Problems in Mathematical Analysis III: Integration [1 ed.] 0821832980, 9780821832981

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Problems in Mathematical Analysis HI Integration

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STUDENT MATHEMATICAL LIBRARY Volume 21

Problems in Mathematical Analysis III Integration W. J. Kaczor M.T. Nowak

#AMS

AMERICAN MATHEMATICA L SOCIET Y

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Editorial Boar d David Bressoud , Chai r Danie

l L . GorofT Car

l Pomeranc e

2000 Mathematics Subject Classification. Primar y 00A07 , 26A42 ; Secondary 26A45 , 26A46 , 26D1 5 , 28A1 2 . For additiona l informatio n an d updates o n this book , visi t www.ams.org/bookpages/stml-21 Library o f Congres s Cataloging-in-Publicatio n D a t a Kaczor, W . J . (Wieslaw a J.) , 1 949 [Zadania z analizy matematycznej . English ] Problems i n mathematica l analysis . I . Rea l numbers , sequence s an d serie s / W. J . Kaczor , M . T. Nowak . p. cm . — (Studen t mathematica l library , ISS N 1 520-91 2 1 ; v. 4) Includes bibliographica l references . ISBN 0-821 8-2050- 8 (softcove r ; alk. paper ) 1. Mathematica l analysis . I . Nowak , M . T . (Mari a T.) , 1 951 - II . Title . III. Series . QA300K32513 200 0 515'.076— dc2199-08703

9

Copying an d reprinting . Individua l reader s o f thi s publication , an d nonprofi t libraries actin g fo r them , ar e permitte d t o mak e fai r us e of th e material , suc h a s to copy a chapte r fo r us e in teachin g o r research . Permissio n i s grante d t o quot e brie f passages fro m thi s publicatio n i n reviews, provide d th e customary acknowledgmen t of the sourc e i s given. Republication, systemati c copying , o r multiple reproductio n o f any materia l i n this publication i s permitted onl y unde r licens e fro m th e American Mathematica l Society . Requests fo r suc h permissio n shoul d b e addresse d t o th e Acquisition s Department , American Mathematica l Society , 20 1 Charles Street , Providence , Rhod e Islan d 02904 2294, USA . Requests ca n also be made b y e-mail t o [email protected] . © 200 3 b y the American Mathematica l Society . Al l rights reserved . The America n Mathematica l Societ y retain s al l right s except thos e grante d t o the United State s Government . Printed i n the United State s o f America . @ Th e paper use d i n this boo k i s acid-free an d fall s withi n th e guideline s established t o ensure permanenc e an d durability . Visit th e AMS home pag e a t http://www.ams.org / 10 9 8 7 6 5 4 3 2 1 0

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Contents

Preface vi

i

Part 1 . Problem s Chapter 1 . Th e Riemann-Stieltje s Integra l 3 §1.1. Propertie s o f th e Riemann-Stieltje s Integra l 3 §1.2. Function s1 o f Bounde d Variatio n

0

§1.3. Furthe r Propertie s o f th e Riemann-Stieltje s Integra 1l

5

§1.4. Prope r Integral s 2 1 §1.5. Imprope r Integral s 2

8

§1.6. Integra l Inequalitie s 4

2

§1.7. Jorda n Measur e 5

2

Chapter 2 . Th e Lebesgu e Integra l 5

9

§2.1. Lebesgu e Measur e o n th e Rea l Lin e 5

9

§2.2. Lebesgu e Measurabl e Function s 6

6

§2.3. Lebesgu e Integratio n 7 1 §2.4. Absolut e Continuity , Differentiatio n an d Integratio n 7

9

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Contents

VI

§2.5. Fourie r Serie s 8

4

Part 2 . Solution s Chapter 1 . Th e Riemann-Stieltje s Integra l 9

7

§1.1. Propertie s o f th e Riemann-Stieltje s Integra l 9

7

§1.2. Function s 1 o1 f Bounde d Variatio n

4

§1.3. Furthe r Propertie s o f the Riemann-Stieltje s Integra1l 2

6

§1.4. Prope r Integral s 4

3

1 §1.5. Imprope r Integral s 6

4

§1.6. Integra l Inequalitie s 20

7

§1.7. Jorda n Measur e 22

8

Chapter 2 . Th e Lebesgu e Integra l 24

7

§2.1. Lebesgu e Measur e o n th e Rea l Lin e 24

7

§2.2. Lebesgu e Measurabl e Function s 26

8

§2.3. Lebesgu e Integratio n 28 1 §2.4. Absolut e Continuity , Differentiatio n an d Integratio n 29

6

§2.5. Fourie r Serie s 3

6

Bibliography - Book s 35 1 Index 35

5

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Preface

This i s a seque l t o ou r book s Problems in Mathematical Analysis I, II (Volume s 4 an d 1 2 i n th e Studen t Mathematica l Librar y series) . The boo k deal s with th e Riemann-Stieltje s integra l an d th e Lebesgu e integral fo r rea l function s o f one rea l variable . Th e boo k i s organize d in a wa y simila r t o tha t o f the firs t tw o volumes , tha t is , i t i s divide d into tw o parts : problem s an d thei r solutions . Eac h sectio n start s with a numbe r o f problem s tha t ar e moderat e i n difficulty , bu t som e of th e problem s ar e actuall y theorems . Thu s i t i s no t a typica l prob lem book , bu t rathe r a supplemen t t o undergraduat e an d graduat e textbooks i n mathematica l analysis . W e hop e tha t thi s boo k wil l b e of interes t t o undergraduat e students , graduat e students , instructor s and researche s i n mathematical analysi s an d it s applications. W e also hope tha t i t wil l b e suitabl e fo r independen t study . The first chapte r of the book is devoted to Riemann an d Riemann Stieltjes integrals . I n Sectio n 1 . 1 w e conside r th e Riemann-Stieltje s integral wit h respec t t o monotoni c functions , an d i n Sectio n 1 . 3 w e turn t o integratio n wit h respec t t o function s o f bounde d variation . In Sectio n 1 . 6 w e collec t famou s an d no t s o famou s integra l inequal ities. Amon g others , on e ca n fin d OpiaP s inequalit y an d Steffensen' s inequality. W e clos e th e chapte r wit h th e sectio n entitle d "Jorda n measure". Th e Jorda n measure , als o calle d conten t b y som e authors ,

vn

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Vlll

Preface

is no t a measur e i n th e usua l sens e becaus e i t i s no t coun t ably addi tive. However , i t i s closely connecte d wit h th e Rieman n integral , an d we hope that thi s section will give the student a deeper understandin g of the idea s underlyin g th e calculus . Chapter 2 deals with the Lebesgu e measur e an d integration . Sec tion 2. 3 present s man y problem s connecte d wit h convergenc e theo rems tha t permi t th e interchang e o f limi t an d integral ; LP spaces o n finite interval s ar e als o considere d here . I n th e nex t section , absolut e continuity an d the relation between differentiation an d integration ar e discussed. W e present a proo f o f th e theore m o f Banac h an d Zareck i which state s tha t a function / i s absolutely continuou s o n a finite in terval [a , b] i f and onl y i f it i s continuous an d o f bounded variatio n o n [a, b], an d map s set s o f measur e zer o int o set s o f measur e zero . Fur ther, th e concep t o f approximate continuit y i s introduced. I t i s worth noting her e that ther e i s a certain analog y betwee n tw o relationships : the relationshi p betwee n Rieman n integrabilit y an d continuity , o n the on e hand , an d th e relationshi p betwee n approximat e continuit y and Lebesgu e integrability , o n th e othe r hand . Namely , a bounde d function o n [a , b] i s Rieman n integrabl e i f an d onl y i f i t i s almost ev erywhere continuous ; an d similarly , a bounde d functio n o n [a , b] is measurable, an d s o Lebesgu e integrable , i f an d onl y i f i t i s almos t everywhere approximatel y continuous . Th e las t sectio n i s devoted t o the Fourie r series . Give n th e existenc e o f extensiv e literatur e o n th e subject, e.g. , th e book s b y A . Zygmun d "Trigonometri c Series" , b y N. K . Bar i " A Treatis e o n Trigonometri c Series" , an d b y R . E . Ed wards "Fourie r Series" , w e foun d i t difficul t t o decid e wha t materia l to includ e i n a boo k whic h i s primaril y addresse d t o undergraduat e students. Consequently , w e hav e mainl y concentrate d o n Fourie r co efficients o f function s fro m variou s classe s an d o n basi c theorem s fo r convergence o f Fourie r series . All the notatio n an d definition s use d i n this volume ar e standard . One ca n find the m i n the textbook s [27 ] and [28] , which als o provid e the reade r wit h th e sufficien t theoretica l background . However , t o avoid ambiguit y an d t o mak e the boo k self-containe d w e start almos t every sectio n wit h a n introductor y paragrap h containin g basi c defi nitions an d theorem s use d i n th e section . Ou r referenc e convention s

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Preface

IX

are bes t explaine d b y th e followin g examples : 1 .2.1 3 or I , 1 .2.1 3 o r II, 1 .2.1 3 , whic h denot e th e numbe r o f th e proble m i n thi s volume , in Volum e I o r i n Volum e II , respectively . W e als o us e notatio n an d terminology give n i n th e first tw o volumes . Many problem s hav e bee n borrowe d freel y fro m proble m section s of journals lik e the America n Mathematica l Monthl y an d Mathemat ics Today (Russian) , an d fro m variou s textbooks an d proble m books ; of thos e onl y book s ar e liste d i n th e bibliography . W e woul d lik e t o add tha t man y problem s i n Sectio n 1 . 5 com e fro m th e boo k o f Ficht enholz [1 0 ] an d Sectio n 1 . 7 i s influence d b y th e boo k o f Rogosinsk i [26]. Regrettably , i t wa s beyon d ou r scop e t o trac e al l th e origina l sources, an d w e offer ou r sincer e apologies if we have overlooked som e contributions. Finally, w e woul d lik e t o than k severa l peopl e fro m th e Depart ment o f Mathematics o f Maria Curie-Sklodowsk a Universit y t o who m we are indebted . Specia l mentio n shoul d b e mad e o f Tadeusz Kuczu mow an d Witol d Rzymowsk i fo r suggestion s o f severa l problem s an d solutions, an d o f Stanisla w Pru s fo r hi s counselin g an d Te X support . Words o f gratitud e g o t o Richar d J . Libera , Universit y o f Delaware , for hi s generou s hel p wit h Englis h an d th e presentatio n o f th e ma terial. W e ar e ver y gratefu l t o Jadwig a Zygmun t fro m th e Catholi c University o f Lublin , wh o ha s draw n al l th e figure s an d helpe d u s with incorporatin g the m int o th e text . W e than k ou r student s wh o helped u s i n th e lon g an d tediou s proces s o f proofreading . Specia l thanks g o t o Pawe l Sobolewsk i an d Przemysla w Widelski , wh o hav e read th e manuscrip t wit h muc h car e and thought , an d provide d man y useful suggestions . Withou t thei r assistanc e som e errors , no t onl y ty pographical, coul d have passed unnoticed . However , we do accept ful l responsibility fo r an y mistake s o r blunder s tha t remain . W e woul d like t o tak e thi s opportunit y t o than k th e staf f a t th e AM S fo r thei r long-lasting cooperation , patienc e an d encouragement . W. J . Kaczor , M . T . Nowa k

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Part 1

Problems

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Chapter 1

The Riemann-Stieltje s Integral

1.1. Propertie s o f th e Riemann-Stieltje s Integra l We star t wit h som e basi c notations , definition s an d theorems . B y a partition P o f a close d interva l [a , b] we mea n a finite se t o f point s xo, xi,..., xn suc h tha t a = XQ < x\ < .. . < x n-\ < The numbe r n(P) = maxjx ^ — xi-\ : raes/i of P.

x n = b.

i = 1 , 2 , . .. ,n} i s calle d th e

For a functio n a monotonicall y increasin g o n [a , b] w e writ e A ^ = a(xi) - a(xi-i). If / i s a real function bounde d o n [a , 6], we define th e uppe r an d lowe r Darboux sum s o f / wit h respec t t o a an d relativ e t o P , respectively , by nn

U(P,f,a) =

Y,MiAa t, L(P,f,a)

=

^m^Aa, ,

where Mi = su

p /(#)

, ^

i = in

f /(#)

• 3

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4

Problems. 1 : Th e Riemann-Stieltje s Integra l

We als o pu t pb rb

/ fda = mfU(P if,a), /

fda = supL(P , f,a),

J aJ

a

where th e infimu m an d th e supremu m ar e taken ove r al l partition s P of [a, 6], and call them, respectively, the upper an d the lower Riemann Stieltjes integral . I f th e uppe r an d th e lowe r Riemann-Stieltje s inte grals are equal, we denote the common valu e by / fda an d cal l it th e Riemann-Stieltjes integra l o f / wit h respec t t o a ove r [a , b]. I n thi s case w e sa y tha t / i s integrabl e wit h respec t t o a , i n th e Rieman n sense, an d w e writ e / G 1Z(a). I n th e specia l cas e o f a(x) = x w e get th e Rieman n integral . I n thi s cas e th e uppe r (lower ) Darbou x sum correspondin g t o a partitio n P , an d th e uppe r (lower ) Rieman n integral ar e denoted , respectively , b y U(P,f) ( L ( P , / ) ) , an d f (fafdx) .

afdx

The Rieman n integra l o f / ove r [a , b] i s denoted b y j a fdx.

Moreover, correspondin g t o ever y partitio n P o f [a , b] w e choos e points t i , t 2 , . . . , t n suc h tha t Xi-\ < t{ < Xi, i = 1 ,2 , . . . , n , an d consider th e su m n

5(P,/,a) = 53/(ti)Aai. 2 =1

We say tha t lim S(PJ,a)

=

A

if, fo r ever y e > 0 , ther e i s S > 0 suc h tha t /i(P ) < S implie s tha t |5(P, /, a) — A\ < e fo r al l admissibl e choices o f U. I n th e cas e whe n a(x) — x w e se t

5(P,/) = £/(*i)(x i -x < _ 1 ). i=l

Throughout thi s section , / i s always assume d t o b e bounde d an d a monotonicall y increasin g o n [a , b]. I n th e solution s w e will often us e the followin g theorem s (see , e.g. , Rudi n [28]) . Theorem 1 . / G 71(a) on [a , b] if and only if for every e > 0 there exists a partition P such that

U(PJ,a)-L(PJ,a) 0 an d (x ifx G [ - o , o ] n Q , [0 i f x G [ - a , a ] \ Q . Find th e uppe r an d lowe r Rieman n integral s o f / ove r [—a , a]. 1.1.5. Sho w tha t th e so-calle d Rieman n functio n 0i f{x) = {l/q i

f x i s irrational o r x = 0 , f x = p/q, p G Z , g r G N , an d p, g are co-prime ,

is Riemann integrabl e o n ever y interva l [a , b]. 1.1.6. Le t / : [0,1] - + R b e define d b y settin g

[0 otherwise . Show tha t J Q f(x)dx = 0. 1.1.7. Sho w tha t / : [0,1] - > R define d b y

\x ~ ix] otherwis

e

is Riemann integrabl e o n [0,1 ] .

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6

Problems. 1 : Th e Riemann-Stieltje s Integra l

1.1.8. Defin e JO if * G [-1,0], / /(x) = < an d a(x \ l i f x G ( 0 , l ] , \l Show tha t / G 7£(a ) althoug h li m S(P,

O if * G [-1,0), )= < iixe [0, 1 ] . / , a) doe s no t exist .

1.1.9. Sho w tha t i f / an d a hav e a commo n poin t o f discontinuity i n fa, 61 , then li m S(P, f.a) doe s no t exist . LJ

'M

(P)-o v

)

1.1.10. Prov e tha t i f li m S(P,

/ , a) exists , the n / G 1 1 (a) o n [a , 6]

/x(P)-0

and lim 5(P ; , /, a) = / / d a .

M(P)-O V

J

a

Show als o tha t fo r ever y / continuou s o n [a , 6], th e abov e equalit y holds. 1.1.11. Sho w tha t i f / i s bounded an d a i s continuous o n [a , 6], the n / G 1 Z(a) i f an d onl y i f li m S(P, / , a) exists . MP)-o 1.1.12. Le t (c i f a < x < x*, a(x) = < [d i f x * < x < 6 , where c < d an d c < a(x*) < d. Sho w tha t i f / i s bounde d o n [a , 6] and suc h tha t a t leas t on e o f th e function s / o r a i s continuous fro m the lef t a t x * an d th e othe r i s continuou s fro m th e righ t a t x*, the n / G 1 1 (a) an d C

f(x)da(x) =

f(x*)(d-c).

Ja

1.1.13. Suppos e that / i s continuous o n [a , 6] and a i s a step function that i s constant o n th e subinterval s (a , ci), (ci, C2),..., (c m , 6), wher e a < c\ < C2 < • - • < c m < 6. Sho w tha t b ™>

/ f(x)da(x) =

/(o)(a(a+ ) - a(a)) + £ /(c

fc )(a(c+)

- a(c fe"))

fc=i

+ /(6)(a(6)-a(fe")) .

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1.1. Propertie s o f the Riemann-Stieltjes Integra l 7 1.1.14. Usin g Rieman n integral s o f suitabl y chose n functions , find the followin g limits : (a) li m ( + + •• • + — ), v; n-o o \n + 1 n + 2 3nJ ini n2 3 + -" + ( b ) Jn-^oo „y n ( ^ r3 T+-l^3 +n ^3 -+ T2 ^ n

(c) h m ^-

j,

k

3

- f n3 / '

> 0,

(d) li m - ij/( n + l)( n + 2)---( n + n), n—>oo 77 , / 77

, 77

, 77

,

(e) li m si n +i2 i 2 n sin H2 + 2 2 n h + ol/n 0. /

1.1.17. Fo r k > 0, calculat e fc fc l i m / l + 3 + -- - + ( 2 n - l )

n—>oo \ 7 7 /

fc

^

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Problems. 1 : Th e Riemann-Stieltje s Integra l

8

1.1.18. Suppos e tha t / i s twic e differentiabl e o n [0,1 ] an d / " i s bounded an d Rieman n integrable . Sho w tha t

^y(£mdx-±±f

2i-i\\ _ r ( i ) - r ( o ^ 2n 2

)

4

1.1.19. Fo r n G N , defin e

1 1

J

_

and 2

2

2n + 1 2

n+3 4

n- 1

Show tha t lim [7 n = li m V n — I n 2. n—>oo n—

->oo

Moreover, usin g th e result s state d i n 1 .1 .1 6 and 1 .1 .1 8 , sho w tha t lim n(In 2 — U n) — - an n-^oo 4

d li

n—>o

m n 2 (ln2 — V n) — —. o 3

2

1.1.20. Sho w tha t i f / i s Rieman n integrabl e ove r [a , 6], the n / ca n be change d a t a finit e numbe r o f point s withou t affectin g eithe r th e integrability o f / o r th e valu e o f it s integral . 1.1.21. Sho w tha t i f / i s monotoni c an d a i s continuou s o n [a , 6], then / G 71(a). 1.1.22. Prov e tha t i f / G 71 (a) an d a i s neither continuou s fro m th e left no r fro m th e righ t a t a point i n [a , 6], then / i s continuous a t thi s point. 1.1.23. Le t / b e Rieman n integrabl e an d a continuou s o n [a , b]. I f a i s differentiabl e o n [a , b] except fo r finitel y man y point s an d a' i s Riemann integrable , the n / G IZ(a) an d pb pb

/ f(x)da(x) =

Jo..

/ f(x)a'(x)dx. Ja

1.1.24. Le t / b e Rieman n integrabl e an d a b e continuou s o n [a , b] except fo r finitel y man y points . I f a i s differentiabl e o n [a , b] except

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1.1. P r o p e r t i e s o f t h e Riemann-Stieltje s Integra l 9 for finitely man y point s an d a' i s Riemann integrable , the n / G 7Z(a) and pb pb

/ f(x)da{x)= f{x)a'{x)dx J a

+

f{a){a{a

+

)-a{a))

J a m

+ £ f(c

k)(a(4)

-

a(c~)) + f(b)(a(b) - a(b~)),

k=l

where c^ , k = 1 , 2 , . . ., m, ar e point s o f discontinuit y o f a i n (a , b). 1.1.25. Calculat e j_ 2x2da{x), wher

e

x+ 2 i

f -2 a , an d pu t V ( / ; a , o o ) = li m V(f;a,b). b—>oo

Show tha t i f V(f; a , 00) < 00 , the n th e finite limi t li m f(x) exists . x—*oo

Does th e opposit e implicatio n hold ? 1.2.15. Fo r / define d o n [a , b] an d a partitio n P — {xo,x'i,.. . , x n } of [a , 6], w e for m th e su m

vup)=x;i/(si)-/(zi-i)iProve tha t i f / i s continuou s o n [a , 6], the n lim V ( / , P ) = V(/:a,6) , that is , fo r an y s > 0 there exist s 6 > 0 such tha t //(P ) < < 5 implie s V(f;a.b)~V(f,P) m > 0 , x G [a , 6]. Sho w t h a t ther e ar e tw o monotonicall y increasin g functions g an d / i suc h t h a t

/?,(x)

1.2.19. Comput e th e positiv e an d negativ e variatio n function s o f (a) f(x)=^-\x\, a : G [ - l . l ] (b) f(x)

=

cosa: , x

(c) / ( ! • ) = J - M . . r

,

€ [0.2TT] ,

e [0.3] .

1.2.20. Assum e t h a t / i s o f bounde d variatio n o n [a , b]. Prove t h a t i f / i s continuou s fro m th e righ t (left ) a t TQ , the n Vf i s als o continuou s from th e righ t (left ) a t TQ. 1 . 2 . 2 1 . Sho w t h a t th e se t o f point s o f discontinuit y o f a functio n / o f bounde d variatio n o n [a J)] i s a t mos t countable . Moreover , i f

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14

P r o b l e m s . 1 : T h e Riemann-Stieltje s Integra l

{xn} i s the sequenc e o f points o f discontinuity o f / , the n th e functio n g(x) = f(x) — s(x), wher e s(a) = 0 an d s(x) = f(a+) - f(a) + £ (/(*+

) - f(x~)) + f(x) - / ( * " )

Xn/ also satisfies th e Lipschitz conditio n with the same Lipschitz constant . 1.2.24. Prov e tha t i f / i s o f bounde d variatio n o n [a , b] and enjoy s the intermediat e valu e property , the n / i s continuous. Conclud e tha t if / ' i s o f bounde d variatio n o n [a , 6], the n f i s continuous . 1.2.25. Prov e tha t i f / i s continuously differentiat e o n [a , 6], the n X

«/(*)= [

\f'(t)\dt.

Ja

1.2.26. Sho w tha t i f / i s continuous an d a monotonicall y increasin g on [a , 6], the n th e functio n F(x) = / f(t)da(t),

[a , b],

xe

./a

is o f bounde d variatio n o n [a , 6]. 1.2.27. I f / O ) = li m f n(x) fo r x G [a, b], the n n—->oo

V(f;a,b)< li

m V(/„;a,b) . n—• oo oo o

1.2.28. Suppos e tha t th e serie s ^T

a

o

n a n d Yl b n ar e absolutel y con -

n=l n=

l

vergent, an d le t {x n} b e a sequenc e o f distinc t point s i n (0,1 ) . Prov e that th e functio n / define d b y

/(0)=0, f(x)=

Yl

a

- + J2

b

^ fo

r

*e(0,l ]

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1.3. Furthe r Propertie s .. .

15

is continuou s a t ever y x ^ x n , n G N , an d f(xn) -

f(x~) =

a n , f(x+)

Prove als o tha t

- f(x n) =

b n.

oo

V(f;0,l) = Y,(K\ + K\). n=l

1.3. Furthe r Propertie s o f th e Riemann-Stieltje s Integral In this section w e consider Riemann-Stieltje s integral s wit h respec t t o functions o f bounded variation . I f a i s a function o f bounded variatio n on [a , 6], and i f a = p — q7 where p and q are monotonicall y increasing , then rb pb

rb

/ f(x)da(x) =

/ f{x)dp{x) - / f(x)dq(x),

Ja Ja

Ja

provided tha t / G Ti(p) an d / G l^{q) (se e th e definitio n i n Sectio n 1.1). Thi s definitio n doe s no t depen d o n a decompositio n o f a int o a difference o f tw o increasin g functions . T h e o r e m 1 . If functions f and a are of bounded variation on [a , b] and one of them is continuous, then f(b)a(b) - f(a)a(a) -

j f(x)da(x) =

f a(x)df(x).

The abov e formul a i s calle d th e partial integration formula. T h e o r e m 2 . Suppose f and Lp are continuous on [a , b] and ip is strictly increasing on [a , b]. If ip is the inverse function of (p, then / f(x)dx

=

/ f{rl>{y))dil>(y).

Ja J

(f(a)

The formula i n Theorem 2 is called the change of variable formula. T h e o r e m 3 . Suppose that either f is continuous and a is of bounded variation on [a,b], or f and a are of bounded variation on [a , b] and a is continuous. Then

I

b|

f(x)da(x)\
0, ^ c

n

< oc , an d defin e

n=\ oo Q ;r

C

( )= ^

np{x-Xn),

where th e functio n p is given b y 0i fx

< 0,

1i f x

> 0.

Prove tha t i f / i s continuou s o n [0,1 ] , the n DC

fda = ] P c

nf(xn

n=l

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1.3. Furthe r Propertie s .. .

17

1.3.5. Suppos e tha t a i s a continuou s functio n o f bounde d variatio n on [a , b] suc h tha t fo r ever y / continuou s o n [a , 6], rb

/

0.

f(x)da(x) =

J a

Show tha t a i s constan t o n [a , 6]. 1.3.6. Le t a b e monotonicall y increasin g o n [0,7r ] an d suc h tha t / sin

xda(x) = a(7r ) — a(0) .

Show tha t Ja(0) i

f XG[0,TT/2)

I CK(TT ) i f x

,

£ (7r/2,7r] .

1.3.7. Fin d a functio n a monotonicall y increasin g o n [0,1 ] an d suc h that

[f{x)Mx) = m±m for ever y / continuou s o n [0,1 ] . 1.3.8. Fin d a functio n / continuou s o n [a , b] an d suc h tha t rb

f(x)da(x) =

a(b) — a (a)

for ever y a monotonicall y increasin g o n [a , b]. 1.3.9. Assum e tha t a i s of bounde d variatio n o n [a , b] an d th e func tions / n , n = 1 ,2,... , ar e Riemann-Stieltje s integrabl e wit h respec t to a ove r [a , b]. Prove tha t i f {f n} converge s uniforml y o n [a , b] t o / , then / i s Riemann-Stieltje s integrabl e an d rb

rb

I f(x)da(x) =

li m / f

n(x)da(x).

1.3.10. Calculat e

f1 nx{\ - x

lim / o

l n

) dx.

n-o c 7Q

1.3.11. Fo r a o f bounde d variatio n o n [0,1 ] , find lim / x

n

da(x).

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P r o b l e m s . 1 : T h e Riemann-Stieltje s Integra l

18

1.3.12. Suppos e tha t {a n} i s a sequenc e o f function s whos e tota l variations ar e uniforml y bounde d o n [a , 6], that is , there i s a positiv e M suc h that V(a n; a , b) < M fo r al l n. Prov e that i f {a n} i s pointwise convergent t o a o n [a , 6], the n fo r ever y / continuou s o n [a , 6], lim / f(x)da n(x) = / f(x)da(x). ^°° J a J a 1.3.13. Suppos e tha t {a n} i s a sequenc e o f function s whos e tota l variations ar e uniformly bounde d o n [a , 6], and tha t {a n} i s pointwise convergent t o a o n [a , 6]. Suppos e als o tha t {fn} i s a sequenc e o f continuous function s uniforml y convergen t o n [a,b] t o / . Prov e tha t n

rb

lim / f

rb

n(x)dan(x) =

«/ a J

/ f(x)da(x). a

1.3.14. Prov e th e followin g Helly selection theorem. Le t {a n} b e a sequence o f function s define d o n [a,b] such tha t |a n (a)| < M an d V(an; a,b) < M fo r n G N . Then {a n} contain s a subsequenc e {a nk} convergent t o a functio n a o f bounde d variatio n o n [a , 6], an d fo r every continuou s functio n / , rb

rb

lim / f(x)da nk(x) = / /(x)da(x) . k -^°°Ja J a 1.3.15. Prov e th e followin g theorem of Helly whic h generalize s th e result i n 1 .3.1 2 . Le t / b e continuou s an d a o f bounde d variatio n on [a , b]. I f th e sequenc e {a n} o f function s o f uniforml y bounde d variation converge s t o a o n a se t A dens e i n [a , b] and suc h tha t a, b G A, the n rb

lim / f(x)da

rb

n(x)

=

/ f(x)da(x).

n—»oo

1.3.16. Prov e th e second mean value theorem. Suppos e / i s mono tonic an d a i s continuou s an d o f bounde d variatio n o n [a , b]. Then there i s a poin t c G [a, 6] such tha t /»6

/»c

rb

/ /(ar)da(a: ) = / ( a ) / da(x) + f(b) / da(x) J a

Ja

Jc

= /(a)(a(c ) - a(a) ) + /(6)(a(6 ) - a{c)). 1.3.17. Prov e th e followin g Bonnet forms of the second mean value theorem .

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1.3. Furthe r Propertie s .. .

19

(a) I f / i s a positiv e increasin g functio n o n [a , b] an d a i s a con tinuous functio n o f bounde d variatio n o n [a , 6], the n ther e i s a poin t c G [a , b] suc h tha t b rb

f(b) / da(x) = f(b)(a(b) - a(c)) .

f(x)da(x) =

(b) I f / i s a positiv e decreasin g functio n o n [a , b] an d a i s a con tinuous functio n o f bounde d variatio n o n [a , b], the n ther e i s a poin t c £ [a , 6] suc h tha t /(a ) / da(x) = f(a)(a(c) -

f{x)da{x) =

a(a)).

1.3.18. Fo r 0 < a < 6, find

inn r^^dx. 1.3.19. Fo r x > 0 , prov e tha t (a) i f F(x ) = / * + 1 sin(* 2)dt, the n |F(a;) | < 1 /x , (b) i f F ( » = / * + 1 sin(e*)cfe , the n |F(x) | < 2/(e x ). 1.3.20. Sho w tha t i f th e function s / , OL\,OL2 bounded variatio n o n [a , 6], the n b rb

a re

continuou s an d o f

pb

/ f(x)a 1 (x)da2{x)+ f(x)a

f(x)d(a1(x)a2{x)) =

JaJ

1 2{x)da

(x).

a

1.3.21. Sho w tha t i f / i s continuou s an d o f bounde d variatio n o n [a, 6], the n fo r a positiv e intege r n , rb r

b

f(x)d({f(x))n) =

nf (f(x))

n

df(x)

Ja

U

((/W)

n+ 1

n1 +

- ( / ( « ) ) n 1+ ) -

1.3.22. Suppos e tha t / i s continuou s o n [0,1 ] . Fin d th e followin g limits: (a) li m \n [ n x n f(x)dx) . n—>oo \ "

(b) li m (nfie-

u

"

J nx

f(x)dx),

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20

Problems. 1 : Th e Riemann-Stieltje s Integra l (c) Jo x

lim

J0 x

n

f{x)dx

n x2

e dx

2n

(d) li m (jnj* f(x)sm

(27rx)dxY

(e) JQ1 f(x)sin 2n(27rx)dx lim n ~*°° Jo e x2sm2n(27rx)dx ' 1.3.23. Prov e th e followin g monotone convergence theorem for the Riemann integral. I f {/ n} i s a decreasin g sequenc e o f Rieman n in tegrable function s o n [a , b] which converge s o n [a , b] to a Rieman n integrable functio n / , the n nb rb

lim / f n(x)dx = / f(x)dx. n -*°° J a J a 1.3.24. Prov e th e followin g monoton e convergenc e theore m fo r th e lower Rieman n integral . I f {f n} i s a decreasin g sequenc e o f bounde d functions o n [a , 6], and i f li m f n{x) = 0 for x E [a , 6], the n 71—•OO

rb

lim / f n(x)dx =

0.

Ja

1.3.25. Prov e th e followin g Arzela theorem. I f {/ n } i s a sequenc e o f Riemann integrabl e function s o n [a , b] whic h converge s o n [a , b] t o a Riemann integrabl e functio n / , an d i f there i s a constant M > 0 such that \f n(x)\< M fo r al l x G [a, b] an d al l n G N, the n b rb rno

no

lim / f n(x)dx = / f{x)dx. n oc -" Ja J a 1.3.26. Prov e th e followin g Fatou lemma for Riemann integrals. I f {/n} i s a sequenc e o f nonnegativ e Rieman n integrabl e function s o n [a, 6] whic h converge s o n [a , b] to a Rieman n integrabl e functio n / , then br rru

I f(x)dx < J a n-*oo

b pO

li m / J

f

n(x)dx.

a

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1.4. Prope r Integral s

21

1.4. Prope r Integral s 1.4.1. Fo r n G N , calculate (a;,) /

f4 \

x

-1 | n

(c) /

\lnx\dx , (d

(e) /

tan

(g) /

n

: : —\dx, (b ) / Jo \x — 2\ + \x — 3 | ,y

Jo

:

j0 sin

2n

n

)/

£cb, (f

)/

n

sm 0 J t

xdx, /

cos o

17

xdx,

_ 9_dx, + cos 2 x

;

dx,

Jo sin " x' +' co " "s ~x

sm x x-h cosn x-ax.

1.4.2. Fo r n E N, use the integral J Q ( l — x 2) dx 1 fn\ 1

fn\ 1

1 VOy 3

/ n\ ,

r1

v

\lj 5\2J

y

t o calculat e /

n

2 n + l V^

1.4.3. Suppos e tha t a functio n / ha s an indefinite integra l (o r antiderivative) o n an interva l I ; tha t is , there i s a differentiable functio n F suc h tha t F'{x) — f(x) fo r x G I. Sho w tha t i f a one-sided limi t of / a t xo G I exist s an d is equal t o a, the n f(xo) — a. 1.4.4. Le t / b e defined b y „ N (sin

± i f x^O,

[c i

f x = 0,

where c G [—1,1]. For which value s o f c does ther e exis t a n antideriv ative o f / ? 1.4.5. Le t xn — 1/y/n fo r n G N. Construct a function / continuou s on (0,1 ] and suc h tha t / > 0 on [x2k,X2k-i], / < 0 on [x2k+i,X2k] and F(x 2k-i) - F{x 2k) = F{x 2k+i) ~ F{x 2k) = 1 /fc , wher e F i s an antiderivative of/. Exten d the function / t o [0,1 ] by setting /(0) = 0. Prove tha t / ha s an antiderivative o n [0,1 ] , bu t | / | doe s not. 1.4.6. Suppos e / i s continuous o n [0,1 ] , Sho w tha t r

\ xf(smx)dx Jo 2

7T r

= — / f(sinx)dx. J0

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22

Problems. 1 : Th e Riemann-Stieltje s Integra l

Using thi s equality , comput e •

I

2r ?

xsm x o sin x + cos 2n x dx, n 2n

G N. 1.4.7. Assum e tha t / i s continuou s o n [—a, a], a > 0. Sho w tha t (a) a ra

/

i

f(x)dx = 2 f(x)dx,

f / i s even ,

-a JO

(b)

pa

/

f(x)dx = 0 , i f / i s odd . 1.4.8. Le t / : R— » R b e continuou s an d periodi c wit h perio d T > 0 . Prove tha t fo r ever y rea l a, —i

J

/ f(x)dx

= / f(x)dx.

Ja

JO

1.4.9. Le t / : R— > R b e continuou s an d periodi c wit h perio d T > 0 . Prove tha t fo r ever y a < 6, lim / f(nx)dx =

-—/

f(x)dx.

1.4.10. Suppos e tha t / G C ( [ - l, 1 ]) . Find th e followin g limits : (a) li m ±Cf(sinx)dx, (b) li m ^ /

n

/(|sinx|)dx,

(c) li m L xf(s'm(27rnx))dx. u

n—>oo

1.4.11. Fo r / G C([a,6]), find th e followin g limits : (a) li m f f(x) cos(nx)dx, li

m f /(# ) sin(nx)dx ,

(b) li m f f(x) si n (nx)dx. n—>oo a

1.4.12. Suppos e / G C([0,oo)) an d se t an = I f(n + x)d x fo Jo

rn

= 0,1 ,

Suppose als o tha t li m a n = a. Fin d th e limi t li m J

Q

f(nx)dx.

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23

1.4. Prope r Integral s

1.4.13. Fo r a functio n / positiv e an d continuou s o n [0,1 ] , comput e

f f(x)+f(l-x)

-dx.

1.4.14. Sho w that, i f / Jo i s continuous an d eve n on [—a , a], a > 0 , the n / " . $ ? * - ! >*• 1.4.15. Sho w tha t i f / i s nonnegative an d continuou s o n [a, b] and

f f{x)dx = 0, Ja

then / i s identicall y zer o o n [a , b]. 1.4.16. Sho w tha t i f / i s continuou s o n [a , b] and fo r eac h a,/3 , a < a< (3 0 , if / * + T f(t)dt = J QT f(t)dt fo r ever y x G R , the n / i s periodic wit h perio d T > 0 .

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24

P r o b l e m s . 1 : T h e R i e m a n n - S t ieltjes Integra l

1.4.20. Comput e

(a) ta

(„./-_**) ,

(b) li m ( n'3 f 2n xdx

\

(c) li m / I n ( x H )n dx, r2n xdx n-*coJ1 n-^oo \3TT J^ arctan(nx) 1.4.21. Fin d th e followin g limits : (a) li m / e-

Rsint

dt,

(b) li m / \/xsinxdx, (c) li m / dx. n >00 - Jo v l+ ^ 1.4.22. Fo r a functio n / continuou s o n [0,1 ] , find n

lim / f(x

)dx.

n-^ooJQ

1.4.23. Sho w that , i f / i s Riemann integrabl e o n [a , &], then ther e i s 0 G [a, b] suc h tha t

/ f(t)dt = [ f(t)dt.

Ja

JO

1.4.24. Le t / b e continuou s o n [a , 6] and le t / f(x)dx = 0.

Ja

Show tha t ther e i s 0 G (a, 6) suc h tha t

f f(x)dx = f(9).

Ja

1.4.25. Le t / G C([a,6]), a > 0 , an d le t / /(x)d z = 0 .

Ja

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25

1.4. P r o p e r Integral s Show tha t ther e i s 0 G (a, b) such tha t

f f{x)dx = 6f{6). Ja

1.4.26. Suppos e / , g G C([a, &]). Sho w tha t ther e i s 0 G (a, b) suc h that g(0) f f(x)dx =

/(0 ) / ^(x)dx .

1.4.27. Suppos e / , # E C([a, 6]). Sho w tha t ther e i s 0 G (a, 6) suc h that (/(0) / f(x)dx Ja

= f(0) [ g(x)dx. JO

1.4.28. Suppos e / an d g are positive an d continuous o n [a , b]. Show that ther e i s 0 G (a, b) such tha t =1. 1.4.29. Le t / b e positiv e an d continuou s o n [0,1 ] . Prov e tha t fo r every n G N there i s 0{n) such tha t e n Ir1 r ()r 1 — / f(x)dx = / f(x)dx + / f(x)dx. n

JO

JO

Jl-O(n)

Find th e limit lir a (n0(n)). n—>oo

1.4.30. Le t / G C^QO, 1]). Show tha t ther e i s a 0 G (0,1) such tha t

f(x)dx = f{Q)+ l-f'{0).

/* 2 Jo 1.4.31. Le t / G C ([0,1]). Sho w tha t ther e i s a 0 G (0,1) such tha t

jf 1 /(x)dx = /(0) + i// (0) + ^/ ,/(fl). 1.4.32. Suppos e / G C^IO, 1]) an d /'(()) ^ 0 . For x G (0,1], le t 0(x) be suc h tha t

r f(t)dt=f(6(x))x. Jo Find th e limi t

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26

Problems. 1 : Th e Riemann-Stieltje s Integra l

1.4.33. Suppos e / i s continuous, nonnegativ e an d strictl y increasin g on [a , b]. Fo r p > 0, le t 0(p) denot e th e uniqu e numbe r suc h tha t

(f(o(p)r = ~ j\f{x)Ydx. b a

~ J

a

Find li m 0(p). p—>oo

1.4.34. Suppos e / i s continuous o n [a , b] an d suc h tha t

f x nf(x)dx = 0 Ja

for n = 0 , 1 , . . . . Prov e tha t / i s identicall y zer o o n [a , b\. 1.4.35. Suppos e / i s continuou s o n [a , b] an d suc h tha t / x nf(x)dx = 0 Ja for n = 0 , 1 , . . . , N. Prov e tha t / ha s a t leas t N -\- 1 zeros i n [a , b]. 1.4.36. Suppos e tha t / e C([-a,a]),a >

0. Sho w tha t

(a) i f I x 2nf(x)dx = J —a

0 fo r n = 0 , l , . . . ,

then / i s odd o n [—a , a], (b) i f 1 I x 2n+ f(x)dx = 0 fo r n = 0 , 1 , . . . , J —a then / i s even o n [—a , a]. 1.4.37. Fo r / continuou s o n R , fin d

i- [ (f(x + h)-f(x))dx.

h^O h J a

1.4.38. Fo r / continuou s o n R an d a < 6, defin e

g(x)= ( f(x + t)dt. Ja

Find th e derivativ e o f g. 1.4.39. Fin d th e followin g limits :

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27

1.4. Prope r Integral s 1 (a) li m - = / I n ( 1 + 4 = ) dt, x->oo , / x Ji /I \ T

Z" 1

Vi

X

C O S

^7

(b) h m x / — f2?r-dt,

, ,T c li

m^

n si n \fidt ^ _ ,

{ I [ x Pit) \ (d) li m I —a / I n d t 1 , wher e a > 1 , an d P an d Q ar e x-^oo y ^ J 0 Q\t) J polynomials positiv e o n R + . 1.4.40. Fin d th e followin g limits : (a) li m f -— / dt)

(b)

^Gr

,

(i+sint)i +t

(c) li m (\ fV

^

dtV

(r 2 \ 1 /(*2) l (d) li m \ e dt) x ^°° \Jo ) 1.4.41. Sho w tha t i f / i s continuous o n [0,1 ] , the n lim (f \f(x)\*dx)

=

ma x \f(x)\.

1.4.42. Suppos e tha t a real-value d functio n f(x,y) i s continuou s o n a rectangl e R = [a , b] x [c , d]. Sho w tha t i{y) = / f(x,y)dx Ja is continuou s o n [c , d]. 1.4.43. Suppos e tha t a real-valued functio n f(x,y) define d o n a rectangle R = [a , b] x [c , d] is Rieman n integrabl e ove r [a , b] for eac h 2/ G [c , d], an d th e partia l derivativ e -g- is continuou s o n R . Prov e that dy

/ f(x,y)dx= /

—(x,y)dx.

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28

P r o b l e m s . 1 : T h e Riemann-Stieltje s Integra l

1.4.44. Le t / b e positiv e an d continuou s o n [0,1 ] . Fin d i/p

lim ( / (f(x)) pdx ' p-

1.4.45. Le t / b e positiv e an d continuou s o n [0,1 ] . Fin d ^ I/P lim j\f{x)fdx p—» —oo

1.4.46. Prov e tha t fo r ever y positiv e intege r N th e equatio n /t

t

e t 1+

Jo l0

' { v. +

2

2N

t + +

^. - (2Ny.]dt =

N

has a solutio n i n th e interva l (TV , 27V). 1.4.47. Le t P b e a polynomia l o f degre e n suc h tha t

f /o

Prove tha t

xkP(x)dx =

0 fo

r fe =

l,2...,n .

Jo

2

f (P(x)) 2dx =

( n + l ) 2 ^ / P(x)d ix a

1.4.48. Sho w tha t i f / i s continuous o n R = [a , b] x [c , d], the n /I

/ f{x,y)dx\dy= /

I

/ f{x,y)dy\dx.

1.4.49. Prov e tha t fo r 0 < a < b, 1

xb - x a , , 1 +6 ax = I n IQ 1 I nx +a

I

1.5. Imprope r Integral s Assume tha t / i s define d o n [a , oo) an d i s Rieman n integrabl e ove r any finite interva l [a , b]. Then w e defin e /•OO

fO

I f(x)dx = li m / f(x)dx, J ab ->°° J a provided tha t thi s limi t exist s an d i s finite. I n thi s cas e w e sa y tha t the imprope r integra l o n th e lef t converges ; otherwise , w e say tha t i t

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1.5. Imprope r Integral s

29

diverges. Th e imprope r integra l J_ f(x)dx

i s define d analogously .

The integra l J_^ f(x)dx i s defined b y + o o pa

r~\-oo

)dx

/

f(x)dx = f(x)dx+ f(x -oo J— oo Ja provided tha t bot h imprope r integral s o n th e righ t converge . Th e definition doe s no t depen d o n th e choic e o f a. For / define d o n [a , b) and Rieman n integrabl e ove r eac h close d subinterval o f [a , 6), the imprope r integra l J f(x)dx i s define d a s rb ro—r/ pb—r] ro

/ f(x)dx = li m / f(x)dx, J a ^o+ J a provided th e limi t i s finite. I f / i s define d o n (a , b] and Rieman n integrable ove r eac h close d subinterva l o f (a , 6], the imprope r integra l f f(x)dx i s defined i n a simila r way . 1.5.1. Fo r n G N and positiv e a, calculat e dx r-l*

pZTV

Jo sin c) /

4[

(b) [

:x + cos 4 x '

x n{l-x)adx, a > - l Jo Jo

, (d

e) Jop - ^ - ^ V

(f) f

x

s)

/

(T

)/

\n(sinx)dx, Jo n

(-\nx)

Jo

dx,

x n ln a xdx,

dx 7i x

dx y/x2 — 1

2

In a:

k) /

ln( l + cosa:)dx ,

(1)

rda:,

f'"K)r

+x

2 *

1.5.2. Fo r 0 < a < 1 , defin e

/a 0*0

rai

r ii

— —a — X Vx\

00+ x - » l

-

lim

/ ( * ) + / ( S ) + " + /( aJT1 )

n—»oo fl

exists a s a finite limit , the n th e imprope r integra l J 0 f(x)dx exists . 1.5.5. Sho w b y exampl e tha t i n th e precedin g proble m th e assump tion tha t on e o f th e one-side d limit s i s finite canno t b e omitted . 1.5.6. Usin g th e resul t i n 1 .5.3 , find (a) li m

n—>-oo fl

,\r

, . 7T

.

2TT

.

( n -1

u (b) li nm \ si n — s i n - - . . . sin 2n 2n 2n

\ fc=i \

)T T

fc=i /

1.5.7. Suppos e tha t th e functio n / : (0,1 ]— > R i s monoton e an d fo r some a G M the imprope r integra l j 0 x af(x)dx exists . Sho w tha t lim x a+1 f(x) = 0 . x-+0+

1.5.8. Verif y whethe r th e followin g imprope r integral s converg e o r diverge:

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1.5. Imprope r Integral s

(a)

31

L ^ ' (b

)

L IT^ '

1

(c) /

r r (-lnx)°dx , a e l , (d Jo Jo o1

l

)/

dx — - — ^ , o,ieR \ mx j

x

,

xdx + x 2 sin 2 x

1.5.9. Suppos e tha t / an d g ar e positiv e o n [a , oo) an d J a°° g{x)dx diverges. Sho w tha t a t leas t on e o f th e integral s / f(x)g(x)dx, Ja Ja

/

dx

-Tr^ f{

X

)

diverges. 1.5.10. Prov e th e followin g Cauchy theorem. I n orde r tha t th e im proper integra l J a°° f(x)dx converge , a necessary an d sufficien t condi tion i s that, give n e > 0, ther e i s a^ > a suc h tha t fo r a2 > a\ > ao, a

r 22 I />a

/(z)da; < £. J a\

1.5.11. Sho w tha t th e imprope r integra l J f(x)dx converge s i f an d only if for every increasing sequence {a n }, an > a , divergin g to infinit y the serie s ^ /*a

(1) }

n

\ J/ / f(x)dx, n = 1

wher

ea

0

= a,

^an_i

converges. Moreover , i n th e cas e o f convergence ,

/ f(x)dx

=

j ; / /(*)£*!;

.

Show als o tha t i f / i s nonnegative, a sufficien t conditio n fo r th e con vergence o f th e imprope r integra l i s tha t ther e i s a n increasin g se quence {a n }, a n > a, divergin g t o infinit y fo r whic h th e serie s (1 ) converges. 1.5.12. Fo r positiv e a, stud y th e convergenc e o f th e integra l

I

o1

dx + x a sin 2 x

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32

Problems. 1 : Th e Riemann-Stieltje s Integra l

1.5.13. Suppos e / i s positive o n [0 , oo) an d f Q f(x)dx exists . Mus t f(x) ten d t o zer o a s x — » oo ? 1.5.14. Suppos e / i s positive, differentiable o n [a , oo), and \f'(x)\ < 2 for x > a. Doe s th e convergenc e o f J°° f(x)dx impl y tha t f(x) tend s to zer o a s x — > oo ? 1.5.15. Prov e tha t i f / i s uniforml y continuou s o n [a , oo) an d th e improper integra l f °° f(x)dx converges , the n li m f(x) = 0 . 1.5.16. Assum e tha t / : [0 , oo)— * [0,oo) i s monoton e decreasing . Prove tha t i f

then li m xf(x) =

f(x)dx
oo

hold; tha t is , th e conditio n li m xf(x) —

0 doe s no t impl y th e con -

x—>oo

vergence o f J 0 f(x)dx. 1.5.17. Assum e tha t / : [1 , oo)— > (e , oo) i s monotone increasin g an d f°° dx

h fix) f °° dx (a) Prov e tha t als o / —•— —— = oo. :ln/(x) (b) Giv e an example of a function / satisfyin g th e abov e assump dx r x l n / ( x ) l x ) )converges . tion fo r whic h / _ ^ £f ^ ^ n n( l_n£f/ („w

1.5.18. Le t / b e a continuou s functio n o n [0 , oo) suc h tha t lim (fix)-]- f f(t)dt] ^°° \ Jo / exists a s a finite limit . Prov e tha t li m f(x) = 0 . x

x—>-oo

1.5.19. Le t / b e a nonnegativ e an d continuou s functio n o n [0 , oo) and Jo Jo

Prove tha t

f(x)dx
0 such tha t f(t)dt
0 ) wit h / 0°° g(x)dx < oo. The n /»oo p\

/ f(x)g(x)dx Jo Jo

R b e a nonnegativ e decreasin g functio n an d let g : [a , b] —> R b e a nonnegativ e an d Rieman n integrabl e functio n such tha t

/

J aJ

g

2{t)dt

a

and rb nb

/ gi (t)dt= / J aJ

g

2(t)dt,

a

Show tha t i f / i s increasing o n [a , 6], the n nb

/»6

/ f(t)

9l(t)dt
f f(t)g 2(t)dt. J aJ

a

1.6.43. Us e the Steffense n inequalit y t o prov e tha t i f / i s contin uously differentiat e o n [a , b] and m < f'(x) < M (m < M) fo r x G [a, b], then +

(M - m)A 2 /(& ) - / ( a ) ( (6-a) 2 " 6a~ (b

M - m)(f r - a - A) - a) 2

2

where A

/(&)-/(a)-m(6-a) M -m

1.6.44. Prov e tha t i f / i s continuousl y differen t iable o n [a , b] and m < f'(x) < M (m < M) fo r x e [a , 6], the n / f{x)dx (b-a)\
0.

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1.7. J o r d a n M e a s u r e

55

1.7.19. Sho w t h a t a bounde d se t A i s Jorda n measurabl e i f an d onl y if ther e exis t tw o sequence s { B n } an d { C n } o f J o r d a n measurabl e sets suc h t h a t B n C A C C n an d li m | B n | = li m | C n | = | A | . n—>oo n—>o

o

1.7.20. Sho w t h a t th e se t A=|(x,|/)Gl

2

: 0 < x < l ,0

< y
0 , consist s o f thre e con gruent leaves , tangen t t o eac h othe r a t th e origin . Fin d th e are a o f one o f th e leaves . 1.7.25. Fin d th e are a o f th e limago n r = a + bcosO, wher e positiv e numbers a an d b ar e given , distinguishin g th e case s i n whic h a > b, a = 6 , an d a < b. 1.7.26. Fin d th e are a o f th e loo p o f th e curv e x 5 + y 5 = 5ax where a > 0 i s given .

2 2

y,

1.7.27. Fin d th e are a o f th e regio n t h a t lie s withi n t h e limago n r = 1 + 2 cos 6 an d outsid e th e circl e r = 2 .

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56

Problems. 1 : Th e Riemann-Stieltje s Integra l

1.7.28. Suppos e / i s nonnegativ e an d continuou s o n [a , b]. A plan e region A / = {(x,y) £ M 2 : a < b, 0 < y < f(x)} i s revolve d abou t the x-axis , generatin g a solid o f revolution V wit h volum e |V| . Prov e that |V| = 7 T / f\x)dx. Ja 1.7.29. Suppos e / i s nonnegative an d continuou s o n [a , b] and 0 < a. Prove tha t th e volum e o f th e soli d V generate d b y th e plan e regio n Af = {(x,y) G M 2 : a < 6, 0 ( 2k + i) 71 "] a b o u t t h

e

x-axis. 1.7.32. Sho w tha t th e lengt h L o f the ellips e

©Mir-

satisfies 7r(a + b) < L < 7r^2{a 2 + b 2).

1.7.33. Sho w tha t th e lengt h o f th e ellips e

is given b y

where e = ~

b

i s th e eccentricit y o f th e ellipse .

1.7.34. Deriv e th e followin g formul a fo r th e lengt h o f a curv e i n th e polar coordinate s (r , 6). I f / i s continuousl y differen t iable o n [a,/?] , then th e lengt h L o f th e curv e r = /(0) , a < 9 < (3, i s give n b y

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57

1.7. J o r d a n M e a s u r e 1.7.35. Fin d th e lengt h o f th e curv e wit h pola r equatio n Q

(a) r = a s i n 3 - , 0 o

< 6 < 3TT.

( b ) r =~i l —E> ~ f < ^ < f K; 1 + co s 0 2 - 2 1.7.36. Fin d th e lengt h o f the curv e wit h pola r equatio n

0 = \(r + iy l 0 there i s a n open se t G suc h that A C G an d ra(G) < ra*(A) +£ . Sho w als o tha t there i s a Q$ se t A 2 suc h tha t A C A 2 an d ra(A 2) = m*(A) . 2.1.12. Prov e that fo r A C R the following statements are equivalent : (i) A i s measurable . (ii) Give n e > 0 , ther e i s a n ope n se t G D A suc h tha t m*(G\A) 0, ther e i s a close d se t F C A suc h tha t m*(A\F) 0 there i s a finite unio n W o f open interval s suc h tha t ?7i* (W A A ) < £ , wher e W A A i s th e symmetri c differenc e fo r W and A , tha t is , W A A = ( W \ A ) U (A \ W) . 2.1.14. Fo r A C R , defin e th e Lebesgue inner measure m*(A ) o f A by settin g ra*(A) = sup{ra(B ) : B e S0t , B c A } , where 9J t denotes th e a-algebr a o f al l measurabl e subset s o f R. Prove : (a) I f A i s measurable , the n m * (A) = ra*(A). (b) I f A i s a subse t o f a bounde d close d interva l I , the n m*(A) = |I | - r a * ( I \ A) . (c) I f ra* (A) = m*(A ) < oo , the n A i s measurable . (d) Fo r an y set s A an d C , m*(A U C) + ??7 * (A H C) > ra*(A) + 7n*(C) . (e) I f A n , n = 1 ,2,... , ar e pairwis e disjoint , the n 777* ( | J A \n = l I

n

J > ^m^An). n

=l

(f) I f M i s o f measur e zero , the n m* (A U M) — m*(A).

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62

Problems. 2 : Th e Lebesgu e Integra l

2.1.15. Prov e tha t A C I, wher e I i s a bounde d an d close d interval , is measurabl e i f an d onl y i f |I| = m * ( A ) + m * ( I \ A ) . 2.1.16. Le t A an d B b e given sets of finite outer measure . Sho w tha t m * ( A u B ) = m*(A ) 4-m*(B) i f and onl y i f there ar e measurable set s Ai an d B i suc h A C A i , B C B i an d ra(Ai n Bi ) = 0 . 2.1.17. Le t A an d B b e given sets of finite outer measure . Sho w tha t if ra*(AUB) = r a * ( A ) + m*(B) , t h e n r a * ( A u B ) = ra*(A) + m*(B) . 2.1.18. Prov e tha t i f {A n } i s a n increasin g sequenc e o f measurabl e sets, the n m I Iv J/ A n -- l // \n=

J = li m m ( A n ) . n,

—»oo

2.1.19. Prov e tha t i f {A n } i s a decreasin g sequenc e o f measurabl e sets an d ra(Afc) i s finite fo r a t leas t on e fc, then m I P i A n J = li m m ( A n ) . \' ' / \n=l J

n—>o

o

Show als o tha t th e assumptio n tha t ra(Afc) < o o fo r som e k canno t be omitted . 2.1.20. Fo r a sequence {A n } o f sets in R, we define th e limi t superio r and th e limi t inferio r o f {A n } b y settin g oo

lim A

n

oo

oo

= p | I ) A n an d li

mA

n ~^°° , -, 7 n-^oc 1 k=ln=fc K= n=k

n

oo

= ( J p | A n.

,

1

,

(a) Sho w tha t i f A n , n G N, ar e measurable , the n m I li m A \n—+oo J

n

1 < li m m(A n ). n—

>-oo

(b) Sho w that if , moreover , m(A n U one n , the n

An + i U . . .) < o o for a t leas t

m l li m A n 1 > li m m ( A n ) . \n—>oo /

n—>-o

o

2.1.^1. W e say that a sequence {A n } of sets in R converges if li m A n—>oo = li m A n , an d w e denot e th e commo n valu e b y li m A *-n* ^

n

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63

2.1. Lebesgu e Measur e o n th e Rea l Lin e (a) Sho w tha t a monotoni c sequenc e o f set s converges .

(b) Sho w that , i f a sequenc e {A n } o f measurabl e sets , A n C B , where ra*(B) < oo , converges , the n m ( li m A n ) = li m ra(A \n—>oo /

n—->o

n).

o

2.1.22. Sho w tha t th e Lebesgu e measur e o f th e se t A define d i n 1.7.12 i s equal t o 1 — a. 2.1.23. Le t A b e th e se t o f points i n [0,1 ] suc h tha t x i s in A i f an d only if the decimal expansio n o f x doe s not requir e th e us e of the digi t 7. Sho w tha t A ha s Lebesgu e measur e zero . 2.1.24. Le t B c M b e th e se t o f al l number s whos e decima l expan sions d o no t requir e th e us e o f th e digi t 7 afte r th e decima l point . Show tha t B ha s Lebesgu e measur e zero . 2.1.25. Le t A b e th e se t o f point s i n [0,1 ] whic h admi t o f binar y expansions with zeroes in all even positions. Sho w that A i s a nowhere dense se t o f Lebesgu e measur e zero . 2.1.26. Fin d th e Lebesgu e measur e o f the se t o f points i n [0,1 ] whic h admit decima l expansion s containin g al l th e digit s 1 , 2 , . . ., 9 . 2.1.27. Wha t i s th e Lebesgu e measur e o f th e se t o f point s i n [0,1 ] which admi t decima l expansion s 0.di^2^ 3 . • • suc h tha t n o sequenc e d^k+i^3/c+2^3/c+3 consist s o f thre e consecutiv e 2's ? 2.1.28. Le t A b e th e unio n o f interval s centere d a t point s o f th e Cantor se t an d eac h o f lengt h 0.1 . Fin d th e Lebesgu e measur e o f A . 2.1.29. Sho w tha t i f A i s a bounde d measurabl e se t o f measur e ra(A) = p > 0 , the n fo r eac h q G (0,p ) ther e i s a measurabl e se t B C A o f measur e q. 2.1.30. Sho w that i f 0 < ra(A) < oo , then for each positive q < ra(A) there i s a measurabl e se t B C A o f measur e q. 2.1.31. Sho w that i f 0 < ra(A) < oo , then for each positive q < ra(A) there i s a perfec t se t B C A o f measur e q. 2.1.32. Sho w tha t an y se t A o f positiv e Lebesgu e measur e ha s th e cardinality o f th e continuum .

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64

Problems. 2 : Th e Lebesgu e Integra l

2.1.33. Sho w tha t an y nonempt y an d close d se t A C R o f Lebesgu e measure zer o i s nowher e dense . 2.1.34. Suppos e tha t A C i i s a nowher e dens e se t o f Lebesgu e measure zero . Mus t it s closur e b e o f Lebesgu e measur e zero? 2.1.35. Sho w tha t i f A C [a , b] an d ra(A) > 0 , the n ther e ar e x an d y in A suc h tha t \x — y\ i s a n irrationa l number . 2.1.36. Doe s ther e exis t a countable collectio n o f nowhere dens e an d perfect subset s o f [0,1 ] whos e unio n i s of Lebesgu e measur e 1 ? 2.1.37. Doe s ther e exis t a nowher e dens e an d perfec t subse t o f [0,1 ] of Lebesgu e measur e 1 ? 2.1.38. Giv e a n exampl e o f a measurable se t A C R wit h th e follow ing property : Fo r eac h interva l (a,/3) , ra(An(a,/?)) >

0 an d ra((R \

A ) n (a,/?) ) > 0 .

2.1.39. Assum e tha t a measurabl e se t A C R ha s th e propert y tha t for eac h 8 > 0, m ( A f l ( 4, 5)) > 0 and 0 £ A . Prov e that ther e exist s a perfec t se t B C A suc h tha t ra(B H (—6, 5)) > 0 for ever y S > 0 . 2.1.40. A measurable se t A C R i s said t o hav e density d at x if th e limit m(An\x-h,x + h\) hm — —— — /i—o+ 2h

exists an d i s equal t o d. I f d — 1, then x i s called a point of density of A, an d i f d — 0 , the n x i s called a point of dispersion of A. Fin d th e points o f density an d point s o f dispersion o f A = (—1 , 0)U(0,1)U{2}. 2.1.41. Give n a £ (0,1 ) , construc t a se t A whos e densit y a t XQ £ R is equa l t o a. 2.1.42. Le t A b e a measurabl e se t suc h tha t 0 £ A i s a poin t o f density o f A . Prov e tha t ther e i s a perfec t se t B C A suc h tha t 0 is a poin t o f densit y o f B . 2.1.43. I f x an d y ar e i n [0,1 ) , we define th e su m x + y(mod 1 ) to b e x - f y, i f x + y < 1 , an d t o b e x + y — 1 , i f x + y > 1 . Fo r A C [0,1 ) and a £ [0,1 ) , w e defin e th e translat e modul o 1 of A t o b e th e se t A - f a(mo d 1 ) = {x - f a (mod 1 ) : x £ A} . Show tha t ?n* (A) = m*( A + a(mo d 1 )) .

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2.1. Lebesgu e Measur e o n th e Rea l Lin e

65

2.1.44. Sho w tha t i f A C [0,1 ) an d a G [0,1 ) , the n th e se t A + a (mod 1 ) i s measurabl e i f an d onl y i f A i s measurabl e an d ra(A + a(mo d 1 ) ) = m(A) . 2.1.45. W e sa y tha t x,y G [0,1 ) ar e equivalent , an d writ e x ~ y , i f and onl y i f x — y i s a rational . Thi s equivalenc e relatio n partition s [0,1) int o a n uncountabl e famil y o f disjoin t equivalenc e classes . B y the axiom of choice there is a set V whic h contains exactly one element from eac h equivalence class . W e call any such a set a Vitali set. Prov e that a Vital i se t i s nonmeasurable . 2.1.46. Sho w that i f A i s a measurable subse t o f a Vitali se t V , the n ra(A) = 0 . 2.1.47. Show , b y example , tha t th e convers e o f th e resul t state d in 2.1 .1 7 i s no t true ; tha t is , ther e exis t set s A an d B suc h tha t ra*(A U B) = ra*(A) + ra*(B) bu t ra*(A U B) < ra*(A) + ra*(B). 2.1.48. Sho w tha t an y se t o f positiv e oute r measur e contain s a non measurable subset . 2.1.49. Giv e a n exampl e o f a sequence {A n } o f pairwise disjoin t set s such tha t

(

oo \

o

c

(J A n c} is measurabl e fo r ever y c G C. 2.2.5. Sho w tha t a real-value d functio n / define d o n a measurabl e set A i s measurabl e i f an d onl y i f / - 1 ( G ) i s measurabl e fo r ever y open G C R . 2.2.6. Sho w tha t i f a real-value d functio n / define d o n R i s measur able, the n / _ 1 ( B ) i s measurabl e fo r ever y Bore l se t B C R . 2.2.7. Prov e tha t continuou s real-value d function s define d o n mea surable set s ar e measurable . 2.2.8. Suppos e extended-value d function s / an d g ar e define d o n a measurable se t A . Prov e tha t i f / i s a measurable functio n an d f — g a.e., the n g is measurable . 2.2.9. Prov e tha t ever y Rieman n integrabl e functio n define d o n [a , b] is measurabl e o n [a , b]. 2.2.10. Sho w tha t eac h functio n o f bounde d variatio n o n [a , b] is measurable. 2.2.11. Defin e th e Cantor function ip : [0,1 ]— > [0,1] a s follows : I f oo

x i s a n elemen t o f th e Canto r se t C an d x — Yl &t with a n = 0 o r 71=1

an = 2 , the n w e pu t

(

oo \

o

o 1

Z^ 3 n J Z-J

n=l /

n=

2

l

2

n

that is , i f a n i s th e nt h ternar y digi t fo r x , the n th e nt h binar y digi t for 0, there i s a measurable se t B C A suc h tha t ra(A\B) < e and / restricte d t o B i s bounded . 2.2.22. Prov e th e followin g Egorov theorem. Le t A C R b e a mea surable se t o f finit e measure . I f {f n} i s a sequenc e o f measurabl e functions whic h converge s t o a real-value d functio n / almos t every where o n A , then , give n e > 0, ther e exist s a measurabl e subse t B o f A suc h that ra(A\B) < e and th e sequenc e {f n} converge s uniforml y to / o n B .

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70

Problems. 2 : Th e Lebesgu e Integra l

2.2.23. Sho w by example that th e assumption m(A) < o o is essential in th e Egoro v theorem . 2.2.24. Suppos e tha t A i s measurabl e an d {f n} i s a sequenc e o f measurable function s whic h converge s t o / almos t everywher e o n A . Prove that ther e is a set B C A suc h that B = I J ^ i B* , ra(A\B) = 0 and th e sequenc e {f n} converge s uniforml y t o / o n ever y B^ . 2.2.25. Le t {V n } b e th e sequenc e o f set s define d i n th e solutio n t o 2.1.45 an d le t {f n} b e th e sequenc e o f function s o n [0,1 ) define d b y fn = X\J°l Vi- Sho w tha t f n— » 0 o n [0,1 ) , bu t th e assertio n o f th e Egorov theore m doe s not hold , that is , there i s e > 0 such that o n any measurable subse t B o f [0,1 ) wit h m([0,1 ) \ B ) < e th e convergenc e is no t uniform . 2.2.26. Construc t a sequenc e {f n} o f measurabl e function s o n [0,1 ] such tha t th e sequenc e converge s everywher e o n [0,1 ] bu t fo r ever y set B C [0,1 ] o f measur e 1 , th e sequenc e fail s t o converg e uniforml y onB. 2.2.27. Prov e th e followin g Lusin theorem: I n orde r tha t a real valued functio n / define d o n a measurabl e se t A b e measurable , a necessary an d sufficien t conditio n i s tha t fo r ever y e > 0 ther e i s a closed se t F C A suc h tha t ra(A \ F ) < e an d / restricte d t o F i s continuous. 2.2.28. Usin g the result i n 2.1.38, construct a function / , measurabl e on M , such tha t fo r an y se t E o f measur e zero , / i s not continuou s a t any poin t i n M \ E . 2.2.29. Le t / : [0, a]—> R b e a measurable function . Sho w that ther e exists a monoton e decreasin g functio n g o n [0 , a] such tha t fo r an y real y, m({x E [0,a ] : f(x) > y}) = m{{x e [0 , a] : g{x) > y}). 2.2.30. Le t {f n} b e a sequenc e o f real-value d function s measurabl e on A . W e sa y tha t {f n} converges in measure t o a measurabl e func tion / i f fo r ever y e > 0 , li m m({x e A : \f n(x) — f(x)\ > e}) — 0 . n—*oo

Show tha t i f a sequenc e {f n} converge s i n measur e t o / an d {f converges i n measur e t o g, the n / = g a.e . o n A .

n}

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2.3. Lebesgu e Integratio n

71

2 . 2 . 3 1 . Prov e th e followin g theorem of Lebesgue. I f 771 (A ) < 0 0 an d {fn} converge s t o / a.e . o n A , the n {f n} converge s i n measur e t o / . 2 . 2 . 3 2 . Sho w b y exampl e t h a t th e assumptio n ra(A) < in th e Lebesgu e theore m state d above .

0 0 i s essentia l

2 . 2 . 3 3 . Giv e a n exampl e o f a sequenc e o f measurabl e function s o n th e interval [0,1 ] t h a t converge s i n measur e o n [0,1 ] bu t i s no t convergen t at an y poin t o f t h a t interval . 2 . 2 . 3 4 . Le t {f n} b e th e marchin g sequenc e define d i n th e solutio n to th e previou s problem . Fin d a subsequenc e o f i t t h a t converge s t o the zer o functio n a.e . o n [0,1 ] . 2 . 2 . 3 5 . Prov e th e followin g Riesz theorem. Ever y sequenc e {/ n } con vergent i n measur e t o / o n A contain s a subsequenc e convergin g t o / a.e . o n A . 2 . 2 . 3 6 . Le t {f n} t> e a sequenc e o f monotonicall y increasin g function s on (a , b). Sho w t h a t i f th e sequenc e converge s i n measur e t o / , the n lim f n(x) — f\x) a t eac h poin t x o f continuit y o f / . n^oo

2 . 2 . 3 7 . Prov e th e followin g Frechet theorem. I f / i s a real-value d measurable functio n o n A , the n ther e i s a n T a se t H suc h t h a t ra(A \ H ) = 0 an d / restricte d t o H i s i n th e firs t Bair e clas s (tha t is, / i s a pointwis e limi t o f a sequenc e o f continuou s function s o n H ) . 2 . 2 . 3 8 . Prov e th e followin g Vitali theorem. I f / i s a real-value d mea surable functio n o n A , the n ther e i s a functio n g i n th e secon d Bair e class (tha t is , g i s a pointwis e limi t o f a sequenc e o f function s i n th e first Bair e clas s o n A ) suc h / = g a.e .

2.3. Lebesgu e Integratio n n

T h e Lebesgu e integra l o f a simpl e functio n r}) = 0 fo r som e rea l numbe r r . I n this cas e w e define th e essential supremum o f / b y H/lloo = inf{ r : m({x € A : \f(x)\ > r}) = 0} . Then th e se t E = {x e A : |/(a;) | > ||/||oo } i s o f measur e zer o an d |/0r)| < H/llo o outsid e E . Thu s | / ( x ) | < | | / | U a.e . We wil l ofte n us e th e followin g theorems : Theorem 1 (Lebesgue' s Monoton e Convergenc e Theorem) . Suppose {f n} is an increasing sequence of nonnegative measurable functions on A . If f(x) = li m f n{x), x E A, then lim / f ndm = / fdm. ^°° J A J A Theorem 2 (Fatou' s Theorem) . Suppose {f n} is nonnegative and measurable functions on A. Then n

/ Ui n fndm < li m / f J A n—>o o n—*oo

a sequence of

ndm.

JA

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73

2.3. Lebesgu e Integratio n

Theorem 3 (Lebesgue' s Dominate d Convergenc e Theorem) . Suppose {f n} is a sequence of measurable functions on A and f(x) = lim f n(x), x G A. If there exists a function g integrable on A and n—>oo

g{x), n = 1 , 2 , . . ., x G A, then

such that \f n(x)\
0 ther e i s S > 0 suc h tha t fo r ever y measurable subse t B o f A , J B \f n\dm < e fo r n = 1 ,2,.. . when ever m (B) < # . Show tha t i f {/ n } i s a convergen t sequenc e o f equi integrable function s o n a set A o f finite measure , the n lim / fndm — / li m f ndm. ~ °° J A J An ^°° 2.3.21. Prov e th e following versio n o f Lebesgue's dominate d conver gence theorem . Suppos e {f n} i s a sequenc e o f measurable function s converging i n measur e o n A t o / . I f ther e exist s a functio n g in tegrable o n A an d suc h tha t |/ n (a:)| < g(x), n = 1 ,2,... , x G A, then n >

lim / fndm — \ fdm. ™-*°° J A J

A

2.3.22. Sho w tha t th e theore m state d i n 2.3.2 0 remain s tru e whe n convergence i s replaced b y convergence i n measure . 2.3.23. Suppos e th e sequence {f n} converge s i n measur e t o / o n a set A o f finite measure , an d \f n(x)\ < C fo r x G A, n = 1 ,2,... . Show tha t i f g is continuous o n [—C , C], the n I™ / g{f n)dm= / g(f)dm. -*°° J A J A 2.3.24. Suppos e tha t {f n} i s a sequence of functions define d o n a set A o f finite measur e tha t converge s i n measure o n A t o /. Sho w tha t n

lim / s'm(f n)dm = / sin(/)d r ^°° J A J A

n

2.3.25. Suppos e / G Lp[a, b], 1 < p < oo . Show that , give n e > 0, there is (i) a simple functio n{x)\ : x e [o,6] } > 1 / 2 for al l step function s ip.

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77

2.3. Lebesgu e Integratio n

2 . 3 . 2 7 . Suppos e / G L p[a,b],1 < p < oo . Sho w t h a t , give n £ > 0 , there i s a continuou s functio n g suc h t h a t L u\f — g\ pdm < e. 2 . 3 . 2 8 . Show , b y example , t h a t th e resul t i n th e previou s proble m i s false i f p = oo . 2 . 3 . 2 9 . Suppos e t h a t g satisfie s a Lipschit z conditio n o n R . Prov e t h a t i f {/ n } i s a sequenc e o f measurabl e function s o n [a , b] t h at con verges i n measur e t o / an d ther e i s a Lebesgu e integrabl e functio n G such t h a t \f n(x)\ < G(x), the n

lim / g{f

n)dm

= / g{f)dm.

J[a,b] ^^JlaM[a,6] J\aM

2 . 3 . 3 0 . Suppos e t h a t 1 < p < oo , an d t h a t / i s measurabl e o n A and suc h t h a t \f\ p i s integrabl e o n A . Prov e t h a t , give n e > 0 , ther e is a continuou s functio n g vanishin g outsid e a finite interva l an d suc h that / \f-g\Pdm > g a.e. Sho w t h a t f ngn— > / ^ i n L p [a,6]. 2 . 3 . 3 3 . Sho w t h a t i f / i s a n essentiall y bounde d functio n o n [a . 6], then

lim If \J7dm

) =!i/lk

-

(Compare wit h 1 .4.41 . )

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78

Problems. 2 : Th e Lebesgu e Integra l

2.3.34. Prov e th e Jensen inequality for Lebesgue integrals (se e als o 1.6.29). Suppos e tha t / i s Lebesgue integrable o n [a , b]. If if is convex on R , the n (f - / fdm \b-aJ[aM J

/

< b

a5}, then 2

2

m(As)>m(A)(l-5) j. 2.4. Absolut e Continuity , Differentiatio n an d Integration A real-valued functio n / define d o n [a , b] i s said t o b e absolutely continuous o n [a , b] if , give n e > 0, ther e i s 5 > 0 such tha t n

J2\f(xk)-f(x'k)\oo x—

»oo

2.4.20. Le t / G Lp[a, 6] , 1 < p < oo , an d pu t f(x) = 0 for x £ [a , b}. For h > 0 , w e defin e fh b y settin g px+h

A M - S £»'«>*• Show tha t fh i s continuous an d \\fh\\p < II/li p (compar e wit h 1 .6.34) . 2.4.21. Prov e tha t i n th e notatio n an d unde r th e assumption s o f 2.4.20 limL||/ft-/|| p = 0 . h—>0

2.4.22. Suppos e tha t / G L l[a,b] an d tha t x G (a , b) i s suc h tha t f{x) ^ ±oo . The n x i s calle d a Lebesgue point for fit f>x+h

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2.4. A b s o l u t e C o n t i n u i t y , D i f f e r e n t i a t i o n . . .

83

Show t h a t i f x i s a Lebesgu e poin t fo r / , the n th e functio n F(x) = J f(t)dt i s differentiabl e a t x an d F'(x) — f(x). 2 . 4 . 2 3 . Sho w t h a t eac h poin t o f continuit y o f / £ L l[a, b] is a Lebes gue poin t fo r / . 2 . 4 . 2 4 . Prov e t h a t almos t ever y poin t o f [a , b] is a Lebesgu e poin t fo r

2 . 4 . 2 5 . Sho w t h a t almos t al l point s o f a measurabl e se t A ar e point s of densit y o f A . (Se e 2.1 .4 0 fo r th e definitio n o f a poin t o f density. ) 2 . 4 . 2 6 . A se t A C R i s sai d t o hav e outer density d at x i f t h e limi t ra*(An \x

lim — h-+o+ 2h

- h,x + h\)

exists an d i s equa l t o d. I f d = 1 , the n x i s calle d a point of outer density of A, an d i f d = 0 , the n x i s calle d a point of outer dispersion of A. Prov e th e followin g generalizatio n o f th e resul t i n th e previou s problem. I f A i s an y se t (measurabl e o r not) , the n almos t al l point s i n A ar e point s o f it s oute r density . A necessar y an d sufficien t conditio n t h a t A b e measurabl e i s t h a t almos t al l point s i n A c ar e point s o f outer dispersio n o f A . 2 . 4 . 2 7 . Assum e t h a t a real-value d functio n / (measurabl e o r not ) i s defined o n [a , b]. W e sa y t h a t XQ £ [a , b] i s a point of approximate continuity of f i f fo r eac h e > 0 , l i m m*([x h-^o 2h

0

- fe,s

0

+ h] H {x G [a , 6] : \f(x) -

f(x 0)\ >

g»=

Q

Prove t h a t a real-value d functio n / i s measurabl e o n [a , b] if an d onl y if almos t al l point s o f [a , b] are point s o f approximat e continuit y o f / . 2 . 4 . 2 8 . Sho w t h a t a Lebesgu e integrabl e functio n o n [a , b] is approx imately continuou s a t eac h o f it s Lebesgu e points . Sho w b y exampl e t h a t th e convers e i s no t true . 2 . 4 . 2 9 . Sho w t h a t i f / i s measurabl e an d bounde d o n [a , 6], the n x G (a , b) i s a Lebesgu e poin t fo r / i f an d onl y i f x i s a poin t o f approximate continuit y o f / .

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84

Problems. 2 : Th e Lebesgu e Integra l

2.4.30. A n alternate definitio n o f approximate continuit y i s that / i s approximately continuou s a t xo provided ther e i s a measurable se t A such tha t xo is a point o f density fo r A an d such tha t th e restrictio n of / t o A i s continuous a t x$. Sho w tha t thi s definitio n i s equivalent to th e one given i n 2.4.27. 2.4.31. Giv e a n example o f a bounded functio n / : [0,1]—> R whic h is not Rieman n integrabl e bu t i s a derivativ e o f some functio n F on [0,1]2.4.32. Suppos e / i s integrable o n [a , b] an d for t £ R defin e G(t) = m({x e [a,b] : f(x) < t}). Prov e tha t pb rOG

/ f(x)dx = / td(G(t)), J — oo

Ja

where oo /*

/

0 /»o

o

td{G{t)) = I td(G(t))

+ / td(G(t))

— oo JO

-oo J

pO nB

= li m / td{G(t))+ li m / td(G(t)). 2.5. Fourie r Serie s For / G L1[—7r,7r], the Fourier coefficients o f / a r e give n by an — — f(x)

cosnxdx, b

^ 7-7 T

n

= — / / ( # ) sinnxdir. ^ 7-7 T

The serie s 1° ° -ao + / J ( a n cos nx-f - 6n sin nx) 2

n=

l

is the Fourier series of f, an d we write ^

oo

/(#) ~ -a o + Y ^ ( a n c o s n x + 6 n sinnx). 2' n=l

We denot e b y s n{x) th e nt h partia l su m of th e Fourie r serie s o f / ; n

that is , sn{x) — \a§ + Yl (dk cos kx + bk sin fcr). fc=i

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2.5. Fourie r Serie s

85

2.5.1. Sho w tha t i f p n = y a ^ + b*^, the n oo

oo

2_"(o>n

c o sn x

+ b n sinnx) = \ J p n sin(n:r + a n ) ,

n=l n=

l

with som e a n. 2.5.2. Sho w tha t i f a n an d 6 n ar e th e Fourie r coefficient s o f / £ L1[—7r, 7r], the n lim a n = li m b n = 0 . n—»oo n—>o

o

2.5.3. Suppos e tha t / i s a 27r-periodic function tha t satisfie s th e Lip schitz conditio n o f orde r a ( 0 < a < 1 ) ; tha t is , f(x + h) — f(x) = 0(\h\a) a s / i— > 0 uniforml y wit h respec t t o x. Sho w tha t i f a n an d bn ar e th e Fourie r coefficient s o f / , the n an = 0(n- a) an

d6

= 0(n _ Q ! ).

n

2.5.4. Sho w tha t i f / i s of bounde d variatio n o n [—7r,7r] , the n a n — 0 ( n _ 1 ) an

db

n

= 0{n~

l

).

2.5.5. Sho w tha t i f / i s a 27r-periodi c functio n tha t i s absolutel y continuous o n [—7r,7r] , the n an = o(n _ 1 ) an d b

n

— o{n~ l).

2.5.6. Sho w tha t i f / i s a 27r-periodi c functio n an d / G Cfc (R), the n an = o(n~ k) an

db

n

— o(n~ k).

2.5.7. Sho w that fo r / G Ll[—TT, 7r], the nth partia l sum of the Fourie r n

series o f / , tha t is , s n(x) = \a$ -f - ^ (a ^ coskx + 6 ^ sin/ex), i s give n fc=i

by 7r t/_7r 2 s m ^ ( t - x

)

2.5.8. Asum e tha t / i s a periodic functio n o n R wit h perio d 27T , and that / i s integrable o n [—7r , TT]. Sho w that i f s n denote s the nth partia l sum o f th e Fourie r serie s o f / , the n

Snix) = - [^ (/(* + t) + f(x - t)) TT J0 2

Sin n+

(

p^.

sin %t

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86

Problems. 2 : Th e Lebesgu e Integra l

2.5.9. Prov e the Dini-Lipschitz test for convergence of Fourier series. Suppose / i s a periodic function o n R with period 2TT that i s integrable on [—7r,7r] . I f / satisfie s th e conditio n \f(xo + t) — /(#o) | < L\t\ a fo r \t\ < 8 with som e positive L , 8 and 0 < a < 1 , then th e Fourie r serie s of / a t XQ converges t o f(xo). 2.5.10. Prov e th e Dirichlet-Jordan test for convergence of Fourier series. Assum e tha t / i s a 27r-periodi c functio n integrabl e o n [—7r,7r] and tha t / i s of bounded variatio n o n [xo — 8, XQ - f 8] . Show tha t th e Fourier serie s o f / a t XQ converges t o |(/(#o~ ) + /(#o~)) 2.5.11. Sho w b y exampl e tha t th e Dirichlet-Jorda n tes t doe s no t include th e Dini-Lipschit z tes t an d tha t th e Dini-Lipschit z tes t doe s not includ e th e Dirichlet-Jorda n test . 2.5.12. Sho w tha t oo

7r — x v-

^ si n nx 71=1

and deduc e tha t 7r ^- ^ sin(2 n — l)x 4_ ^ 2 n -1 '

0 < X < 7T ,

71 = 1

IT

1

4

= 1

1

- 3

2 ^5

1 +

5 - 7

+

7 1 11

- ' 3

2.5.13. Fo r x G (0, 27r), find th e sum s o f th e followin g series : OO

E

OO

cosnx Y

n=l n =

2

n ^

^ sinn x n

3

l

2.5.14. Sho w tha t i n 2.5. 4 bi g O' s canno t b e replace d b y littl e o's . 2.5.15. Le t / b e a 27r-periodic function suc h that f(x) = E ^- fo r x G [0, 27r), and le t s n(x) denot e th e nt h partia l su m o f it s Fourie r serie s (see 2.5.12). Sho w that i f x n i s the smalles t positiv e numbe r a t whic h sn(x) attain s it s loca l maximum , the n li m s n(xn) = L ^^dx.

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2.5. Fourie r S e r i e s

87

2 . 5 . 1 6 . Sho w t h a t fo r a ^ 0 , e2an e—

~~ n

_ 1 / 1 — i / i„ \2a

a

2

V^/ / ^

n aT T ,

^

12

Z

a7T 7

e**=

a c ao scos n nxx — — nn si n nx

,0

< X < 27T ,

+ n xaCOSnX a2

T

_ \ ( 1 _ (1 _ )nea7r^__ ^( 7T ^ a2 + n 2 n=l

_ ^ n2 '

) < X < 7T .

Find th e sum s o f th e abov e serie s whe n x = 0 2 . 5 . 1 7 . Fo r 0 < x < 2TT and a ^ O , find t h e sum s o f t h e followin g series: n si n n a a co s n x a2 + n 2 ' a2 + n 2

E

E

71=1

n=l

2 . 5 . 1 8 . Sho w t h a t O

^,

OO

\~^ / -,\n

C0SnX

n=l

Using thi s equality , sho w t h a t

SA = ?an d E n=l n =

Mn+1 ^

2 12'

l

2 . 5 . 1 9 . Suppos e / G L 1 [—7r, 7r] and a n , 6 n ar e th e Fourie r coefficient s of / . Sho w t h a t (a) i f / ( x + 7r ) = f{x) fo r x G [-7r , 0] , the n a 2 n - i = b

2n-i

=

0,

(b) i f f(x + 7r ) = —/(# ) fo r x G [—7r , 0], the n a 2 n = &2 n = 0 . 2 . 5 . 2 0 . Sho w t h a t fo r 0 < x < TT , cosx

8 v~ ^ n s i n 2 n x 2 7T ^ 4 n -l n=l

24 and s i n x = —z 4 Jinx = > 7T7

v^- ^^ cc oo ss 2z nn xx —z . T ni L=-l ' 4 n 2 - 1

2 . 5 . 2 1 . Sho w t h a t In 2 sin x

~E

cosnx

for X ^ 2 / C T

T (fceZ)

,

n=l

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Problems. 2 : Th e Lebesgu e Integra l and In 2 c o s | U ^ ( - l )

n + 1

^ ^ fo

r x^(2k

+

G Z) .

l)7r (k

n=l

2.5.22. Suppos e a 27r-periodi c functio n / i s i n L l[—7r,7r] and a are th e Fourie r coefficient s o f / . Sho w tha t fo r x G [0, 27r], [x (£/±\ *

" \-^ v

a

n,bn

^ a n s i n n x + ^ n (l - cosnx )

=

x (^-2 °r § »

•

2.5.23. Prov e tha t unde r th e assumption s o f th e previou s problem , oo

the serie s ^2 n converges . Us e thi s resul t t o sho w tha t ^

l^ f*

s

not a Fourie r serie s o f an y integrabl e function . 2.5.24. Prov e tha t i f / G L2[-7r,7r], the n -. o

n

r

oi

(This inequalit y i s known a s Bessel's inequality.) 2.5.25. Us e th e result s give n i n 2.5.2 2 an d 2.5.2 4 t o prov e tha t i f / G L2[—7r,7r], then fo r an y measurabl e se t A C [—7r,7r] , lim / (f(x)-s n(x))dx = 0, -^°°JA where s n denote s th e nt h partia l su m o f th e Fourie r serie s o f / . n

2.5.26. Sho w tha t i f f,g G L2[-7r,7r], the n lim / g(x)(f(x)-s

n(x))dx

=

0,

where s n denote s th e nt h partia l su m o f th e Fourie r serie s o f / . 2.5.27. Sho w tha t i f f,g G L 2[-7r,7r] an d 1

o

o

f(x) ~ ~a o + y ^ ( a n c o s n x - f b 2

n=

Io

nsinnx),

l o

g(x) ~ - a 0 + Y ^ ( an co s nx + f3 n sin nx), n=l

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2.5. Fourie r Serie s

89

then - / f(x)g(x)dx 2.5.28. Sho w tha t i f / e cients, the n °°1 b

=

-a 0a0 + y^(a nan +

L 2 [0,2TT]

/*

& n/?n)-

an d a n, 6 n ar e it s Fourie r coeffi -

27r

2.5.29. Sho w tha t i f / , g G L 1 [—7r,7r] hav e th e sam e Fourie r series , then f = g a.e . o n [—7r,7r] . 2.5.30. Us e Parseval' s formul a (a) t o su m th e serie s l ^1

E ^ 4 ' 2-

^j ( a2 + n 2 ) 2 ' ^ lv

n=l n=

y

z

>n n=

2

(a lv

+ n 2)2'

7

(b) t o comput e 7T /»7

/

T

cos6 xcb an

d/

-IT J

cos

8

xdx.

—TV

2.5.31. Sho w tha t i f / G L2[—7r,7r] an d a nybn ar e th e Fourie r coeffi cients o f / , the n th e serie s Yl \ an\+\ n\

conver

ges

71=1

2.5.32. Sho w tha t i f a 27r-periodi c functio n / i s continuousl y differ ent iable o n R , the n max \f(x) - s n(x)\ = o(n~

1 /2

).

2.5.33. Prov e th e followin g Bernstein theorem. I f / i s 27r-periodi c and satisfie s th e conditio n \f(x + h) — f(x)\ < L\h\ a wit h som e L > 0 OO

and a > 1 /2 , the n th e serie s J2 V an + ^n converges . n=l

2.5.34. Sho w tha t i f / i s 27r-periodi c an d satisfie s th e Lipschit z con dition I / O + h) - f(x)\ < L\h\ a wit h som e L > 0 an d 0 < a < 1 , OO

then th e serie s J2 (\ an\^ + \b n\^) converge s fo r (3 > n=l

2a+l'

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Problems. 2 : Th e Lebesgu e Integra l

90

2.5.35. Suppos e tha t / i s 27r-periodi c an d satisfie s th e conditio n \f(x + h) — f(x)\ < L\h\ a wit h som e positiv e L an d a. Sho w tha t if / i s o f bounde d variatio n o n [0 , 27r], the n th e serie s J ^ y ! a \ + b\ n=l

converges. 2.5.36. Giv e a n exampl e o f a continuou s an d 27r-periodi c functio n whose Fourie r serie s fail s t o converg e o n R . 2.5.37. Sho w tha t i f a 27r-periodi c functio n / i s bounded o n R , the n |s n (x)| = 0 ( l n n ) . 2.5.38. Le t L n b e th e Lebesgue constants give n b y

Ln

~^L

sin (n + \) x\

dx. n

G N.

Show tha t L n = £ In n + 0(1 ) . 2.5.39. Sho w tha t i f a 27r-periodi c functio n / i s continuou s o n R , then |s n (x)| = o(lnn) . 2.5.40. Assum e tha t / i s 27r-periodi c o n R an d / G Ll[—7r,7r]. Us e the formul a give n i n 2.5. 8 t o sho w tha t i f s0(x) + si(x ) H hs n _i(x) n (which i s called th e nth Cesaro mean of the Fourier series for f), the n crn(x) =

1 f n sin 27rn Jo sin

2 z

-nt |t

2.5.41. Sho w tha t i f / i s 27r-periodi c an d m < f(x) < M fo r al l x , then m < o~ n(x) < M fo r al l x G R an d n G N. Sho w als o tha t i f for som e n an d x w e hav e cr n(x) = M (cr n(x) = m) , the n /(x ) = M (resp. /(x ) — m ) a.e . 2.5.42. Sho w tha t i f / i s 27r-periodic , m < f(x) < M fo r al l x an d n\an\ < A, n\b

n\

< B, n

= l,2,... ,

then fo r al l value s o f n an d x , m - A - B < s n(x) oo

uous, the n li m s n(x) — f(x) uniforml y o n R . n—>oo

2.5.53. Prov e th e followin g Wiener theorem. A 27r-periodi c functio n of bounde d variatio n o n [0, 2n] i s continuou s i f an d onl y i f -. n

lim -YkJa

2 k

+

b

2

k=0.

k=l

Conclude tha t a 27r-periodi c functio n o f bounde d variatio n o n [0 , 2ir] for whic h a n = o(n~ 1 ) i s continuous . 2.5.54. Prov e tha t i f th e Fourie r serie s o f / G L1[—7r,7r] i s lacunary , that is , CO

f(x) ~ ^ ( a n f c cosn kx + b nk smn

kx),

k=i

where ! ^±±1 - > q > 1 for k = 1 , 2 , . . ., the n li m s Hfc

k—

nk

(x) — f(x) almos t

>oo

everywhere.

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2.5. Fourie r Serie s

93

2.5.55. Assum e tha t / i s 27r-periodi c o n R an d / G Lp[—7r,7r], 1 < p < oo. Prov e that th e sequenc e o f Cesaro means of the Fourie r serie s for / converge s i n th e LP norm t o / , tha t is , lim | | < r n - / | | p = li m ( f \a

n(x)

- f(x)fdxY =

0.

2.5.56. Suppos e tha t {b n} i s a monotonicall y decreasin g sequenc e oo

converging t o zero . Prov e tha t ^ b

nsmnx

i s uniforml y convergen t

n=l

on [0 , 2TT] i f an d onl y i f li m nb n = 0 . n—->oo

2.5.57. Suppos e tha t {6 n } i s a monotonicall y decreasin g sequenc e oo

converging t o zero . Prov e tha t ] P b n sinnx i s the Fourie r serie s o f a 71=1

continuous functio n i f an d onl y i f li m nb n = 0 . n—>oo

2.5.58. Suppos e tha t {b n} i s a monotonicall y decreasin g sequenc e oo

converging t o zero . Prov e tha t ^ b

n

sin nx i s the Fourie r serie s o f a

n=l

bounded functio n i f an d onl y i f nb n — 0(1). 2.5.59. Le t {a n} b e a monotonically decreasin g sequenc e convergin g to zer o an d suc h tha t a n + i < a " + ^ + 2 for n G N . Prov e tha t —+ > a 2

n=

n

co s nx

l

is th e Fourie r serie s o f a functio n nonnegativ e an d integrabl e o n [—7r,7r]. Deduc e tha t th e serie s cos E. Icosnx n

n=2

is th e Fourie r serie s o f suc h a function . (Compar e wit h th e resul t i n 2.5.23.)

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Part 2

Solutions

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http://dx.doi.org/10.1090/stml/021/03

Chapter 1

The Riemann-Stieltje s Integral

1.1. Propertie s o f th e Riemann-Stieltje s Integra l 1.1.1. Le t £ > 0 be given . B y Theore m 1 , t o sho w tha t / e TZ(a) i t suffices t o fin d a partitio n P suc h tha t

U{P,f,a)-L(P,f,a) 0 such tha t \a(x) — a(c)\ < e/2 i f | x — c | < 5. If w e tak e a partitio n P whose mes h i s les s tha n S an d suc h tha t #;_ i < c < xi fo r som e i , then U(P, / , a) - L(P , /, a) = a(xi) - a(xi_i ) < e. The sam e reasonin g ca n b e applie d t o th e case s c — a an d c = b. Clearly, J a fda = su p L(P, f, a) = 0 . 1.1.2. Le t a = x 0 < • • • < x n — b be a partitio n o f [a , 6], an d le t z G { 1, 2 , . . ., n} b e suc h tha t a^_ i < c < Xi. The n U(PJ,a)= su

p f(x)

= Mt

xE:[Xi-i,Xi]

and L(P,f,a)= in

f /(x

) = m* .

97

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

98

Since / i s continuous o n [a, 6], taking partition s whos e mes h tend s to zero, w e see that sap L{P,f, a) = / ( c ).

iniU(P,f,a) =

1.1.3. Sinc e the sets Q and M\Q are dense in M, we see that L(P , /) = 0 fo r any partition P o f [a , b]. Consequently , fdx = 0.

/ Ja

On th e other han d

nn

U(P,f) =

x

x

x

Y^ i( i ~

-{b 2 - a 2 ) ,

^2 X1 -^XiXi-i >

i-i) =

i=l 2= 1 i=l

because 2xiXi-i < (xf + x 2_x). I t i s worth notin g her e tha t fdx=l-{b2-a2)

/ Ja

Indeed, takin g the sequence of partitions P n wit h x^ — a + (b — a)i/n, i = 0 , 1, . .. ,n , we see that li m U(P n,f) = hb 2 - a 2). 1.1.4. Choos e a partition P o f [—a , a] suc h tha t —a = xo < • • • < Xj-i = 0 < Xj < - - - < xn — a. Then, a s in the previous problem , w e get n_1

U(P, f) = Y^

2

a

x

x

k+ii k+i ~

x k) > —

fc=j-i

and j 2

L P

( J) =

2

~ a ^ xk{xk+i -x

k)


0. The n there are at most 2k^ subinterval s [a^-i , Xi] containin g at least on e of the rational s mentione d above . O n the other subinterval s f we hav e Mj — rrij < 1/N. Consequently ,

U(P,f)-L(PJ) 0, we take N > 2(b — a)/e. The n fo r any partition o f mesh S < £/(4/CJV) w e get, in view o f the above,

U(PJ)-L(P,f) 0 be given . Ther e i s a positiv e intege r n o such tha t 1/n < e/2 fo r n > UQ. Choos e a partition P determine d b y 0 = xo < x\ —

X2 < • • • < Xk = 1


0 be given . Ther e i s a positiv e intege r n o such tha t 1/n < e/2 fo r n > no - Choos e a partition P determine d b y 1 1 0 = x 0 < xi = — < x 2 < • • • < x n> = — n0 + 1 ° n 0 < X n' + ! < • • • < X

nt

=


—

1

such tha t xi — Xi-\ < e/(4no), i > 2. Then -, n

o

U(P, f) - L(P, f) - — — + T(Ml - m 2)(x, - Xi -!) 710 + 1

7^2

no—2 n

fc+i

+ ^2 ILJ

i Mi ~ mi)ixi ~

x

i-l)

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

100

So, w e hav e prove d tha t / i s Riemann integrabl e o n [0,1 ] . 1.1.8. Sinc e fo r ever y partitio n P suc h tha t Xk = 0 fo r som e k w e 6

have U(P, / , a) = L(P , /, a) = 0 , w e se e tha t / fda = 0 . No w t o se e a

that li m 5(P , /) doe s not exist , conside r partition s wit h /x(P )— » 0 Ax(P)^0

and suc h tha t fo r som e fc, x^-i < 0 < #& . I f w e selec t t ^ = 0 an d £'fc G (0, arfc], then w e get £(P , /) = 0 and 5(P , /) = 1 , respectively . 1.1.9. Suppos e tha t c , a < c < 6, is a commo n poin t o f discontinuit y of / an d a. The n ther e i s £ > 0 suc h tha t fo r an y 6 > 0 ther e exis t x' an d x" suc h tha t |x ' — c\ < 5, \x" — c\ < 6, and (1) \f(x')

- f(c)\ > s, \a(x")-a(c)\>s.

Assume, contrar y t o ou r claim , tha t li

m 5(P , /) exists . I t the n

follows tha t ther e i s 5' > 0 such tha t /i(P ) < 5' implie s tha t 2

(2) \S(PJ,a)-S'(PJ,a)\l + £ ; x/P Our tas k i s no w t o sho w tha t i f a > /? , the n V ( / ; 0,1) < +oo . Le t P = { x o , x i , . . . ,x n} b e a partitio n o f [0,1 ] . Tak e n s o larg e tha t n-i/f3 < X l j a n ( j ^d |- 0 t n e p a r tition P point s 1 1 1 1 1 nV/3' ( n _ i/2)V/3 ' ( n _ i ) i / / 3 ' ' ' " ' 2 V£ ' ( 2 - 1 /2) 1 //? * Note als o tha t / i s monotoni c o n eac h interva l [ i - 1 ^ , (i — l / 2 )- 1/ ^ ] , % = 2 , 3 , . . ., n , an d [( i + l ^ ) " 1 / ^ - 1 ^ ] , z = 1 , 2 , . .. ,n - 1 . Thu s ~

nn

o

o 1

£i/(*«)-/te-i)i a suc h that y ( / ; a , a) - V ( / ; a, x') = V ( / ; z', *) < e whenever x > x' > M. Thu s

|/(x')-/(a;)| 0. a

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1.2. Function s o f Bounde d Variatio n

123

1.2.23. Indeed , suppos e tha t \f(x')-f(x")\ < [a, b]. If x r = XQ < x\ < - • • < x n = x" , the n

L\x f -x"\ fo r x',x " G

nn

^ |/(a; 0 - f(xi- x)\ < £L{ X l -

Xi

-{) = £(*" - x').

Taking th e supremu m ove r al l partition s o f [x^x" ] yield s vf(x") -

v f(x') = V(f; x' , x") < L(x" - x') .

1.2.24. B y Theore m 1 , / = p — q, wher e p an d q ar e monoton e o n [a, b]. So, th e one-side d limit s o f / exis t a t ever y poin t o f [a , b] (see, e.g., II , 1 .1 .35) . Suppose , contrar y t o ou r claim , tha t x 0 i s a poin t o f discontinuity o f / . I f f(x^) — /(XQ) = d > 0 , the n b y th e definitio n of one-side d limit s ther e i s 5 > 0 such tha t f(x) < f(xn) + - fo

rx

G (x0 - 6, x0)

f(x)>f(x$)-- fo

rx

G (x0 ,x 0 + £) .

and

Then, i f y G [/(x 0~) + d/3,/(x+ ) - d/3 ] \ {/(*) } C [/(m) , f(x 2)} \ {f(xo)} w ^ h X i G (x o — 5,x 0 ) an d X 2 G (XO,X Q + 6), ther e i s n o x G (xi,X2) suc h tha t /(x ) = y, a contradiction . I f / ( x j ) = /(x^) ^ /(xo), analogou s reasonin g ca n b e applied . Since th e derivativ e / ' enjoy s th e intermediat e valu e propert y (see, e.g. , II , 2.2.31 ) , th e othe r statemen t follow s immediately . 1.2.25. Fo r x G [a, 6], let P = { x o , x i , . . . , x n } b e a partition o f [a,x] . Since /' i s continuous on [a , 6], the results in 1.2.1 5 and in 1.1.26 imply that n x

M ) = i™ n y i i/(xfc) - /(zfc-i)i

iim y

(t)dt

= ^ mnEi//^)i(^-^-i)= f V (*)|dt.

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Solutions. 1 : The Riemann-Stieltjes Integra l

124

1.2.26. Clearly , ther e i s M > 0 such tha t |/(x) | < M fo r x € [a , b}. Moreover, fo r a = XQ < x\ < • • • < xn = 6, using the firs t mea n valu e theorem (see , e.g. , 1 .1 .26) , w e get

J2\F(Xi) - F(x^)\

da

2=1

i

da

=E i rw ^ - r w w £

f' f(t)da(t)

2=1

= X)l/te)(«( a:oo

pose tha t li m V ( / n ; a , b) = £ < -foo . Then , give n e > 0 , there i s a 71—>00

subsequence {n/ J suc h that V(f nk; a, 6) < Z+£ . Thus for any partitio n a = xo < x\ < • • • < x n = 6, n 2=1

Letting / c tend t o infinity, w e get n

£ | / ( * i ) - / ( * i - i ) | < i + e, and th e desired inequalit y follows . oo

1.2.28. Sinc e the serie s ^2 a n

an

oo

d ^ C bn are absolutel y convergent ,

n=l n=

l

given e > 0, there i s no such tha t y ^ |a n | < e an d ^

|6

n|

< e.

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1.2. Function s o f Bounde d Variatio n

125

If x / Xk, k — 1,2,..., the n ther e exist s a 8 > 0 suc h tha t ther e i s no x n wit h n < no i n (x — J, x + 8). Consequently , fo r x — 8 < y < x,

\m-f(x)\< Y, K I + £ K\ < 1 . Summin g up ,

a(x)

a(0) i

fx

= 0,

a(0) + | i

f X G ( 01, )

a(0) + l i

fz

,

= l.

1.3.8. Conside r th e functio n a(x) =

a(a) i f x

G [a,it),

a(6) i f x

G (w, &],

with a(a) < a(b). The n f{x)da{x) =

f(u)(a(b) -

a(a)) ,

which combine d wit h ou r assumptio n give s f(u) = 1 fo r u G (a , 6). Since / i s continuou s o n [a , 6], we see tha t / ( a ) = /(& ) = 1 . 1.3.9. Sinc e a = p — g, where p, g are monotonicall y increasing , i t i s enough t o sho w tha t / G H(p) i f f n £ ^GP) > n = 1 , 2 , . . ., an d / f(x)dp(x) =

li m / f

n(x)dp(x).

The unifor m convergenc e o f {/ n} mean s tha t sup \f n{x) - f(x)\ = d n >0

x£[a,b] n

. -^°°

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1.3. Furthe r Propertie s .. .

131

Thus, give n e > 0, ther e i s no suc h tha t dn(p(b) - p{a)) < - fo o

rn

> n 0.

By Theore m 1 in Sectio n 1 .1 , ther e i s a partitio n P suc h tha t U(PJno,p)-L(P,fno,p) no, rb

/ /

rb

n

I

/>6

( x ) ^ ( x ) - / /(x)dp(a: ) < / \f

n{x)-

f(x)\dp(x)

fb £ < / d ndp(x) < - . 1.3.10. Observ e that fo r x G [0,1] we have li m nx(l — x2)n = 0 , bu t n—>oo

the convergenc e i s not unifor m o n [0,1 ] an d th e resul t i n th e previou s problem canno t b e applied . W e hav e f1 n 1 lim / nx(l — x 2)ndx = li m — =

n ^ o o j 0 n-^o

o 2( n + 1 ) 2

- .

1.3.11. A s i n th e foregoin g problem , w e canno t appl y th e resul t i n 1.3.9 becaus e {x n} doe s no t converg e uniforml y o n [0,1 ] . Let a = p — q, wher e p an d q ar e monotonicall y increasin g o n [0,1]. Fo r e e (0,1 ) , w e hav e 0 < / x ndp(x) < e n(p{e)-p(0)) > Jo n

0 ^°°

and /x

Jl-s

n

dp(x) J x

n

dp(x)

> ( 1-— ) \P(1 )-P1 Consequently, lettin g n — » o o give s p{l)-p{\-)< li

m/ x n-+oo JO

n

dp(x) < li m / x

n

n

dp(x) < p(l)-p(l-e).

~*°° Jo

Since e € (0,1 ) ca n b e arbitraril y small , n

lim / x

dp(x)=p(l)-p(l-).

The sam e reasonin g ca n b e applie d t o q. Thu s lim / x

n

da(x) =

a(l)-a(l-).

n->ocJQ

1.3.12. B y assumption , fo r an y partitio n a = x$ < x\ < • • • < # m = 6, we hav e ^2 \ an(Xk) ~ OJ n(Zfc-l)| < M. k=l

Letting n — > oo , we see tha t m

^|a(xib)-a(xife-.i)| < M , fc=i

which mean s tha t a i s o f bounde d variatio n o n [a , 6]. Moreover , w e have f f(x)da n(x) -

[ f(x)da(x)\
0 , ther e i s S > 0 suc h tha t i f the mes h o f the partitio n i s less than 6, then \f(x) — f(xk)\ < e/(3M) for x G [xk-i,xk]. Consequently , b y Theore m 3 , x) ~ f(x

/

and

m

k))da(x)

( / ( •


o

n(x),

o

which show s tha t th e sequenc e {p n} decrease s t o zer o o n [a , 6]. B y the monoton e convergenc e theore m fo r th e lowe r Rieman n integra l stated i n th e las t problem , rb rb

0 < li m / f

n(x)dx




rb

(/(x ) - / n ( i ) ) + d x + / /

/ /(x)cfe < li m / Ja

f

n (x)dx

j

v)dx, n(x)

a

where th e las t equalit y follow s fro m I , 2.4.1 9 .

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1.4. Prope r Integral s

143

1.4. P r o p e r I n t e g r a l s 1.4.1. (a ) W e hav e / i ^7—r—^dz = / - ^ -d x + / - ^ -d x J0 \x-2\ + \x- 3 | J o2 x- 5 A -2 x + 5

+J\ x-i)dx + 12 J3

££z± t ^

^— 5

93 = 2 + - l n 3 - -In5 . 44 (b) Usin g th e substitutio n t = 7r/ 2 — x, w e se e tha t / sin n xdx = / cos Jo Jo Now w e sho w b y inductio n tha t (i) / %

n

xdx.

. 2 n 7 l - 3v . . . ( 2 n -Jl )T T (v 2 ny - l ) ! !T T snT xdx = i 2-4...(2rc) 2 (2n)! !2 Jo

and 2 - 4 . . . ( 2 n ) (2n)! ! Psin s i n 2 "2 n+1+1 ;xd x l 3 . . . ( 2 n + l ) ( 2 n + l ) ! ! ' Jo It i s eas y t o chec k tha t bot h formula s ar e tru e fo r n — 1. Fo r n > 1 , integration b y part s give s (2) /

In = sin ./o i

n

xdx = — / s i n o

n1 _

xdcosx

= / cosxdsin n1_ x = ( n — 1 ) / cos Jo Jo = ( n - l)/ n-2 - ( n - l)/ n.

2

x s i nn _ 2 d x

Hence n— 1 = ^n-2 n

(3) I

-

So, if ^2n

_ ( 2 n - l ) ! !T T (2n)!! 2 '

we ge t fro m (3 ) tha t 2n+2

_ 2 n + l ( 2 n - l ) ! ! 7 r _ ( 2 n + l ) ! !T T " " 2 n + 2 ' (2n) H ' 2 ~ ~ (2n + 2)! ! ' 2 "

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

144

This complete s th e proof o f (1 ) . Formul a (2 ) ca n b e prove d analo gously. (c) W e hav e / |lnx|d x = — / Inxd x + / Inxdx . ee

Integrating b y parts , w e get re

2

/ |lnx|d x = 2 - - . J± e e

(d) Substitutin g t = ir — x give s / =

/^ xsin

/ 7 1r

-

0

=

x7

77

Z*

dx=

+; — cosrz ~x y

(7r — x ) s i n x 7

dx

x / 1~ r ~ ;+—c o2s— 0

fn 7rsin x / ^ xsin x x da; / 7 ^ 2~^ ~ / 7 7 7~ Jo1 + cos ^ x j + cos ^ x 01

So, 1 _

f^

7TSin

2

X 7T

i = - / ^—a x= — . 2 J 01 + cos 2 x 4 (e) W e hav e hn = / tan

2n

xd x =1 / (

Jo Jo

V

cos x

2n_2

) tan2 n ~ 2 xdx

J

= / tan xd(tanx ) - I 2n-2 = ~ Jo 2 n- 1 Consequently, on e ca n sho w b y inductio n tha t 1l hn = (-ir(~-(i-Ul---^(-ir3 5 4V 2nv

hn-2-

l

(f) W e hav e sinx -dx = / ^ -rz. -dx ( r/ 0 sin x + co s x" 7 Jo0 si s i n (n ( |f - x ) + co s (J - x ) CO S X — Sin X , 7 T 1 , \/ 2 ox = — i — In—— /o 2cos x8 2 2 (g) Observ e tha t sin"x /"" 2 cos"x n n -ax = / ——r n: ax n sin x -f cos x J sin x + cos x . 0 0

I

4

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1.4. Prope r Integral s

145

It the n follow s tha t 7T

1.4.2. Usin g th e substitution x = cost an d 1 .4. 1 (b) , we get f\l-x2)ndx = Jo J On th e other hand ,

r'\m 2n+xtdt= J o (2n+l)!

2n)!

'„. !

n\* - i

1 t

kr '

2A ; + 1

(-1)

1.4.3. Assume , fo r example , tha t f(x^) = a. W e have f(xo) = F(x 0)= h

m.

x-+x0

x — Xo

On th e other hand , usin g l'Hospital' s rule , w e get ,. F(x) hm = X—+XQ X

~ F(x Q) F'(X) h m= XQ

X

1 -+X+

h m j{x) = a. ^,x+

x

2

1.4.4. Defin e F(x) = x cos \ - 2 f* t cos \dt fo r x ^ 0 and F(0) = 0 . Then F'(x) — f(x) fo r x e R. Thu s F i s an antiderivative o f / i n the case whe n c = 0. If c ^ 0 , then a n antiderivativ e G of f woul d equa l F(x) + c\ fo r x > 0 an d F(x ) + C 2 for x < 0 , wit h som e constant s C\ and C2 - Moreover, sinc e G mus t b e continuou s a t zero , c\ = c 2 . Consequently, G'(0 ) = 0 . Thi s prove s tha t a n antiderivativ e o f / exists i f and only i f c = 0 . 1.4.5. W e will sho w tha t ^sin^ri [0 i

f x G [a^fc+i^fc-i] , fz = 0

has th e desired properties . Indeed , sinc e / i s continuous o n (0,1 ] , 7T [

i ^ - i ) " F{x 2k) =

X2k

1 ~l

7

--/ —

T

sin — dx

X2k

(2fc-l)7T -

2k J22/CT , T

.

sin tdt = - . ^

Similarly, ^(>2fc+l) - ^ ( ^ 2 / c ) = T -

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

146 Now w e pu t

\oi

fx

= 0.

Calculation show s tha t F(x) = J " ^ ( C 0 S ^ + \0 i

1

)i

f x

[ x2fc+i^2fc-i], fi = 0. e

Then F'(x) — f(x) fo r x G (0,1]. Moreover , fo r x G [a^fc+i^fc-i] ?

- — i — < - ^ b^(f(b-2e))». Since

' (f(*)) pdx-> I' (f(b-s)y>dx, Jb-e

we see tha t b ir ,. , ,i ' * f(,-))Pdx>1— I (f(b-c))"d.r b - a J a ' b-a J h c(f(b-s))" >(/(&-2r))". b-a Hence 0(p) must , satisf y 0(p) > b — 2c, fo r p > po. whic h mean s I hat lim 0{p) = b. P—>OC

1.4.34. Clearly , rb

P(x)f(x)dx =

0

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

158

for ever y polynomia l P. B y th e approximatio n theore m o f Weier strass (see , e.g. , II , 3.1 .33) , ther e i s a sequenc e o f polynomial s {P n} uniformly convergen t o n [a , b] t o / . Thu s b y th e resul t i n 1 .3.9 , ob pb

/ f 2(x)dx = J an

li m / P - > °° J a

n(x)f(x)dx

=

0,

which give s f(x) = 0 in [a , b\. 1.4.35. B y assumption , b

0

P{x)f(x)dx =

for an y polynomia l P o f degre e les s tha n o r equa l t o TV . Suppose , contrary t o ou r claim , tha t a < x\ < X2 < • • • < x m < 6 , m < iV, ar e all zero s o f / . No w choos e thos e point s xi x,..., Xi r wher e / change s its sign . Withou t los s o f generalit y w e ca n assum e tha t f{x) > 0 i n [a,x^], f(x) < 0 in [x^,Xi 2 ], an d s o on. Pu t r

P(x) = l[(x

lk-x).

k=l

Then P(x)f(x) > 0 in [a , b] an d P(x)f(x) > (a,xi), ( x i , x 2 ) , . . . , (x m ,6). Therefor e

0 in each o f the interval s

rb

P(x)f(x)dx > 0, / although P i s a polynomia l o f degre e les s tha n o r equa l t o iV , a contradiction. 1.4.36. Suppos e tha t / G C{[~a,a\). (a) Not e firs t that , b y assumption , [* {f(x) + f(-x))x 2ndx= r J—a

f{x)x

2n

dx+ r

J —a

/(-x)(-a;)

2n

J—a

dx

2n

dx = 0.

= 2 f f(x)x J —a

Moreover, f (fix)

J—a

+ f(-x))x 2n+1 dx =

f f(x)x J—a

J

1 2n+

dx

i

f{-x){-x)2n+1dx =

0.

J —I

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159

1.4. Prope r Integral s Thus, w e se e tha t + f(-x))xndx =

/ (f{x)

0 fo r n = 0 , 1 , 2 , . . . ,

J— a

and th e desired resul t follow s fro m 1 .4.34 . (b) Th e proo f run s a s in (a) . 1.4.37. B y the first mean valu e theore m w e get rb r f^O

b+h r fO-\-n

/ (f(x + h)- f(x))dx =

b rO

- / f(x)dix ix

/ f{x)dx

J aJ

a-\-h Ja ra rb+h

= / f(x)dx+

/

f(x)dx

Ja+h Jb

= -hf(a + 6h) + hf(b + 6'h), where # , 6' G [0,1]. It then follow s fro m th e continuity o f / tha t Urn \ [ (/( * + h) - f(x))dx =

f(b) - / ( a ) .

1.4.38. W e hav e pb+x g(x) = / f(u)du. J a+x

Thus g'(x) = f(b + x)-f(a +

x).

1.4.39. Usin g l'Hospital' s rule , w e obtai n

h

^=11^i)=1

X—*00 yJX V

X—>0 2y/x

0 —i-

=

(b) il^dt lim —— \ =

li m cos a; = 1.

(c)

J ? si n y/idt 2xsin lim — = x^Q+ X

3

x2

li m3x—

x^0+

2

3

(d) X

lim (— [ x^oo \x

a

J

0

In ^\dt) =

Q(t)

J

li m x-^o

In P{x)-]nQ(x)

= 0.

o

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

160

1.4.40. Usin g l'Hospital' s rule , w e ge t (a)

,. f i r

i

x

,\

hm - — / at — hm= x^oo^lnxJo V I + t 5 / ^ ^ V l + a; 5

1 .

(b) = lim( l + sinx) 1 / x = e ,

lim [ - / ( 1 + smt)idt J because _. ln( l + sinx ) 1 hm = x->o x x-+o

cos x h m= 1 + sin x

1 .

(c) lim (\ ( x-^o+ \x 2 J

X 0

t^dt] = li m ^ = \. J x^o+ 2 2

(d) x x2 1f 2 e lim —2 In / e l dt — li m x — x-^oo x 7 0 a?-+o o 2.x J0 e * d t

2xex2 ^-oo 2 / 0 V d £ + 2xe x2

= l i m —-r~— 2-0 7

= li

m~

e x2 + 2x 2 e x2 ~

x-^oo 2e x

So li m ( ^

- f 2x 2 e x "

= 1

.

l/(x 2 )

1.4.41. I f A = \f(x 0)\ =

max{|/(x) | : x G [0,1], the n fo r p > 0,

[ imr)'" < (£*>)'" - A. On the othe r hand , th e continuit y o f / implie s that, give n e > 0, ther e is a n interva l [a,/3] C [0,1 ] suc h tha t \f(x)\ > A - e/2 fo r x G [a,/?]. Consequently,

' |/(*)lp) " > I [ l/(*)l P] > = (,4-£/2)(/?-a)

1 /p

( £V - e/2)" ) >,4-£

for sufficientl y larg e p.

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1.4. Prope r Integral s

161

1.4.42. Le t yo G [c,d\ an d e > 0 b e chose n arbitrarily . Sinc e / is uniformly continuou s o n R , ther e i s 5 > 0 such tha t l/(zi>2/l) ~ f(x2,V2)\
0 such tha t fo r x G [a, b] an d |/i | < (5, ^(x,y +

h)-/ y(x,y)\
0.

fc=0

If w e se t 2

F ( x ) = c t i + t t+ +t

+

2iV

r " ( ^ ^ "" (^j*'

then, b y th e abov e inequality , r2N /

.

,2

J.2N

F(2N)= /JO e - ' [ 1l + l !+2 i!r + -+ (2iV) v

7

^ !) d t

On th e othe r hand ,

Fm=i

fN (

t

t

2

t

2N

N

\ f

.-( 1 + _ + _ + ... + _) t f < > |; lli(=w .

Since F is continuous, th e desired resul t follow s fro m th e intermediat e value property . 1.4.47. Pu t P(x) = anxn + an _ i x n _ 1H h

a0.

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163

1.4. P r o p e r I n t e g r a l s By assumption , (P(x))2dx = /o 7

a 0 P{x)dz o

On th e othe r hand , r-l

a0

(i) a

:1

+ /c

" fc +

+ ^4 + •

1 "

ifc + 1

fc = 0 , l ,

fc + r

n+ k

^n ^n— 1

n +A ; + 1 n Observe no w tha t

a0

+ -^-4 +

/ x f c P(x)dx= ^ 7o n +/c + l and, b y assumption ,

0 fo r k

— 1, . . ., n.

Q(k) (k + l)...{n +

k + l)

,

where Q i s a polynomia l o f degre e a t mos t n. Sinc e Q{k) = 0 fo r fc=l,2,...,n, ther

e i s a constan t C suc h t h a t Q(k) = C(k-l)(k-2)...(k-n).

P u t t i n g k = 0 i n (1 ) w e se e t h a t

Jo n

v

+1 1 n

;

rc + 1

Now, multiplyin g (1 ) b y A ; -f - 1 an d nex t settin gA : = —1 , w e obtai n a 0 = ( - l ) n ( n + l ) C . So , a 0 = ( n + l ) 2 / P(x)dx,

Jo

which implie s th e desire d equality . 1.4.48. I f I(y) = J a }'{x,y)dx, the n b y 1 .4.42 , I(y) i s continuou s o n [c, d]. Thu s th e integra l o n th e left-han d sid e o f th e equalit y t o b e proved exists . Moreover , i f g(z) = I(y)dy

an

d h(z)

= I

f(x,y)dy\dx,

then g'(z) = I(z) fo r z G [c , d], an d b y 1 .4.42 , th e functio n F(x,z) = /

f(x,y)dy

is continuou s wit h respec t t o x . I t i s clea r t h a t ^(x,z) = f(x,z) i s continuous o n R . Henc e b y 1 .4.43 , /i'(2 ) = I(z). S o /i'(2: ) = g'(z) =

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Solutions. 1 : The Riemann-Stieltje s Integra l

164

I(z) fo r z G [c, d], and therefor e h(z) = g(z) + C. Sinc e h(c) — g(c) = 0, w e see tha t C = 0, whic h end s th e proof . b_ a

b

_a

0 and li m x , x =

1.4.49. Sinc e li m x , x =

b — a, the proper

integral exists . B y 1.4.48, /X

'^dx =

f x ydy\ dx=

f (

f ( f x ydx) dy

1.5. Imprope r Integral s 1.5.1. (a ) We hav e 1 /•27T -j /•27T

/

_/ »/ 7 r /

-T~ 4

Jo si

> 7 r 44 1 /

7 ~ ^=

8

n x -f- cos4 x J

/

4

rn/4 ^

_o

-

^

/ 2

Jo cos = s

T

n xx + + cos cos44 xx n

o si o si

/ 4

r

i

J0 cos

(2x) + |sin 2 (2x

( 2

0

dx_

(2x) V I + | tan 2 (2x)7 .

/.TT/ 4

Jo l

dx

1

+ ^tan 2 (2x) J- -t- 2 ^

= 2TIV 2

(b) Observ e firs t tha t r-n/2 PTT/2

/ ln(sina;)da Jo J Thus 1 />7r/ / InfsinxW Jo

/.TT/

: = / ln(cos(7r/ oJ

2

2 — x))dx = I \n{cosx)dx. o

/ 'r / 2sin2 1 x x= - / I n dx * Jo 2

i r /2

i r7T si n a; = - / I n —dx 4 J0 2

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1.5. Imprope r Integral s

165

Consequently, r 7r/2

I hi{smx)dx Jo 2

— — — In 2.

(c) W e prov e b y inductio n tha t 1 n]

oV

(

a + l)( a + 2)---( a + n + l )

For n = 0 and a > — 1 we hav e 1 f1 Jo a-\-l

0 a

! +1

Assuming th e equalit y t o hol d fo r n , w e integrat e b y part s an d get

xn+1(l-x)adx =

f

xn+1

(~^~X^1 ) dx

,x Oi + 1 Jo

n

(l-x)a+ldx

(n+1)! (a + l)( a + 2 ) . . . (a + n + l ) ( a + n + 2) ' (d) Integratin g b y part s give s i:

In= (-lnx) o

n

dx =

n (-lnx)

1 n

~ dx =

nl

n_i.

Since I\ — 1, we se e tha t 7 n = n! . (e) Th e substitutio n £ = 1 — x/n give s n

1 - (l-x/n) n , ^ ^-ct x Jo

Z* 1 l - tn , e = / dt 1 -t = [\l + t + t 2 + --- + t Jo 1 1 1 23 n

n1

- )dt

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166

Solutions. 1

: T h e R i e m a n n - S t i e l t j e s Integra l

(f) Integratin g b y part s n times , w e ge t 1

nn n x ln xdx =

1

f

/

n

lnn~lxdx

x

o

= ••

•=

(g) I f n = 1 , the n f™ f Xf^2\n — x — t an t give s ^oo

7

/,

(1 -

)

TT/2. I f n > 1 , the n th e substitutio n

/.TT/ 2

,-

9

/

2

J0 ( 1 + x ) - 7

/.TT/ 2

2

cos

" £ c o s "2 tdt = / c o s

o7

2n 2

- tdt

o

_ l - 3 . . . ( 2 n - 3 ) ix ~ 2-4...(2n-2 )'

^'

where th e las t equalit y follow s fro m 1 .4.1 (b) . (h) Th e substitutio n t = x — \ / x2 — 1 yield s

i;

dx f 2

1

2

X \JX

17

At

o ( l + ^ 2):

rdt = 1 .

(i) Observ e firs t t h a t oo

0

x

l n x Z" 1 l n x f r 52 +^ aax2- 7 - -- r—a o- « j^ = / -T 0X

00 ln x h / i— -dxa ^ ^ Z+

and t h a t b o t h integral s o n th e righ t sid e o f th e equalit y converge . Furthermore, a f°° l n x , f \nx / -7> z Tjdx = / -1 1 z3

J0 x

+a J

0x

l

f°° ln x7 -d£ + / — 2-dx

+ a2

7a

x -f a 2

7 r^x + / , r^.To ^ o ( - T T ) dt 0 x2+a2 Ja (a2/£)2 + a 2 V ^2 a

a f 21 na-ln t. dx + — — dt 2 2 2 0x + a J o^ + a a 2 In a 7rln a

In

x,

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1.5. I m p r o p e r I n t e g r a l s

167

(j) W e hav e I n a: f°°°° lnx _ 1 f°° m a + m^ J0 (xi + a^dX-^J0 ~(T^W r

hit + t 2)'

7rlna

1

4a 3 a

3

-dt

Int ;dt, 7o ( 1 + t 2)2 '

where th e las t equalit y follow s fro m (g) . Substitutin g t — t a nx give s at = co

s im(tanx)aa :

o ( 1 + *2)2^ X

p l + cos2 x m(tanx)a x Jo 2 1p = - I cos2xln(tanx)(ix , 2 Jo x = 0 (se e t h e solutio n t o (b)) . Finally , inte =/

because J 0 l n ( t a n x ) d gration b y part s give s

2

sin2 x

1

• dx 2 ta n x cos (k) Usin g th e substitutio n t = T T — x , w e se e t h a t / cos2xln(tanx)(i Jo Jo

/= /

x= 0 — /

l n ( l + cosx)(i x = / Jo Jo

== 2

x2

. *

l n ( l — cost)dt

So, b y th e symmetr y o f s i n x an d b y (b) , 21 = / l n ( l - cos Jo Jo

2

x)dx = 2 / l n ( s i n x )

d

rn/2

/ ln(sinx) ln-+ln

n ^ \x fc=l

n

n-

J\

7

\

-

>I

n^ n

)

fc=l

n - + In

Hence b y 1 .5. 3 an d 1 .5.1 (d) , lim a n—\ (lnx)

n

-*°° Jo

2

\Jo

dx — I / Inxdx ) = 2 — 1 = 1 .

)

1.5.7. W e firs t observ e tha t th e Cauch y theore m (see , II , 1 .1 .37 ) implies th e followin g criterio n fo r convergenc e o f imprope r integrals : For g : (a , b]—> R , th e imprope r integra l J bg{x)dx converge s i f an d only if, give n e > 0 , there is S > 0 such tha t

a C 29(x)dx < e whenever

a < a\ < a + 5 and a < a2 < a + S. Consequently, b y assumption , 2x rrZX

lim / t x^0+ J

a

f{t)dt = 0.

x

Clearly, w e ca n assum e tha t ther e i s a positiv e XQ such tha t / i s either positiv e o r negativ e o n (0,xo) . I f / i s positiv e o n (0 , #o) an d decreasing, the n \f(2x){2x)ax i

/

fa

< 0.

Likewise, i f / i s positive o n (0 , XQ) an d increasing , the n

l

„ ,/ x ,I

f(x)xax i fa > 0, )/(x)(2x)ax i f a < 0 .

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

170

It follow s fro m th e abov e inequalitie s tha t li m / ( x ) x a + 1 = 0 . Anal ogous reasonin g ca n b e applie d i n th e cas e whe n / i s negativ e o n (0,x 0 ). 1.5.8. (a ) W e hav e f°° dx , /— =

x->o

J 2 Xn\X

., , h m m i n x — m m 2 = +oo . o

(b) Sinc e 2 sin 1 x 1 + ir 2 ~ 1

+ x 2'

the integra l converges . (c) Th e substitutio n —h\x = t give s pi poo

pi a

/ (~hix) Jo Jo Since

dx= /

poo

a

t Jo

e~ldt= /

t

a

Ji

e-tdt+ /

t

a

e'ldt.

tae~l lim —3^ - = 0 ,

the las t integra l converge s fo r al l rea l a. Moreover , sinc e tae~t lim = a t-o+ t

1 ,

the integra l J Q t ae~tdt converge s i f an d onl y i f th e integra l J 0 t adt converges, tha t is , i f an d onl y i f a > — 1. (d) A s i n (c) , J0 x

dx {-lnx)b

t-be-(l-a)tdt

a

0

It i s eas y t o chec k tha t i f a > 1 , the n th e las t integra l diverge s t o infinity fo r an y b. Sinc e fo r a < 1 pOO pOO

h l / t- be-^-a^dt = (1 - a ) 6 "1 / r e- dt, Jo Jo it follow s fro m th e solutio n t o (c ) tha t ou r integra l converge s i f an d only i f a < 1 and b < 1 .

(e) Th e divergenc e o f th e integra l follow s fro m th e inequalit y /z Jo1

o — dx > / Tydx. + x 2 su r x J o1 +r

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1.5. Imprope r Integral s

171

1.5.9. Sinc e f(t) + l / ( / ( t ) ) > 2 , we see tha t

J* f(t)g(t)dt + jT ^j|dt > 2 £ 0(*)d* . Letting x — > oo , we get th e desire d result . 1.5.10. Th e assertio n follow s immediatel y fro m th e definitio n o f a n improper integra l an d th e Cauch y theore m give n i n II , 1 .1 .37 . 1.5.11. Suppos e tha t th e integra l J a°° f(x)dx converge s an d {a n }, an > a, i s a n increasin g sequenc e divergin g t o infinity . The n n

poo pa

/ f(x)dx

pa

n

= li m / f(x)dx =^ / n=l ^

k

= li m V ^ / f(x)dx

f(x)dx. a

n-i

On th e othe r hand , i f the las t serie s converge s fo r ever y increasin g se quence {a n} divergin g t o infinity , the n th e limi t li m f f(t)dt exist s and i s equa l t o th e su m o f thi s series . If / i s nonnegative , the n th e functio n F(x) = J f(t)dt i s mono tonically increasing. Moreover , i f there is an increasing sequence {an }, an > a , divergin g t o infinit y fo r whic h th e serie s (1 ) converges , the n F i s bounded abov e b y the su m o f this series. So , the limi t li m F(x) x—>oo

exists. 1.5.12. B y the resul t i n the previou s problem , th e integra l converge s if an d onl y i f J^inTr l

+ x a s i n 2 x ~ ^Jo l

+ ( x + n7r) a sin 2 x < °°

'

Now observ e tha t o1

dx f < + ( ( n + l)7r) a sin 2 x Jo

n

dx

1

~ Jn 1

+ (x + n?r) a sin 2 dx -f (n7r) a sin2 x

Moreover, w e hav e

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

172

Substituting y = tanx , w e se e tha t ^ 2 dx

dy 2

2

5

+ {b 2 + l)y2 2>/6

+ b sm x~ J 01

J/o0 l

TT"

It the n follow s tha t th e integra l converge s fo r a > 2 and diverge s fo r 0 < a< 2 . 1.5.13. No . Tak e f(x) = e~ x + g{x), x e [0 , oo), wher e

9{x)

0

if x

6 [0,7/4] ,

1

if s

= 2,3,... ,

if x

e [n — n~ 2,ri\, n

2

n {x - n) + 1 2

—n (x - n) + 1

if x G [ n , n - f n "

0

otherwise.

2

], n

= 2,3,... , = 2,3,... ,

1.5.14. Yes , i t does . Indeed , i f i t i s no t tru e tha t li m f(x) = x—>oo

0,

then ther e exis t e > 0 an d a sequenc e {x n } divergin g t o infinity , xn+i—xn > 2e, and such that f(x n) > e. B y the mean value theorem , f(x) = f(x n) + f'(£ n)(x ~

x n),

Xn < £ n < X < X n + e/4 .

Consequently, £

f(x)>e-2-

2'

and

f(x)dx > J2 / /(*)

&

* > E / f^

dx

^

o

8

-foo ,

a contradiction . 1.5.15. Assume , contrar y t o ou r claim , tha t li m f(x) = 0 doe s no t x—>oo

hold. The n ther e exist a positive number e and a n increasing sequenc e {xn} divergin g t o infinit y suc h tha t eithe r f(x n) > £ fo r al l n o r f(xn) < — £ for all n . Suppose , fo r example , tha t f{x n) > £ for al l n. By th e unifor m continuit y o f / , ther e i s 5 > 0 such tha t i f \t — s\ < e/2, an d therefor e rxn+5

(1) /

f{x)dx

> eS.

J xn — 5

On th e othe r hand , sinc e the imprope r integra l J a°° f(x)dx converges , by th e Cauch y theore m (se e 1 .5.1 0) , ther e i s a o > a suc h tha t fo r a2 > a i > ao , < eS,

f(x)dx\ i

I

CL-L

which contradict s (1 ) . Additionally, w e observ e tha t i f / i s continuou s o n [a , oo) an d lim f(x) — 0 , then / i s uniformly continuou s o n [a , oo) (see , e.g. , II , x—»oo

1.5.7(b)). It i s als o wort h notin g her e tha t th e resul t i n 1 .5.1 4 is containe d as a specia l cas e i n thi s problem . 1.5.16. Le t {x n} b e a n increasin g sequenc e divergin g t o infinity . Then, give n e > 0, ther e i s no suc h tha t /•OO

/ f(x)dx Jxno


no, pX-n POO

(xn - x

no)f(xn)


a contradiction .

(b) Le t en

an = e e , n = 0,1 ,2 , We defin e / b y settin g f(x) = a n fo r a n _i < x < a n , n = 1 , 2, Then, b y 1 .5.1 1 , oo

dx \-^ e

a n- a

~ ar, /(-) n 1 =

Je

n _i

.

because li m ^-^- — 0 . O n th e othe r hand , n—>-oo

(y, Tan dx "

dx ^__

f°° dx

xe e e " KS*J p '

E

- p'»

e—

*•

e

1 *-! O

O

-,

^1

Y

n=l n=

l

1.5.18. Se t F{x) = J* f(t)dt. The n F'(x) = / ( x ) , an d b y assump tion, lim (F'{x) + F(x)) = Z , ! G R .

X—>00

It follow s fro m II , 2.2.3 2 tha t li m F(x) = / , an d therefore , x—+oo

lim /(x ) = li m F\x) = :r—>oo x—>o

0.

o

1.5.19. Pu t F(x) — J Q f(t)dt. Then , b y assumption , F i s increasing , and integratio n b y part s give s

ir i

r i

- / xf(x)dx

=

r

- / xdF{x)

)dx.

= F{n) - - F(x

We kno w tha t li m F(n) = L f(x)dx
oo n

u

m+

n—>o

-+

o

F(n-l) < 1

T


0 such that |/(s ) — f{t)\ < 1 i f \t — s\ < 5. Now suppose , contrar y t o ou r claim, tha t / i s no t bounded . The n ther e i s a sequenc e {a n} suc h that a n + i > a n + S and \f(a n)\ > n. B y assumption ,

/ f(t)dt\

> / f(t)dt\

f(t)dt

- /

f(t)dt -M. Moreover, w e hav e \f(t) — f(a n)\ < quently,

1 for t G [a n — 5/2 yan]. Conse -

f(t)dt >(\f(an)\-l)S->(n-l)Sand

f{t)dt >{n-l)~(

M,

Ja

a contradiction . 1.5.21. Integratio n b y part s give s

(1) f

f(t)f"(t)dt = /(*)/' (x) - /(a)/'(a ) - f (/'(*)) t/a J

S

dt.

a

By th e inequalit y (/(x)) 2 + tf"{x)) 2 > 2\f(x)f"{x)\ an d b y assump tion, th e integra l J a°° f(i)f"{t)dt converges . Clearly , f*(f'(t)) 2dt ei ther converge s o r diverge s t o infinit y a s x — » oo . In th e latte r case , b y (1), li m f(x)f f(x) = +oo , an d sinc e /2(*)-/2(a)

r Ht)f(t)dt, Ja

we se e tha t li m f 2(x) = 2

x—>oo

J^°(f(x)) dx. S

+oo , whic h contradict s th e convergenc e o f

o the convergenc e o f f^°(f'(t)) 2dt i

s proved .

1.5.22. I t follow s fro m th e Cauch y theore m (se e 1 .5.1 0 ) tha t th e improper integra l J^° f(x)g(x)dx converge s if and onl y if, give n e > 0 , there i s CLQ > a suc h tha t fo r ci2 > a\ > ao,

f

f(x)g(x)dx


0 such tha t \g(x)\ < C fo r al l x G [a, oo). By (1 ) , the imprope r integra l J a f{x)dx exists, s o th e Cauch y theore m implie s tha t ther e is ao > a such tha t / f(x)dx

an

ao. Thu s G2

/ f{x)g(x)dx

)dx + l^2)|

< l5(ai ) 1,

and th e functio n x — i > l/(x a) monotonicall y tend s t o zer o a s x — • oo, the convergenc e o f J x § ^^dx follow s fro m th e Dirichle t test . (b) I f th e imprope r integra l f™ * sl"x*dx converged , the n J^° ^^dx would als o converg e (becaus e sin 2 x < | sinx|). Thi s woul d giv e (i)

If 1

cos 2x

-dx < 00 .

2/1"

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1.5. Imprope r Integral s

177

As i n (a) , on e ca n sho w tha t J^° CQ ^2xdx converges . Thi s combine d with (1 ) woul d impl y th e convergenc e o f J^° ^ , a contradiction . (c) Th e substitutio n x = yfi give s OO rC

sint dt. 2y/t

/

s'm(x2)dx = j Thus, b y (a) , th e integra l converges .

(d) On e ca n appl y th e Dirichle t tes t wit h g(x) = l/(x a) an d f(x) = esin x gm 2 X, becaus e

/

2esinx(sinx-l)

e s i n x sin(2x)dx


1 . (e) Pu t lnax g(x) = an

d f(x)

= si n a:.

Then g\x) =

In"" 1 x

(a - l n x ) < 0

for sufficientl y larg e x. Thu s g i s monotonicall y decreasin g t o zero , and th e Dirichle t tes t fo r convergenc e ca n b e applied . 1.5.25. I f J a a + T f(x)dx = 0 , the n als o J^ kT f(x)dx = 0 fo r k G N. For 6 > a put fc = [ ^ ] . Then , b y th e resul t i n 1 .4.8 , pb pb

/ f(x)dx\ = Ja \Ja+kT

—kT

pb

\ f(x)dx\ la+kT

=

\ f(x)d.

\Ja

nb-kT rb-kT pa+T

< / \f(x)\dx< JaJ

/

\f(x)\dx

= C.

a

Now the firs t statemen t i s an immediat e consequenc e o f the Dirichle t test. To prov e th e secon d statement , se t f* f(x)dx consider th e integra l

=

I ^ 0 an d

/(*) - ^ ) g{x)dx. Ja

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

178

It follow s fro m wha t w e hav e alread y prove d tha t thi s integra l con verges. Consequently , J^° f(x)g(x)dx converge s i f an d onl y i f th e integral J g(x)dx converges . 1.5.26. (a ) Sinc e r27T

sm(smx)ecosxdx cos:E

= / sin(sinx)e

^+ / sin(sinx)e

JO

cosj:

(ix = 0 ,

JTT

the give n integra l converges , (b) W e hav e / sin(sinx)e JO

sm;r

(ix = / sin(sinx)e

sm:r

Jo /o

sm:r;

Gfo + / sin(sinx)e

JO

(ix

JTT

(esinx - e~ s'mx) sm(smx)dx >

0.

Thus th e integra l diverges . x

1.5.27. Observ e first tha t li m ^ x-+o+

x

— exist

+

smx

s a s a finite limi t fo r al l

positive a. S o w e ca n confin e ourselve s t o studyin g th e convergenc e of oo sinx : ax. a x + sm x We hav e 2 sin x si n x sin x a a a a x + sin x x x (x + sinx ) We kno w (se e 1 .5.24(a) ) tha t fo r a > 0 th e integra l J 2°° § ^-dx con verges. So , i t i s enoug h t o stud y th e integra l

s:

sin2 x J2 x a(xa + sinx ) dx. poo

Since '' x

2 sm 2 x sin x a a a a (x + 1 ) " " x (x + sinx) ~ x a(xa -

1 ) '

we see that integra l (1 ) converge s fo r a > 1 /2 . Applyin g th e resul t i n 1.5.25, we find that th e integra l J* 2°° w ^ + D^ diverge s f° r a — V^So, i t follow s fro m (2 ) tha t th e integra l (1 ) als o diverge s i f a < 1 /2 . For a > 1 / 2 th e integra l converges .

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1.5. Imprope r Integral s

179

1.5.28. Sinc e ,oo 1

f(x)dx= /

0/0))-cb , J aX

the assertio n follow s fro m th e Abe l tes t fo r convergenc e o f imprope r integrals (se e 1 .5.22) . 1.5.29. Assume , fo r example , tha t / i s decreasin g o n [0 , oo). Then , since f n°° f(x)dx exists , w e se e tha t li m f(x) = 0 . Hence , b y 1 .5.1 1 , />oo °

/ f(x)dx

° «(

= Y, /

n+l)/i

°

°

f(*)dx

h^2f(nh) J® n =

So,

=

h^2f(nh)-hf(0).

l n=0 0 0

POO

0 < / i V / ( n / i ) - / f(x)dx 1 , the n li m e

x

xa x

— 0, s o i t suffice s t o sho w th e conver -

gence of ^~ldx _.

r

X

/ ex with som e XQ > 0 . Sinc e li m e x /2x1a

— 0, ther e i s x$ suc h tha t

re—•oo

xa-ie-x/2


XQ . Henc e /•oo

/>co

x a l

/ e~

x - dx
1 , the n Cn+1 n

-

cn a

'

i + I'i > i ,

+ n V n

y

a

because ( l + ±) > 1 + g (see , e.g., II , 2.3.7(a)). Thu s i n this case the sequence {c n} eithe r converge s o r diverge s t o infinity . I f 0 < a < 1 , then ( l + £) a < 1 + % (see , e.g. , II , 2.3.7(b)) , an d consequently , the sequenc e {c n} i s monotonicall y decreasing . T o en d th e proo f i t is enoug h t o appl y th e resul t give n i n II , 3.3.2 1 (whic h ca n als o b e extended t o th e cas e whe n A = -hoc) .

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1.5. I m p r o p e r Integral s

181

1.5.32. W e hav e r}. a _ 1 n.!

r ( a ) = lin->oo m a (a + 1 ) . . . (a + n — 1 ) n.a—1 n.

lim ™ a ( a + l)(f + l ) . . . ( ^I + l ) a fc= = h m n1 -e

71-1

lim

a e

a/f

(lnn-(1 +i+-+^r))IF T e "

n-+oo a

,-* • \1

c

+?

1.5.33. B y 1 .5.24(a ) we know that th e integra l f 0°° ^^dx converges Therefore

.

n7r/2 / 2 r ° sin1 x, sin x , v r sinn x 7 B f /a x = li m / ax = li m / ax. n n Jo x ^°°Jo x ^°°Jo x Hence / 2 f°° sin x , , . r sin(2 n + l) x , / ax = li m / ax . n Jo ^ ->°°Jo x We wil l sho w tha t r / 2 s i n ( 2 n + l ) xT T 1/ — ^ -dx = - , n = l,2,... ,

and (2) li

mf f n ^°° \Jo

/2 x

M ^ + D x ^_ Jo

r/ » sin(2n +l), \ sin xy

=

Q

To sho w (1 ) , recal l tha t fo r x E (0, 27r), 1 sin(±(2 n + l)s ) - 4 - cos x + co s 2x + • • • + co s nx = ^——. 22 sm | Hence sin(2n + l) x ^ ^ —— = 1 + 2cos2 x + 2cos4x H + sinx

. 2cos2nx , x

x

e (0,7r) .

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

182 This give s

r 7r/ 2

sin(2n - f l)x . n ax — x2 //o sin To prove (2) , observeJOtha t b y 1 .4.1 1 (a) , ^/2M2n+ sin(2 )xn + l)x , f 1

n

ton r Q x

**-TQ s

j Jo

^ ° ° V Jo x

f ' smx — x . .

= h m / sm(2 n-+oo j0 xsin

n/2

x

_

sin(2 n + l)x dx mx .

n + l)xdx = 0 .

1.5.34. I t follow s fro m th e inequalit y l_

x

2 < e - 21
ao pO rb

b' rrD

f{x,y)da x ix\ < e

/ f{x,y)dx- / J a

J a

for al l y G A. Takin g y — > yo , w e ge t b y (1 ) 6 r/»0

/.&' no

/ (p(x)dx — / (p(x)da Ja

0 be chose n arbitrarily . The n there exist s ao > a such tha t fo r a b > ao,

and

r

f(x,y)dx

f

0, there i s 770 > 0 such tha t for 0 < 77 , rjf < 77 0 pb—r) no — rjrb—r}'

no

f(x,y)dx- / / f{x,y)dx-

jf(x,y)dx

«/ a t

|< £

/a

for al l I / G A. Takin g 2 /—» yo, we get by (1 ) t>—77 pb—r}'

(p(x)dx — \ (p(x)dx

0 be chosen arbitrarily . The n ther e exist s 77 0 > 0 such that fo r an 0 oo

^ = = // (p(x)dx = 0 .

Since th e resul t i n 1 .5.4 0 ca n b e applie d t o th e integra l J 2°° e xV dx, the proo f i s complete . 1.5.55. Pu t fn(t)

{{l-^Y^-1 i I0i

f 0oo

theorem (see , e.g., II , 3.1 .1 6) , {f n} converge s to / uniforml y o n [0 , a]. Since | 0 °° f(t)dt = T(x) < o o (se e 1 .5.30) , give n e > 0 , •>o o

/•o o

/ fn(t)dt

< / f{t)dt

JaJ

< e

a

for sufficientl y larg e a. Therefor e th e resul t i n 1 .5.4 0 can b e applied . 1.5.56. W e firs t sho w tha t 1 W

2

smnx dx = n o sinx We kno w (se e (1 ) i n th e solutio n o f 1 .5.33 ) tha t (i)

I

1x12

sin(2 n + l)x , T T dx = —. sinx 2

Moreover, sinc e n-1

^sm(2k +

1 — co s 2nx 2sinx

l)x

k=0

sin nx

X ^ 0 , ±7T , ±27T, . . . ,

equality (i ) follows . Substitutin g y — nx i n (i ) give s ^ / 2 /sin V2

2

2//^

dyy = —. 2

sin(2//n)

2/

Now defin e for x

1

/nW = { ( ^ ) ( ; E & ) ^ 2

0 fo

= 0,

(Xa^mr/2 rx

> nir/2.

for x

= 0,

,

We hav e 1

lim / n ( z ) v

( ^ j fo

r x>0

,

and th e convergenc e i s unifor m o n eac h interva l [0,a] . Recal l (see , e.g., II , 2.5.28(a) ) tha t fo r 0 < x < TT/2 , smx 2 > - . X IX

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1.5. Imprope r Integral s

195

So, „ , N , ' s m x\ 7r fn{x) < [ —^j —

2

,£

>0 ,

and since J 0°° ( ^ ^ ) dx < oo, the uniform convergenc e of J0°° f follows fro m 1 .5.35 . Hence , b y 1 .5.40 , - = li m / f I n-+ooJ

Q

n(x)dx

=

J

/ li Q

mf

n(x)dx

n(x)dx

n^oo

1.5.57. W e hav e 00 f°° x a-x , f 1 x a~l , f x*" 1 , T T / dx = / a x+ / a x= ii+i 2Jo1 +^ J o H x J 11 +x

For 0 < x < 1 , a —1 °

°

and th e serie s converge s uniforml y o n [77, 1 — 77'] , wher e 0 < 7 7 < \ — T)' < 1 . Moreover ,

0 < sn(x) = Vc-i)*^*-1 = ^ d - ( - * ) ") < ^-^ Since J Q x a~1 dx = satisfied an d

1 /a , w e se e tha t th e assumption s o f 1 .5.5 3 ar e

n=0 u

n=

0

Furthermore, th e substitutio n x = 1 /y give s r1

7|

-a r

io H y 7

l 0

7l

(l-a)-l

i+ y ~

i-

a

where th e las t equalit y follow s fro m th e above . Henc e / -, Jo1

dx= +s a

- + > (-l) Mr + v aa -f / c a f- f \ + ^

— fc/

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

196

which i s th e firs t equalit y t o b e proved . T o prov e th e othe r on e w e start wit h th e identit y (see , e.g. , I , 3.8.37 )

=*n( i -

smx

n=l

Thus i f x j^ kn, k G Z, the n oo

= l ^ In In I sin £ I = I n \x\ -fn Y

Note tha t th e derive d serie s 2

^x

— n 27r2

converges uniforml y o n eac h compac t interva l disjoin t fro m th e se t {kn : k G Z}. Henc e w e ge t cosx 1 ^ 2 x COt X = = - + > —z — z -z smx x ^ x — n 7T n=l

1 + niT J

—' \ x — nil+x

x*

n=l

It the n follow s tha t tanx = — cot(x — TT/2)

£

^ \x

1

- (2n - 1 )TT/ 2 x+

+ (2 n - l)?r/ 2

Finally, usin g th e identit y 1 1 / x x\ —= - co t - + ta n - , smx 2 V 2 2 / we obtai n 1 1 sinx x

^

/

1

i

\

+ y; ( -i)M—i_ + _ i _ ), n=l x

7

which implie s th e desire d equality .

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1.5. Imprope r Integral s

197

1.5.58. W e hav e 6 ] l>oo xa-l _ -^ xb-l

Jo 1

-dx z fl x a~l -x h~l , f°° x a~x - x h~l , = I dx - f / dx Jo l x J 1 -x x -

(!/x)a-l _ ( l / x ) ^ 1

rl x a-l _ xb-l rl

~ Jo l x f1 x a~x -x~ a , = / dx

Jo l

+

dx

7 o (l-(l/x))x* f 1 x b~l -x~ b , — / — dx

~x Jo

l

~x

= Ji-/ 2. As i n th e solutio n o f th e precedin g problem , w e ge t 1

W 1 hr = ~ + > — a ^-^

\a + n a

—n

= 7 r cot 7ra.

n—l

1.5.59. (a ) Th e functio n ln(l - x) n=l

can b e continuousl y extende d t o [0,1 ) an d th e powe r serie s converge s uniformly o n each interval [0 , 6], 0 < b < 1 . Moreover, i f S n(x) denote s the nt h partia l su m o f th e series , the n \Sn(x)\ = S n(x) 1 I

/ tan Jo J

a

rl

x d x = / y a(l-y2)-%-^dy= o^ 1„ 1 fa

2V

2 2

a 1

'2

2

t%~* Jo \7

(1 - £ ) ~ ^ dt

T

/ 2cos(7ra/2)

'

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203

1.5. Imprope r Integral s where th e las t equalit y follow s fro m 1 .5.64 . (b) Sinc e

r dx

- co s x V2Jo

Jo y/3

dx

r

i

H

+ sin 2 (x/2)

the substitutio n y — sin(x/2 ) yield s

f,

dx

= v~2 f «L= = ^ f ,-3/ 4(1 _ t)-^dt \J\-y A 4

Jo V 3 - co s x Jo

J0

-—5(1/4,1/2)-—

r(3/4

)

.

Now usin g th e identit y F(a)F(l — a) = -^—, 0 < a < 1 , we se e tha t T(l/2) = v^ F an d T(3/4 ) = y/2n/(T(l/A)), whic h give s th e desire d equality. (c) A s i n (a) , w e ge t / sin"" Jo 2

1

xdx=- t*' Jo

1 2

- ^ - ty

= l -B(a/2M2) =

2

xl2

dt

a 2

- B(a/2,a/2) 1

where th e las t equalit y follow s fro m th e duplicatio n formula . 1.5.70. Ther e ar e man y proof s o f thi s formula . Her e w e presen t a proof du e t o W . Felle r [Amer . Math . Monthly , 74(1 967) , 1 223-1 225] . Put (1) A

n

= I n n\-(n + 1 /2 ) I n n + n.

Our ai m i s to sho w tha t lim A

n

= In V2TT.

n—*oo

Let &k

1

rk

rk

r

r

-Ink— lnxdx= 1 1/2 Jk-

/2

hi(k/x)dx,

1 2 Jk- 1 /2 Jk- /2 1 rk+l/2

bk = / lnxdx—-lnk Jk 2

1 rk+ /2

= J

\n(x/k)dx. k

Then

«fc — 6jf e = 1 + lnfc + Tfe — - j I n (fe — -J — ( f c + 2 ) ^ ( f c + 2 Copyright 2003 American Mathematical Society. Duplication prohibited. Please report unauth Thank You!

204

Solutions. 1 : Th e Riemann-Stieltje s Integra l

Consequently, (2) ] T ( a , e - ^ ) = n + lnn ! + - m - - (n + - j I n (n + - j = B n. oo

Now w e show tha t th e serie s ] T (ak — b k) converges . Sinc e k=l 1 , , / r l / 2-

2

(3)

ak = I n— dt an d bk = / I n (1 + t/k) dt, — 1 Jo V& Jo we se e tha t ak > bk > ak+i > 0 . Moreover , li m ak = li m bk — 0. fc—>oo fc—+00

Thus th e serie s ai - fei + a 2 - b 2 H +

a n - b n + .. .

converges b y th e Leibni z test . Thi s mean s tha t li m B n exist s an d i s n—>oo

oo

a

equal t o th e su m o f the serie s Y2 ( k — bk)- I t follow s fro m (3 ) tha t fe=i

ak-bk =

1/2 2 - / f I ( n ( t1 - — \I dt.

Thus 1 oo . /

2o o,

1

2

\

5>-«~jL EH "*)* because th e serie s o n th e right-han d sid e converge s uniforml y o n [0,1/2]. So , "1/2 a oo

oo « ! /

! > - «=

/ oo ^

f2

2\

) dt

-/ ^ ( i - p

/• 1 / 2 , sinTr t 7 -- / I n dt, Jo Kt where th e las t equalit y follow s from , e.g. , I , 3.8.37 . Usin g th e resul t in 1 .5.1 (b) , w e ge t f1/2^ sinnt 1 l n

-J0 m - ^ * = - ( m . - i ) . By (1 ) an d (2) , l i m ( A n - B n ) = li m ( n+ i ) In ( 1 + -* - j + ]• In 2 = ^ ( 1 + l n 2 ) . n—>oo n—»o

o\ Z

/V

Zn

I I

Z

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1.5. Imprope r Integral s

205

Finally, w e arrive a t lim A n = li m B n + - ( 1 + I n 2) = I n y/2n. 1.5.71. W e first not e tha t T i s infinitely differentiabl e o n (0 , oo). In deed, th e imprope r integra l f Q x a~1 \i±xe~xdx converge s uniforml y on eac h interva l [c , d], 0 < c < d, becaus e eac h o f the integral s /x

a1

~ \nxe~xdx an

d/

x

1 a

~ \nxe~xdx

converges uniforml y o n [c,d]. So , applyin g 1 .5.44 , w e see tha t />oo

1 a

~ \nxe~xdx.

r'(a)= / x Jo The formul a T(n\a)= /

a x 1 - ]nnxe-xdx Jo /o .5.655 , can b e obtaine d inductively \. Now ,, bbyy 11 .5.6 r(o)r(6) T(b)b r( T(b)-B{a,b) = r(b) r(o + b) T(a + b)

o + 6)-r(a )

r ( 6 + l ) T(a + & ) - ! » T(a + b) Letting b —» 0 + w e ge t T'(a) T(a) 6^0

lim ( r ( 6 ) - S ( a , 6 ) ) . +

Since (se e th e solutio n o f 1 .5.65 )

f

X

.6-1

B(a,b) — I — (l +——rdx, x) we obtai n

(i) m

= , im r

X

6-i1 fe-

^

^

Moreover, th e imprope r integral s

iy'{'"-(n^h r^'( e--(iT^)

dx

converge uniformly o n [0 , bo]. Thu s the result i n 1 .5.4 3 can be applied , and w e ca n switc h th e limi t an d th e integra l i n (1 ) . Thi s complete s the proof .

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206

Solutions. 1 : Th e Riemann-Stieltje s Integra l

1.5.72. (a ) W e firs t sho w tha t th e functio n = ScTjx + 1 ,2) ' T{x + l)

V

is monotonicall y increasin g fo r x > 1 . T o this end , observ e tha t (ln [X))

*

+

2x

+ 1 )-

r(x+±) T(x

By th e precedin g problem ,

rr(x (x ++1) i) r'(r ' x(x++|)\)f r°°_J__i yr+ r(x + i) r( x + i) ~y 0 (i + t)*+i'

t -1

/ o ( 1 + 0-+ 1 v T + 7 + 1 JO

< ~ /

2j0 ( 1 "

7^

r-7T *

l + t)*+ i

2a; '

;

which shows that (lnF(x)) > 0 . Consequently , F increase s on (1 , oo). So li m F(x) = li m F(n) . T o find thi s limit , w e wil l us e th e dupli cc—>oo n—->o

o

cation formul a an d th e Stirlin g formul a give n i n 1 .5.6 8 an d 1 .5.70 , respectively. Henc e lim F(n) = li m ^ ( n + 1 /2 ) = ^ ^ oo r ( n + l ) n - . o o r ( n ) r ( n + l ) 2 mf(2n-1 ) ! 2 1 n i ^ o ( n _ i)! n !2 ^-

^

1 2n

-

,. Q ^ ( 2 n - l ) 2 - -1 / 2 e - ( 2 n -1 ) = h m Pn—7^ ; r = 1 , n-oo M v / 2 n ( 2 n - 2 ) 2 r i 1- e - ( 2 n - 2 ) where {f3 n} i s a sequenc e convergin g t o 1 . (b) Not e firs t tha t i f a = 1 , then b y 1 .5.65 ,

x"B(a,x)=xB(l,x) =

X 1

^™ 1 =

1 [x 4- 1) for al l x > 0 . So , i n thi s case , th e limi t t o b e foun d i s 1 . W e wil l show tha t th e functio n F a(x) — xaB(a,x),x > 0 , i s monotonicall y

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207

1.6. Integra l Inequalitie s

increasing i f a > 1 , an d monotonicall y decreasin g i f 0 < a < 1 . W e have linear;J - ^ +

r (

^^

+

a )

00 1 1 /

_ a Z*

\

dt

+ t)x ~ {i + ty+*j J'

~x~j0 \(i If a > 1 , then r°° / I

I

\

dt _ r°° I +1 — ( i +1) 1 -^ dt 1 /•°° , a < a / T ^ x— rrdt = - ,

where th e las t inequalit y follow s fro m th e inequalit y ( 1 + t) 1 a > 1 -j - ( 1 - a)t (se e II, 2.3.7(a)) . Thi s show s tha t (\nF a{x))' > 0 , whic h means tha t F a i s increasing . Similarly , i f 0 < a < 1 , the n th e in equality ( 1 4- * )1_ 0 < 1 + ( 1 - a)t (se e II , 2.3.7(b) ) implie s tha t F a is decreasing . Consequently , t o find th e desire d limi t i t i s enoug h t o calculate li m F a(n). W e ge t n—>oo a

lim n B(a,n)= h

m -^oo r ( a + n )

naI»(n-l)! ^oo (a + n - l)( a + n - 2 ) . . . (a + l)aT(a) na~ln\ , N — hm — — -— - = r(a) , n-oo ( a + n - l)( a + rc - 2 ) . . . (a + l) a where th e las t equalit y follow s fro m 1 .5.31 .

1.6. Integra l Inequalitie s 1.6.1. I t i s clear tha t

i [la

U{X)g{V)

~ !{V)9{X))2dX) dV ~ ^

Hence 2 f f\x)dx f J aJ

g a\

2

(x)dx - 2 ( I f(x)g(x)dx) >

0,

J aJ

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208

Solutions. 1 : Th e Riemann-Stieltje s Integra l

which give s the desire d inequality . Now , i f / an d g are continuou s o n [a, b] an d th e equalit y holds , the n / (f(x)g(y)

-

f(y)g(x))

2

dx = 0

Ja

for ever y y G [a, b]. Consequently, f(x)g(y) — f(y)g{x) = 0 fo r ever y x G [a, b] an d y G [a, 6]. So if there i s a yo G [a, 6] such tha t g(yo) ^ 0 , then f(x) — ( \g(x). I f g i s identicall y zer o o n [a , 6], the n w e ca n take A i = 0 and A 2 = 1 . 1.6.2. I t i s a n immediat e consequenc e o f the Schwar z inequality . 1.6.3. B y th e Schwar z inequality , 9

(b Moreover, sinc e

(f(x)-m)(f(x)-M) „ ^ - ^ ;;

r

V
0

J f (x)dx-(J f(x) Jo V o

On th e othe :rr hand hand , D(f, f) = (M i - F)(F -

mi

)-

[ (M i - f(x))(f(x) Jo

mi)dx ,

which implie s tha t D(f,f) -

feAj0 ~

IT

.

The infimu m i s equa l t o 4/-7 T and i s attaine d fo r f(x) = ^aXx

2

)'

1.6.7. Th e inequalit y follow s fro m

(M-f(x))(f(x)-m)dx>0. 1.6.8. Se t

F(t)=^f(x)dx)j -j\f(*)Y

dx, i e [ 0 , l ]

.

Then F\t) =

f(t)(2J*f(x)dx-(f(t))

2

y

and, i f G(t) = 2f*f(x)dx-(f(t)) 2, the n G'{t) = 2f(t)(l-f'(t)) > 0. Consequently, G(t) > G(0 ) = 0 , whic h give s F'(t) > 0 . So , F(t ) > 0 and, i n particular , F(l) > 0. Moreover, i f F ( l ) = 0 , the n F(t) = 0 fo r t e [0,1 ] , an d therefor e F'(t) = f(t)G(t) = 0 . This , in turn, implie s G'{t) = 2f{t)(l-f(t)) = Oand 1 - / ' ( * ) = 0 forte (0,1 ) . 1.6.9. I t follow s fro m th e solutio n o f 1 .2.2 2 that th e functio n 1 g{x) = f(t)dt x— a is monotonicall y increasin g o n (a , 6]. Thu s g(x) < g(b). A s i n th e solution o f 1 .2.22 , one ca n sho w tha t th e functio n rb

o-x J

x

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1.6. Integra l Inequalitie s

211

is monotonicall y increasin g o n [a , 6), and therefor e h(a) < h(x). 1.6.10. Assum e first tha t bot h th e function s ar e monoton e i n th e same sense . W e hav e

f n\f(x)g(x)

-

f(x)g(y))dx) dy

pb pb

= (b-a) /

pb

f(x)g(x)dx -

/ f{x)dx /

Ja Ja

g(x)dx

Ja

and

J (j

aU(y)g{y)-f{y)g(x))dx\dy pb pb

pb

= (b-a) f(x)g(x)dx-

f(x)dx

JaJ

g(x)dx.

aJ

a

Therefore pb pb

(b-a) /

pb

f{x)g(x)dx -

/ f(x)dx /

Ja Ja

= \f a ( /

g(x)dx

Ja 6

( / W - f{y)){g{x) - g{y))dx\ dy > 0,

because, b y assumption , (f(x) — f(y))(g(x) —

g{y)) > 0 for al l x an d

1.6.11. Th e proo f i s analogou s t o tha t o f 1 .6.1 0 . 1.6.12. B y th e Chebyshe v inequalit y (se e 1 .6.1 0) , pa -

/ f(a Jo a

i pa

pa

— x)g(x)dx < - f(a J0 7

— x)dx I g{x)dx o

= - / f(x)dx «7o Jo

/

g(x)dx

< / f(x)g(x)dx. Jo 1.6.13. I t i s enoug h t o appl y th e generalize d Chebyshe v inequalit y (see 1 .6.1 1 ) wit h p(x) = q 2{x), f(x) — x an d g{x) = l/(q(x)).

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Solutions. 1 : Th e Riemann-Stieltje s Integra l

212

1.6.14. B y th e convexit y o f / , w e get / f(x)dx =

(b — a) / ( (

Ja

l — x)a + xb)dx

JO

jr^'-^m

1.6.15. Le t f(x) = ^ , x > 0 . Sinc e f"{x) = (21 n x - 3)/x 3 , w e se e that / i s strictly conve x on (e 3 / 2 , oo) and strictl y concav e o n (0 , e3//2 ). Hence i f y > x > e 3 / 2 , w e hav e 1 Tlnt

AV

2 ;

y - x j

x

^ In

t

2

y- In 2 x In 2(2/-x) L

G '

which gives A L < G A. Analogou s reasonin g ca n b e use d t o sho w tha t AL>GA i f 0 < x < y < e 3 / 2 . 1.6.16. Le t c G [a , 6] b e suc h tha t f(c) = ma x / ( # ) . Then , b y x£[a,b]

assumption, nb

rC

rb

/ f(x)dx — \ f(x)dx + / f(x)dx Ja

Ja

Jc

= (c — a) / f((l — x)a + xc)dx Jo + (b-c) f((l-x)c Jo

+

xb)dx

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1.6. Integra l Inequalitie s

213

a) f

>{c-a) ((l-x)f(a)+xf(c))dx Jo + (b-c)J «l-x)f(c)

+

Jo

xf(b))dx

>l(b-a)f(c). 1.6.17. Integratin g b y parts yield s fag(x)f'(x)dx+ f Jo /O Jo

g'(x)f(x)dx ^0

= g(x)f(x)\ aQ- [

a g'(x)f(x)dx+ f Jo Jo

g'(x)f(x)dx

= f(a)g(a) + f g'(x)f(x)dx Ja

> f(a)g(a) + f g'{x)f{a)dx =

f(a)g(b).

Ja

The case o f equality follow s immediatel y fro m th e above . 1.6.18. Substitutin g u = f~ l(t) an fX Pf(x)

d integrating b y parts give s

pX

/ f(t)dt+ / f-\t)dt=

Jo Jo

Jo

rX

/

f(t)dt+ / uf'(u)du

Jo = xf(x).

1.6.19. Assum e firs t tha t f(a) yr\y) -(y-

{y) = &/ _1 (y) = ba.

l

b)r

1.6.20. Applyin g th e Youn g inequalit y wit h f(x) = ln( l + x), we get pa pb

/ ln( l + x)dx + / (e Jo Jo which give s th e desire d inequality .

x

- l)dx > ab,

1.6.21. I f there is xo such that g~ l{x$) > f(xo), the n by assumption , xog^ixo)^ /

/

f{x)dx+ Jo Jo

g(x)dx

r^o

< / g~ Jo Jo =x

1 0g"

pg~ l

(x0)

{x)dx + / g(x)dx

(x0),

where th e las t equalit y follow s fro m 1 .6.1 8 . A contradiction . 1.6.22. I t follow s fro m th e Schwar z inequalit y tha t i f / G A, the n (2 + 36) 2 = (f f(x){x

+ b)dxj
ma x —i2 , — = 12. bem 3b + 3 6 + 1

The equalit y i s attaine d a t f(x) = 6x

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1.6. Integra l Inequalitie s

215

1.6.23. Le t / G A. The n b y th e Schwar z inequality , (1) (J

( 1 - x)f'(x)dx)
- 1,

XX

which implie s tha t f{x)>l —

x. Likewise .

f(x) - /(2 ) /(x x-2 x-

)- 1 2

r(^2) < i,

which give s f(x) > x — 1 . Consequently , /(x ) > | x — 1 | . Thu s / f{x)dx> 7o Jo

[

\x-l\dx =

1.

Since / i s continuous, th e equalit y hold s i f and onl y i f f(x) = \x — 1 | , x G [0,2] . Bu t thi s functio n i s no t differen t iable a t x — 1 . S o th e answer t o ou r questio n i s no. 1.6.25. B y th e mea n valu e theorem , f(x) = f'(0 1 )(x-a) an

d f(x)

= f'(9

2)(x-b).

If M = ma x |/'(:r ) |, the n xG[a,6]

|/0r)| < M ( a ; - a ) an

d \f(x)\ ° > t h en I baty f^ dz = J a + f{ z)dz < J a f(x)dx. I f y < 0 , analogou s reasonin g ca n b e ap plied. Assum e now that / i s an arbitrary continuou s function , an d se t I rX-\-h A 0*0 2h Jx-h 1 /(01 ^ - I t follow s fro m wha t w e hav e prove d tha t 7 Jrab fh{x)dx < J a \f(x)\dx. Moreover , w e hav e -i px+h

- 1 px+h

\h(x)\ = r r / f(t)dt\