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Princdoo

Uni~tnily

Fi.-, Ilall. Was!linglOO Rood

Princeroo. NJ 08j.44..1000

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98765

Preface

.""""-

The present volume contains aU the exercises and their solutions lOr Lang's second edition of Undergraduate Analysis. The wide variety of exerdllell, which range from cornputatiolUl1 to mote conOOl1uallUld which are of varying difficulty, cover the folJowing subjects and more: real numbers. (jmltll, continuous functioos, diJferentiation and elementary illtegration, nonoed vect.oT spaees, eornpa.ctncsa, series, IntegraUon In one \'lI.riabIe, ImJX'OPCl" integrals, convolutions, Fourier series and the Fourier Inkgrsl, funcUons in nspace, derlvatlVfJa In vector spaces, the IIM!l'8e and Impiclt mapping theorem, ordinary differential equations, multiple Integrsls, and dilferenUal Jonns. My ob!ectiw Is to offer those learning and teaching analyllis at the uode:rgradua.te level II Ia.q:e number of oompl: Funct. . . aod LimitlI _ . Limita with J.o.fi.l:Uty ContinlOUll F"nd""" . . . . . . . . . . .

_ . . _ _ _ _ . . ..

1922

2429

DifferentiationPiopettiellOfthe DmYltlveMeau

55_.......~

Ul2 IIL3

Value TI>eorem . . .. Invene F'un.ctioulI' ..

38 39

II

Cootenta

IV IV.l IV.2 IV.3 IV.4V V.2V.3

Elementary FunctionsExponential. . . . Ulgarithm. . . . . Sine and Coeine Complex Nwnbel'll. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ..

43 43 5L 65 7173 7380

The Elementary Real IntegralPropertiell of the Integr&l 'nt,ylor'. fOrmula. . . . . . . . . . . . . . . . .. Asymptotic Estimates and Stirling'lI Formula. . . . . .NOI"J'I>f:ld Vector Spacell

VA VI Vl2 Vl3 VI.4 VI.S VII VII.l VII.2 VII.3

8491

Normed Vector Sp&eeS . . . . . . . . . . . . 91 n-Spao:::e and fUnction Sp&eeS. . . . . . . . . .. 96 Completenells. . . . . . . . . . . . . . . . . .. 99 Open and CLoeed Sets . . . . . . . . . . . .. 104Limits Basic Ptopertiell . . . . . . . . . . . . . . ContinllOU5 Maps . . . . . . . . . . . . .. Limitll in Function Sp&eeS . . . . . . . .111

HI H3 120

vmVIII.I VIII.2 VIn.4

1211 &sic Propertiell of Compact Sets . . . . . . 125 OmtinuoUll Maps on Compact Sets . . . . . .. 126 Relation with Open Coverillgll L29

Compact_

IX 1X.2 1X.3IX.S IX.6 IX.7

133 SerIes of Po6ItMl NumbeI'II . . . . . . . . . . . . 133 Non-Abeolute Convergenci:l 146 Absolute and Unifwm Convergence . . . . . . . . 150 Power Series . . . . . . . . . . . . . . . 156 Differentiation and fnUoration of Seriell. . . . . . . . . . 160The Integral in One Variable Approximation by Step Maps Properties of the Integral Relation Between the Intqral and the Derivative . . . 1611 165 170 179

Bedell

X X.3 X.4 X.6 XI XI.l XI.2

ApproxLrnaUon with Clmvolutions 183 Dirac Seqoencea . . . . . . . . . . . . . . . . . . . .. 183 The Wciel'lltriWl Theorem. . . . . . . . . . . . . 185 189 Hennlti&ll Producta and Orthogonali\y 189 Trigonometric Polyoomillls lIS a Thta1 Family. . . . . 199 Fourier Sericll

xnX11.1XII.2

XU.3

xn.4

Explicit Uniform Approximation . . . . . . . 203 Pointwise Convergence . . . . . . . . . . . . . . . 208

XDIXIII.l XUI.2

Improper IntegrM

217

xnl.3XIV XIV.l

Definition . . . . . . . . . . . . . . . . . . . 217 Criteria ror ConvergellCll . . . . . . . . . . . . . . 219 1l1tetdlangiDg Derivatives and IDtegra1B 2~

XN.2 XN.3XVXV.l XV.2 XV.3 XV.4

The fburler Integral The Schwartz Space . . . . . . . . . . . . . . The Fourier InveT8ion Formula. . . . . . . . An Example of Fourier Transrorm Not in the Schwartz Space . . . . . .

243 243 247250

F\mctJoll8 on n-Space 253 Partial Derivatives . . . . . . . . . . . . . . . 253 Differentiability and the Cbaln Rule . . . . . . . . . . 262Potential Funct.ioos . . . . . . . . . . . 266 Curve Integrals . . . . . . . . . . . . 267 Taylor'll Formula . . . . . . . . . . . . . . . Maxima and the Derivative. . . . . . . . . . . . . . .

XV.SXV.6

m m

XVIXVI.2

The Winding Numberand Global Potential Functions 281' The WindIng Number 1100 HomoJogy 287 The Homotopy Form of the Integrability Theorem. . . . 288

XVI.SXVI.6

Mom 011 Homotopies . 290DerlvativeB in Vector Spaces 293 The Space of Conti1U1OUll Linear MIlp8 . . . . 293The Derivative as 11 LineM MIlP . . . . . . . . . 29S

XV" XVII.1XVII.2

XVII.3 XVlL4 XVIt.s XVII.6XVDl XVIII.1 XVUI.2 XVllL3

Propertiell ohbe Derivative . . . . . . . . . . . 296Mean Value Theorem . . . . . . . . . . . . . . . . . .. 297 The Second Derivative . . . . . . . . . . . . . . . 298 Higher Derivatives and Taylor'll Formu1a . 30i

Inverse Mapping Theorem The Shrinking Lemlllll Invetlle Mapplngll, Linear Case ....... " The Inverse MIlpplng Theorem. . . . . . . . XVlU.S Product DecompoaitlOO8 ......

303 303310 318 320

XIX)(1)(.1 )(1)(.3

Ordinary Differentlal Equations 327 Local Existence and UniquenC8ll . ., 327Linear Differential EquatioOlS . . . . . . . . . . . .. 331

xxJCX.I

xx...XJC..5

xx.2 XU

MultlP'e IDtegrU E'JeDlCllte:y MuitipIl! I~ Criteria for- Ad.miMibiIity . . ~ ~. . . . . . . . ~ of Variab&M . . . \o'eI:toI" F" oekb OIl Spbera DUrereDti&l ForIn.[)efjnjll

SST. . . . . . . 337

343 . . . . . . . . . . . . . .. 3-tS. . . . .

. .. 346 . .. 358

XXI XXLI XXl3 XXU

SSlil

35lil loYerae ~ of Form . . . . . . 362 ~ . FonauIa lOr S;mpliaw ... 363

oSets and Mappings

0.2

MappingsSn(TuT') .. (SnT)u(SnT').

Exercise 0.2.1 Ld S, T, T' be uu. Show th4t

IjTIo ,Tn are leU. .how tholSn (TI U UT.. ) = (Sn T I ) U U (SnT..).

Solution. Assume that S, T, and T' are all non-empty (ifoot!.be equality 18 trivial). Suppose 1Iw ~ E Sn (TUT'). TheD ~ belongs neeesnrily to S and 10 at least one of tbe sets TorT'. Thus % belongs to al least one of the !IetII 5 n TorS n T'. Heace

Sn(TUT')

c (SnT}U(SnT').

To get t.be revene blclusloo, note that T C (TuT') so (snT) c Sn(TuT') and similarly, (SnT') c Sn (TUT'). Therefore(SnT)u(SnT') c Sn(TuT').For 1 < j

< n, let l'j = T 1 U .. U Tj _ Then by our previous argument snv" = sn(v..-, UT.. ) = (Snv..- I ) U(SnT..).

Repeating this p.t'ocell8 n - 1 times we find

Sn (T1 U'" UT..) .. (sn VI) U (SnT,) u u (SnT..).But VI = T, 110 tbia pI"OYUI the equality.

2

O. 5etI 6Ild Mappinp

ExerdllCl 0.2.2 Show that the equoJitiu 01 &erciM I remain trw: in~rm:titm lind 1ItIion ligns n 1100 U IIni intm:hGnged.

if tM

SoIu&lon. We want tD IIhow that S U (T n T) = (S U T) n (S U T). Suppoee :i: E S U (T n T), then :i: belong!l to S or T and T. But since S C (SUT) n (SUT') and (TnT) c (Su T)n (su1") we mll8t have :i: E (suT)n(SuT). Conversely, If:i: E (SUT)n(SU11, then:i: belongs to (SU T) and (SUT'). If:i: docs not belong to S, then it must lie In T and T', thllll tiel in S U (T n 1") M waa to be ahown. The same argumm 811 In Exercillc 1 with unkln and lntcracctioo Bigos i~ shows that if TI , ,T,. arc seta, then SU (T1 n nT,.) = (SU T1 ) n n (SuT.. ).E:.wrclee 0.2.3 Ld A,B ~ .nIbldl 010 let S. Denole by A" the corrtpte. JI'IeI'It 0/ A in S. ShOlJl that the cornplDnent: 0/ the inUratcllori " the W1wno/lht~,i.~

(AnBr=A"Unc

cnJd (AU B)" = A"nB".

Solution. Suppose:i: E (A n B)e, 1Il):i: is not in both A and B, thal. III :i: tI. A OT:i: tI. B, lhllll (A n B)" C (AC U B"). ConVCI'llCly, if:i: E (A" U 8"), then thwl :i: E (A n B)". Hence:i:

tI. A Of :i: t/. B

Ill)

certainly,

:i:

t/. A n

B,

(AeuB') C (AnBy.

For the IlCallId mmu1a, IlUppose:l' E (AUB)", then:i: tI. AuB, 1lO:i: t/. A and:l' t/. B, thllll :i: E A" n 8". Conversely, If:i: tI. A and :i: t/. B, then:i:

t/. A U B 1Il):i: E (A U By.

Exercise 0.2.4 1/ X, Y, Z

IIni

1dI, Mow that

(XUY< z= (X x Z)U(Y x Z),(Xn Y 0 .such &hlIt lor all inkgerl q,p, end 0 '" 0 we ha~

=($

1qa-P1> ~.{Nou: TM same c dwuld UIQli: />-

J9Q+PI

310lq

q'

Ir 10 - pfql2:: 101, then

Th1lI c:ouctude8 the exetd8e.

from below (i.e. there U

nOl'lmtpty let oj i ~ S which it bounded 30me integer m 31lch that m 0. ShoUl tMt tkre ui.. ~ m. n with n > ,ueh lOOt Irna - nl < f. (6) In fPet, given p po6itive in~ N, IhoU1 tJwt there e:zW iflUgrr, rn, n, Pnd < rn.:!> N mch lhnt Imo -nl < lIN. (e) Ld w be lI1Iy rmrn6er Pnd ( > 0. SIloUl thnt tkre eNt tnt(fer3 q,p,ueh

+ Sf = R(%) + R(y) + n", + n~. By IL9SUmption, R(z) + R(y) < I and since "a- + n, E Z we conclude that R(~ + 11) = R(%) + R(Sf).

ut

"'"

lqa-p-Ull

% - n. = :! < I. Divide 'he Interva!..l0,11 In N Interva1s liIN,(j + I)/N], i = 0, ... ,N -I. Then consider ka for k = O,I, ... ,N. The number a ill irrational, there are N Interva1s ami N + I numbers ka, therefore lW ROme k"k3 the numbers k1a and k,:o belong to the same interval. Thus

Ik l () -

~

"I""

=

I(k, -

k,:)a - fIj"o

I + ",,00 1< N'

(e) Let

f

> 0. Sdeet.lwgers m,n 8UCh that0< Irna-nl

Assume without 10118 of generality that < rna - n < f. Let "0 be the greatel!t lovrer bound of the set of integers k such that UI .:!> k(rna - n).

("0 which ImpUe!l that.

I)(rna - n)

< UI .:!> "o(rna -

n),

lkoma - "on - wi < f.

u '"'" Coml'~ Axinm

17

&ercbe 1.4.1 l.d 5 be 0 non-empty Itt 01 real numben, end Ed b be 0 ~t upper 00und lor 5. Ld. -5 dDloU tM .d. oloU ntmlben -z, tvith z e 5. Show thol -b"o greate.tllotou bowJd /oT -5. Show t1ult om-hal/ of &he compltkm# lUiom i~ tM otMr h4lf.

Solution. Since z :s; b we have -b :s -z for aU z E 580 -b Is a Iowet bound for -5. Suppose there exists c E R such that -b < c :s II for all II E -5. Then b > -c ~ z for all z E 5. Thls contradicts the act that b Isa least upper bound for 5. The first half of the completeuess Axiom Implies the olber half. Indeed, 6UppolIe 5 Is bounded from below. Then -5 Is bounded from 800w, thus -5 bas 8 leIllIt upper bound. I:Ience 5 has 8 greatest lower bound. Use the MIlle klod of argument to prove that the 8eCOOd half of the com}ietenellS axiom Implies the first half.

&ercl3e U.s Given Clfl real number ~ 0, .Jww that it haJ CI

~

root.

Solution. Assume that 0 > 0 (the CBBe CI ... 0 is obvious). lof!t 5 = {z E R such that O:s z BOd -l So}. Since 5 is lIOo-emJ:t)', SUP(S) exists, call It II. Then, proceed as in the proof of ProposltWn 4.2. Suppose /il < 0. If n > (2b + 1)/(0 - /il), then

1)2 1 I 2+_+_I?-->C1 ( n n n n/l-2b I 2/l2

because d our choice for n. Thill contradiction concludce the exercl6e.

&erellle L4.9 kt Zit ... ,%n be real numlltn. Show._~

lhat:z:f +... +~ "

Solution. Since

shows that :z:f+" has a square root..

~ ~

0 for all i "" I. _.,n, 8ZI easy Induction ugument .+~ ~ O. The previous exercise implies that :z:f+" .+~

IILimits and Continuous Functions

III

Sequences of Numbers

Determine in each case whether the given &eq1l(!JlC(l hall a Ii.mit, and If It docs, prove that your lItated wlue is a limit. Exercise ILLl2:..

=~.

Solution. Given f > 0 choose N 8Uch that N Il/nl < f, 110 {~} OOJIVerges to O.

> IlL 1'ben for all n 0 choose N 80 t.hat. N > 2/t. Tben roe all .. ~ N, we have 1z..1 < t. Hence the sequence {r,,} converges to 0.Exercise 11.1.5 r" = slnrtll".Solution.. For aU JKlIlltlve 11lteger&", r .. = 8lnrnr = 0, thus {r,,} OOllvetgeli lo O.

Exerd88 11.1.6 7 .. = ain (T)

+ coemr.

Solutkln. Let n be a poeitIve Integer. We haver ...... ain

(4~) + coe411" =

I,

7'....1 = ain

(2nlr + ~) + coe(4rtll" + 11") = 1- I = O.

ThUll {7..} has at least two distinct points of aceumulatlon (actually It has

exactly three, -2, 0, and I) and therefure the sequence (r,,) does DOt. have a limit.Exerci!te ILl.? r .. = ..:/'+1'Solution. Since n' + I

> n' ror all n we have

1:r,,1 < ~ ... n'80

n

I

(r,,)

oonverges

to O.

Exerd88 11.1.8 r .. =

k.

Solution. Foe all .. ~ 1 ~ have

I:r,,-II =80

In'~ I -

::::1n3

=

I~-~ II- n'~1 S ~, S~,

{r,,} converges to 1.

Excrcbe 11.1.9 :t" -

"f:,.~(I+n\) = I+J,. ~2'n n

Solution.. For all n ~ I, ~ have I + 1/..' s 2. ThUll:t,,=

Given any poaIUve real number M, 7 .. ?: M whenever n > 2M, thus {r..} does not have a limit,Exerdae ILl.l0 :t" = ~::;:~.

11.1

5eQuellcell

of Numbers

21

Solution. Mt all n ~ I, we have 0 ~ n' - n ~ n' and n3 ~ n 5 + I, thus

l%.. I~n3=nThis implies that {:I;,} corm::rgcs to O.

n'

1

ExerclllC D.t.ll Let S be a bourukd u! 0/ real number& ld A be 1M set 0/i/3 poinU 0/ accumulation. That if, A ~ 0/ aU I'ItlJllbeNo E R.wdl /hot 0 if the point 0/ O c - [.Solut5on. (a) 11lete exilltll a point of ~latlon d of S at distance less than [/2 of 6. The open b&U of ramUlI [/2 centered at d contains infinitely roany elemenl.ll of S. Hence the open ball of ramUlI [centered at 6rontains infinitely many elements of S. TItiB proves that 6 ill a point of accumulaUon of S. (b) Suppose that c = limllUpS. Theu given any [> 0, part (a) implies that there exiats infinitely many xES auch that x > c - ~ 1 for lIOme eo there exiatB infinitely U\II.fIy xES IIUCh that x > c + [0, then the WeierstrassBolumo theorem impliea that S baa a point d aecumulaUon 8trIctly greater than c, which is a contradiction. Convenely, suppose that c ill a number that satisfies both properties. Given [ > 0, there exista Infinitely many xES such that z > c - [ and there e>dst8 only finitely many xES such that x > c + [ 80 there are infinitely many elements d S in the open ba1I centered at c d radiUll 21. Hence c Is a point of 8Ccum.tlation of S. Now lIUppose that 6 Is a point of accumulation oi S such that c < 6. Then ill is small, say [ < (6 - c)/2 we know that there exista Infinitely many elemen1ll in S at dlst.ance < [ from 6. Thill implies that there Wsts infinitely UIlll\Y elements in S that are > % + [, wbi c - IE.

22

IL Limitll and CoWnUOUll f\.",ctioN

te) lj {an} ami {bn} are till I, there uim N 6lldlthm if 1'1 > N, Ikm d" > B. {llint.: Wrik d = I +6 with 6 > O. ~n d" = 1 +flb+:::: l+n6.1 Solution. Write d = 1 + 6 with 6 > O. By the binomial formula we get

4" =(I+W=

t(:)lf' .~

= l+n6+. 2: 1 +'16.

U.2 FuncUo... And Limits

23

So given B > 1 chooee N such that. N > (B - 1)/6. Tben 101" all n > N, we have d" ~ 1 + nil > B, 88 W88 to be shown.Excrciae 11.2.2 Pro11e that i/O < c,-~

< 1,

then

lim r!' = o.

Whllt i/ -1 < e S. Of {lfinl: Wrik e == -l/d wWl d>

1./

Solutkm. Write e == l/d with d > I. Execcise 1 Implie8 that. given f > 0 there eldsw N 1IO that lor all n > N we have d" > l/f. Then lOt all n > N, we get r!' < f. lienee lim--. r!' == O. If c = 0 the result is tdvl81. If -1 < c < 0, then 0 < lei < 1 and 100.._"" let' = 0, 1IO Ii.m,._"" e" = 0. Exercise 11.2.3 Show thatlor ClfI1I number x

f< 1 we haw.

l+x++z....

x.... l - l

.-,

lflcl < 1, Ihow that..-00

Urn (l+c+ ... +c") = - ' -. 1-e

Solutklo. We simply expand

(x-l)(Z.. +Z....I++ z +l) = %,,+1+%"++x-x"- ... _x_l == z..+1_1.When

lei < 1, consider the differencel+c+"'+c"--== 1-e1 1 - c.. +1 - 1 lc

-c"+1 =-.':::::Ie

So

sf,"I." '11 I/e. Then fot aU n 2: N

so /(%)

=0

O.

[1.3

Limits with Infinity

Exerclse lL3.1 fbnnulctte COfJIJIldeIy the rtdu for limiLt oj produ.cU. at 0, oM conrider 8rl il'lkrval 0 < 6:5 % :5 2A - 6. mm 0 oomtant 0, ond lor t.aeh po.ritive il'lttger n, thm e:risU 0 polpomiol P" lIuch that lor oU z in tM iftten1oI, om hoIItlog(z) - P,,(z)l:5 0/f1.

lHim: Write % = A + (% - A)logz ""

110

thm

kJgA + 10 (1 + z ~ A) .J1 ,

Solution. Let e = (A - 6)/A. Then 0:5 e < I, and -c:5 (% - A)/A:5 elf and only If 6 :5 % :5 2A - 6. The estim&telJ for the remainder In 11Ieotem 3.4 show that if -c:S X :5 (l, tben1R,,(X)I:5 ;; 1- c'

so ifQ"_I(X) = X -(X'/2)+. .+(-I)"-'X"-I/(n-I), the polynomialsP,,(%) = IOCA+Q,,_1

(% ~ A)

aatisfy thedeslred property with 0 = c/(1-e). Note, that we can tab more terms In the expaneiOll of the klgarithm 80 that we get a better uniIorm approximation. However, here we have shown that we can c:hooee P" to be of degree n - I.

V.4 Asymptotic Estimates and Stirling's FormulaExercise V.4.1 lntegrotmg ~ r-tI, prtlVe the loilMuing lonnvl4ll. (0) f sin" %Ib: = --: 610,,-1 %00liI% +.!!.:! f 610,,-2 %dz. (b) f 00liI" %Ib: = -: 00lI"-1 %s1nz + ,,;r f 00liI,,-2 zdz.

Solution. (a)

I~ by

part8 we have

f

sin,,-I ztlin%1b: = -lIin,,-1 %OOliIZ + (n - I)

f

OOliI'z!lin"-' zdz,

but OOliI' %mn"-' %= (l_sln2 %)sIn"-2% = sIn,,-2 %-610" %which Implle8 the deslred reIIUIt.

26

It Limit. and ContIJlUOOJ3 FunctioruI

Solution. If lei :S. I, the result ill trivial. Suppoee lei > I. Since c ill a root oil WI:! have -c!' .. On_Ic"-1 + ... +ao, thus lei" :S. 1o,,-llIcI,,-1 + ... + laolDividing by

Icr- '

impies

lei :S. lOn-II + ... + leI,,-I'but elnee 0 0 6UCh thot Ix"y,,1 < C lor all n.

Solution. Let x" = (-I)"/n and lin = n. Then xnJl" = (-1)" and Ix"y,,1 :S. 1 Coxall n 2: 1.Exerclse IL3.7 Ld

I(x) = G"r"++ao g(x) = b",r'" +... + bo

IL1 Umlb with Infinity

27

be polynQl'llials, with an, b", >F 0, 10 01 ~ n, m ruplitlelJl. Auvme that a", b", > O. IrwenigQk tM limitlim /(x) ,,--. 9(X) ,dininguitllmg !he al!$Q

> m, Q = m, aJUI n < m.

Solution. For large values of x ~ can write

/(x)

an:!:" 1 + ... + ;;;;:..),

9(X)=b",X"'(I+"'+~

where (1 + ... + G()/anx") and (I + .. + bo/b",%"') ..... 1 8IJ z ..... lXI. ThUll we have the three CllSeS

{

*, =-00 = lXI, . ~ n = m * Iim-oo : = Il,./bm , Q .-ne number a, ld 9 be defind lor all numbenl > 60me l'lllmber b, and cmtme that 1(%) > II lor oil % > a. SIJf11N)U that~_oo

lim /(X)=lXI and

"'-0

tim 9(X) =00.

-~

lim g{/(% = lXI.

Solution. For all x > a, 9(/(%) is defined. Let B > O. Oloose M~ > II IlUCh that for all x > M~ we have 9{Z) > B. Choose M, > a such that or all % > M, we have I(x) > M~. Then for all % > M, we have g{/(x)) > B.

Exercise 11.3.9 I'Jvve; ut 5 be a ut of numbeN, and Id a be to S. Lei I be defined UI'l S and Q$4Ivme

~rmt

La 9 be dejinallor aU 8Uffieientlrl tarye numberl, arwl a.vume

-

11m I(x) = lXI.

. O. C'ltolJ&> M lluch that if 11 :> M, then jgW) - Lj < (. Select 6 :> 0 6UCb that whenever z E S arnllz -"I < 6 _ bll\ltl /(z) :> M. Clearly for all z e S and Iz - aJ < 6 we have III 0 eboo6e A and B such that 11:> B l~ies 19(11)-1 < f andsuch that z :> A impUes/(z) :> B (z E S). Then for all z :> A we have III 0, choose A and B such that for all 11 :> B we have 9(11) :> M and sucl1 that z :> A implies fez) :> B. Then II M whenever z :> A.

Exer'dse 11.3.12 Find tM following IimiU a.t n ..... 00: {oJ~. (6) v'n - vn + 1. (e) ~. (d)~i/z#O. (e)v'n-Vn+l0.Solution. (a) TIle limit is O. To see this write

2 -+ nn2 n'Since l/n2 0 and l/n ..... 0 IIIl n ..... 00 we have lim" 00(1 + n)Jn2 = O. (b) The limit is O. To see this, write.....

1+n i l

n-(n+l) C r:-:--o ~ 1 C' vn+vn+1 2v n (e) The limit ill I. Indeed, we ean write1 vn+1 = ../I+I/n'(d) The limit is O. For n large, _ ha\ltl

r- .~ Ivn-v n +.,=

I

I

,rn

and nlzl- 1 ..... 00 IIIl n .....

ll:nzl~nlzt00.

1

(e) 1be limit ill 0 because we have the bound

1v'n-vn+l01

=1

n-(n+IO) I 90. To be precise we let

The

8Ct ~

diBt(1:0,

S~) = ~ {diet (:ta.'.:!O~"'90

n }.

Then dist(:ta.S~) > 0 becauIle 1:0 illlrraUonal. so Ilelect 6 aoch that0< 6 < min{l. min ditit(1:o, S~)}. Then 11: -

zol < IJ implies I/(z) -/(zo)1 < l.

Exetdsc IL4A Show tho! a polyoomicll of odd d~ with mil coeificim" h& a root.Solution. Suppll'lll we have a polynomial p(z) = am1:""

+ ... + 00.

where m Is odd a.nd am '" O. We 0 (If not, consider -P(1:. 1beu we can write p(1:) = 1:"" [am

+

0.;' + ... + ;]."--00lim p(z) = -00.

From this expr

Pl. it is clear that~

lim p(1:) = 00

and

Since " Is cowmous, the intermediate value theorem Implies that " IeIl8t one real root..

has at

11.4 Cont.inuouo f\ulctiona

31

Exercise 0.4.5 For z

'F -I 6huw tOOt 1M lollowing limit uUl.t:I(z)", _UrnI' ("" -I)'z~+

W1W CITe 1(1)./(~),/(2)r WMt ill Iim~_I/(z)? W1W ill lim",__ I/(z)1' For lDhich wlue& 01 z '" -I ill I amtil\1lOU&l'l& ill JIllUi6/(! to 1(-1) in 8uch a _~ thllt I ill amtiJl1lOt I, then

-.)' = ('-'1,,")' ("" z"+1 I+I/z"so I(z) = I. If Ixl < I, then

,,_cc

11m

.. -I =1 (z"-I + )'()'Z" I I

.

(a) The above argUlDem shows that 1(1) = 0, 1(1/2) = 1, and 1(2) = 1. (b) Note that I(z) = I CO!" all % IiUCh that IzJ 'F I, but 1(1) 18 defined and 1(1) = 0. So lim.._I/(z) does not exist, but_1",l'!1

Hm

J(%) = 1.

(c) SIml1atIy, lim__ I/(z) does tm exist., but:0:--1..-1'1-1

Um

1(%) = 1.

(d) The CUnd.lon I is conl.inU0U8 at aU % 'F 1,-1. Howe\l'Cf,l can be extended continuously 88 -1 by defining 1(-1) = 1.Exercise 11.4.6 Let

,.

. _l+r'" (0) It'7IoI ill 1M doImJin of de(lfIitiDn all, i.e-lor which numbers z does 1M timil exiIItl' (6) Give uplicidlllM vow I(z) 011 lar 1M various z in tM domain 01

I(z) = Hm

""

fc) For whidl % in tM dom 0 iBsmell we conclude that f(c) The IilIllle wgumcut appIil)8 when c < x.

~

I(x)+< as WlllI to be shown.

Excn:be 11.4.10 Let I,g !Ie conllU vpllIllrd and GUIIJ7Ie thm ~ in10ge 01 I if conlainerf in ~ inttn,r(j/ 01 tkfinitiDn of g. A&nlI7Ie 0l6t 9 if an incTeaftng fundion, thm it if x < V. ~n g(x) g(y). Show t1Ult 9 0 I i.f

s:

Solution. We have the foUowing

~--

go 11 - t)1I + tb)

s: g((l- t)/(lI) + tf(b)) s: (1 h(x) = max(J(x),g(x))

t)go 1(11) + tgo feb).~

Exercise II.4.lI Let I,g !Ie functioN tkfimd on tM rrJ3.X(J,g) to !Ie ~ lunction h ItJdl thm

ut s. lJtJim

and Mmihlrlv. defim the minimum 0/ ~ two fundioN, mln(J,g). Let I,g be defimd on II ut 01 numberl. Show t1Ult if /,g are continuow, ~n fDIIJ{(J,g) and min(J, g) are continVOUl.Solution. Sincem/lJt.(J,g) =~rcl6e

~(J + 9 + 1/- gllI

and

min(f, g) =

~(f + g-I/- gl),

12 Implies lhe coutlnuity of max(f,g) and min(/,g).~ ~

Exercise II.4.12 fA

luncion

at

be tkfimd Otl II -'d 01 nvrn/la" and let III be tM x it I/(x)l. If I if conlinuow, &how Mil! III it

contmuow.

Solution. The result Inllows from lhe inequality

Jl/(xll-l/(xv)1I ~ J/(x) - }(xo)l.

IIIDifferentiation

I1ll

Properties of the Derivative

Exercise m.I.1 Let Q ~ om imltitmlll nul'l'lbtr hoving the following J1RlPerty. There ui8U a number c > 0 MldIlh/lt for any flltional number pIll (in 101llut form) with q > 0 _ have

IQ-~I>; , fqa-pl>-(a) 1.6 I be lhe jundilm defind for alllltlmbtn (l$ folw$. If Z" .., not a rationGl number, then I(z) = O. 1/ % U II miaMI nul'l'lbtr, which CllJI be written 4.f 1I fraclilm pIll, with rmegen fl,P and if thU jrndiqn u iJllowut form, q > 0, tMn fez) = IIt1. Show t1uIll it differentiGble at Q. (6) Let 9 be the function defin6J for fill nurnbenf tu/ollowf. 1/z .., imlUonal, then g(z) = O. 11 Z .., flltional, written lU II fradion pIll in lowest form, q > 0, therl g(z) '"' l/q. 1 ~ the diff~ 0/9 at the nt 0 IIelect 6 I/q < ct. Tben

> 0 such that lllx - 01 < 6 and plq -= % E Q, then

,,..Urn 1(%)Q~

1(0) '" O.00

%

If % is IrratJonal, then t.he Newton quotient ill 0, 8(1 lis differentiable at aDd 1'(0) '" O. (b) For plq '" % E Q, the Newton quotient of 9 at 0 becomes

By Exercise

4, d Chapter 1 we know tbet given N > 0 there integet'l!PN,qN such tbat

6.

l

o(.)-o(.)I~%0

1

exillt$

IPqal

IPN

1qNO I 0 we have.~

lim 1(0+ h) h

J(O) = lim ~ = I,.~,

and II h < 0, thm11m I(O+h) h_ h

1(0) = 11m -h =-1 h_h 'we get

whence lis not differentiable at O. (b) If % > 0, then I(x) = %' aDd il h > 0

h_

lim 1(0 + h~ - J(O) = lim h = O.

h_

If % < 0, then J(%) = -%' and the Newton Quotient at 0 tends to 0 as h ..... 0 with h < O. Thus I is d1fft'l'mtiable lOT all % and for % > 0 its derivaUw is 2% and lor % < 0 its derivatiw is -2:1:.Exercl9c 111.1.3 FOf' a poritive inkger k, let I(t) denote the k-lh deriw. tiVe of J. Let p(x) = ao +01%+'" +Qnz" be II polynomial. Shaw thflt for all k,

IILI Propertiel 01 the Derivative

37

Solution. We prove by Induction thB.t. for 0 :S k :S n we haVll the formula

P"'(z)=k

'lair

+(k+l)l +(k+2)1 ...2+ .. (k+n-k)1 -,,-/r 11 al 2, /(2) = -3, and /(1) = 2. Thus the Inverse function g: [-3,a) ..... [J273, 21 of / is well defined and

0'(2) = 1'(1) = -1.Exercise III.3.S J(x) =

2xS + 1; find tI(21).

Solution. Restrict / to 10.31 because Cor all x In thia interval, I'(x) .. 6.:t' 2: O. Furtbcnnore, /(0) I, /(3) 55. and /((IW) :021. Then the inverse function 9: 11.55) ..... [0,31 of/is weU defined and

=

=

9

'(21)-

- J'({IW)

1

=

1 6.10~/3

Excrcille 111.3.6 Let / be G oontinU0U6 functiofl on 1M intenIallo.b1. AI81I71Ie that / u twice diff~ on 1M opm intemJI 0 < % < b. ond thGt I'(x) > 0 ond I"(x) > 0 on UW inknIot. Ld 9 be the i"mJtnle fundion 0/

/.

(G) Find ml aprestion /or the KC:ond lkritrGtive oj 9. M ShofII that 9"(11) < 0 on ill mterwl 0/ dqinitiDn. ThUl 9 u

the OJ!1J(UiU dirediofl

all.

rontre:I: in

Sohdiorl. (a) Since 9'(x) = 1/I'(o(x, the c:hain rule and the rule b dlffcr'eOOllbng QuoticntlI apply, leading to the following expresion for g":

~(%I = - r(g(x9'(%)If

U'(g(x))I'

(b) ForG < X < lowe havel'(x) >Oand j"(x) > 0. Thereforewe_rom the Ilblwc expreseion of g"(%), that g"(x) < 0 on its interval of definition.

ItO Invnae Functions

41

Exerdlle 111.3.7 In T1teorem 3.2, prove thaI if 1 i, 01 cianO' withp ~ 1, then iU intteru jundion 9 i, also 0/ cln. (a> The inequality NI ttue lor n ... I. Indeed,d Ik (e'"-(I+%"'~-I oand ~-(I+O) "" 0,.0 ~-(I+%) 0, let % = I/v. Tb8n11m P"l(%) - pnl(o) = lim uP,,(u)e- = 0 % 0 ..-00

z-O.80

P"+II(O) = 0. By Induction we conclude that the functlon differentiable at 0 and fl")(O) = 0 for all n. (b) The graph of the functiOll I is

I

is Infinitely

o

Exercise IV.1.6 (a) (Bump Functions). Ld a.b lie number" a Ld I be the /urn:tian 6Udi that/(%} = 0 if %:5 a or % 2: b, and(a) 1(%) = e-I!(z-..}(I>-.-)or

< b.

(b) 1(%) = e-l!{z- 0 be 80 If1WIlIIMl 0; + 6 < b -

6. Show thol there uWs 0 CO"

6tJdi that: g(%) = 0 if x :s: a, and g(z) = 0 if z ';? II; g(x) = 1 MIlo + 6,11 - 6\; IIfld 9 u Mridly increa.ring on [0.,11 + 61, ll!ld gricUy decrta.ring on (II - 6,IIJ. Skdch ~ gTQ:fIIu 0' F OM g.e-l/(~-"K-).

Solution. (8) We t8ke I(z) =

Fm 0 < z

< II we have

fez) '"

(6 -:r) 0) e-1/1z-G)(6-,,). (x 0)2(%_ liP

ex -

TbuB Ilslncree.alng on (4,(1+11)/2) and decreasing on 0+6)/2,11). Just 8lI in Exen:ise 5, UlIe induction to show that there exista a &equellCe cf polynomials {POI} and 8. sequellOEl cI positive ~crs {k,,} such that

r"l(z) _

- lfx

p.. (z) (1)(6

xli""

e-If(,,-..)(~....)

.

A Jinea.r change ot variable and the limlt.s computed In Exercise 5 pI"OV(! that I i.s In/lnlte1y differentiable at both 0; and b. (b) Le\ I = I(l)dt = Coo /(t)dt. Since / is contiDIIO\I8, non-negatlve. and not ldentlca1ly zero. we know that 1 '" 0, 80 we can define

t

F(z) =

j [00 /(t)dt.

Then F(z) = 0 If x:s: 0; and F(z) = 1 If oX ~ II and for all z we haWl

d dz F(z) .. I(z).So F Is strictly Increasing on lo,bj and F is CO". (e) By (b) we can C lIe. Show tOOt the inverse jvnclion 9 uiW. $Ilow th4t one con write log, g(y) = loglogl/l!r(Y), where

limw-oo l!r(y) = 1.

Solution. Since I 18 oont.Inuous and increasing on (1/e, 1 ond z > O.

S1ww thol

!f!" -I ~ o(z-I).(II) La "'9 ~ ntlrl'lllen ~ 1 6uch tholl/V + l/q .. 1. IJ:f ~ I, Mww l1wIt

" 1 :fl IP:s;-+-.

Solution. (a) The function h defined by h(:f) = :r/' - 1 - o(:f - 1) Is cootiPUOU8 on Rl!:o and dilfe~labl.e on R>o with h'(:f) = O(:r/'-I - I). Thus h Is deo'essing on (0,1), increasing on (1,00) and A(I) ;: O. lienee h(:f) ~ 0 for all z > O. (b) Let /(z) = Zl/, and g(z) = zl,,+ I/q. Then /(1) = g{I) .. 1 and for :I' > I we have f(z) .. !ztl-p)Jp and 9'(:1') = !. But if V> 1 and z > I, then zt l-,)/, < I which implies that I'(:f) < g'(:f) wheoevt!r:f> I. Hence /(z) :5 g(:I') and /(z) = g(z) if and only if z '" I, IlS WII5 to be ahown. Exercise IV.2.10 (0) Let u, v ~ poritive numbeY'r, ond let p,q ~ 011 in Exerci8e 8. Show that Ul/pv l/ ' < ! +~.

-

(6) Let u, v

~

po6itive numbenr, ondO < t

..

< 1.

Show that

IV.2 Loltarlthm

51

SoIutkm. (a) Let x = u/l) in the inequality 01 Exerd8e 9(b) and PJUhlply both s1de8 of the inequality by I). (b) Fix t III (0, 1), and Ietp = lIt and q = 1/(I-t). Then 8il"lC(ll/p+l/q = 1 and p,q > I, the prev\oUll exercise implie8 that Ull)l-l ~ til + (I - t)l). Clea.rly, if II = I) tlte equality holds. Conventely, II the equality holds, then u/l) ~ I because 811 we have IIl!eIl in Exercise 9(b), !(x) = g(x) If ...d ooly ifx .. l.Exercise IV.2.II k~ u be u number > O. Find the minimum und maximum 01 the IllfI{;twn I(x) at:l:~I~. Skmh the graph oll(z).

Solution. Since

.!.- ("') ~ 2%u~ thu"

a~

z:2u~ Iogu

the function attaln8 its minimum 0 at 0. U a > I, then 1im".__.... /(x) = 00 ...d II u < I, then lIrn z 'co 1(:1:) = 00. U u = I, then I(z) =

r.

."

,:.

.-

Exercise IV.2.12 U~ng the mean value thearem,

Generaliu btl replocing

i

--

find the limit

lim (n l {3 - (n + W/3)

hll 11k lor _II i~ k ~ 2.

Solution. Let I.(z) = x l / . Tboo/Hz) = l/kz(I-.lI., so by tlte IMall value theorem _ know that 8iven n there exiSlJl a number en,. IlUch that n e",. ~ n + I and

.s

Therefore

0< (n+l)I/_ n l/. < 1 - knt-I/Helice Iim,,--.(n + 1)1/.1< - nilit = 0 fo any integel" It ~ 2-

Exercise 1V.2.13 Find tM limit(I +h)l/a -1 ~ h . .Solution. We rec+l

+ z). TheflflJ~(z)

S I(z) S 1\.,+I(z).

Indeed, flJ~(O) = 1(0) = f\.,(0) = 0, I'(x) = 1/(1 + z), and

P'21>(z) = 1- z+z' - .. - ~-I and P''''+I(Z)'' 1- z+r _ .. +:?'.However we can 6UID the first m terms of a geometric series and obtain the fomwJa

1 - i=I")~-::+~""c.::'C' l-x+z'1 -.+(-I)"'z'" = 1- (_I)"'+lz"'+l = I+z l+x l+x'80 that P:z..(z) :$.I'(z) :$. P:z..+1(z) which implies that f1m(z) I(z) :$. I\,,+I(Z). We 8lso have the inequality z/(I +x) Iog(I + x) for all z?: 0. Indeed, mnsider I(x) = z/(1 + z) -1og(1 + z). TIicti

:s

I'(x) = (1 + z)' :$. 0.Since 1(0) = 0, the inequality follows. (d) Notation being as in part (c) we uve flJ~+l(z)-l\",(z)..... 0 Il8 n ..... 00 for all z E {0,11, hencelog(1 + x) = lim P2tl(z) = lim &'+I(Z)~~-""

-.

:s

for all z E (0,11. If L,.(z) = z /2 + ... + (-I)~+Iz"/n, then for all poeiLive integers n we uve L".(z) = I\",(z) and L:z~+I(Z) = P"*I(Z) which impliesfor all z E [0, II.

r

Exen:lae 1V.:U6 SIwv! t1ult for ~,mitiw inkger k one 11M

I)' \I+%n=lm% 11.-0

Exponentiating yields the desired limit.

Exercise IV.U8 Let (a,,), (b,,) be ~. oj puitive numbenI. De. /iM thue Iefl'encu k> be equivalent, llrad writl! a" ;;; b" Jqr n ..... 0 n+ n+1 ,

IV.2 Logarithm the Iaat Inequality coming rl'lXIl. Exerdse 21 (a). (b) ~ IIimply take the difference and get

63

b,.+l - b,. = an+! - an -

->+.! n+1 n

= -1oe: (I +.!) +.!.

n

n

>0

the last inequality coming rrom Exerei3e 21(a). (c) Let f > O. Choose an integer N > IlL Then fur all m ~ n > N we baveo s: an - 0".

s: lin -

b", + I/n - 11m s: I/n < f.

o s: b", - a.. s: nm - a.. + I/n -

11m s: I/n < f.

So both {an} and (b,.) are Cauchy sequences and tbey COlM!rges to the _limit beco.!!'" lim,,__ (a.. - b,.).Exercbe 1V.2.23 lfO ~ z s: 1/2, IMw that 1oe:(1- z) ~ -z-r .tNou.: When UOO 1Kwe Thf,'lor', lormu4 lind aerita kder, JII'tI am au MlU1og(1- z) - -z- 2' -"3 _ ....The point i, lMt-z UII good ~ tolog(l-z)..mnz i, 811WJl1.}

'" '"

Solution. Let I(z) -1og(1- z) and let g(z) = -z - r. Then f(z) = -1/(I+z) and g'(z) ... -I-h. We wish to show that f(z) ~g'(z). ThIs Inequality is equivalellt to I ~ (1- z)(1 + 2z) or 0 z - 2r = z(l- 22:). Clearly, thia 1ll5l. inequality is true wbeoever 0 :oS z ~ 1/2 so we have f(z) ~ 9'(z) on this interval. F'mally, 1(0) .. g(0) = 0 60 we IIlUat ha~ I(z) ~ 9(Z).

s:

Exercbe 1V.2.24 (0) Let, be a mrmber. de...

Define IN bimomlal coeffi-

(~)=I

and

(:)=B(n,')='('-I)('-2)"'('-fl+I)/nl/orn~2.limlupIS(n,.)r1/n::s I.

Prove !he utimde IS(n,.)1 ~ I'Jelol(n- 1)lol/n . In pnr1ictdnr,

Note IMt !he abot1e utimate npplje, lIS well if. U ~ (b) II. Lt not 0Jl WlUger ~ 0, aOOw thot limIB(n"W/n '" 1. Solution. (a) Since

B(n,.) =

!.(.- (~- ... (-'n 2 n-I

-i)

1)4

IV. Elementary FUnctIoN= 0 and

0;:-=

....,

0+0+1n+1

(n+l)"

0"

=

(

1+;;+1

0 ) ( 0 )" ;;+I .

'The inequalitk's deduced In Exercises 14 and 21(11.) imply that fOI' wBicieutly lIuge n

tog(l+ n:l) S n:1 < Olog(1

+;;}

ExponcnUating, we find that On+l/On < I fOI' all large n, and since On > 0 we see that {c,,} is rnoootonka1ly decreasing for 8Ufficiently large values or

o.

IV.3 Sine and CosineExcrdse IV.S.l Define tanZ"'" 6InZ"/CC6%. Skdch 1M gmph ofUn%.

Pm,

,- Unh.11m

Solution. 'The tangent is DOt defiued Cor % = ,../2+h when k e Z, and is periodie of period ,... We have

.I

d -Un%=

dz

1 ~ >0 = 0 on -_/2 < x < ff/2. DefiJle Ute mver.e function, roUed Ute arcsine. stetc1l Ute groph, cmd Ihow that Ute derivatiorl 0/ aresiox it 1/';1 z:l. Solution. By symmetry with respect. to the line)' '" z "'-C get the graph of araIinx. To compute its derivatlve 00 the given illterv&l, _ simply use the formula fOl' the derivative or the inverse map, so that

.s

-(arcslDx) = = dz COl'I(aralinx) JI-Sln'(ataIIlZ)

d.

1

1

1 = .

vi

z:l

""" "" ,,1\ ~

.

&erclge IV.3.3 &.too the anine functwm to the il'\terwal 0 .s x s "'. Show Ul4t the inver function m.tI. It at mlkd the arcoosine. Sketdl iU

lV.3 Sine and Co8ine

61

graph. and Wow th I. Prove UIGt,,-00

lim

,=0. n.

."

Concludt that the remainder

in the Taylor upanaiona lor the rine, corint, IIfld uponeotiallur.dw.n tcr.d !(. 0 ClJ n Unda to in,fif'lity.ternI.1

82

V. Tbe EIemeutaty Real

(~ral

Solution. L6 Un~ ~

...

_

a"/rd, and select an integer N > 20-1. Fbr all nu..+1 1 . - -a< u"

~

N

n+l

2'

thus for all k ~ 0 we bBve UN+l-:S (1/2)~UN, hence lillln_a"/rd = O. Clearly, thUllimlt is also true when a :S I. Form the ~ given after the formula llOnle coeine and the exponential Wll deduce at once that the remainder ol these runcUons tends to 0 lIS n ..... 00.

rot

Exerdse V.3.5 (a) Provt thlIt log 2 = 1og(4/3)+1og(3/2), or even betkr, 1082 - 710"9 - 2 log 24 + 310g 8lf1025

81

CQl'lt1ll''YfIJIt ezpruMn (J/Iog 2 (I,f ~ alUrnatmg 6erie1.

(b) Find CI roliotwl numbtr ~iIlg 1og2 to fiw decimOO, and prow l/l(lf il dou 10. The abooe trid; u much more fI~rIt them ~

dow'"

Solution. (a) The first. equality foIiows rom the ract that 2 = (4/3)(3/2). Fm- the sewnd equality, note that the right-hand side is equal tolog

('0)' (")'(81)S9 2S 80

=

log

212'3~31~5121~3u5T

=

1og2.

(b) We&ee that 10/9. 2Sf24, and 81{80 are all cloBe to 1. We UIIe 11Ieorem 3.4 to cstimat.e e&1I term. Let P,,(z) = z - (r/2) + ". + (-I)"-Iz"/n. Then f 0, oM conrider 8rl il'lkrval 0 < 6:5 % :5 2A - 6. mm 0 oomtant 0, ond lor t.aeh po.ritive il'lttger n, thm e:risU 0 polpomiol P" lIuch that lor oU z in tM iftten1oI, om hoIItlog(z) - P,,(z)l:5 0/f1.

lHim: Write % = A + (% - A)logz ""

110

thm

kJgA + 10 (1 + z ~ A) .J1 ,

Solution. Let e = (A - 6)/A. Then 0:5 e < I, and -c:5 (% - A)/A:5 elf and only If 6 :5 % :5 2A - 6. The estim&telJ for the remainder In 11Ieotem 3.4 show that if -c:S X :5 (l, tben1R,,(X)I:5 ;; 1- c'

so ifQ"_I(X) = X -(X'/2)+. .+(-I)"-'X"-I/(n-I), the polynomialsP,,(%) = IOCA+Q,,_1

(% ~ A)

aatisfy thedeslred property with 0 = c/(1-e). Note, that we can tab more terms In the expaneiOll of the klgarithm 80 that we get a better uniIorm approximation. However, here we have shown that we can c:hooee P" to be of degree n - I.

V.4 Asymptotic Estimates and Stirling's FormulaExercise V.4.1 lntegrotmg ~ r-tI, prtlVe the loilMuing lonnvl4ll. (0) f sin" %Ib: = --: 610,,-1 %00liI% +.!!.:! f 610,,-2 %dz. (b) f 00liI" %Ib: = -: 00lI"-1 %s1nz + ,,;r f 00liI,,-2 zdz.

Solution. (a)

I~ by

part8 we have

f

sin,,-I ztlin%1b: = -lIin,,-1 %OOliIZ + (n - I)

f

OOliI'z!lin"-' zdz,

but OOliI' %mn"-' %= (l_sln2 %)sIn"-2% = sIn,,-2 %-610" %which Implle8 the deslred reIIUIt.

(b) Since

J~1aDd

zem;t,d% _c:m"-I zidDz+ (n-l)

J

ein'zeos"-':r;dz

u' z _

1 - em' z, \be eQU*lity c1r 0 select $() Buch that z > ~ implie8 hex) < (. Then

[/h= 1% Ih+

/h$.M LI.+fM

h" Ih. Consequently, r, Ih

iff!: 1:'+('h< 2f:

Sioce I ..... 00 BB Z ..... 00 we c:an make the exprt'SSion on the right for all z suf5cieIltI,y large, thereby proving tbal. f(t)dt = o(F(z.

J:

Exerdse V.4.10 A.uume Uult f,h are continuaw plMiti". th4t fez) ..... 00, and tM! h(z) ..... 0 lU z ..... 00. $hm8 thct[ J(f)h(f)dt = o(F(z00,

lor z ..... 00.0080

Solution. Irlilllz-oo I(z) = to Exercise 9.

then lim..__ F(z) =

we are reduced

Exercise V.4.11 S ~ th4l Cf'eQ4ing) lII'Id th4l

Ju~

po6ititle (~g or de-

/(%) = o(F(z))

Jor z ..... 00.

V.4 AAympt R by the lonnula

e S. dqine the

Is(y) = 01(%,,).

(6) Show tOOt d(%,,) = II.. - J.I. (c) Frt an dement a 01 S. Let 9s as Is -

s.

(a) GWen two points %,01n S 6how that Is - I. U a bounded func.!ion on

la. 8IIottI that the mill'

U II dutance pre.!lerring embedding (i.e. i1Ijtit!e map) of S into the normed tIector JtIIOU 01 bounded fundw", on S. with the sup nonn./q Me metrie on S origi'llltly wa.f bounded, you am me Is ~ 01 !}s.] ThU e%en:Ue Mows thot the generoJity 01 marie IpOl:U U ilI~. In applicQtilU, metrie IpOl:U UIUClUy orUe naturolly ll8 ~ 0..' l'IOJ'Illed vector 6plIce..

Vl.2 Notmed V""'tor SJNlCf!I

!l3

SolutJon. (a) The triangle InequalIty immediately lmpllEe-d(~z) :$1z(1I) - '~(II) = d(x, II) - d(4,II)

s d(~x).

(b) By (a) we know tbat ,~ -

III Is bounded and thatfw(z)1 s dCl/.z).

1/~(z) -

Letting z = x or Z = II we _ tbat I/~ - , 11 1 = d(v,x). (e) Let 'I' be the given map, tbat ill 'PCz) = rh. The function 'I' ill il\iecti~ because if 'PCz,) = 'P(x,), then!h, (II) = g.... CI/) for all II. and putting II = XI we get 0 = d(XI,XI) = d(xI,~) bencc Xl = fUrthermOt"e,!h ia bounded and l'PCx) - 'PCII)M = Hg~ - 9111 = d(x,II).

X,.

Exerciae VI.2.4 Ld H be a norm 01'1 a vector 1IpG~ E. Ld a be a number > O. Show thal the jvnclilm x .... aJxl it aL!O (I norm on E.Solution. We bavt! a > 0 110 atzl ~ 0 and = 0 if and only If Ixl = o. The second axiom ooids because a\bzl = a!bllxl = !bI(llxl. Finally the tria.llgle Inequality follows rom

alx + III :$ a(lxl + IYI) S aJxl + 0111IExercise VI.2.:> Ld Ih and 11, be nornu 01'1 E. Show !hat IMluncliom X .... ,xll + Ixl, and x .... lIlaX(lxh, Ixb) ctre Ill7mU on E.Solution. (I) '": ba~ Izh + IxJ, 2: 0 and = 0 If and only If!xh = Ixl, = 0. Clearly, Ja%h + laxb = G(lxh + Ixl,). Fillllily, the triangle Inequallty

Iz + IIh + Ix + 111, :$ Ixll + IlIh + Ixb + II/I,(ii) For the lleCOOd function only the tria.ngle inequality requires a little~k

max(1x + I/Ii, Ix + 1/12) S ma.x(f:rh + 111I" Ixb + 11112) :$ ma.x(lzII, 171,) + ma.x(IIIII, 11/I,).E one meatI$ (I jvnclicn '11'" ,>.,. are numl1eN 2: 0, tlttnt'I t!t.Ctor IIpGce. 811 a0fI

ExerclSfl VI.2.6 Let E be

semlnorm

>'1 0 we can find ~ e S and l? e T such that Iz, - VIis ~ (and Iz2 - v.zlr S (. By the dEfinition clUie Illlp norm It follows that if ~ = (~,v.z) S xT, then Iz- ~Is"r < (. So z e ~ u was to be shown.Exercise Vl.5.2 A boundary point 01 S is 0 poin.I tI e E such thot aJeflI open set U whidi conWirll tI oUo contmn.1 on dement 01 S ond an dement 01 E whidi u not in S. The let 0/ boundarr point, u coiled tM boundary 01 S, ond u denoted br 8S. (0) ShOll1 thGt 8S u dosed.

I

"

(IJ) Show t1ult S .. doftd if ornI only if S Qlntaim all it" boundary pomu. (e) Shot t1ult the boundary of S .. eqool to the 00un4arv oJ't" compltme1Il.Solution. One eould limply note that 85 >= ~ n ~ n E - S. We can also argue stralght from the definitions. SUPpo:ll! x E 8S and IISSWDe without lOllS of generality that XES. 11Ien for some e > 0 the open hall B.(x) dC>ell DOt intersect the complement of S, so B.(x) Is COlltained III the complement lJS. (b) SUppolle S is dosed. Let x E E - S, theD there exists e > 0 eucb that B.(x)nS III empty. So x Ie not 8 blundaly poIDt of S. Coovetsely, llUppo9C S contll.ins e.11 or Itll bounde.ly pointe.. Then givf!n x E E - S there exlste. e > 0 llUch that B,(x) ns Ie empty, otherwise x wouJd be 8 bounde.ly point or S. Thus S Ie dolled. (c) The fact that the boundary of S eque.Js the bounde.Jy ofiu complemellt follows 8t (lIlCe from the IlYmmetry of the definition of 8 bounde.Jy point. Indeed, x EEls 8 blunde.ry poillt of S If e.nd only If x Is 8 boundary point of the complement of S.

se _

_...

Exumee VI.6.S An element u oJ S if caUed on InterioI' point oJ S if tMre cUt" an open boU B 'IIlernt at u .tudl t1ult D .. Qlntoimd in S. The .et oJ intenor point" oJ S .. ~ btllnt(S). It u oInPiowly open. It u immediate t1ult the inter3tdion oJ lnt(S) and 8S u empty. Prove the~=

lnt(S) U 85.

In j:)llrlieulor, a c~ .fel .. the umon. oJm interior poinU and iu bo:MIdarv point". Jj S, T OTt .w-t" oJ normtd vector .paOll.l, thm also Mow Ut 0 the open ball B.XI,l/l)) is contained In S x T. Suppose that IXI - z,1 < e, then the sup norm ltnplies

I(xl, Vi) - (z"YI)ls> 0, the opeD ban 8,,(x) cootaina points of s< and re because (S u T)c = s< n '1"'. IT x Is not 8 bouDdary pointo! SorT, then there ex.1st8 41 > OllUCh that B (x)n(sUT) Is empty. Cootradlctlon. (b) Every open baU centered at 8 point of 8(S n T) inteJSCCts SandT. Since (S n 7Y = s< U'1'" we CODClude that every open ball centered at a point of IJ(Sn1') intersects s< Dr 7"'. Cooclude. (e) We have lJ(S - T) = 8(Sn'1"') C [}Su8'I'" but 1Ir _ BT, 80 the n!5Ult drops out. (d) Suppose (x, y) E IJ(S x 1'). Then since every DJl(IJ ball centered at (X,II) intersects S x T the definitloo of the sup DOnn implies that x E-s and yET. 1 SandT are subaets of E we see that (S x 7y = (SC x E) u(E x80

re),

we have 8(S x T) C (OS x T) x ~ x 8T). Conversely, If (x, y) belongs to [}S )( T Dr"B x BT, then by the definition of the sup norm we see that (x, II) E [}(S )( T).

,.,.....

Exercise Vl.5.5 LaS kll~tOJononnt.dvt(."Ior~E. An dement v oj S .., collt.d 580Iated (in S) if there mm on open boll oenkffil CIt v auch tIUJt v .., the onIJi element oj S in thu open boll. An ~nt x oj E it culkd an accumulation poim (or point of accumulation) oj S l/ x bdong& If) the cltmn oj the $d S - (xl. (0) Show !hoi x .., tJlIhermt If) S if and only if x it eilhu an O O. Let W be the let of 011 elementt t% with z E U. S1ww th4l W if qpen. Prove 0 &imUar Itatemml oWvt dNed xlf.Solution. Let til = tz where z E U. There exist$ r > 0 5UCh that 8,.(z) is cootaioed In U. Then 8 ..(z) is contained In W. H U is closed, then W ill closed because !( E - U) = E - W. Exdne VUi.8 Show that the~ctionRxR ..... Rgitrmby(z,V) .... z i, coNmU01ll. FitwJ on uample of a cW.ed .wbIel A of R x R IftlCh thcl the projection of A on the jim foetor" not eloted FitwJ on e:mrnpk of on open ut U in R' tvho4e projection if d08ed, and U:#

a'.

Soludon. The projection map is continuous because

lz - zol ~ n(z -

ZO.V -lAIll

Let. A = {(z,v) E R' : V = lIz with z E (0. I]}. Then A ill closed in R' and the projection of A is (O.lJ, wbicb is not cloned in R. Let U t= {(z, V) E R' : V > OJ. Then U ill tho upper balf plane which is open in R', and the projection of U ill R which is cloned. Exel'clne VI.S.D Prow the mnork before Theorem 5.6.SoIut;lon. FlrstllUPpose that S isopcn in E. Suppose X ill open in 5. Tben tbere exIrt.B an open lItt U In E such that X = S n U. The lntet'9OCt.loo oJ open sets is open, 110 X is open in E. Conversely. If X is open In E, llOte that X = snx 8Ild therebce Wi! can tab U = X in the definition, thereby proving that X ill open in S. Now DUppose 5 ill closed in E. Suppose that X is closed In E. Then there exlst.8 aclo8ed set. Z in Eouch that X = snz. The Interseetlon of two closed nets is closed, 80 X is closed. Conversely, if X ill closed in E, Wi! llOte that X = sn X 80 that Wi! can t.aJoo Z = X In the delinitioJl, therefore X Is closed in E.

108

VI. Normed

~

S~

Exercise VU~.10 Let U be open if! (I normtd be open .... (I nonned vuWr 6JIClce F. Ld

spoce E and let V

/:U ..... Vbe crmtinwlu 1rnIplI whkhCl1'e

(lnd g:V ..... U

imra' 0/ eGC1l otkr. !hat isand

/og=idv

go/=idv.

whert id mtmlI the identity map. Shm thal ijUI U open in U. thm /(UI ) i, open in V. and Ihat the opera ,l 0, Olen [or some m we have Iz - z... l. fIe, so Iz - z... ~ S f, hence z ill a point o[ accumulation fA {z,.} roc II~. (d) Suppose J ill oomlnoous lit. Zo with respect. to Ih. Let f > 0, tOOn there exists 6 > 0 such that IJ(z) - J(:to)11 < fIe, whenever Iz - zol. < 61CI. 'Then, iJ' Iz - zol, < 6 we lnunedl8tcly conclude that IJ(z) - J(zo)l~ < f, 110 J Is OOlIUllUousllt. :to with respect to II~. (e) By Execcise 3, 8S 'at "S - Int(S) and by (a) and (b) we know that the clollure 01 S ill the &arne with respect to equivaleut norms IIlld similarly fur too itlterioc S, 110 the boundary cI S ill also the same with respect to equivalent norms. (f) By Exerci8e :J. if follows that the clOBUl"e 01 a set ia the B1\rne set. with reBpect to equivalent DOI"IDlI.

s.

or

Exercllle Vl.5.13 fAt 111 andlb be two rwrm.tOllO vedor~E, and ~lJIlPO'e thot there ui4tI 0 conJtam C > 0 ItIdlthat for tilt z E E we have IZh CIz~I. Let bIz) "" IziJ. Prove in detail: Giwn f, there ezilIu 6 ~uch that i/ z, y e E, and lz - yl2 < 6, thtfllb(z) -b(y)1 < f./Rem4rl.. Sin 11 is reoI tIlIlued, the ~ oj the Iign~ 11 denc~~ the lIbIolute valtu! on R.I In porticular, 11 is ccntiJlUQU,f for the norm II~.

.s

wt

Solution. Given e > 0, let 6 = (IC. Then Iz - yl~ < 6 impliclliz -yh < f. The triaugle inequality for the norm impliell Izh -1fII. s. Iz - 1111 and ITIII -Izl. S.lz - 1111 110 IIzll -Iyhl s.lz - TIll end tOOrefore

Ilt(z) - 1I(Y) .s Iz - ylt < f,88 W88 to be shown.VL~.14

Exercise

Ld BS ~ncte the spau oj tilt 'eqvencu oj numbeN

Iz..1.s

which

lin':

bounded, i.e. there emu C > 0 (depending on X) 8uch that C for aU n E Z+. Then BS if a ~PeCW aue oj &ample 2 oj l,

~the qaa oJbotmded ~8(Z+,R). ForX e BS, the 8IIP norm

WXI=sup)z..l,i.e. IXU if the ~f Iqtper bound oj till ~ me" oj the COl'IIponenb. (a) fAt Eo be the m oj till Hquencu X ItIdl thotz,. IS 0 for till but a finite number oj n. Show that Eo if a S1lNpaCfl oj BS. (b) I, Eo deme in BS'! Prove ~ auertion. INote. In 71tec-rem $.1 oj COOpter VII, it will be Mown that BS is complde.1Solution. (a) Additioll and scalar multiplication bdefinedCOlllponentwise. Suppose X IIlld Y belong to Eo, ~i':d that ,HInd q are the number ofnomero /

110

Vl. Normed Veet.ar S _

elements in X and Y, respectively. Thm X + Y has at ~ p + q nonzero elemmts, hence X +Y bclODgll to Eo. Similarly, aX bdon.gs to Eo whenever o E R and X E 0. Clearly, (0,0, ) belongs to Eo, hence we conclude IbM Eo is a wbspace BS. (b) We cootend that Eo is not dense \f BS. ~ X =< (1,1, l, ...). Then we ace that lot all Y E Eo ,he vector X - Y h&8 lit least one component equal to I. In fad, X - Y h8a()llly finitely many components '" I, therefore KX - YU ~ l and this proves our contention.

or

VIILimits

VIl!

Basic Properties$p 0, thef"t uisu z e 5 "loee a sequence d pointe {z,,} web that z" 2: n. Then (f(z,,). ill a Caucby sequence because ifn,m > N, lbm I/(z..) - f(Zm)1 < t. The comp1etenesll of R (Of F) Implies that (flz..)} comrerges to a limit point, say w. Tboo fot all z 2: N and z.. 110 large that %" > N aud 1/(z,,) - UIj.:s; f we bave

IJ(z) -

Ull :s: I/(z) -

J(z,,)1 + l/(z..) -

wi < 2r:.

Exorcise VII.l.3 Let F be 8 normed IJedor 6Pll' WE be 8 vector Ipcu;e (not nonJKd lid) ClII41let L : E -0 F be 8 iimar map, that ~ ~flfin9 L(z+v) L(z)+L(,) and L(u) cL(z) for till c E R, z" E E. AslUl'l\e thl%l L is injective. For each z e E, define I:tl = IL(z)l. Show thl%lthe /unction z .... lzl U 8 reorm ore E.

=

=

Solution. Since L Is iI\)octlve, L(z) = 0 if and only If z "" 0, 110 lzl = 0 if and only if z = O. Clearly,

Finally we cbeclc the triangle inequality

and th18 concludes the prooC.

Exerclee VII.1.4 Let ~ be the IJedor spau oj pdynomial /unctiorll 01 tkgree :s: 5 on the intenrollO,I]. Show thl%l ~ U ciMed In the Ipcu;e oj all bol UwIt Ihe /unction /(x) = Bin(I/x) " not um-

/0fTn1Jl continuow on

Ihe interval

0, pkk ~ > 0 such that U(.%) -/(v)1 < ( whenever 1.% -III < 26 and .%, V e S. Utoo e 3', then fm- aU .%,v E Sand 1.% - tool < 6, 111- tool < 6 we have 1/(.%) -/(v)1 < . Tbeotem l.2 ImPiee the existence oCli~ 1(.%), .% E S.Exerclao VlL:U Ld S. T be cwKd ftdIKtt 01 a nonned vector 'JX!U. and lei A = SUT. Ld/: A _ F be a map into 8lmle nomJlldvectorfP/lU. ShG'W that I u COfltitzucw on A if and only if it" rutrictiofu on SandTare contintlOUl.

Solution. Suppose I is eontinuouB 0lI A. Let .% E S and let {.%.,.} he a IIllquence in S converging to.%. '!ben since {.%,,) is a sequeute iII A converging to.% it follOWll that Iim/(.%.. ) = 1(.%), thereby pwving that I ill coc&1l1UOU8 on S. A similar argument ptOvell that I is continuoua on T. Conversely, suppose that I is eontlnuoua on SandT. Let .% e A and let (> O.If.% rt T, then.% ill not adherent to T, 80 \W can choose a baU of 8lI\3U radius eentered at. .% wblch does not Interse 0. There ex1sts 6 > llUCh that lL(v)1 < ( whenever Ivl < 6. If Iz -1I1 < 6, then lL(z) - L(1I)1 = lL(z -1I)1 < (. This proves the uniform COllthUlity of L.

Exercise VII.2.13 Ld L: E ..... F be a continuow lifIeor map. Show that tM tJOtuu 01 L on tM do6ed ball 0/ rndilU 1 ~ bou1lded. If r u a number > 0, Mow thai tM voluu 0/ L on any do3ed ball 0/ mdiw r ore 1Iounded. (7'M do3ed balls ~ cmlemIlll tM origin.) Soow that the fmtlge tlndn" L 0/ a boundd l Q bounded. ButJtUe 01 Uem.1e 10, a continuolU linellr map L u abo called bounded. II C '" a number 6ucIa that 1l..(:l:)1 :s CI:l:1 for atl z E E, lhen we caU C abound lor L.

Solution. Since L ill contJn\I(JW at 0, there exists 8 number C > IlUclI that IL(z)1 :s: Glzi fw all z e E.lf Izi :s r, then !L(z)1 :s: GIrl which provelI that the values of L on the clo6ed ball of radiWl r centered at the Ol"lgill areGiven II bounded set 5, thereexi8ta II closed ball such that S Is contained in the clo6ed ball. 11Jerefote, the values of L on S are bounded.

""""""'.

Exen:1ae vn.2.14 Ld L be a COI1tintlOW Iintor map, and Jet ILl denou tM grmtuC Io1uer bound of atl nt O. Suppose that

If -

911 < (, then

Ir.(n -1:(1)1 :5....bence

II~(J - g)[ S [I/(X) -

g(x)ldt = 1/- gh < (,the Schwan

r: is oontlnuoWl or the L.-nonn. (b) The 1Dkgr81 is a.1Bo contiOUOUll or the L'norm becauseinequality giveill

1/ - gIl = ; : I/(z) - g(z}ldt

< ([ 1,,"Exetei8e VIL2.16 Let X k

r ([

Ifl') - *)1',,"

r

S CI/- 012'II; romplete metric 8poa sotUJving the semiparolldogrnm lInu. (Cf. Eun:iu 5 oj C1ulpter VI, .I.) Let S k II; eloH.d subut, and Q;UtlIJIl! tJUlt given ZhZ2 e 5 the midpoint between ZI and X2 U abo in S. Let v EX. Prove t1uJt tMre trim an e~1 w e 5 sueh that

d(v,w) = d(v,S).Solution. Let d = d(v,S). We 8S8IlDlt tbat v tt otherwise cboaJe v = w. Let {x n } be a sequence of polnt.8 in S such that d(v,z,,} C(lllverges to d. We DOW show that (xn ) Is a Cauchy sequence. Let z",n be the midpoint of X m and x n . Applyillg the eemlparallelogram law with z = v we get

s,

d(xm , x n )' :5 2d(xm , v) + 2d(xn , v)' - 4d(z",." v)'.By 8MIIlIIptioD,

thet'em d(zm,Xn)' :5 2d(x m,v) + 2d(xn v)' - 4JR,z",.,

e S so 4d(z",."

v)' ~ 4 I lire have Ilm,,-.oo I.,(z) = I and In(l) = 1/2 for all n 80 the sequence {I.,} does II 0 such lJuIt g(1:) ~ G fur all z E S. Shew that the uqumct9n'"' l+ng

unifomUy to tile con"to.r\t juneticn I. Pro'Ot the 8Ilmt tiling GUumplion it t1wt I9(z)1 ~ 11 lor GIl z E S.~

if the

Solution. We prove the stronger result when 19(z)1 ~ 11 > O. 'The uniIonn conva-gence follows from the fact that fo alIlatgtl n we have

Ig.,(z) - II = II + ng{1:)1

,

< nlg(z)1

'-:":-' 1 I), and eacb I., is odd, 110 1/.,(z)l,5; I for aU z and all n. By Iymmetry we can reduce the proof of uniform eoo.vergenoo fOT z ~ 0~

then have

,-,,", f,.(z) = (I + w)"

110 I.. attains it3 maximum 1/(2,fii) at 1: = 1/,fii, tbw 1/..(z)l,5; J/(2,fii) provq; that the lIeljUCIlCf' converges to 0 unlfc:>nnlJr on R.

VlI.3 Limits in f\ulct.jon Specea

121

toionSF (c) Let 0 < c < l. SluM ehtlt I u uniformly contintlOtll on 1M inknIal 1O.e). and ehtlt Ma ~ {Pn} ~ uniJormJy to Ion ehU mUrool.Solution. (a) The function I is no\ unUonnly ooDtillllOUS because if it were, tbeu there would exist 0 < 6 < 1/2 such thai I/(z) - 1(,)1 < I wbenever Iz -111 < 6. Let 11 = I - 6 and suppose 11 < z < 1. Then for values oh close to I we see thu J/(z) - lUi)1 ~ I. (b) We know thllt p,,(z) = (1- zn+I)/(I_ z) 110

Exerdee VlLS.4 Let S be 1M rnurwl 0 ~ z < 1. Ld I be 1M fvm;lion tkfintd on S ~ I(z) _ 1/(1- z). (0) Deknnine ~tMr I u vnl/ormltJ rontinvoUl. M Let Pn(Z) = I +z+.+z". Doe. 1M Ie9ItenCe {Pn} converge unifonnllt

f.'or fixed n we bsve 1:&1....1/11 - zl .... 00 11II Z .... I 110 tOO sequence {Pn} does Dot. convtqe unifoI'mly to I on S. (c) On IO.e), the function I has a bounded derivative, 80 I Is uniIotmly continuoll8. Furthermore. we bave the estimata

lPn(z)-/(z)1 ~ I-c' 110 {p,.} converges uniformly to I on (O,e).

cn+1

Exerciee VII.:U Let fn(z) = r/(1 + nr) u.quenu {In} Clm1/t1yU uniformlv on R.Solution. For each n 2: 0 the function liable OIl R with

lor all rml z. SIIow ehtlt 1M

I ..

ill even, positive, and dilfel'en-

,%~

In(z)" (I +nx2)"and In(z} I.enlb to lIn 11II z .... go. Therefore JI.. (z)l.s. lIn whicb proves that {I.. } OOIIVerges uniformly to 0 11II n .... 00.Exercise VD.3.6 Conrider 1M fvndiMI defined

Find

~

1M valuu 011 at rationcU and irrDtioncll n1l71lben.

Solution. Suppose z = plq, then for large m, m 2: 21ql. hence min = 21rk for IlOIIle k E Z thll8/(z) = 1. If z is irrational, then IC06(mhrz)1 < I for aU poeitive integers m 110 I(z) = o.

122

VIL Limlta

Exerdee VIL3.1

A. in Exercise 5 01 It, Id

J(%,V)~1. I

I O. there exlsts positive iDtegerll N and M such that for all 1'1, m > N we have Jz" - %",1 < E and such that or all k > M we have Ix,.. - xl < t. Select k such that k > M and 1'1" > N. Then for 8111'1 > N we have

1x" -xl:S Iz.. -%.... 1-Ix... -xl < 2(.Exercise VIn.l.2 M La Sa. . .. lie 0 Jiniu number oJ COtl1pOct tI in E. Show that tM ul'liorl 51 u u 8m i$ compact. (b) ut {So}iEI 6e /I /4mi1y 01 compact leU. Show thlll the ~ion (\-EI S; U romped. OJ CQtlf, it may lie emply.

s...

Solution. Ca) We prove the result for two eet8. Let S be WI infinite IIUbset of SIUS,. As!iU.me that SnSt is infinite (iCoot, tOOn sns, DlUlIt be infinite and the argument ill the MII'Ie) IlO that Sn 8 1 baa II. point ol e.caunuIatlon in 51_ Therefore S hll4 8 point of e.ccumulatioa iJI 8\ us,. By induction weconclude that 81 U , u 8m ill compact. (b) Let S be any member of the family. Then S aDd ill therefore compact by Theorem 1.2.

fke, 50 ill a cloeed subset of

1211

VIII.

~

Exercise VIII.1.3 SlwlJl th4t a denume-robk lIrlion 01 CQl'Jlpad lIeU need not be compact.Solution. Each Interval!" - l-n, I'll 18 compact but the union u"1,, ill unbounded and therefore not compact.

Exerclse VIII.l.4 Ld {z,,} be II ~equenoe in a normed vector ~ E f'Itdl tAot {z,,} ~ kJ v. Ld S be the let ~iflg of all z" oruI v. Show tAot 5 U CQl'JlpoctSolution. Let {v,,} be a eequellC(l In 5. If there are Infinltltely IlUl.fOI k'. such that v" Is equal to the same ckment of 5, tben {v,,} has&. converging subecquellC(lln 5. If one elanent 01 5 Is not repeated infnitely IlUl.fOI times in {tit}. then we can find &. subecquellC(l of {v,,} which 18 a su~ of {z,,} and whkh therefore converges to v In 5.

VIII.2 Continuous Maps on Compact SetsExerciH VIIL2.1 Ld 5 c T be MMU 0/11 normedtledor ~ E. Ld I: T ..... F be II IIl 0 ,ud! I1ult q %, 'II E 5 Ofld ~ - '/II < r. !hen % Ofld '/I are tonC4ined in U, lor

_i.

Soluticm. Suppoee that sueh a Ilumber r does DOt exisl& Then roc eaeh n there exlats s point :l:n E 5 sueb that B1/n(%n) is not COI:ltaioed ill Uj or any i. There exllIts 118ubsequeoce {:l:n.} of {:,,} wWeb converges to %. Pick m'such that % E V..... Since V"" ill open, there exlsts II positive Integer N eueh that BI/N(%) C U..... We ean find l' auch that roc all k > l' we have ~ ... - %] < I/2N. Then for k > maxCP.2N) we have B I/... (%...) C U:", II

contradiction.

&erdee VIllA.' Let {Bi}lEl be llfomilfl olc.ompad Midi of 0 normed tredor Ipaee E. SlJP1'O'e tk irIterIeelWn (\el 50 i, tmPt'll' Prom tOOt ~ i, II finiU nt. nns.,.ThU

i, empl'/l.

u tM"duol" 1'1 operl.1I of the finite 00\Itring properl.lI

Solution. Let 5 be one member of the family. Let Vi = Sf, 80 that Vi Is open. Then by byptt.hesia, the family {U,} rovers S, whence II finite number Vi" .. U... toVen 5 by I1teOI'em 4.2, that is 5 cU" U U V.... 'Th.Idng comp)ementa II!l 0 00 C. ConsIder the function

I +ng'

This fmct.km belongs to I and Exercise 2, 3, of Chapter 1 implies thal. ng/(I+ ng) tends uniformly to lone. Thull

which belongs to J tends unifonnly to

I on C, 110 for 811 large n we have

/ -/1 :~,~I < t. 1

VIU.4 Rl!1at.Xm with Opeu

Cow:rin&s

131

On the complement we have the estimate

It

-11 ~gngl:s III +

It :nglI

< 2(.

IXSeries

IX.2 Series of Positive NumbersExerc::lte 1X.2.1 (0) Prove the ~ a/tAt, uriu E 1/"(108n)I+.Jor~(>O,

Dcu tM,mu I:l{nJog"~'Prooff (c) Dcu tAt, ~6 I: Ijn(logn)(loglogn) ~r Prooff W/l6t i/~ .did; an aponmt 0/1 H to the (Ioglogn)fSolutloa. m > N_haveO 5 E~I 0l S (lB. Furthermore, we have Io"bnt S Blo"l, 110 for all n > m > N we have

t I

Olb.\.

s.

Il=m+l

_+1

t

10lbll

< (.

11IU:Ilhe part1al8Ulll8 of Eo.,bn form a Caucb,y sequence, as was to be ahown. Exc!rclae IX.2.4 Shew that E(logn)/n2 ~.J. II a > I, dou E(logn)3/n conwryeV Given 0 positiw 1nUgerd, dou E(logn)"/n con-

Solution. Let a '" 21n the text. Now we prove the general n'8U1" Let. d be a positive integer. and write a = 1+ 2c wllh (> O. Then(Iogn)"nl+2 N we bave

~)ZnJ -z.",jl' 0 auch that 10 - p/ql > C/q /Of' all inlergen p.q with q > O. {Hint: II not, there t:rist8 II ~ 1 < VI 0 such that

We now ahow

that. we can choo5e the Ill'a end! that )J!UI!ly

< ql < q:z < .... To

do 80, It 8Ilffis to prove the following lemma:Lemma. For each i, there Is infinitely eholeea of q > 0 such that

IX.2

smes of Positive Numben

1015

q 'q la-~I 0 8UCb that If 10 - p/Ill < 6, then

for

IlOlIll! "

(which depends on q).

10-~I~;q'Select k eum that k > i and 1/2" < 6. By assumption, there exists Y.t/ with t/ > 0 eoch tbat

80

10 -

y /til ~ 1/('tt(). We choo8e k > i

o.

_.

EC_I)"z3;n

_ .... Jd~in~60wtf ~ h.tdoanot_.ut,ge.tllo-

h 0 chOOlle k auch that E:'d+IIa,,1 < E. Then choose N such that; ~ Nimplies

Ibm,.t - btl < Elk,Ibm,.3 Then for all i O?: N m haveExuclse IX.5.8

b:t[ < Elk, . .. , 1b"'1,/1: -

boll < Elk.

IPm I - PI < 3. This concludes the exercise.

be tM complete notTl'Ied vecCor IfIIJf% 0/ COfItifitiow Junetiofll on ~.2lf) with tM Itlp w.mn. iOra = {a,,} ifI {I, let

ut F

1M

lX. Sere

L(a) = LII.. c:oIIfIZS1ww that L '" II CQfltinu.oUI linea" map 0/1.1 into F, and that lall lor all a E 1. 1Solution. Since [cosflZl:S; I we have

.",

lL(a)l5:

lL(a)1 :s; L [0.,.1- Rani'U fJ -= (b.. ) E 1.1 and c is a number we getL(a+fJ) = LII.. cosflZ+b.. cosflZ=

-, .._1

La..coeflX+ L

..",I

-

b.. COlIfIZ

L(oo) =

L ea"COSflZ = L II..C..-I "",I

-

= L(a) + L(J1)

-

COlIfIZ '"

cL(a) .

So L is linear. Flnally, Lis oootimwus because

lL(a) - L(.8)n :s; la -lJIh.Exerdse 1X.5.9 For % E C (eompb rtIlI'Ilbera) and 1%[ #- I, tJIuntJ that the following limit eriIU Imd give the totJluu;

/(z) = lim ,,_ %"+1

(%" -I).I itt 8l/dl1I WlIl/ to make /

-,

III it potlrible to

tk/im /(%) tuMn 1%1 =

contin-

Solutkm. SUppoIIe

1%1 < 1, then

I~ ~ ~88 " ..... OIl,tIO

-(-1)1

=

Iz~ 11- 0

/(z) = -I. If 1%1 > I, then

I~~:-II=IZ"~II-O88 n __ 011,tIO

/(%) when

Izi

/(z) = I. From thelle results we IIee that we canoot define = I tIO 88 to make / comimlOUS.

1:

n_

,"

O. H 1"1

> I, theo

n ..... 00. We now lnvestl.gate ",bat bllflPl!llS 011 the unit cirde. Let " = e 18 with o 5 < 2lr. Then 1 + %" = 1 + eMf, BO if = 0 we illllned!Mely get 1(1) => 1/2. Jf e 'I< 0, then&II

," I I =1%.. +11 .....:"+1-1

1

0

e

e

e....

1

1(%)= I+eilil= I+e-"-'

and aloce e-.. /I goes around the cirde, we cannot define lilt the pointa 18 with (J 'I< 0. So the doulain of definition of I is the aet % E C 6IlCh %= ethat

1"1'1< lor" .. I.

Exercl8e 1X.5.11 (a) For" compla, Mov t/wlt the aerie.!

ond ,tlbtract I in 1M numerotor, and we 0 partial /mdWn dec.ornpomion, pwhing 1M thing through algdmriaIlly, before ~ tuOr'T1I oboul ~UNl parti4l11unu./

u moat4r a quution of algebrn. F'orowlly, ladQr out 1/", thetl atjirn fIdd I

~ to 1/(1-.11)' 10rl%1

< 1 and to 1/.11(1-.11)' forl.llj > t./Him: ThU

(b) ~ that 1M conttetymOl! ond 1%12= b > I in the lleCOtId.

U lI.niform lor lai 5 c < 1 in 1M jim CMe,

Solution. (a) Let U.. = %"/(1- "")(1- :"+1) and let D(,,) = (1-%")(1,,'*1). Then

u" =

.3"(1 - %) D(,,)(I- ,,) =

i"="'i

1

[%.. - %,,+1]D(%)

1 = 1-"

[(z" -I) + (1- ,,"+I)]

1- 1 -1-2 L 1-%"+1 + 1

r

(1-,,")(1

".. +1)1

% ..

1 .

156

IX. Series

We get. a telescopic IIUm, wbel\C(l" U,,-I [ -I- L - I - z I-z I

~,

and theremz")(l

"'-1

z"+1)" z(1

1['

z) ~ - I

I]Z,,+1

u Izl

I, then 1/{1- Z....I) ..... I all n ..... 00 and therefore s.,{z) ..... -+ 00. U lzl > I, then l/{l- z"+I) -+ 0 118 n ..... 00 80 5,,(1:) -+ I/of(l- r). (b) Suppose Izi $ C < I. A little algebra and part (a) imply that

1/(1_ .11 2) 118 n

I, then

and ",,+I .....

00,

80 the OOllVfl"getlCe 18 uniform m tbe reglon

Izi ~ b > I.

._.(,) r:(d)

IX.6 Power SeriesDekrm~ tk

Exetclse lX.a.)

radii 0/ ~ 0/ the following potr

(0) Em;"

E::.

(e) E2"~" In'~3)' (h) Dsinn".)z".

(lin: n2r>.

E -; . (/) ;:..(c)

Solution. (a) TIle radiUll 01 comrergcncc ill I becauMl n l / ....... 1 118 n ..... 00. (b) The radiUll 01 OODvergellC(l ill I becauBe n2/ n I 118 n -+ 00. (e) The rtLdill8 of convergmce ill I because I/n'/n I 118 n -+ 00. (d) The radius 01 OOllvergellC(l is 00 because lIn 0 118 n ..... 00. (e) The rtLdius of OOllvergentil is 1/'1..

IX.6 Pm.. !lorrW

157

(f) The radiU8 of coovergeuce Is 2. (g) The radiU8 of c:om>ergenoe is 1 ~ n~/n(1+2/1'1)l/n ..... 1 ea 1'1 ..... eo. (b) The radiU8 or COII\IegeDCe ill 00 because ain(mr) = 0 ror all 1'1 hence the IIerie8 ill identica1ly zero.

Exerelac lX.6.2 Detennim the rodii 01 CQI'I~ 01 the tc'lowing .e-

""-

(0) Dlogl'I}%n. (/I) E~%"o (c) E;;Jon%". (d) E n(lOin)'%" (e) E 1":~1W (J) E ~%".

Solution. (a) The radiu8 of coovergence ill 1 because 1 S (Iogn)l/n S n l/n for 1'1 :2: 3. (b) The radiU8 of coovergenc:e is 1 because (logn)l/n ..... 1 and I'Il/n ..... 1 ea

(e) The radiU8 of coovergence is 1 because

&8

(d) The radiU8 of coovergence Is 1 becr.U8e (logn)l/n ..... 1 and 1'I1/n ..... 1. (e) The radiU8 of coovergence Is 00 bec&U8e of Example 1 in the text and the ineqt.L8lity 1/(4n - I)! SolIn!. (f) Tbe radiU8 of convergence is 00 because fOI" aoy r > 0.

1'1 ..... 00.

2""'"1,......1(2(1'1 + 1) + 1)1ean_eo.

(2'1'1 + 1)1 ..... 02"T"

Exercise lX.6.3 SvppoIe tho/. E on.:n 1wu 0 1'Odiu.l 01 com.oergence r > O. Sht'll.ll that gWen A > l/r there !%i6U C > 0 6UC/l tho/.

10nl S CAn

for oll n.

SoIutkm. Since A > l/r = Umauplanl 1/n the ineqt.L8lity 10nP/" S A holds for &11 but finitely IIWI,Y n. Hence lanl S An for all but finitely IIWI,Y n 80 we can choose C 80 large that lanl S CAn for &11 n..

Exercise IX.6.4 Let {an} be o'equenee of poIitive numbaa, ond G.mIme that limlln-f.l/an = A ~ O. Sht'll.ll that Umo~/n = A.Solutton. SUppo&e first A > 0 for simplicity. Given ( > 0, let no be eocb that A - (S On+l/On S A +l if n :2: no. Without Joss of generality. we Cll.D assume l < A 80 A - l > 0. Write

158

IX. SerIs

tben by Induction, there exilIts constanta CI(f) and Gi(t:) sucb that

Ct(f)(A -

t:)"-~

s: lin S G;(f)(A + f)tl-~.

Pu\ c;,(t:) = CI(f)(A - f)-~ and c;(f) = C2(f)(A +f)-..... Then

C:(f)t/"(A _ f)

s: a~/" s: c;(f)I/"(A +t:).

and similarly

n..

A - f + lit(n)(A - f)

s: all" s: A +f +6:l(n)(A + f).

Thill shows that lal/" - AI 2f and rondudes the proof wben A > 0. If A = 0 one ean simply write the WIllS on the lef~ of the inequalitil as 0 throughout.Exdse 1X.6..6 IJtkrmim the nuliw ()/ ~11 ()/ the /oUouMg .e-

s:

(a)E:~z". (6)EI;:'~~t".

Solution. (a) By Cbapw- IV, ~ know that n';;; nnt" fo.- 'I ..... 00, ... here lin;;; bn means that tbere exists a aequertee {u,,} sudi that b" = Una" and lim = I. See Exerci5e 18, 2, of Chapter IV. So

u:/"

and the radius of ronvttgenoo of the Beril is e. Note that we can a1so use Exercise 4 because

(n+I)1 'I" (n+I)>>+1 nI~

. (l+k)" -1(3n)3"e-3n

I e'

(b) In Exercise 18, 2, of Chapter IV, ~ proved that (3n)1 ;;; (3nt" e-3n

....

. (n1)3)'" ~(3n)1

=

( nS"e-3n )1 / I " 27' ,,~~=

lienee the power IIetie8 ht.3 a ~illt' of ronvtrgence equal to 27.

1X.6 Power Series

159

Exercise 1X.6.6 ut {a,,} ~ 0 ~ ~u 0/ poIUiw numbers appoaching O. Prwe that 1M power 6eriu Ea"z" "'lI7Ii/~~ on tM domain 0/ ~ % 6UCh thot

1%ISI and 1%-110:::6, where 6 > o.&~ 6UmrI\alion

fly pmiI.

Solution. Let. T,.(t) ... ~-oat~ and S..(t) = by parte formula gives

I::& la..lf...JA

Solution. We construct. the COO fullCtion lIS follows: Let II = (/2 and 6 = f, and Ie\ h be the funct.1on (bump fUl'lC\ion) defined to be equal to 0 If t:5 0Ot~6and

h(t) = e-I/(-)(I-.)

otbenolDe. Then

h 18 a COO funct.ion (Exerc1ee 6, Sl, of Chapter IV) with all derlvaUves equal to 0 at both (I and b. Let

g(%) '" 1- C

L

h(t)dl,

wbere IIC .. h(t)de. Then '(0) ". l,g{b) .. 0, 0 S g(%) S 1, and 9 is COO with e.ll derivatives equa.1 to 0 at. both 1I and b. For 0 .:S z S 1 define~z)

fa

_

{ o

ifOSz.:S(/2, I g(z) if (/2 S % s: (,

Ir(s%.

For -1 S z:S 0 define 'P by ~zl '" ~-z).

Th get /, le\ 1Ioll(:z) = .,,(z) and tt-t1(z) = ~(t)dt. Then let 7 If En = 1/(n 1a,,1 + I), then Ela..lf.. convergos. Finally, let g(z) =

J:

f = "".

.

L a,,1....~(z) .~

~

For Ie :2: 0, the &el"ies Eo"D"I....A converges uniformly OIl as we split the sum

1%1 s: 1 becauseA

Lo.. D'1...... = L:a"D~/..... + L..:Sot";:0:&+1En

o"D"I....

aod if n 2: Ie + I, we have IDl/....J S

by eonsuuctkJo. Hence

Ie < n we have D"(J....A) = 0 by (1) In the blot and if Ie > 1'1, then we a1Bo have D" I......{O} = 0 '"8l1se 'f ill locally con8tAllt. near the origin. Since D..' ......(O) _ I we ~ thet D""g{i;):r. Qt. as was to be shown.When

164

rx. So!riM

Exet"clee IX.,'T.6 Gitrtn 0 COO /vndion 9: 10,61 ..... R from 0 clcKtl interval, 6ht:NJ thot 9 Clm ~ t:de1ukd tc 0 CO" junction dqined Of! 0" ope7I irlltf'wl rontllireitlg [0,61.

Solution. Let 0" = gl"l(o) 8.Dd b" = 9("1(6). Then by the previous exercise we can extend 9 is some open neighborhood of 0 and III some open neighbodlOOd of b such that thiIl exteruIioo is COO on all inWval containing

~"IExen:1ae 1X.7.7 Let" 2 0 belin

inUger. SJww that tM

~~

J,,(%) =

L

co

~

(_I)....%tt+u 2....-ukl(n + k)!

V'+-v'+ -) y=O. I ( 1"

,

"

Solution. The absolute value of the k.th term is.:s l:tl,,+u/kl IlO the 81!riee converges absolutely b: every % and the convergence Is uniform on every c\otied and bounded interval A similar arg~ shows that we can differentiate term by term to get f/ and f!'. Let

(-1)""Then the coefficient of the tefm ZM2Ir-2 in the 8UIll

f!'+-f/+ I ( 1" ) 11=0

,

"

is equal to

0.. . (" + 2k)(" +2k -1) +0.. . (" +2k) +0.. . _1

_'1

20....

= (4kn +4k2)o.... +0.. . - 1.

But 0 .... _1/0.. . = -4k(n + k) so we conclude that II = J,,(%) is a soIullon to Bessel's equation.

XThe IntegralIII

One Variable

X.3 Approximation by Step MapsEXOf'eI8o X.3.1 III u a COfltinuow mll Vlued jundicm. on [">bI .oow lAat one Clll'l CIJlPl"ll%'im6l:t /16li/0f'l'II4I ~ mp juIIctitJ'M flitcH trllluu 8I'"e' leu Chon or tqlUIl to th04e 011. and abo ~ q jundioM whole wJuu

Chan or~ tothcu oj I. TM inUgroLl olthuemp~ are !Mn 1M Mandord lower and upper Riemann awn"anlgt'l!llJer

Solution. Let e > O. Choolle a ~it.ion of (a, b] lIS in 11leorem 3.1, that 18, a partition ofsize < 6 where 6 ill ChollellllO thal U(~)-/ 0 chooae. 8tep map h tiUCh tbat Ihl Is alao a step map and the inequality

1/- hi < . Then

1I/(lll-lh(t)1I:S I/(t) - h(t)l,

.....

implies tbat KlIl-jhll < (SO III ill ~ted. From t.hilI analysisweextract tbe fact. t.bat.1/1 can be uniformly approximated by poe.itive sup m&p6. so

I1I1 '"'" 1o"#1/1:::?: 0.Furt.ltetrnOlt!, If _ COlillider the IlI.ep map wbich ia 0 01\ (0,6) and 1 at II, theD 11:8 nonn Is O.1f e ill. DUmber and {h,,} ill a sequence of etep m8pll convergIl:\C to I, t.bcn theeequcoce {Ih"I} 00I.lVerge6 to III and tbesequence {lch,,1} COllY(lrges to Iell. The Integral of IlI.ep mape Is a linear functJoo so

r.(I""Il-I~r.(Ih"I)hence Menl "" lell/ll' Fmally,lf I and 9 are ~ and {f"}, {g,,} are aequCl.JOO8 of step IJlIlPlI mV&glng to I and 9, respectlY(lly, tben {fn +g,,} converges to I + 9 and {IIn + g,,1} convergce to II +91. Since

Ifn(t) + 9n(l)1 S I/n(t)1 + 19,,(t)1we cooclude that

II + 9Ut sUI" + .9Uh Unl! I nl

ia. seminOfOl.

X.ll Apprnxlm&l.lnll by Step Map.

U;;"

Exerciae X.3.4 LdF be the wdorlptJa olreolwlued~lIkdfundiqnl on an inknlal[a.bj. We h 0 there exiBt8 (I step map

II -

'PII

< (/(6 -

a).

Y/- 'PII '5 (6 - (1)1/- 'PK < (,

So the Rep functlonll aro dense III F for too LI-semIoonn. 10 prove that the step maps aro Il1so dense for the L'-eemillOl"m, choose a step map 6Ilcb that

[II-'PI' oc:; (6- a)n/-:.re:

170

X. The

~

In One Variable

'f

,

".

I

..L

.,

Let 9 be the fuDCtion resulting (rom modifying 'II 00 each 1;. By eonstructioo, 9 is in CO" and the same estimate as in Exerdse 4 sbOWlJ that

(b) The argument is thesame as In part (a) except that _lISe the ootation and estimal.ell of Exercise 5. We collClude tbat CO" is L~-delllle in F.

W'II- gil 0, then uim 6 6UCh that i{8ize IP) < 6, IhenIS(P;c,f)-LI aueh that

I, cbool:Ie a lltep uuctlon

D/-cpI. By linearity of the integral we can write

0, ,p) belongs to It ""+11, then lP is CODllt.a~ an thillinterva.l and we get.

For each k (k =- 0, .. ,n - 1), look at. the Interval!t.,tHI!' If ron a; (i =

1'' ' 1'' '

I(c.) - f{>(t)dtl S

lt~ a (t.+I -

t.).

U for some Inl.qter i (i = 1, .. ,p) the point a; belnnp to It.,t.''IJ, then If/>(t) - l(c.)1 S 2B

fOl" all t e It.,t.+.] and therefore I(c.) - CP(t)dtl S 2B(t.+I - t.) S 2Bsize(P).DlOllt two

Since each as belongs to at.

Intervals [t.,t....) we conclude that

S(P, c, J) - [

f{> S aI (t.,t.~+.II, IUe upper and lower 8UIlU, and MunsJ thot the di1fmme i8 oWk1ll tuMn the rile 01 the porlitWn i, muJlll The above limit i8 wuolll' kMteJ by

L-

r

Idh.

Solution. Given a partition P of [a,bI we define the upper RiemaJm. Stieltjes 8III1l to be

3'(P, f) =

...

.-. L.-,

1(c.r.)[h(l.+I) - h(t.)),

whore CJ: = max... 9:'ilHl I(t). Similarly, we define the lower RImwmStleltjeB 8III1l to be

B.(P, f) =where

4 = mm,.SISI /(l). C\carty, we a1way8 haveB.(P, f) S 3'(P, f).

...

L 1(4)[h(tk+l) -

h(t.)),

In fact, If P' ia a ~6nement til P, that adding finitely rua.oy paints, tiff! _ tbe.t

Is.

If P' ia obt.alned from P by

X.4 Properties olthe IntEf;fal

113

5.(P,J) $. 5.(P',J) $. S(P',/) $. "S(P,J).This is nbviclua when 1" is Ilbtained by adding llJle pUDt, b 0 choose a poal.tive DUmber Q IIlICb a that 2Ba U such that J~ -III 0, there exista a step UllIp If' on ItIobi such that UI - 'PI < elO. Then), o If' Is also II step map on (o,bl and for all :t E10, ill we have

I), o/(:t) -),0 .,,(:t)I5. Ol/(:t) - .,,(:t)1 < e.80 I), 0

I - ) , 0 'PI < e, thereby proving that), 0 I is regulated. Consider a sequence {y;>,,} 01 step mapll whicll COJ1\'el'gC8 uniformly to I. Tb 080 P(z)th = O. Since 2: 0, W(l IJ1UIIt have P = 0 and. therelOre I = 0-

r

Is COnliDUOUll and

Exerdlte XJ.2.2 PnM UUIt if lUG continuow ftmctioo. then

IImll."~-o

_I

h2 h 2/ (z)d%- 7f /(0). +z

Solution. Suppoae that lis t:OtltinltO\l8 on 1-1, 1)' Then W(l can extend I to a eontinUOUll and bounded runetloo on R. Call tb18 extension I again, and let M be a bound ror I. Tben ~ 10, S, ofCha.pterXm lmpllt"s

."

lim!'[ h'l h p!{:z:)dx= I{O), ~""'II" -co +

'The second limit fuDows from

:S: M= M = M

== M(arct.an(Blh) - ardan{l/h.Let B ..... 00 and t.heo h ..... 0 to OODCtude.

" I. " I. I.I I

h'l +v

h

dx

1 I hl+(%/h)'lUI I

B1h

II"

+u'

dx

Exercise Xl.2.3 (An Integral Operator) d. K .. K(%, tI) be 8 eonMUOW function on the ndGngk ~ !lrt the ~litiu

For I

E COllc,dI),

define the/unctirm TI = TKJTK/(:Z:)"" [

6y theJOTmWa

K(:z:.l/l!C.,)dJ/.

with tM .up oonn on both IpIlCU. M ProtJe that TK i, a ccmtinuow linear f1\l)P with the L'-nmm on both."",...

5011\tlon. ;:..,);...) ...

(I

and if n 'F m we baWl

(xn,Xm>j" ejn~-lioIud:l: =. 1 [e't..-m)~l = O. _. I(n m) -.The ortbogoMJity of the ftmctlons 'AI, '/!ft. "'.. was studied in Exercise 3, 1, 01 Chapter IX.Exercise XD.l.2 On the 'JIllUC" c.msi~ting olallvutor- Z = (ZII'" ,z..) IlJId w = (w" ... , w,,) uohere Zj, Wi E C, define &he fl"O(fud

(x,w) - zllih + ... + .:."tii.". Show t1ult th1$ it a 1lermition product, and that (ZIZ) =Z=

o.

(I

q lind only if

190

X1L Fourier Seriell

Solution. The verification of all the properties ill routine. We have

(:r,to) = L:rotoi = LtDil'i'" (to,:r).'-I ;_1

Then we also haV(!

(tm,v) =

L au;{ii ... 0 L tijl1i '" o(u, v)

and aL'Kl

(u,at =Finally, we haV(!

L

tijlro; =

15"L

lI;ii,

= 15"(u,v).

{v, v) = Llvil~ ~ O.

Now IiUppoee (:r,:r) = 0. Then E l:ro[' = 0 80 l:ro[ = 0 for all i, hence.1: = 0. Conver8cly, II .1: = 0. then it ill clear that (.1:,.1:) = 0.Exei">:Ue xn.l.3 Ld (l ~ tM xl 01 all (c,,} 01 compla n!mllrers IUCh t1utt E Ie"r conve'iU. Sho1JJ that r " a vector qaa, and tMt if {o.. }, {P,.} o.rt! etemenu oll~, then tM product

,equt'I'ICt'

(l... },{p.))- La,J!." II herI'IIUiIm product $UCh thaI (0,0) = 0 if and on/II if 0 = O. (9Iof/J t1utt 1M 8eriu on tM righl ~', tiring the Schwan inequality /or each

partial .ftmI. Ux tM Mime meUwd to prove the jim ,tatemtnt.) Prot1e t1utt (2 Ii' COfIlplele.

Solution. (i) Suppose {o..}, {,B,,} are In (2. Then

L 1o. + "'I' N we have

D...,-.....,F< ...J

lienal, oJ: ellCh ;, 1(J",j - o",,jr < ( and tberefore {O..,J):'I ill a Cauchy &equcnce in C for each j. Let oJ ~ its Iimlt e.nd Jet- X "" {OJ}j:,t. We assert that X beloDgll to P and that X .. ..... X as n ..... OQ witb respect to the norm given by the hermitian product studied at the bq9nning of the exercise. For each positive integer M and all f1i, m > N we have

i .. 1

L 10n.J - omJf

M

l

rwe Ute paral1eklgram law: For ntl v, WEE we ~

Iv + wi' + Iv - wi' = 21vl' + 2Dw1l 2 Solution. Using the properties 01 tt1s hermitian product we find

...

Iv +w1 2 = (v+w,v+w) =

(v,V) + (V,w) + (w,v) + (w,w),

Iv - '111 2 = (v - w, v- w) = (v,v) - (v, w) - (w, v) + (w, w), hence nv + wU 2 + Iv - wB 2 = 2nvll 2 + 211wn 2 , as WlIIJ to be shawn.Exen:1.se XII.l.12 Let E be II tutor space with 0 hermitia>l produd whldl is porititlt Ikfinih, UllIt is if gvl = 0, IMri v = O. Let F be 0 complete .w6plll:e 01 E. Let vEE lind ld0""",E~'

in!

Dz - vII.

b ..) if Cauchy, wing Ute JIllf'lIlIehJgrum laID on

mm lin tlel'l'll:nt ZO E F such thllt II = IIv - zol {Hint: Let {lI..} be 0 SBqUCIU in F such thllt ny" - vi converyes w o. ~ thlltProve thlIt then!

y" -v... = ~ -z) -

Wm -z).J

Solution. LeI. {y,,} be Il. sequence in F 1IUcl1 that The pareUelogram law implies

Iy" -v! ..... 1I as n ..... 00-

2Wl/.. - v8' + 211l1m - vg 2 = III.. - Yml' + Iy" +Ym - 2toM'.

B",

106

len Four\@rs..riH

II~ + lim - 2111' =41~(~ +lIm) because F is 8 wbllpaoo. So

-r

2: 40'

II~Given e

-lImM' :S 21t~ - vii' + 21l1m - \lY' - 40'.

> 0 tbete exlst8 N well t.hal if A 2: N, then U~ - til' < 0 2 + e. Then IW all A, m 2: N we haveB~

- u...8' < 4t,

110 lhe sequence {~} is Cauchy and therefore hall II. limit 2:0 E E. Since F Is cloaed, %0 E F, aod Gxo- \II '" a. Remark. In this exercbe we assume that F ill II. sub8pace. From this assumption. we concluded tbet H~ + y",) E F wbeoover ~,lIm In F. So If F Is ooIy MSumed to be II. dosed c:onvex set. the concluslon of die exaclse &Ill boJds. For II. 8tmllar result, see &erci5e HI, 2, ol Chapter

vn.

Exercise X1I.1.13 Not4tion llf in ~ preceding eureUe, ClUUfl'\l!! that F ." E. Show that tMre aUU II "tdor ~ E E VJhidl h perpendicular tc F and z.';' O. {Hinl: Ld \I E E, \I f F. Ld xo be Q8 in ~ It, lind let If. = \I - ZO. ChIJnging \I br 0 tromlGtion, ~ ma~ Q8.wme thot ~ = \I,

~

"""

l:rl' :S N:r + xII' Jor aU x E F.

YOtI om _ tllJ(l met1uKk ~ of tMm is to con.ri4er ~ + tQz, uth $7lI/Jll plUim lOlUuu oJ t, CIf1d ,UitGbk (T E C. 'I'M otMr is to un ~ '

........,~

Solution. We wish to 8how thatWrite _~

~

is ~lcular to all x in F.

Method 1. Suppoee nxl ." 0 find let. e be the component of :r aJoll& z.-

ex + ex 110 by Pythagonus 1%0' = Iz - exU' + Rexl' 2: 11% - exl'U~M' = I~ -

By theclloioe of ~ we have l:rft' So "z + ~U' for aU ~ E F. Heuce we obtain

exl' whJcb implies that lex.' .. 0 and therefou e = 0, &8 was to be Ilbown. Method 2. Consider l:rl' $. "z + ul' lor amaIl values ol t. Expandingy"","

(~,z.) $. (:r,z) + 2!Re(z,x) +I'(x,:r)...hich implies

o:S 21:Re(~,:r) + t'(:r, :r).

we pick t smaD oCtile oppoelte sign olRe(z.,:r) to get II. COIltradietwn whenR.e(:r,:r) .;. O. Using ix Insttad ol:r proves t.hal Im(:r,x) = O. Thill concludeslhe proof.

XII.I Hermitian ProdIlCtl and 0rth0c0na111,y

197

Exerelse XILl.14 Not{Jliqn G, In ~ II! and 13, ~)., E-+ C lie II "01ltinuow Iimor mop. S/Iol# th4t ~ ui.W ~ E E ,uch th4t >.(:r;) = (:r;.II) for 4U % e E. {Hint: Ld F lie tM NlIpllCt' of all % e E IUCh th4t ).(%) "" O. Show fMt F u cWed. lj F " E, we ~ 13 to gd lUI

dement:

e E.: tJ. F,:;' 0, IUCh Mot: u ~ to F. ShOtlJ th4tI07IIe

norrr.dIIo = X\%)/U:U

there trim

comple:r; 0 Iuch th4t cu = II2 ./

MJluM'

tM require~tI,

Solution. The subepace F is closed because If %" e F and %" ..... %, then 0- ).(%,,) -+ >.(%) by continuity. SO >'(%) = 0 and %e F. ~ F" E and choose.1: as io Exereise 13. 1'beD Jet II = az where 0'" >.(.1:)/I.1:ft2. Any % io E C&II be expcell88d as%=%-

>'(%) >.(%) ).(.1:):+ ).(.1:).1:.

But u = % - ().(%)/>.(.1:.1: be\00g8 to F becauae

).(11) = >.(%) So

~:: ).(.1:) "" 0.

).(%) >'(%) (%.11) - (u + ).(.1:) z.oz) = li ).(z) (z.1:) "" >.(z).

Exercise XII.I.Ui Let E be II tIedor 6JIlIU overC with II humWlUl prod. uct whk1l u JIlUititle definit.e. Ld VI V.. lie demmt.J oj E. 000 GSI'II7IIoe thlIt t1leII are Uneor4r in4ependent. Thu mam.J.' ijclvi + .. '+e",v,. = 0 with c; e C, then c; "'" 0 for all j. Pnwe thlIt jur mdi k = 1.. n there tri.st thmenu III..... ,lilt wllkA ore oflmgth 1. mvtually ~ (thol u (lUi.1II1) ... 0 i), and generote tM MJme IOOIpoce 01 Vt , Vt. Th_ dementi ore UI'IiqIIe l.'/(%).VM the L~-f1OmI on the 'fIaCe E oj continvow functions on la.b!. Prove lhIIt if h., I .. /II"l! in E, mulutJllll orthogonal, lind 0/ L~-nonn equal to J, and an rigmJunctitmf tuith rupect to 1M _ number >. '! 0, then n is bounded bIJ II ntm'Iber deperwIing on'" on K and >.. (Hint: Ulle 1'IIeomn 1.5.)Solution. Flx % withCI

S % S b. Ld.4 '" U~.J.) =

(K,A)

(K

.J..

)

Theorem I.Slmpliell that Lh.II4FI S IKn~ 110

" E>.2IM%)fl S.IKI~~.

XII.2 'nigonomeuic Polypom1ah as a 'l'btaI Family

1M

If B ill a bound for K on lhe lIQuare

..

[a.61 x la.liI, we 8ee that.

L>.2IMr)1 2 :s: B 2 (b_a),

,

1'hiIl last Inequality ill true for all x with 0

respect to x and lI6ing the fad that "",II~ = I, we get

:s: :r :s: b, 80 integrating with

>.2 n :s: dl(b- a)2.

XII.2 Trigonometric Polynomials as a Total FamilyExercise XD.:U Ld 0 be an irTatiornU number. Ld I be a amtinvow l~iOfl (complu 11ahsed, ola real tlGriabk), periodU: oj perWd I. Show thai

H_

1 11m N

L I(no) = -,H

/.' ,

J(x)dz.

IHint: First, lei I(x) "" ~ lor lOme integer k. 1/ k " 0, then ~ can com~ upliatJl/ tM .urn Of} lhe left. and one Itt.f al once that the geometric 6Ul'Jl8

are bovtukd, whence tM auerlion lollmtM. 1/ k _ 0, it U even more trivial. Second, prove that 1/ tM relalioNhip u tnle lor IUI(l fundioru, 111m it u tnle for a linear comllinalion ollhue functWm. Hen illhe relatioruhtp iI true for a fami11/ of ~ of 0 vector IplICe oflunctioN, then it b tnle for aU elementt of l1tU vector 1plICe. Third, prollt' that 1/ the rdaliONhip b true for a aequm 0. choose k such that 11/- hi 0 because O.:s T < I. DIR 2 is abo verified. Indeed, each Pr is cor4luuoue bee&use the aerIea OOllverge& unifonnly (tomparison to the geometric aeriea). Furthermore, we

can integrate the series term by teml, 80

[P.(9)d8=But lin O,'cr e""dO = 0 80

2~tJ~rl"'ewdO.-~

[

P.(l1)d8 =

~ J~ 1d8 = I.

Finally, we verify DIR 3. U 6::; O.:s '11", then CCIIO::; CC11680

,> , P.(9)d8::; , [ /. 1_2Tcos6+~d8.But 1-2TCCII6+r2 -. 2(I-cos6) and I_T2 -. 0 as r -. I, andBince Pr Is even, there exists < TO < 1 such tbat if TO < T < I, then

-'P.+[P. 0. Then DIR 2 implies tba.t

CPr - J)(z) - I(z) = [V(Z - 0) - I(z)/ P.(6)d8.1be function /18 uniformly eontinuoue 00 5 SO there exlsts 6 > 0 8UCh tbM I/(z - 0) - l(z)1 < f whenever 101 < 6. Let B be a boond Cor J, and select To. 0 < TO < I Ruch that 11 TO < T < I, then

[ -. P,+[P' 0 for alIlalge n we have

ID.. _/(r) - /(:1:)1 < (.Lel6 bulso < r. So the middle integre.Iln D.. _g(x)_g(z) 8lId D.. _/(r)_ I(z) are equal. The RIemann-Lebesgue lemma lmpliea tha1 the fim and third integralOOOVi!fgC to 0, 80 weoonelude that for1arge n. D.. -g(Z)-9(Z) issmall. Therefore, the Fourier aeriesoC 9 OOllwrge8 to 9(Z). Conversely, the Fourier eeriee of I oonvelle8 lo I(z) at:l: irthc Fourier aeries of 9 converges to 9(Z) at:l:.

Exercise XII.4.4 Let F be the ~k nonned mctor 8ptJ 01 rontinuow periodit: fundi0n8 on (-"-,,,-J with tM 8Ufl norm. Let I be the W'dt>r ~ of all mil ~Ht ., {a,,} (n ., 1,2, ...) 8uch thol EIa,,1 am11tJ'pU We Ikfine, 0.1 in Eureiu 8 of CIlapkr JX, S, tM norm

101. ~ L ""'I

-.~

Let B be the do8ed tmlt 6aU 01 ro4il''''''') = ".~.

11. J:./(x)coefl.:rth. Then

Ibn -~I:S: -

Now we prove that 13 E B. U not, then for &Orne N and c > 0, we haveN

. If. -

I/(x) -/,,(x)llcosfl.:rld:r:S: -1/- M.

'" .

-,

LIb,.,IZ:I+C.U/- AR < c/(4N). Then

The hypothe&ea implies that fOt ecme k we have

1~1-1~-r+b,.,IZ:~I-I~-b,.,IZ:Ib,.,I-2!/-A.~Ib,.,I-2~' Lla~1 z: LI6,,1..... 1N N

n_1

N; ~ 1+~.

But 10k", :s: 180 we ~ a contr'adittlo