an infimum over all D, we get

I ∗ (f ) + I∗ (g) ≥ bf (b) − af (a)

Similarly

s(D, f ) + S(D 0 , g) = bf (b) − af (a)

so

I∗ (f ) + I ∗ (g) ≤ bf (b) − af (a). Rb R f (b) But I ∗ (f ) = I∗ (f ) = a f (t) dt and I ∗ (g) = I∗ (g) = f (a) g(s) ds so Z f (b) Z b g(s) ds = bf (b) − af (a). f (t) dt + f (a)

a

Z

0

(ii) Without loss of generality, assume Y ≥ F (X). Then Z Y Z X Z f (X) Z X f (x) dx + g(s) ds = f (x) dx + g(s) ds + 0

0

0

Y

g(s) ds f (X)

≥ XF (X) + (Y − F (X))X = Y X

with equality if and only if Y = f (X). (iii) Take f (x) = xp 1/q X = Y 1/p .

−1 −1

, g(y) = y q

−1 −1

. Equality if and only if

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.

146

K134 Suppose γ ∈ [a, b]. Then given 1 > ² > 0 we can find a δ > 0 such that |f (s) − f (γ)|, |f (g(s)) − f (g(γ))|, |g 0 (s) − g 0 (γ)| < ²

for all s ∈ [a, b] with |s − γ| < δ.

Now suppose [α, β] ⊆ (α − δ, α + δ). Without loss generality we suppose g(β) ≥ g(α). Then ¯Z Z g(β) ¯ g(β) ¯ ¡ ¢ ¯ ¯ f (s) ds − g(β) −g(α) f (g(γ)) ≤ |f (s) − f (g(γ))| ds ¯ ¯ g(α) g(α) ¡ ¢ ≤ g(β) − g(α) sup |f (s) − f (g(γ))| s∈[g(α),g(β)]

¡

¢

≤ g(β) − g(α) sup |f (g(s)) − f (g(γ))| ¡

t∈[α,β]

¢

≤ g(β) − g(α) ²

≤ sup |g 0 (t)|(β − α) t∈[α,β]

≤ (|g 0 (γ)| + ²)(β − α)² We also have, by a simpler argument ¯Z β ¯ ¯ ¯ ¡ ¢ ¯ f (g(t)) dt − g(β) − g(α) f (g(γ))¯¯ ≤ (β − α)². ¯ α

Thus ¯ ¯ Z β ¯Z g(β) ¯ ¯ ¯ f (s) ds − f (g(t)) dt¯ ≤ (1 + |g 0 (γ)| + ²)(β − α)² ¯ ¯ g(α) ¯ α We can make ² as small as we want.

≤ (2 + |g 0 (γ)|)²(β − α).

147

K135* No comments.

148

K136 The formula sup x∈[xj−1 ,xj ]

f 0 (t)(xj − xj−1 ) ≥ f (xj ) − f (xj−1 ) ≥

inf

x∈[xj−1 ,xj ]

f 0 (t)(xj − xj−1 )

is a version of the mean value inequality. Consider a dissection of [a, b] D = {x0 , x1 , . . . , xN }

with a = x0 < x1 < · · · < xN = b. By summing the formula of the first paragraph we get so

S(D, f 0 ) ≥ f (b) − f (a) ≥ s(D, f 0 ) I ∗ (f 0 ) ≥ f (b) − f (a) ≥ I∗ (f 0 )

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 R1 as N → ∞. Thus f is Riemann integrable and −1 f (t) dt = 0.

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

151

K139 (i) n X r=1

n X (r + 1)r(r − 1) − r(r − 1)(r − 2)

r(r − 1) =

3

r=1

n X

r=

r=1

n X

2

r =

r=1

n X (r + 1)r − r(r − 1)

2

r=1

n X

r(r − 1) +

r=1

n X r=1

r=

=

=

(n + 1)n(n − 1) . 3

(n + 1)n 2

n(n + 1)(2n + 1) . 6

(ii) n X r=1

n3 =

n2 (n + 1)2 . 4

(iii) If f (x) = xm S(D, f ) = and s(D, f ) =

n X

1 Pm (n) 1 r + m+1 → n−1 ( )m = n m+1 n m+1 r=1

n−1 X

r 1 Pm (n − 1) 1 n−1 ( )m = (1 − n−1 )m+1 + → n m+1 nm+1 m+1 r=0

so I ∗ f = I∗ f = (m + 1)−1 . (iv) If f (x) = xm and F (x) = xm+1 /(m + 1) then F 0 = f and Z b f (x) dx = F (b) − F (a). a

152

K140 (v) We have |g(t) − g(−1)(t + 1)| ≤ K(t − 1) so ¯Z 1 ¯ Z 1 ¯ ¯ ¯ ¯ g(t) dt − g(−1)¯ ≤ K (t − 1) dt = 2K. ¯ −1

Scaling, we get ¯Z ¯ ¯ ¯

−1

a−rh

a−(r−1)h

0

¯ ¯ f (t) dt − f (a − (r − 1)h)¯¯ ≤ Kh2

if |f (t)| ≤ K for t ∈ [a, b] and N h = b − a. Thus, summing, ¯Z b ¯ ¯ ¯ ¯ ¯ ≤ K(b − a)h. f (t) dt − S (f ) h ¯ ¯ a

Taking a = 0, b = π, N h = π and F (t) = sin2 (nt), we see that the result cannot be substantially improved.

153

K141 (ii) Let s = yt. (viii) We have tn+1 1 − (1 + t + · · · + tn ) = . 1−t 1−t So, integrating both sides from 0 to x, we have µ ¶ Z x n+1 x2 t xn+1 − log(1 − x) − x + = + ··· + dt → 0 2 n+1 0 1−t

as n → ∞ for all |x| < 1. Thus

∞ X xr+1 log(1 − x) = − r+1 r=1

for |x| < 1. [The result also holds if x = −1 but the argument above does not prove this without extra work.] (ix) By (viii) log

µ

1+x 1−x

¶

= log(1 + x) − log(1 − x) =

∞ X (−1)r xr+1 r=1 ∞ X

=2

r=1

for |x| < 1.

r+1

x2r+1 2r + 1

+

∞ X xr+1 r+1 r=1

If y > 0 and we seek an x with y= we obtain

1+x 1−x

y−1 y+1 so |x| < 1 and substitution in the formula of (ix) gives the desired result. x=

154

K142 Power series can be multiplied term by term within their radius of convergence. We observe that ¶ n−1 µ n−1 X (−1)r (−1)n−r (−1)n X 1 1 (−1)n 2Sn−1 × = + . = r n−r n r=1 r n − r n r=1 The result is valid for |x| < 1.

155

K143 (i) Write λ= and observe that 1 > λ > 0, so

x3 − x 2 x3 − x 1

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 , then f (y2 ) − f (y1 ) f (y3 ) − f (y2 ) f (y3 ) − f (y1 ) f (y3 ) − f (y1 ) ≥ and ≥ . y3 − y 1 y2 − y 1 y3 − y 2 y3 − y 1 (ii) By (i), if h > 0, then f (c) − f (d) f (c + h) − f (c) ≥ h c−d where d is fixed with c > d > a. Since a non-empty set bounded below has an infimum, σf (c+) exists. By definition, given ² > 0 we can find a δ > 0 such that f (c + δ) − f (c) . δ If h > δ > 0 then (looking at the last paragraph of our discussion of (i)) σf (c+) + ² ≥

so

σf (c+) + ² ≥

f (c + h) − f (c) f (c + δ) − f (c) ≥ ≥ σf (c+) δ h f (c + h) − f (c) → σf (c+) h

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) ≥ σf (c+) h

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−).

156

(v) Observe that αg(t) ≥ tg(t) ≥ βg(t)

and integrate to obtain c ∈ [α, β]. so

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

Now integrate.

(f (t) − f (c))g(t) ≥ B(t − c)g(t).

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

whence Ã n ! n n n X X X X λj f (xj ) − f λj xj = λj (f (xj ) − f (c)) ≥ λj (xj − c) = 0. j=1

j=1

j=1

j=1

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 a0 , b0 so that a < a0 < c < b0 < b Observe that, E = {(x, y) ∈ R2 : y ≥ f (x), x ∈ [a0 , b0 ]}

is a closed convex set. Applying Exercise 32 (iii) with y = (c, f (c) − n−1 0 we see that there exist an , bn and cn such that bn x + an y ≤ cn whenever (x, y) ∈ E

and bn c + an (f (c) − n−1 ) > cn . By considering what happens when y is very large, we see that an ≤ 0 and the conditions given are incompatible if an = 0. Thus an < 0 and, setting Bn = bn /an ,Cn = cn /an , we obtain Bn x + y ≥ Cn whenever (x, y) ∈ E

and Bn c + (f (c) − n−1 ) < Cn . In particular we have

Bn x + f (x) > Bn c + (f (c) − n−1 )

for all x ∈ [a0 , b0 ]. Thus

f (x) − f (c) > Bn (x − c) − n−1

for all x ∈ [a0 , b0 ]. If δ > 0 is such that [c − δ, c + δ] ⊆ [a0 , b0 ] then

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

157

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 ∈ [a0 , b0 ].

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

f (x) − f (c) ≥ Bn (x − c)

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

f (x) − f (c) ≥ B(x − c)

158

K144 (ii) Suppose that γ < β. If 0 < h, k < (β − γ)/2 then f (β − h) − f (γ + h f (γ + k) − f (γ) f (β) − f (β − h) ≥ ≥ h β−γ−h−k k

so

σf (β−) ≥ σf (γ+). Thus N X ¡ j=1

σf (cj +) − σf (cj −)

¢

= σf (cN +) −

N X ¡ 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 the integer part of x). Thus E = {a, b} ∪

∞ [

Ej

j=1

is countable. If c ∈ / E, then σf (c−) = σf (c+) = σf (c), say, and f (c + h) − f (c) → σf (c) h 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 P Thus, using the comparison test, ∞ n=1 fn (x) converges everywhere to f (x), say.

159

Noting that, by convexity, fm (x + h) + fm (x − h) − 2fm (x) ≥ 0 for all m when x, x − h, x + h ∈ (−1, 1), we have f (qn + h) + f (qn − h) − 2f (qn ) f (qn + h) − f (qn ) f (qn ) − f (qn − h) − = h h h fn (qn + h) + f (qn − h) − 2f (qn ) ≥ h −2 = 2n 9 0 as h → 0+, so f is not differentiable at qn

160

K145 (i) Given ² > 0, we can find a δ > 0 such that |f (x) − f (c)| < ² for |x − c| < δ. If [an , bn ] ⊆ (c − δ, c + δ), then ¯ ¯ ¯ ¯ Z Z ¯ ¯ 1 ¯ ¯ 1 ¯ ¯ ¯ f (t) dt − f (c)¯ = ¯ f (t) − f (c) dt¯¯ ¯ |In | |In | In In Z 1 ≤ |f (t) − f (c)| dt |In | In Z 1 ≤ ² dt = ² |In | In (ii) The following is one generalisation (interpreting the notation appropriately). Let B(c, r) be the ball centre c then (provided the integrals exist) Z 1 f (t) dV (t) → f (c) Vol B(c, r) B(c,r)

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 that F,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 under the integral to obtain Z y F,1 (x, y) = g,1 (x, y) dx. 0

If (x0 , y0 ) is fixed Take R ≥ 2 + |y0 | + |x0 |. Since g,1 is continuous on K = [−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 |g,1 (x, t) − g,1 (x0 , t)| + |y − y0 | |t|≤|y|

≤R

sup (s,t),(s0 ,t)∈K,|s−s0 |≤|x−x

+ |y − y0 |

0|

sup (s,t),(s0 ,t)∈K

sup

|t|≤max(|y|,|y0 |)

|g,1 (x0 , t)|

|g,1 (s, t) − g,1 (s0 , t)|

|g,1 (x0 , t)| → 0

as k(x, y) − (x0 , y0 )k → 0. Thus F,2 is continuous and F is once differentiable. (iii) The chain rule shows that G is differentiable and Z x 0 G (x) = F1 (x, x) + F2 (x, x) = g,1 (x, t) dt + g(x, x). 0

(iv) We have H(x) = F (x, h(x)) so the chain rule gives Z h(x) 0 0 H (x) = F1 (x, h(x)) + F2 (x, h(x))h (x) = g,1 (x, t) dt + h0 (x)g(x, h(x)). 0

163

K148 Observe that differentiating under the integral, using the symmetry of partial derivatives and the fundamental theorem of the calculus

E 0 (t) =

Z

=2

1 0

Z

Z

∂ ∂t

Ãµ

∂u (x, t) ∂t

¶2

+

µ

∂u (x, t) ∂x

¶2 !

dx

1

u,2 (x, t)u,22 (x, t) + u,1 (x, t)u,12 (x, t) dx 0 1

−u,2 (x, t)u,11 (x, t) + u,1 (x, t)u,21 (x, t) dx Ãµ ¶2 µ ¶2 ! Z 1 ∂u d ∂u (x, t) + (x, t) = dx ∂t ∂x 0 dx "µ ¶2 µ ¶2 #1 ∂u ∂u = =0 (x, t) + (x, t) ∂t ∂x =2

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 Hδ (t) = cδ

for t ≤ 0

for t ≥ 2δ

and 0 ≤ Hδ (t) ≤ cδ for all t, where cδ is a strictly positive number. Let Lδ (t) = c−1 δ (Hδ (t) − Hδ (t + 1 − 2δ). If 0 < δ < 1/4 then Lδ (t) = 0

Lδ (t) = 1 and 0 ≤ Lδ (t) ≤ 1 for all t.

for t ∈ / [0, 1]

for t ∈ [2δ, 1 − 2δ]

Now let kn (x) = (−1)[2nx] L²/20 (2nx−2[nx]) and follow Exercise 8.66.

165

K150 We seek to minimise Z b F (x, y(x), y 0 (x)) dx with F (u, v, w) = g(u, v)(1 + w 2 )1/2 . a

The Euler-Lagrange equation gives d 0 = F,2 (x, y(x), y 0 (x)) − F,3 (x, y(x), y 0 (x)) dx d g(x, y)y 0 2 = g,2 (x, y)(1 + y 0 )1/2 − dx (1 + y 0 2 )1/2

g,1 (x, y)y 0 + g,2 (x, y)y 0 2 + g(x, y)y 00 g(x, y)y 0 2 y 00 + . = g,2 (x, y)(1 + y ) − (1 + y 0 2 )1/2 (1 + y 0 2 )3/2 Multiplying through by 1 + y 0 2 gives the required result. 0 2 1/2

We seek to minimise 2π

Z

b

x ds.

a

We take g(x, y) = x and obtain −y 0 −

xy 00 =0 1 + y02

or ¶ xy 0 =0 (1 + y 0 2 )1/2 (observe that g(x, y) has no direct dependence on y and recall Exercise 8.63). Thus xy 0 =c (1 + y 0 2 )1/2 whence c y0 = 2 (x − c2 )1/2 and x y + a = c cosh−1 c for appropriate constants a and c. d dx

µ

166

K151 Precisely as in the standard case, we consider Z b G(η) = f (x, y(x) + ηg(x), y 0 (x) + ηg 0 (x), y 00 (x) + ηg 00 (x)) dx a

and require 0

0 = G (0) =

Z

b

f,2 (x, y, y 0 , y 00 ) + g 0 (x)f,3 (x, y, y 0 , y 00 ) + g 00 (x)f,4 (x, y, y 0 , y 00 ) dx a

for all well behaved g with g(a) = g 0 (a) = 0, g(b) = g 0 (b) = 0. Integrating by parts once and twice gives µ Z b d g(x) f,2 (x, y(x), y 0 (x), y 00 (x)) − f,3 (x, y(x), y 0 (x), y 00 (x)) 0= dx a ¶ 2 d + 2 f,4 (x, y(x), y 0 (x), y 00 (x)) dx dx for all g as above, so we get an Euler-Lagrange type equation d 0 = f,2 (x, y(x), y 0 (x), y 00 (x)) − f,3 (x, y(x), y 0 (x), y 00 (x)) dx d2 + 2 f,4 (x, y(x), y 0 (x), y 00 (x)). dx In the case given this becomes 0 = −24 +

d2 00 y (x) dx2

that is to say y 0000 (x) = 24 so y = A + Bx + Cx2 + Dx3 + x4 . The boundary conditions give us A = 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) + ²x2 (1 − x)2 sin N x, we can make small changes in 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−1 2−n−3 , 0 ≤ r ≤ n, 0 otherwise.

169

K154 Second sentence done exactly as the case h = 1 (Lemma 9.6). R∞ If 1 f (t) dt converges then, since Z Nh N −1 N −1 X X f (nh), f (t) dt ≥ h f (nh) ≥ h n=0

0

we have (allowing N → ∞) Z ∞ X f (nh) ≥ h n=0

and so

as h → 0+.

n=1

∞ 0

f (t) dt ≥ h

∞ X

f (nh),

n=1

¯ ¯ ∞ Z ∞ ¯ ¯ X ¯ ¯ f (t) dt¯ ≤ hf (0) → 0 f (nh) − ¯h ¯ ¯ n=1 0

Without loss of generality, suppose 1 > ² > 0. Choose K > 8N ²−1 and set ( 1 − K|x − rN −1 | if |x − rN −1 | < K, 0 ≤ r ≤ n, g(x) = 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 Z Z x 0 f (t) dt ≤ f (n) + f (x) ≤ f (n) + n

x 0

n

|f (t)| dt ≤ f (n) +

for n ≤ x ≤ n + 1 and so Z n+1 Z f (x) dx ≤ f (n) + n

Z

n+1 n

|f 0 (t)| dt

n+1

|f 0 (t)| dt.

n

Similarly

Z

n+1 n

f (x) dx ≥ f (n) −

Z

n+1 n

|f 0 (t)| dt.

It follows that, ¯ ¯ ¯Z ¯ Z m+1 m ¯X ¯ ¯ m+1 ¯ ¯ ¯ ¯ f (r)¯ ≤ ¯ f (x) dx¯¯ + |f 0 (x)| dx ¯ ¯ ¯ n n r=n R∞ Pn and, if 1 f (x) dx converges, r=0 f (r) is a Cauchy sequence and converges by the general principle of convergence. P∞ similar argument shows that R nConversely, if r=0 f (r) converges, a P f (x) dx tends to a limit. Further, if ∞ r=0 f (r) converges we have 1 f (n) → 0 and the arguments above show that Z n+1 Z n+1 |f 0 (x)| dx → 0 |f (x)| dx ≤ |f (n)| + n n R∞ as n → ∞ and so 1 f (x) dx converges. (ii) If f is decreasing, positive and has a continuous derivative Z X Z X 0 f 0 (x) dx = f (1) − f (X) ≤ f (1). |f (x)| dx = − 1

1

(iii) False. We consider a continuously differentiable function g such that g(x) =

(

(2n)−1 2−2n −(2n + 1)−1 2−2n−1

if 22n + 2−1 ≤ x ≤ 22n+1 − 2−1 , n ≥ 0, 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 ], increasing on [22n − 2−1 , 22n + 2−1 ] and g(2m ) = 0 for all m.

171

K156

Z

n+1/2 n−1/2

log x dx − log n =

Z

=

Z

n+1/2

(log x − log n) dx +

n n+1/2

log(x/n) dx + n 1/2

Z

n

Z

n n−1/2

(log x − log n) dx

log(x/n) dx n−1/2

¶ µ ¶¶ t t + log 1 − dt log 1 + = n n 0 making substitutions like x = t + n and x = n − t. Z

µ

µ

But by the mean value inequality, ¯ ¯ ¯ µ ¶ µ ¶¯ ¯ 1 ¯ ¯ ¯ t t 1 −1 ¯≤2 ¯ ¯log 1 + sup ¯¯ + log 1 − − ¯ ¯ n n n − t¯ t∈(0,1/2) t + n 2t 4 = 2−1 sup ≤ 2. 2 2 3n t∈(0,1/2) n − t Thus by the comparison test ¶ µ ¶¶ µ ∞ µ X t t log 1 + + log 1 − n n n=1

is absolutely convergent and so convergent to c, say. Thus Z N +1/2 log x dx − log N ! → c. 1/2

But, integrating by parts, Z N +1/2 Z N +1/2 log x dx = [x log x]1/2 − 1/2

= (N + 12 ) log(N +

Thus

N +1/2

dx 1/2 1 ) − 21 ) log 21 2

− (N + 1).

(N + 12 ) log(N + 12 ) − N − log N ! → c0

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

172

K157 With the notation of the previous question Áµ ¶2 n! 2n! −2n −n e e → C −1 (2n + 1/2)(2n+1/2) (n + 1/2)(n+1/2) where C is the constant of K156. Thus µ ¶ 1 2n ) 2n 1/2 2n (1 + 4n → C0 1 2n 2 n n (1 + 2n )

and

µ ¶ 2n → C 00 4 n n as n → ∞ for appropriate constants C 0 and C 00 . In particular, we can find constants K1 and K2 with K1 > K2 > 0 such that µ ¶ 2n −n −1/2 ≥ 4−n n−1/2 . K1 4 n ≥ n Thus ¯ µ ¶ ¯ ¯ µ ¶ ¯ ¯ α 2n n ¯ ¯ 2n n ¯¯ ¯n z ¯¯ → 0 if |z| < 1/4 and ¯¯nα z ¯ → 0 if |z| > 1/4 ¯ n n ¡ ¢ P α 2n n z has radius of convergence 1/4. so ∞ n=0 n n ¡ ¢ n z | ≤ K nα−1/2 for all |z| = 1/4 so, If α < −1/2 we have |nα 2n P∞ n α ¡2n¢ n 1 z converges absolutely and so by the comparison test, n=0 n n converges everywhere on the circle of convergence. ¡ ¢ n P∞ α ¡2n¢ n If α ≥ 1/2 we have |nα 2n z z | ≥ K for all |z| = 1/4 so 2 n=0 n n n diverges everywhere on the circle of convergence. ¡ ¢ α 2n If α ≥ −1/2 then n ≥ K2 nα−1/2 and the comparison test tells P∞ α ¡2n¢ n n us that n=0 n n z diverges when z = 1/4. n 1/2

The only remaining case is |z| = 1/4, ¡ ¢z 6= 1/4 and 1/2 > α ≥ −1/2. −n α 2n Let us write w = 4z and un = 4 n n . We know that un → 0 so, if we can showPthat un > un+1 for all sufficiently large n, Abel’s test will n tell us that ∞ n=0 un w . But 1 ) (1 + 2n un+1 = 1 1−α < 1 un (1 + n )

β so the conditions of Abel’s test are satisfied. (Observe that ¢≥ P∞ (1 +α ¡x) 2n n 1 + βx for β ≥ 0 and x ≥ 0.) If 1/2 > α ≥ −1/2 then n=0 n n z converges everywhere on the circle of convergence except the point z = 1/4.

173

K158 (i) Observe that Z

g(X)

f (s) ds = 0

Z

X

f (x)g 0 (x) dx 0

and that g(X) → ∞ as X → ∞ and, if g(h(Y )) → ∞ as Y → ∞, then h(Y ) → ∞ as Y → ∞. Only one limit so no problem of interchange. (ii) (sin x)/x → 1 as x → 0 (by considering differentiation, if you wish). Observe that 0 ≤ (−1)n+2

sin(x + nπ) sin(x + (n + 1)π) ≤ (−1)n+1 x + (n + 1)π x + nπ

so 0 ≤ (−1)

n+2

Z

π 0

sin(x + (n + 1)π) dx ≤ (−1)n+1 x + (n + 1)π

and, if we write vn = (−1)

n+1

Z

(n+1)π nπ

Z

π 0

sin(x + nπ) dx x + nπ

sin x dx, x

the vN are a decreasing sequence with Z (n+1)π 1 dx ≤ (nπ)−1 → 0 0 ≤ vn ≤ x nπ as n → ∞. Thus, by the alternating series test, Z nπ n−1 X sin x dx = (−1)r vr x 0 r=0

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

we have

as X → ∞.

Z

(n+1)π nπ

¯ ¯ ¯ sin x ¯ −1 ¯ ¯ ¯ x ¯ dx ≤ (nπ) → 0,

Z

X

0

sin x dx → L 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/π 2 for 0 ≤ x ≤ π/2 will do. Observe that ¯ ¯ Z nπ ¯ n−1 Z (r+1)π ¯ X ¯ sin x ¯ ¯ sin x ¯ ¯ ¯ ¯ dx = ¯ ¯ x ¯ dx ¯ x ¯ π r=1 rπ n−1 Z (r+1)π X x − rπ ≥ dx r rπ r=1 Z π n−1 X 1 g(x) dx = →∞ r 0 r=1 as n → ∞. R∞ x (v) If α > 0, then the arguments above show that 1 sin dx conxα verges. If α ≤ 0, then ¯ ¯Z ¯ (n+1)π sin x ¯ Z (n+1)π ¯ ¯ | sin x| dx 9 0 dx¯ ≥ ¯ α ¯ ¯ nπ x nπ R∞ x so 1 sin dx does not converge. xα However, if 0 ≤ x ≤ 1 we have x ≥ sin x ≥ 2x/π so sin x x1−α ≥ α ≥ 2x1−α /π x and comparison shows that Z 1 sin x dx xα ² converges as ² → 0+ if and only if α < 2. R∞ x Thus 0 sin dx converges if and only if 2 > α > 0. xα

175

K159 (i) Observe that n n X X sin((n + 12 )x) 1 − ei(2n+1)x = 1+2 cos rx = eirx = e−inx 1 − eix x r=1 r=−n

for |x| ≤ π.

(ii) Change of variable t = λx. (iii) Observe that the integrals Z (r+1)π/(n+1/2) rπ/(n+1/2)

¡ ¢ sin (n + 12 )x dx sin 12 x

alternate in sign and decease in absolute value as r increases from 1 to n and as r decreases from −! to −n. Observe further that if ² ≤ α ≤ β ≤ π and α − β ≤ π/(n + 1/2) then ¯ ¯ ¯ ¯Z ¯ β sin ¡(n + 1 )x¢ ¯ ¯Z −alpha sin ¡(n + 1 )x¢ ¯ π ¯ ¯ ¯ ¯ 2 2 dx¯ = ¯ dx¯ ≤ →0 ¯ 1 1 ¯ ¯ −β ¯ (n + f rac12)| sin ²| ¯ α sin 2 x sin 2 x as n → ∞. (iv) Write

2(sin 12 x − 21 x) 1 2 = − x sin 21 x x sin 12 x and use L’Hˆopital or Taylor expansions. (v) Given any η > 0 we can find an ² > 0 with η > ² such that ¯ ¯ ¯2 ¯ 1 ¯ − ¯ 1 ¯ x sin x ¯ < η 2

for |x| < ² and so ¯ ¯Z ² ¯ 2 sin((n + 21 )x) ¯¯ sin((n + 12 )x) ¯ dx − dx¯ ≤ 2²η ≤ 2η 2 . ¯ sin x x −²

2

Thus, allowing n → ∞ and using (ii) and (iii) we have ¯ ¯ Z ∞ ¯ ¯ sin x ¯2π − 2 ¯ ≤ 2η 2 . , dx ¯ ¯ x −∞

Since η was arbitrary

and we are done.

Z

∞

−∞

sin x , dx = π x

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)) and forbids f (n) ∼ g(n). f (n) = Ω(g(n)) permits f (n) = O(g(n)), permits f (n) ∼ g(n) and forbids f (n) = o(g(n)). f (n) ∼ g(n) implies f (n) = O(g(n)), implies f (n) = Ω(g(n)) and forbids 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 is bounded. (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) ≤ Aj gj (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) ≤ Aj gj (n) with Aj ≥ 0 then

0 ≤ f1 (n) + f2 ≤ (A1 + A2 ) max(g1 (n), g2 (n)). 2

(iv) True. n! ≤ nn ≤ (2n )n = 2n . (v) True. Use L’Hˆopital 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)2 x5

and (sin x)2 − x2 = O(x4 ). (vi) False. Use L’Hˆopital or (I think better) observe that cos x − 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)x6

where |γ1 (x)|, |γ2 (x)| ≤ 10 for |x| small (we can do better but we do not need to) so cos x − 1 + sin2 x/2 − sin4 x/4! + x4 /6 = O(x6 ).

178

K162 (i) If −1 < α ≤ 0, then xα is decreasing so Z n+1 n n Z r+1 n+1 X X X α α α r ≥ x dx = x dx ≥ rα . r=1

r=1

r

1

r=2

Since

Z

this gives

n

xα dx = 1

nα+1 − 1 , α+1

¯ ¯ P ¯ (α + 1) nr=1 rα ¯ 1 + 2nα ¯ ¯≤ →0 ¯ ¯ nα+1 nα+1

as n → ∞.

If α ≥ 0, then xα is increasing so Z n Z r+1 n X X α α x dx = r ≤ r=1

r=1

r

n+1 1

and much the same argument works.

xα dx ≤

n+1 X r=2

(ii) If α = −1, then much the same argument gives n X 1 ∼ log n. r r=1 (iv) No such modification, since

P∞

r=n

r−1 diverges.

rα

179

K163 Argument similar to that in K156. We have |g 00 (x)| ≤ Ax−λ for x ≥ R say. If |t| ≤ 12 and n ≥ R + 1, then applying the mean value theorem twice gives |(g(t + n + 12 ) − g(n + 12 )) + (g(−t + n + 12 ) − g(n + 12 ))| = |g 0 (s + n + 12 ) − g 0 (−s + n + 12 )|

= |2sg 0 (u + n + 12 )| ≤ A(n − 12 )−λ

for some s and u with such that

1 2

> s > 0 and |u| < s. Thus we can find a B

|(g(t + n + 21 ) − g(n + 12 )) + (g(−t + n + 12 ) − g(n + 12 ))| ≤ Bn−λ

for n ≥ R and |t| ≤ 21 . We have ¯Z ¯ ¯ ¯

n+1 n

¯ (g(x) − g(n + 21 ) dx¯ ¯ ¯ ¯ ¯Z 1 ¯ ¯ 2 1 1 1 1 ¯ (g(t + n + 2 ) − g(n + 2 )) + (g(−t + n + 2 ) − g(n + 2 )) dt¯¯ =¯ ¯ ¯ 0 ≤

so

Z

1 2

0

|(g(t + n + 12 ) − g(n + 12 )) + (g(−t + n + 12 ) − g(n + 12 ))| dt

≤ Bn−λ Z

n+1

(g(x) − g(n + 21 )dx = O(n−λ ).

n

We have ¯Z ¯ ¯ ¯

n+1 n

g(x) dx − g(n +

¯ ¯

1 ¯ ) 2 ¯

≤ Bn−λ

for n ≥ R so adding and using K162 or direct argument ¯Z ¯ n ¯ n+1 ¯ X Bn−λ+1 ¯ ¯ g(x) dx − g(r)¯ ≤ C + ¯ ¯ 1 ¯ −λ + 1 r=1 R n+1 for some constant C. Dividing through by 1 g(x) dx gives the required 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 R π/2 when x is small and positive, so, by comparison, 0 log(sin x) dx converges. (Similar arguments apply to all the integrals of this question.) Using the change of variables formula with t = 2x and symmetry we have Z π/2 Z Z π/2 1 π log(sin 2x) dx = log(sin t) dt = log(sin x) dx. 2 0 0 0 But, using symmetry again, Z π/2 Z π/2 log(sin 2x) dx = log(2 sin x cos x) dx 0

0

=

Z

π/2

log 2 dx + 0

Z

π/2

log(sin x) dx + 0

Z π/2 π = log 2 + 2 log(sin x) dx. 2 0 Combining our formulae gives the result.

Z

π/2

log(cos x) dx 0

181

K165

so

Ra Suppose that 0 f (x) dx converges. Since f is decreasing Z a Z a Z a N a X ³ ar ´ f (x) dx f f (x) dx → f (x) dx ≥ ≥ N r=1 N 0 a/N 0 Z a N ³ ar ´ X f f (x) dx. → N 0 r=1

Ra Suppose that 0 f (x) dx diverges. Then given any K we can find an ² > 0 such that Z a f (x) dx ≥ K + 1. ²

However, since f is continuous Z a ³ ar ´ X f → f (x) dx N ² ²N ≤r≤N

so we can find an N0 such that Z a ³ ar ´ X → f (x) dx − 1 f N ² ²N ≤r≤N

and so

N ³ ar ´ X ≥K f N r=1

for all N ≥ N0 . Thus

N ³ ar ´ X f →∞ N r=1

as N → ∞.

(ii) Consider a = 1, ( 2n (1 − 24n |x − 2−n |) if |x − 2−n | ≤ 2−4n , f (x) 0 otherwise. (iii) Observe that (1 − z) so, dividing by z − 1,

N −1 X r=0

N −1 X r=0

n

n

z =1−z

z =1−z

N

N

=

=

N −1 Y r=0

N −1 Y r=1

(z − ω r )

(z − ω r )

182

for z 6= 1. Since a polynomial of degree N − 1 which vanishes at N points vanishes identically we have N −1 X r=0

for all z.

n

z =1−z

N

=

N −1 Y r=1

(z − ω r )

In particular, setting z = 1, we have N=

N −1 Y r=1

r

(1 − ω ) = τ

1+2+···+(N −1)

r=1

where τ = π/N . Thus N =τ

N (N −1)/2

(−2i)

N −1

N −1 Y

N −1 Y r=1

(τ −r − τ r )

N −1 Y rπ rπ N −1 sin =2 sin . N N r=1

(iv) Observe that ÃN −1 ! N ³ ³ πr ´´ ³ πr ´ X π π X log sin sin log = 2N r=1 N 2N N r=1

N π log N −1 2N 2 log N π N −1 log 2 + π → − log 2. = −π 2N 2N 2 =

183

K166 Observe that Ek = {x : f (x) = k}

is the union of a finite set of disjoint intervals (of the form [a, b)) of total length (9/10)k−1 /10. Thus (writing IA for the indicator function of A) Ã ! X X kIEk + X 1 − I Ek fX = 1≤k≤X

1≤k≤X

is Riemann integrable with ! Ã Z 1 X X (9/10)k−1 . k(9/10)k−1 + X 1 − 9−1 fX (x) dx = 10−1 0

1≤k≤X

1≤k≤X

Now writing [X] for the integer part of X, Ã ! X (9/10)k−1 = X(9/10)[X] ≤ ([X] + 1)(9/10)[X] → 0 0 ≤ X 1 − 10−1 1≤k≤X

and using the Taylor expansion (1 − t)

−2

=

∞ X

(j + 1)tj

j=0

(Newton’s binomial expansion) we have Z 1 fX (x) dx → 10−1 (1 − 9/10)−2 = 10. 0

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

gX (x) − fX (x) = 0

for all x ∈ k=1 Ek so, given any ² > 0, we know that gX (x)−fX (x) = 0 on the union of a finite set of disjoint intervals of total length 1 − ²/2 and so IfX − X² = I∗ fX − X² ≤ I∗ gX ≤ I ∗ gX ≤≤ I ∗ fX + X² = IfX + X².

Since ² 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 Z Z b fR,S (x) dx ≥ ab−η f (x) dx a

for all R, S ≥ R0 (η), and if R and S are fixed, we can find an η0 (R, S) > 0 such that Z Z b b−η fR,S (x) dx a f (x) dx ≥ a

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 both f+ and f− . There is this no loss of generality in assuming that f is positive. 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], the apparent singularity at x = 0 must be a trick of the notation.) Z X −ax Z XZ b e − e−bx e−tx dt dx dx = x 0 a 0 Z bZ X = e−tx dx dt a 0 Z b 1 − e−tX = dt. t a Now e−tX e−aX 0≤ ≤ t t for t ∈ [a, b] so µ ¶ Z b −tX e b −aX →0 0≤ dt ≤ e log t a a as X → 0 so Z b Z b −tX Z X −ax 1 e e − e−bx b dx = dt − dt → log x t a a t a 0 as x → ∞.

186

K169 By the fundamental theorem of the calculus ¶ Z t µZ d Z d d f (u, v) dv du = f (t, v) dv. dt a c c By differentiating under the integral sign and using the fundamental theorem of the calculus ¶ µZ t ¶ Z d Z d µZ t Z d d ∂ f (u, v) du dv = f (u, v) du dv = f (t, v) dv. dt c a a c c ∂t Thus µZ t µZ d ¶ ¶ ¶ Z d µZ t d f (u, v) dv du − f (u, v) du dv = 0, dt a c c a and by the constant value theorem ¶ Z t µZ d Z d µZ t f (u, v) dv) du − f (u, v) du dv a c c a ¶ ¶ Z a µZ d Z d µZ a = f (u, v) dv du − f (u, v) du dv = 0. a

c

c

a

Z It is weaker because we say nothing about the existence or value of f (x) dA. R

187

K170 (i) We have Z

for all x and Z

so Z

1 0

Z

1

f (x, y) dx = 0 0

1

f (x, y) dy = 0

(

2 if 2−1 < y < 1, 0 otherwise,

1

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

0

Z

1 0

Z

1

f (x, y) dy dx. 0

(ii) We have

for all x and Z

Z 1

g(x, y) dy = 0

1

g(x, y) dx = 0 0

(

2u(2y − 1) if 2−1 < y < 1, 0 otherwise,

so

Z

1 0

Z

1

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

0

Z

1 0

Z

1

g(x, y) dy dx. 0

Theorem 9.20 demands continuity everywhere. (The underlying problem is that g is not bounded.) (iii) We have Z

1 0

1+y 2

y(w − 2y 2 ) dw w3 y2 ¸1+y2 · 1 y y3 = − + 2 2 w w y2 −y = 2(1 + y 2 )2

1 h(x, y) dx = 2

Z

so that Z

1 0

Z

1 0

·

1 h(x, y) dx dy = 4(1 + y 2 )

Observe that h(y, x) = −h(x, y).

¸1 0

= −1/8.

188

K171 (ii) Observe that, if R ≥ 1, then ¯ Z ¯Z ∞ Z R ¯ ¯ ¯ f (x, y) dx¯¯ ≤ f (x, y) dx − F ¯

A

(1 + + y2) − tan−1 R =A 1 + y2 ≤ A( π2 − tan−1 R)

0

0

∞ R π 2

x2 )(1

dx

Thus given ² > 0 we can find an R such that ¯Z ∞ ¯ Z R ¯ ¯ ¯ ¯ ≤ ²/4 f (x, y) dx − f (x, y) dx ¯ ¯ 0 0 RR for all y. Since 0 f (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 ¯ ¯Z R Z R ¯ ¯ ¯ ≤ ²/2 ¯ f (x, y ) dx f (x, y) dx − 0 ¯ ¯ 0

0

and so

¯ ¯Z ∞ Z ∞ ¯ ¯ ¯ f (x, y0 ) dx¯¯ < ². f (x, y) dx − ¯ 0 0 R∞ If follows that 0 f (x, y) dx is a continuous function of y. (This can be done better using uniform convergence.) R∞ Since 0 f (x, y) dx is continuous in y and ¯Z ∞ ¯ ¯ ¯ ¯ ¯ ≤ Aπ/2 , f (x, y) dx ¯ ¯ 1 + y2 0 R∞R∞ comparison shows that 0 0 f (x, y) dx dy exists. A similar but simR∞RR pler argument shows that 0 0 f (x, y) dx exists Equation F shows that ¯Z ∞ Z ∞ ¯ Z ∞Z R ¯ ¯ ¯ ¯ ≤ ( π − tan−1 R) πA . f (x, y) dx dy − f (x, y) dx dy 2 ¯ ¯ 2 0 0 0 0 We also know that ¯Z ∞ Z R Z ¯ ¯ f (x, y) dx dy − ¯ 0

so that ¯Z ∞ Z ¯ ¯ ¯ 0

0

∞

0

f (x, y) dx dy −

R

0

Z

Similarly ¯Z ∞ Z ∞ Z ¯ ¯ f (x, y) dy dx − ¯ 0

0

R 0

R 0

Z

R

0

Z Z

R 0

R 0

¯ ¯ πA f (x, y) dx dy ¯¯ ≤ ( π2 − tan−1 R) 2

¯ ¯ f (x, y) dx dy ¯¯ ≤ ( π2 − tan−1 R)πA.

¯ ¯ f (x, y) dy dx¯¯ ≤ ( π2 − tan−1 R)πA,

189

so, using Theorem 9.20 (Fubini on products of intervals), ¯Z ∞ Z ∞ ¯ Z ∞Z ∞ ¯ ¯ ¯ ¯ ≤ 2( π − tan−1 R)πA. f (x, y) dy dx − f (x, y) dx dy 2 ¯ ¯ 0

0

0

0

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 Riemann length so [−3, 3] \ U is closed, bounded and does not have Riemann length.

193

K175* No comments.

194

K176 (i) Observe that f (tj ) ≤ f (b). Recall that an increasing sequence bounded 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 strictly decreasing 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˜ is right 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 and only 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( sup f (x) − inf f (x)) t→0, t<0 x∈(t,0]

x∈(t,0]

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 , 0} we 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 P∞ set of points xj and and that, writing λj = u(xj ) − supt<xj u(t), j=1 λj converges (indeed ∞ X j=1

λj ≤ lim f (t). t→∞

Without loss of generality we may suppose λ1 ≥ λ2 ≥ . . . . Define un (t) so that n X H(x − xj ). u(t) = un (t) + j=1

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 observe that 0 ≤ inf un (t) − un (s) ≤ λn s<x
and 0 ≤ un+1 (t) − un+1 (s) ≤ un (t) − un (s)

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

inf U (t) − U (s) = 0

s<x
and U is continuous. A very much simpler argument shows that for all x.

Pn

j=1

H(x−xj ) converges

(ii) Suppose that f and g are bounded increasing left continuous functions with 0 ≤ f (t) − g(t) ≤ ² for all t and g is Riemann-Stieljes integrable with respect to G. Then I ∗ (f, G) − I∗ (f, G) ≤ ²( lim G(t) − lim G(t)). t→∞

t→−∞

By (i), we can write f (t) = v(t) +

∞ X j=1

˜ − xj ) λj H(t

P ˜ ˜ where H(t) = 0 for t ≤ 0, H(t) = 1 for t > 0, λj > 0 and ∞ j=1 λj converges. By the first paragraph we need only show that the function fN given by fN (t) = v(t) +

N X j=1

˜ − xj ) λj H(t

197

is Riemann-Stieljes integrable. We know that v is Riemann-Stieljes ˜ − xj ) is a Riemannintegrable so we need only show that t 7→ H(t Stieljes integrable function and this follows much the same pattern as in K177. (iii) f will be Riemann-Stieljes integrable if and only if no discontinuity 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 (G(t) − G(s)) inf (f (t) − f (s)) > 0 s<x0
s<x0
so f is not Riemann-Stieljes integrable with respect to G. If f and G have no points of discontinuity in common, the argument of (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 theorem fails. Suppose G is strictly increasing. If f is bounded continuous positive function with f (x0 ) 6= 0, observe that there exists an ² > 0 such that |f (t) − f (x0 )| ≤ f (x0 )/2 so f (t) ≥ f (x0 /2 for |x0 − t| ≤ 2². Now Z (G(x0 + ²) − G(x0 − ²))f (x0 ) f (x) dG(x) ≥ > 0. 2 R (iii) Same condition as (ii). Suppose G is strictly increasing. If f (x0 ) 6= 0 there exists an ² > 0 such that |f (t)−f (x0 )| ≤ |f (x0 )|/2 for |x0 −t| ≤ ². Choose g continuous so that g vanishes outside (x0 − ², x0 + ²), g(t)f (x0 ) ≥ 0 for all t and g(X0 )f (x0 ) = 1. By part (ii), g(t)f (t) = 0 for all t which is impossible. Thus f is identically zero.

199

K180 (i) Observe that n n X X |g(xj ) − g(xj−1 )| = g(xj ) − g(xj−1 ) = g(b) − g(a). j=1

j=1

(ii) Observe that n n X X |f (xj ) − f (xj−1 )| |(f + g)(xj ) − (f + g)(xj−1 )| ≤ j=1

j=1

=

n X j=1

|g(xj ) − g(xj−1 )|.

(iii) Observe that |f (x)| ≤ |f (a)| + |f (b) − f (x)| + |f (x) − f (a)|. ¡ (iv) Observe that, setting x0 = 0, xN +2 = 1 and xj = (N + 4 − j + ¢−1 1 )π , we have 2 N X j=1

as N → ∞.

|f (xj ) − f (xj−1 )| ≥ π

−1

N +5 X j=5

j −1 → ∞

200

K181 (ii) Given ² > 0, we can find a = x0 ≤ y0 ≤ x1 ≤ . . . xn ≤ yn = t such that n X F (yj ) − F (xj ) ≥ F+ (t) − ² j=1

and a =

x00

≤

y00

0 = t such that ≤ x01 ≤ . . . xm ≤ ym n X j=1

F (x0j ) − F (yj0 ) ≥ F− (t) − ².

Let a = z0 ≤ z1 ≤ . . . zN = t with

{z0 , z1 , z2 . . . zN } ={x0 , x1 , x2 . . . xn } ∪ {y0 , y1 , y2 . . . yn }

0 } ∪ {x00 , x01 , x02 . . . x0m } ∪ {y00 , y10 , y20 . . . ym

Then F (t) − F (a) =

N X j=1

F (zj ) − F (zj−1 ) X

=

F (zj )−F (zj−1 )>≥0

F+ (t) ≥

X

F (zj ) − F (zj−1 ) +

F (zj )−F (zj−1 )>≥0

X

F (zj )−F (zj−1 )<0

F (zj ) − F (zj−1 ),

F (zj ) − F (zj−1 ) ≥ F+ (t) − ²

and F− (t) ≥ −

X

F (zj )−F (zj−1 )>≤0

F (zj ) − F (zj−1 ) ≥ F− (t) − ²

so |F (t)−F+ (t)+F+ (t)| ≤ 2². Since ² 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 X

G+ (yj ) − G+ (xj ) +

j=1

≥ = ≥

n X

j=1 n X

j=1

G+ (xj ) − G+ (yj−1 )

G+ (yj ) − G+ (xj )

(f (yj ) − f (xj )) + (G− (yj ) − G− (xj ))

j=1 n X j=1

n X

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 is impossible by (i)). (iii) Write G+ (t) =

Z

t 0

a

max(F (x), 0) dx, G− (t) = −

Z

t

min(F 0 (x), 0) dx. a

Then, by the fundamental theorem of the calculus, G+ is differentiable with G0+ (t) = max(F 0 (t), 0) so G+ is increasing. Similarly G− is decreasing. Further G0+ (t) − G0− (t) = F 0 (t), so using the fundamental theorem of the calculus, F = F (a) + G+ − G− . Now let ² > 0. Since F 0 is continuous on [a, t], it is uniformly continuous so we can find an integer N such that |F 0 (x) − F 0 (y)| < ² 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 0 (x)| > ². Let us call the other integers r with 1 ≤ r ≤ n ‘bad’. Observe that, if r is good, then F 0 is single signed on x ∈ [(r − 1)/N, rN ]. Thus, using the mean value theorem N X X Z xj |F (xj ) − F (xj−1 )| ≥ |F 0 (x)| dx j=1

j good

so that VF (t) ≥

Z

≥

Z

≥

Z

xj−1

t

|F 0 (x)| dx −

a t a

X

j bad

²(t − a)/N

|F 0 (x)| dx − ²(t − a)

t a

|F 0 (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 is about 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+2β ²(|x|)

with ²(|x|) → 0 as x → 0.

If α + β > 1, fα,β is differentiable at 0 with derivative 0. We check 0 that fα,β is continuous at 0 so fα,β is everywhere differentiable with continuous derivative (so of bounded variation). If 1 ≥ α + β > 0, fα,β is not differentiable at 0 but is continuous 0 there (so everywhere continuous). By inspection of fα,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 of bounded variation. (B) If β = 0. fα,0 = |x|α sin 1. The function is differentiable everywhere with continuous derivative if and only if α > 1, and continuous everywhere if and only if α > 0, of bounded variation if and only if α ≥ 0. (C) If β < 0 then the function differentiable at 0 so everywhere if and only if α > 1. Derivative is then 0. Now 0 fα,β (x) = αxα−1 sin xβ + βxα+β−1 cos xβ

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 if α > 0. If α ≤ 0, inspection shows that fα,β is not of bounded variation. If α > 0 then by differentiating (or rescaling) we see that in each interval In = [((n + 1)π)1/β , (nπ)1/β ] the function fα,β either increases and then decreases or decreases and then increases. The total variation Vn of fα,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 P α/β so fα,β is of bounded variation if and only if ∞ converges, i.e. n=1 n 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)

so

I ∗ (f, F + G) = I ∗ (f, F ) + I ∗ (f, G) and similarly I∗ (f, F + G) = I∗ (f, F ) + I∗ (f, G). (ii) Rewrite as Z Z Z Z f (x) dF1 (x) + f (x) dG2 (x) = f (x) dF2 (x) + f (x) dG1 (x). R

R

R

R

and observe that by part (i) Z Z Z f (x) dF1 (x) + f (x) dG2 (x) = f (x) d(F1 + G2 )(x) R

R

R

and

Z

R

f (x) dF2 (x) +

Z

R

f (x) dG1 (x) =

Z

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 Taylor theorems, (1 − (1 − s)2 )−1/2 = (2s − s)−1/2

= 2−1/2 s−1/2 (1 − s/2)−1/2

2−1/2 s−1/2 (1 − ²(s)s) R1 with ²(s) → 0 as ² → 0. Thus y (1−t12 )1/2 dt converges and Z 1−y Z 1 1 1 dt = − ds 2 1/2 (1 − (1 − s)2 )1/2 y (1 − t ) 0

= 21/2 (1 − y)1/2 + (1 − y)3/2 δ(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 that the extended function sin is differentiable at ω with sin0 (ω) = 0. If ω>u>0 sin0 (ω − u) = −(1 − (sin ω − u)2 ) and

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

sin0 (ω + u) = − sin0 (ω + u) = −u − β(u)u

with β(u) → 0 as u → 0+. Thus sin is twice differentiable at ω with sin00 (ω) = −1 = − sin ω.

208

K188 The points at issue are very similar to those of K187.

209

K189 (iv) Observe that Z R Z 2 π f (x) dx = π 1

R

x−2 dx = π(1 − R−1 ) → π

1

but

2π

Z

R 0

2 1/2

f (x)(1 + f (x) ) 1

dx ≥ 2π

as R → ∞.

Z

R 1

x−1 dx = 2π log R → ∞

(v) We have Vol K = π but 0

Vol(K \ K) = π

Z

∞

Z0 ∞

Z

∞ 0

x−1 dx = ∞

(x−1/2 + x−1 )2 − (x−1/2 )2 dx

(x−2 + 2x−3/2 ) dx 0 ¡ ¢ = π lim (1 − R−1 ) + 4(1 − R−1/2 ) = 5π. =π

R→∞

210

K190 We have Z

f (x) ds =

γ1

= = =

Z

Z

f (x) ds γ2

f (x) ds γ3

Z

f (x) ds

γ4 Z 1

cos(2πt)(−2π sin(2πt), 2π cos(2πt)) dt Z 1 (− cos(2πt) sin(2πt), cos2 (2πt)) dt = 2π 0 Z 1 (sin(4πt), 1 + cos(4πt)) dt =π 0

0

= (0, π)

Also

and

Z

γ5

Z

γ6

f (x) ds = −

Z

f (x) ds = 2

Z

γ1

f (x) ds = (0, −π) f (x) ds = (0, 2π).

γ1

211

K191 (ii) We have p(n − k) un (k + 1) = un k (1 − p)k

so un (k + 1) > un k if p(n − k) > (1 − p)k, i.e. if np > k and un (k + 1) < un k if np < k (and, if np = k, then un (k + 1) = un k). We take k0 so that np ≥ k0 − 1 and np ≤ k0 . (ii) If k ≥ n(p + ²/2) then

p(n − k) p(1 − p − ²/2) un (k + 1) = ≤ un k (1 − p)k (1 − p)(p + ²/2)

Pn Now 1 = j=0 un (j) ≥ un (k) so, writing k1 = k1 (n) for the least integer greater than n(p + ²/2), we see that ¶k−k1 µ p(1 − p − ²/2) un (k) ≤ (1 − p)(p + ²/2)

for all k ≥ k1

If k2 = k2 (n) is the greatest integer less than n(p + ²/2) we have X X µ p(1 − p − ²/2) ¶k−k1 un (k) ≤ (1 − p)(p + ²/2) k≥np+n² k≥k2 ¶k2 (n)−k1 (n) µ ¶−1 µ p(1 − p − ²/2) p(1 − p − ²/2) 1− ≤ →0 (1 − p)(p + ²/2) (1 − p)(p + ²/2) as n → ∞. Similarly X

k≤np−n²

so

un (k) → 0

Pr{|Nn − np| ≥ ²n} = as n → ∞.

X

|k−np|≥n²

un (k) → 0

(iii) We have EXj = p1 + (1 − p)0 = p and EXj2 = p1 + (1 − p)0 = p so var Xj = p − p2 = p(1 − p).

212

The sum of the expectations is the expectation of the sum so n n X X EXj . Xj = ENn = E j=1

j=1

For independent random variables, the sum of variances is the variance of the sum so and Thchebychev gives

var Nn = np(1 − p)

Pr{|Nn − np| ≥ ²n} ≤ as n → ∞.

var Nn p(1 − p) = →0 2 (²n) ²2 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)0 to obtain (B). (C) now follows from (B) and (C)0 . (ii) Setting d0 (x, y) = −d(x, y), we see that d0 is a metric space. If d(x, y) ≥ 0 for x, y ∈ X then X has exactly one point.

214

K193 By (A)0 , 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 ∼ x0 , y ∼ y 0 gives

d(x, y) = d(x0 , x) + d(x, y) + d(y, y 0 ) ≥ d(x0 , y 0 ) and d(x0 , y 0 ) ≥ d(x, y) so d(x, y) = d(x0 , y 0 ). Thus d˜ is well defined.

215

K194 (i) The result is true with A1 = 1. We use induction on n to prove it. 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 0 = σ −1 τ we see that τ 0 fixes m + 1 so is generated by at most m elements of the form (ij) (transpositions). Thus τ = στ 0 is is generated by at most m + 1 transpositions. 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 + 1 from 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 ar b = 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 little relationship between size and diameter.

216

K195 (i) We wish to prove the statement P (N ). If |x| ≥ 4−k for some integer k. and x=

N X j=1

xj with |xj | ≤ 4−n(j) for some integer n(j) ≥ 0 [1 ≤ j ≤ N ],

PN

2−n(j) ≥ 2−k . P If x = 1j=1 xj then x = x1 so P (1) is true.

then

j=1

Suppose now that P (r) is true for all 1 ≤ r ≤ N − 1. If |x| ≥ 4−k and N X

x=

j=1

xj with |xj | ≤ 4−n(j)

then if n(j) ≤ k for any j we are done. Thus we need only consider the case n(j) ≥ k + 1 for all 1 ≤ j ≤ N . Let M be the smallest m such that m X |xj | ≥ 4−k−1 . j=1

We know that M X j=1

so

|xj | ≤

N X

M +1

M −1 X j=1

|xj | + 4−k−1 ≤ 2 × 4−k−1

|xj | ≥ |x| −

M X j=1

|xj | ≥ 2 × 4−k−1

. Applying the inductive hypothesis to 0

x =

M X j=1

00

|xj | and x =

N X

M +1

|xj |

we see that N X j=1

2−n(j) =

M X j=1

2−n(j) +

N X

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= 0 then |x| ≥ 4−k for some k and so, by (i), d(x, 0) ≥ 2−k .

217

P (iii) If 4−k+1 > |x| ≥ 4−k then taking x = 1j=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 . Thus 4|x|2 ≥ d(x, 0) ≥ 4−1 |x|2 . (iv) I think the graph is quite complex with f constant on intervals of the form [4−k (1 − η), 4−k ] (for some η > 0) and on other intervals which 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 sophisticated. Here are longer solutions. (ii) True. If x ∈ X1 ∪ X2 then, without loss of generality, we may suppose x ∈ X1 . Thus there exists a δ1 > 0 such that, if z ∈ X and d(x, z) < δ1 , we have z ∈ X1 . Since f |X1 is continuous, we know that, given ² > 0 there exists a δ2 (²) > 0 such that d(x, z) < δ2 (²) implies ρ(f (x), f (z)) < ² for all z ∈ X1 . Now d(x, z) < min(δ1 , δ2 (²)) implies ρ(f (x), f (z)) < ² for all z ∈ X. (iii) True. If x ∈ X \ Xj , the argument of (ii) shows that f is continuous at x. If x ∈ X1 ∩ X2 , then given ² > 0 there exists a ηj (²) > 0 such that d(x, z) < ηj (²) implies ρ(f (x), f (z)) < ² for all z ∈ Xj and d(x, z) < minj=1,2 ηj (²) implies ρ(f (x), f (z)) < ² for all z ∈ X.

Set f (x) = g(x) when kg(x)k < 1, f (x) = kg(x)k−1 g(x) otherwise. Let X1 = {x ∈ X : kg(x)k ≤ 1} and X2 = {x ∈ X : kg(x)k ≥ 1}

and 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 with f (c) = 1/2. (iv) If A is closed, then {t ∈ [0, 1) : (1 − t)a + tb ∈ A} is closed. If A 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 find ¯ Thus A¯ ∈ F. xm ∈ A with d(xm , ym ) < 1/m so xm → y and y ∈ A. ¯ 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 closed sets, A¯ ∈ F and automatically F ⊇ A¯ whenever F ∈ F so A¯ satisfies (iii). Suppose A¯ satisfies (iii). If xn ∈ A and xn → x then xn ∈ A¯ so since ¯ ¯ Suppose x ∈ A, ¯ then (writing B(x, 1/n) for the A is closed x ∈ A. open ball centre x radius 1/n) if B(x, 1/n) ∩ A = ∅ A¯ \ B(x, 1/n) is a strictly 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 have xn ∈ A and xn → x. (iii0 B ◦ is an open set with B ⊇ B ◦ such that, if U is open and B ⊇ U , we have B ◦ ⊇ U . Int Q = ∅, Int Z = ∅, Int{1/n : n ≥ 1} = ∅, Int[a, b] = Int(a, b) = Int[a, b) = (a, b).

Cl Q = R, Cl Z = Z, Cl{1/n : n ≥ 1} = {0} ∪ {1/n : n ≥ 1}, Cl[a, b] = Cl(a, b) = Cl[a, b) = [a, b]. ¯ ⊆ f (A) for every A. If xn → x but f (xn ) 9 f (x), Suppose f (A) 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 ≥ 1}

to obtain a contradiction (x ∈ A¯ but f (x) ∈ / f (A)). Thus f is continuous. ¯ then we can find xn ∈ A Conversely, if f is continuous and x ∈ A, ¯ ⊆ such 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 Cl U ⊇ Int(Cl U )) and Cl U is closed so that Cl U ⊇ Cl(Int(Cl U )).

However, since U is open, and Cl U ⊇ U , we have Int(Cl U )) ⊇ U so Cl(Int(Cl U )) ⊇ Cl U

and Cl(Int(Cl U )) = Cl U .

It follows that, If B is any set, Cl(Int(Cl(Int B))) = Cl(Int B) and by taking complements, if C is any set Int(Cl(Int(Cl C))) = Int(Cl C) Since Cl Cl A = Cl B and Int Int A = Int B, the only possible distinct sets that we can produce are A, Int A, Cl A, Cl Int A, Int Cl A, Int Cl Int A and Cl Int Cl A. 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, kxk))−1 , for example. (iii) Since Γ is absorbing we can define n(x) = inf{t ≥ 0 : t−1 x ∈ Γ}.

We note that n(x) ≥ 0. Our object is to show that kxk = n(x) defines the required norm. If λ ≥ 0 the definition gives n(λx) = λn(x). The fact that Γ is symmetric now gives for all λ ∈ R.

n(λx) = |λ|n(x)

Condition (B) tells us that 0 ∈ Γ, and then (D)0 together with convexity tells us that, if x 6= 0, there exists a µ > 0 such that tx ∈ /Γ for all t ≥ µ. Thus n(x) = 0 if and only if x = 0. Suppose that x, y 6= 0. Then, if ² > 0, ¡ ¢ (1 + ²)−1 n(x)−1 x, (1 + ²)−1 n(y)−1 y ∈ Γ

and so, by convexity, −1 −1 1 −1 −1 n(x) n(y) (x + y) (x + y) = (1 + ²) (1 + ²) n(x) + n(y) n(x)−1 + n(y)−1 n(y)−1 n(x)−1 −1 −1 = (1 + ²) n(x) x + (1 + ²)−1 n(y)−1 y ∈ Γ. n(x)−1 + n(y)−1 n(x)−1 + n(y)−1 Since ² 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 the norm k kC .

223

K201 Observe that but that

xn = (1, 2−1 , 3−1 , . . . , n−1 , 0, 0, . . . ) ∈ E

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 | ≤ ka − a(j)k → 0

as j → 0 so a2m = 0 for all m. Thus a ∈ F .

Suppose that (V, k k) is a normed space and E is a finite dimensional subspace. Let e1 , e2 , . . . , eN be a basis for E. Then ° N ° N °X ° X ° ° xj ej ° = |xj | ° ° ° j=1

∗

j=1

is a norm on E. But all norms on a finite dimensional space are Lipschitz equivalent so there exists K ≥ 1 with Kkxk ≥ kxk∗ ≥ K −1 kxk. If yn ∈ E, y ∈ V and kyn − yk → 0, then yn is a Cauchy sequence in (V, k k) so a Cauchy sequence in (E, k k) so, by Lipschitz equivalence, a Cauchy sequence in (E, k k∗ ) so, since (E, k k∗ ) is complete, converges in (E, k k∗ ) to some z ∈ E. Since kyn − zk∗ → 0, it follows, by Lipschitz equivalence, that kyn − zk → 0. By the uniqueness of limits, y = z ∈ E. Thus E is closed.

224

K202 (i) If λ0 y + e0 = λy + e with λ, λ0 ∈ R and e, e0 ∈ E then

(λ0 − λ)y = e − e0

so, applying T to both sides,

(λ0 − λ)T y = 0

so λ = λ0 and e = e0 . Also

x−

Tx y ∈ E. Ty

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 ky − f k < δ

and (setting e = −λf ) this is equivalent to saying that for some e ∈ N and λ 6= 0.

kλy + ek < δλ

(iii) If T is continuous then since the inverse image of a closed set under a continuous function is closed N is closed. If N is closed pick a y ∈ / N and observe that there exists a δ > 0 such that kλy + ek ≥ δ|λ

for all λ and e ∈ N .

By part (i), if x ∈ E we have

x = λy + e

with e ∈ N , so, by (ii), Thus

kxk ≥ δ|λ| whilst T x = λT y. kT xk ≤ δ −1 |T x|kxk

for all x and T is continuous.

(iv) We have N = {a ∈ s00 :

P∞

j=1

aj = 0}.

225

(a) If we use the norm k k∞ , then N not closed. Take   if j = 1 1 −1 ej (n) = −n if 2 ≤ j ≤ n + 1  0 otherwise.

Then e(n) ∈ N , but e → (1, 0, 0, . . . ) ∈ / N . Thus N is not closed. P −1 (b) If we use the norm k kw , N not closed. Observe that ∞ j=2 j diverges so (since j −1 ≤ 1) we can find N (n) such that N (n)

n≤ Set A(n) =

PN (n)

X j=2

j −1 ≤ n + 1.

j −1 . Take   1 ej (n) = −A(n)−1 j −1  0

j=2

if j = 1 if 2 ≤ j ≤ N (n) otherwise.

Then e(n) ∈ N but e → (1, 0, 0, . . . ) ∈ / N . Thus N is not closed. (c) If we use the norm k k1 , N is closed. If e(n) ∈ N and ke(n) − ek1 → 0 then (remember all sums over a finite number of non-zero terms) ¯ ¯∞ ¯ ¯∞ ∞ ¯ ¯X ¯ ¯X X ¯ ¯ ¯ ¯ ej (n)¯ = ke(n) − ek1 → 0 ej − ej ¯ = ¯ ¯ ¯ ¯ ¯ ¯ j=1 j=1 j=1 P∞ so j=1 ej = 0.

(d) If we use the norm k2 , N not closed. Consider the same example as (a). (e) If we use the norm k ku then N is closed. We can use much the same proof as (c). [To see T not continuous in (a), (b) and (d) we can consider the norms of e − e(n) and T (e − e(n)). In (c) and (e) simple inequalities give kT k ≤ 1, indeed kT k = 1.]

226

K203 The key observation is that, since 21/2 is irrational, x + y21/2 = x + y 0 21/2 for x, y, x0 , y 0 ∈ Q if and only if x = x0 and y = y 0 . 0

If pn , qn are positive¡integers with pn /qn → 21/2 (in the usual Eu¢ clidean metric) then N (−pn , qn ) → 0 but k(−pn , qn )k → ∞. On the other hand ¡ ¢ N (x, y) = |x + y21/2 | ≤ |x| + 2|y| ≤ 2k(x, y)k

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) → 0 as n → ∞. Since x ∈ Fm for each m so T∞ xn ∈ Fm for all n ≥ m and Fm isTclosed, ∞ x ∈ 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 T 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 for m 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 strictly increasing. 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 may suppose that f is not left continuous and there exist x1 < x2 < . . . with xj < x and xj → x together with a δ > 0 such that f (x) − δ ≥ f (xj ) for all j. Since N X j=1

we have

P∞

ρ(xj+1 , xj ) = f (xN +1 ) − f (x1 ) < f (x) − f (x1 ),

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 > 0 for all j so ρ(t, xj ) 9 0. Thus the xj form a non-convergent Cauchy sequence for ρ. If f (t) 9 ∞ then a similar argument shows that, if xj = j, the xj form a non-convergent Cauchy sequence for ρ. A similar argument covers 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+1 g(x) ≥ 2n=1 g(2−n−1 ) =

n+1 2X

j=1

g(2−n−1 ) ≥ g

Ã2n+1 X j=1

2−n−1

!

= g(1)

so g(x) ≥ 2−n−1 g(1) ≥ x2−1 g(1). Take K = 2−1 g(1).

If we take g(x) = x1/2 we see that no relation d(x, y) ≤ L|x − y| will hold in this case. (v) Since g is increasing and g(t) ≥ 0, g(t) → l as t → 0+ for some l ≥ 0. If l > 0 then d(x, y) > l for all x 6= y so any Cauchy sequence for d is eventually constant and so converges.

230

If l = 0 then, since g is decreasing, given any δ > 0 we can find and ² > 0 such that ² > g(s) ≥ 0 implies |s| < δ. Thus if xn is Cauchy for d, it follows that xn is Cauchy for the Euclidean metric so there exists an 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 the discrete 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 and so

kaj ej k ≤ kak |aj | ≤ kakkej k−1 .

232

K208 We have N X j=1

Ã

N X

Ã

∞ X

(aj + bj )2 ≤ 

≤

a2j

j=1

a2j

j=1

!1/2 !1/2

+

+

Ã

N X

Ã

∞ X

a2j

j=1

j=1

!1/2 2 

!1/2 2  , a2j

so, sequence bounded above converges we have P∞since an increasing 2 j=1 (aj + bj ) convergent. Moreover ∞ X j=1

Ã

(aj + bj )2 ≤ 

∞ X

a2j

j=1

!1/2

+

Ã

∞ X j=1

a2j

!1/2 2

 .

(ii) To show completeness, suppose a(n) a Cauchy sequence in l 2 . We have |aj (n) − aj (m)| ≤ ka(n) − a(m)k2 so aj (n) is Cauchy with respect to the usual Euclidean distance and there exists an aj ∈ R such that |aj (n) − aj | → 0. Now observe that, ÃM !1/2 Ã M !1/2 Ã M !1/2 X X X a2j (n) ≤ (aj − aj (n))2 + aj (n)2 j=1

j=1

≤

Ã

M X

≤

Ã

M X

j=1

j=1

j=1

(aj − aj (n))2

!1/2

+ ka(n)k2

(aj − aj (n))2

!1/2

+ sup ka(r)k2 r≥1

→ sup ka(r)k2 r≥1

as n → ∞. (Recall that a Cauchy sequence is bounded.) Thus Ã for all M and so a ∈ l2 .

M X j=1

a2j

!1/2

≤ sup ka(r)k2 r≥1

233

Ã

Finally, if n, m ≥ N !1/2 !1/2 Ã M !1/2 Ã M M X X X (a2j (n) − aj (m))2 (a2j − aj (m))2 + (a2j − aj (n))2 ≤ j=1

j=1

j=1

≤

Ã

M X

≤

Ã

M X

j=1

j=1

(a2j − aj (m))2

!1/2

+ ka(n) − a(m)k2

(a2j − aj (m))2

!1/2

+ sup ka(s) − a(r)k2 r,,s≥N

so, allowing m → ∞, ÃM !1/2 X (a2j − aj (n))2 ≤ sup ka(s) − a(r)k2 r,s≥N

j=1

and, allowing M → ∞,

ka − a(n)k2 ≤ sup ka(s) − a(r)k2 r,s≥N

for all n ≥ N . Thus as n → ∞.

ka − a(n)k2 → 0

(iii) We have (by Cauchy-Schwarz in N dimensions) !1/2 !1/2 Ã N Ã N N X X X ≤ kak2 kbk a2j |aj bj | ≤ a2j j=1

j=1

j=1

so, P∞since an increasing sequence bounded above converges, we have j=1 aj bj absolutely convergent and so convergent.

234

K209 (ii) we have − log

µ

x y + p q

¶

1 1 ≤ − log x − log y p q

for all x y > 0. Setting x = X p , y = Y Q , multiplying by −1 and taking exponentials gives the inequality. (iii) By (ii), |f (t)|p |g(t)|q + . |f (t)g(t)| ≤ p q Now integrate.

235

K210 (i) Observe that (p − 1)q = p, so, taking g(t) = |f (t)|p−1 , we have ¶1/q µZ b ¶1/q µZ b Z b p (p−1)q p , |f (t)| dt =A |f (t)| |f (t)| dt ≤ A a

a

a

whence the result. (ii) We have Z

b a

|(f (t) + h(t))g(t)| dt ≤ ÃµZ ¶1/p b

≤

p

a

|f (t)| dt

Z

b a

+

|f (t)g(t)| dt + µZ

b p

a

|h(t)| dt

Z

b

|h(t)g(t)| dt

a ¶1/p ! µZ b a

q

|g(t)| dt

¶1/q

.

Now use (i). (iii) To see that (C([a, b]), k kp ) is not complete follow the proof that (C([a, b]), k k1 ) is not complete. (iv) We work in C([a, b]). kf gk1 ≤ kf kp kgkq . If kf gk1 ≤ Akgkq for all g then kf kp ≤ A.

236

K211 It may be helpful to look at K209. (iv) If k k∗ is derived from an inner product, then Taking

ka + bk2∗ + ka − bk2∗ = 2(kak2 + (kak2 ). a = (1, 0, 0, 0 . . . ), b = (0, 1, 0, 0, . . . ),

we see that this fails for k kp unless p = 2. In the case C([0, 1], k kp ) we look at (for example) Observe that

f (x) = g(1 − x) = max(1 − 4x, 0). Z

1 p

f (x) dx = 0

Z

1/4(4t)p dt. 0

(v) We consider C([a, b]). It is easy to check that kf gk1 ≤ kf k1 kgk∞

and (by setting g = 1) that if

kf gk1 ≤ Akgk∞

for all continuous g then kf k1 ≤ 1. If f ∈ C([a, b]) and there exists an x ∈ [a, b] with |f (x)| ≥ A + ². Without loss of generality, we may suppose f (x) ≥ A + ². By continuity we can find a δ > 0 such that f (t) ≥ A = ²/2 for t ∈ [a, b] ∩ (x − δ, x + δ). Choose a non-zero g ∈ C([a, b]) with g(t) = 0 for all t ∈ / [a, b] ∩ (x − δ, x + δ). Then |f gk1 > (A + ²/2)kgk1 .

Thus H¨older and reverse H¨older 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 differentiable except at (0, 0). Since fp (x, 0) = sgn x 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) = sgn h 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) 1 if h > 0, = h −1 if h < 0, so f∞ is not differentiable at (x, x) if x > 0 and a similar argument shows that f∞ is not differentiable at (0, 0). Similar arguments show that f∞ is differentiable at the points (x, y) with x 6= y and only there. (iii) If kxks = 1 then |xj | ≤ 1 and |xj |s ≤ |xj |r so 1 = kxkss ≤ kxkrr

so kxkr ≥ 1. x = (1, 0, 0, . . . , 0). (iv) We have n X j=1

|xj |r ≤

Ã

n X j=1

Now set rp = s (so q(s − r) = s).

|xj |rp

!1/p

n1/q .

(vi) If s > r we can find an α with r −1 > α > s−1 . If bj (N ) = j −α for j ≤ N , bj (N ) = 0 otherwise, then kb(N )ks /kb(N )kr → ∞ 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 kf ks ≤ (b − a)1/s kf kr .

239

K213 (ii) First look at maps from (Rn , k k∞ ) to (Rm , k k∞ ) Observe that ¯ n ¯ n ¯X ¯ X ¯ ¯ kT xk∞ = max ¯ aij xj ¯ ≤ max |aij |kxk∞ 1≤i≤m ¯ ¯ 1≤i≤m j=1

j=1

and that, if we set yj (i) = sgn aij we have kT y(i)k∞ = Thus kT k = max1≤i≤m

Pn

n X j=1

j=1

|aij | =

n X j=1

|aij |ky(i)k∞ .

|aij |.

Now look at maps from (Rn , k k1 ) to (Rm , k k∞ ). Observe that ¯ n ¯ n ¯X ¯ X ¯ ¯ kT xk∞ = max ¯ aij xj ¯ ≤ max |aij ||xj | ≤ max max |aij |kxk1 1≤i≤m ¯ 1≤i≤m 1≤j≤n ¯ 1≤i≤m j=1

j=1

and that, if we set yj (k) = 1 if j = k, yj (k) = 0, otherwise, then kT y(k)k∞ = max |aik | = max |aik |ky(k)k1 1≤i≤m

1≤i≤m

Thus kT k = max1≤i≤m max1≤j≤n |aij |.

Now look at maps from (Rn , k k1 ) to (Rm , k k1 ). Observe that ¯ n ¯ m ¯X ¯ X ¯ ¯ kT xk1 = aij xj ¯ ¯ ¯ ¯ i=1

≤ = =

j=1

n m X X

i=1 j=1 n X m X j=1 i=1 n X j=1

|xj |

≤ max | 1≤j≤n

|aij ||xj |

|aij ||xj | m X

i=1 m X i=1

|aij |

|aij |kxk1

and that, if we set yk (j) = 1 if k = j, yk (j) = 1, otherwise then kT y(j)k1 = Thus kT k = max1≤j≤n |

Pm

m X i=1

i=1

|aij | =

|aij |.

m X i=1

|aij ky(j)k1 .

240

If we look at maps from (Rn , k k∞ ) to (Rm , k k1 ) we have the obvious bound m X n X kT k ≤ |aij | i=1 j=1

but looking at the maps T1 µ 1 1

and T2 with matrices ¶ ¶ µ 1 1 1 and −1 1 1

we see that, in the first case the bound is attained but in the second case it is not since µ ¶µ ¶ µ ¶ 1 1 x x+y = 1 −1 y x−y and |x + y| + |x − y| ≤ 2 max(|x|, |y|). However, if we really need a to find kT k we can obtain it as a solution of a linear programming problem.

241

K214 Suppose a, b ∈ X. If ² > 0, we can find a y ∈ E such that d(a, E) ≥ d(a, y) − ².

Thus

d(b, E) ≤ d(b, y) ≤ d(a, b) + d(a, y) ≤ d(a, b) + d(a, E) + ².

Since ² 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 so, by definition,

d(k, M ) ≤ d(k, l) + d(l, M )

d(k, M ) ≤ d(k, l) + ρ0 (L, M )

so, allowing l to range freely,

d(k, M ) ≤ d(k, L) + ρ0 (L, M )

and so, by definition again, so

d(k, M ) ≤ ρ0 (K, L) + ρ0 (L, M ) ρ0 (K, M ) ≤ ρ0 (K, L) + ρ0 (L, M ).

It follows at once that ρ obeys the triangle inequality. Also ρ is symmetric and positive. If ρ(E, F ) = 0 then ρ0 (E, F ) = 0 and, if e ∈ E, we can find fn ∈ F such that d(fn , e) → 0. Since F is closed e ∈ 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 and bounded. If |fn (x) − f (x)| < ² then ρ0 (f (Kn ), F (K)), ρ0 (f (K), F (Kn )) < ².

No. Let K = [−π, π] and fn (x) = sin nx.

242

K215 The intersection of closed bounded sets is closed and bounded. Choose xn ∈ Kn . Since K1 is closed and bounded, the theorem of Bolzano– Weierstrass says that we can pick n(j) → ∞ such that xn(j) → x for some x ∈ K1 . Since xn(j) ∈ Km for j sufficiently large and Km is closed x ∈ 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 the first paragraph we can find n(j) → ∞ and x ∈ K with d(xn(j) , x) → 0 which 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{kz − yk : y ∈ E} > 0. If 1 > ²0 > 0 we can find y0 ∈ E such that kz − y0 k > r(1 − ²0 )

Set x0 = z − y0 . Then kx0 k < r(1 + ²0 ) and for all e ∈ E.

kx0 − ek = kz − (y0 + e)k ≥ r

If we set x = kx0 k−1 x0 k then kxk = 1 and

kx − ek ≥ kx0 k−1 r = (1 + ²0 )−1 > 1 − ²

for all e ∈ E if we choose ²0 appropriately.

Let x1 be any vector of norm 1. Let E1 be the subspace generated by x1 . Now proceed inductively choosing xn+1 of norm 1 with kxn+1 −ek ≥ 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. If un(j) (j) = δ, uk (j) = 0 if k 6= n(j), then u(j) ∈ E but the sequence has 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 to find 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 ¡ ¡ ¢ ¢ kx(n(r, r)) − xk ≤ max max |xj n(r, j) − xj |, sup |κj | 1≤j≤r

j≥r+1

−r

≤ max(2 , sup |κj |) → 0 j≥r+1

so x(n(r, r)) → x as r → ∞. P PN (j) (ii) If ∞ κ diverges, we can find N (j) such that n n=1 n=j κn > 1. PN (j) If uk (r) = κk /( 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 ∈ X such that xn(j) → x. Now there must exist a U ∈ U with x ∈ U . Since U is open there exists a δ > 0 such that B(x, δ) ⊆ U . But we can find a J such that n(J) > 2δ −1 and kxn(J ) − xk < δ/2 so B(xn(J) , 1/n(J)) ⊆ B(x, δ) ⊆ U

which contradicts our initial assumption.

S (iii) By Lemma 11.22 we can find y1 , y2 , . . . yM such that M m=1 B(ym , δ) = X and by (ii) we can find Um ∈ U with B(ym , δ) ⊆ Um . Thus SM m=1 Um = X.

247

K220 (i) Take contrapositives. (ii) If the space has property (B) then, taking U to be the collection of S B(y, δy ), we can find y1 , y2 , . . . , yM such that X = M m=1 B(ym , δym ). Now set N = max1≤m≤M Nm . We have xN ∈ / B(ym , δym ) for each 1 ≤ m ≤ M and this gives a contradiction. (iii) Combine the results of this question with that of K219.

248

K221 k kB is not a norm since k1kA = 0 but 1 6= 0. The rest can be checked to be norms. Recall that Lipschitz equivalent metrics are either both complete or neither complete. kf k∞ ≤ kf kA ≤ 2kf k∞

so k k∞ and k kA are equivalent. Looking at fn (x) = (n−2 + (x − 12 )2 )1/2 we see that fn is Cauchy. If it has a limit f in the uniform norm then / C 1 ([0, 1]). it is a pointwise limit of fn so f (x) = |x − 21 |. But f ∈ kf kD ≤ kf kC ≤ 2kf kD

so k kC and k kD are equivalent. If fn is Cauchy in k kC then both fn and fn0 are Cauchy in k k∞ converge uniformly to continuous f and F and standard theorems tell us that f is differentiable with f 0 = F . Thus f ∈ C 1 ([0, 1]) and kfn − f kC → 0. By considering fN (x) = sin N x we see that k kD is not Lipschitz equivalent to k k∞ or k k1 . By considering fN (x) = xN we see that k k∞ is not equivalent to k k1 (or observe that one norm is complete and the other is not).

249

K222 (i) By the inverse function rule d 1 sin−1 x = dx (1 − x2 )1/2 for |x| < 1. By the binomial expansion (or by some rigorous Taylor expansion) n ∞ Y X (2r − 1) n 1 = 1 + t (1 − t)1/2 2r n=1 r=1

for |t| < 1. Thus

∞ Y n X d (2r − 1) 2n −1 sin x = 1 + x dx 2r n=1 r=1

for |x| < 1. But we may integrate term by term within the radius of convergence so (since sin−1 0 = 0) µ ¶ ∞ Y ∞ n X X 2−2n 2n 2n+1 (2r − 1) x2n+1 −1 sin x = x + x = 2r 2n + 1 n=0 2n + 1 n n=1 r=1

for |x| < 1.

(ii) The power series has radius of convergence 1 so the only other points to consider are x = ±1. Taking logarithms (cf K96) shows that ¶ n n µ Y 1 (2r − 1) Y 1− = → n−1/2 2r 2r r=1 r=1 so ∞ n X 1 Y (2r − 1)2r 2n + 1 n=1 r=1

converges by comparison with n−3/2 so we actually have (by the Weierstrass M-test) uniform convergence on [−1, 1] so the power series is continuous on [−1, 1] so, since sin−1 is also continuous on [−1, 1], we have 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 k k∗ then both fn and fn0 are Cauchy in k k∞ converge uniformly to continuous f and F but then we know that f is differentiable with f 0 = F . Thus f ∈ C 1 ([0, 1]) and kfn − f kC → 0.

251

K224 Let hn : [−1, 1] → R be given  2n+3 (x + 1)    1 hn (x) =  −2n+3 x    −h(−x)

by for for for for

Let gn : [−1, 1] → R be given by gn (x) =

Z

x ∈ [−1, −1 + 2−n−3 ], x ∈ [−1 + 2−n−3 , −2−n−3 ] x ∈ [−2−n−3 , 0] x ∈ [0, 1]. 1

hn (t) dt −1

By the fundamental theorem of analysis, gn is differentiable with continuous derivative hn so |gn0 (x)| ≤ 1 for all x and |gn (x)| = 1 outside three 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) = 2−1 gn (2(x − 1)) for x ∈ [−1, 1] and define fn : [0, 1] → R by fn = n−1 kn (nx) for x ∈ [0, 1]. Observe that fn ∈ A, kfn k∞ → 0 as n → 0 but I(fn ) → 0 and I(0) = 1. Thus I is not continuous with respect to the uniform norm. However if gn , g ∈ A and kgn −gk∗ → 0 then gn0 (x) → g(x) uniformly, so (1 − (gn0 (x))4 )2 → (1 − (g 0 (x))4 )2 uniformly, so (1 − (gn0 (x))4 )2 + gn (x)2 → (1 − (g 0 (x))4 )2 + (x)2

uniformly as n → 0 so I(gn ) → I(g) and I is continuous with respect to k k∗ . (iii) Done in (i). (iv) Since I is continuous it follows that, if fN → f with respect to kR k∗ then Ifn → If , so If = 0 which is impossible. (If If = 0 then 1 f (x)2 dx = 0 so f = 0 (since f is continuous) but I(0) = 1.) −1

252

K225 (i) If fN,² (x) = ² sin 2πN x then Z 1 ¡ I(fN,² ) = 1 − 2²(2πN )4 (cos 2πN x)4 0

¢ + ²2 (2πN )8 (cos 2πN x)8 + ²2 (sin 2πN x)2 dx

with

= 1 − A²N 4 + B²8 N 8 + C²2 A = 2(2π) B = (2π) C = (2π)

8

2

Also

4

Z

Z

Z

1

(cos 2πx)4 dx > 0

0 1

(cos 2πx)8 dx > 0 0 1

(sin 2πx)2 dx > 0. 0

kfN,² k∗ = (1 + 2πN )².

If we take gN = fN,N −4/3 then kgN k∗ → 0, and (when N is sufficiently large) I(gN ) < 1 = I(0). If we take hN = f1,N −1 , then khN k∗ → 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 zeros at 1 and 2. Thus f (t) = 2t3 − 9t2 + 12t has stationary points at 1 and 2 so (from our knowledge of cubics or by looking more closely) has a unique zero at 0. Setting P (t) = f (t)2 gives a function of the required type. (iii) There exists a δ > 0 such that P (t) > 0 for all 0 < |t| < δ. Thus, if |f 0 (t)| < δ 1/2 , the function h : [0, 1] → R defined by h(x) = P (1 − f 0 (x)2 ) + f (x)2 − P (1)

is continuous and positive so

J(f ) − I(0) =

Z

1 0

h(x) dx ≥ 0

with equality only if h(x) = 0 for all x ∈ [0, 1] i.e. only if f = 0. Thus if kf k∗ < δ 1/2 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 function u : R → R with 0 < u(x) < 1 for all t and inf x∈R u(x) = 0, supx∈R u(x) = 1. Set v(x) =

1 2

+ 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 continuous functions form a closed subspace of B(E) and that the uniform limit of continuous functions is continuous since the proofs do not use the completeness of (Y, ρ). Suppose that Y is not complete, so we can find a Cauchy sequence y(n) in Y which does not converge. Let E = e and fy (e) = y. (Note B(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 convergence fails.

254

K227 Observe that, if 0 < v < u, then evx e−ux xγ−1 → 0 as x → 0 (exponentials beat powers) so we can find an A such that 0 ≤ e−ux xγ−1 ≤ Ae−vx R∞ for x ≥ 1 so, by comparison, 1 e−ux xγ−1 dx converges. Also so, by comparison, converges.

R1 0

0 ≤ e−ux xγ−1 ≤ xγ−1

e−ux xγ−1 dx converges. Thus

R∞ 0

e−ux xγ−1 dx

Our theorem on differentiating under a finite integral show that Z n Z n d −ux γ−1 e−ux xγ dx. e x dx = du 1/n 1/n

The argument of our proof on differentiating under an infinite integral now shows that Z ∞ Z ∞ d −ux γ−1 e x dx = e−ux xγ dx. du 0 0 Now integrate by parts, to get Z X Z X £ −1 γ −ux ¤X −1 −ux γ e−ux xγ−1 dx + γu e x dx = −u x e ² ²

²

and, taking limits,

φ0γ (u) = γu−1 φγ (u). Thus ¢ d ¡ −γ u φγ (u) = 0 du and, by the constant value theorem, φγ (u) = Aγ uγ . (ii) We have φn (u) =

1 uγ . (n − 1)!

(iii) The arguments including the proofs of convergence for the appropriate integrals are similar to but simpler than those of (i). Z ∞ h i∞ 2 2 0 ψ (u) = − xe−ux e−x /2 dx = −e−ux e−x /2 − uψ(u) = −uψ(u) −∞

0

so

and ψ(u) = Ae−u

2 /2

¢ d ¡ 2 ψ(u)eu /2 = 0 du for some constant A.

255

(iv) To obtain (i) make the substitution s = ux. To obtain (iii) make the substitution y = x − u.

256

K228 Since f (n+1) → g uniformly on the closed interval with end points a and x, we have Z x g(t) dt = g(x) − g(a). a

Since g is continuous, the fundamental theorem of analysis gives g(x) = g 0 (x) so ¢ d ¡ −x e g(x) = 0 dx so, by the constant value theorem, g(x) = Cex for some C. No, f (x) = 1 + ex has the property.

257

K229 (i) Since exponentials beat polynomials, fn (x) → 0 for x 6= 0. But fn (0) = 0 → 0, so fn → 0 pointwise for all α, β > 0. (ii) We observe that (if x ∈ (0, 1))

¡ ¢ fn0 (x) = nα xβ−1 (1 − x)n−1 β − (β + n)x

so, examining the sign of fn0 , we see that there is a unique maximum at x = β/(β + n). Thus µ ¶ ¶β µ ¶n µ β β β α−β =n 1− 0 ≤ f (x) ≤ f β+n 1 + β/n β + nr

so, remembering that (1 + γn−1 )n → eγ , we see that fn → 0 uniformly if and only if β > α. (iii) We have Z 1 0

fn (x) dx ≥

Z

≥

Z

2/n

fn (x) dx 1/n 2/n α

n (n ) 1/n

=n so

R1 0

−1 β

α−β−1

µ

2 1− n

µ

2 1− n

¶n

¶n

dx

fn (x) dx 9 0 if α ≥ β + 1.

Now suppose α < β + 1. Choose γ with α . 1>γ> β+1 We have Z 1 Z fn (x) dx = 0

≤

as n → ∞.

n−γ

fn (x) dx + 0

Z

n−γ

nα xβ dx +

Z

1

n−γ Z 1 n−γ

0

fn (x) dx nα (1 − x)n dx

nα−(β+1)γ nα = + (1 − n−γ )n+1 β+1 n+1 ¡ nα nα−(β+1)γ γ ¢n(1−γ) + (1 − n−γ ) (1 − n−γ )n → 0, = β+1 n+1

[An alternative approach is to show that, when α = β + 1, Z 1 Z 1 ∞ > lim sup fn (x) dx ≥ lim inf fn (x) dx > 0 N →∞

0

N →∞

0

258

and deduce the other cases.]

259

K230 (i) Suppose that gn → g uniformly and xn → x. Let ² > 0. By the continuity of g at x, we can find a δ > 0 such that |g(t)−gn (t)| < ²/2 for |x − t| < δ and t ∈ [a, b]. Now there exists an N such that |xn − x| < δ for n ≥ N and an M such that kgn − gk∞ < ²/2 for n ≥ M . Thus if n ≥ max(N, M ) |gn (xn ) − g(x)| ≤ |gn (xn ) − g(xn )| + |g(xn ) − g(x)| < ².

If gn 9 g uniformly then we can find a δ > 0 and n(j) → ∞ such that |gn(j) (xn(j) ) − g(xn(j) )| > δ.

By Bolzano–Weierstrass we can find j(k) → ∞ such that xn(j(k)) → y say. 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)|

we have gr (yr ) 9 g(y).

≥ δ − |g(xn(j) ) − g(y)| → δ

(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 for x > 0. Then gn → g uniformly but g(1/n) 9 g(0).

However, the ‘if’ part still works. Suppose gn 9 g uniformly as before and define δ > 0 as before. Now define n(j) and m(j) inductively as follows. Set m(0) = 0. For each j ≥ 1 choose n(j) > m(j − 1) and xn(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)) → y say. 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 (even if gn is continuous). Let (a, b) = (0, 1) and gn (x) = max(1 − nx, 0). If

260

x ∈ (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 that xn → x implies gn (xn ) → g(x). However gn does not converge uniformly on [0, 1].

261

K231 (i) Since f is continuous on [0, 1 − δ], it attains its bounds so there exists a y ∈ [0, 1 − δ] with f (y) ≥ f (x) ≥ 0. Since 0 ≤ f (y) < 1, we have Z 1−δ f (x)n dx ≤ nf (y)n → 0. n 0

Observe that (if f 0 (1) 6= 0) f 0 (1) > 0 since f (1) > f (1 − x) for x ∈ (0, 1). Given any ² with f 0 (1) > ² > 0 we can find a 1 > δ > 0 such that |f (x) − 1 − f 0 (0)(1 − x)| = |f (x) − f (1) − f 0 (1)(1 − x)| ≤ ²(1 − x) and so 1 − (f 0 (1) − ²)(1 − x) ≤ f (x) ≤ 1 − (f 0 (1) − ²)(1 − x) for x ∈ [1 − δ, 1]. Thus Z 1 Z 0 n (1 − (f (1) − ²)(1 − x)) dx ≤ 1−δ

≤

Z

1

f (x)n 1−δ 1 1−δ

(1 − (f 0 (1) − ²)(1 − x))n dx.

Now n

Z

1 n

1−δ

(1 − A(1 − x)) dx = n =

Z

δ 0

(1 − At)n dt

n (1 − (1 − Aδ)n+1 ) → A−1 A(n + 1)

so, using the first paragraph, Z 1 n f (x)n dx → 0

1 . f 0 (1)

(ii) Suppose g(y) = 1 for some y ∈ (0, 1). By the argument of Rolle’s theorem, g 0 (y) = 0, so given any ² > 0, we can find a δ > 0 with [y − δ, y + δ] ⊆ (0, 1) such that g(t) > 1 − ²|y − t| for all t with |y − t| ≤ δ. Thus Z 1 Z y+δ n n g(x) dx ≥ n (1 − ²|y − t|)n dt 0

y−δ

=

2n (1 − (1 − Aδ)n+1 ) → ²−1 . ²(n + 1)

262

Since ² was arbitrary, n as n → ∞.

Z

1 0

g(x)n dx → ∞

263

K232 Observe that, if η is fixed with 1/3 > η > 0, Z 1 Z η1/2 /2 Sm,η (x) dx ≥ Sm,η (x) dx −η 1/2 /2

−1

≥ but, if |x| ≥ 2η 1/2 , then

Z

η 1/2 /2

−η 1/2 /2

µ

3η 1+ 4

¶

dx → ∞

|Sm,η (x)| ≤ (1 − 3η)n → 0

as m → ∞. Thus Tm,η (t) → 0 uniformly on [−1, −2η 1/2 ] ∪ [2η 1/2 , 1] as m → ∞.

(vi) If F : [−1, 1] → R is continuous, then it is uniformly continuous and, given any ² > 0, we can find an integer N such whenever |t − s| ≤ N −1 and t, s ∈ [−1, 1] we have |F (t) − F (s)| < ²/4. Thus if h is the simplest piecewise linear function on [−1, 1] with h(r/N ) = F (r/N ) for N ≤ r ≤ N we have |h(t) − F (t)| < ² for all t ∈ [−1, 1].

264

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 polynomial Pn with Pn (t2 ) = 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.

265

K234 Write Xn (t) = Y1 + Y2 + · · · + Yn where Yj = 1 if the jth trial is a success and Yj = 0 otherwise. Observe that (since the expectation of the sum of random variable is the sum of the expectations, and the variance of the sum of independent random variables is the sum of the variances) EXn (t) = var Xn (t) =

n X

j=1 n X j=1

EYj = nt 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 ² > 0 there exists a δ > 0 such that |f (t)−f (s)| ≤ ² whenever t, s ∈ [0, 1] and |t−s| ≤ δ. By Tchebychev’s inequality ¯ ¶¯ µ¯ µ ¶ ¶ µ¯ ¯ Xn (t) ¯ ¯ Xn (t) Xn (t) ¯¯ ¯ ¯ ¯ Pr ¯ − t¯ ≥ δ = Pr ¯ −E ¯≥δ n n n

var(Xn (t)/n) δ2 t(1 − t) = nδ 2 1 ≤ 2 nδ so, writing IA for the indicator function of A, µ µ ¶ µµ µ ¶ ¶ ¶¶ Xn (t) Xn (t) Xn (t) Ef − f (t)E f − f (t) I[t−δ,t+δ]∪[0,1] n n n µ µµ µ ¶ ¶ ¶¶ Xn (t) Xn (t) +E f − f (t) I[0,1]\[t−δ,t+δ] n n ¯ ¯ ¶ µ¯ ¶ µ¯ ¯ Xn (t) ¯ ¯ ¯ Xn (t) − t¯¯ ≤ δ + 2M Pr ¯¯ − t¯¯ > δ ≤ ² Pr ¯¯ n n 2M ≤ ²+ 2. nδ Allowing n → ∞ and observing that ² is arbitrary gives the required result. =

We have n n ³r´ ³ r ´ µn ¶ X X pn (t) = Pr(Xn (t) = r) = f f tr (1 − t)n−r , n n r r=0 r=0

so pn is polynomial of degree n.

266

Since

and

µ ¶ n µ ¶ X n r n r n−r t (1 − t) ≥ 0, t (1 − t)n−r = 1 r r r=0 n µ ¶ X n r r=0

r n

tr (1 − t)n−r = t

we can use Jensen’s inequality (or we could just quote K144 (iv)).

267

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 such that fn (xn ) > f (x) + δ. By the theorem of Bolzano–Weierstrass we can find a subsequence n(j) → ∞ and an x ∈ I such that xn(j) → x. Since fn (x) → f (x) we can find an N such that f (x) + δ/2 > fN (x) Since f − 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| < η we 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 1 pn+1 (t) − pn (t) = pn (t) − pn−1 (t) + (pn (t)2 − pn−1 (t)2 ) 2¶ µ pn (t) + pn (t) (pn (t) − pn−1 (t)) = 1− 2 ≤ (1 − |t|)(pn (t) − pn−1 (t) ≥ 0

and

t2 + |t| |t|(|t| + 1) = ≤ |t|. 2 2 The result follows by induction. pn+1 (t) ≤

Since pn (t) is increasing and bounded above, pn (t) → g(t) say. We have 1 0 = pn (t) − pn−1 (t) − (t2 − pn−1 (t)) 2 1 2 1 → g(t) − g(t) − (t − g(t)) = − (t2 − g(t)) 2 2 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 metric space satisfying the Bolzano–Weierstrass condition.

268

K236 (ii) Proceed inductively. Write pn (x) = xn . Set s0 = 1. When s0 , s1 , . . . , sn−1 have been defined set sn = p n −

n−1 X hpn , sj i j=0

hsj , sj i

.

(This is the Gramm-Schmidt technique.) (iii) To prove uniqueness observe that n X

bj xj =

n X

aj x j

j=0

j=0

implies bn = an and use induction. A similar induction proves existence. P j Observe that Q = n−1 j=0 cj x .

(iv) Write Pn for the set of polynomials of degree or less. Since the polynomials are uniformly dense, we have, using (iii), ° ° n °X ° ° ° infn+1 ° aj sj − f ° = inf kP − f k2 P ∈Pn ° ° a∈R j=0

2

≤ (b − a)1/2 inf kP − f k∞ → 0 P ∈Pn

as n → ∞. (v) We have k

n X j=0

bj s j −

f k22

= =

kf k22 kf k22

+ +

n X

j=0 n X j=0

(b2j ksj k22 − 2bj hf, sj i) (bj ksj k2 −

2 hf, sj iksj k−1 2 )

n X (hf, sj iksj k−1 − 2 ) j=0

° °2 ¶2 µ n n ° X X hf, sj i ° hf, sj i ° ° 2 = °f − sj ° + ksj k2 bj − ° ksj k22 ° ksj k22 j=0 j=0 2 ° °2 n n ° ° X X ° ° ˆ = °f − f (j)sj ° + ksj k22 (bj − fˆ(j))2 ° ° j=0

2

j=0

The result of the first sentence can now be read off.

The result of the second sentence follows from part (iv).

269

(vi) Observe that sn (t)q(t)w(t) is single signed (that is always positive or always negative) and only zero at finitely many points. Thus Z b sn (t)q(t)w(t) dt 6= 0 a

and, by part (iii), q must be a polynomial of degree at least n. Thus k ≥ 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.

270

K237 Observe (by Leibniz rule or induction or otherwise) that, if n > r ≥ dr 0, then (x2 − 1)n has (x2 − 1) as a factor and so vanishes when dxr x = ±1. Thus, if n ≥ m ≥ 0, integrating by parts n times gives Z 1 n m d 2 nd (x − 1) (x2 − 1)m dx n m dx −1 dx ¸1 · n−1 Z 1 n−1 n m+1 d d 2 nd 2 n 2 nd − (x − 1) (x − 1) (x − 1) (x2 − 1)m dx = n−1 n n−1 m+1 dx dx dx −1 dx −1 Z 1 n−1 m+1 d 2 nd (x − 1) (x2 − 1)m dx =− n−1 m+1 dx dx −1 = ... Z 1 dm+n n = (−1) (x2 − 1)n m+n (x2 − 1)m dx. dx −1 Thus Z

1 −1

m dn 2 nd (x − 1) (x2 − 1)m dx = dxn dxm

(

0 if n 6= m R1 2 n (2n)! −1 (1 − x ) dx if n = m.

Making the substitution x = sin θ, we have Z 1 Z π/2 2 n (1 − x ) dx = cos2n+1 θ dθ. −1

If In =

R π/2

−π/2

In =

Z £

−π/2

cosn θ dθ, and n ≥ 2 then integration by parts gives

π/2

cosn−1 θ cos θ dθ −π/2

= cos

n−1

= (n − 1)

θ sin θ Z

π/2 −π/2

¤π/2

+ (n − 1) −π/2

Z

π/2

cosn−2 θ sin2 θ dθ −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 Y j=0

2n − 2j 2n+1 (n!)2n = . 2n + 1 − 2j (2n + 1)!

271

By expanding (x2 − 1)n and differentiating n times we see that pn is a polynomial of degree n with leading coefficient (2n)!/(n!). Take cn = (n!)/(2n)!. There are several ways of doing the last sentence including integrating by parts along the lines of the first part of the question.

272

K238 P (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 polynomial.

aj =

Z

1

ej (t) dt. −1

(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 degree k can be written as p = sq + r where s has degree at most k − s if k ≥ s and is zero, otherwise, and r has degree at most k − 1). P P Observe that nj=1 aj Q(xj )pn (xj ) = nj=1 0 = 0. Q (iv) Let p(t) = nj=1 (t − yj ). Then p is polynomial of degree exactly n with leading coefficient 1 such that Z 1 n X bj p(yj )q(yj ) = 0 p(t)q(t) dt = 1

j=1

whenever q has degree at most n−1 (and so, in particular, when q = pr with 0 ≤ r ≤ n − 1). Thus p = pn .

273

K239 (i) The argument of K236 shows that sn + λsn−1 must change sign at least n − 1 times in (a, b). It changes sign only at zeros of odd order and has exactly n zeros if multiple roots are counted multiply so, by counting, it must have n simple zeros in (a, b). (ii) Let the common root be y. We have P (x) = (x − y)p(x) and Q(x) = (x − y)q(x). If p(y) = 0, then P has a multiple root at 0 and we 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 µ both non-zero (since sn and sn−1 do not have multiple roots) such that µsn +λsn−1 and so sn +µ−1 λsn−1 has multiple roots which is impossible. (iii) Without loss of generality we may suppose that P has no real roots in the open interval (x, y). Thus both P and Q are single signed and, without loss of generality, we may suppose that P and Q are positive in the open interval (x, y). Since a continuous function on a closed bounded interval is bounded and attains its bounds we can find s1 , 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 a supremum λ0 say.We observe that, for each n ≥ 1 there exists an wn ∈ [x, y] such that P (wn ) − (λ0 − 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) − λ0 Q(w) = 0.

We note that this implies w ∈ (x, y). Since

P (t) − λ0 Q(t) ≤ 0

for all t ∈ [x, y] we see that w is a multiple root. Set λ = −λ0 . Last sentence much as for (ii).

274

K240 Fix M ≥ 1 temporarily. Observe that (if |x| < M/2) we have ¯ ¯ ¯ ¯ 4 1 ¯ ¯ ¯ (x − n)−2 ¯ ≤ n2 P 2 so by Weierstrass M-test N n=M (x − n) converges uniformly. We write FM,N (x) =

N X

n=M

1 1 + . (x − n)2 (x + n)2

Since the uniform limit of continuous functions is continuous, fM is continuous. ¯ ¯ ¯ ¯ ¯ −2 ¯ ¯ −2 ¯ 8 ¯ ¯ ¯ ¯ ¯ (x − n)3 ¯ + ¯ (x + n)3 ¯ ≤ n3

0 so by the Weierstrass M test fM,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 integer M such that |y| + 2δ < M/4. Thus since the sum of two differentiable functions is differentiable M −1 X 1 1 + F (x) = FM (x) + (x − n)2 (x + n)2 n=0

is well defined, continuous and differentiable on B(y, 2δ). Since y was chosen arbitrarily, we are done. (iii) Fix x ∈ / Z. 0=

N X

n=−N

N +1 X 1 1 − ¡ ¢2 → F (x) − F (x + 1) 2 (x − n) (x + 1) − n n=−N +1

as N → 0, so F (x) = F (x + 1).

275

K241 (i) Since g is continuous on [0, 1] it is bounded, with |g(x)| ≤ K say for 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 set h(x) = x for all x then h is continuous but not bounded. (ii) Observe that 1 |g(x)| = 4 and repeat.

µ¯ ³ ´¯ ¯ µ ¶¯¶ x + 1 ¯¯ K ¯ x ¯ ¯¯ ≤ ¯g ¯ + ¯g ¯ 2 2 2

(iii) Use L’Hˆopital or power series manipulation π2 π2 = (sin πx)2 (πx − (πx)3 /3! + ²1 (x)x3 )2 1 = 2 x (1 − (πx)2 /6 + ²2 (x)x3 )2 µ ¶ 2(πx)2 1 3 = 2 1+ + ²3 (x)x x 6

=

1 π2 + + ²3 (x)x x2 3

where ²j (x) → 0 as x → 0. Observe that g(x) → g(0) as x → 0 so g is continuous at 0. Now use periodicity. (iv) Observe that, if 2x ∈ / Z, then µ ¶ ³x´ x+1 +g g 2 2 µ ¶ ∞ X 1 1 = + x − π 2 (cosec2 (πx/2) − cosec2 (π(x + 1)/2) x 1 2 2 ( 2 − n) (2 − n + 2) n=−∞ ¶ µ ∞ 2 2 X 4 2 2 cos (πx/2) − sin (πx/2) + (x − 2n + 1) − π = (x − 2n)2 cos2 (πx/2) sin2 (πx/2) n=−∞ = 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 ∞ X

1 = π 2 cosec2 (πx) 2 (x − n) n=−∞

276

and looking at x = 0, ∞ X 1 π2 = f1 (0) = 2 . 3 n2 n=1

277

K242 (i) fn0 (x) = x(x2 + n2 )−1/2 . Thus   if x ∈ (0, 1], 1 0 fn (x) → 0 if x = 0,  −1 if x ∈ [−1, 0).

We also have

¯ ¯ ¯fn − |x|¯ =

uniformly on [−1, 1].

1 n−2 ≤ →0 2 2 1/2 |x| + (x + n ) n

−1 R x(ii) We could take gn (x) = n max(1 − |x − (2n) |, 0) and fn (x) = g (t) dt. fn tends pointwise to a (continuous from below at 0) Heav0 n iside function.

(iii) Demand uniform convergence of the derivatives.

278

K243* No comments.

279

K244 Observe that |t|−1/2 v(t) = |t|1/2

v(t) − v(0) → 0 × v 0 (0) = 0. t

(ii) Standard results (chain rule, product rule etc) show that g is continuous except possibly at points (x0 , 0). Observe that since u is non-zero only a bounded closed set we can find a K such that |u(t)| ≤ K for all t. Thus, if y 6= 0 |g(x, y) − g(x0 , 0)| = |g(x, y)| ≤ K|y −1/2 v(y)| → 0

as y → 0. Thus g is everywhere continuous.

(iii) Standard results (chain rule, product rule etc) show that g has continuous 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| > 3}

and so, trivially, has continuous partial derivatives there. The only point not in A ∪ B is (0, 0). But g(x, 0) = 0 for all x and g(0, y) = 0 for all y so partial derivatives exist at (0, 0) (iv) If y 6= 0, and |y| ≤ 1/9 Z 1 G(y) = g(x, y) dx 0 Z 1 −1/2 = |y| v(y) u(x|y|−1/2 ) dx 0

= v(y)

Z

y 1/2

u(s) ds = v(y),

0

and G(0) = 0 = v(0) so G0 (0) = v 0 (0). Since (x, 0) ∈ B for x 6= 0, we have g,2 (x, 0) = 0 for x 6= 0 (and for the reasons given in the last sentence of (iii), g,2 (0, 0) = 0 as well). R1 Thus 0 g,2 (x, 0) dx = 0.

We do not have g,2 continuous at the origin. (More vividly, look at the behaviour of g(x, y) when x is fixed and small and y runs from x2 /9 to x2 .)

280

K245 R∞ By comparison, 0 |fn (t)| dtR converges and so, since absolute con∞ fn (t) dt converges. Given ² > 0, we vergence implies convergence, 0 R∞ can find an X such that X g(t) dt ≤ ²/2 and then ¯Z ∞ ¯ Z ∞ Z ∞ ¯ ¯ ¯ |fn (t)| dt ≤ g(t) dt ≤ ²/2. fn (t) dt¯¯ ≤ ¯ X X X R∞ Similarly 0 f (t) dt converges and ¯Z ∞ ¯ ¯ ¯ ¯ ¯ ≤ ²/2. f (t) dt ¯ ¯ X

Thus ¯ ¯Z ∞ Z ∞ ¯ ¯ ¯ ¯ f (t) dt f (t) dt − n n ¯ ¯ 0 0 ¯ ¯ ¯Z ∞ ¯Z ∞ ¯ ¯Z ∞ Z ∞ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ f (t) dt¯¯ fn (t) dt¯ + ¯ ≤¯ fn (t) dt − fn (t) dt¯ + ¯ X X X ¯ ¯ZX∞ ¯ ¯ fn (t) − f (t) dt¯¯ + ² → ² ≤ ¯¯ X

as n → 0 and so, since ² was arbitrary, the result follows.

The proof above works under the conditions of the last sentence.

281

K246 (ii) If follows by comparison and (using Dini’s theorem) dominated convergence (K245). R∞ To prove only if, Robserve that ifR any of the 0 fn (t) dt fails to con∞ ∞ verge then so does 0 f (t) dt. If 0 fn (t) dt converges to an then we can find Xn such that Z Xn fn (t) dt ≥ an − 2−n 0

and so

Z

Xn

f (t) dt ≥ an − 2−n 0 R∞ R∞ . Thus if 0 f (t) dt converges so do all the 0 fn (t) dt and the values an of the integrals form an increasing sequence bounded above so converging to a limit A. We have Z ∞ Z X Z X A≥ fn (t) dt ≥ fn (t) dt → f (t) dt 0 0 0 R∞ by part (i). Since X is arbitrary f (t) dt converges. The fact that 0 R∞ R∞ fn (t) dt → 0 f (t) dt now follows from the ‘if’ part. 0

282

K247 (i) Observe that 0≤

2−j dj (x, y) ≤ 2−j 1 + dj (x, y)

so the sum converges and d is well defined. To prove the triangle inequality, observe that, if d is metric, d(y, z) d(x, z) d(x, y) + − 1 + d(x, y) 1 + d(y, 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 2j d(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 X j=1

N X

−j

2 dj (xn , z) +

j=1

∞ X

2−j

j=N +1

2−j dj (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 eventually constant (and then converges to that constant value) or is a subsequence 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 Cauchy then there is an fj such that f (j) → fj . Now use the theorem on interchanging derivative and limit.

283

(v) Could take d(f, g) =

∞ X 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)|

284

K248* No comments.

285

K249* No comments

286

K250 (i) Suppose that the result is false for some j. Then we can find (j) (j) gk ∈ D such that kgk k ≤ 1/k but kgk k∞ ≥ 1. Since kgk k → 0 but gk does not converge uniformly to 0 as k → ∞ we have a contradiction. (ii) Use the fact that we have a norm.

287

K251* No comments.

288

K252* No comments.

289

K253 (v) Observe that, given any ² > 0, we can find 0 = x0 < y0 < x1 < y1 < · · · < xn−1 < yn−1 < xn < yn = 1 P such that f (yj−1 ) = f (xj ) for 1 ≤ j ≤ n and nj=0 (yj − xj ) ≤ ². Thus, if |f (x)−f (y)| ≤ K|x−y| for all x, y ∈ [0, 1], we have 1 = f (1)−f (0) ≤ K² which is impossible if ² is chosen sufficiently small.

290

K254 (i) Observe that rn (x) = xn is a strictly increasing, differentiable function. It thus has a differentiable inverse r1/n whose derivative is given by the rule for differentiation of inverses rn−1 (x) rn−1 (x) 1 = = . (rn−1 )0 (x) = 0 −1 rn (rn (x)) rn (nrn−1 (x)) nx (ii) Take nn0 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 .

291

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) ≥ 0 and so for all 0 < x ≤ R.

0 ≤ rα (x) ≤ rN (x) = xN ≤ RN

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 and combining this with the previous paragraph we obtain ¡ ¢ F |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) → 1 as γ → 0 through values of γ ∈ Q (proof by e.g. observing that if 1/n ≥ γ ≥ 0 and x ≥ 1 then r1/n (x) ≥ rγ (x) ≥ 1). (ii) Observe that rα0 n → rα0 uniformly on [a, b].

292

K256 (v) Since f 0 (y) = y 1−1/n − y 1−1/(n+1) ≥ 0

for y ≥ 1, f is increasing on [1, ∞). But f (1) = 0 so f (y) ≥ 0 for y ≥ 1. Thus n(y 1/n − 1) ≥ (n + 1)(y 1/(n+1) − 1) ≥ 0

and so, since a decreasing sequence bounded below tends to a limit, n(y 1/n − 1) → L. (iii) If 1/n ≥ x ≥ 1/(n + 1) we also have P (x) − P (0) P (1/(n + 1)) − P (0) ≥ x x n P (1/(n + 1)) − P (0) ≥ × n+1 1/(n + 1) as n → ∞.

→1×L=L

(viii) To deal with 1 > a > 0 repeat argument or use chain rule on the maps t 7→ −t and t 7→ at . If La = 0 then Pa is constant, if La > 0 then Pa is strictly increasing and 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 λ t d λt (a ) = a = λLa aλt = λLa (aλ )t dt dt so Laλ = λLa . −1

L2 Choose λ = L−1 . 2 and set e = 2

(x) We step outside the context of the question and observe that et = exp t so La = Lelog a = log aLe = log a.

293

K257 (i) Observe that ex + e−x eix + e−ix 2 2 1 ¡ 21/2 ωx 1/2 3 1/2 5 1/2 7 + e2 ω x + e2 ω x + e2 ω x ) = e 4

cosh x cos x =

with ω = eiπ/4 .

(ii) The coefficient of x2n is ¶ n µ n X 1 X 2n (−1)r = (−1)r (2r)!(2n − 2r)! (2n)! 2r r=0 r=0

1 (1 + i)2n + (1 − i)2n = = (2n)! 2

(

0 22n (2n)!

if n is odd if n is even.

(iii) Set f (x) = cosh x cos x. 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 n µ ¶ X n (n) cosh x ≤ 2n cosh R |f (x)| ≤ r r=0 for |x| ≤ R. This gives an estimate for the remainder (2R)n cosh R |Rn (f, x)| ≤ (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 (0) = y 00 (0) = y (3) (0) = 0. Remember to check the radius of convergence of the resulting power series.

294

K258 P∞ n n=0 an z has radius of convergence at least R, we know that PSince ∞ n n=0 an z converges absolutely. Thus if |h| < R − |z0 | we know that µ ¶ ∞ X n |an |(|z0 | + |h|)n . |z0 |n−r |h|r = |an | r n=0 r=0

n ∞ X X n=0

¡ ¢ Write cn,r = |an | nr |z0 |n−r |h|r if r ≤ n, cn,r = 0 otherwise. By Fubini’s theorem for sums, ∞ X ∞ X

cn,r =

n=0 r=0

∞ X ∞ X

cn,r

r=0 n=0

so ∞ X

n ∞ X X

µ ¶ n n−r r z h an an (z0 + h) = r 0 n=0 r=0 n=0 µ ¶ ∞ ∞ X X n n−r r an z h = r 0 r=0 n=r ∞ µ ¶ ∞ X X n n−r r z0 h = r n=r r=0 n

=

∞ X

br h r

r=0

with

br =

¶ ∞ µ X n+m

m=0

Thus P ∞

P∞

r=0 br z r

r=0 br h

.

r

m

z0m .

converges for |z| < R − |z0 | and

P∞

n=0

an (z0 + h)n =

295

P r Let 2δ = R − |z0 |. Since ∞ r=0 br δ converges we can find an M such −r that |br | ≤ M δ . It follows that, if |w| ≤ δ/2, ¯ ¯ ¯¯X ¯ ¯ ¯ ¯∞ ¯ g(z0 + w) − g(z0 ) r−1 ¯ ¯=¯ ¯ b w − b ¯ r 1 ¯ ¯ ¯ ¯ w r=2 ≤ |w|

∞ X k=0

≤ |w|M δ

bk−2 |w|k

−2

≤ |w|M δ −2 −2

as |w| → 0.

∞ X

k=0 ∞ X

(w|δ −1 )k 2−k

k=0

≤ 2M δ |w| → 0

296

K259 ¡ (i) We could set A0 = 4, B0 = 1, C0 = π/2, Bn = 2n+1 (n!)2 + 1 + Pn−1 (n) n−1 and take Cn so that sin(n) Cn = 1 for n ≥ 1. r=0 |fr (x)|, An = Bn Observe that

|fn(r) (x)| ≤ 2−n−1

P (r) for all n > r and all x so, by the Weierstrass M-test, ∞ n=r+1 fn (x) converges uniformly and ¯ ∞ ¯ ¯X ¯ ¯ ¯ fn(r) (x)¯ ≤ 2−r . ¯ ¯n=r+1 ¯ P P∞ (r) Thus ∞ f (x) converges uniformly for each r and f = n n=1 n=1 fn is infinitely differentiable with ∞ X (r) f (x) = fn(r) (x). n=1

Also

f

(r)

(0) ≥ f

(r)

(0) −

r−1 X n=1

|fn(r) (0)|

−

∞ X

n=r+1

|fn(r) (0)| ≥ (n!)2 .

297

K260 (i) Just observe that ∞ X n=0

n

|an z | ≤

∞ X n=0

|an |(R − ²)n .

if |z| ≤ R − ².

(ii) Since |nan z n | ≥ |an z n |, the radius of convergence does not increase. observe that, if |z| < R and we choose r = (2R+|z0 |)/3, we PNow ∞ have n=0 an rn convergent so an rn → 0 so we can find an M ≥ 0 with |an rn | ≤ M for all n and so |an | ≤ M r−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 |nan z n | ≤ K(|z|/ρ)n and, by comparison, P ∞ n n=0 nan z converges absolutely and so converges. P P∞ n n−1 Thus ∞ nz n=0 nan z has radius of convergence PR∞ and so n=1 na n−2 folhas radius of convergence R. The result for n=2 n(n − 1)an z lows. (iii) and (iv). Observe that µ

n−2 n(n − 1) j

¶

n(n − 1)(n − 2)! n! = ≥ = j!(n − 2 − j)! j!(n − j)!

µ ¶ n . j

so ¯ n−2 µ ¶ ¯ ¯X n ¯ ¯ n n n−1 j n−j ¯ |(z + h) − z − nz h| = ¯ z h ¯ ¯ ¯ j j=0 n−2 µ ¶ X n ≤ |z|j |h|n−j j j=0 ¶ n−2 µ X n−2 2 ≤ n(n − 1)|h| |z|j |h|n−j−2 j j=0 ≤ n(n − 1)(|z| + |h|)n−2 .

Suppose δ > 0 and |z| + |h| < r − δ. By parts (ii) and (i), there exists an A(δ) independent of h and z such that ∞ X n=2

n(n − 1)|an |(|z| + |h|)n ≤ A(δ).

298

Thus ¯Ã N ¯ ! N N ¯ X ¯ X X ¯ n n n−1 ¯ a (z + h) − an z −h nan z ¯ ¯ ¯ n=0 n ¯ n=0 n=1 ¯ ¯ N ¯ ¯X ¡ ¢ ¯ ¯ =¯ an (z + h)n − z n − nz n−1 h ¯ ¯ ¯ n=0 ≤ |h|2

N X n=2

n(n − 1)|an |(|z| + |h|)n

≤ A(δ)|h|2

and, allowing N → ∞, ¯ ¯Ã ∞ ! ∞ ∞ ¯ ¯ X X X ¯ n n−1 ¯ n nan z ¯ ≤ A(δ)|h|2 an z an (z + h) − −h ¯ ¯ ¯ n=0 n=1 n=0 (v) Divide by h and allow |h| → 0.

299

K261 P j We try, formally, y = ∞ j=0 aj x obtaining, on substituting in the Legendre equation, ∞ ∞ ∞ X X X 2 j−2 j−1 (1 − x ) j(j − 1)aj x − 2x jaj x + l(l + 1) aj xj = 0. j=2

j=1

j=0

Rearranging gives ¶ ∞ µ X ¡ ¢ − j(j − 1) − 2j + l(l + 1) aj + (j + 2)(j + 1)aj+2 xj = 0 j=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) aj . (j + 2)(j + 1)

This gives y(x) = a0

∞ Qr X

− 2k)(l + 2k + 1) 2r x (2r)! ∞ Qr X k=0 (l − 2k − 1)(l + 2k + 2) 2r+1 x + a1 (2r + 1)! r=0 k=0 (l

r=0

where a0 and a1 can be chosen freely. When l is not a positive integer we observe that (j + l + 2)(j − l) (1 + (l + 2)j −1 )(1 − j −1 ) = →1 (j + 2)(j + 1) (1 + 2j −1 )(1 + j −1 ) so careful application of the ratio test to the sums of even and odd powers shows that the power series has radius of convergence 1 and justifies our formal manipulation for |x| < 1. If l is positive integer we see that one of the two infinite sums in the first paragraph actually terminates to give a polynomial of degree l. The other does not terminate and the argument above shows it has radius of convergence 1, If v(x) = (x2 − 1)n then v 0 (x) = 2nx(x2 − 1)n−1 so (x2 − 1)v 0 (x) = 2nxv(x)

as required. Using Leibniz’s rule we have

dn+1 (1 − x2 )v 0 (x) dx = (1 − x2 )v (n+2) (x) − 2(n + 1)xv (n+1) (x) + n(n + 1)v (n) (x)

300

and dn+1 2nxv 0 (x) = 2nxv (n+1) (x) + 2n(n + 1)v (n) (x). dx Thus differentiating the final equation of the first sentence of this paragraph n + 1 times gives (1 − x2 )vn00 (x) − 2xvn0 (x) + n(n + 1)v(x) = 0

as required. Observe that vn is the pn of K237.

301

K262 (this is a special case of Bessel’s equation.) P j We try, formally, w = ∞ j=0 aj z obtaining, on substituting in the given equation, z

2

∞ X j=2

j(j − 1)aj z

j−2

+z

∞ X j=1

jaj z

j−1

2

+ (z − 1)

∞ X

aj z j = 0

j=0

that is to say a0 +

∞ X ¡ j=2

Thus a0 = 0 and

¢ − (j − 1)(j + 1)aj − aj−2 z j = 0 aj = −

aj−2 (j − 1)(j + 1) j

(−1) for j ≥ 2. It follows that a2j = 0 for all j ≥ 0 and a2j+1 = a1 22j (j!) 2 (j+1) . Thus ∞ X (−1)j w(z) = a1 z 2j+1 2j 2 2 (j!) (j + 1) j=0

Since 1 →0 (j − 1)(j + 1) the ratio test shows that our power series has infinite radius of convergence so our solution is valid everywhere. We note that we can only choose one constant a1 freely. Observe that setting z = 0 in the original equation shows that automatically w(0) = 0. P j Let us try and find a solution w = ∞ j=0 bj z for the equation z2

d2 w dw + z + (z 2 − 1)w = z 2 . dz 2 dz

We obtain b0 + (3b2 − b0 − 1)z

3

∞ X ¡ j=3

¢ (j − 1)(j + 1)bj − bj−2 z j = 0

so, equating coefficients, b0 = 0, b2 = 1/3 and bj = −

bj−2 (j − 1)(j + 1)

302

for j ≥ 3. Thus w(z) = b1 W (z) +

X j=1

with W (z) =

X j=0

(−1)j z 2j Qj−1 2 (2j + 1) k=1 (2k + 1)

(−1)j z 2j+1 . 22j (j!)2 (j + 1)

(We can talk about a particular integral plus a complementary function.) As before the solution is valid everywhere. P j If we try and find a solution w = ∞ j=0 cj z for d2 w dw +z + (z 2 − 1)w = z, 2 dz dz looking at the coefficient of z gives 0 = 1 which is impossible. z2

303

K263 By Lemma 11.81 we have (provided |x| < 1) (1 + x)1/2 =

∞ X

an x n

n=0

with n−1

1 Y 1 an = (−1) ( − j). n! j=0 2 n

We observe that (if z 6= 0) | 12 − n||z| |an+1 z n+1 | = → |z| |an z n | n P∞ n so, by the ratio test, n=0 an z has radius of convergence 1. Since power series can be multiplied within their circles of convergence Ã

∞ X

an z

n=0

n

!2

=

∞ X

cn z n

n=0

for |z| > 1 with 1+x=

∞ X

c n xn

n=0

for all |x| < 1 and x real. By the uniqueness of power series (or we could use K268 below) we have c0 = c1 = 1 and cj = 0 otherwise so Ã

∞ X

an z

n

n=0

!2

=1+z

has radius of convergence infinity. Take K > 1. Observe that (1 + z)−1/2 has power series radius of convergence 1 (proof much as above). Since (1 + K −1 z)1/2 has power series has radius of convergence K the function (1+z)−1/2 (1+K −1 z)1/2 has power series radius of convergence at least 1. Since power series are continuous within their circle of convergence and (1 − z)−1/2 (1 + K −1 z)1/2

304

P n −1/2 (1 + is not continuous at −1, the power series ∞ n=0 cn z of (1 − z) −1 1/2 K z) has radius of convergence exactly 1. Now ∞ ∞ X X n cn z n = (1 + z)1/2 (1 + z)−1/2 (1 + K −1 z)1/2 an z | n=0

n=0

= (1 + K −1 z)1/2 ∞ X an K −n z n = n=0

for |z| < 1. So we have an example of two power series of radius of convergence 1 whose product has radius of convergence K for any K > 1. An example with radius of convergence infinity was given in the first paragraph, An example with radius of convergence 1 is given by considering (1 +P z)−1/2 (1 + z)−1/2 . General result obtained by rescaling −j j (i.e. looking at ∞ j=0 bj R z for appropriate R. [For enthusiasts. Observe that (if we ignore convergence) to obtain ∞ ∞ ∞ X X X j j uj z vj z = wj z j j=0

j=0

j=0

we merely have to solve

n X

un−j vj = wn

j=0

Thus u2n = (2n)! and v2n+1 = (2n + 1)! (so P∞ we j can ensure P∞ that j u z and v z certainly have radius of convergence 0 and j j=0 j Pj=0 ∞ j j=0 wj z 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+K −1 z)1/2 has power series with radius of convergence 1 and (1+ z)(1 + L−1 z)1/2 has power series with radius of convergence L and their product has power series with radius of convergence K. On the other hand, if K > L > 1 then (1+z)−1 (1+L−1 z)−1/2 (1+K −1 z)1/2 has power series with radius of convergence 1 and (1 + z)(1 + L−1 z)1/2 has power series with radius of convergence L and their product has power series with radius of convergence K. If K > 1 then (1 + z)−1 (1 + K −1 z)−1/2 has power series with radius of convergence 1 and (1 + K −1 z)−1/2 has has power series with radius of convergence K and their product has power series with radius of convergence K. Remaining cases (infinite radius of convergence) dealt with similarly. −1

305

K264 If we write Sn (t) =

n X

aj (t),

j=1

then q X

λj aj (t) =

j=p

q X j=p

λj (Sj (t) − Sj−1 (t))

= λq+1 Sq (t) +

q X j=p

(λj − λj+1 )Sj (t) − λp Sp−1 (t)

and so ° ° q q ° °X X ° ° (λj − λj+1 )K + λp K = 2λp K λj aj (t)° ≤ λq+1 K + ° ° ° j=p

j=p

for all P q ≥ p ≥ 1 and so, by the general principle of uniform convergence, ∞ j=1 λj aj (t) converges uniformly.

Observe that, since xn is a decreasing sequence, ¯ ¯ ¯ ¯ q ¯ q ¯ q ¯ ¯ ¯ ¯X ¯X ¯X ¯ ¯ ¯ j¯ p j¯ j¯ bj x ¯ ≤ 2x sup ¯ bj x ¯ ≤ 2 sup ¯ bj x ¯ → 0 ¯ ¯ ¯ ¯ ¯ q≥p ¯ q≥p ¯ j=p j=p j=p P j as p → ∞, so by the general principle of uniform convergence, ∞ j=1 bj x converges uniformly. We may thus integrate term by term to get ! Z 1 ÃX ∞ ∞ Z 1 ∞ X X bn n n bn x dx = bn x dx = . n+1 0 n=0 n=0 0 n=0

P∞ P bn xn /(n+1) If ∞ n=0 n=0 bn /(n+1) converges, then the power series P∞ n has radius of convergence at least 1 so the P∞powernseries n=0 bn x has radius of convergence at least 1 and so n=0 bn x converges uniformly on [0, 1 − ²] and ! Z 1−² ÃX ∞ 1−² Z 1 ∞ X X bn (1 − ²)n+1 n n bn x dx = dx = bn x n+1 0 n=0 n=0 0 n=0 for all 1 > ² > 0. By the result of the last paragraph but one, P ∞ j+1 /(j + 1) converges uniformly on [0, 1]. The function j=1 bj x ∞ X

xj+1 bj j+1 j=1

306

is thus defined and continuous on [0, 1] so ∞ ∞ X X bj bj xj → j+1 j+1 j=1 j=1

as x → 1−. Thus

as ² → 0+.

Z

1−² 0

Ã

∞ X n=0

bn x

n

!

∞ X bj dx → j+1 j=1

bj = (−1)j we know,Pby the alternating series test, that PTaking ∞ ∞ j −1 for 0 ≤ x < 1. j=0 bj /(j + 1) converges and j=0 bj x = (1 + x) Also Z 1 1 dx = [log(1 + x)]10 = log 2. 0 1+x j Taking bP series 2j = (−1) , b2j+1 = 0 we know, P by thej alternating ∞ 2 −1 test, that j=0 bj /(j + 1) converges and ∞ b x = (1 + x ) for j=0 j 0 ≤ x < 1. Also Z 1 £ −1 ¤1 π 1 dx = tan x 0 = . 2 4 0 1+x

Not good methods. Rate of convergence painfully slow. Observe that log a loga b = log b ∞ ∞ X X j+1 (−1) log2j+1 e = j=0

j=0

log 2 (−1)j+1 = = 1. (j + 1) log 2 log 2

307

K265 Can be done in several ways. Here is one. Write pn = Pr(X = n). Observe that, since pn ≥ 0 and 1 − tn = 1 + t + · · · + tn−1 ≤ n 0≥ 1−t for 1 > t > 0, we have PM PN N M j X X 1 − tj j=0 pj − j=0 pj t pj jpj ≤ EX ≤ ≤ 1−t 1−t j=0 j=0 for M ≥ N and so N X j=0

pj

1 − tj φX (1) − φX (t) ≤ ≤ EX 1−t 1−t

for all N ≥ 0 and all 0 < t < 1. Allowing t → 1− we have N X j=0

jpj ≤ lim inf t→1−

φX (1) − φX (t) φX (1) − φX (t) ≤ lim sup ≤ EX. 1−t 1−t t→1−

Allowing N → ∞ gives the result.

308

K266 Just as in K82, if m ≥ n, ¯ n ¯ m ¯Y ¯ Y ¯ ¯ ¯ (1 + aj (z)) − (1 + aj (z))¯ ¯ ¯ j=1 j=1 ! ! !Ã Ã m Ã n X X |aj (z)| − 1 |aj (z)| exp ≤ exp j=n+1

j=1

≤ exp

Ã

∞ X j=1

Mj

!Ã

exp

Ã

∞ X

j=n+1

Mj

!

−1

!

→0

as n → QN∞ so, by the general principle of uniform convergence, it follows that j=1 (1 + aj (z)) converges uniformly.

309

K267 (i) We have |z 2 n−2 | ≤ R2 n−2

for all |z| ≤ R so, by the previous question, SN (z) converges uniformly on D(R) = {z : |z| < R}. Since the uniform limit of continuous functions is continuous, S is continuous on D(R) for each R > 0 and so on all of C. (ii) Observe that (if z is not an integer) Q (N !)2 N SN (z + 1) r=−N (r − (1 + z)) = Q SN (z) (N !)2 N r=−N (r − z)

N +1−z 1 − (z − 1)/N = → −1. −N − z −1 − z/N Thus S(z + 1) = −S(z) for z not an integer. Extend to all z by continuity. P RecallQfrom K82 that, if ∞ −1 for all n=1 |an | converges, and an 6= Q ∞ n then ∞ (1 + a ) = 6 0. Thus, if N > |z| we know that n n=1 n=N (1 − Q N −1 2 −2 2 −2 z n ) 6= 0 and so S(Z) = 0 if and only if z n=1 (1 − z n ) = 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 z so, if the guess is correct, A = π −1 .]

310

K268 P∞ n (i) Take R > r > 0 As usual we note that n=0 an r converges, n −n so an r → 0 and there exists an M such that |an | ≤ M r . Thus, if |z| < r/2, ¯ ¯ ¯ q ¯ q ¯ ¯ ¯X ¯X ¯ ¯ n¯ n¯ a z = a z ¯ ¯ ¯ ¯ n n ¯ ¯ ¯ ¯ n=0

n=N

≥ |aN z N | −

= |z|

N

≥ |z|N N

Ã

q X

n=N +1

|aN | − |z|

Ã

|an z n | q X

n=N +1

|an z

|aN | − M |z|r −N −1

≥ |z| (|aN | − 2M |z|r

−N −1

n−N −1

q X

(|z|r)n−N −1

n=N +1

).

Thus if we set δ = min(r/2, |aN |rN +1 /4M ) we have ¯ ¯ q ¯ |a ||z|N ¯X ¯ N n¯ a z ¯≥ ¯ n ¯ ¯ n=0 2 for all q ≥ N and so

¯∞ ¯ ¯X ¯ |a ||z|N ¯ N n¯ . an z ¯ ≥ ¯ ¯ n=0 ¯ 2

|

! !

311

K269 (i) Proof of K268 works. (ii) Exercise 5.91. (iii) sin x =

∞ X (−1)n x2n+1 n=0

(2n + 1)!

.

312

K270 (i) Observe that µ ¶ ¶ µ ¶ µ n! 1 n 1 n = + + r (r − 1)!(n − r)! n − r + 1 r r−1 µ ¶ n! n+1 = (n + 1) = . (r!(n + 1 − r)! 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)n y ¶ µ ¶¶ n µµ X n n n+1 xr y n+1−r + y n+1 + =x + r r − 1 r=1 ¶r n+1 µ X n+1 = y n+1−r x r=0

so the result is true by induction.

(iii) Result true for n = 0. If true for n then µ ¶Y r−1 n n−r−1 n Y Y X n (x − j) (x + y − q) = (x + y − n) (y − k) r q=0 r=0 j=0 k=0 µ µ ¶ ¶ n n n−r−1 n−r r r X n Y X n Y Y Y (y − k) + = (x − j) (y − k) (x − j) r j=0 r j=0 r=0 r=0 k=0 k=0 µµ ¶ µ ¶¶ n n−1 r n−r n Y X Y Y Y n n = (x − j) + + (x − j) (y − k) + (y − k) r r − 1 r=0 j=0 j=0 k=0 k=0 ¶ µ n−r r−1 n+1 Y X n+1 Y (x − j) (y − k) = r r=0 j=0 k=0

so the result is true by induction.

313

K271 (i) Since |w| < 1, we can find a ρ > 1 such that ρ|w| < 1. We can now find an N ≥ 1 such that ¯ ¯ ¯x − j ¯ M + j ¯ ¯ = 1 + M j −1 < ρ ¯ j ¯≤ j

for all |x| ≤ M and all j ≥ N − 1. We observe that ¯∞ ¯ ¯X 1 n−1 ¯ Y ¯ ¯ (x − j)¯ ≤ (N + M )N = K, say ¯ ¯ n=0 n! ¯ j=0 for all |x| ≤ M and all n ≤ N . Thus ¯ ¯∞ ¯ ¯X 1 n−1 Y ¯ ¯ (x − j)w n ¯ ≤ K(ρw)n ¯ ¯ ¯ n=0 n! j=0

for all |x| ≤ M and all n ≥ 1 so by the Weierstrass M-test, ∞ n−1 X 1 Y (x − j)w n n! j=0 n=0

converges uniformly. Since the uniform limit of continuous functions is continuous, f is continuous on [−M, M ]. Since N is arbitrary, f is defined and continuous everywhere. Note that we have shown that the sum is absolutely convergent. (ii) By Cauchy’s lemma (Exercise 5.38) or considering the power series in w we have ∞ X cn f (x)f (y) = j=0

with

cn =

n µ ¶Y r−1 X n r=0

r

j=0

(x − j)

n−r−1 Y k=0

(y − k).

Exercise K270 now gives f (x)f (y) = f (x + y). Q (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 intermediate value theorem, f (x) > 0 for all x. (iv) Since f (x) > 0, we can define g(x) = log f (x) and, since f is continuous, g is. Thus by K101, g(x) = ax and f (x) = eax = bx for some b. (v) Since b = f (1) = 1 + w, f (x) = (1 + w)x and we are done.

314

K272 (i) We prove that AN ≥ |aN | inductively. The result is trivial for N = 0. If the result is true for N = k − 1, we observe that kak =

k−1 k−1 X X

cn,m Pn,m (a1 , a2 , . . . , ak−1 )

n=0 m=0

and kAk =

k−1 k−1 X X

Kρn+m Pn,m (A1 , A2 , . . . , Ak−1 )

n=0 m=0

where the Pn.m are multinomials with all coefficients positive. Thus k|ak | ≤ ≤

k−1 X k−1 X

n=0 m=0

k−1 X k−1 X

|cn,m |Pn,m (|a1 |, |a2 |, . . . , |ak−1 |) Kρn+m Pn,m (A1 , A2 , . . . , Ak−1 ) = kAk

n=0 m=0

and the induction can proceed. (ii) Formally ∞ X ∞ X dw =K (ρt)n (ρw)m dt n=0 m=0 Ã∞ !Ã ∞ ! X X =K (ρt)n (ρw)m n=0

=

m=0

K . (1 − ρt)(1 − ρw)

To solve the differential equation we set, informally, (1 − ρw)dw = obtaining w(1 − ρw/2) = −

K dt 1 − ρt

K log(1 − ρt) + C. ρ

Since w(0) = 0 this gives w(1 − ρw/2) = −

K log(1 − ρt) ρ

and w(t) =

−1 + (1 − k 2 (log(1 − ρt))2 )1/2 . ρ

315

(iii) If we now fix w as in the previous paragraph then Exercise K79 tells us that w has a power series expansion ∞ X Aj tj w(t) = j=0

with radius of convergence η > 0. (Note A0 = 0 since w(0) = 0.) Since w satisfies FF our standard methods show that the P An satisfy ∞ n F. By the arguments of (i), |an | ≤ An , so, by comparison, n=0 an t P∞ has radius of convergence at least η. If we set u(t) = n=0 an tn , then, since a0 = 0, we have u(0) = 0. By continuity, we can find δ with 0 < δ ≤ η and |u(t)| < ρ for |t| < δ. The required results can now be read off.

316

K273 (i) Observe that, since f is continuous on [0, 4π], it is uniformly continuous on [0, 4π]. Since f is 2π periodic it is thus uniformly continuous on R. Similarly g is bounded. Thus ¯ Z π ¯ ¯1 ¯ |f ∗ g(t + u) − f ∗ g(t)| = ¯¯ (f (t + u − s) − f (t − s))g(s) ds¯¯ 2π ≤ kgk∞

−π

sup |f (x) − f (y)| → 0

|x−y|≤|u|

as u → 0. (iii) Observe, for example, that making the substitution x = t − s we have Z π Z t+π 1 1 f (t − s)g(s) ds = f (x)g(t − x) dx = g ∗ f (t). f ∗ g(t) = 2π −π 2π t−π Fubini gives f ∗ (g ∗ h)(t) = = = = =

¶ µZ π Z π 1 g(s − u)h(u) du ds f (t − s) 2π −π −π Z πZ π 1 f (t − s)g(s − u)h(u) du ds (2π)2 −π −π Z πZ π 1 f (t − s)g(s − u)h(u) ds du (2π)2 −π −π Z π Z π−u 1 f (t − u − v)g(v)h(u) dv du (2π)2 −π −π−u Z π 1 f ∗ g(t − u)h(u) du = (f ∗ g) ∗ h(t). 2π −π

(iv) If we set G(s, t) = f (t − s)g(s), then G has continuous partial derivative G,2 (s, t) = f 0 (t − s)g(s). We may thus apply our theorem on differentiating under the integral to show that f ∗ g is differentiable and (f ∗ g)0 = f 0 ∗ g. (v) We have ¯ Z π ¯ ¯1 ¯ |un ∗ f (t) − f (t)| = ¯¯ un (s)(f (t − s) − f (t)) ds¯¯ 2π ¯ Z−π ¯ ¯1 ¯ π/n ¯ ¯ =¯ un (s)(f (t − s) − f (t)) ds¯ ¯ 2π −π/n ¯ Z π/n 1 un (s) ds ≤ sup |f (u) − f (v)| 2π −π/n |u−v|≤π/n =

sup

|u−v|≤π/n

|f (u) − f (v)| → 0

317

as n → ∞. (vii) Take un to be a three times continuously differentiable function in (v).

318

K274 (i) Fubini gives µ Z π ¶ Z π 1 1 −int e f (t − s)g(s) ds dt f[ ∗ g(n) = 2π −π 2π −π Z πZ π 1 e−int f (t − s)g(s) ds dt 2 (2π) −π −π Z πZ π 1 e−int f (t − s)g(s) dt ds = (2π)2 −π −π µ Z π ¶ Z π 1 1 −ins −in(t−s) = e g(s) f (t − s)e dt ds 2π −π 2π −π gˆ(n)fˆ(n). (iii) We have |fˆ(n) − gˆ(n)| = |f[ − g(n)| ≤ kf − gk∞ . (v) If e ∗ f = f for all f then

fˆ(n) = e[ ∗ f (n) = eˆ(n)fˆ(n)

for all f and all n. Take f (t) = eimt to see that eˆ(m) = 1 for all m contradicting the Riemann-Lebesgue lemma.

319

K275 (i) We have 1 fˆ1 (n) = 2π

Z

π

−π

g1 (t)

(eit − e−it ) −int gˆ1 (n + 1) − gˆ1 (n − 1) e dt = 2i 2i

Thus, using Riemann-Lebesgue, n X gˆ1 (j + 1) − gˆ1 (j − 1) Sn (f1 , 0) = 2i j=−n =

gˆ1 (n + 1) + gˆ1 (n − 1) − gˆ1 (−n + 1) − gˆ1 (−n − 1) → 0. 2i

(ii) If we set  f2 (t)  if t/π is not an integer,  sin t 0 g2 (t) = f2 (0) if t = 2nπ,  −f 0 (π) if t = (2n + 1)π, 2

then g2 is automatically continuous except possibly at points nπ. But, f2 (t) − f2 (π) t−π ¡ ¢ → −f20 (π) = g2 (π) g2 (t) = t−π − sin(t − 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 Z π 1 ˆ f4 (n) = f3 (t/2)e−int dt 2π −π Z Z π 1 1 2π −i2ns f3 (s)e ds f3 (s)e−i2ns dt = fˆ4 (2n). = π −2π 2π −π

Observe that f4 obeys the conditions of (iii). (iv) Translate.

320

K276 (i) Let S(z) = z − f (z). We have S 0 (z) = 1 − f 0 (z) so |S 0 (z)| ≤ 1/2 for |z| ≤ δ and |T z| ≤ |w| + |S(z) − S(0)| ≤ |w| + |z| sup |S 0 (u)| ≤ δ/2 + δ/2 = δ |u|≤δ

for all |z| ≤ δ. (ii) Observe that, if z, u ∈ X,

|T z − T u| = |Sz − Su| ≤ |z − u|/2.

Thus T is a contraction mapping and there is a unique z0 ∈ X with T z0 = z0 i.e. with f (z0 ) = w. (iii) Observe that the continuity of F 0 at 0 gives us a δ > 0 such that |F 0 (z) − 1| ≤ 1/2 for |z| ≤ δ. (iv) Let F (z) = (g(z) − g(0))/g 0 (0) and apply (iii).

(v) No. g = 0 is a counterexample. [However, in more advanced work, it is shown that this is the only counterexample!] (vi) No. If g(z) = z 2 , the equation g(z) = reiθ has two roots r 1/2 eiθ/2 and −r1/2 eiθ/2 .

321

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 0. (iv) and (vi) True. For (vi), observe that we can find a δ with 1/6 > δ > 0 such that |F 0 (z) − 1| ≤ 1/3 for |z| ≤ δ 0 . Let X = {z : |z| ≤ δ}. Set δ = δ 0 /3. If |w| ≤ δ write Sz = F (z) − 1 and T z = Sz + |z|1/2 − w. Using the mean value inequality, we have 0

|T z| ≤ |w| + |Sz| + |z|2 ≤ δ 0 /3 + sup |S 0 (u)||z| + |z|/3 ≤ δ 0 |u|<δ 0

whenever |z| ≤ δ 0 and

|T z − T u| ≤ |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 ∈ X with T z0 = z0 i.e. w = F (z0 ) + |z0 |2 .

322

K278 (i) Observe that f (x)f 00 (x) f 0 (x) f (x)f 00 (x) = + f 0 (x) f 0 (x)2 f 0 (x)2 so, by the mean value inequality, T 0 (x) = 1 −

|T a − T b| ≤ |λ||a − b|.

Since R is complete under the usual metric, the contraction mapping theorem applies. (ii) Much as in (i), |T x| = |T x − T 0| ≤ |x| sup |T 0 (w)| ≤ |λ||x| |w|≤a

so, if |x| ≤ a, we see, by induction that |T n x| ≤ |a| and |T n x| ≤ |λ|n |x| → 0. Observe that T 0 (x) =

F (x)F 00 (x) F 0 (x)2

and F 00 (x) F (x)F 000 (x) 2F (x)(F 00 (x))2 + − , F 0 (x) F 0 (x)2 F 0 (x)3 so T (0) = T 0 (0) = 0 and T 00 (0) = F 00 (0)/F 0 (0). Thus ¯ ¯ 00 ¯ ¯ F (0) 2 ¯ = |T x − T (0) − T 0 (0)x − 1 T 00 (0)| ≤ ²(x)x2 ¯T x − x 2 ¯ 0 2F (0) ¯ T 00 (x) =

with ²(x) → 0 as as x → 0. Choose η so that ²(x) ≤ 1 for |x| ≤ η.

(iii) We have quadratic convergence (roughly doubling of correct decimal places) with Newton rather than linear convergence (increasing the number of correct decimal places by the same amount) as is implied by the contraction mapping theorem. There may be problems if F 0 (0) is small.

323

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 f g = gf = f , f has only one fixed point but g has three. Let X = {−1, 1}, f (t) = t, g(t) = −t. Then f g = gf = f , f has two fixed points but g has none.

324

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 ∈ X such that that T xn+1 = xn (this uses surjectivity). We observe that so

d(xn+1 , xn ) = d(T xn+2 , T xn+1 ) ≥ Kd(xn+2 , xn+1 ) K −1 d(xn+1 , xn ) ≥ d(xn+2 , xn+1 )

and precisely the same argument as in the contraction mapping theorem shows that xn → z for some z ∈ X. Now z = T w for some w and d(z, w) ≤ d(xn , z) + d(xn , w) ≤ d(xn , z) + Kd(T xn , T w) = d(xn , z) + Kd(xn−1 , z) → 0

as n → ∞. Thus d(z, w) = 0 so z = w and T z = z.

325

K281 (i) If T (a) = a and T (b) = b, then kT (a) − T (b)k = ka − bk so a = b. (ii) Uniqueness much as in (i). Existence not implied see e.g. second paragraph of K281. (iii) Let E = {−1, 1} be a subset of R. The mapping T : E → E given by T x = −x preserves length but has no fixed point. The map T : E → E given by T x = x preserve length and has two fixed points.

(The reader may prefer the example of a circle E in R2 . Non-trivial rotations have no fixed points, reflection in a diameter has two.) (iv) Observe that, since T decreases distance, it is continuous. The map x 7→ kx − T xk

is thus continuous and, since E is closed and bounded in Rn , thus attains a minimum at x0 , say. By definition so T x0 = x0 .

kx0 − T x0 k ≤ kT x0 − T 2 x0 k

(v) Let E ∗ be the closure of the set {T n c : n ≥ 0}. Then E ∗ is is closed bounded set and T |E ∗ is a distance decreasing map from E ∗ to itself. By (iv), T |E ∗ and so T has a fixed point.

326

K282 Observe that, if X = [0, ∞) with the usual metric, the map f given by x 7→ x + 1 is an isometry but we cannot find an isometry g : X → 0 such that f ◦ g(x) = x. We now restrict ourselves to bijective isometries. Since the bijective maps on X form a group under composition we need only show we have 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. The identity 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 the group 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 have different length. (iii) Take X to be a circle. (iv) Take X to be a regular n-gon and T rotation about the centre of symmetry through 2π/n.

327

K283 (i) Observe that, since (X, d) has the Bolzano–Weierstrass property, we can find a sequence m(j) → ∞ of strictly positive integers and an α ∈ X such that d(am(j) , α) → 0. Take a sequence j(r) → ∞ of strictly positive integers such that m(j(r + 1)) > 3m(j(r)) + 1. Then setting n0 (r) = m(j(r + 1)) − m(j(r)) we have d(an0 (r) , a) ≤ d(f m(j(r) an(r) , f m(j(r) a) = d(am(j(r+1)) , am(j(r)) )

as r → ∞.

≤ d(am(j(r+1)) , α) + d(am(j(r)) , α) → 0

Taking n(j) to be subsequence of the n0 (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(f n(r) (a), f n(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 implies x = y, f is injective. To show surjectivity take any a ∈ X. As before we can find a subsequence 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 and r(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.

328

K284 Observe that g ◦ f : X → X is expansive so, by K283, g ◦ f is bijective 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 property and g is an isometric bijection. If X = Y = {x ∈ R2 : kxk = 1} with the usual metric, then, taking f = g to be rotation through π/4, we see that f and g need not be inverses. 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 distance preserving and g is distance preserving but not a bijection.

329

K285 We note that x 7→ f (x) and x 7→ kf (x)k are continuous maps on a closed bounded subset of Rn . Thus there exists an M with kf (x)k ≤ M for all x ∈ A. By convexity, g maps A into A. We have kg(a) − g(b)k = (1 − ²)kf (a) − f (b)k ≤ (1 − ²)ka − bk so g is a contraction mapping and so, since A is closed and we are dealing with a complete metric space, g has a fixed point z, say. We observe that kf (z) − zk = kf (z) − gzk

≤ ²kf (x)k + ²ka0 k

≤ (M + ka0 k)². Thus

inf{kf (a) − ak : a ∈ A} = 0. But x 7→ kf (x) − xk is a continuous map on a closed bounded subset of Rn so attains its infimum. Thus kf (y) − yk = 0 for some y ∈ A which is thus a fixed point. Let V = R, A = [−2, −1] ∪ [1, 2] and T x = −x. Then A is closed and bounded and T is distance non-increasing with no fixed point. Let V = R, A = (0, 1) and T x = x/2. Then A is convex and bounded and T is distance decreasing with no fixed point. Let V = A = R and T x = x + 1. Then A is closed and convex and T is distance non-increasing with no fixed point. In final paragraph take A = {q ∈ R : n

n X i=1

qi = 1 and qi ≥ 0 for 1 ≤ i ≤ n}

˜ with and f (q) = q q˜j =

n X i=1

qi pij .

330

We then have A closed, convex and bounded, f maps A to A and ¯ n ° n ¯X n n X ° X X ¯ ° kf (a) − f (b)k1 = |ai − bi |pij ≤ ¯ (ai − bi )pij ° ¯ ° =

=

j=1 i=1 n n X X

i=1 j=1 n X i=1

|ai − bi |pij

|ai − bi | = ka − bk1 .

j=1 i=1

331

K286 (i) Note that |d(x, T x) − d(y, T y) ≤ d(x, y) + d(T x, T y) ≤ 2d(x, y)

so S is continuous. Thus since our space has the Bolzano–Weierstrass property S must attain a minimum at z say. If T z 6= z then ¡ ¢ S(T N (z,T z) z) = d T N (z,T z) z, T (T N (z,T z) z) ¡ ¢ = d T N (z,T z) z, T N (z,T z) (T z) < d(z, T z) = S(z)

which is absurd. Thus T z = z. (ii) Write fn (x) = d(T n x, z). Then fn is continuous (by direct proof or by composition of continuous functions) so Un = fn−1 ((−², ²)) is open. Since d(T n+1 x, z) = d(T (T n x), T z) ≤ d(T n x, z) we have Un ⊆ Un+1 . Observe that F is a closed subset of (X, d) and so inherits the Bolzano–Weierstrass property. If T x ∈ U then T x ∈ Un for some n so x ∈ Un+1 ⊆ U . Thus, if F 6= ∅ we can apply (i) to T |F : F → F to obtain w ∈ F with T w = w contradicting uniqueness. (iii) Thus U = X and, given x ∈ X, we can find an n such that x ∈ Un so d(T m x, z) < ² for m ≥ n. Since ² was arbitrary, the required result follows.

332

K287 Observe that if q ∈ X then m X i=1

qi pij ≥ 0 and

m X m X

qi pij =

j=1 i=1

m X m X

qi pij = 1.

i=1 j=1

(i) Note all norms on Rm Lipschitz equivalent. X is a closed (e.g. because the of closed sets of form Ei = {x : xi ≥ 0} and Pintersection m E = {x : i=1 xi = 1}) and bounded (if x ∈ E then kxk ≤ 1). (ii) We have

¯ m ¯ m ¯X ¯ X ¯ ¯ kT u − T vk1 = ¯ (ui − vi )pij ¯ ¯ ¯ j=1

≤ =

i=1

m m X X

j=1 i=1 m X m X i=1 j=1

|ui − vi |pij

|ui − vi |pij = ku − vk1 .

(iii) Show that (n+1) pij

=

m X

pnik pkj

k=1

and use induction. (n)

Probabilistically, pij 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 there must exist I and I 0 such that uI − vI and uI 0 − vI 0 are non-zero and have opposite signs. Thus ¯ ¯ m m ¯ X ¯X ¯ N¯ |ui − vi |pN (u − v )p < ¯ i i ij ¯ ij ¯ ¯ i=1

i=1

for each j.

(vi) T π = π gives π1 = π2 so π1 = π2 = 1/2. T n q → π if and only (2n+1) (2n) if q = π. Observe p11 = 0 and p12 = 0 for all n so there is no n with pnij 6= 0 for all i, j.

(vii) T π = π for all π ∈ X. We have T n q = q for all q ∈ X. (n) Observe that p12 = 0 for all n. (viii) T π = π has as solutions all π ∈ X with π1 = π2 and π3 = π4 .

333

Given q ∈ X set π1 = π2 = (q1 + q2 )/2 and π3 = π4 = (q3 + q4 )/2. Then, T q = π and (if n ≥ 1) T n q = π → π.

334

K288 We have ∞ X ∞ X i=1 j=1

|xi pij | =

∞ X ∞ X i=1 j=1

|xi |pij =

P∞

∞ X i=1

|xi | = kxk1

so Fubini’s P∞ theorem tells us that i=1 xi pij converges for each j. Further i=1 |xi |pij converges and ¯ ¯∞ ∞ ∞ X ∞ ¯X ¯ X X ¯ ¯ |xi |pij xi pij ¯ ≤ ¯ ¯ ¯ j=1

i=1

=

j=1 i=1 ∞ X ∞ X i=1 j=1

|xi |pij = kxk1 .

Thus T x is a well defined point of l 1 and kT xk1 ≤ kxk1 . (Note that this only shows that kT k ≤ 1.) Since αn → α and βn → β imply λαn + µβn → λα + µβ, it is easy to check that T is linear. Further, using Fubini again, if q ∈ X ∞ X ∞ ∞ X ∞ X X qi pij = qi pij = 1 j=1 i=1

and, trivially,

P∞

j=1 qi pij

i=1 j=1

≥ 0 so T q ∈ X.

(ii) If u, v ∈ X and u 6= v then we can find I and I 0 such that hat uI − vI and uI 0 − vI 0 are non-zero and have opposite signs. If k = max{I 0 , I} then as in (iv) of K287, d(T N (k) u, T N (k) v) < d(u, v). (iii) If T u = u and T v = v then (ii) shows that u = v. If T h = h and hi ≥ 0 for all i then either h = 0 or setting k = khk−1 h we have k ∈ X and T k = k. If T π = π has a solution in X then the required solutions are λπ with λ ≥ 0. If not, the only solution is 0. (iv) We must have πI > 0 for some I. Take k = max{I, i} ∞ X πi = πr pN (k) pri ≥ πI pN (k) pIi > 0. r=1

(v)(a) We have for i ≥ 2

πi = 21 πi+1 + 14 πi + 41 πi−1 .

so 2πi+1 − 3πi + πi−1 = 0

and πn = Aαn + Bβ n with α and β roots of the auxiliary equation 2θ2 − 3θ + 1 = 0. Thus πn = A + B(1/2)n .

335

Treating the previous paragraph as purely exploratory we observe that 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 B giving 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 B giving π ∈ X.

336

K289 (i) Observe that πJ > 0 and set h = πJ−1 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 ke(k) − e(l)k1 = 2 for k 6= l.) Thus X does not have the Bolzano– Weierstrass property. However (see Exercise K218) Y = {y : hi ≥ |yi | for i ≥ 1}

does so, since X is closed, so does XJ = X ∩ Y . Observe that π = khk−1 1 h ∈ XJ .

Also if q ∈ XJ then T q ∈ X and ∞ X (T q)j = qi pij i=1

≤

∞ X

hi pij = hj

i=1

so T q ∈ XJ . Thus by Exercise K186 T n q → π for all q ∈ XJ . (iii) In particular T n e(J) → π. But J is arbitrary.

P Now if q ∈ X and ² > 0 we can find M such that M r=1 qr > 1 − ². PM Write u = q − r=1 qr e(r). Then °Ã ° ° ! ° M M M ° ° X ° ° X X ° ° ° ° qr π ° + kT n uk1 kT n q − πk1 ≤ °T n qr e(r) − qr π ° + ° 1 − ° ° r=1 ° ° r=1 r=1 1

≤

≤

M X r=1

M X r=1

→ 2²

1

qr kT n e(r) − πk1 + ² + kuk1

qr kT n e(r) − πk1 + 2²

as n → ∞. Since ² was arbitrary, kT n q − πk1 → 0 as required.

337

K290 For uniqueness. Suppose T x = T y. Then T n x = T n−1 T x = T T y = T n y so, by the contraction mapping theorem, x = y. n−1

For existence. By the contraction mapping theorem we can find an x0 with T n x0 = x0 . Now T n (T x0 ) = T (T n x0 ) = T x0 so, by uniqueness, T x0 = x0 . In question K282, T n has a fixed point but is not a contraction mapping. Desired result trivially true for n = 0. If true for n then ¯ ¯Z t ¯ ¯ n+1 n+1 ¯ |(T h)(t) − (T k)(t)| ≤ ¯ φ(h(s), s) − φ(k(s), s) ds¯¯ a Z t ≤ |φ(h(s), s) − φ(k(s), s)| ds a Z t ≤ |M (h(s) − k(s))| ds a Z t M n+1 ≤ kh − kk∞ (s − a)n ds n! a n+1 M kh − kk∞ (t − a)n+1 . ≤ (n + 1)! The induction can proceed. We thus have kT n h − T n kk ≤

1 n M (b − a)n kh − kk∞ . n!

Since 1 n M (b − a)n → 0 n! (observe that un+1 /un → 0) we can find an N such that 1 kT N h − T N kk ≤ kh − kk∞ 2 for all h, k ∈ C([a, b]). Thus there exists a unique f ∈ C([a, b]) with Z t f (t) = g(t) + φ(f (s), s) ds un =

a

and so (if g(t) = c) a unique solution of

f 0 (t) = φ(f (t), t) with initial condition f (a) = c.

338

K291 Consider Rn with its usual Euclidean norm and recall that it is complete. Set n X T x = y, with yi = aij f (xj ) + bi j=1

then, using Cauchy-Schwarz inequality and the mean value inequality, we have  !2 1/2 Ã n n X X aij (f (uj ) − f (vj )  kT u − T vk =  j=1

i=1

≤

Ã

≤

Ã

Ã n n X X Ã

n n X X

a2ij

!Ã

j=1

= M ku − vk with K ≥ 0 and

!Ã

j=1

i=1

i=1

a2ij

= Kku − vk 2

Ã

n X j=1

K =M

2

n X (f (uj ) − f (vj ))2 j=1

n X (M |uj − vj |2 ) j=1

a2ij

!1/2

n X

a2ij .

j=1

If K < 1 the contraction mapping theorem applies.

!!1/2

!!1/2

339

K292 A little fiddling around (we could write y −α dy = dx obtaining, in a purely exploratory, non-rigorous manner (1 − α)y 1−α = x + A) shows that, 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 need s ≥ 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) = 0 for all t is unique.

340

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 constant on [0, c], then y(t) = 0 for t ∈ [0, c]. Otherwise, since a continuous function on a closed bounded interval is bounded and attains its bounds, y must 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 0 (α) = f (α, y(α)) > 0 which is impossible. Thus y(t) = 0 for t ∈ [0, c] for all c > 0 so y(t) = 0 for t ≥ 0 and, similarly, for all t. Thus F has at most one solution. Since y = 0 is a solution it is the unique solution. (ii) If E = {(x, t) : k(x, t)k ≥ δ} with δ > 0 then E is closed and the restriction of f is continuous on each of the three closed sets, and

E1 = {(x, t) ∈ E : x ≥ t2 }, E2 = {(x, t) ∈ E : |x| ≤ t2 } E3 = {(x, t) ∈ E : x ≤ −t2 }

S and so the restriction f |E is continuous on E = 3i=1 Ei . Thus f is continuous at every (x, t) with k(x, t)k > δ with δ > 0 and so at every (x, t) 6= (0, 0) Since |f (t, u)| ≤ 2 max(|x|, |t|), f is continuous at (0, 0) (Lots of other ways to do this, many probably better.) Now y1 (t) =

Z

t 2

f (s, s ) ds = 0

Z

t 0

(−2s) ds = −y0 (t)

for t ≥ 0 and a similar calculation applies when t < 0. Thus yn = (−1)n y0 .

341

K294 Suppose, if possible, that y(τ ) ≤ 0 for some τ > 0. Let σ = inf{s ≥ 0 : y(s) ≤ 0}.

By continuity, y(σ) = 0. By the mean value theorem, we can find an s with 0 < s < σ such that |y(s)|β ≤ y 0 (s) = s−1 (y(s) − y(0)) < 0

which is absurd. Thus y(t) > 0 for all t, so y 0 (t) > 0 for all t, so y is strictly increasing, so y 0 (t) ≥ y(0)β = 1 so y(t) ≥ 1 + t for all t ≥ 0. On (tj , tj+1 ), we have y 0 (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 so

tj+1 − tj ≤ 2j(1−β) .

Thus tr =

r−1 X j=0

for all r and so

tj+1 − tj ≤

a≤

r−1 X j=0

2j(1−β) ≤

1 1 − 21−β

1 . 1 − 21−β

(ii) Observe that, if y 0 = 1 + y 2 , then y 0 ≥ y 2 . R (iii) Recall that (w log w)−1 dw = log log w + a. We have y(t) = exp(exp t) as a solution.

342

K295 Observe that y((r + 1)h) − y(rh) ≈ y 0 (rh). h Observe that |f (t, u) − f (t, v)| ≤ sup |f,2 (t, s)||u − v| ≤ K|u − v| s

so we have a (global) Lipschitz condition. By the chain rule y 00 exists and y 00 (t) = f,1 (t, y(t)) + y 0 (t)f,2 (t, y(t)) = f,1 (t, y(t)) + f (t, y(t))f,2 (t, y(t)). Thus |y 00 (t)| ≤ L for all t and so |y(nh + s) − y(nh) − y 0 (nh)s| ≤

Ls2 2

whence |y((n + 1)h) − y(nh) − f (nh, y(nh))h| ≤

Lh2 2

and |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 − y n+1 + yn |

≤ |y(nh) − yn | + |y((n + 1)h) − y(nh) − f (nh, y(nh))h| + |f (nh, y(nh))h − f (nh, y n )h| Lh2 + Kh|y(nh) − yn | 2 Lh2 . = (1 + Kh)|y(nh) − yn | + 2

≤ |y(nh) − yn | +

Thus if |y(nh) − yn | ≤ it follows that

¢ Lh ¡ (1 + Kh)n+1 − 1 , 2K

¢ Lh2 LhN ¡ n+1 |y((n + 1)h) − yn+1 | ≤ (1 + Kh) (1 + Kh) −1 + 2K 2 ¢ Lh ¡ (1 + Kh)n+2 − 1 . ≤ 2K Since the result is true for n = 0 it is true inductively. In particular |y(N h) − yN | ≤

¢ Lh ¡ (1 + Kh)N +1 − 1 . 2K

343

K296* No comments.

344

K297 As in K295, |y((n + 1)h) − y(nh) − y 0 (nh)h| ≤

Lh2 , 2

and then, much as in K295, |y((n + 1)h) − y˜n+1 | ≤ (1 + Kh)|y(nh) − yn | +

Lh2 +² 2

so, by induction, |y(nh) − y˜n | ≤ Thus

µ

² Lh + 2K Kh

|y(a) − y˜N | ≤

µ

¶

¡

¢ (1 + Kh)n+1 − 1 .

² Lh + 2K Kh

¶

(eaK − 1).

The error due to machine inaccuracy increases directly with the number of computations. (ii) and (iii) Recall that µ ¶ µ 1/2 ¶2 A A 1/2 1/2 −B h + 2A1/2 B 1/2 + Bh = h h1/2

takes its minimum value 2A1/2 B 1/2 when h = (A/B)1/2 . We want h = (2²/L)1/2 giving minimum error (2L)1/2 K −1 ²1/2 .

345

K298 (i) Set f (t, y(t)) = g(t), M = L. Either redo calculations with K = 0 or observe that ¯ ¯Z N −1 ¯ ¯ a X LhN eaK − 1 ¯ ¯ [hN ] g(rh) = |y(a) − y | ≤ g(t) dt − h ¯ ¯ N ¯ ¯ 0 2 K n=0

→

aM h aLhN = 2 2

as K → 0+. Take κa = M a/2. (ii) We have Z Z a N −1 X G(rh) = G(t) dt − h 0

n=0

π 0

1 + sin 2N t π π dt = = kGk∞ . 2 2 2

(iii) No. Consider f (t, u) = G(t) in (ii).

346

K299 (i) The result is trivially true for n = 0. If it is true for n then (uv)(n+1) (x) = ((uv)0 )(n) (x) = (u0 v + uv 0 )(n) (x) n µ ¶ n µ ¶ X X n (n+1−r) n (n−r) (r) = u (x)v (x) + u (x)v (r+1) (x) r r r=0 r=0 ¶¶ µ n µµ ¶ X n n (n+1) u(n+1−r) (x)v (r) (x) + u(x)v (n+1) (x) + =u (x)v(x) + r − 1 r r=1 µ ¶ n+1 X n+1 = u(n+1−r) (x)v (r) (x) r r=0

so the induction can proceed. (ii) Differentiating we get

x(1 + x2 )−1/2 y(x) + (1 + x2 )1/2 y 0 (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 0 (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) + n2 y (n−1) (0) = 0

Since y(0) = 0 and y 0 (0) = 1 (by setting x = 0 in the first two displayed equations in the statement of (ii)) we have y (2n) (0) = 0 and y (2n+1) (0) = 2n n!. Set un = y (n) (0)/n!. Then |u2n+1 x2 | x2 |u2n−1 | 2n + 1 which tends to zero if |x| ≤ 1 but diverges otherwise. Thus the Taylor series for y has radius 1. P n Now ∞ term by term with in its circle j=0 un x may be differentiated P∞ n of convergence so u(x) = j=0 un x is a solution for the given differential equation with the same value as y at 0. Since the Lipschitz conditions apply, y(t) = u(t) for |t| < 1 so ∞ X y (n) (0) n y(x) = x n! j=0 for |x| < 1.

347

K300 Since K is continuous on the closed bounded subset [0, 1]2 of R2 , it is bounded so there exists an M with M ≥ K(s, t) for (s, t) ∈ [0, 1]2 . If f ∈ C([0, 1]), set (T f )(t) = g(t) + λ

Z

1

K(s, t)f (s) ds. 0

We observe that T f ∈ C([0, 1]) since K is uniformly continuous (because K is continuous on the closed bounded subset [0, 1]2 ) and so ¯ ¯Z 1 ¯ ¯ ¯ |(T f )(u) − (T f )(v)| = ¯ (K(u, s) − K(v, s))f (s) ds¯¯ 0

≤ kf k∞

as u → v.

sup

kx−yk≤|u−v|

|K(x) − K(y)| → 0

Further, if |λ| < M −1 , T is a contraction mapping on the complete normed space (C([0, 1]), k k∞ ) since ¯ ¯Z 1 ¯ ¯ ¯ K(s, t)(f (s) − h(s) ds¯¯ |(T f )(t) − (T h)(t)| = |λ| ¯ 0 Z 1 ≤ |λ| |K(s, t)(f (s) − h(s)| ds 0

≤ |λ|M kf − hk∞

so kT f − T hk∞ ≤ M kf − hk∞ . The contraction mapping theorem now gives the required result.

348

K301 (ii) Observe that ¯ ¯ u (t) d ¯¯ 01 u1 (t) dt ¯¯ 00 u1 (t) ¯ 0 ¯ u (t) ¯ 1 = ¯¯ u01 (t) ¯u00 (t) 1

u2 (t) u02 (t) u00 2 (t) u02 (t) u02 (t) u00 2 (t)

¯ u3 (t) ¯¯ u03 (t) ¯¯ ¯ u00 3 (t) ¯ ¯ u03 (t) ¯¯ ¯¯ u1 (t) 0 u3 (t) ¯¯ + ¯¯u00 1 (t) ¯ ¯u00 (t) u00 3 (t) 1

u2 (t) u00 2 (t) u00 2 (t)

¯ ¯ ¯ u1 (t) u2 (t) u3 (t) ¯¯ ¯ 0 0 0 ¯ u2 (t) u3 (t) ¯¯ = 0 + 0 + ¯ u1 (t) ¯u000 (t) u000 (t) u000 (t)¯ 1 2 3 ¯ ¯ u (t) 1 ¯ u01 (t) = ¯¯ ¯−a(t)u00 (t) − b(t)u0 (t) − c(t)u1 (t) 1 1 ¯ 0 ¯ ¯ u (t) u0 (t) u0 (t) ¯ 2 3 ¯ 01 ¯ 0 0 ¯ = −a(t) ¯ u1 (t) u2 (t) u3 (t) ¯¯ ¯u00 (t) u00 (t) u00 (t)¯ 1

0

2

3

so W (t) = −a(t)W (t).

¯ ¯ u3 (t) ¯¯ ¯¯ u1 (t) 00 u3 (t)¯¯ + ¯¯ u01 (t) ¯ ¯u000 (t) u00 3 (t) 1

u2 (t) u02 (t) u000 2 (t)

¯ u3 (t) ¯¯ 0 u3 (t) ¯¯ ¯ u000 3 (t)

u2 (t) u02 (t) 0 −a(t)u00 2 (t) − b(t)u2 (t) − c(t)u2 (t)

¯ ¯ u3 (t) ¯ ¯ u03 (t) ¯ 00 0 −a(t)u3 (t) − b(t)u3 (t) − c(t)u3 (t)¯

349

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 /g is well defined continuous function on [a, b] which is differentiable on (a, b). By Rolle’s theorem there exists a c ∈ (a, b) with µ ¶0 f f 0 (c)g(c) − g 0 (c)f (c) W (c) 0= (c) = = . g (g 0 (c))2 (g 0 (c))2

Thus W (c) = 0 contrary to the hypothesis. Thus there must be a zero of g between any two zeros of f and, similarly, a zero of f between any two zeros of g. [Moreover the zeros must be simple, if f (c) = f 0 (c) = 0, then W (c) = 0.]

350

K303 (i) Without loss of generality suppose that λ1 6= 0. Then without loss of generality we may suppose λ1 = 1 and set λ2 = λ. Then µ ¶ f1 f2 W (f1 , f2 )(x) = det f1 f20 µ ¶ f1 + λf2 f2 = det 0 f1 + λf20 f20 ¶ µ 0 f2 = 0. = det 0 f20 However, if g1 (x) = (x − 1)2 for x ≥ 1 and g2 (x) = g1 (−x) (we could play the same trick with infinitely differentiable functions) then g1 and g2 are continuously differentiable and their Wronskian vanishes everywhere but they are not linearly dependent. (ii) By Exercise 12.24 applied to the vectorial equation ¶ ¶ µ µ d y(t) v1 (t) , = −a1 (t)v(t) − a2 (t)y1 (t) dt v(t) it follows that if a1 and a2 are continuous (so bounded on [a, b]), the equation y 00 (x) + a1 (x)y 0 (x) + a2 y(x) = 0 has exactly one solution with y(a) = A and y 0 (a) = B. If W (f1 , f2 )(x) never vanishes then setting ¶ µ ¡ ¢−1 f1 (x) f2 (x) a1 (x) = W (f1 , f2 )(x) det 00 f1 (x) f200 (x)

and

µ 0 ¶ ¡ ¢−1 f1 (x) f20 (x) a2 (x) = W (f1 , f2 )(x) det 00 f1 (x) f200 (x)

we have a1 and a2 continuous. Thus the equation F

y 00 (x) + a1 (x)y 0 (x) + a2 y(x) = 0

has exactly one solution with y(a) = A and y 0 (a) = B. But F 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 λ1 f1 (a) + λ2 f2 (a) = y(a) λ1 f10 (a) + λ2 f20 (a) = y 0 (a)

351

have a solution, so, if y is a solution, the uniqueness result of the previous paragraph shows that y(x) = λ1 f1 (x) + λ2 f2 (x) 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 00 (x) + b1 (x)y 0 (x) + b2 y(x) = 0 then if W (f1 , f2 ) vanishes at one point it vanishes everywhere.

352

K304 Observe that 0 = (u00 v + 2v 0 u0 + v 00 u) + p(u0 v + v 0 u) + qvu = v(u00 + pu0 + qu) + v 00 u + v 0 (2u0 + pu) = v 00 u + v 0 (2u0 + pu) so, writing w = v 0 we have †

w0 (x)u(x) + w(x)(2u0 (x) + p(x)u(x)) = 0

Observe that † is linear so, if w is a solution, so is Bw. Thus µ ¶ Z x y(x) = u(x)v(x) = u(x) A + B w(t) dt 0

is a solution. Since it contains two arbitrary constants we would expect it to be the general solution. (ii) We try y(x) = v(x)(exp x) obtaining v 00 (x) = (2x − 1)v 0 (x)

so v 0 (x) = B(exp(x2 − x)) and

y(x) = A exp(x) + B exp(x)

Z

t 0

exp(x2 − x) dx.

(iii) y(x) = (A + Bx) exp x. (iv) If u is a solution, try y = uv, obtaining 0 = (u000 v + 3v 0 u00 + 3v 00 u0 + v 000 u) + p1 (u00 v + 2u0 v 0 + v 00 u) + p2 (u0 v + v 0 u) + p3 vu = w00 u + w0 (3u0 + p1 u) + w(3u00 + 2u0 p1 + p2 ) a second order equation for w = v 0 .

353

K305 (i) We have f = (u001 y1 + 2u01 y10 + u1 y100 + p(u01 y1 + u1 y10 ) + qu1 y1 ) + (u002 y2 + 2u02 y20 + u2 y200 ) + p(u02 y2 + u2 y20 ) + qu2 y2 = 2u01 y10 + u001 y1 + pu01 y1 + 2u02 y20 + u002 y2 + pu02 y2 . (ii) If we differentiate ††, then we get

0 = u001 y1 + u01 y1 + u002 y2 + u02 y2

so † becomes

f = u01 y10 + u02 y20 .

In other words, we have the pair of equations u01 y10 + u02 y20 = f u01 y1 + u02 y2 = 0 so, since W never vanishes, u01 = −W −1 f y2 , and u02 = W −1 f y1 . (iii) Performing the integration we get y(x) = A1 y1 (x) + A2 y2 (x) Z a Z −1 + y1 (x) W (s)f (s)y2 (s) dx + y2 (x) x

x

W −1 (s)f (s)y1 (s) dx b

and, if y1 and y2 are chosen as in our discussion of Green’s function, we obtain our standard Green’s function solution plus the general solution of 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 u01 (x) = −

ex ex e−x −x 0 = + e − 1 and u (x) = . 2 1 + ex 1 + 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) = A1 ex + A2 e−x + (e−x − ex ) log(1 + ex ) − ex + 1. (v) Set y = y1 u1 + y2 u2 + y3 u3 subject to 0 = y1 u01 + y20 u02 + y3 u03 0 = y10 u01 + y20 u02 + y30 u03 .

354

We get f = y100 u1 + y200 u2 + y300 u3 . Since the Wronskian never vanishes, we can solve the last three equations to find u01 , u02 and u03 . Integrating now gives u1 , u2 , u3 and thus the solution.

355

K306 Observe that since K is continuous on the closed bounded set S it is uniformly continuous. Thus ¯Z 1 ¯ ¯ ¯ ¯ |(TK (f ))(s) − (TK (f ))(t)| = ¯ K(s, y) − K(t, y)f (y) dy ¯¯ 0

≤ kf k∞

sup

kz−wk≤|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)| ≤ kf k∞ kKk∞

so TK is continuous and kT k ≤ kKk∞ .

(i) True, since kTKn − TK k = kTKn −K k ≤ kKn − Kk∞ .

(ii) False. If Kn (x, y) = max(0, 1 − n2 (x2 + y 2 )) then kKn k∞ = 1 for all n but Z 1/n |(TKn (f ))(s)| ≤ |f (y)| dy ≤ kf k∞ /n 0

so kTKn k ≤ n

−1

→ 0.

(iii) True. Since TK − TL = TK−L , it suffices to show that TK = 0 implies K = 0. Suppose K 6= 0. Then we can find (x, y) ∈ S such that K(x, y) 6= 0. Without loss of generality suppose K(x, y) = m > 0. By continuity we can find a δ > 0 such that (s, t) ∈ S and |s−x|, |t−y| ≤ δ implies K(s, t) ≥ m/2. Set f (t) = max(0, 1 − 2δ −1 |t − y|). Then TK f (x) > 0 so TK 6= 0. (iv) False. Let T f (x) = f (1/2) for all [x ∈ [0, 1]. Then T is linear and continuous with kT k ≤ 1. We claim that T 6= TK for all K. For consider fn (x) = max(0, 1 − n|x − 12 |). T fn = 1 for all n but, if n ≥ 2, Z 1 +1/n 2 |TK fn | ≤ kKk∞ ds = 2kKk∞ /n → 0 as n → ∞.

1 −1/n 2

356

K307 (ii) We apply Exercise 12.24 to the vectorial equation µ ¶ µ ¶ d y1 (t) v1 (t) = −a(t)v1 (t) − b(t)y1 (t) dt v1 (t) to obtain the existence and uniqueness of y1 . R1 Rt (iii) We have (noting that t u(x) dx = − 1 u(x) dx) Z 1 Z t 0 ˜ ,2 (s, t)f (s) ds ˜ t)f (t) + H H,2 (s, t)f (s) ds − H(t, y (t) = H(t, t)f (t) + t 0 Z t Z 1 = H,2 (s, t)f (s) ds + H,2 (s, t)f (s) ds 0

t

and

t

Z

˜ ,2 (t, t)f (t) + y (t) = H,2 (t, t)f (t) + H,22 (s, t)f (s) ds − H 0 Z t Z t ˜ ,22 (s, t)f (s) ds. H H,22 (s, t)f (s) ds + = f (t) + 00

Z

t

˜ ,22 (s, t)f (s) ds H

0

0

0

Summing gives

y 00 (t) + a(t)y 0 (t) + b(t)y(t) Z t = f (t) + (b(t)H,2 + a(t)H,2 (s, t) + H,22 (s, t))f (s) ds 0 Z t ˜ ,2 + a(t)H ˜ ,2 (s, t) + H ˜ ,22 (s, t))f (s) ds + (b(t)H 0

= f (t).

Further y(0) =

Z

and similarly y(1) = 0.

1 0

˜ 1)f (s) ds = H(s,

Z

1

0 ds = 0 0

357

K308 We seek a twice continuously differentiable function y(x) = G(x, t) which is four times differentiable (using left and right derivatives at end points) on [0, t) and (t, 1] and satisfies y(0) = y 0 (0) = 0, y (4) (x) − k 4 y(x) = 0 for x ∈ [0, t)

and y(1) = y 0 (1) = 0,

y (4) (x) − k 4 y(x) = 0 for x ∈ (t, 1]

whilst This gives

y 000 (t+) − y 000 (t−) = 1.

y(x) = a sinh kx + b sin kx + c cosh x + d cos x 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 condition on the discontinuity for the in the next paragraph) ¡ ¢ ¡ ¢ M A B C D = 0 0 0 1 so

where

¢  ¡ ¡ sinh kt − sin kt ¢  k cosh kt − cos kt ¡ ¢ M =  k2 sinh kt − sin kt ¡ ¢ 3 k cosh kt + cos kt

¡ ¢ ¡cosh kt − cos kt ¢ k ¡ sinh kt + sin kt ¢ k2 ¡ cosh kt − cos kt¢ k3 sinh kt + sin kt

and second derivatives and the third gives (subject to the check ¡

A B C D

¢T

¡ ¢T = M −1 0 0 0 1

¡ −¡ sinh k(1 − t) + sin k(1 − t)) k cosh k(1 − t) + cos k(1 − t)) ¡ −k¡2 sinh k(1 − t) − sin k(1 − t)) k3 cosh k(1 − t) − cos k(1 − t))

¡ ¢  −¡ cosh k(1 − t) − cos k(1 − t)¢ k ¡sinh k(1 − t) + sin k(1 − t) ¢ .  −k 2¡ cosh k(1 − t) + cos k(1 − t) ¢ 3 k sinh k(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

358

we must have and y(1) = y 0 (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

µ¡ ¢ ¡ ¢¶ sinh k − sin k cosh k − cos k ¢ ¡ ¢ det ¡ 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 so C = D = 0). If k = nπ then, by direct calculation, the Wronskian of our four functions (which is a multiple of det M ) vanishes at 1 so det M = 0 everywhere and our method fails. (The method will also fail for simpler reasons in the case k = 0.) [In fact we are in a situation like the last example of Chapter 12 and there is no solution.]

359

K309 We seek a continuous function y(x) = G(x, t) which is twice differentiable (using left and right derivatives at end points) on [0, t) and (t, 1]) and satisfies y(0) = y 0 (0) = 0, y 00 (x) + 2βy 0 (x) + ω 2 y(x) = 0 for x ∈ [0, t)

and

y 00 (x) + 2βy 0 (x) + ω 2 y(x) = 0 for x ∈ (t, ∞)

whilst

y 0 (t+) − y 0 (t−) = 1.

This gives

y(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 first derivative gives 1 = αA. Thus G(x, t) =

(

0 if x ≤ t −1 −β(x−t) α e sinh(α(x − t))

so, if β > ω, the general solution on [0, ∞) of

y 00 (x) + 2βy 0 (x) + ω 2 y(x) = f (x)

is y(t) =

Z

∞

G(s, t)f (s) ds = α

0

−1

Z

0 2

t

f (s)e−β(t−s) sinh(α(t − s)) ds

with α the positive square root of β − ω 2 . 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 ω 2 − β 2 . Arguing much as above we get B = 0, A = λ−1 and the general solution on [0, ∞) of y 00 (x) + 2βy 0 (x) + ω 2 y(x) = f (x)

as y(t) = λ

−1

Z

t 0

f (s)e−β(t−s) sin(λ(t − s)) ds.

360

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) .

Arguing much as before we get B = 0, A = 1 and the general solution on [0, ∞) of y 00 (x) + 2βy 0 (x) + ω 2 y(x) = f (x)

as y(t) =

Z

0

t

(t − s)f (s)e−β(t−s) ds.

If β is large the pointer takes a long time to return to close to equilibrium. If β is small the pointer overshoots and we have violent oscillation so it also takes a long time before the pointer remains close to equilibrium. As a rule of thumb β close to ω is preferred.

361

K310 (i) Observe that, if m ≥ n m m X X kT kn+1 r →0 kSn − Sm k ≤ kT k ≤ kT kr ≤ 1 − kT k r=n+1 r=n+1

as n → ∞, so the sequence Sn is Cauchy. Thus we can find an S ∈ L(U, U ) with kSn − Sk → 0. Now kSk ≤ kSn − Sk + kSn k ≤ kSn − Sk + ≤ kSn − Sk + (1 − kT k)

−1

n X r=0

kT kr

→ (1 − kT k)−1

so kSk ≤ (1 − kT k)−1 . A similar calculation gives kI − Sk ≤ kT k(1 − kT k)−1 .

(ii) Observe that kS(I − T ) − Ik ≤ kSn (I − T ) − Ik + k(S − Sn )(I − T )k = kT n+1 k + k(S − Sn )(I − T )k

≤ kT kn+1 + kS − Sn kkI − T k → 0

as n → 0, so kS(I − T ) − Ik = 0 and S(I − T ) = I. Similarly (I −T )S = I. Thus I −T is invertible with inverse S. Setting T = I −A gives the result. (iii) Observe that kA − Ik = kB −1 B − B −1 Ck ≤ kB − CkkB −1 k < 1

so A is invertible. Thus

A−1 B −1 C = I and CA−1 B −1 = BB −1 CA−1 B −1 = BB −1 = I so C is invertible with inverse A−1 B −1 . We observe that kC −1 k ≤ kB −1 kkA−1 k ≤ kB −1 k(1 − kI − Ak)−1 = kB −1 k(1 − kI − Ak)−1

but so

1 > kB − CkkB −1 k ≥ kA − Ik ≥ 0 kC −1 k ≤ kB −1 k(1 − kB − CkkB −1 k)−1 .

Similar arguments give the second inequality. (iv) If B ∈ E, then part (ii) shows that

Γ = {C ∈ L(U, U ) : kB − Ck < kB −1 k−1 } ⊆ E

362

so E is open. If C ∈ Γ then

kΘ(C) − Θ(B)k ≤ kBk−1 kkB − Ck(1 − kBk−1 kB − Ck)−1 → 0

as kC − Bk → 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 −1 H))−1 = (I + B −1 H)−1 B −1 = (I − B −1 H + ²(B −1 H)kHk)B −1 = B −1 − B −1 HB −1 + η(H)kHk

= Θ(B) + ΦB (H) + η(H)kHk. where

kη(H)k = kB −1 ²(B −1 H)B −1 k ≤ kB −1 k2 k²(B −1 H)k → 0

as kHk → 0.

If n = 1, this reduces to the statement that d cx−1 = −cx−2 . dx

363

K311 (ii) If kAk = 0, then A = 0 and there is nothing to prove so we assume kAk 6= 0. ∆ exists since any non-empty set of real numbers bounded below has an infimum. Now choose ² > 0. We can find a j such that kAj k1/j ≤ ∆ + ². Thus ¢1/(jk+r) ¡ kAjk+r k1/(jk+r) ≤ kAj kk kAkr ≤ (∆ + ²)jk/(jk+r) kAkr/(jk+r) → ∆ + ²

for all 0 ≤ r ≤ j − 1. Thus

∆ ≤ lim inf kAn k1/n ≤ lim sup kAn k1/n ≤ ∆ + ² n→∞

n 1/n

and since ² > 0 kA k

n→∞

→ ∆.

(iii) and (iv) If ρ(A) < 1, set λ = (ρ(A) + 1)/2. There exists an n N such that kAn k1/n ≤ λ and so kAkP ≤ λn for n ≥ N . The kind ∞ of arguments used in K310 show that j=0 Aj converges and that the limit is the inverse of I − A.

n 1/n If ρ(A) > 1 then there exists an N such ≥ 1 for all P∞ thatj kA k n ≥ N so, by the easy part of the GPC, j=0 A fails to converge.

364

K312 (i) k(λα)n k1/n = |λ|kαn k1/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 Pm eigenvectors ej with Q eigenvalues λj are linearly independent. (If m µ e = 0 apply j=1 j j j6=k (α − λj ι) to both sides.) They thus form a basis. If the standard basis is uj , let θ be the unique linear map with θej = uj . Observe that k(θαθ−1 )n k1/n = kθαn θ−1 k1/n ≤ kθ−1 k1/n kαn k1/n kθk1/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 the standard basis, φ has norm equal to the absolute value of the largest diagonal entry so ρ(θαθ −1 ) = max |λj |. (iv) If u1 and u2 form the standard basis let α, β be the linear maps given by αu1 = u2 , αu2 = 0, βu2 = u1 , βu1 = 0.

365

K313 (i) This is just the formula for change of basis. (ii) Here and elsewhere we use the fact that m2 sup |aij | ≥ kαk ≥ sup |aij | 1≤i,j≤m

1≤i,j≤m

when (aij ) is the matrix of α with respect to the matrix. (In fact we merely use K sup |aij | ≥ kαk ≥ K −1 sup |aij | 1≤i,j≤m

1≤i,j≤m

for some K and this follows from the fact that all norms on a finite dimensional 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 distinct eigenvalues and kβn − βk → 0. Let αn = θ−1 βn θ. We know that αn has the same eigenvalues as βn and kα − αn k = kθ−1 (β − βn )θk ≤ kθ −1 kkβ − βn kkθk → 0

as n → 0.

(iv) If α has m distinct eigenvalues we know that we can find a basis of eigenvectors. With respect to this basis, α is a diagonal matrix D with entries the eigenvalues so m Y χα (t) = det(tI − D) = (t − λj ) j=1

and χα (D) = 0 the zero m × m matrix. Thus χα (α) = 0. 2

(v) We know that the map from L(Cm , Cm ) to C m given by α 7→ (aij ) is continuous so the map α 7→ χα (A) is continuous. But the inverse map (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 kα − αn k → 0 with χαn (αn ) = 0, so, by continuity, χα (α) = 0 (vi) We need αβ − βα non singular. One (among many ways) of doing this is as follows. Take e1 , e2 , . . . , em to be a basis of Cm . Let α be the linear map with αej = αej+1 for 1 ≤ j ≤ m − 1, αem = αe1 and β be the linear map with βej = jαej for 1 ≤ j ≤ m.

366

K314 (i) We know that, given any ² > 0, we can find a(²) ≥ 1 and b(²) ≥ 1 such that kαr k ≤ a(²)(ρ(α) + ²)r and kβ r k ≤ b(²)(ρ(β) + ²)r

for all r ≥ 0. Thus

° n µ ¶ ° °X n ° ° ° k(α + β)n k = ° αj β n−j ° ° ° j j=0 n µ ¶ X n kαj kkβ n−j k ≤ j j=0 n µ ¶ X n ≤ a(²)b(²) (ρ(α) + ²)j (ρ(β) + ²)n−j j j=0 = a(²)b(²)(ρ(α) + ρ(β) + 2²)n

so, taking n-th roots and allowing n → ∞,

ρ(α + β) ≤ (ρ(α) + ρ(β) + 2²).

Since ² > 0 this gives the result.

(ii) Observe that, given an upper triangular matrix A, we can find a diagonal matrix B with entries in absolute value less than any give η > 0 such that the diagonal entries of A + B are all distinct. Observe that 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 kαn e1 k/ke1 k = |λ1 |n , we have ρ(α) ≥ |λ1 |. However, if λ > |λ1 | we have ° ° ¯ m m ¯ ° X ° X ¯ λj ¯ ° ° n n ¯ ¯ λ °α xj ej ° ≤ ¯ λ ¯ kej k → 0 ° ° j=1

j=1

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, kβr k ≤ 1/r and α and βr commute. Observe that, by (i), and

ρ(α + βr ) ≤ ρ(α) + ρ(βr ) ≤ ρ(α) + kβr k = ρ(α) + 1/r ρ(α) = ρ(α + βr − βr ) ≤ ρ(α + βr ) + 1/r

so ρ(α + βr ) → ρ(α) and the desired result follows.

367

(iv) If α is represented by A with respect to one basis and by B with respect to another, we can find an invertible θ such that θαθ −1 is represented by B with respect to the first basis. Now k(θαθ−1 )n k = kθαn θ−1 k ≤ kθkkαn kkθ−1 k,

so, taking n-th roots and letting n → ∞, we have ρ(θαθ −1 ) ≤ ρ(α) and similarly ρ(α) = ρ(θ −1 θαθ−1 θ) ≤ ρ(θαθ −1 )

so ρ(θαθ−1 ) = ρ(α) and there is no ambiguity.

368

K315 (i) Observe that n

xn = α x0 +

n−1 X

αj b

j=0

−1

so xn → (ι − α) b.

(ii) Observe that the map τ : Cm → Cm given by τ x = b + αx satisfies kτ n x − τ n yk = kαn (x − y)k ≤ kαn kkx − yk.

For large enough n we have τ n a contraction mapping so xn → c where c is the unique fixed point given by τc = c i.e. by αc = b. P j Suppose ρ(α) > 1. If n−1 j=0 α b diverges, then take x0 = 0. If not, take x0 to be an eigenvector corresponding to a largest (in absolute value eigenvalue). Pn−1 j Suppose ρ(α) = 1. If j=0 α b diverges, then take x0 = 0. If not, then either all of the largest in absolute value eigenvalues are 1 and we consider x0 = ±e with e an associated eigenvector to get two fixed points (with necessarily different limits) or one of the largest in absolute value eigenvalues, is not 1. Take x0 = e with e an associated eigenvector to get a non-convergent sequence.

369

K316 We require ρ(I − A) < 1. If kI − Ak is small, convergence will be rapid with the error roughly multiplied by kI − Ak at each step.

Jacobi needs ρ(D −1 (U +L)) < 1 and Gauss needs ρ((D−L)−1 U ) < 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 of its diagonal elements we have det H = (det(D − ωL))−1 det((1 − ω)D + ωU ) = (det D)−1 det((1 − ω)D) = (det D)−1 (1 − ω)n det D = (1 − ω)n .

Now the determinant of a matrix is the product of its eigenvalues (multiple roots counted multiply) (consider the coefficient of t0 in det(tI − A)) so, if | det H| > 1, we have ρ(H) > 1. Thus the scheme fails for ω < 0 or ω > 2.

370

K317 From K310 we know that the map α 7→ α−1 is continuous on the open subset of L(U, U ) where the inverse is defined. Thus thus composition of the maps is continuous.

x 7→ Df (x), α 7→ α−1

371

K318 (i) Observe that, if m ≥ n, ° ° m m n °X α r X r° X kαkr α ° ° − →0 °≤ ° ° r=0 r! r! ° r=n+1 r! r=0

P Pn r r (since m r=0 (x /r!) converges for all x) so, by completeness, r=0 α r! converges.

(ii) Observe that ° n ° ° ° n n °X α r X ° r s° r° r r s X X X α β ° (α + β) ° ° β α β ° − − ° °=° ° ° r=0 r! r=0 r! ° ° r! r! s! 0≤r+s≤n r,s≥0 r! s! ° r=0 0≤r,s≤n ° ° ° r s° X β α ° ° =° ° ° ° r! s! r+s>n, 0≤r,s≤n ≤

≤

kαkr kβks r! s! 0≤r,s≤n

X

r+s>n,

kαkr kβks r! s! r,s≥0

X

2n≥r+s>n,

2n X 1 = (kαk + |βk)k → 0 k! k=n+1

[Or use results from Chapter 3.] (iii) Observe that, if 0 < khk < kαk/2 we have ° ¯ ¯ ° n n °X ¯ °X r¯ r hα hα ° ¯ ° ¯ − ι − hα¯ = ° ° ¯ ° ° ¯ r! r! ¯ r=2 r=0 ≤

n X |h|kαkr r=2

2

r!

≤ |h| kαk

2

∞ X r=0

2−r−1 = |h|2 kαk2 .

Thus exp(hα) = ι + hα + h2 θα (h)h2 where kθα (h)k ≤ kαk2 for 0 < khk < kαk/2. Similarly exp(hβ) = ι + hβ + h2 θβ (h)h2 where kθβ (h)k ≤ kβk2 for 0 < khk < kβk/2.

372

We thus have kh−2 (exp(hα) exp(hβ) − exp(hβ) exp(hα)) − (αβ − βα)k ¡ = kh−2 (ι + hα + h2 θα (h)h2 )(ι + hβ + h2 θβ (h)h2 ) ¢ − (ι + hβ + h2 θβ (h)h2 )(ι + hα + h2 θα (h)h2 ) − (αβ − βα)k

≤ 2|h|kαkkθβ (h)k + kβkkθα(h) k + 2h2 kθα(h) kkθβ (h)k → 0

as h → 0. Thus, if αβ 6= βα, we have exp(hα) exp(hβ) 6= exp(hβ) exp(hα) for h small. If exp α exp β = exp(α + β) and exp β exp α = exp(β + α) then exp α exp β = exp β exp α. (iv) Observe by considering the various terms in the expansion or by induction that n µ ¶ X n n n k(α + κ) − α k ≤ kκkj kαkn−j j j=1 ¶ µ n X n−1 kκkj kαkn−j ≤ n j − 1 j=1 ¶ n−1 µ X n−1 = kκk n kκkr kαkn−1−r r r=0 = nkκk(kαk + kκk)n−1

(We shall do something similar in K319.)

Thus ° N ° N N °X (α + κ)n X (α)n ° ° ° X 1 − k(α + κ)n − αn k ° °≤ ° n=0 ° n! n! n! n=0 n=1 ≤ kκk = kκk so that

N X n (kαk + kκk)n−1 n! n=1

N −1 X n=0

1 (kαk + kκk)n n!

= kκk exp(kαk + kκk) k exp(α + κ) − exp(α)k ≤ kκk exp(kαk + kκk) → 0

as kκk → 0. Thus exp is continuous.

373

K319 (ii) If α ∈ L(U, U ) and β ∈ L(U, U ) write

Φn (α)(β) = βαn−1 + αβαn−2 + α2 βαn−3 + · · · + αn−1 β.

It is easy to check that Φn (α) : L(U, U ) → L(U, U ) is a continuous linear function with kΦn (α)k ≤ nkαkn−1 , The next paragraph (not surprisingly) uses K260 (iii). Observe by considering the various terms in the expansion or by induction that n µ ¶ X n n n k(α + κ) − α − Φn (α)(κ)k ≤ kκkj kαkn−j j j=2 ¶ µ n X n−2 kκkj kαkn−j n(n − 1) ≤ j−2 j=2 ¶ n−2 µ X n−2 2 kκkr kαkn−2−r = n(n − 1)kκk r r=0 = n(n − 1)kκk2 (kαk + kκk)n−2

Thus Θn is differentiable at α with derivative Φn (α). (iii) Since kΦn (α)k ≤ nkαkn−1 we have N X kΦn (α)k j=1

PN

Φn (α) j=1 n!

n!

≤

N X nkαkn−1 j=1

n!

≤ ekαk

so 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 N N °X (α + κ)n X (α)n X Φn (α) ° ° X 1 ° − − κ° ≤ k(α + κ)n − αn − Φn (α)(κ)k ° ° ° n=0 n! n! n! n! n=0 n=1 n=1

N X 1 ≤ n(n − 1)kκk2 (kαk + kκk)n−2 n! n=2

= kκk2 2

so

N −2 X n=0

1 (kαk + kκk)n n!

≤ kκk exp(kαk + kκk) k exp(α + κ) − exp(α) − Φ(α)κk ≤ kκk2 exp(kαk + kκk)

and exp is differentiable at α with derivative Φ(α).

374

K320 Use the fact that m2 sup |aij | ≥ kαk ≥ sup |aij | 1≤i,j≤m

1≤i,j≤m

when (aij ) is the matrix of α with respect to the matrix. (Or merely use K sup |aij | ≥ kαk ≥ K −1 sup |aij | 1≤i,j≤m

1≤i,j≤m

for some K and this follows from the fact that all norms on a finite dimensional space are Lipschitz equivalent.) If A is upper triangular then Ar is upper triangular with (j, j) th entry the rth power of ajj the (j, j) th entry of A. Thus exp A is upper triangular with (j, j) th entry exp ajj . Thus Y X det(exp A) = exp ajj = exp ajj = exp Trace A

and, since any α has an upper triangular representation and det and Trace depend on α and not the representation chosen, det(exp α) = exp Trace α.

375

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 is absurd. (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 = ±I or a + d = 0 and, either a2 = 1 in which case bc = 0 and ¶ ¶ µ µ 1 0 1 β or A = ± A=± β −1 0 −1 for some β or a2 6= 1 and A=

µ

a b b−1 (1 − a2 ) −a

¶

for some a and b. Observe that 1 = det I = det A2 = (det A)2 so det A = ±1. By inspection the only such matrices with det A = 1 are ±I and the only such matrices with diagonal entries are ¶ µ ±1 0 . A= 0 ±1 (iv) Observe that S(I) = I and that S has a continuous derivative and that DS(I)(C) = 2C so DS(I) is invertible. The result now follows from the inverse function theorem. (v) The result fails for Y = 0 since µ ¶ 1 0 β −1 is a solution for all β.

376

We show that DS(Y ) is not invertible so the the hypotheses of the inverse function theorem fail. Observe that if µ ¶ u v H= w x then

DS(Y )H = Y H + Y H ¶ ¶ µ µ u −v u v + = w −x −w −x ¶ µ u 0 =2 0 −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, unless B = ±I then, if B 2 = I, there exist Cn with Cn 6= B, kCn − Bk → 0 and Cn2 = I. If B = −I then the argument of (iv) (or argument from the result of (iv)) shows that G and H exist.

377

K322 We have hαx, xi = hx, αT xi = −hx, αxi = −hαx, xi.

So hαx, xi = 0.

If α2 x = λx, then λhx, xi = hλx, xi = hα2 x, xi = −hαx, αxi ≤ 0.

Since a real cubic has at least one real root, the characteristic equation det(tι − α) = 0 must have at least one real root. Thus α has an eigenvector e3 of norm 1. By the first paragraph αe3 = 0. We observe that e3 is an eigenvector of α2 and so since α2 is symmetric we can find e1 and e2 such that the ej form an orthonormal basis. We know that αe1 is perpendicular to e1 . Also hαe1 , e3 i = he1 , −αe3 i = 0

Thus αe1 = µe2 for some µ ∈ R.

Similarly αe2 is perpendicular to e2 and e3 whilst so αe2 = µe1 .

hαe2 , e1 i = he2 , −αe1 i = −µ

With respect to the basis ej the linear map exp α has the representation P∞ (−1)r µ2r+1  n  P∞ (−1)r µ2r ∞ 0 µ 0 r=0 r=0 (2r+1)! (2r)! X 1 P P∞ (−1) r µ2r (−1)r µ2r+1 −µ 0 0 =  − ∞ r=0 (2r+1)! r=0 (2r)! n! n=0 0 0 0 0 0   cos µ − sin µ 0 =  sin µ cos µ 0 0 0 1

so exp α is a rotation through µ about e3 .

matrix 0



 0 1

378

K323 We have α = (α + αT )/2 + (α − αT )/2.

The first term is symmetric the second antisymmetric. Note that the decomposition is unique (if α = θ + φ with θ symmetric and φ antisymmetric then (α + αT )/2 = θ). If α is as stated k(α − αT )/2k ≤ k(α − ι)/2k + k(α − ι)T /2k = kα − ιk < ²

so α = ι + ²β + φ with β antisymmetric, kφk symmetric and kβk ≤ 1. Now ι = ααT = (ι + ²β + φ)(ι − ²β + φ) = ι + ²(βφ − φβ) − ²2 β 2 + 2φ + φ2

so

But so

2φ = −φ2 + ²(φβ − βφ) + ²2 β 2 . k(α + αT )/2k ≤ k(α − ι)/2k + k(α − ι)T /2k = kα − ιk < ² 2kφk ≤ kφk2 + 2²kφkβk + ²2 β 2 ≤ ²2 + 2²2 + ²2 = 4²2

so kφk ≤ 2²2 . [We can reuse this estimate to show that kφk = ²2 /2 + O(²3 ).]

379

K324 We call matrices U with U U T orthogonal. If, in addition, det U = 1 we say it is special orthogonal. (i) Observe that 1 = det U U T = det U det U T = (det U )2 . (ii) Observe that !T Ã N X Bj j=0

j!

=

N X (B T )j j=0

j!

=

N X (−B)j j=0

j!

so, since the map A 7→ AT is continuous, (exp B)T = exp(−B) = (exp B)−1 . (Recall that, if C and D commute, exp(C+D) = exp C exp D.) (iii) The maps h 7→ hB, A 7→ exp A and C 7→ det C are continuous so their composition θ is. Note if B T = −B we have (hB)T = −hB so exp hB is orthogonal so det(exp hB) = ±1. Since h 7→ det(exp hB) is continuous, the intermediate value theorem tells us that that the map is constant so θ(1) = θ(0) and det(exp B) = 1. (iv) By (ii) and (iii) exp hB is special orthogonal so A(exp hB) = (exp hB)A. But, as in K318, we have kh−1 (I − hB − exp hB)k → 0

so so

kA(h−1 (I − hB − exp hB)) − (h−1 (I − hB − exp hB)Ak → 0 kAB − BAk → 0

i.e. AB = BA.

(v) If n ≥ 2, we can have 1 ≤ I, J ≤ n with I 6= J. Considering B with matrix given by bIJ = −bJI = 1, bij = 0 otherwise, we see that aII = aJJ and aIJ = −aJI for all I 6= J. If n ≥ 3, we can have 1 ≤ I, J ≤ n with I, J and K distinct Considering B with matrix given by bIJ = −bJI = 1, bJK = −bKJ = 1, bij = 0 otherwise, we see that n n X X aIs bsK = aIK . bIr arK = aJK = r=1

s=1

On the other hand, if we consider B with matrix given by bIJ = −bJI = 1, −bJK = bKJ = 1, bij = 0 otherwise, we see that n n X X aIs bsK = −aIK . bIr arK = aJK = r=1

s=1

380

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 is isotropic. (vi) When n = 2 we have shown that µ ¶ a b A= −b a

Let k be the positive square root of a2 + b2 . Then ¶ µ cos θ sin θ , A=k − sin θ cos θ

that is A is product of a dilation and a rotation. But the special orthogonal matrices are precisely the rotations so the isotropic matrices are 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 det U = 1, we get a stronger condition so the new isotropic matrices are a subset of the old. If n ≥ 3 or n = 1 this makes no difference, by inspection. If n = 2, we observe that µ ¶µ ¶µ ¶ µ ¶ 1 0 cos θ sin θ 1 0 cos θ − sin θ = 0 −1 − sin θ cos θ 0 −1 sin 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).

381

K325 The area of the inscribed n-gon is given by n n X θk r2 X θk 2 = A= sin θk r sin cos 2 2 2 k=1 k=1

subject to

n X

θk = 2π.

k=1

We form the Lagrangian

n n X r2 X θk = 2π sin θk − λ L= 2 k=1 k=1

and observe that, at a stationary point, ∂L 0= = cos θk − λ. ∂θk Thus θ1 = θ2 = · · · = θn . Inspection (this is methods question) shows that we have the maximum. The area of the inscribed n-gon is given by n X θk A= r2 tan 2 k=1

subject to

n X

θk = 2π.

k=1

We form the Lagrangian n n X X θk 2 θk = 2π L= r tan − λ 2 k=1 k=1

and observe that, at a stationary point, ∂L 1 0= = ∂θk 2 cos2

θk 2

−λ

Thus θ1 = θ2 = · · · = θn . Inspection (this is methods question) shows that we have the maximum.

382

K326 Observe that the set E = {x ∈ Rn : xj ≥ 0 for all j and

X

xpj = c}

j=1

P p is closed (observe that the map x 7→ j=1 xj is continuous) and bounded (if x ∈ E then c1/p ≥ xP j ≥ 0.) Thus the continuous function f : E → R given by f (x) = j=1 xj yj attains a maximum. This maximum does not occur if any of the xj = 0 (for, if xJ = 0, then if delta is very small and positive and we take x0J = δ and take ² so that taking x0r = xr − ² if xr 6= 0, x0r = 0 if xr = 0 and r 6= J we have x0 ∈ E, we see that ² = O(δ p ) and f (x0 ) > f (x)). Thus, if the Lagrange method gives us a unique stationary point with xj 6= 0 for all j, it will indeed be the maximum. Form the Lagrangian L=

n X j=1

xj yj − λ

n X

xpj .

j=1

At a stationary point, 0=

∂L = yj − λpxjp−1 , ∂xj

so we have indeed got the maximum. Rewriting the last equations we obtain yj = λpxjp−1 so xpj = kyjq ´−1 ³P P n q for some constant k. Since nj=1 xpj = c, we have k = c y j=1 j and !−1/p !Ã n Ã n n X X X 1+q/p xj yj = c1/p yj yjq j=1

j=1

=

Ã

=c

c1/p

1/p

n X

j=1

yjq

!Ã

yjq

!1/q

j=1

Ã

n X j=1

n X j=1

.

yjq

!−1/p

383

Thus c1/p Pn

n X

Thus

n X j=1

tpj

tpj

!1/q

≥

n X

t j yj

j=1

= c with equality if and only if tpj = kyjq .

!Ã

with equality if and only if

yjq

j=1

p j=1 tj

whenever tj ≥ 0 and Ã

Ã

n X j=1

=

yjq

!1/q

kyjq .

≥

n X

t j yj

j=1

We assumed that all the yj where non-zero but a little reflection shows that if tj ≥ 0 and yj ≥ 0 (and t, y 6= 0) then !1/q Ã n !Ã n n X p X X q ≥ tj t j yj yj j=1

with equality if and only if

j=1

j=1

tpj

=

kyjq .

Thus if t, y 6= 0 then Ã n !Ã n !1/q n X X X p q |tj | |yj | ≥ |tj yj | j=1

j=1 p

j=1

q

with equality if and only if |tj | = k|yj | .

384

K327 (iii) The sum of the squares of the four sides of a parallelogram equals the sum of the squares of the two diagonals. (iv) Observe that kxk ≤ kx − yk + kyk

so

kxk − kyk ≤ kx − yk.

Now interchange x and y. Observe that

4hx, yi = kx + yk2 + kx − yk2

≤ (kxk + kyk)2 + (kxk − kyk)2

= 4kxkkyk

(v) Observe, for example, that the parallelogram law fails for the uniform norm when we consider f (x) = max(1 − 4x, 0) and g(x) = f (1 − x).

385

K328 (ii) We have ku + v + wk2 + ku + v − wk2 = k(u + v) + wk2 + k(u + v) − wk2 = 2ku + vk2 + 2kwk2 .

Thus 4(h(u + w, vi + h(u − w, vi)

= ku + w + vk2 − ku + w − vk2 + ku − w + vk2 − ku − w − vk2

= 2ku + vk2 + 2kwk2 − 2ku − vk2 − 2kwk2

= 8hu, vi so (1) holds.

Setting w = u, we obtain (2). Now set u = (x + y)/2 and w = (x − y)/2 to obtain hx, vi + hy, vi = hu + w, vi + hu − w, vi = 2hu, vi

= h2u, vi

= hx + y, vi. (iii) A simple induction now gives hnx, yi = nhx, yi for n a positive integer. The remark that h−x, yi = 4−1 (kx − yk2 − kx + yk2 = −hx, yi

now gives the result for all integer n. Now we have so

nhmn−1 x, yi = hmx, yi = mhx, yi hmn−1 x, yi = mn−1 hx, yi

for all integer m and n with n 6= 0.

Now observe that the Cauchy-Schwarz inequality holds (see K327 (iv)) so |hλx, yi − hλk x, yi| = |h(λ − λk )x, yi| ≤ k(λ − λk )xkkyk → 0

as λk → 0. Thus allowing λk → λ through rational values of λk gives the full result. (vi) Observe that if a norm obeys the parallelogram law on a dense subset it must obey it on the whole space. (vii) In parts (ii) and (iii) start by proving the identities for
386

K329* No comments.

387

K330 (i) Since |d(e, t) − d(e, s)| ≤ d(s, t), the function fe is continuous. Since |d(e, t) − d(e0 , t)| ≤ d(e0 , e) it is bounded. (ii) We have and

|fu (t) − fv (t)| = |d(u, t) − d(v, t)| ≤ d(u, v)

so kfu − fv k = d(u, v).

|fu (u) − fv (u)| = d(u, v)

(iii) Since C(E) is complete, the closed subset Y is complete under the uniform norm. We now observe that θ is a distance preserving ˜ such that θ(E) is dense in Y . mapping from (E, d) to (Y, d)

388

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 particular M (0, 1) = M (0, 2), which is incompatible with group laws (ab = cb implies a = b). (iii) As for (ii). (iv) If xn is Cauchy for d, then d(xn , 1) is bounded so there exist A > 1 such that A−1 ≤ xn ≤ A. Observe that (by the mean value theorem) there exists a K > 1 such that K −1 |x−y| ≤ d(x, y) ≤ K|x−y| for all |x|, |y| ∈ [A−1 , A]. Take M (x, y) = xy.

389

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 = inf{x ∈ E : x > 0}. If α = 0 then we can find xn ∈ E with 0 < 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 find k ∈ 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 = S 1 or E = {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 kxn k → 0, or (B) not. In case (A) either (A)0 There exists an N such that all the xn lie in a one dimensional subspace or (A)00 not. In case (A)00 , E will be dense in R2 and so E = R2 . In case (A)0 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{kxk : x ∈ E \ U }.

If α = 0, E will be dense in R2 and so E = R2 . If α 6= 0 then since U is a subgroup of E, ky − zk ≥ α whenever y ∈ E \ U and z ∈ U . Thus, since E is closed, we can find v ∈ E \ U with kvk = α. Suppose e ∈ E. By simple geometry we can find x ∈ R and k ∈ Z such that ke − xu − kvk ≤ kvk/2 so e − xu − kv ∈ U so e − kv ∈ U . Thus E = {xu + kv : x ∈ R, k ∈ Z}.

390

In case (B) let u be an element of smallest norm E \ {0}, write U = {lu : l ∈ Z}

and let v an element of smallest norm in E \ U . If e ∈ E then by simple geometry we can find integers l and k such that e − lu − kv lies in the parallelogram with vertices (±u ± v)/2 and so is 0. Thus E = {lu + kv : l ∈ Z, k ∈ Z}.

391

K333 Second paragraph. Let X = [−1, 1] with the usual metric d. Let E = 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 a dense subset of X then if (Y, ρ) is a complete metric space any uniformly continuous 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 can find xn ∈ E with d(xn , x) → 0. Since f is uniformly continuous on E, given any ² > 0 we can find a δ(²) > 0 such that d(u, v) < δ implies ρ(f (u), f (v)) < ². Choose N such that d(xn , x) < δ/2 for n ≥ N . Then d(xn , xm ) < δ and ρ(f (xn ), f (xm )) < ² for n, m ≥ N . Thus f (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 ² > 0, u, v ∈ X and d(u, v) < δ(²)/3. We can find un , vn ∈ E with d(un , u), d(vn , v) < δ(²)/3 and d(un , u), d(vn , v) → 0. By the triangle inequality d(un , vn ) < δ(²) so ρ(f (un ), f (vn )) < ² and ρ(f˜(u), f˜(v)) ≤ ². Thus f˜ is uniformly continuous.

392

K334 Observe that f (x) − f (y) →0 x−y as y → x. Thus f is everywhere differentiable with derivative zero and so is constant. If we replace R by Q, then f is uniformly continuous on Q and so can be extended to a continuous function f˜ on R. If x, y ∈ R and x 6= y choose xn , yn ∈ Q with xn → x and yn → y. Since we have so, allowing n → ∞,

|f (xn ) − f (yn )| ≤ (xn − yn )2

|f˜(xn ) − f˜(yn )| ≤ (xn − yn )2 |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.

393

K335 This is the statement that a complete metric space is totally bounded (see e.g. Section 11.2). S S If Fn is finite set such that x∈Fn B(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 no convergent subsequence, (|xn − xm | ≥ 1 for m 6= n), Z does not have the Bolzano–Weierstrass property. However Z is a countable dense subset of itself and since any Cauchy sequence is eventually constant it is complete. The space Q with the usual metric is not complete but is a countable dense subset of itself. If we give R the discrete metric (d(x, y) = 1 if x 6= y) then B(x, 1/2) = {x} so the only dense subset of R is R itself which is uncountable. Since any Cauchy sequence is eventually constant R with the discrete metric is complete.

394

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 , then put n ∈ Γ. Since U is open we can find a δn0 > 0 such that d(xn , y) > δn0 for all y ∈ / U . If U = X set δn = 1. Otherwise inf y∈U / d(xn , y) is a well defined strictly positive number. Set δn = inf y∈U / d(xn , y)/2. S We claim that Sn∈Γ B(xn , δn ) = U . Observe that if n ∈ Γ then B(xn , δn ) ⊆ U so n∈Γ B(xn , δn ) ⊆ U . Now suppose x ∈ U . Since U is open we can find a δ > 0 with 1 > δ such that B(x, δ) ⊆ U . Since the sequence xn is dense we can find an N such that d(xN , x) < δ/4. Automatically inf y∈U / d(xN , y) > 3δ/4 so δN ≥ 3δ/8 and x ∈ B(xN , δN ). S Thus n∈Γ B(xn , δn ) = U . (iii) The space R with the discrete metric d(x, y) = 1 if x 6= y is complete. Any open ball of the form B(x, r) with r > 1 is R itself. Any open ball of the form B(x, r) with r < 1 is {x}. Thus the countable unions of open balls are countable subsets of R and R itself. But every set is open so any uncountable set (e.g. {x : x > 0}) which is not the whole space is a counterexample. S −n ¯ )r). Since the count(iv) Observe that B(x, r) = ∞ n=1 B(x, (1 − 2 able union of countable sets is countable, part (ii) shows that every open set Sj U is the countable union of closed balls L1 , L2 , . . . say. Set Kj = r=1 Lr .

395

K337 (i) If x ∈ U , we can find a δ > 0 such that (x − δ, x + δ) ⊆ U . Thus x ∼ 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 at α = 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 and α − δ ∼ α + δ. Thus x ∼ α + δ contradicting the definition of α. By reductio 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 ) ⊂ U with a, αn ∈ (an , cn ). Thus ∞ ∞ [ [ {y ∈ [x] : y ≥ x} ⊇ [x, cn ) ⊇ [x, αn ) = [x, α). n=1

n=1

Thus

{y ∈ [x] : y ≥ x} = [x, α). If [x] is unbounded above we can find αn ≥ n with αn ∼ x so we can find (an , cn ) ⊂ U with x, αn ∈ (an , cn ). Thus ∞ ∞ [ [ {y ∈ [x] : y ≥ x} ⊇ [x, cn ) ⊇ [x, ∞) n=1

n=1

so

{y ∈ [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 C contains a rational which belongs to no other element so C is countable. (Or say that the map f Q → C given by f : (q) = [q] is surjective.) If D is an open disc {x : kx − x0 k = r} and kx − x0 k = r then any open disc containing y intersects D but y ∈ / D.

396

K338 (i) Observe that

n X j=1

|(f + g)(xj ) − (f + g)(xj−1 )| ≤

n X j=1

|f (xj ) − f (xj−1 )| +

n X 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 ) = kfp − fq kBV so fn (t) is Cauchy in R for each t. Thus fn → f (t) for some f (t). We now show that f ∈ BV . Since any Cauchy sequence is bounded we can find a K with K ≥ kfn kBV for all n. Now suppose 0 = x0 ≤ x1 ≤ · · · ≤ xN = 1, Then N X j=1

|f (xj ) − f (xj−1 )| ≤ ≤

N X j=1

N X j=1

→K

|(f − fn )(xj ) − (f − fn )(xj−1 )| +

N X j=1

|(f − fn )(xj ) − (f − fn )(xj−1 )| + K

|fn (xj ) − fn (xj−1 )|

397

as n → ∞. Thus f ∈ BV . Finally we use an ‘irrelevant m argument’ to observe that |(f − fn )(0)| +

N X j=1

≤ |(f − fm )(0)| + +

N X j=1

|(f − fn )(xj ) − (f − fn )(xj−1 )| N X j=1

|(f − fm )(xj ) − (f − fm )(xj−1 )| + |(fm − fn )(0)|

|(fm − fn )(xj ) − (fm − fn )(xj−1 )|

≤ |(f − fm )(0)| + ≤ |(f − fm )(0)| +

N X j=1

N X j=1

|(f − fm )(xj ) − (f − fm )(xj−1 )| + kfn − fm kBV |(f − fm )(xj ) − (f − fm )(xj−1 )| + sup kfp − fq kBV p,q≥n

→ sup kfp − fq kBV p,q≥n

as m → ∞. Thus kfn − f kBV → 0 as n → ∞. (ii) If f (t) = 0 for 0 ≤ t ≤ 1/2, f (t) = 1 for 1/2 < t ≤ 1 then fP∈ BV but f ∈ / C. If g(t) = t cos πt then g ∈ C but, examining n / BV . j=1 |g(1/j) − g(1/(j + 1))|, we see that g ∈

C ∩ BV is the intersection of two vector spaces and so a vector subspace of BV . If f ∈ BV ∩ C then arguments like those in (i) show that kf k∞ ≤ kf kBV . Thus if fn ∈ BV ∩ C and f ∈ BV and kfn − f kBV → 0 we know that fn is Cauchy in (BV, k kBV ) and so fn is Cauchy in (C, k k∞ ). Thus fn converges uniformly to some g ∈ C. We now observe that |fn (t) − f (t)| ≤ kf − fn kBV → 0 and |fn (t) − g(t)| ≤ kg − fn k∞ → 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, k kBV ).

The mean value inequality shows that CR1 is a subspace of BV . We t observe that if g is continuous and G(t) = 1/2 g(x) dx then |G(xj ) − G(xj−1 )| ≤

sup

t∈[xj−1 ,xj ]

|g(t)||xj − xj−1 |.

398

Thus if we take   if 0 ≤ x ≤ 2−1 − 2−1 n−1 , −1 gn (x) = 4n(x − 1/2) if 2−1 − 2−1 n−1 < x < 2−1 + 2−1 n−1 ,  1 if 2−1 + 2−1 n−1 ≤ x ≤ 1, Rt and set Gn = 1/2 gn (x) dx, G(t) = |t − 2−1 | we have Gn ∈ C 1 , G ∈ BV and kG − Gn kBV → 0 as n → ∞ but G ∈ / C 1.

(iii) Since g ∈ C 1 there exists an M such that |g 0 (t)| ≤ M for all t. Let ² > 0. By the definition of V (g). 0 = x0 ≤ x1 ≤ · · · ≤ xn = 1 such that n X j=1

|g(xj ) − g(xj−1 )| ≥ V (g) − ².

Let m be a strictly positive integer. By the staircase property of the function f we can find a finite collection I of disjoint closed sets of total length 1 − (2/3)m such that f is constant on each I ∈ I. We can choose 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 X r=1

|g(yr ) − g(yr−1 )| ≥

n X j=1

|g(xj ) − g(xj−1 )| ≥ V (g) − ².

Let us say that r ∈ A if (yr−1 , yr ) ∩ I = ∅ for all I ∈ I. We have X X X |(f + g)(yr ) − (f + g)(yr−1 )| ≥ |f (yr ) − f (yr−1 )| − |g(yr ) − g(yr−1 )| r∈A

=

r∈A

r∈A

X

X

r∈A

(f (yr ) − f (yr−1 )) −

=1− ≥1− ≥1−

X r∈A /

(f (yr ) − f (yr−1 )) −

X r∈A

X r∈A

r∈A

|g(yr ) − g(yr−1 )| X r∈A

|g(yr ) − g(yr−1 )|

|g(yr ) − g(yr−1 )| M |yr − yr−1 |

≥ 1 − M (2/3)m = V (f ) − M (2/3)m

399

and X r∈A /

|(f + g)(yr ) − (f + g)(yr−1 )| ≥ ≥

Thus

X r

|g(yr ) − g(yr−1 )| −

≥ V (g) − ² − M (2/3)m .

V (f + g) ≥

X r∈A

X r∈A

X r∈A /

|g(yr ) − g(yr−1 )| −

|g(yr ) − g(yr−1 )| −

|(f + g)(yr ) − (f + g)(yr−1 )| +

≥ V (f ) + V (g) − ² − 2M (2/3)m .

X r∈A /

X r∈A /

X r∈A /

|f (yr ) − f (yr−1 )|

(f (yr ) − f (yr−1 ))

|(f + g)(yr ) − (f + g)(yr−1 )|

Since ² and m can be chosen freely, V (f + g) ≥ V (f ) + V (g) and so V (f + g) = V (f ) + V (g). In particular, if g ∈ C 1 we have 1

kg − f kBV ≥ V (f − g) ≥ V (f ) + V (−g) ≥ V (f ) = 1

so C is not dense in BV .

400

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 J is a dense subset of {z ∈ C : |z| = 1} which is complete. The Cauchy sequences xn = 1 − 1/n and yn = −1 + 1/n have no limits in I. But |xn − yn | 9 0 so they can not have the same limit in the completion.

401

K340 (i) Suppose, if possible, that f −1 is not uniformly continuous. Then we can find an ² > 0 and un , vn ∈ Y such that ρ(un , vn ) → 0 but d(f −1 (un ), f −1 (vn )) > ². By the Bolzano–Weierstrass property of X, applied twice to obtain a subsequence and then a subsequence of that subsequence, 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 ² < 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–Weierstrass property of X 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 uniformly continuous on [−2, 2] so f is uniformly continuous on R. However f −1 (x) = x3 so f (x + δ) − f (x) ≥ 3xδ 2 → ∞ 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.

402

K341 Since (X, d) has the Bolzano–Weierstrass property, there exists a K such that d(x, y) ≤ K for all x, y ∈ X (direct or use total boundP∞ proof −j edness), thus, by the Weierstrass M-test, j=1 (2 d(x, xj ))2 converges and f (x) ∈ l2 . Observe that kf (x) − f (y)k22 = ≤ so f is continuous.

∞ X

(2−j (d(x, xj ) − d(y, xj ))2

j=1 ∞ X j=1

(2−j d(x, y))2 ≤ d(x, y)2

If x 6= y then d(x, y) > 0. Set 4k = d(x, y). We can find an N with d(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) is a bijective continuous function and, by K339 (i) is a homeomorphism.

403

K342 (i) We can have (X, d) complete but (Y, ρ) not. Let X = R with the usual 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) with the usual metric d and y = [−1, 1] with the usual metric ρ. Take f (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 −1 ρ(f (xn ), f (xm ) = ρ(yn , ym ),

the xn are Cauchy so we can find x ∈ X with d(xn , x) → 0. By continuity, ρ(yn , f (x)) = ρ(f (xn 0, f (x)) → 0 as n → ∞.

We can have (Y, ρ) complete but (X, d)) not. Let Y = R with the usual metric ρ, and X = (−π/2, π/2) with the usual metric d f (x) = tan(x). Using the mean value theorem, |f (x) − f (y)| ≥ |x − y|.

404

K343 Observe that α4β ∪ β4γ ⊇ α4γ

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 know that 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 → a for some a. Similarly bn → b for some b. Since bn − an > c we have b − 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, then we can find n(j) → ∞ such that bn(j) − an(j) → 0. If θ is any open interval of length |θ|, d(θ, αn(j) ) → |θ|. Thus (X, d) is not complete.

However if we take X ∗ = 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∗ (∞, ∞) = 0 then d∗ is a metric and a Cauchy sequence which consists from some point on of intervals of length greater than some fixed c converges by the arguments above and (since a Cauchy sequence with a convergent subsequence converges) one which does not converges to ∞. [The reader may prefer to replace ∞ by ∅.]

405

K344 (ii) Observe that the open unit ball E centre 1 consists of the odd numbers. 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 closed and {n} is not open.

(iv) d(r, s) ≤ 2−k implies r ≡ s mod 2k so r2 ≡ s2 mod 2k and d(r2 , s2 ) ≤ 2−k . Thus f is continuous. k

(v) d(n + 2k , n) → 0 as k → ∞ but d(2n+2 , 2n ) = 2−n 9 0. Thus g is nowhere continuous. P (vi) Let xn = nr=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 sequence xn can not converge. However d(xn , xm ) ≤ 2−2r for all n, m ≥ r so the sequence is Cauchy.

406

K345 Recall that we can find E : R → R infinitely differentiable with E(x) = 0 for x < 0 and E strictly increasing on [0, ∞). Thus, if we consider F (x, y) = E(x + 1)E(−x + 1)E(y + 1)E(−y + 1), we have F infinitely 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) =

∞ X

m=1

−1 −1 (y − ym )). n−1 F (4δm (x − xm ), 4δm

For each (u, v) 6= 0 we can find a δ > 0 and an n such that −1 −1 F (4δm (x − xm ), 4δm (y − ym )) = 0

for all m 6= n and all |u − x|, |v − y| < δ. Thus G is well defined −1 −1 everywhere (observe that F (4δm (−xm ), 4δm (−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 so G is continuous at (0, 0). Since no line through the origin can intersect both An and Am for n 6= m, G(λt, µt) = 0 for sufficiently small |t|, so G has directional derivative zero at the origin. However, n−1 E(1)4 |G(xn , yn ) − G(0, 0)| = ≥ nE(1)4 /2 → ∞, k(xn , yn )k k(xn , yn )k so G is not differentiable at the origin.