REAL ANALYSIS STUDY MATERIAL. B.Sc. Mathematics VI SEMESTER CORE COURSE. (2011 Admission) UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION

REAL ANALYSIS STUDY MATERIAL BSc Mthemtics VI SEMESTER CORE COURSE ( Admissio) UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION THENJIPALAM, CALICUT UNIVERSITY PO, MALAPPURAM, KERALA - 67 65 57

UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION Study Mteril BScMthemtics VI SEMESTER CORE COURSE REAL ANALYSIS Prepred y : SriNdkumr, Assistt Professor, NAM College, Kllikkdi, Kur Scrutiised y : Dr Ailkumr V Reder, Deprtmet of Mthemtics, Uiversity of Clicut Type settigs d Ly out : Computer Sectio, SDE Reserved Rel Alysis Pge

Cotets CONTINUOUS FUNCTIONS ON INTERVALS 5 BOLZANO S INTERMEDIATE VALUE THEOREM 8 UNIFORM CONTINUITY 7 4 THE RIEMANN INTEGRAL - PART I 4 5 THE RIEMANN INTEGRAL - PART II 5 6 FUNDAMENTAL THEOREMS OF CALCULUS 6 7 POINTWISE AND UNIFORM CONVERGENCE 68 8 UNIFORM CONVERGENCE AND CONTINUITY 79 9 SERIES OF FUNCTIONS 8 IMPROPER INTEGRALS OF FIRST KIND - PART I 89 IMPROPER INTEGRALS OF FIRST KIND - PART II IMPROPER INTEGRALS OF SECOND AND THIRD KINDS PART I 8 IMPROPER INTEGRALS OF SECOND AND THIRD KINDS PART II 8 4 BETA AND GAMMA FUNCTIONS 4 Rel Alysis Pge

Rel Alysis Pge 4

CONTINUOUS FUNCTIONS ON INTERVALS Defiitios A fuctio f : A is sid to e ouded o A if there eists costt M such tht f () M for ll A Tht is, fuctio f : A is ouded o the set A if its rge f () A is ouded set i f : A is uouded o A, if f is ot ouded o A ie, f is uouded o A if give y M, there eists poit A M f () M M such tht Emple We ow give emple of cotiuous fuctio tht is ot ouded Cosider the fuctio f defied o the itervl A (,) y f () is ot ouded o A ecuse for y M we c tke the poit M i A to get M f () M M M However, eig the quotiet two cotiuous fuctios d, f is M cotiuous o A We ow review some sic defiitios d theorems Cluster Poit: Let A A poit c is cluster poit of A if for every there eists t lest oe poit A, c such tht c ie, A poit c is cluster poit of A if for every -eighorhood V ()( c c,) c of c cotis t lest oe poit of Adistict from c The cluster poit c my or my ote e memer of A, ut eve if it is i A, it is igored whe decidig whether it is cluster poit of A or ot, sice we eplicitly require tht there e poits i V () c A distict from c i order for c to e cluster poit of A Emple We ow show tht A {,, }, the set cosistig of three elemets, d, hs o cluster poit The poit is ot cluster poit of A, sice choosig gives eighorhood of tht cotis o poits of A distict from The sme is true for the poit d Also, rel umer other th, d cot e cluster poit For, if c with c, c, d c, the choose mi c, c, c The V () c A Hece A hs o cluster poit Theorem A umer c is cluster poit of suset A of if d oly if there eists sequece () i A such tht lim() c d c for ll Rel Alysis Pge 5

Emple Let A e the ope itervl A (, ) All the poits of A re cluster poits of A Also the poits, re cluster poits of A, though they do ot elog to A the closed itervl [, ] is cluster poit of A Hece every poit of Emple We ow show tht A fiite set hs o cluster poit Let F e fiite set No umer c is cluster poit of F s it is ot possile to fid sequece () i F such tht lim() c d c for ll Emple The ifiite set hs o cluster poit No umer c is cluster poit of s it is ot possile to fid sequece () Emple The set B : i such tht lim() hs oly the poit s cluster poit c d c for ll Emple If I [, ], the the set C I cosists of ll the rtiol umers i I It follows from Desity Theorem ( If d y re rel umers with y, the there eists rtiol umer r such tht r y ) tht every poit i I is cluster poit of C Defiitio Let A, d let c e cluster poit of A For fuctio f : A, rel umer L is sid to e limit of f t c if, for every umer, there eists correspodig umer such tht if A d c, the () f L I tht we write lim() f L c Sice the vlue of usully depeds o, it my e deoted y () emphsize this depedece The iequlity c is equivlet to syig c isted of to Theorem If f : A d if c is cluster poit of A, the f c hve t most oe limit t c Theorem Let f : A d let c e cluster poit of A The the followig sttemets re equivlet (i) lim() f L c (ii) Give y -eighorhood V () L of L, there eists -eighorhood V () c if c is y poit i V () c A, the f () elogs to V () L Importt Limits of Fuctios () lim k k () lim c c c of c such tht () lim c (4) lim if c c c c The followig importt formultio of limit of fuctio is i terms of limits of sequeces Sequetil Criterio for Limits: Let f : A d let c e cluster poit of A The the followig sttemets re equivlet (i) lim() f L c (ii) For every sequece () i A tht coverges to c such tht c for ll, the sequece (()) f coverges to f () c Divergece Criterio: Let A, let f : A d let c e cluster poit of A Rel Alysis Pge 6

() If L, the f does ot hve limit L t c if d oly if there eists sequece () i A with c for ll such tht the sequece () coverges to c ut the sequece (()) f does ot coverge to f () c () The fuctio f does ot hve limit t c if d oly if there eists sequece () i A with c for ll such tht the sequece () coverges to c ut the sequece (()) f does ot coverge i Remrk The ssertio () i theorem eles us to show tht certi umer L is ot the limit of fuctio t poit, while () sys tht the fuctio does ot hve limit t poit We list some emples, which works o the sis of Divergece Criteri The detils re voided s it is ot metioed i the syllus Emple If f ()( ), the lim() f lim doesot eist i The sigum fuctio sg is defied y The limsg() limsi(/) for, sg() for, for doesot eist We lso ote tht sg() doesot eist i for Properties of Limits The followig rules hold if lim() c f L d lim() c g M (L d M rel umers) Sum Rule: lim[()()] f g L M c ie, the limit of the sum of two fuctios is the sum of their limits Differece Rule: lim[()()] f g L M c ie, the limit of the differece of two fuctios is the differece of their limits Product Rule: lim()() c f g L M ie, the limit of the product of two fuctios is the product of their limits 4 Costt Multiple Rule: lim() kf kl (y umer k) c ie, the limit of costt times fuctio is tht costt times the limit of the fuctio () 5 Quotiet Rule: lim f L, M c g () M ie, the limit of the quotiet of two fuctios is the quotiet of their limits, provided the limit of the deomitor is ot zero 6 Power Rule: If m d re itegers, the lim[()] f c m m L, provided m L is rel umer ie, the limit of y rtiol power of fuctio is tht power of the limit of the fuctio, provided the ltter is rel umer Rel Alysis Pge 7

Defiitio Let A, let f : A, d let c A We sy tht f is cotiuous t c if, for every umer there eists such tht if is y poit of A stisfyig c, the ()() f f c Defiitio If f fils to e cotiuous t c, the f is discotiuous t c Similr to the defiitio of limit, the defiitio of cotiuity t poit c e formulted very icely i terms of eighorhoods Theorem A fuctio f : A is cotiuous t poit c A if d oly if give y - eighorhood V (()) f c of f () c there eists -eighorhood V () c poit of A V (), c the f () elogs to V (()), f c tht is f (())(()) A V c V f c of c such tht if is y If c A is cluster poit of A, the compriso of Defiitios d show tht f is cotiuous t c if d oly if f () c lim() f c () Thus, if c is cluster poit of A, the three coditios must hold for f to e cotiuous t c : (i) f must e defied t c (so tht f () c mkes sese), (ii) the limit of f t c must eist i (so tht lim() f (iii) f () c lim() f c c mkes sese), d If c A is ot cluster poit of A, the there eists eighorhood V () c of c such tht AV () c { } c Thus we coclude tht f is cotiuous t poit c A tht is ot cluster poit of A Such poits re ofte clled isolted poits of A They re of little prcticl iterest to us, sice they hve o reltio to limitig process Sice cotiuity is utomtic for such poits, we geerlly test for cotiuity oly t cluster poits Thus we regrd coditio () s eig chrcteristic for cotiuity t c Similr to Sequetil Criterio for Limits (Theorem 4), we hve the Sequetil Criterio for Cotiuity Sequetil Criterio for Cotiuity: A fuctio f : A is cotiuous t the poit c A if d oly if for every sequece () f () c i A tht coverges to c, the sequece (()) f coverges to Discotiuity Criterio: Let A, let f : A, d let c A The f is discotiuous t c if d oly if there eists sequece () i A such tht () (()) f does ot coverge to f () c Emple f () is ot cotiuous t The sigum fuctio sg is ot cotiuous t coverges to c, ut the sequece Defiitio Let A, d let f : A If B is suset of A, we sy tht f o the set B if f is cotiuous t every poit of B is cotiuous Emple The followig fuctios re cotiuous o Rel Alysis Pge 8

The costt fuctio f () The idetity fuctio g() h() eist Emple The fuctio f () Emple f () is cotiuous o A { : } Emple The fuctio f () si(/) for is ot cotiuous t s limsi(/) si(/) for is ot cotiuous t s f () Emple Give the grph of f(), show elow, determie if f() is cotiuous t,, does ot is ot defied t Solutio To swer the questio for ech poit we ll eed to get oth the limit t tht poit d the fuctio vlue t tht poit If they re equl the fuctio is cotiuous t tht poit d if they re t equl the fuctio is t cotiuous t tht poit First f ( ) The fuctio vlue d the limit re t the sme d so the fuctio is ot cotiuous t this poit This kid of discotiuity i grph is clled jump discotiuity Jump discotiuities occur wheree the grph hs rek i it s this grph does Now The fuctio is cotiuous t this poit sice the fuctio d limit hve the sme vlue Filly d lim() f f () d f () d does t eist lim() f so tht lim() f () f lim() f Rel Alysis Pge 9

The fuctio is ot cotiuous t this poit This kid of discotiuity is clled removle discotiuity Removle discotiuities re those where there is hole i the grph s there is i this cse Emple The fuctio si(/) for, F() for, is cotiuous t Comitios of Cotiuous Fuctios: Let A d let f d g e fuctios tht re defied o A to d let c A costt d f d g re cotiuous t c, the (i) f g, f g, fg, d kf re cotiuous t c (ii) f / g is cotiuous t c, provided g() for ll A If k is The et result is immedite cosequece of Theorem, pplied to every poit of A Theorem Let A d let f d g e cotiuous o A to If k is costt d f d g re cotiuous o A, the (i) f g, f g, fg, d kf re cotiuous o A (ii) f / g is cotiuous o A, provided g() for ll A Emples A polyomil fuctio is cotiuous o Rtiol fuctio r defied y r() p(), q() s the quotiet of two polyomils p() d q(), is cotiuous t every rel umer for which it is defied The trigoometric fuctios sie d cosie fuctios re cotiuous o The trigoometric fuctios t, cot, sec, d csc re cotiuous where they re defied For istce, the tget fuctio is defied y si t cos provided cos (tht is, provided ( ) /,) Sice sie d cosie fuctios re cotiuous o, it follows tht the fuctio t is cotiuous o its domi Theorem Let A, let f : A, d let f e defied y f ()() f for A () If f is cotiuous t poit c A, the f is cotiuous t c () If f is cotiuous o A, the f is cotiuous o A Emple Let g() for The g is cotiuous o Cotiuity of Composites: If f is cotiuous t c, d g is cotiuous t f () c, the g f is cotiuous t c Rel Alysis Pge

Emple Let g() si for d f () for The, eig the compositio of the cotiuous fuctios g d f, ()() g f si(/) is cotiuous t every poit c Theorem If X () is coverget sequece d if for ll, the lim() Sequetil Criterio for Cotiuity A fuctio f : A is cotiuous t the poit c A if d oly if for every sequece () f () c Hece if f is cotiuous t, the lim lim()() f f Equivletly, Squeeze Theorem: Suppose tht such tht d tht lim() i A tht coverges to c, the sequece (()) f lim lim()(lim) f f X (), Y () y, d Z () z y z for ll, lim() z The Y () y is coverget d lim() lim() ylim() z coverges to re sequeces of rel umers Nested Itervls Property: If I [, ],, is ested sequece of closed d ouded itervls, the there eists umer such tht I for ll Chrcteriztio Theorem for Itervls: If S is suset of tht cotis t lest two poits d hs the property if, y S d y the [, y] S, () the S is itervl Discotiuity Criterio: Let A, let f : A, d let c A d oly if there eists sequece () i A such tht () The f is discotiuous t c if coverges to c, ut the sequece (()) f does ot coverge to f () c The Supremum Property (or Completeess Property) : Every oempty set of rel umers tht hs upper oud hs supremum i The logous property of ifim c e deduced from the Supremum Property The Ifimum Property of : Every oempty set of rel umers tht hs lower oud hs ifimum i Theorem (Boudedess Theorem) Let I [, ] e closed ouded itervl d let f : I e cotiuous o I The f is ouded o I To prove this suppose tht f is ot ouded o I The, for y there is umer such tht f () Sice I is ouded, it follows tht the sequece X () I is ouded Beig ouded sequece of rel umers, y the Bolzo-Weierstrss Theorem, X hs susequece X () r tht coverges to umer, sy, Sice I is closed d sice the elemets of the susequece X elogs to I, it follows, from Theorem, tht I Rel Alysis Pge

As f is cotiuous o I, I implies f is cotiuous t Now f is cotiuous t, together with () r coverges to f () coverges to implies, y Sequetil Criterio for Cotiuity, tht f Sice coverget sequece of rel umers is ouded, the coverget sequece must e ouded sequece But this is cotrdictio sice f () r r r for r Therefore the suppositio tht the cotiuous fuctio f is ot ouded o the closed ouded itervl I leds to cotrdictio Hece f is ouded This completes the proof The coclusio of the Boudedess Theorem fils if y oe of the hypotheses is reled The followig emples illustrte this Emple The fuctio f () for i the uouded, closed itervl A [,) is cotiuous ut ot ouded o A Emple The fuctio g() () r f () r for i the hlf-ope itervl B (,] is cotiuous ut ot ouded o B Emple The fuctio h defied o the closed itervl C [, ] y whe h() whe is discotiuous d uouded o C Emple Let I [, ] d sice f : I e cotiuous fuctio such tht f () I We prove tht there eists umer such tht f () for ll I for ech To prove this, defie g : I y g() f () for I Sice f () for ll I, d sice f is cotiuous o I, it follows tht g is cotiuous o the closed ouded itervl I Hece, y Boudedess Theorem, g is ouded o I there eists such tht As f () g() for I, the ove implies g() f () for I Hece Hece f () for I Asolute Mimum d Asolute Miimum Defiitio Let A d let f : A We sy tht f hs solute mimum o A if there is poit * A such tht Rel Alysis Pge

* f ()() f for ll A We sy tht f hs solute miimum o A if there is poit * A such tht f ()() f for ll A * We sy tht is solute mimum poit for f o A, d tht * is solute miimum poit for f o A, if they eist Emple We ow give emple to show tht cotiuous fuctio o set A does ot ecessrily hve solute mimum or solute miimum o the set The fuctio f () hs either solute mimum or solute miimum o the set A (,) (Fig ) There c e o solute mimum for f o A sice f is ot ouded ove o A f c hve o solute miimum s there is o poit t which f ttis the vlue if{() f : } A The sme fuctio hs either solute mimum or solute miimum whe it is restricted to the set (, ), while it hs oth solute mimum d solute miimum whe it is restricted to the set [, ] I dditio, f () hs solute mimum ut o solute miimum whe restricted to the set[,), ut o solute mimum d o solute miimum whe restricted to the set (,) Emple We ow give emple to show tht if fuctio hs solute mimum poit, the this poit is ot ecessrily uiquely determied Cosider the fuctio g() defied for A [, ] hs two poits givig the solute mimum o A, d the sigle poit yieldig its solute miimum o A Emple The costt fuctio h() for is such tht every poit of is oth solute mimum d solute miimum poit for h Theorem (Mimum-Miimum Theorem) Let I [, ] e closed ouded itervl d let f : I e cotiuous o I The f hs solute mimum d solute miimum o I To prove the result, let f () I {() f: } I ie, f () I is the oempty set of vlues of f o I Sice f is cotiuous d I is closed d ouded, usig Boudedess Theorem, we hve f () I is ouded suset of The f () I is ouded ove d ouded elow Hece y the completeess property of, supremum d ifimum of the set eist Let s * We clim tht there eist poits sup() f I d s* if() f I * d * i I such tht s f () d s * f () * * * * We oly estlish the eistece of the poit, s the proof of the eistece of * is similr Sice s * sup() f I, if, the the umer s is ot upper oud of the set f () I Cosequetly there eists umer I such tht * Rel Alysis Pge

* * s f () s for ll () By this wy we oti sequece X () with memers i I Also, sice I is ouded, the sequece X () is ouded Therefore, y the Bolzo-Weierstrss Theorem, there is susequece to I [, ], we hve Tht is, X () r of X tht coverges to some umer * Sice the elemets of X elog r d hece it follows from Theorem A tht lim r * d hece * I Therefore f is cotiuous t follows tht lim()() f r * f Sice it follows from () tht * * s f () r s for ll r, we coclude from the Squeeze Theorem tht Therefore we hve We coclude tht r lim(()) f r * * r * s f () lim() f sup() s f I * d, y Sequetil Criterio for Cotiuity, it * is solute mimum poit of f o I This completes the proof * * Remrk With the ssumptios of the theorem, s f ()() f I d s * f ()() * f I Theorem (Loctio of Roots Theorem) Let I [, ] d let f : I e cotiuous o I If f () () f, or if f () () f, the there eists umer c (,) such tht f () c To prove the result, we ssume tht f () () f We will geerte sequece of itervls y successive isectios I [, ], Let where,, d let p e the midpoit p () If f () p, we tke c p d we re doe If f () p, the either f () p or f () p If f () p, the we set, p, while if f () p, the we set p, I either cse, we let I [, ] ; the we hve I I d f (),() f Rel Alysis Pge 4

We cotiue the isectio process Suppose tht the itervls I, I,, Ik hve ee otied y successive isectio i the sme mer The we hve f () k d f () k, d we set k k pk If f () pk, we tke c pk d we re doe If f () pk, we set, p, k k k k while if f () pk, we set p, I either cse, we let k k k k I [, ]; k k k the I k I d k f () k,() f k If the process termites y loctig poit p such tht f () p, the we re doe If the process does ot termite, the we oti ested sequece of closed ouded itervls I [, ] such tht for every we hve f () d f () Furthermore, sice the itervls re otied y repeted isectios, the legth of elogs to I is equl to () / It follows from the Nested Itervls Property tht there eists poit c tht I for ll Now c I for ll implies c for ll, d hece we hve d () c, () / c Hece, it follows, y Squeeze Theorem, tht lim()()lim c d lim()()lim c Tht is, c lim() d lim() c Hece lim() clim() Sice f is cotiuous t c, the fct tht () for Cotiuity) tht f lim() f c f lim()() c The fct tht f () for ll implies tht f () c lim() f d () c implies (y Sequetil Criterio Rel Alysis Pge 5

Also, the fct tht f () for ll, f () c lim() f Thus, we coclude tht f () c implies tht Cosequetly, c is root of f This completes the proof Remrk The ove theorem c e restted s follows: Let I [, ] d let f : I e cotiuous o I If f ()() f there eists umer c (,) such tht f () c, the RemrkThe ove result is clled loctio of roots theorem s it is the theoreticl sis for loctig roots of cotiuous fuctio y mes of sig chges of the fuctio The proof of the theorem provides lgorithm, kow s the Bisectio Method, for fidig solutios of equtios of the form f (), where f is cotiuous fuctio (This method is discussed i detils i the tet Numericl Alysis Sith semester core) Now usig loctio of roots theorem, we emie whether there is rel umer tht is oe less th its fifth power? The umer we seek must stisfy the equtio 5 5 ie, I other words we re lookig for zero of the fuctio By tril, we hve f 5 () f () d f () Thus f () d f () Hece, y loctio of roots theorem, there eists umer c (,) ie, c such tht f c c c 5 () 5 Hece c c Therefore c is the required umer which is oe less th its fifth power Assigmets Show tht every polyomil of odd degree with rel coefficiets hs t lest oe rel root Let I [, ] d let f : I e cotiuous fuctio o I such tht for ech i I there eists y i I such tht f ()() y f Prove tht there eists poit c i I such tht f () c Show tht the polyomil these roots to withi two deciml plces 4 Show tht the polyomil p() hs t lest two rel roots Use clcultor to locte 4 p() 7 9 hs t lest two rel roots Use clcultor to locte these roots to withi two deciml plces 5 Let f e cotiuous o the itervl [,] to d such tht f ()() f poit c i [, ] such tht f () c f c Prove tht there eists [Hit: Cosider g()()() f f ] Coclude tht there re, t y time, tipodl poits o the erth s equtor tht hve the sme temperture Rel Alysis Pge 6

6 Show tht the fuctio f () hs root i the itervl[, ] Use the Bisectio Method d clcultor to fid the root with error less th 7 Show tht the equtio cos hs solutio i the itervl [, ] Use the Bisectio Method d clcultor to fid pproimte solutio of this equtio, with error less th 8 The fuctio f ()( )( )( )( 4)( 5) hs five roots i the itervl [, 7] If the Bisectio Method is pplied o this itervl, which of the roots is locted? 9 The fuctio for g()( )( )( 4)( 5)( 6) hs five roots o the itervl [, 7] If the Bisectio Method is pplied o this itervl, which of the roots is locted? If the Bisectio Method is used o itervl of legth to fid p with error determie the lest vlue of tht will ssure this ccurcy p c 5, Rel Alysis Pge 7

BOLZANO S INTERMEDIATE VALUE THEOREM The et result is geerliztio of the Loctio of Roots Theorem It ssures us tht cotiuous fuctio o itervl tkes o (t lest oce) y umer tht lies etwee two of its vlues Bolzo s Itermedite Vlue Theorem: Let I e itervl d let f : I e cotiuous o I If, I d if k stisfies f ()() k f, the there eists poit c I etwee d such tht f () c k To prove the result, suppose tht d let g()() f k The f ()() k f implies g() () g By the Loctio of Roots Theorem there eists poit c with c such tht ()g c ie, such tht () f c Therefore f () c k k If, let h()() k f so tht h() () h Therefore there eists poit c with c such tht ()()h c k f c, d hece f () c k This completes the proof Corollry Let I [, ] e closed, ouded itervl d let f : I e cotiuous o I If k is y umer stisfyig if() f I sup() k f I the there eists umer c I such tht f () c k It follows from the Mimum-Miimum Theorem tht there re poits c * d c * if()() f I Hece y the ssumptio if()()() f c * d f c * () sup() * f I sup() f c* k f c f I f I i I such tht Hece y Bolzo s Itermedite Vlue Theorem, there eists poit c I etwee c * d * c such tht f () c k This completes the proof Theorem Let I e closed ouded itervl d let f : I e cotiuous o I The the set f () I {() f: } I is closed ouded itervl To prove this, we let m if() f I d M sup() f I m d M elog to f () I Moreover, we hve f () I [ m, M] (), the y the Mimum Miimum Theorem, If k is y elemet of [ m, M ], the it follows from Corollry to Bolzo s Itermedite Vlue Theorem tht there eists poit c I such tht k f () c Hece, k f () I [ m, M ]() f I (4) d we coclude tht From () d (4), we hve f () I [ m, M] Sice [ m, M ] is closed d ouded itervl, f () I is lso closed d ouded itervl (Also, we ote tht the edpoits of the imge itervl re m d M, resp, the solute miimum d solute mimum vlues of the fuctio) This completes the proof Rel Alysis Pge 8

The previous theorem sttes tht the imge of closed ouded itervl uder cotiuous fuctio is lso closed ouded itervl The edpoits of the imge itervl re the solute miimum d solute mimum vlues of the fuctio, d ll vlues etwee the solute miimum d the solute mimum vlues elog to the imge of the fuctio If I [, ] is itervl d f I is cotiuous o I, we hve proved tht f () I is the itervl[ m, M ] We hve ot proved (d it is o t lwys true) tht f () I [(),()] f f is the itervl The precedig theorem is preservtio theorem i the sese tht it sttes tht the cotiuous imge of closed ouded itervl is set of the sme type The et theorem eteds this result to geerl itervls However, it should e oted tht lthough the cotiuous imge of itervl is show to e itervl, it is ot true tht the imge itervl ecessrily hs the sme form s the domi itervl Emple We ow give emples to show tht the cotiuous imge of ope itervl eed ot e ope itervl, d the cotiuous imge of uouded closed itervl eed ot e closed itervl If f () for, the f is cotiuous o It is esy to see tht if (,) f () I,, which is ot ope itervl Also, if which is ot closed itervl (Fig 4) I, the I is the closed itervl give y I, the (),, f I, Preservtio of Itervls Theorem Let I e itervl d let f : I e cotiuous o I The the set f () I is itervl To prove the result, let,() f I with The there eist poits, I such tht f () d f () Further, it follows from Bolzo s Itermedite Vlue Theorem tht if k (,) the there eists umer c I with k f ()() c f I Hece,[, ]() f I, showig tht f () I possesses property () of Chrcteriztio Theorem for Itervls Therefore f () I is itervl d this completes the proof Emple Let I [, ] d let f : I d g : I e cotiuous fuctios o I We show tht the set E { I :()()} f g is with the property tht if () E d, the E To prove this, we ote tht sice f d g re cotiuous, f g : I is lso cotiuous Also, sice E { I :()() f g}, we hve E f g As {} is closed set d f () {} g is cotiuous, () f {} g is closed suset of I ie, E is closed suset of I Hece if we tke sequece () i E with limit, the its limit must elogs to the set E Solved Emples Verify the followig two importt limits: Rel Alysis Pge 9

() lim () lim k k (k costt) Aswers ) Let e give We must fid implies The implictio will hold if = lim ) Let e give We must fid such tht for ll, implies k k Sice k k, we c use y positive umer for d the implictio will hold This proves tht lim k k such tht for ll or y smller positive umer This proves tht Emple Evlute the followig limit Sice we kow tht epoetils re cotiuous everywhere we c use the fct ove Emple For the limit lim 5, fid tht works for Aswers We hve to fid such tht for ll 5 Step : We solve the iequlity to fid itervl out 5 o which the iequlity holds for ll 9 The iequlity holds for ll i the ope itervl itervl s well (, ), so it holds for ll 5 i this Step : We fid vlue of tht plces the cetered itervl 5 5 iside the itervl (, ) The distce from 5 to the erer edpoit of (, ) is If we tke or y smller positive umer, the the iequlity 5 will utomticlly plce etwee d to mke 5 Emple Prove tht lim() 4f f if Rel Alysis Pge

, f (), Aswers Our tsk is to show tht give there eists such tht for ll () f 4 Step : We solve the iequlity () f 4 to fid o ope itervl out o which the iequlity holds for ll For, we hve f () 4 4 4 4 4 4 here we ssume tht 4 4 4 this is ope itervl out tht solves the The iequlity f () 4 holds for ll i the ope itervl 4, 4 Step : We fid vlue of tht plces the cetered itervl (, ) iside the itervl ( 4, 4) : Tke to e the distce from to the erer edpoit of 4, 4 I other words, tke mi 4, 4,, d the iequlity to solve is 4 : 4 iequlity i the ope itervl to the erer edpoit of the miimum (the smller) of the two umers 4 d 4 If hs this or y smller positive vlue, the iequlity will utomticlly plce etwee 4 d 4 to mke () f 4 For ll, This completes the proof () f 4 Emple Show tht hs root somewhere i the itervl [-,] Wht we re relly skig here is whether or ot the fuctio will tke o the vlue somewhere etwee - d I other words, we wt to show tht there is umer c such tht d However if we defie d ckowledge tht d we c see tht these two coditio o cre ectly the coclusios of the Itermedite Vlue Theorem So, this prolem is set up to use the Itermedite Vlue Theorem d i fct, ll we eed to do is to show tht the fuctio is cotiuous d tht is etwee d (ie or d we ll e doe Rel Alysis Pge

To do this ll we eed to do is compute, So we hve, Therefore is etweee d d sice is polyomil it s cotiuous everywhere d so i prticulr it s cotiuous o the itervl [-,] So y the Itermedite Vlue Theorem there must e umer so tht, Therefore the polyomil does hve root etwee - d Emple If possile, determie if itervl [,5] f () si( )cos tkes the followig vlues i the () Does? () Does? If possile, suppose the fuctio tkes o either of the two vlues ove i the itervl [,5] First, let s otice tht this is cotiuous fuctio d so we kow tht we c use the Itermedite Vlue Theorem to do this prolem Now, for ech prt we will let M e the give vlue for tht prt d the we ll eed to show tht M lives etwee d If it does the we c use the Itermedite Vlue Theorem to prove tht the fuctio will tke the give vlue So, sice we ll eed the two fuctio evlutios for ech prt let s give them here, Now, let s tke look t ech prt () I this cse we ll defie d we c see tht, So, y the Itermedite Vlue Theorem there must e umer such tht () I this prt we ll defie e We ow hve prolem I this prt M does ot live etwee d So, wht does this me for us? Does this me tht i [,5]? Rel Alysis Pge

Ufortutely for us, this does t me ythig It is possile tht i [,5], ut is it lso possile tht i [,5] The Itermedite Vlue Theorem will oly tell us tht c s will eist The theorem will NOT tell us thtc s do t eist I this cse it is ot possile to determie if Theorem i [,5] usig the Itermedite Vlue Theory: Let s tke look t the followig grph d let s lso ssume tht the limit does eist Wht the defiitio is tellig us is tht for y umer tht we pick we c go to our grph d sketch two horizotl lies t d s show o the grph ove The somewhere out there i the world is other umer, which we will eed to determie, tht will llow us to dd i two verticl lies to our grph t d Now, if we tke y i the regio, ie etwee d, the this will e closer to th either of d Or, If we ow idetify the poit o the grph tht our choice of gives the this poit o the grph will lie i the itersectio of the regios This mes tht this fuctio vlue f() will e closer to L th either of d Or, So, if we tke y vlue of i the regio the the grph for those vlues of will lie i the regio Rel Alysis Pge

Notice tht there re ctully ifiite umer of possile δ ss tht we c choose I fct, if we go ck d look t the grph ove it looks like we could hve tke slightly lrger δ d still gotte the grph from tht regio to e completely cotied i the regio Also, otice tht s the defiitio poits out we oly eed to mke sure tht the fuctio is defied i some itervl roud ut we do t relly cre if it is defied t Rememer tht limits do ot cre wht is hppeig t the poit, they oly cre wht is hppeig roud the poit i questio Oky, ow tht we ve uderstd it let s see how gotte the defiitio out of the wy d mde ttempt to it s ctully used i prctice These re little tricky sometimes d it c tke lot of prctice to get good t these so do t feel too d if you do t pick up o this stuff right wy We re goig to e lookig t couple of emples tht work out firly esily Emple Use the defiitio of the limit to prove the followig limit I this cse oth L d re zero So, let e y umer Do t worry out wht the umer is, is just some ritrry umer Now ccordig to the defiitio of the limit, if this limit is to e true we will eed to fid some other umer so tht the followig will e true Or upo simplifyig thigs we eed, Ofte the wy to go through these is to strt with the left iequlity d do little simplifictio d see if tht suggests choice for We ll strt y rigig the epoet out of the solute vlue rs d the tkig the squre root of oth sides Now, the results of this simplifictio looks wful lot like with the eceptio of the prt Missig tht however is t prolem, it is just tellig us tht we c t tke So, it looks like if we choose we should get wht we wt We ll et eed to verify tht our choice of will give us wht we wt, ie, Rel Alysis Pge 4

Verifictio is i fct pretty much the sme work tht we did to get our guess First, let s gi let e y umer d the choose Now, ssume tht choosig to stisfy this we will get, We eed to show tht y To strt the verifictio process we ll strt with d the first strip out the epoet from the solute vlues Oce this is doe we ll use our ssumptio o, mely tht Doig ll this gives, Or, upo tkig the middle terms out, if we ssume tht the we will get, d this is ectly wht we eeded to show So, just wht hve we doe? hve, We ve show tht if we choose the we c fid so tht we d ccordig to our defiitio this mes tht, Assigmets Let I [, ], let f : I e cotiuous o I, d ssume tht f (),() f Let W { I :() f }, d let w supw Prove tht f () w of Loctio of Roots Theorem) Let I [, / ] d let f : I e defied y (This provides ltertive proof f () sup{,cos } for I Show there eists solute miimum poit I for f o I Show tht is solutio to the equtio cos Suppose tht f : e cotiuous o d tht lim f d lim f Prove tht f is ouded o d ttis either mimum or miimum o Give emple to show tht oth 4 Let f : e cotiuous o d let Show tht if is such tht f (), the there eists -eighorhood U of such tht f () for ll U Rel Alysis Pge 5

5 Emie which ope [respectively, closed] itervls re mpped y f () for oto ope [respectively, closed] itervls 6 Emie the mppig of ope [respectively, closed] itervls uder the fuctios g () /( ) d h() for 7 Give emple of ope itervl tht is ot mpped y f () for oto ope itervl 8 If f [,] is cotiuous d hs oly rtiol [respectively, irrtiol] vlues, must f e costt? Prove your ssertio 9 Let J (,) d let g : J e cotiuous fuctio with the property tht for every J, the fuctio g is ouded o eighorhood V () of Show y emple tht g is ot ecessrily ouded o J Let I [, ] d let f : I e (ot ecessrily cotiuous) fuctio with the property tht for every I, the fuctio f is ouded o eighorhood V () Defiitio Let A, let f : of (i the sese of the A, d let c e cluster poit of A We sy tht f is ouded o eighorhood of c if there eists eighorhood V () c of c d costt M such tht such tht f () M for ll A V () c ) Prove tht f is ouded o I Rel Alysis Pge 6

UNIFORM CONTINUITY A fuctio f is uiformly cotiuous if it is possile to gurtee tht f () d f () u s close to ech other s we plese y requirig oly tht d u re sufficietly close to ech other; ulike ordiry cotiuity, the mimum distce etwee d u cot deped o d u themselves Every uiformly cotiuous fuctio is cotiuous Uiform cotiuity, ulike cotiuity, relies o the ility to compre the sizes of eighourhoods of distict poits of give spce Let A d let f : A By the defiitio of cotiuity t poit it follows tht the followig sttemets re equivlet (i) f is cotiuous t every poit u A (ii) Give d u A, there is (,) u such tht for ll such tht A u (,) u, the f ()() f u I the ove ote tht depeds, i geerl, o oth d u A For emple, cosider f () si(/) for f chge its vlues rpidly er certi poits d slowly er other poits Hece depeds o u Emple If g() / for A { : }, the If u u g()() g u () u A is give d if we tke (,) u if{, u u }, () the if u (,) u, we hve u u so tht u u u so tht u u, d hece it follows tht Thus, if u u, the equlity () yields the iequlity u g()() g u u () u Hece if u (,) u, the () d () imply tht g()()() g u u u We hve see tht the selectio of (,)u e d y the formul () works i the sese tht it eles us to give vlue of tht will esure tht g()() g u whe u d, u A We ote tht the vlue of (,)u give i () certily depeds o the poit u A If we wish to cosider ll u A, formul () does ot led to oe vlue () u, sice if {(,) : u u} tht will work simulteously for ll Rel Alysis Pge 7

There re other selectios tht c e mde for (F or emple we could lso tke (,) u if{, u u } ; however, we still hve if{(,) : u u} ) I fct, there is o wy of choosig oe vlue of tht will work for ll u for the fuctio g() / see Uiform Cotiuity, s we shll Now it ofte hppes tht the fuctio f is such tht the umer c e chose to e idepedet of the poit u A d to deped oly o For emple, if f () for ll, the f ()() f u u, d so we c choose (,) u / for ll, u Here is idepedet of u Defiitio Let A d let f : A We sy tht f is uiformly cotiuous o A if for ech there is () such tht if, u A re y umers stisfyig u (), the f ()() f u We ote tht () i the defiitio ove is idepedet of u Also, it is cler tht if f is uiformly cotiuous o A, the it is cotiuous t every poit of A I geerl, however, the coverse does ot hold, s is show y the fuctio g() / o the set A { : } Emple Show tht the fuctio Now f () Let, y e y two poits of [,] The Hece, for, y [, ] f ()() f y y y y we ote tht y y, sice, y [,] f ()() f y y Hece for y, f ()() f y is uiformly cotiuous o [,] wheever y Hece tkig, we hve for y d y, y [,] f ()() f y wheever y This shows tht the give fuctio is uiformly cotiuous Emple Show tht the fuctio f () / is ot uiformly cotiuous o A { : } Sice is cotiuous o A (,] d for ll A, / is cotiuous o A If possile, let f () / is uiformly cotiuous o A The correspodig to, there eists such tht Rel Alysis Pge 8

wheever y y By Archimede Property of rel umers [ If, the there eists such tht ], correspodig to this, we c fid turl umer m such tht m The m Sice m, oth m d re elemets of (,] m A d m m m m But f f m m m m m / m / m This is cotrdictio to the iequlity Hece our ssumptio is wrog ie, f is ot uiformly cotiuous o A Emple If g() / for B [/, ], the u g()() g u () u Sice, u [/, ], u / / 4 4 u u So if u,()() f f u 4 if / 4 u Here does t deped o d u It works for y pir of umers i the itervl [/, ] with u Hece g() / is uiformly cotiuous o B [/, ] Emple Show tht If possile let f () cos, is ot uiformly cotiuous o f () cos,, there eists, such tht for y, y, is uiformly cotiuous o The correspodig to f ()() f y wheever y By Archimede Property, correspodig to this we c fid turl umer such tht Hece Now let ( ) d y The But y ( ) ( ) ( ) f ()() fcos y cos cos( y ) cos ( )( ) Rel Alysis Pge 9

This is i cotrdictio with iequlity (6) Hece our ssump tio is wrog ie, f is ot uiformly cotiuous o It is useful to formulte coditio equivlet to syig tht f is ot uiformly cotiuous o A We give such criteri i the et result Nouiform Cotiuity Criteri: Let A d let f : A The the followig sttemets re equivlet: (i) f is ot uiformly cotiuous o A (ii) There eists such tht for every there re poits, u i A such tht u d f ()() f u (iii) There eists f ()() f u for ll d two sequeces () d () u i A such tht lim() u d Emple Show tht g() / is ot uiformly cotiuous o A { : } We pply Nouiform Cotiuity Criterio For, if / d u /( ), the we hve lim() u, ut g()() g u ( ) u / /( ) for ll Now choose By the Nouiform Cotiuity Criterio, g is ot uiformly cotiuous We ow preset importt result tht ssures tht cotiuous fuctio o closed ouded itervl I is uiformly cotiuous o I Uiform Cotiuity Theorem: Let I e closed ouded itervl d let f : I e cotiuous o I The f is uiformly cotiuous o I To prove the theorem, we pply Cotrpositive Method (which works o the logic p q q p ) for provig the Theorem If f is ot uiformly cotiuous o I, the, y the Nouiform Cotiuity Criterio, there eists d two sequeces () d () u i I such tht u d f ()() f u for ll Sice I is ouded, the sequece () is ouded; d hece y the Bolzo-Weierstrss Theorem there is susequece () k of () closed, the limit z elogs to I, y Theorem A i chpter Sice u z u k k k z k z k k tht coverges to elemet, sy, z Sice I is d sice d k z s k, we hve the susequece () u k z k lso coverges to Rel Alysis Pge

Now if f is cotiuous t the poit z, the, y the sequetil criterio for cotiuity, s () k z d () u k z, we hve oth of the sequeces f () k d f () u k coverge to f () z But this is ot possile sice f ()() f u for ll Thus the hypothesis tht f is ot uiformly cotiuous o the closed ouded itervl I implies tht f is ot cotiuous t some poit z I Cosequetly, if f is cotiuous t every poit of I, the f is uiformly cotiuous o I This completes the proof Emple Show tht g() / is uiformly cotiuous o A { : } To show this, we ote tht sice A [, ] is closed d ouded d g is cotiuous o A, y Uiform Cotiuity Theorem, g is Uiformly Cotiuous Lipchitz Fuctios If uiformly cotiuous fuctio is give o set tht is ot closed ouded itervl, the it is sometimes difficult to estlish its uiform cotiuity However, there is coditio tht frequetly occurs tht is sufficiet to gurtee uiform cotiuity Defiitio Let A d let f : A If there eists costt K such tht f ()() f u K u (**) for ll, u A, the f is sid to e Lipchitz fuctio (or to stisfy Lipchitz coditio) o A Geometric Iterprettio The coditio ( **) tht fuctio f : I o itervl I is Lipchitz fuctio c e iterpreted geometriclly s follows If we write the coditio s f ()() f u K,, u I, u, u the the qutity iside the solute vlue is the slope of lie segmet joiig the poits,() f du,() f u Thus fuctio f stisfies Lipchitz coditio if d oly if the slopes of ll lie segmets joiig two poits o the grph of y f () over I re ouded y some umer K Theorem If f : A is Lipchitz fuctio, the f is uiformly cotiuous o A To prove this we ote tht if f is Lipchitz fuctio, the there eists K such tht f ()() f u u ie, such tht K,, u I, u, f ()() f u K,, u, u I u If is give, we c tke / K If, u A stisfy u, the f ()() f u K K Therefore f is uiformly cotiuous o A This completes the proof Emple If f () o A [, ], where, the f ()() f u u u u for ll, u i[, ] Thus f ()() f u u for ll, u [, ], u Rel Alysis Pge

Thus f stisfies the Lipchitz coditio (4) with K o A, d therefore f is uiformly cotiuous o A Of course, sice f is cotiuous d A is closed ouded itervl, this c lso e deducted from the Uiform Cotiuity Theorem ( Attetio! The fuctio f i the ove emple f does ot stisfy Lipchitz coditio o the itervl[,) ) Coverse of the Theorem is ot true, i geerl Not every uiformly cotiuous fuctio is Lipchitz fuctio The followig emple shows this Emple Show tht g() fuctio If possile let g() o [, ] is ot Lipchitz fuctio, ut uiformly cotiuous, for [,] K such tht g()() g u K u for ll, u[,] ie, u k u for ll, u [,] The, u u u u K u u is Lipchitz fuctio The there eists costt for ll, ua By Archimede Property, correspodig to the rel umer K, we c fid turl umer, such tht K Now cosider 4 d u 6 Clerly, oth, u [,] But u, sice K 4 4 K This is cotrdictio Hece our ssumptio is wrog Hece g() is ot Lipchitz fuctio Sice [, ] is closed d ouded d sice g is cotiuous, y Uiform Cotiuity Theorem g is uiformly cotiuous Altertively, There is o umer K such tht g() K for ll I The reso is this: If there is such K the, i prticulr for, K, which is ot possile / s the set { / : } is uouded Therefore, g is ot Lipchitz fuctio o I The followig emple illustrtes tht Uiform Cotiuity Theorem d Theorem c sometimes e comied to estlish the uiform cotiuity of fuctio o set Emple We ow verify the uiform cotiuity of the fuctio g defied y g() set A [,) o the The uiform cotiuity of the cotiuous fuctio g o the closed ouded itervl I [,] follows from the Uiform Cotiuity Theorem If J [,), the if oth, u re i J, we hve Rel Alysis Pge

u g()() g u u u u Thus g is Lipchitz fuctio o J with costt K, d hece, y Theorem, g is uiformly cotiuous o[,) Now A I J, d for give, uiform cotiuity of g o I gives, deoted y I () d uiform cotiuity of g o J gives, deoted y J () Now tke () if{,(),()} The for give there eists such tht I g()() g u for, u A d u Hece g is uiformly cotiuous o A J The Cotiuous Etesio Theorem We hve see emples of fuctios tht re cotiuous ut ot uiformly cotiuous o ope itervls; for emple, the fuctio f () / o the itervl (, ) O other hd, y the Uiformly Theorem, fuctio tht is cotiuous o closed ouded itervl is lwys uiformly cotiuous So the questio rises: Uder wht coditios is fuctio uiformly cotiuous o ouded ope itervl? The swer revels the stregth of uiform cotiuity, for it will e show tht fuctio o (, ) is uiformly cotiuous if d oly if it c e defied t the edpoits to produce fuctio tht is cotiuous o the closed itervl We first estlish result tht is of iterest i itself Theorem If f : A is uiformly cotiuous o set A of d if () is Cuchy sequece i A, the f () is Cuchy sequece i To prove this we Let () e Cuchy sequece i A d let e give Choose such tht if, u i A stisfy u, the f ()() f u Sice () defiitio of Cuchy sequece, there eists H () such tht for ll,() m H By the choice of, this implies tht for,() m m f ()() f m H, we hve Therefore, y the defiitio of Cuchy sequece, the sequece is Cuchy sequece, y the f () is Cuchy sequece This completes the proof Defiitio A sequece X () of rel umers is sid to e Cuchy sequece if for every there is turl umer H () such tht for ll turl umers,() m H, the terms, stisfy m m Emple Usig Theorem 4 show tht f () / is ot uiformly cotiuous o (,) We ote tht the sequece give y / i (,) is Cuchy sequece, ut the imge sequece, where f (), is ot Cuchy sequece Sequetil Criterio for Limits Let f : A d let c e cluster poit of A The the followig sttemets re equivlet Rel Alysis Pge

(i) lim() f L c (ii) For every sequece () i A tht coverges to c such tht c for ll, the sequece (()) f coverges to f () c Cuchy Covergece Criterio A sequece of rel umers is coverget if d oly if it is Cuchy sequece Cotiuous Etesio Theorem: A fuctio f is uiformly cotiuous o the itervl (,) d oly if it c e defied t the edpoits d such tht the eteded fuctio is cotiuous o [, ] Proof () This directio is trivl () Suppose f is uiformly cotiuous o (,) We shll show how to eted f rgumet for is similr This is doe y showig tht lim() f Rel Alysis Pge 4 L if to ; the eists, d this is ccomplished y usig the sequetil criterio for limits If () is sequece i (,) lim() with, the it is Cuchy sequece, d y the precedig theorem, the sequece f () is lso Cuchy sequece, d so is coverget y Cuchy Covergece Criterio Thus the limit lim() f L eists If () u is y other sequece i (,) lim() u, so y the uiform cotiuity of f we hve f u f u f f lim() lim()() lim() tht coverges to, the L L Sice we get the sme vlue L for every sequece covergig to, we ifer from the sequetil criterio for limits tht f hs limit L t If we defie f () L, the f is cotiuous t The sme rgumet pplies to, so we coclude tht f hs cotiuous etesio to the itervl [, ] This completes the proof Emple Emie the uiform covergece of the fuctios f () si(/) d o (, ] for ll g() si(/) Sice the limit of f () si(/) t does ot eist (Fig B), we ifer from the Cotiuous Etesio Theorem tht the fuctio is ot uiformly cotiuous o (, ] for y O the other hd, sice lim si(/) eists, the fuctio g() si(/) o (, ] for ll is uiformly cotiuous Approimtio I this sectio we descrie the wy of pproimtig cotiuous fuctios y fuctios of elemetry ture Defiitio Let I e itervl d let s : I This s is clled step fuctio if it hs oly fiite umer of distict vlues, ech vlue eig ssumed o oe or more itervls i I Emple The fuctio s :[,4] defied y

s(),,,,,,,,,, 4, is step fuctio We will ow show tht cotiuous fuctio o closed ouded itervl I c e pproimted ritrrily closely y step fuctios Theorem Let I e closed ouded itervl d let f : I e cotiuous o I If, the there eists step fuctio s : I such tht f ()() s for ll I Proof Sice f is cotiuous o the closed ouded itervl I, y the Uiform Cotiuity Theorem, the fuctio f is uiformly cotiuous o I Hece it follows tht, give there is umer () such tht if, y I d y (), the f ()() f y Let I [, ] d let m e sufficietly lrge so tht h () We ow divide m I [, ] ito m disjoit itervls of legth h; mely, I [, h] d Ik (( ) k, h ] kh for k,, m Sice the legth of ech suitervl I is h (), the differece etwee y two vlues of f i Ik is less th We ow defie s ()() f kh for Ik, k,, m, so tht s is costt o ech itervl I k (I fct the vlue of s o I k right edpoit of I k Cosequetly if Ik, the f ()()()() s f f kh Therefore we hve f ()() s for ll I This completes the proof k is the vlue of f t the The proof of the precedig theorem lso estlishes the followig result Corollry Let I [, ] e closed ouded itervl d let f : I e cotiuous o I If, there eists turl m such tht if we divide I ito m disjoit itervls Ik h () / m, the the step fuctio s defied y stisfies s ()() f kh for Ik, k,, m, f ()() s for ll I hvig legth Step fuctios re etremely elemetry i chrcter, ut they re ot cotiuous (ecept i trivil cses) Sice it is ofte desirle to pproimte cotiuous fuctios y elemetry cotiuous fuctios, we ow shll show tht we c pproimte cotiuous fuctios y cotiuous piecewise lier fuctios Rel Alysis Pge 5

Defiitio Let I [, ] e itervl The fuctio g : I is sid to e piecewise lier o I if I is the uio of fiite umer of disjoit itervls I,, Im, such tht the restrictio of g to ech itervl Ik is lier fuctio Remrk It is evidet tht i order for piecewise lier fuctio g to e cotiuous o I, the lie segmets tht form the grph of g must meet t the edpoits of djcet suitervls Ik,( Ik, k, m) Theorem Let I e closed ouded itervl d let f : I e cotiuous o I If, the there eists cotiuous piecewise lier fuctio g : I such tht f ()() g for ll I To prove this we ote tht eig cotiuous o closed ouded itervl, f is uiformly cotiuous o I [, ] Hece correspodig to, there is umer () such tht if, y I d y (), the f ()() f y Let m e sufficietly lrge so tht h () /() m Divide I [, ] ito m disjoit itervls of legth h; mely let I [, h], d let I [( ) k, h ] kh for k,, m k O ech itervl Ik we defie g to e lier fuctio joiig the poits (( ) k,(( h f))) k h d ( kh,()) f kh The g is cotiuous piecewise lier fuctio o I If I the (( ))() k h h, so tht f ()(( f)) k h Similrly, k f ()() f kh ie, for Ik the vlue f () is withi of f (( )) k h d f () Hece for I k, g ()(( f)) k h f ()(( kh)) f k h h ((( ))) k h Hece for I k, f ()()()(( g )) f f k h s (( )) k h h f ()(( kh)) f k h h h, f ()(( f))()(( k )) h f kh f k h f ()(( kh))(( f ))() k h f k h f f ())() kh f, y Trigle Iequlity kh Tht is, for I k, f ()() g for ll Ik ; therefore this iequlity holds for ll I This completes the proof Net is the importt theorem of Weierstrss cocerig the pproimtio of cotiuous fuctios y polyomil fuctios Rel Alysis Pge 6

Weierstrss Approimtio Theorem Let I [, ] d let f : I e cotiuous fuctio If is give, the there eists polyomil fuctio p such tht f ()() p for ll I Remrk I order to oti pproimtio withi ritrrily pre-ssiged, we hve to choose polyomils of ritrrily high degree There re umer of proofs of Weierstrss Approimtio Theorem Oe of the most elemetry proofs is sed o the followig theorem, due to Serge Berstei, for cotiuous fuctios o [,] Give f :[,], Berstei defied the sequece of polyomils: k k k B ()() f k k The polyomil fuctio B is clled the th Berstei polyomil for f; it is polyomil of degree t most d its coefficiets deped o the vlues of the fuctio f t the eqully spced poits, /, /,, k /,, d o the iomil coefficiets Emple If () k f () B Here!( )( ) k k!() k k, for [,] f () for [,] Berstei polyomils for f re give y k k k B ()(,)() f, C,, k k k k The B ()(,)() f C k k k C(,)() k k k k k k B ()(,)() f C k k k k C k k k 4 (,)(), clculte the first few Berstei polyomils for f Show tht k k () 4 k k B ()(,)() f C k k k k C k k k 9 (,)() k () () 4 9 9 9 9 Rel Alysis Pge 7

[ ] [ ] 4 I geerl for y,, k k B ()(,)() C k k k k! k [( k k ) k]() k!()! k! k ( k )!()! k k () k k! k k () k ( k )!()! k ( )! k ( )() k ( k )!()! k ( )! k k () k ( k )!()! k ( ) [()] [ ()] ( ) Berstei s Approimtio Theorem Let f :[,] e cotiuous d let There eists such tht if, the we hve f ()() B for ll [,] The Weierstrss Approimtio Theorem c e derived from the Berstei Approimtio Theorem y chge of vrile Specificlly, we replce f :[, ] y fuctio F :[,], defied y F()(()) t f t for t [,] The fuctio F c e pproimted y Berstei polyomils for F o the itervl [, ], which c the yield polyomils o [, ] tht pproimte f Assigmets Show tht the fuctio f () is uiformly cotiuous o A [,), ut tht it is ot uiformly cotiuous o B (,) Show tht the fuctio f () is uiformly cotiuous o the set A [,), where is positive costt Use the Nouiform Cotiuity Criterio to show tht the followig fuctios re ot uiformly cotiuous o the give sets ) f (), A [,) ) g() si(/), B[,) k Rel Alysis Pge 8

4 Show tht the fuctio f () /() for is uiformly cotiuous o 5 Show tht if f d g re uiformly cotiuous o suset A of, the f g is uiformly cotiuous o A 6 Show tht if f d g re uiformly cotiuous o A d if they re oth ouded o A, the their product fg is uiformly cotiuous o A 7 If f () d g() si, show tht oth f d g re uiformly cotiuous o, ut tht their product f g is ot uiformly cotiuous o 8 Give emple to show tht oudedess of f d g is ecessry coditio for the uiform cotiuity of the product 9 Prove tht if f d g re ech uiformly cotiuous o, the the composite fuctio f g is uiformly cotiuous o If f is uiformly cotiuous o A, d f () k for ll A, show tht / f is uiformly cotiuous o A Prove tht if f is uiformly cotiuous o ouded suset A of, the f is ouded o A If g() for [,], show tht there does ot eist costt K such tht g() K for ll [,] Coclude tht the uiformly cotiuous g is ot Lipchitz fuctio o [,] Show tht if f is cotiuous o [,) d uiformly cotiuous o [,) costt, the f is uiformly cotiuous o [,) for some positive 4 Let A d suppose tht f : A hs the followig property: for ech eists fuctio g : A such tht g is uiformly cotiuous o A d f ()() g there for ll A Prove tht f is uiformly cotiuous o A 5 A fuctio f : is sid to e periodic o if there eists umer p such tht f ()() p f for ll Prove tht cotiuous periodic fuctio o is ouded d uiformly cotiuous o 6 If f () for [,], clculte the first few Berstei polyomils for f Show tht they coicide with f [Hit: Biomil Theorem sserts tht 7 Let f :[] 8 If f () k k k () ] k is cotiuous d hs oly rtiol vlues Show tht f must e costt for [,], clculte the first few Berstei polyomils for f coicide with f Show tht they Rel Alysis Pge 9

4 THE RIEMANN INTEGRAL - PART I The Riem itegrl, creted y Berhrd Riem, ws the first rigorous defiitio of the itegrl of fuctio o itervl For my fuctios d prcticl pplictios, the Riem itegrl c e evluted y usig the fudmetl theorem of clculus or (pproimtely) y umericl itegrtio The Riem itegrl is usuitle for my theoreticl purposes Some of the techicl deficiecies i Riem itegrtio c e remedied with the Riem Stieltjes itegrl, d most dispper with the Leesgue itegrl Prtitio d Tgged Prtitios If I [, ] is closed ouded itervl i, the prtitio of I is fiite, ordered set P (,,,,) of poits i I such tht The poits of P re used to divide I [, ] ito o-overlppig suitervls I [, ], I [, ],, I [, ] Ofte we will deote the prtitio P y the ottiop {[, ]} i i i We defie the orm (or mesh) of P to e the umer P m{,,, } Thus the orm of prtitio is merely the legth of the lrgest suitervl ito which the prtitio divides[, ] Clerly, my prtitios hve the sme orm, so the prtitio is ot fuctio of the orm If poit ti hs ee selected from ech suitervl Ii [ i, i ], for i,,,, the the poits re clled tgs of the su-itervls I i A set of ordered pirs {([, ],)} t P i i i i of suitervls d correspodig tgs is clled tgged prtitio of I (The dot over P idictes tht tg hs ee chose for ech suitervl) The tgs c e chose i wholly ritrry fshio; for emple, we c choose the tgs to e the left edpoits, or the midpoits of the suitervls, etc Note tht edpoit of suitervl c e used s tg for two cosecutive suitervls Sice ech tg c e chose i ifiitely my wys, ech prtitio c e tgged i ifiitely my wys The orm of tgged prtitio is defied s for ordiry prtitio d does ot deped o the choice of tgs Defiitio If P is the tgged prtitio give ove, we defie the Riem sum of fuctio f :[, ] correspodig to P to e the umer Rel Alysis Pge 4

S( f ;)()() P f ti i i i Nottio We will lso use this ottio whe etire prtitio P deotes suset of prtitio, d ot the Fig Riem sum of positive fuctio Defiitio If the fuctio f is positive o [, ], the the Riem sum () is the sum of the res of rectgles whose ses re the suitervls Ii [ i, i ] d whose heights re f () t i (Fig) Riem Itegrl We ow defie the Riem itegrl of fuctio f o itervl [, ] Defiitio A fuctio f :[, ] is sid to e Riem itegrle o [, ] umer L such tht for every there eists such tht if P prtitio of [, ] with P, the S( f ;) P L if there eists The set of ll Riem itegrle fuctios o [, ] will e deoted y R [, ] Remrk It is sometimes sid tht itegrl L is the limit of the Riem sums S( f ;) P orm P However, sice S( f ;) P is ot fuctio of P is y tgged s the, this limit is ot of the type tht we hve studied efore First we will show tht if f R [, ], the the umer L is uiquely determied It will e clled the Riem itegrl of f over[, ] Isted of L, we will usully write Rel Alysis Pge 4

L f or () f d It should e uderstood tht y letter th c e used i the ltter epressio, so log s it does ot cuse y miguity Theorem If f R [, ], the the vlue of the itegrl is uiquely determied Proof Assume tht L d L oth stisfy the defiitio of Riem itegrl of f over [, ], d let The there eists / such tht if P is y tgged prtitio with P /, the S( f ;) P L / (*) Also there eists / such tht if P is y tgged prtitio with P / S( f ;) P L / (**) Now let mi{ /, / } d let P e tgged prtitio with P P / d P /, the (*) d (**) gives S( f ;) P L / d S( f ;) P L /, d hece it follows from the Trigle Iequlity tht L L L S( f ;)( P ;) S f P L L S( f ;)( P ;) S f P L / / Sice is ritrry, it follows tht L L Method of Verifyig f R [, ] usig Defiitio, the Sice oth If we use oly the defiitio, i order to show tht fuctio f is Riem itegrle we must (i) kow (or guess correctly) the vlue L of the itegrl, d (ii) costruct tht will suffice for ritrry The determitio of L is sometimes doe y clcultig Riem sums d guessig wht L must e The determitio of is likely to e difficult I ctul prctice, we usully show tht f R, y mkig use of some of the theorems tht will e discussed fter cosiderig some emples sed o the Defiitio Emple Every costt fuctio o [, ] is i R [, ] Let f () k for ll [, ] If P {[, ],)} i i ti i is y tgged prtitio of [, ], the it is cler tht S( f ;)()()()() P f t k k i i i i i i i i i i k() k() Rel Alysis Pge 4

k() Hece, for y, we c choose so tht if P, the S( f ;)() P k Sice is ritrry, we coclude tht f R [, ] d f k () Emple Let g :[,] e defied y g() for d g() for Fid g A prelimiry ivestigtio, sed o the grph of g, suggests tht we might epect tht g 8 Let P e tgged prtitio of [,] with orm ; we will show how to determie i order to esure tht S( g;) P 8 Let P e the suset of P hvig its tgs i [,] where g(), d the let P e the suset of P with its tgs i (,] where g() tht we hve S( g;)( P ;)( S;) g P S g P It is ovious Sice P, if u [, ] d u [ i, ], the i so tht i i, d hece the tg t [,] Therefore, the itervl [, ] is cotied i the uio of ll suitervls i i P with tgs t [,] Similrly, this uio is cotied i [, ] Sice g() t for these i tgs, we hve ()( ;) S() g P A similr rgumet shows tht the uio of ll suitervls with tgs ti (,] cotis the itervl [,] of legth, d is cotied i [, ] of legth Therefore, ()( ;) S() g P Addig these iequlities d usig equtio () we hve 8 5( ;)( S ;)( g P ;) 8S g5 P S g P d hece it follows tht S( g;) P 8 5 To hve this fil term, we re led to tke /5 Mkig such choice (for emple, if we tke / ), we c retrce the rgumet d see tht S( g;) P 8 whe P Sice proved tht g R[,] d g 8, s epected i is ritrry, we hve Rel Alysis Pge 4

Emple Let h() for [,] Show tht h R[,] We will employ trick tht eles us to guess the vlue of the itegrl y cosiderig prticulr choice of the tg poits Ideed, if { I } i i is y prtitio of [, ] d we choose the tg of the itervl Ii [ i, i ] to e the midpoit q () i i i, the the cotriutio of this term to the Riem sum correspodig to the tgged prtitio Q {( Ii,)} qi i is h()()()()() qi i i i i i i i i If we dd these terms d ote tht the sum telescopes, we oti i i i i S( h,)()( Q ) Now let P {( Ii,)} ti i e ritrry tgged prtitio of [,] with P i so tht for i,, Also let Q hve the sme prtitio poits, ut where we choose the tg i q i to e the midpoit of the itervl I i Sice oth t i d t q Usig the Trigle Iequlity, we deduce i S( h;)( P ;)()() S h Q t q i i i i i i i i t q () i i i i i i () i i () Sice S( h;) Q, we ifer tht if P is y tgged prtitio with S( h;) P Therefore we re led to tke coclude tht h R [,] d h d If we choose q i elog to this itervl, we hve P, the, we c retrce the rgumet to 4 Emple Let F() for,,,, d F() elsewhere o [, ] Show tht F R [,] 5 5 5 5 d F Here there re four poits where F is ot, ech of which c elog to two su-itervls i give tgged prtitiop Oly these terms will mke o zero cotriutio to S( F;) P Therefore we choose /8 If P, let P e the suset of P 4 with tgs differet from,,,, d let P 5 5 5 5 e the suset of P with tgs t these poits Sice S( F;) P, it is see tht Rel Alysis Pge 44

S( F;)( ;)( S ;) F S F P P P S( F;) P Sice there re t most 8 terms i S( F;) P d ech term is, we coclude tht ( S;)( F ;) P 8 S F P Thus F R [,] d F Emple Let G() /(), d G() elsewhere o [,] Show tht G R [,] d G Give, let E e the (fiite) set of poits where G(), let e the umer of poits i E, d let /() Let P e tgged prtitio such tht P Let P e the suset of P with tgs outside of E d let P e the suset of P previous emple, we hve ( S;)( G;)() P S G P with tgs i E The, s i the Sice is ritrry, we coclude tht G R[,] d G Emple Suppose tht c d re poits i[, ] If :[, ] stisfies () for [ c, d] d () elsewhere i[, ], prove tht R [, ] d tht () d c For tgged prtitio P {([, ],)} i i ti i, S(,)()() P t i i i i Riem sum is givig y Give, let / 4 The if P, the P / 4, d the the uio of the suitervls i [ c, d ] Therefore d hece P with tgs i [ c, d] cotis the itervl [ c, d ] d is cotied i ( d c ) S( ;) P ( d c ), S( ;)() P d c 4 Hece y the Defiitio of Riem itegrility, R [, ] d tht () d c Some Properties of the Itegrl The difficulties ivolved i determiig the vlue of the itegrl d of suggest tht it would e very useful to hve some geerl theorems The first result i this directio eles us to form certi lgeric comitios of itegrle fuctios Theorem Suppose tht f d g re i R [, ] The: ) If k, the fuctio kf is i R [, ] kf k f ) The fuctio f g is i R [, ] d () f g f g d Rel Alysis Pge 45

c) If f ()() g for ll [, ], the f g To prove (), we ote tht if P {([ i, i ],)} ti i is tgged prtitio of [, ], the S( kf ;)()()()()() P kf t k f t i i i i i i i i k f ()()( t ;) i i i ks f P i Now the ssertio () follows S( f g;)( P ;)( S ;) f P S g P Give, we c use the rgumet i the proof of the Uiqueess Theorem to costruct P, the oth umer such tht if P is y tgged prtitio with Hece S( f ;) P f / d S( g;) P g / (4) S( f g;)( P ;)( ;) f g S f P S g P f g S ( f ;)( P ;) f S g P g / / Sice is ritrry, we coclude tht f g R[, ] d tht its itegrl is the sum of the itegrls of f d g To prove (c), we ote tht S( f ;)( P ;) S g P d lso (4) implies f ( ;) S f P d ( ;) S g P g If we use the fct tht S( f ;)( P ;) S g P, we hve But sice Theorem If elogs to R [, ] f g is ritrry, we coclude tht f g This completes the proof f,, f re i R[, ] d if d i i f k f i k,, k, the the lier comitio Hit for the Proof Use mthemticl Iductio d prt () d () of Theorem ove Theorem If f R[, ] d f () Tke g() M for [, ] M for ll [, ], the f M () f i i k f i Rel Alysis Pge 46

d use prt (c) of Theorem Boudedess Theorem If f R [, ], the f is ouded o [, ] Proof Assume tht f is uouded fuctio i R[, ] such tht if P is y tgged prtitio of [, ] with which implies tht S( f ;) P L, P S( f ;) P L (***) Now let Q {[, ]} i i e prtitio of [, ] with with itegrl L The there eists, the we hve Q Sice f is ot ouded o,, the there eists t lest oe suitervl i Q, sy [ k, k ], o which f is ot ouded for, if f is ouded o ech suitervl [ i, i ] { M,, M } y i M, the it is ouded o [, ] y m We will ow pick tgs for Q tht will provide cotrdictio to (***) We tg Q y t i i for i k d we pick tk [ k, k ] such tht f ()() t ()() L f t k k k i i i ik From the Trigle Iequlity (i the form A B A B ), we hve S( f ;)()()()() Q f t f t L, which cotrdicts (***) k k k i i i ik Remrk I view of the ove theorem we hve the followig: A uouded fuctio cot e Riem itegrle We ow cosider emple of fuctio tht is discotiuous t every rtiol umer d is ot mootoe, ut is Riem itegrle Emple We cosider Thome s fuctio h :[,] defied y h() if [,] irrtiol, h() d y h() / if [,] is the rtiol umer m/ where m, hve o commo iteger fctors ecept We clim tht h is cotiuous t every irrtiol umer d discotiuous t every rtiol umer i[,] If is rtiol, let () e sequece of irrtiol umers i A tht coverges to The lim(()) h lim(), while h() Hece, i view of Sequetil Criterio for Cotiuity, h is discotiuous t the rtiol poit is Rel Alysis Pge 47

If is irrtiol umer d, the (y the Archimede Property) there is turl umer such tht ie, such tht There re oly fiite umer of rtiols with deomitor less th i the itervl (, ) Hece c e chose so smll tht the eighorhood (,) cotis o rtiol umers with deomitor less th It the follows tht for d (,), we hve h is cotiuous t the irrtiol umer We will ow show tht h R [,] h()()() h h The Let The the set E { [,]:() h } is fiite set We let e the umer of elemets i E d let /(4) If P is tgged prtitio with P, let P e the suset of P hvig tgs i E d P e the suset of P hvig tgs elsewhere i[,] We oserve tht P hs t most itervls whose totl legth is / d tht () ht i for every tg ip Also the totl legths of the suitervls i P is d h() ti / for every tg ip Therefore we hve P P P S( h;)( ;)( S;) h ( S/ ) h Sice is ritrry, we ifer tht h R[,] with itegrl Assigmets If I [, 8], clculte the orms of the followig prtitios: ) P (,,,4,5,6,7,8) ) P (,,4,6,8) c) P (,,6,8) d) P4 (,5,8) If f () for [,8], clculte the followig Riem sums where P i prtitio poits s i Assigmet, d the tgs re selected s idicted ) P with the tgs t the left edpoits of the suitervls ) P with the tgs t the right edpoits of the suitervls c) P with the tgs t the left edpoits of the suitervls d) P with the tgs t the right edpoits of the suitervls hs the sme Show tht f :[, ] is Riem itegrle o [, ] if d oly if there eists L such tht for every there eists such tht if P is y tgged prtitio with orm P, the S( f ;) P L 4 Let P e tgged prtitio of [,] Rel Alysis Pge 48

) Show tht the uio U of ll suitervls i U [, P ] P with tgs i [,] stisfies [, P ] ) Show tht the uio U of ll suitervls i P with tgs i [,] stisfies [ P, P ] U [ P, P ] 5 Let P {( Ii,)} ti i e tgged prtitio of [, ] d let c c ) If u elogs to suitervl Ii whose tg stisfies c ti c, show tht c P u c P ) If v [, ] d stisfies c P v c P, the the tg t i of y suitervl cotis v stisfiesti [ c, c] 6 ) Let f () if d f () if Show tht f R[,] d evlute its itegrl ) Let h() if, h() d h() if Show tht h R[,] d evlute its itegrl f R d if 7 If [, ] Ii tht P is y sequece of tgged prtitios of [, ] such tht P, prove tht f lim( S ;) f P 8 Let g() if [,] is rtiol d g() / if [,] is irrtiol Epli why g R [,] However, show tht there eists sequece P of tgged prtitios of [, ] such tht P d lim( S;) g P eists 9 Suppose tht f is ouded o [, ] d tht there eists two sequeces of tgged prtitios of [, ] such tht P d Q, ut such tht lim( S;) f Plim( ;) S f Q Show tht f is ot i R[, ] Cosider the Dirichlet fuctio, defied y f () for [,] rtiol d f () [,] [,] irrtiol Use the precedig eercise to show tht f is ot Riem itegrle o Suppose tht f :[, ] d tht f () ecept for fiite umer of poits c,, c i [, ] Prove tht f R[, ] d tht f If g R[, ] d if f ()() g ecept for fiite umer of poits i[, ], prove tht f R[, ] d tht f g Let, let ech i, let Q() for [, ] qi e the positive squre root of ) Show tht i () i i i i q stisfies i qi i d let P {[, ]} i i i e prtitio of [, ] For for Rel Alysis Pge 49

) Show tht Q()()() q i i i i i c) If Q is the tgged prtitio with the sme suitervls s P d the tgs q i, show tht S( Q,)() Q d) Show tht Q R [, ] d () Q d 4 If f R [, ] d c, we defie g o [ c, c] y g()() y f y c Prove tht g R [ c, c] d tht c c 5 Let d m, let M () [, ] For ech i, let g f The fuctio g is clled the c-trslte of f qi e the positive m for [, ] mth () m m m m i i i i i i m ) Show tht qi stisfies i qi i ) Show tht M ()()() q d let P {[, ]} i i i e prtitio of root of m m i i i m i i c) If Q is the tgged prtitio with the sme suitervls s P d the tgs q i, show tht S( M ;)() Q m d) Show tht M R [, ] d () m m m m m m M d Rel Alysis Pge 5

5 THE RIEMANN INTEGRAL - PART II Cuchy Covergece Criterio for Sequeces: A sequece of rel umers is coverget if d oly if it is Cuchy sequece Cuchy Criterio for Itegrility: A fuctio f :[, ] elogs to R [, ] if d oly if for every there eists such tht if P d Q re y tgged prtitios of [, ] with P d Q, the S( f ;)( P ;) S f Q Proof () If f R [, ] with itegrl L, let e such tht if P d Q re y tgged prtitios of [, ] with P d Q, the S( f ;) P L / d S( f ;) Q L / Therefore we hve S( f ;)( P ;)( S ;)( f Q ;) S f P L L S f Q S( f ;)( P ;) L L S f / / () For ech, let e such tht if P d Q re tgged prtitios with orms the S( f ;)( P ;) S / f Q Evidetly we my ssume tht for ; otherwise, we replce mi{,, } For ech, let P e tgged prtitio with P Clerly, if m the oth P m P hve orms, so tht S( f ;)( P ;) S f P m for m (*) Cosequetly, the sequece S( f ;) P m is Cuchy sequece i Therefore, y Cuchy m Covergece Criterio for Sequeces this sequece coverges i d we let A lim( S;) f P m m Pssig to the limit i () s m, we hve S( f ;) P A/ for ll Q, y d Rel Alysis Pge 5

To see tht A is the Riem itegrl of f, give, let K stisfy K / If Q is y tgged prtitio with Q K, the S( f ;)( Q ;)( A ;)( S;) f Q S f P K S f P k A / K / K Sice is ritrry, the f R [, ] with itegrl A This completes the proof Emple Let g :[,] e defied y g() for, d g() for We hve see tht if Hece if P is tgged prtitio of [,] with orm P 8 5( ;) S 8g P 5 Q is other tgged prtitio with 8 5( ;) S 8g Q 5 Q If we sutrct these two iequlities, we oti S( g;)( P ;) S g Q, the, the I order to mke this fil term, we re led to employ the Cuchy Criterio with / The Cuchy Criterio c e used to show tht fuctio f :[, ] is ot Riem itegrle To do this we eed to show tht: There eists such tht for y eists tgged prtitios P d S( f ;)( P ;) S f Q Q with P d Q such tht there Emple Show tht the Dirichlet fuctio, defied y f () if [,] is rtiol d f () if [,] is irrtiol is ot Riem itegrle Here we tke If P is y prtitio ll of whose tgs re rtiol umers the S( f ;) P, while if Q is y tgged prtitio ll of whose tgs re irrtiol umers the S( f ;) Q Sice we re le to tke such tgged prtitios with ritrrily smll orms, we coclude tht the Dirichlet fuctio is ot Riem itegrle The et result will e used to estlish the Riem itegrility of some importt clsses of fuctios Squeeze Theorem: Let f :[, ] The f R [, ] if d oly if for every there eist fuctios d such tht d i R[, ] with ()()() f for ll [, ], (**) () (***) Proof () Tke f for ll Rel Alysis Pge 5

() Let Sice d elog to R [, ], there eists such tht if P tgged prtitio with P the S( ;) P d S( ;) P It follows from these iequlities tht S( ;) P d ( ;) P S I view of iequlity (**), we hve S( ;)( P ;)( S f;) P S P, d hece If S( f ;) P Q is other tgged prtitio with Q S( f ;) Q, the we lso hve If we sutrct these two iequlities d use (***), we coclude tht S ( f ;)( P ;) S f Q is y () Sice is ritrry, the Cuchy Criterio implies tht f R [, ] This completes the proof Clsses of Riem Itegrle Fuctios A fuctio :[, ] is step fuctio if it hs oly fiite umer of distict vlues, ech eig ssumed o oe or more suitervls of [, ] We egi with result see i the previous chpter Lemm Suppose tht c d re poits i[, ] If :[, ] stisfies () for [ c, d] d () elsewhere i[, ], prove tht R [, ] d tht () d c Proof For tgged prtitio {([, ],)} t, Riem sum is givig y P i i i i S(,)()() P t i i i i Give, let / 4 The if P, the P / 4, d the the uio of the suitervls i [ c, d ] Therefore d hece P with tgs i [ c, d] cotis the itervl [ c, d ] d is cotied i ( d c ) S( ;) P ( d c ), S( ;)() P d c 4 Rel Alysis Pge 5

Hece y the Defiitio of Riem itegrility, R [, ] d tht () d c Lemm If J is suitervl of [, ] hvig edpoits c d d if () for J J d J () elsewhere i[, ], the J R[, ] d J d c Proof If J [ c, d] with c d, the the proof of Lemm, we c choose / 4 A similr proof c e give for the three other suitervls hvig these edpoits Altertively, we oserve tht we c write [ c,) d [, c ] d [, d ]( d, ] c d [, ] c d [, ] c c d ( c,) d [,) c d [, ] c c Sice, ll four of these fuctios hve itegrl equl to d c This completes the proof [ c, c] Riem Itegrility of Step Fuctios Step fuctios of the type pperig i the previous lemm re clled elemetry step fuctios A ritrry step fuctio c e epressed s lier comitio of elemetry step fuctios: where m k j j J j hs edpoits c j d j Theorem If :[, ] is step fuctio, the R[, ] J j Proof A ritrry step fuctio c e epressed s lier comitio of elemetry step fuctios: where The m k (4) j J j hs edpoits c j d j j J j m k j J j j j k j J j k j J j j, usig prt () of Theorem of the pervious chpter, usig prt () of Theorem of the pervious chpter j Hece R [, ] k j d j c j, usig Lemm We will ow use the Squeeze Theorem to show tht ritrry cotiuous fuctio is Riem itegrle Theorem If f :[, ] is cotiuous o [, ], the Rel Alysis Pge 54

f R [, ] Proof Beig cotiuous fuctio o closed d ouded itervl, it follows tht f is uiformly cotiuous o [, ] Therefore, give there eists such tht if u, v [, ] d u v, the we hve f ()() u f v Let P { I } i i e prtitio such tht P, let ui Ii e poit where f ttis its miimum vlue o I i d let vi Ii e poit where f ttis mimum vlue o I i Let e the step fuctio defied y ()() f ui for [ i,)( i i,, ) d ()() f u for [, ] Let e defied similrly usig the poits vi isted of the u i ie, ()() f vi for [ i,)( i i,, ) d ()() f v for [, ] The we oti ()()() f for ll [, ] Moreover, it is cler tht f vi f u j i i ()(()()() i () i i i Therefore it follows from the Squeeze Theorem tht f R[, ] This completes the proof Mootoe fuctios re ot ecessrily cotiuous t every poit, ut they re lso Riem itegrle Before provig this result, we recll Chrcteriztio Theorem for Itervls Chrcteriztio Theorem for Itegrls: If S is suset of tht cotis t lest two poits d hs the property if, y S d y the [, y] S, the S is itervl Theorem If f :[, ] is mootoe o [, ], the f R [, ] Proof Suppose tht f is icresig o the itervl[, ], If such tht f ()() f h q is give, we let q e Let yk f () kh for k,,, q d cosider sets Ak f ([ yk,)) yk for k,, q d A f ([ y, y ]) The sets A re pirwise disjoit d hve uio, The q q q k Chrcteriztio Theorem implies tht ech A k is either (i) empty, (ii) cotis sigle poit, or (iii) is odegeerte itervl (ot ecessrily closed) i [, ] We discrd the sets for which (i) holds d relel the remiig oes If we djoi the edpoits to the remiig itervls Rel Alysis Pge 55

k A, we oti closed itervls q I k It c e show tht releled itervls kk q wise disjoit, stisfy [, ] k Ak d tht f () [ yk, y] k for Ak We ow defie step fuctios d o [, ] y settig () yk d () yk for Ak for Ak It is cler tht ()()() f for ll [, ] d tht q ()()() yk yk k k k q h ()() k k h k A re pir Sice is ritrry, the Squeeze Theorem implies tht f R [, ] This completes the proof Additivity Theorem: Let f :[, ] d let c (,) The f R[, ] restrictios to [, c] d [ c, ] re oth Riem itegrle I this cse c f f f (*) c if d oly if its Proof () Suppose tht the restrictio f of f to[, c ], d the restrictio f of f to [ c, ] re Riem itegrle to L d L, respectively The, give there eists such tht if P is tgged prtitio of [, c] with P, the S( f;) P L / Also there eists such tht if P is tgged prtitio of [ c, ] with P S( f ;) P L / If M is oud for f, we defie mi{,, / 6 M} d let P e tgged prtitio of [, ] with Q We will prove tht S( f ;)() Q L L (**) (i) If c is prtitio poit of Q, we split Q ito prtitio Q of [, c] d prtitio Q of[ c, ] Sice S( f ;)( Q ;) S f Q S( f ;) Q, d sice Q hs orm d Q hs orm, the iequlity (**) is cler (ii) If c is ot prtitio poit i Q {( I,)} m k tk k, there eists k m such tht c ( k,) k We let Q e the tgged prtitio of [, c] defied y Q {( I,),,( t I,),([ t, ],)} c c k k k d Q e the tgged prtitio of [ c, ] defied y Q {([ c, ],),( c I,),,( t,)} I t k k k m m A strightforwrd clcultio yields the Rel Alysis Pge 56

S( f ;)( Q ;)( S ;)()()()() f Q S f Q f tk k k f c k k (()())() f tk f c k k, d hece it follows tht S( f ;)( Q ;)( S ;) f Q () S f Q M k k But sice Q d Q, it follows tht S( f ;) Q L d S( f ;) Q L, from which we oti (**) Sice is ritrry, we ifer tht f R [, ] d tht (6) holds () We suppose tht f R [, ] d, give, we let stisfy the Cuchy Criterio Let f e the restrictio of f to [, c ] d let P, Q, e tgged prtitios of [, c] with P d Q By ddig dditiol prtitio poits d tgs from[ c, ], we c eted P d Q to tgged prtitios P d Q of [, ] tht stisfy dditiol poits d tgs i [ c, ] for oth Sice oth P d S( f ;)( P ;)( S ;)( f Q ;) S f P S f Q P d P d Q Q, the If we use the sme Q hve orm, the S( f;)( P ;) S f Q Therefore the Cuchy Coditio shows tht the restrictio f of f to [, c] is i R [, c] I the sme wy, we see tht the restrictio f of f to [ c, ] is i R [ c, ] The equlity (*) ow follows from the first prt of the theorem This completes the proof Corollry If f R [, ], d if [ c, d] [, ], the the restrictio of f to [ c, d] is i R [ c, d] Proof Sice f R [, ] d c [, ], it follows from the theorem tht its restrictio to [ c, ] is i R [ c, ] But if d [ c, ], the other pplictio of the theorem shows tht the restrictio of f to [ c, d] is i R [ c, d] This completes the proof Corollry If f R [, ] d if c c cm, the the restrictios of f to ech of the suitervls [ ci, ci ] re Riem itegrle d m i f f ci i c Defiitio If f R [, ] d if, [, ] with, we defie f f d f Theorem 7 If f R [, ] d if,, re y umers i[, ], the f f f, (***) i the sese tht the eistece of y two of these itegrls implies the eistece of the third itegrl d the equlity (***) Rel Alysis Pge 57

Proof If y two of the umers,, re equl, the (***) holds Thus we my suppose tht ll three of these umers re distict For the ske of symmetry, we itroduce the epressio L(,,) f f It is cler tht ( ***) holds if d oly if L(,,) Therefore, to estlish the ssertio, we eed to show tht L for ll si permuttios of the rgumets, d We ote tht the Additivity Theorem implies tht L(,,) whe But it is esily see tht oth L(,,) d L(,,) equl L(,,) Moreover, the umers re ll equl to L(,,) L(,,), L(,,) d L(,,) poits This completes the proof Therefore, L vishes for ll possile cofigurtios of these three Eercises Cosider the fuctio h defied y h() for [,] rtiol, d h() [,] irrtiol Show tht h is ot Riem itegrle Let f :[, ] Show tht f R [, ] if d oly if there eists such tht for every there eist tgged prtitios S( f ;)( P ;) S f Q Let H () k for /() k k d H () P d Q with P / d Q / for such tht elsewhere o [,] Use Eercise to show tht H is ot Riem itegrle 4 If () d () d if ()()() f for ll [,], does it follow from the Squeeze Theorem tht f R[,]? 5 If J is y suitervl of [, ] d if () for J d () elsewhere o[, ], we J sy tht J is elemetry step fuctio o[, ] Show tht every step fuctio is lier comitio of elemetry step fuctios 6 If :[, ] tkes o oly fiite umer of distict vlues, is step fuctio? 7 If S( f ;) P is y Riem sum of f :[, ], show tht there eists step fuctio :[, ] such tht S( f ;) P 8 Suppose tht f is cotiuous o[, ], tht f () for ll [, ] d tht f Prove tht f () for ll [, ] 9 Show tht the cotiuity hypothesis i the precedig eercise cot e dropped If f d g re cotiuous o [, ] d if f g, prove tht there eists c [, ] such tht f ()() c g c J Rel Alysis Pge 58

If f is ouded y M o[, ] d if the restrictio of f to every itervl [ c, ] where c (,) is Riem itegrle, show tht f R [, ] d tht f f c s c Show tht g() si(/) for (,] d g() elogs to R [,] Give emple of fuctio f :[, ] tht is i R [ c, ] for every c (,) ut which is ot i R [, ] 4 Suppose tht f :[, ], tht c c cm d tht the restrictios of f to [ ci, ci ] elog to R [ ci, ci ] for i,, m Prove tht f R [, ] d tht the formul i Corollry holds 5 If f is ouded d there is fiite set E such tht f is cotiuous t every poit of [, ]\ E, show tht f R [, ] 6 If f is cotiuous o[, ],, show tht there eists c [, ] f m i c c i i such tht we hve f f ()() c This result is sometimes clled the Me Vlue Theorem for Itegrls 7 If f d g re cotiuous o [, ] d g() for ll [, ], show tht there eists c [, ] such tht f g f () c g g() Show tht this coclusio fils if we do ot hve (Note tht this result is etesio of the precedig eercise) 8 Let f e cotiuous o[, ], let f () for ll [, ] lim() M sup{() : f [, ]} 9 Suppose tht d tht f R[, ] ) If f is eve (tht is, if f ()() f for ll [, ] ), show tht, d let / M f f f f ) If f is odd (tht is, if f ()() f for ll [, ] ), show tht f Show tht Suppose tht f :[, ] d tht Let P e the prtitio of [, ] ito suitervls hvig equl legths, so tht i ;() / i for i,,, Let L ()( f ;) S f P, l d R ()( f ;) S f P, r, where P, l hs its tgs t the left edpoits, d P, r hs its tgs t the right edpoits of the suitervls[ i, i ] ) If f is icresig o[, ], show tht L ()() f R f () ()()(()() R f L f f f d tht ) Show tht f ()()()()()() L f f R f f c) If f is decresig o[, ], oti iequlity similr to tht i () d) If f R[, ] R () f is ot mootoe, show tht f is ot ecessrily etwee L () f d Rel Alysis Pge 59

If f is cotiuous o[,], show tht Emie certi Riem sums] / f (cos) d / (si)(si) f d f d [Hit: If f is cotiuous o[, ], show tht f () d() f d Aswers If the tgs re ll rtiol, the S( h;) P, while if the tgs re ll irrtiol, the S( h;) P Let P e the prtitio of [,] ito equl suitervls with t / d Q e the sme suitervls tgged y irrtiol poits 5 If c,, c re the distict vlues tke y, the j jr j () c j { J,, J } of disjoit suitervls of [, ] We c write 6 Not ecessrily is the uio of fiite collectio rj c j k 8 If f () c for some c (,), there eists such tht f ()() f c for c The f c f ()() f c If c is edpoit, similr rgumet pplies c Use Bolzo s Itermedite Vlue Theorem (stted i chpter ) Let c () M d c () M for [,) c d ()()() w f for [ c, ] Apply the Squeeze Theorem for c sufficietly er ] Ideed, g() d is cotiuous o every itervl [ c,] where c The precedig eercise pplies Let f () / for (,] d f () 6 Let m if() f d M sup f The we hve m() f M () By Bolzo s Itermedite Vlue Theorem, there eists c [, ] such tht f () c f 9 () Let P e sequece of tgged prtitios of [, ] with P d let correspodig symmetric prtitio of [, ] Show tht Let i( / ) for i,,, The we hve tht i f (cos)(si) i i f k k Note tht f () is eve cotiuous fuctio c c j j jk P * * S( f ;) P ( ;) S f P f e the Rel Alysis Pge 6

6 FUNDAMENTAL THEOREMS OF CALCULUS I this chpter we will eplore the coectio etwee the otios of derivtive d itegrl The First form of Fudmetl Theorem of Clculus: If f is cotiuous t every poit of [, ] d F is y tiderivtive of f o [, ], the f ()()() d F F The First Form of the Fudmetl Theorem provides theoreticl sis for the method of clcultig itegrl It sserts tht if fuctio f is the derivtive of fuctio F, d if f elogs to R [, ], the the itegrl f c e clculted y mes of the evlutio F()() F A fuctio F such tht F()() f for ll [, ] is clled tiderivtive or primitive of f o[, ] Thus, whe f hs tiderivtive, it is very simple mtter to clculte its itegrl I prctice, it is coveiet to llow some eceptiol poits c where F() c does ot eist i, or where it does ot equl f () c It turs o ut tht we c permit fiite umer of such eceptiol poits We recll Me Vlue Theorem Me Vlue Theorem: Suppose tht f is cotiuous o closed itervl I [, ], d tht f hs derivtive i the ope itervl (,) The there eists t lest oe poit c i (,) such tht f ()() f f ()() c The followig is result eeded i the comig sectio Theorem If f : I hs derivtive t c I, the f is cotiuous t c Fudmetl Theorem of Clculus (First Form ): Suppose there is fiite set E i [, ] fuctios f, F :[, ] such tht: ) F is cotiuous o[, ], ) F()() f for ll [, ]\ E, The we hve f F ()() F () Proof We will prove the theorem i the cse where E {, } The geerl cse c e otied y rekig the itervl ito the uio of fiite umer of itervls Let e give By ssumptio (c), f R [, ] The there eists such tht if P is y tgged prtitio with P, the S( f ;) P f () d Rel Alysis Pge 6

If the suitervls i P re[ i, i ], the the Me vlue Theorem pplied to F o [ i, i ] implies tht there eists ui ( i,) i such tht F()()()() i F i F ui i i for i,, If we dd these terms, ote the telescopig of the sum, d use the fct tht F()() u i f u i, d oti i i f ()() ui i i i i F()()(()()) F F F Now letp {([, ],)} u i i ui i, the the sum o right equls S( f ;) P u F()()( F;) S f P u ito (), we coclude tht F()() F f If we sustitute But, sice is ritrry, we ifer tht equtio () holds This completes the proof Remrk If the fuctio F is differetile t every poit of [, ], the (y Theorem C) hypothesis () is utomticlly stisfied If f is ot defied for some poits c E, we tke f () c Eve if F is differetile t every poits of [, ], coditio ()c is ot utomticlly stisfied, sice there eist fuctio F such tht F is ot Riem itegrle (This is illustrted i the followig Emple 5) Emple If F() for ll [, ], the F() for ll [, ] Further, f F is cotiuous so it is i R[, ] Therefore the Fudmetl Theorem (with E ) implies tht ()()() d F F Emple If G() Arc t for [, ], the G() /( ) for ll [, ] ; lso G is cotiuous, so it is i R[, ] Therefore the Fudmetl Theorem (with E ) implies tht d Arc t Arc t Emple If A() for [,], the A() if [,) d A() for (,] The we hve A() sg() for ll [,]\{}, where for sg() for for (sg is clled the sigum fuctio) Sice the sigum fuctio is step fuctio, it elogs to R [,] Therefore the Fudmetl Theorem (with E {} ) implies tht sg()()( d ) A A Emple If H() for [, ], the H is cotiuous o [, ] d H() / for (, ] Sice h H is ot ouded o (, ], it does ot elog to R[, ] o mtter how we defie h () Therefore, the Fudmetl Theorem does ot pply Emple Let K() cos(/) Rule d the Chi Rule, tht for (,] d let K() It follows, pplyig the Product Rel Alysis Pge 6

() cos(/)(/)si(/) for (,] K Further, we hve K()() K cos() K() lim lim lim cos() Thus K is cotiuous d differetile t every poit of [, ] Sice the first term i K is cotiuous o [, ], it elogs to R [, ] However, the secod term i K is ot ouded, so it does ot elog to R [, ] Cosequetly KR [,], d the Fudmetl Theorem does ot pply to K The Fudmetl Theorem (Secod Form) We ow discuss the Fudmetl Theorem (Secod form) i which we wish to differetite itegrl ivolvig vrile upper limit Defiitio If f R [, ], the the fuctio defied y F() z z f for z [, ] () is clled the idefiite itegrl of f with se poit We will show tht if f R [, ], the its idefiite itegrl F stisfies Lipschitz coditio; hece F is cotiuous o[, ] We recll result i the chpter Riem Itegrle Fuctios Additivity Theorem: Let f :[, ] d let c (,) The f R[, ] restrictios to [, c] d [ c, ] re oth Riem itegrle I this cse c f f f c Theorem The idefiite itegrl F defied y F() z z f for z [, ] is cotiuous o[, ] I fct, if f () M for ll [, ], the F()() z F w M z w for ll z, w [, ] Proof The Additivity Theorem implies tht if z, w[, ] d w z, the d hece we hve z w z z, F()() z f f f F w f w w F()() z F w f w Now if M f () M for ll [, ], the Theorem D implies tht implies implies z z z M f M w w w z z z M f M w w w M ()() z w f M z w d hece it follows tht z w z if d oly if its Rel Alysis Pge 6

F()() z F w f M z w z w This completes the proof We will ow show tht the idefiite itegrl F is differetile t y poit where f is cotiuous Fudmetl Theorem of Clculus (Secod Form): Let f R [, ] d let f e cotiuous t poit c [, ] The the idefiite itegrl, defied y F() z z f for z [, ], is differetile t c d F()() c f c Proof We will suppose tht c [,) d cosider the right-hd derivtive of F t c Sice f is cotiuous t c, give there eists such tht if c c, the f ()()() c f f c (4) Let h stisfy h The Additivity Theorem implies tht f is itegrle o the itervls[, c ], [, c h] d [ c, c h] d tht ch F()() c h F c f c Now o the itervl [ c, c h] the fuctio f stisfies iequlity (4), so tht (y Theorem D) we hve ch c ch ch ch (())(()) f c f f c c c c Sice f F()() c h F c, the ove implies (())()()(()) f c h F c h F c f c h If we divide y h d sutrct f () c, we oti F()() c h F c h f () c which implies F()() c h F c h f () c But, sice is ritrry, we coclude tht the right-hd limit is give y F()() c h F c lim() f c h h It is proved i the sme wy tht the left-hd limit of this differece quotiet lso equls f () c whe c (, ] This completes the proof If f is cotiuous o ll of [, ], we oti the followig result Theorem If f is cotiuous o [, ] the the idefiite itegrl F, defied y F() z z f for z [, ], is differetile o [, ] d F()() f for ll [, ] Remrk The ove Theorem c e summrized s follows: Rel Alysis Pge 64

If f is cotiuous o[, ], the its idefiite itegrl is tiderivtive of f I geerl, the idefiite itegrl eed ot e tiderivtive (either ecuse the derivtive of the idefiite itegrl does ot eist or does ot equl f () ) This is illustrted i the followig emples Emple If f () sg o[,], the If, the If, F()() f d ( ) the ` F()() f f f [( )] [ ] Thus f R [,] d hs the idefiite itegrl F() with the se poit However, sice F() does ot eist, F is ot tiderivtive of f o[,] Emple If h deotes Thome s fuctio h :[,] defied y defied y h() if [,] is irrtiol, h() d y h() / if [,] is the rtiol umer m/ where m, hve o commo iteger fctors ecept The its idefiite itegrl H () h is ideticlly o[, ] Here, the derivtive of this idefiite itegrl eists t every poit d H() But H()() h wheever [,], so tht H is ot tiderivtive of h o[,] Sustitutio Theorem The et theorem provides the justifictio for the chge of vrile method tht is ofte used to evlute itegrls This theorem is employed (usully implicitly) i the evlutio y mes of procedures tht ivolve the mipultio of differetils Sustitutio Theorem: Let J [, ] d let : J hve cotiuous derivtive o J If f : I is cotiuous o itervl I cotiig ()J, the () f (())()() t t dt f d (5) Proof Defie u () () F()() u f d for u I, d H ()(())()() t F t F t fort J The H()(())() t f t t for t J d tht () () f ()(())()(())() d F H f t t dt The hypotheses tht f d re cotiuous re restrictive, ut re used to esure the eistece of the Riem itegrl o the left side of (5) Emple Evlute the itegrl 4 si t dt t Here we sustitute ()t t for t [,4] so tht () t /() t is cotiuous o[,4] If we let f () si, the the itegrd hs the form ()f d the Sustitutio Theorem implies tht the itegrl equls si d cos (cos cos ) Rel Alysis Pge 65

4 si t Emple Cosider the itegrl dt Sice ()t t does ot hve cotiuous t derivtive o[,4], the Sustitutio Theorem is ot pplicle, t lest with this sustitutio Eercises Eted the proof of the Fudmetl Theorem to the cse of ritrry fiite set E If d tht H for [, ], show tht the Fudmetl Theorem implies () /( ) d () /( ) If g() for d g() g()()( d ) G 5/ G 4 Let B() for d B() 5 Let f :[, ] d let C Wht is the set E here? for for d if () If :[, ] is tiderivtive of f o[, ], show tht C ()() C is lso tiderivtive of f o [, ] () If d re tiderivtive of f o[, ], show tht is costt fuctio o [, ] 6 If f R [, ] d if c [, ], the fuctio defied y F () z G(), show tht Show tht d B ()() B idefiite itegrl of f with se poit c Fid reltio etwee 7 Thome s fuctio is i R [,] c c z f for z [, ] is clled the F d F c with itegrl equl to C the Fudmetl Theorem e used to oti this coclusio? Epli your swer 8 Let F() e defied for y F()( )( ) / for [,), Show tht F is cotiuous d evlute F() t poits where this derivtive eists Use this result to evlute d for, where [ ] is clled the gretest iteger fuctio, for emple, [7] 7, [ 8] ) 9 Let f R [, ] d defie F() () Evlute G() () Evlute H () (c) Evlute S() c f for [, ] f i terms of F, where c [, ] f i terms of F si f i terms of F deotes the gretest iteger i (The fuctio [ ], Let f :[, ] e cotiuous o [, ] d let v [ c, d] e differetile o [ c, d] with v([ c, d]) [, ] If we defie G() Fid F() v () whe F is defied o [,] y: f, show tht G()(())() f v v for ll [ c, d] Rel Alysis Pge 66

) F()() t dt ) F() Let f :[,] e defied y f () Oti formuls for differetile? Evlute F() If f : F() t dt for, f () for d f () t ll such poits is cotiuous d c, defie g : y g()() differetile o d fid g () 4 If f :[,] is cotiuous d f for d sketch the grphs of f d F Where is F c c f t dt Show tht g is f f for ll [,], show tht f () for ll [,] 5 Use the Sustitutio Theorem to evlute the followig itegrls ) t t dt ) / t () t dt 4 t 4 cos t c) dt, d) dt t t 6 Sometimes the Sustitutio Theorem cot e pplied ut the followig result, clled the Secod Sustitutio Theorem is useful I dditio to the hypotheses of Sustitutio Theorem, ssume tht () t for ll t J, so the fuctio :() J iverse to eists d hs derivtive (()) t /() t The To prove this, let () f (())()() t dt f d () G()(()) t t f s ds for t J, so tht G()(()) t f t Note tht K()(()) G is differetile o the itervl ()J K()(())() G f (())()()() f Clculte G()(()) K i two wys to oti the formul 7 Apply the Secod Sustitutio Theorem to evlute the followig itegrls 9 ) dt t ) dt ( ) t t c) 4 tdt t d) 4 dt Arct() Arct(/ ) t ( t 4) d tht 8 Epli why Sustitutio Theorem d/or Secod shiftig Theorem cot e pplied to evlute the followig itegrls, usig the idicted sustitutio ) c) 4 tdt () t t ) t ()t dt t t d) cos tdt () t t t dt () t Arcsi t t 4 Rel Alysis Pge 67

7 POINTWISE AND UNIFORM CONVERGENCE Let A e give d suppose tht for ech there is fuctio f : A ; we sy tht () f is sequece of fuctios o A to Clerly, for ech A, sequece of fuctios gives rise to sequece of rel umers, mely the sequece (()) f, () otied y evlutig ech of the fuctios t the poit For certi vlues of A the sequece () my coverge, d for other vlues of A this sequece my diverge For ech A for which the sequece () coverges, there is uiquely determied rel umer lim(()) f I geerl, the vlue of this limit, whe it eists, will deped o the choice of the poit A Thus, there rises i this wy fuctio whose domi cosists of ll umers A for which the sequece () coverges Defiitio follows: Let () f e sequece of fuctios o A to, let A A, d let f : A We sy tht the sequece () f coverges o A to f if, for ech A, the sequece (()) f coverges to f () i I this cse we cll f the limit o A of the sequece () f Whe such fuctio f eists, we sy tht the sequece () f is coverget o A, or tht () f coverges poitwise o A Ecept for possile modifictio of the domi A, the limit fuctio is uiquely determied Ordirily we choose A to e the lrgest set possile; tht is, we tke A to e the set of ll A for which the sequece () is coverget i If the sequece () f coverges o A to f, we deote it y f lim() f o A, or f f o A Sometimes, whe d f re give y formuls, we write f f () lim() f for A, or f ()() f for A We eed the followig result Theorem Let Y () y e sequece of rel umers tht coverge to y d let The the sequece coverge to y ie, Y () y lim() y lim() y Emple Show tht lim( /) for For, let f () / d let f () for For, it follows tht lim(()) f lim( /) lim(/),, s lim(/) usig Theorem A Rel Alysis Pge 68

Hece for ll, lim( /) Emple Discuss the poitwise covergece of the sequece of fuctios () Let g () for, Clerly, if, the the sequece (()) g is the costt sequece () d hece coverges to It follows from the result if, the lim() tht lim() for d it is redily see tht this is lso true for If, the g ( )( ), d sice (( )) is diverget, the sequece (( g )) is diverget Similrly, if, the the sequece () is ot ouded, d so it is ot coverget i Hece if for, g() for, the the sequece () g coverges to g o the set (,] Reso: If this sequece is coverget, it must e ouded (y Theorem B), which is ot the cse Theorem A coverget sequece of rel umers is ouded Emple Let The Show tht lim(() /) for (Fig ) h ()() / for,, d let h() lim(()) h lim lim lim(), usig Theorem A for We c write () h lim, usig Theorem A d otig tht for fied, the limit of the costt sequece () is The (())() h h for ll Emple Show tht lim(()si()) for (Fig 4) Let F ()()si() for,, d let F() for Sice si y for ll y we hve F ()() F si() () for ll Therefore it follows tht lim(()) F () F for ll It should ote tht, give y, if is sufficietly lrge, the F ()() F for ll vlues of is simulteously! Rel Alysis Pge 69

Gettig ituitio from the remrk ove, we hve the followig reformultio of the defiitio of poitwise covergece Lemm A sequece () f of fuctios o A to coverges to fuctio f : A o A if d oly if for ech d ech A there is turl umer K(,) such tht if K (,), the f ()() f () Emple Let f () for [,),,, We show tht () f f where f () for ll [,) For, f () For fied (,), it follows tht coverges poit wise to lim(()) f lim, usig property of Limit of sequece of rel umers lim() lim(( /)), s lim(/), s lim(/) Hece () f coverges poit wise to f where f () for ll [,) The ove c e doe s follows lso: For y [,),() f s Now, we let f () Let e give We hve to fid K(,) such tht if K (,), the Now implies f ()() f implies implies Now if we choose Rel Alysis Pge 7

k(,) itegrl prt of (), whe, whe ie, k(,) s the smllest positive iteger greter th or equl to, the for (,), k(,) Hece, y Lemm, () f coverges poit wise to f where f () for ll [,) Uiform Covergece Defiitio A sequece () f of fuctios o A to coverges uiformly o A A to fuctio f : A if for ech there is turl umer K() (depedig o ut ot o A ) such tht if K(), the f ()() f for ll A (4) I this cse we sy tht the sequece () f is uiformly coverget o A Sometimes we write f f o A, or f ()() f for A It is immedite cosequece of the defiitios tht if the sequece () f is uiformly coverget o A to f, the this sequece lso coverges poitwise o A to f i the sese of Defiitio of poitwise covergece Tht the coverse is ot lwys true is see y creful emitio of erlier Emples i tht sectio; other emples will e give elow It is sometimes useful to hve the followig ecessry d sufficiet coditio for sequece () f to fil to coverge uiformly o A to f The proof of the followig result requires oly tht tke the egtio of Defiitio of uiform covergece Lemm A sequece () f of fuctios o A to does ot coverge uiformly o A A to fuctio : f A if d oly if for some sequece () k i A such tht f ()() f for ll k k k k Emple Let f (), d lim(()) f for ll Let () there is susequece () f k of () f d We hve oted i erlier emple tht f for ll The f coverges to f poitwise If k we let k k d k k, the f () k k so tht f ()() k k f k Let The, k y the lemm ove, the sequece () f does ot coverge uiformly o to f Emple Let g (),(, ], We hve see i erlier emple tht g coverges to g o the set (, ], where g() for for Rel Alysis Pge 7

If k k k k, the k k k d / g ()() g k k k k g (), so tht Let 4 (,] to g The, y the lemm ove, the sequece () g doesot coverge uiformly o Emple Let () h, We hve see i erlier emple tht () h ll, where h() for ll If k k d k Let k h ()() k k k k, the h () k k d h() k k so tht h ()() k k h k k k The, y the Lemm ove, the sequece () h doesot coverge uiformly o to h Emple Show tht the sequece (()) f where () f, is uiformly coverget i the closed ouded itervl [, m], whtever m my e, ut ot i the itervl Here for y give, f () lim() f lim lim ( /) = The for give, f ()(), f if ie, if ie, if ie, if Let K(,) the smllest iteger greter th (/ ) Also, we ote tht K(,) s icreses d teds s such tht for icreses Hece it is ot possile to choose positive iteger K() f ()() f for every K() d for every vlue of i [, ] Hece the covergece is ouiform i [,) Rel Alysis Pge 7

If we cosider y fiite itervl [, m] where m is fied umer, the the mimum vlue of (/ ) o [, m] is m(/ ) so tht if we tke The K() y iteger greter th m(/ ), f ()() f for every K() d for every i [, m ] This shows tht the sequece (()) f coverges uiformly i the itervl [, m] where m is y fied positive umer Emple Show tht, where Solutio Here is poit of ouiform covergece of the sequece (()) f i f () 4 / f () lim(()) f lim lim 4 4 (/) = for For give, we hve if ie, if ie, if ie, if f ()() f, 4 4 4 4 Also () shows tht if, so tht it is ot possile to choose K() such tht f ()() f for every K() d for every [, ] Hece the sequece is ouiformly coverget i [, ] But if we cosider the itervl [ m, ] where m, the covergece is uiform sice i tht cse it is possile to tke 4 K() iteger just greter th Hece is the poit of ouiform covergece of the give sequece i [, ] Emple Test for uiform covergece the sequece e for Let f () e The lim(()) f lim() e whe whe () Rel Alysis Pge 7

whe Let f () whe The for ech [,), lim (())() f f Hece () f coverges poit wise to f i [,) Now for y d (,), if ie, if f ()() f e log e ie, if log Hece choosig K(,) = the smllest iteger greter th (/)log(/) we get, for ll (,) s ()() S for ll K(,) Here, oviously, K(,) depeds o We ote tht K(,) teds to ie, K(,) icreses d teds to s is ot ouded fuctio of o (,) Hece it is ot possile to fid turl umer K(), depedig oly o, such tht s ()() S wheever K() d [,) Hece () f is ot uiformly coverget o [,) If, however, we cosider the itervl [,) where is y fied rel umer greter th zero, however smll, the K(,) is ouded fuctio o [,) The mimum vlue of K(,) o [,) the is the iteger just greter th (/)log(/) K() the iteger just greter th (/)log(/), f ()() f wheever K() d [,) Hece if we tke Hece the sequece () f is uiformly coverget i [,) where, coverget i [,) ut ouiformly Emple Show tht if g () for, the () g coverges uiformly o [,) / For y, lim(()) g lim lim / Hece () g coverges poit wise to g where g() for o [,) Now for y d (,), if g ()() g Rel Alysis Pge 74

ie, if ie, if Tke K(,) the positive iteger just greter th The for ll (,), g ()() g for ll k(,) Here K(,) = the positive iteger just greter th is ouded ove y d is fuctio of o (,) The mimum vlue of K(,) o (,) is the iteger just greter th / Hece if we tke K() iteger just greter th /, the g ()() g for ll K() d (,) But whe,() g for ll, d hece we hve g ()() g for ll K() d [,) Hece the sequece () g is uiformly coverget o [,) The Uiform Norm I discussig uiform covergece, it is ofte coveiet to use the otio of the uiform orm o set of ouded fuctios If A d : A is fuctio, we sy tht is ouded o A if the set ()A is ouded suset of If is ouded we defie the uiform orm of o A y sup{() : } A (6) A Note tht it follows tht if, the () for ll A (7) Lemm A sequece () f d oly if f f A A of ouded fuctios o A coverges uiformly o A to f if Proof () If () f coverges uiformly o A to f, the y the Defiitio of uiform covergece, give y there eists K() such tht if K() d A the f ()() f From the defiitio of supremum, it follows tht f f wheever K() A Sice is ritrry this implies tht f f A () If f f, the give there is turl umer H () such tht if H () A the f f It follows from (7) tht f ()() A f for ll H () d A Therefore () f coverges uiformly o A to f This completes the proof We ow illustrte the use of Lemm s tool i emiig sequece of ouded fuctios for uiform covergece Rel Alysis Pge 75

Emple Show tht lim( /) for Is the covergece uiform o? Is the covergece uiform o A [, ]? Epli For, let f () / d let f () for By Emple, for ll, lim( /), so tht the sequece of fuctios () f coverges poit wise to f I prticulr, for, ( /) coverges to To emie the whether the sequece of fuctios is uiformly coverget o, we cot pply Lemm sice the fuctio f ()() f is ot ouded o But we hve verified i erlier emple tht this covergece is ot uiform o Now we emie the uiform covergece o A [, ] To see this, we oserve tht (for fied ) f sup : sup : f A so tht f f s Therefore, y Lemm, () f is uiformly coverget o A to f A We coclude tht lthough the sequece ( /) does ot coverge uiformly o to the zero fuctio, the covergece is uiform o A Emple Let g () for A [, ] d, d let g() for d g() The fuctios g ()() g re ouded o A d, we hve for y, Sice g g A g for g sup A for = does ot coverge to, we ifer, y pplyig Lemm, tht the sequece () g does ot coverge uiformly o A to g Emple We cot pply Lemm to the sequece i Emple sice the fuctio h ()() / h is ot ouded o Isted, let A [,8] d cosider 64 h h sup : 8 A Therefore, the sequece () h coverges uiformly o A to h Emple We hve see from () tht F F Hece () F coverges uiformly o to F First Derivtive Test for Etrem Let f e cotiuous o the itervl I [, ] d let c e iterior poit of I Assume tht f is differetile o (,) c d ( c,) The: () If there is eigourhood ( c,) c I such tht f () for c c d f () for c c, the f hs reltive mimum t c () If there is eigourhood ( c,) c I such tht f () for c c d f () for c c, the f hs reltive miimum t c Rel Alysis Pge 76

Emple Let G()() for A [,] The the sequece (()) G coverges to G() for ech A To clculte the uiform orm G G G o A, we fid the derivtive d solve G ()(( )) to oti the poit This is iterior poit of [,], d it is esily verified y usig the First derivtive Test tht G ttis mimum o [,] t Therefore, we oti G G ()() A which coverges to () e Thus we see tht covergece is uiform o A By mkig use of the uiform orm, we c oti ecessry d sufficiet coditio for uiform covergece tht is ofte useful Cuchy Criterio for Uiform Covergece of Sequece of Fuctios: Let () f e sequece of ouded fuctios o A The this sequece coverges uiformly o A to ouded fuctio f if d oly if for ech there is umer H () i such tht for ll m,() H, the f m f A Proof () If f f o A, the give K() the f f Hece if oth m,() K A f ()()()()()() f f f f f m m there eists turl umer, the we coclude tht K such tht if for ll A Therefore fm f, for m,()() K H A () Coversely, suppose tht for there is H () such tht if m,() H fm f Therefore, for ech A A we hve f ()() f f f for m,() H (8) m m A, the It follows tht (()) f is Cuchy sequece i ; therefore, eig Cuchy sequece, it is coverget sequece We defie f : A y f () lim(()) f for A From (8), we hve for ech A f ()() f for m,() H m Now fi m H () d let, the we oti for A, f ()() m f Hece for ech A, we hve f ()() m f for m H () Therefore the sequece () f coverges uiformly o A to f This completes the proof Eercises Show tht lim( /()) for ll, Rel Alysis Pge 77

Show tht lim( /()) for ll Evlute lim( /()) for, 4 Evlute lim( /()) for, 5 Evlute lim((si) /()) 6 Show tht lim(arct)( / ) sg for, 7 Evlute lim() e for, 8 Show tht lim() e for, 9 Show tht lim() e d tht Show tht lim((cos)) Show tht if, itervl [, ], Show tht if, itervl [, ], Show tht if, itervl [, ], for, where sg is the sigum fuctio e for, lim() eists for ll Wht is its limit? the the covergece of the sequece i Eercise is uiform o the ut is ot uiform o the itervl [,) the the covergece of the sequece i Eercise is uiform o the ut is ot uiform o the itervl [,) the the covergece of the sequece i Eercise is uiform o the ut is ot uiform o the itervl [,) 4Show tht if, the the covergece of the sequece i Eercise 4 is uiform o the itervl [, ], ut is ot uiform o the itervl [, ] 5Show tht if, itervl [,), the the covergece of the sequece i Eercise 5 is uiform o the ut is ot uiform o the itervl [,) 6Show tht if, the the covergece of the sequece i Eercise 6 is uiform o the itervl [,), ut is ot uiform o the itervl (,) 7Show tht if, the the covergece of the sequece i Eercise 7 is uiform o the itervl [,), ut is ot uiform o the itervl [,) 8Show tht the covergece of the sequece i Eercise 8 is uiform o [,) 9 Show tht the sequece () e Show tht if, the the sequece coverges uiformly o [,) ut tht it does ot coverge uiformly o the itervl [,) () e coverges uiformly o the itervl [,), Show tht if (),() f g coverge uiformly o the set A to f, g, respectively, the () f g coverges uiformly o A to f g Show tht if f () / d f () for, the () f coverges uiformly o to f, ut the sequece () f does ot coverge uiformly o (Thus the product of uiform ly coverget sequeces of fuctios my ot coverge uiformly) Let (),() f g e sequeces of ouded fuctios o A tht coverge uiformly o A to f, g, respectively Show tht () f g coverges uiformly o A to fg 4 Let () f e sequece of fuctios tht coverges uiformly to f o A d tht stisfies f () M for ll d ll A If g is cotiuous o the itervl[ M, M ], show tht the sequece () g f coverges uiformly to g f o A Rel Alysis Pge 78

8 UNIFORM CONVERGENCE AND CONTINUITY We first list some emples which shows tht, i geerl, limit of sequece of cotiuous fuctios eed ot e cotiuous Similrly, limit of sequece of differetile (resp, Riem itegrle) fuctios eed ot e differetile (resp, Riem itegrle) I this tet we discuss the limit of sequece of cotiuous fuctios oly We will cosider results tht revel the importce of uiform covergece tht llows iterchge of limits which mkes the limit of sequece of cotiuous fuctio lso cotiuous Let g () for [,] fuctio for for g() Although ll of the fuctios d The, the sequece () g coverges poitwise to the g re cotiuous t, the limit fuctio g is ot cotiuous t Recll tht it ws show i Emple 9 i the previous chpter tht this sequece does ot coverge uiformly to g o[,] Ech of the fuctios g () for [,] d hs cotiuous derivtive o[,] For g [, ],() However, the limit fuctio g does ot hve derivtive t it is ot cotiuous t tht poit Let f :[,] e defied for y for f ()() for for It is cler tht ech of the fuctios f, sice is cotiuous o[,] ; hece it is Riem itegrle We ote tht for [,], f () s Also, for, descried s elow: / / ()() / / f d d d d / / () / for f () d This c e Rel Alysis Pge 79

If [, ] d, we hve so tht f () s Hece the limit fuctio f is such tht f () for [, ] Beig costt fuctio, f is cotiuous o [, ] d hece is Riem itegrle o [, ] d I this emple, f () d lim() f d lim() f dlim(), f d so tht iterchge of limits is ot possile Emple Cosider the sequece () h Sice h H, where H () e h ()()() d H H e f () d of fuctios defied y, the Fudmetl Theorem gives For [,], h () s Hece if we tke h() lim(()) h e h() lim(()) h for ll [,] Hece h () d lim() h d Tht is iterchge of limits is ot possile i this cse lso Iterchge of Limit d Cotiuity Therefore we hve h () e for [,],, we hve Theorem Let () f e sequece of cotiuous fuctios o set A d suppose tht () f coverges uiformly o A to fuctio f : A The f is cotiuous o A Proof By hypothesis, give there eists turl umer H H () such tht if H the f ()() f for ll A Let c A e ritrry; we will show tht f is cotiuous t c By the Trigle Iequlity we hve Sice f ()()()()()()()() f c f fh fh fh c fh c f c H f ()() f c H H f is cotiuous t c, there eists umer (, c,) f H such tht if c d A, the f ()() f c Therefore, if c d A, the we hve f ()() f c H H Sice is ritrry, this estlishes the cotiuity of f t the ritrry poit c A Hece f is cotiuous t every poit i A Hece f is cotiuous o A This completes the proof Although the uiform covergece of the sequece of cotiuous fuctios is sufficiet to gurtee the cotiuity of the limit fuctios, it is ot ecessry This is illustrted i the followig emple Emple Rel Alysis Pge 8

Let f () for ll [,],,, The / lim(()) f lim lim for ll [, ] Let us defie f :[,] y f () for ll [, ] The () f coverges poitwise to f o [, ] Here clerly ech f d f re cotiuous o [,] But f f sup()() f : [,] [,] f sup : [,] () To fid sup : [,] ie, the mimum vlue of o the set [, ], we c use the Differetitio Method By the method, the fuctio y, tti etremum whe y ie, whe () () ie, whe ie, whe ie, whe Hece y ttis mimum i [,] t d the mimum vlue is Hece from (), we hve f f [,] Hece f f [,] s f f does ot ted to zero s teds to ifiity, () f is ot uiformly coverget to Sice [,] f o[,] Eercises Show tht the sequece (( /()) the limit fuctio is ot cotiuous o[,] Let f :[,] e defied for does ot coverge uiformly o [,] y showig tht y Rel Alysis Pge 8

for / f ()() for / / for / Prove tht the sequece i () f is emple of sequece of cotiuous fuctios tht coverges o-uiformly to cotiuous limit Costruct sequece of fuctios o [,] ech of which is discotiuous t every poit of [,] d which coverges uiformly to fuctio tht is cotiuous t every poit 4 Suppose () f is sequece of cotiuous fuctios o itervl I tht coverges uiformly o I to fuctio f If () I coverges to I, show tht lim(())() f f 5 Let f : e cotiuous o d let f()( /) f for Show tht () f coverges uiformly o to f 6 Let f () /() for [,] Fid the poitwise limit f of the sequece () f o[,] Does () f coverges uiformly to f o[,]? 7 Suppose the sequece () f coverges uiformly to f o the set A d suppose tht ech f is ouded o A (Tht is, for ech there is costt Show tht the fuctio f is ouded o A 8 Let f () /() for A [,) Show tht ech f poitwise limit f of the sequece is ot ouded o A Does () f A? M such tht f () M for ll A ) is ouded o A, ut the coverge uiformly to f o Rel Alysis Pge 8

If () f 9 SERIES OF FUNCTIONS is sequece of fuctios defied o suset D of with vlues i, the sequece of prtil sums () s of the ifiite series f is defied for i D y s ()() f s ()()() s f s ()()() s f I cse the sequece () s series of fuctios f of fuctios coverges o D to fuctio f, we sy tht the ifiite coverges to f o D Nottio We will ofte write f for f to deote either the series or the limit fuctio, whe it eists If the series () f coverges for ech i D, we sy tht f is solutely coverget o D If the sequece () s of prtil sums is uiformly coverget o D to f, we sy tht f is uiformly coverget o D, or tht it coverges to f uiformly o D Oe of the mi resos for the iterest i uiformly coverget series of fuctios is the vlidity of the followig results which give coditios justifyig the chge of order of the summtio d other limitig opertios Theorem If f is cotiuous o D to for ech d if f coverges to f uiformly o D, the f is cotiuous o D We ote tht Theorem is direct trsltio of the followig Theorem for series Theorem A Let () f e sequece of cotiuous fuctios o set A d suppose tht () f coverges uiformly o A to fuctio f : A The f is cotiuous o A Emple We ow show tht the series 4 4 4 4 4 4 4 4 4 () ()() is ot uiformly coverget o [, ] Discuss the cotiuity of the sum fuctio Let f () () 4 4 d N N s ()() f () N 4 4 Rel Alysis Pge 8

Hece s N () N () 4 4 4 () 4 N () 4 N () 4 4 4 N () 4 N if () Whe, f () for ll d hece s () Hece s () s N For, usig (), sn () () 4 4 N 4 N s N 4, Hece if f () the, () s N coverges poit wise to f d hece f coverges poitwise to f o [, ] Ech f () is cotiuous fuctio o [, ] But its limit f () is ot cotiuous t Hece (with the id of Theorem ) f does ot coverge uiformly o [, ] Emple We ow show tht the series ( ) [( ) ] is uiformly coverget o y itervl [, ],, ut oly poit wise o [, ] For ech, let f () ( ) [( ) ]( ) ( ) The the sequece of prtil sums of the series f () Whe N sn ()() f, () N ( ) N whe N is give y (s other terms, occur i pirs with sme solute vlue ut with opposite sigs d hece, get ccelled) f for d hece () N s for ll N Comiig the ove, we hve Rel Alysis Pge 84

, if sn () N, if If lim(()) sn lim N d lim() s N Therefore () s N d hece f coverges poit wise to f where, f (), For ech, f () is cotiuous o [, ] ut f () is ot cotiuous o [, ] Hece, pplyig Theorem, f does ot coverge uiformly o [, ] Now let J [, ], where The s f sup {()() S : [ f, ]} N J N sup : [, ] N, sice sup : [, ] N s N N Hece () s N coverges uiformly to f o [, ] Hece the ssocited series f coverges uiformly to f o [, ] (where Tests for Uiform Covergece We ow preset few tests tht c e used to estlish uiform covergece We first recll Cuchy Criterio for Series of rel umers d the descrie Cuchy Criterio for Series of Fuctios Cuchy Criterio for Series: The series coverges if d oly if for every there eists M () such tht if m M (), the m lso Cuchy Criterio for Series of Fuctios: Let () f e sequece of fuctios o D to The series f is uiformly coverget o D if d oly if for every M () such tht if m M (), the f ()() fm for ll D Proof Let () s N e the sequece of Nth prtil sums of the series f The there eists Rel Alysis Pge 85

N s ()()()() f f f for ll D N i N i By defiitio of uiform covergece of series, the ifiite series fi coverges uiformly o D if d oly if () s N coverges uiformly o D Now, y Cuchy Criterio for Uiform Covergece of Sequece of Fuctios, () s N coverges uiformly o D if d oly if for every, there eists turl umer M () ie, ie, m sm s D for ll m M () (depedig oly o ) such tht s ()() s for ll m M () d D m i i i f ()() f for ll m M () d D i m i i f () for ll m M () d D ie, f ()()() f fm for ll m M () d D Weierstrss M-test: Let () M e sequece of positive rel umers such tht f() M for D, If the series M is coverget, the f is uiformly coverget o D Proof If m, we hve the reltio f ()() f M M for D () m m Now y Cuchy Criterio for Series of Rel Numers (Theorem A), the series coverget implies for every there eists M () such tht if m M (), the Hece from (), for m M () M M m f ()() fm for D Hece y Cuchy Criterio for series of Fuctios (Theorem ), the series coverget o D Emple Test for uiform covergece the series f M is is uiformly!!! To discuss the uiform covergece of the give series of fuctios, we cosider the series eglectig The!!! the th term of the remiig series is f ()! Rel Alysis Pge 86

The f ()!! for Notig tht! d the series is coverget, we hve, with the id of compriso test of series of rel umers, the series is lso coverget! Thus y Weierstrss s M-test, it follows tht the give series is uiformly coverget for ll vlues of [, ] Emple We ow show tht p q is uiformly coverget for ll vlues of if p Here we hve f () for ll vlues of p q p It is kow tht the p-series p is coverget if p Now tke M The y the discussio ove, p M is coverget if p We coclude from Weierstrss s M- test tht the give series is uiformly coverget for ll vlues of, if p Emple We ow show tht the series 5 ( ) ( )!! 5! is uiformly coverget i every itervl i [, ] Let M e y positive umer greter th ech of d so tht for every [, ], we hve ( ) M ( )!( )! Now cosider the series 5 M M M ()! 5! Sice M M M e M,!! d M M M e M,!! Rel Alysis Pge 87

we hve M M 5 e e M M M ()! 5! The idetity () shows tht the series () is coverget, whtever the positive costt M my e Hece y Weierstrss s M-test, the give series ( ) ( )! is uiformly coverget i [, ] Remrk The series of fuctios give ove is the importt sie series I the previous emple we hve verified the uiform covergece of the sie series 5 si! 5! i [, ] Emple We ow show tht the series cos cos cos 4 cos 4 coverges uiformly, d lso, give the itervl of uiform covergece Here () cos u for ll fiite vlues of So tkig M d otig tht M is coverget, with the id of Weierstrss s cos M-test, we hve the uiform covergece of i y fiite itervl The itervl of uiform covergece is where d re y fiite uequl qutities Eercises Discuss the covergece d the uiform covergece of the series f give y ) () ) ()( ) c) si( /) d) ( )( ) e) /( )( ) f) ( )()( ) Aswers/Hits for selected Eercises () Tke M / i the Weierstrss M-test, where f () is (c) Sice si y y, the series coverges for ll But it is ot uiformly coverget o If, the series is uiformly coverget for (d) If, the series is diverget If, the series is coverget It is uiformly coverget o [,) for However, it is ot uiformly coverget o (,) Rel Alysis Pge 88

IMPROPER INTEGRALS OF FIRST KIND - PART I Evlutio of the defiite itegrls of the type f () d is required i my prolems So fr you my hve ee see those defiite itegrls tht hve the followig two properties: The domi of itegrtio, from to, is fiite The rge of the itegrd is fiite (i other words, the itegrd is defied t every poit i the domi) l Now cosider the defiite itegrls d d d Property is violted i the first defiite itegrl while Property is ot stisfied t y the secod itegrl I this chpter we cosider the methods for the evlutio of such defiite itegrls A improper itegrl is defiite itegrl which does ot stisfy Property or give ove ie, itegrls with ifiite limits of itegrtio d itegrls of fuctios tht ecome ifiite t poit withi the itervl of itegrtio re improper itegrls Whe the limits ivolved eist, we c evlute such itegrls We discuss this i this chpter i detil Improper Itegrl of First Kid The defiitio or evlutio of the itegrl f () d does ot follow from the discussio o Riem itegrtio sice the itervl [,) is ot ouded Such itegrl is clled improper itegrl of first kid The theory of this type of itegrl resemles to gret etet the theory of ifiite series We defie f () d s follows: If f R [, s] for every s >, the f () d is defied to e the ordered pir f, F s where F()()() s f d s Itegrls of the type discussed i this sectio re sometimes clled improper itegrls of the first kid Rel Alysis Pge 89

Covergece of Improper Itegrl We sy tht f is coverget to A if lim() F s s does ot coverge, we sy tht f is diverget Emple We ow show tht the itegrl d is coverget s With the usul ottio, we hve F() s d, d the F() s d hece lim() F s Thus s s d A I this cse we write f A If f Emple We ow show tht the itegrl The itegrl diverges sice d d s F() s d ( s ) lim() F s does ot eist s d diverges l Emple We ow vlute d, if it eists s Here F() s d Recll the itegrtio y prts formul, uv uv u v ie, u ()()()()()() dv u v du v Tkig u() l,() dv d, we oti s l ()(l) s F s d d l s s s s Rel Alysis Pge 9

Hece l s s s l s l s lim() F slim lim s s s s s s L lim / s, pplyig L hospitl rule s Hece l lim() d F s s Emple We ow Ivestigte the covergece of d d d d lim d liml s l s s Hece the itegrl s d s is diverget Now, d lim d lim + = s s s Hece the itegrl d is coverget d its vlue is RESULT: If f d oth coverge d c R, the () () f () cf g, coverges d coverges d g () f g f g cf c f Tests for Covergece d Divergece I prctice, most of the improper itegrl cot e evluted directly So we tur to the twostep procedure of first estlishig the fct of covergece d the pproimtig the itegrl umericlly The pricipl tests for covergece re the direct compriso d limit compriso tests Direct Compriso Test Let f d g e cotiuous o [,) d suppose tht ()()f g for ll The Rel Alysis Pge 9

f () d coverges if g () d coverges g () d diverges if f () d diverges Emple By defiitio, We ow vestigte the covergece of e d lim e d We cot evlute the ltter itegrl directly ecuse it is o-elemetry But we c show tht its limit s is fiite We kow tht e e d d is icresig fuctio of Therefore either it ecomes ifiite s or it hs fiite limit s It does ot ecome ifiite: For every vlue of we hve e e, Also, lim lim lim e d e d 6788 e e e e Hece y Direct compriso Test, e d lim e d coverges to some defiite fiite vlue We do ot kow ectly wht the vlue is ecept tht it is somethig less th 7 Emple We o ivestigte the covergece of si d We kow tht si o [,) d d coverges Hece y Direct compriso test, si d coverges Emple We ow Ivestigte the covergece of d diverges ecuse o [,) d d d diverges RESULT: The improper itegrl d diverges Proof For y iteger N we hve N N N d d d N N k (4) k Agi, sice s N the right side of (4) diverges to ifiity, we see tht lim s d s does ot eist This proves the theorem Rel Alysis Pge 9

RESULT: d, where p is costt d, coverges if p d diverges if p p Proof But Hece p s d lim d lim, p s p s p provided p p p p lim s s lim s p s p p p if p, lim s s if p s, provided p provided p d is coverget whe p d diverget whe p p Also, whe p, Therefore, Emple Show tht But s d lim d liml s l p s s d coverges if p d diverges if p p e t d where t is costt, coverges if t d diverges if t t s t t e e d lim e d lim s s t e t st t st lim e e e lime s t t s ts if t, lim e s if t Hece t coverges if t d diverges if t Whe t, Therefore, s e d d lim d lim() s s s e t coverges if s t d diverges if t Limit Compriso Test/Quotiet Test: If the positive fuctios f d g re cotiuous o [,) d if Rel Alysis Pge 9

f () lim() g () L L the f () d d() g d oth coverge or oth diverge Emple Usig d d the Limit compriso Test, we ow discuss the covergece of d Also compre the vlues of ech itegrl With f () / d() g /(), we hve f () lim lim / g() /() = lim lim, positive fiite limit Therefore, d coverges ecuse d coverges However, the itegrls coverge to differet vlues By erlier Emple, d Also, Emple We kow tht g() : e 5 s d lim d s lim t s s 4 4 We ow show tht d coverges e 5 d e lim t t coverges Now we use Limit Compriso Test with f () lim lim / e lim e 5 g() /( e 5) e = lim 5, e positive fiite limit As, e 5 e y Limit Compriso Test, we coclude tht d coverges e 5 f () e d Emple We ow test for covergece the itegrl Let f () The d g() d Rel Alysis Pge 94

f (), s g() f () Sice lim, fiite o zero umer, y Quotiet Test, g() coverge or diverge together But g() d d d f () d coverges f () d d g() d coverges Hece Emple We ow test for covergece the itegrl Let The Sice f () 5 d g() 5 d 5/ f () s g() 5 5 / f () lim, fiite o zero umer, y Quotiet Test g() coverges or diverge together But g() d d d / f () d is diverget 5 Emple We ow test for covergece the itegrl f () d d g() d is diverget s e d Hece Sice is ot poit of ifiite discotiuity, we hve to emie the covergece t oly e d e d e d () Sice first itegrl o the right is proper itegrl we eed oly test the covergece of e d Now, Hece e e, for ll rel for ll Sice d coverges, y Compriso Test, we hve eig the sum of two coverget itegrls, e d coverges e d coverges Hece usig (), Emple We ow test for covergece the itegrl log d Rel Alysis Pge 95

Let The log f () d g() f () log log g() / s log Here g() d d diverges Hece y Quotiet test, () f d d diverges Defiitio If f R [, s] for every s d if () f d coverges, we sy tht f () d coverges solutely RESULT: If s F() s () f d d if F is ouded (ove) o [,), the lim() F s eists d hece f () d coverges solutely RESULT: If f () d coverges solutely, if g [, s] for every s, d if () g () f (), the g() d coverges solutely Proof s s s G() s () g () d () f d f d Hece G is ouded ove o[,) d the solute covergece of g() d follows from the precedig theorem RESULT: If () f d coverges solutely, the f () d coverges Proof f () is either f () or f () I either cse f ()() f d () () f () f f Hece () f () () f f (), () d sice (y ssumptio) () f f () d coverges implies lim() F s d coverges s eists where F()() s s f d Beig coverget prtil itegrl F is ouded (fuctio) Hece from iequlities i (), s s f ()()()() f d f f d is ouded Rel Alysis Pge 96

Hece y the theorem ove, lim()() f f d s eists d hece f ()() f d coverges solutely It follows tht f () () f d coverges solutely Sice f () () f this mes simply tht f () () f d f d () () () () coverges But, sice () coverges, it follows (y sutrctio) tht f f d f d f d itself coverges This completes the proof cos Emple Prove tht d coverges cos cos We hve for ll sice cos d cos The, sice coverges, y compriso test, it follows tht d coverges, ie, cos d coverges solutely Sice, every solutely coverget itegrl coverges, cos d coverges If f () d coverges ut does ot coverge solutely, we sy tht f () d coverges coditiolly si Emple Show tht d is coditiolly coverget improper itegrl Also fid the si vlue of d si To show tht d coverges, we hve for y s (usig itegrtio y prts) s s si cos s cos d d () s cos Now () s Sice d cos coverges (solutely) it follows tht d coverges solutely d hece coverges Thus s s ll terms o the right of () pproch limits Thus Rel Alysis Pge 97

s si lim d s si eists, which proves tht d coverges Lettig s, () gives si cos d d Note tht the secod itegrl is solutely coverget while the first itegrl (s we shll ow show) is ot si Now we show tht d does ot coverge solutely For y N, we hve N Now si d N ( )( ) N si d ( ) N si() u du si() u si cos u cos si u si cos u si d Sice cos this shows tht si() u si u Hece if u, the si() u si u Thus N N si d Sice the series k N N si u du k () k is diverget, the right side of () c e mde s lrge s we plese y k tkig N sufficietly lrge This d () show tht si lim s d s si cot eist Hece d does ot coverge solutely Eercises I Eercises -, discuss the covergece of the followig improper itegrl of the first kid d d d 4 4 4 d si 5 d 6 d log 4 7 d 8 4 e d d 9 4 5 d Rel Alysis Pge 98

4 d cos d d (log) ` 4 d () 4 5 d 6 d () log cos 7 d 8 d 9 d 6 d 4 6 d d 4 d 4 5/ si d 5 Show tht d d ()() si 6 Show tht d is coverget d deduce tht d cos d is coverget 7 Show tht () d 4 8 Show tht the itegrl 9 True or flse? If f is cotiuous o[,) d if si Show tht d is coverget si p d where p is solutely coverget If f () d coverges d if lim() f, L prove tht L= Give emple of cotiuous fuctio f such tht f () (), d such tht ut f () f () d coverges diverges Give emple of cotiuous fuctio f such tht f () () d such tht f () d coverges f () d coverges, the lim() f Rel Alysis Pge 99

ut f () diverges si 4 Show tht, p d p is coverget ut ot solutely coverget si 5 Discuss solute covergece of d 6 If f () (), if f is oicresig o[,), d if f () d coverges, the show tht lim() f cos 7 Show tht d is coditiolly coverget 8 Let f e cotiuous fuctio o [,), such tht, if F()() f t dt (), the f is ouded o (,) Let g e fuctio o [,) such tht g is cotiuous o [,), g() t for t, d such tht lim() g t f ()() t g t dt coverges t 9 Use the precedig eercise to show tht si t dt coverges logt 4 Show tht cosu du d 4 Evlute d 4 Evlute / d 4 Evlute 44 Evlute e d d 45 Evlute 46 Show tht e d is coverget is coverget Prove tht Rel Alysis Pge

IMPROPER INTEGRALS OF FIRST KIND - PART II Itegrl Test for Series We hve just used the divergece of ifiite series to estlish the divergece of improper itegrl It is more usul to use itegrl i the ivestigtio of series This is kow s the Itegrl Test for Series RESULT Let f e oicresig fuctio o [,) such tht f () diverges f () will coverge if f () d Proof For y we hve f ()()( f ) f ( ) coverges, d sice f is oicresig Itegrtig from to we the hve or or Thus, for N If f ()()( d ) f d f d f ()()( d ) f d f d f ()()( ) f d f we hve N N N N k f ()()( )() f d f f k (5) f () d coverges to A, the, y (5), N k The prtil sums of f ()() k f d A N k f () k f () will diverge if () The f () d re thus ouded ove ie, the sequece of Nth prtil sums of the series is ouded d hece the sequece of the Nth prtil sum is coverget, d hece the series If f () d k f () k coverges d hece diverges, the divergece of k f () f () k coverges usig the left-hd iequlity i (5) This completes the proof Emple Usig itegrl test for series estlish the covergece of If f () /(), the f () Sice f () d d is coverget, it follows tht f () coverges my e estlished i similr fshio, Rel Alysis Pge

Emple Show tht [/( log)] diverges If g() /( log), the g is oegtive d o-decresig o [,) Sice G()() g where G() log log, the d hece diverges g() s g () d log log s log log, diverges It follows from Itegrl Test for Series tht Emple Usig the Itegrl Test, show tht the p-series () p p p P p (p rel costt) coverges if p, d diverges if p Cse If p, the p p f () is positive decresig fuctio of for p p d d lim s p lim ( ) p s p s p p Hece d p s, sice p s s s for p Now, coverges d hece, y the Itegrl Test, the give series coverges Cse If p, the p d lim( p ), d s s lim s p for p p p s s Hece, y the Itegrl Test, the series diverges for p If p, we hve the hrmoic series, which is kow to e diverget, Hece we coclude tht the series coverges for p ut diverges for p [/( log)] Itegrls of the form f () d A itegrl of the form f () d my e treted y the sme methods s those used o itegrls of the form f () d Thus, we sy tht f () d coverges to A if lim() f d s s A Rel Alysis Pge

The chge of vrile u will chge prolem ito Emple Emie the covergece or divergece of For y s we hve s d ( ) du du s s u u Sice for u, it follows, from the fct u u du u lim s s d d prolem is diverget, tht does ot eist Hece lim d does ot eist, which proves tht s s coverge (I this prolem the divergece of d ws deduced from the divergece of du ) u d does ot Emple Test for covergece We write 6 d d d d 6 6 6 Now ()() y y d () dy 6 6 () [Puttig y ] y () y y dy 6 y y y y y dy dy y y 6 6 y y Tkig f () y d g() y, we hve 6 y y f ()() y y / y y y lim lim lim, 6 6 g() y y / y y y 6 dy y o zero fiite umer Hece, y Quotiet Test, d dy y diverge together Sice dy coverges, y Sice y 6 y y 6 y y dy is proper itegrl, it follows tht y dy coverges 6 d coverges coverge or Rel Alysis Pge

6 d Similrly, we c show tht coverges Hece, (eig sum of two coverget improper itegrls) is coverget 6 d Emple Test the covergece of ( ) d lim ( )( ) d lim lim ( ) ( ) lim Hece, the give itegrl is coverget d hs vlue Emple Test the covergece of cosh d cosh d lim cosh d lim sih e e lim sih sih lim Hece, the give itegrl is diverget Emple Evlute d d d d lim d lim lim t lim t d lim t t lim t t t t, sice lim t d lim t Hece, the give itegrl is coverget d hs vlue Emple The cross sectios of the solid hor i Fig perpediculr to the -is re circulr disks with dimeters rechig from the -is to the curve y e, l Fid the volume of the hor Rel Alysis Pge 4

Solutio For typicl cross sectio, rdius is y d re is A()(rdius) y e 4 Now the volume of the hor i the Fig is s V of this portio is l A() d ie, the volume of the hor is the limit of the volume V of the portio from to l By the method of slicig, the volume l V A() l 4 d e d e 8 l 4 ()(4) e e e 8 8 As, e d( V /8)(4 ) / Hece l A() d lim V ie, the volume of the give hor is / l Emple Test for covergece Let y The, Now, e y y e d y e for ll y () dy dy coverget Hece y compriso test, e d lso coverges Eercises I Eercises -, discuss the y e y e d e y y d, y Emple ove with y e dy y e y y dy coverges Thereforee y d t, we hve e dy covergece of the followig improper itegrl of the first kid is Rel Alysis Pge 5