UNIT I. Definition and existence of Riemann-Stieltjes

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1 1 UNIT I Defto d exstece of Rem-Steltjes Itroducto: The reder wll recll from elemetry clculus tht to fd the re of the rego uder the grph of postve fucto f defed o [, ], we sudvde the tervl [, ] to fte umer of sutervls, sy, the k th sutervl hvg legth x k, d we cosder sums of the form f ( t k k ) x 1 k where t k s some pot the k th sutervl. Such sum s pproxmto to the re y mes of rectgles. If f s suffcetly well ehved [, ] cotuous. The two cocepts, dervtve d tegrl, rse etrely dfferet wys d t s remrkle fct deed tht the two re tmtely coected. If we cosder the defte tegrl of cotuous fucto f s fucto of ts upper lmt, sy we wrte F ( x) f ( t) dt. The F hs dervtve d F'(x) = f(x). Ths mportt result shows tht dfferetto d tegrto re, sese, verse opertos. I ths ut we study the process of tegrto some detl. Actully we cosder more geerl cocept th tht of Rem mely Rem-Steltjes tegrl, whch volves two fuctos f d α. The symol for such tegrl s f ( x) d ( x) or somethg smlr, d the usul Rem tegrl occurs s the specl cse whch α(x) = x. Whe α hs cotuous dervtve, the defto s such tht the Steltjes tegrl ecomes the Rem tegrl. However, the Steltjes tegrl stll mkes sese whe α s ot dfferetle or eve whe α s dscotuous. Prolems physcs whch volve mss dstrutos tht re prtly dscrete d prtly cotuous c lso e treted y usg Steltjes tegrls. I the mthemtcl theory of prolty ths tegrl s very useful tool tht mkes possle the smulteous tretmet of cotuous d dscrete rdom vrles. Ojectve: The Rem-Stelges Itegrl s sed o the defto of Rem Itegrl whch we hd studed prevous clsses for the ske of coveece we re gvg the defto d prelmres of Rem Itegrls. Accordgly, we eg y dscussg tegrto of relvlued fuctos o tervls The Rem-Steltjes Itegrl Deftos: Let [, ] e gve tervl. The set P = {x 0, x 1,, x -1, x } of [, ] such tht = x 0 x 1 x -1 x = s sd to e Prtto of [, ]. The set of ll prttos of [, ] s deoted y P([, ]). The tervls [x 0, x 1 ], [x 0, x 1 ], [x 1, x 2 ],, [x -1, x ] re clled the sutervls of [, ]. Wrte x = x x -1 s clled the legth of the tervl [x -1, x ] ( = 1,, ) d mx x s clled the orm of the prtto P d s deoted y P or (P). A prtto Q of [, ] such tht P Q s clled the refemet or fer of the prtto P. Suppose f s ouded rel vlued fucto defed o [, ] d x

2 2 M = sup f(x), m = f f(x) (x -1 x x ) for ech P P( [, ]). The U(P, f ) = M x d L(P, f ) = m x re clled the Upper d Lower Rem sums =1 =1 or Upper d Lower Droux sums of f o [, ] wth respect to the prtto P. Further wrte - f dx = f U(P, f) d f dx = sup L(P, f) - where the f d the sup re tke over ll prttos P of [, ] re clled the Upper d Lower Rem tegrls of f over [, ], respectvely. If the upper d lower Rem tegrls re equl, we sy tht f s Rem-tegrle o [, ] d we wrte f R[, ] d we deote the commo vlue of these tegrls y f d(x), -.e., f dx = f dx = f dx. - Aother defto of Rem Itegrl : A ouded fucto f : [, ] R s Rem Itegrle f for > 0 there exsts > 0 such tht for y prtto P = {x 0, x 1, x 2, x } wth P < d [x -1, x ] f( )( x x -1 ) - f < Lemm. If f : [, ] R s ouded fucto the the upper d lower Rem tegrls of f re ouded. Sce f s ouded, there exst two umers m d M such tht m f(x) M ( x ). Hece, for every prtto P of [, ] we hve m m M M m x m x M x M x, = 1, 2, 3,,. m( ) L(P,f) U(P,f) M(-), so tht the umers L(P,f) d U(P,f) form ouded set. Therefore y the defto of lower d upper Rem tegrls ths shows tht the upper d lower tegrls re defed for every ouded fucto f re ouded lso. The questo of ther equlty, d hece the questo of the tegrlty of f, s more delcte oe. Here we stte some other vestgtos of Rem tegrls Lemm. If f : [, ] R s ouded fucto, P s y prtto of [, ] d P* s the refemet of P, the L(P, f) L(P*, f) d U(P*, f) U(P, f) Lemm. If f : [, ] R s ouded fucto d P 1, P 2 re y two prttos of [, ] the L(P 1, f) U(P 2, f) d L(P 2, f) U(P 1, f).

3 Lemm. If f, g : [, ] R re ouded fuctos d P s y prtto of [, ] the () L(P, f + g) L(P, f) + L(P, g) () U(P, f + g) U(P, f) + U(P, g) Theorem. If f : [, ] R s ouded fucto the - f dx f dx Theorem (Droux). If f : [, ] R s ouded fucto the for > 0 there exsts > 0 such tht - U(P, f) < f + d L(P, f) > f dx Theorem. If f : [, ] R s ouded fucto s Rem Itegrle f the osclltory sum <,.e. (P, f) = U(P,f) L(P,f) <, for > 0 d y prtto P of [, ] Theorem. Every cotuous fucto f : [, ] R s Rem Itegrle Theorem. Every mootoe fucto f : [, ] R s Rem Itegrle. Studets you studed the propertes gve ove d other propertes of Rem Itegrls prevous clsses therefore we re ot terested to vestgte these here. However we shll mmedtely cosder more geerl stuto Defto. Let f : [, ] R s ouded fucto d e mootoclly cresg fucto o [, ]. Let P = {x 0, x 1,, x -1, x } such tht = x 0 x 1 x -1 x = e y Prtto of [, ]. We wrte =(x ) - (x -1 ), = 1, 2, 3,,. By the defto of mootoe fucto () d () re fte therefore s ouded o [, ], lso sce s mootoclly cresg fucto the clerly 0, = 1, 2, 3,,. Let M = sup f(x), m = f f(x) (x -1 x x ) for ech P P( [, ]). We defe U(P, f, ) = M, d L(P, f, ) = m, =1 =1 re clled the Upper d Lower Rem Steltjes sums respectvely. Further we defe - f d = f U(P, f, ) d f d = sup L(P, f, ), - where the f d the sup re tke over ll prttos P of [, ], re clled the Upper d Lower Rem Steltjes tegrls of f over [, ], respectvely. If the upper d lower Rem Steltjes tegrls re equl, we sy tht f s Rem Steltjes tegrle o [, ] d we wrte f R() d deote the commo vlue of these tegrls y f d or f(x) d(x).

4 4 Ths s the Rettm-Steltjes tegrl (or smply the Slelljes tegrl of f wth respect to over [,]. If we put (x) = x we see tht the Rem tegrl s the specl cse of the Rem- Stetjes tegrl Lemm If f: [, ] R s ouded fucto d e mootoclly cresg fucto o [, ]. Let P e y Prtto of [, ].The the upper d lower Rem-Stetjes tegrls of f wth respect to re ouded. Proof. Sce f s ouded, there exst two umers m d M such tht m f(x) M ( x ). Hece, for every prtto P of [, ] we hve m m M M m m M M, = 1, 2, 3,,. m[() ()] L(P, f, ) U(P, f, ) M[() ()], so tht the umers L(P, f, ) d U(P, f, ) form ouded set. Therefore y the defto of lower d upper Rem-Stetjes tegrls ths shows tht the upper d lower tegrls re defed for every ouded fucto f re ouded lso Lemm. If P* s refemet of the prtto P of [, ], the L(P, f, ) L(P*, f, ) d U(P*, f, ) U(P, f, ). Proof. Let P = {x 0, x 1,, x -1, x } such tht = x 0 x 1 x -1 x = e y Prtto of [, ] d P* the refemet of P cots just oe pot X* more th P such tht x -1 < x* <x where x -1 d x re two cosecutve pots of P. Let m, m, m re the fmum of f(x) [x -1, x ], [x -1, x*] d [ x*, x ] respectvely the clerly m m d m m. Therefore L(P*,f,) - L(P,f, ) = m [(x*) - (x -1 )] + m [(x ) - (x*)] - m [(x ) - (x -1 )] = m [(x*) - (x -1 )] + m [(x ) - (x*)] - m [(x ) - (x*) + (x*) - (x -1 )] = (m - m )[(x*) - (x -1 )] + (m - m )[(x ) - (x*)] 0. Hece L(P, f, ) L(P*, f, ). If P* cots k pots more th P the y repetg the sme process we rrve t the sme result. The proof of U(P*, f, ) U(P, f, ) s logous Lemm. If P 1, P 2 re y two prttos of [, ] the L(P 1, f, ) U(P 2, f, ) d L(P 2, f, ) U(P 1, f, ). Proof. We hve P 1, P 2 re y two prttos of [, ]. Let P = P 1 P 2, the P s the commo refemet of P 1, P 2. The L(P 1, f, ) L(P, f, ) ; U(P, f, ) U(P 1, f, ), d L(P 2, f, ) L(P, f, ) ; U(P, f, ) U(P 2, f, ). Hece L(P 1, f, ) U(P 2, f, ) d L(P 2, f, ) U(P 1, f, ) Theorem: f dα f dα - Proof. Sce f P s the commo refemet of two prttos P 1 d P 2 of [, ]. The

5 5 L(P 1, f, α) L(P*, f, α) U(P*, f, α) U(P 2, f, α). Therefore for ll prttos P 1 d P 2 of [, ], sup L(P 1, f, α) f U(P 2, f, α), mples tht - f dα f dα Theorem. f R(α) o [, ] f d oly f for every ε > 0 there exsts prtto P such tht U(P, f, α) - L(P, f, α ) < ε.... () Proof. Let for prtto P of [, ], () holds. The y the defto of Rem Stetjes tegrls - L(P, f, α ) f dα f dα U(P, f, α ). - Ths mples tht - 0 f dα - f dα U(P, f, α) - L(P, f, α ) < ε - - Hece f dα = f dα.e., tht s f R(α). - - Coversely suppose tht f R(α), the f dα = f dα = f dα. Now f ε > 0 e gve y defto of Rem Stetjes tegrls there exst prttos P 1 d P 2 such tht f dα < L(P 1, f, α ) + ε/2 d U(P 2, f, α ) - ε/2 < f dα. Let P = P 1 P 2 e the commo refemet of P 1 d P 2. The U(P, f, α ) U(P 2, f, α ) < f dα + ε/2 < L(P 1, f, α ) + ε L(P, f, α ) + ε. Yelds, U(P, f, α) - L(P, f, α ) < ε Theorem. () If for every ε > 0 there exsts prtto P such tht U(P,f, α) - L(P, f, α ) < ε holds the t s holds (wth the sme ) for every refemet of P. () If for every ε > 0 there exsts prtto P = {x o, x 1,..., x } such tht U(P, f, α) - L(P, f, α) < ε holds d f s, t, re rtrry pots [x -1, x ], the f(s ) f(t ) I. =1 (c) If f R() d the hypotheses of () hold, the f(t ) f d. =1 Proof: () Let P* e y refemet of the prtto P, the we kow tht L(P, f, ) L(P*, f, ) d U(P*, f, ) U(P, f, ), therefore U(P*, f, ) - L(P*, f, ) U(P, f, ) - L(P, f, ) < ε. Sce P* s rtrry, hece the cocluso holds for ll refemets of P. () Let M = sup f(x), m = f f(x) x -1 x x the for rtrry pots s, t [x -1, x ], m f(s ) M d m f(s ) M, so tht f(s ) - f(t ) M - m. Thus f(s ) f(t ) M m = U(P, f, ) - L(P, f, ) < ε.

6 6 =1 (c) Sce f f R() the y the ovous equltes L(P, f, ) f(t ) I U(P, f, ) d L(P, f, ) f d U(P, f, ), mples tht f(t ) f d U(P, f, ) - L(P, f, ) < ε. = Theorem. If f s cotuous o [, ] d s mootoc cresg o [, ] the f R() o [, ]. Proof. Let > 0 e gve. Choose η > 0 so tht () - () < ε/ η. Sce f s cotuous the tervl [, ] so t s uformly cotuous o [, ], there exsts δ > 0 such tht f(x) - f(t) η ; x - t δ x, t [, ]. (*) If P = {x 0, x 1,, x -1, x } prtto of [, ] such tht P < δ d M = sup f(x), m = f f(x) (x -1 x x ) the x < δ for ll, therefore y (*), we hve M - m η ( = 1,, ). Yelds tht U(P, f, ) - L(P,f,) = (M I - m ) η = η[() - ()] <. =1 =1 Hece f R() Theorem. If f s mootoc o [, ] d f s cotuous o [,], the f R(). (We stll ssume, of course, tht s mootoc.) Proof. Let > 0 e gve. Sce s cotuous o [, ] d so t tkes ll vlues etwee () d () lo t s mootoclly cresg. Therefore for postve teger there exts prtto P = {x 0, x 1,, x -1, x } of [, ] such tht =[() - ()] / ( = 1,, ). We suppose tht f s mootoclly cresg the M = supf(x) = f(x ) d m = f(x) = f(x -1 ) (x -1 x x ) so tht U(P, f, ) - L(P, f, ) = (M m ) = [{() - ()} /] {f (x ) f(x -1 ) } =1 =1 Hece f R(). = [{() - ()}{f() f()}] / < s s tke lrge eough Theorem. Suppose f s ouded o [,], f hs oly ftely my pots of dscotuty o [, ], d s cotuous t every pot t whch f s dscotuous. The f R(). Proof. Let > 0 e gve. Let M = sup f(x), x d let E e the set of pots t whch f s dscotuous. So E s fte d s cotuous t every elemet of E, we c cover E y ftely my dsjot tervls [u j, v j ] [, ] such tht the sum of the correspodg dffereces (v j ) - (u j ) s less th, m.e., [(v j ) - (u j )] <. =1

7 7 Furthermore wthout loss geerlty we my plce these tervls such wy tht every pot of E (, ) les the teror of some [u j, v j ],.e., x E [u j, v j ] x s teror pot of [u j, v j ]. m Let K = [, ] - (u j, v j ) the the set K s compct. Hece f s uformly cotuous o K so =1 there exsts δ > 0 such tht f(s) - f(t) < ; s - t < δ, s, t K. (1) Cosder prtto P = {x o, x,..., x } of [, ] such tht () ech u j, v j P () ll x (u j, v j ) x P () f x -1 u j the x < δ for ll. The M m = M m M + m 2 M = 2 sup f(x) 2 sup f(x) M 2M. Now y (), we hve M m f x [, ] d x -1 u j. Hece U(P, f, ) - L(P, f, ) = (M m ) = 1 (M m ) (x = u j ) + 2 (M m ) (x u j ) =1 =1 =1 2M M + [() - ()] =. Hece f R(). =1 =1 Questo: Is f f d hve commo pot of dscotuty, the f R()? Theorem. Suppose f R() o [, ], m f M, s cotuous o [m, M], d h(x) = (f(x)) o [, ]. The h R() o [, ]. Proof. Choose > 0. Sce s uformly cotuous o [m, M], there exsts δ > 0 such tht δ < d (s) - (t) < f s - t δ s, t [m, M]. Further sce f R(), there s prtto P = {x o, x 1, x 2,, x } of [, ] such tht U(P, f, ) - L(P, f, ) < δ 2 () Let M, m d M *, m * re the supremum d fmum of f d h respectvely [x -1, x ]. Cosder two clsses : A f M - m < δ, M *- m * d B f M - m δ, M *- m * 2K, where K= sup (t), m t M. By (), we hve δ (M - m ) I < δ 2 mples tht < δ. It follows tht B B B U(P, h, ) - L(P, h, ) = (M *- m *) + ( M * - m *) B B 1.2. Propertes of Rem-Stetjes Itegrls [() - ()] + 2Kδ < [() - () + 2K] =. Hece h R().

8 Theorem () lf f 1, f 2 R() o [, ], the f 1 + f 2 R(), cf R() for every costt c, d (f 1 +f 2 )dα = f 1 dα + f 2 dα ; cf dα = c f dα. () If f 1 (x) f 2 (x) o [,], the f 1 dα f 2 dα. (c) If f R() o [, ] d f < c <, the f R() o [, c] d o [c, ], d c f dα + f dα = f dα. c (d) If f R() o [, ] d f f(x) M o [, ], the f dα M[() - ()]. (e) If f R( 1 ) d f R( 2 ), the f R( ) d f d( ) = f dα 1 + f dα 2 ; If f R() d c s postve costt, the f R(c) d f d(cα) = c f dα. Proof. () Sce f 1, f 2 R() the for gve > 0 there re prttos P j (j = 1, 2) such tht U(P j, f j, ) - L(P j, f j,) < /2, j = 1, 2. () If P = P 1 P 2 the () rems true for P, U(P j, f j, ) - L(P j, f j,) < /2, j = 1, 2. () Now f f = f 1 + f 2, M, M, M, m, m, m re the supremum d fmum x [x -1, x ] of f, f 1, f 2 respectvely the M M I + M, m m + m so tht U(P, f, ) = M (M + M ) = M + M =1 =1 =1 =1 U(P, f 1, ) + U(P, f 2, ). Smlrly, L(P, f, ) L(P, f 1, ) + L(P, f 2, ). Therefore U(P, f, ) L(P, f, ) [U(P, f 1, ) + U(P, f 2, ) {L(P, f 1, ) + L(P, f 2, )}] [U(P, f 1, ) - L(P, f 1, )] + [U(P, f 2, ) - L(P, f 2, )}] <, y () whch proves tht f = f 1 + f 2 R() so tht Further for y > 0 d prtto P of [, ] - f d = f dα = f dα. - U(P, f 1, ) < f 1 d + /2 d U(P, f 2, ) < f 2 d + /2 the o ddg these, we hve

9 9 U(P, f 1, ) + U(P, f 2, ) < f 1 d + f 2 d + U(P, f, ) < f 1 d + f 2 d + - Sce fd = fd < f 1 d + f 2 d + fd f 1 d + f 2 d () O replcg f 1 d f 2 () y - f 1 d -f 2, we hve fd f 1 d + f 2 d (v) From () d (v), we hve (f 1 + f 2 )d = f 1 d + f 2 d. Smlrly t c e prove tht f = f 1 - f 2 R() d (f 1 +f 2 )dα = f 1 dα + f 2 dα ; cf dα = c f dα. Next for the secod prt f c = 0 the the proof s trvl, ut f c 0, the sy c > 0 d let f R() the for gve > 0 there s prtto P of [, ] such tht U(P, f, ) L(P, f, ) < /c, (v) Also we defe (cf)(x) = cf (x) so tht, sup(cf)(x) = c supf (x), d f(cf)(x) = c f f(x),x -1 x x of [, ]. Whch mples tht, U(P, cf, ) = c U(P, f, ) d L(P, cf, ) = cl(p, f, ). Therefore U(P, cf, ) - L(P, cf, ) = c[u(p, cf, ) - L(P, cf, )] <, y (v). Hece cf R().Further f c < 0 the U(P, cf, ) - L(P, cf, ) = (-c)[u(p, cf, ) - L(P, cf, )] <, y (v). Hece cf R(). - Therefore (cf)dα = (cf) dα = (cf) dα d for l c 0, (cf) dα = (cf)dα = f U(P, cf, α) = c f U(P, f, α) = c f dα = c f dα.. () Let P e y prtto of [, ] d f(x) 0 for ll x, s mootoclly cresg fucto o [, ] the = (x ) - (x -1 ) 0 lso f m = f f(x) x -1 x x, the L(P, f, ) 0. Cosequetly fd = sup L(P, f, ) 0. Therefore f f 1 (x) f 2 (x) o [,] the f 2 (x) f 1 (x) = (f 2 f 1 )(x) 0 mples tht 0 (f 2 f 1 )d = f 2 d - f 1 d f 1 dα f 2 dα. (c) Sce f R() the for gve > 0 there exst prtto P = {x 0, x 1,, x -1, x } of [, ] such tht U(P, f, ) - L(P, f, ) < (v) where U(P, f, ) = M, d L(P, f, ) = m,, M = sup f(x), m = f f(x)(x -1 x x ) =1 r =1 Let x r = c, the U(P, f, ) = M, + M, d L(P, f, ) = m, + m. r

10 10 Therefore =1 r =r+1 =1 =r+1 U(P, f, ) - L(P, f, )= (M m ), + (M m ) <, y (v) r =1 =r+1 whch mples tht (M m ), < d (M m ), <. =1 =r+1 Hece f R() o [, c] d o [c, ]. Further sce f R() o [, c] d o [c, ] therefore for > 0, we hve c f(t ) - fd < /3, f(t ) - fd < /3 d f(t ) - fd < /3. [, c] [c, ] c [, ] Therefore, c c fd - fd - fd = f(t ) - fd - [f(t ) - fd]- [f(t ) - fd] c [, ] [, c] [c, ] c Lettg 0, we hve (d) Sce c f(t ) - fd + f(t ) - fd + f(t ) - fd <. [, ] [, c] [c, ] c c fd = fd + fd. c fd f d (we wll prove ths result ext theorem). If f(x) M o [, ], the fd Md = M[() - ()]. (e) We hve gve tht f R( 1 ) d f R( 2 ) the for gve > 0 there exsts prttos P 1 d P 2 of [, ] such tht U(P, f, ) - L(P, f, ) < /2, j = 1, 2 () If P = P 1 P 2 the U(P, f, ) - L(P, f, ) < /2, j = 1, 2 () Suppose tht = the, we c wrte M [(x ) - (x -1 )] = M [( )(x ) ( )(x -1 )] =1 =1 = M [ 1 (x ) - 1 (x -1 )] + M [ 2 (x ) - 2 (x -1 )] =1 =1.e., U(P, f, ) = U(P, f, 1 ) + U(P, f, 2 ), smlrly L(P, f, ) = U(P, f, 1 ) + L(P, f, 2 ). So, U(P, f, ) - L(P, f, ) = [U(P, f, 1 ) - L(P, f, 1 )] + [U(P, f, 2 ) - L(P, f, 2 )] < y (). Hece f R( ). Further f f R( = ) d c s y postve costt the for prtto P of [, ], we hve U(P, f, ) - L(P, f, ) < cu(p, f, ) - L(P, f, ) < c U(P, f, c) - L(P, f, c) < c f R(c) Further f d(c) = fu(p, f, c) = c fu(p, f, ) = c f dα.

11 Theorem. If f R() d g R() o [, ] the () fg R() ; () f R() d f dα f dα. Proof. () Sce f f R(), g R() o [, ] the f + g, f - g R() o [, ]. lso f theorem f we tke (t) = t 2 the h(x) = (f(x)) [f(x)] 2 R() o [, ]. Therefore y detty 4fg = (f + g) 2 - (f - g) 2 R() or fg R(). () Sce f R() the for gve > 0 there s prttos P such tht U(P, f, ) - L(P, f, ) <, () (M - m ) <, where M, m re the supremum d fmum of f x [x -1, x ]. =1 Now f M, m re the supremum d fmum of f x [x -1, x ]. The for x r, x s [x -1, x ] we hve f(x r ) f(x s ) f(x r ) f(x s ) mples tht M - m M - m (M - m ) (M - m ) <, =1 =1 U(P, f, ) - L(P, f, ) <. Hece f R(). Also sce M M therefore M M f dα f dα.. =1 =1 Note : Here we ote the verse of ths theorem 1.2.2() s ot true lwys Defto. (step fucto) The ut step fucto I s defed y I(x) = 0, (x 0), 1, (x > 0) Theorem. If < s <, f s ouded o [, ], f s cotuous t s d (x) = I(x - s), the f dα = f(s). Proof. Let P = {x 0, x 1, x 2, x 3 ), where x o = d x 1 = s < x 2 < x 3 = e y prtto of [, ]. The M = supf(x) = f(s) d m = ff(x) = f(s), x [x -1, x ], = 1, 2, 3 s f s cotuous t s. 3 3 Therefore U(P, f, α) = M = M [(x ) - (x -1 )] = M 1 [(x 1 ) - (x 0 )] + M 2 [(x 2 ) - (x 1 )] =1 =1 + M 3 [(x 3 ) - (x 2 )] = M 1 [I(x 1 - s) I(x 0 s)] + M 2 [I(x 2 - s) I(x 1 s)] + M 3 [I(x 3 - s) I(x 2 s)] = M 2. Smlrly L(P, f, α) = m = m 2. Sce f s cotuous t s so tht M 2 d m 2 coverge to f(s) s x 2 s. Therefore - f dα = f U(P, f, α) = f(s) d f dα = supl(p, f, α) = f(s). Hece

12 f dα = f dα = f dα = f(s) Theorem. Suppose c 0 for = 1, 2, 3,..., c coverges, {s } s sequece of dstct pots (, ), d (*) α(x) = c I(x - s ). Let f e cotuous o [, ]. The =1 f dα = c f(s ). =1 Proof. Sce c I(x - s ) c, the y the comprso test the seres (*) s coverges for every x. Also ts sum α (x) s evdetly mootoc d α() = 0, α() = c. Also sce c coverges the for gve > 0 choose N so tht c <, N Put α 1 (x) = c I(x - s ) d α 2 (x) = c I(x - s ). =1 N+1 By theorem & 1.2.3, we hve f dα 1 = c f(s ) d f dα 2 M. =1 where M = sup f(x). Sce α = α 1 + α 2 N fdα - c f(s ) M. =1 Lettg N, we hve f dα = c f(s ). =1 N+1 the α 2 = α α 1, t follows tht Theorem.(Relto etwee Rem tegrl d Rem-Stelges tegrl) Assume α creses mootoclly d α R[, ]. Let f e ouded rel fucto o [, ]. The f R(α) f d oly f f α R[, ]. Moreover f dα = f (x) α(x)dx () Proof. Let > 0 e gve, sce α R[, ] therefore there s prtto P = {x 0,..., x } of [, ] such tht U(P, α') - L(P, α') < () Now y me vlue theorem for rel vlued fucto for t [x -1, x ] such tht α (x ) - α (x -1 ) = α'(t )(x - x ) or α (x ) = α '(t ) x, for = 1,...,. () Ag sce equto () s true for s [x -1, x ], the α '(s )- α '(t ) x < (v) =1 If M = sup f(x), the y (), we hve f(s ) α = f(s ) α '(t ) x t follows tht =1 =1

13 13 f(s ) α - f(s ) α '(t ) x = f(s ) α '(t ) x - f(s ) α '(t ) x =1 =1 =1 =1 = f(s )[α '(t ) - α '(t )]x f(s ) α '(t ) - α '(t ) x =1 =1 sup f(x ) α'(t ) - α'(t ) x = M α'(t ) - α'(t ) x < M [y(v)] =1 =1 I prtculr, f(s )α U(P, f, α') + M s [x -1, x ]. The =1 U(P, f, α) U(P, f, α') + M or U(P, f, α) - U(P, f, α') M. Smlrly y the sme rgumet we hve U(P, f, α ') U(P, f, α) + M or - [U(P, f, α) - U(P, f, α ')] M. Thus U(P, f, α) - U(P, f, α') M (v) Here we ote tht () d (v) rems true f P s replced y y refemet. We coclude tht - - f dα - f(x)α'(x)dx M. - - Sce s rtrry. Hece f dα = f(x)α'(x)dx (v) Smlrly, f dα = f(x)α'(x)dx (v) - - From (v) d (v) we hve f dα = f(x)α'(x)dx Theorem.(chge of vrle). Suppose s strctly cresg cotuous fucto tht mps tervl [A, B] oto [, ]. Suppose α s mootoclly cresg o [, ] d f R(α) o [, ]. Defe d g o [A, B] y (y) = α ((y)), g(y) = f((y)) () The g R() d B g d = f d. A Proof. Let to ech prtto P = {x 0,...,x } of [, ] correspods prtto Q = {y 0,,y } of [A,B], the x = (y ). All prttos of [A, B] re oted ths wy. Sce the vlues tke y f o [x -1, x ] re exctly the sme s those tke y g o [y -1, y ], we see tht U(Q, g, ) = U(P, f, ) d L(Q, g, ) = L(P, f, ). Sce f R(α) the for gve > 0, U (P, f, ) - L(P, f, ) <. Hece for the prtto Q U(Q, g, ) - L(Q, g, ) <. Whch shows tht g R(). Furthermore B g d = f d. A

14 14 Note: Cosder the followg specl cse: Tke (x) = x, the =. Assume R [A, B]. The y theorem 1.2.5, we hve B f(x)dx = f((y)) (y) dy. A 1.3. Itegrto d Dfferetto. I ths secto we dscuss the fmous the fudmetl theorem of clculus, whch stted tht tegrto d dfferetto re, cert sese, verse opertos. We shll mde ths study for Rem tegrls Defto. Itegrl Fucto. If f s Rem tegrle fucto o [, ] the fucto x F(x) = f(t) dt s clled tegrl fucto. Further f f(x) s dfferetle o [, ] d F(x) = f(x), the F(x) s clled the prmtve or t-dervtve of f o [, ]. Here we ote tht the prmtve of f(x) s ot uque lso tegrle fucto s ot ecessrly cotuous ut the fucto ssocted to f s lwys cotuous s show the followg theorem x Theorem. Let f R[, ]. For x, put F(x) = f(t) dt. The F s cotuous o [, ] Furthermore, f f s cotuous t pot x 0 of [, ], the F s dfferetle t x 0, d F'(x 0 ) = f(x 0 ). I other word The tegrl of Rem tegrle fucto s cotuous d s dfferetle f f s cotuous. Proof. Sce f R[, ] so y defto f s ouded therefore there exsts postve costt M such tht f(t) M for t. Now for x y, we hve y x y y F(y) - F(x) = f(t) dt - f(t) dt = f(t) dt + f(t) dt = f(t)dt M(y - x). x x Suppose tht for gve > 0 such tht y x < /M. The we hve F(y) - F(x) <. Ths proves cotuty (d, fct, uform cotuty) of F. Next suppose f s cotuous t x 0. The for gve > 0, choose δ > 0 such tht f(t) f(x 0 ) < ; f t x 0 < δ, where t. Hece, f x 0 δ < s x o t < x o + δ d s < t, we oserve tht t {1/(t - s)} dt = 1. Therefore s [{F(t) F(s)}/(t - s)] f(x 0 ) = {1/(t - s)} [f(u)du - {1/(t - s)} f(x 0 )du s s t t t

15 15 {1/ t s } f(u) - f(x 0 ) du < s It follows tht F I dfferetle t x 0 d F'(x 0 ) = f(x 0 ). Sce here x 0 s rtrry so tht here we oserve tht the cotuty of f mples the dfferetlty of F. Some tme ths theorem s clled the frst fudmetl theorem of tegrl clculus Theorem (The fudmetl theorem of clculus). If f R[, ] d f there s dfferette fucto F o [, ] such tht F ' = f, the f(x) dx = F() - F(). Proof. Let > 0 e gve, sce F' = f R[, ] therefore there exsts P = {x 0,..., x } of [, ] wth P < such tht F '(t ) x - F '(x)dx < where t [x -1, x ] () Now y Lgrge s me vlue theorem for pot t [x -1, x ] such tht F(x ) - F(x -1 ) = F '(t )x for = 1,...,. Therefore F '(t )x = [F(x ) - F(x -1 )] = F() - F(). =1 =1 Therefore y equto (), we hve F() - F() - F '(x)dx <. Whch mples tht F() - F() = F '(x)dx = f(x)dx or f(x)dx = F() - F() Theorem (tegrto y prts). Suppose F d G re dfferetle fuctos o [, ], F ' = f R[, ] d G' = g R[, ]. The F(x)g(x)dx = F()G() - F()G() - f(x)g(x)dx. Proof. Put H(x) = F(x)G(x), sce F d G re dfferetle so H(x) s lso dfferetle d H'(x) = F(x)G'(x) + G(x)F '(x) R[, ]. The usg fudmetl theorem of tegrl clculus, we hve H'(x) dx = H() - H() = F()G() - F()G(). F(x)G'(x) + G(x)F '(x) dx = H() - H() = F()G() - F()G(). F(x)g(x) = H() - H() = F()G() - F()G() - G(x)f(x) dx Itegrto of vector- vlued fuctos.

16 Defto Let f 1,...,f k e rel fuctos o [, ], d let f = (f,...,f k ) e the correspodg mppg of [, ] to R k. If creses mootoclly o [,], we sy tht f R() f f j R() for j = 1,..., k. I ths cse, we hve f dα = ( f 1 dα,, f k dα)..e., f dα s the pot R k whose j th co-ordte s f j dα Theorem () lf f, F R() o [, ], the f + F R(), cf R() for every costt c, d (f + F)dα = f dα + F dα ; cf dα = c f dα. () If f R() o [, ] d f < c <, the f R() o [, c] d o [c, ], d c f dα + f dα = f dα. c (c) If f R( 1 ) d f R( 2 ), the f R( ) d f d( ) = f dα 1 + f dα 2 ; If f R() d c s postve costt, the f R(c) d f d(cα) = c f dα. Proof. Sce y the defto of vector vlued fucto these propertes holds good. We smply pply the results (), (c), d (e) of Theorem to ech co-ordte of f Theorem. () Let α creses mootoclly d α R[, ]. Let f e ouded rel fucto o [, ]. The f R(α) f d oly f f α R[, ]. Moreover f dα = f (x) α(x)dx x () Let f R[, ]. For x, put F(x) = f(t) dt. The F s cotuous o [, ]. Furthermore, f f s cotuous t pot x 0 of [, ], the F s dfferetle t x 0, d F'(x 0 ) = f(x 0 ). Here f dα = ( f 1 dα,, f k dα) = (F 1, F 2, F 3, F k,) = F(x) Proof. If we pply the theorems d o ech co-ordte of f, the the results re vld. Therefore y the defto of vector vlued fucto these results holds good.

17 If f d F re s theorem 1.4.2() f R() d f F s dfferette fucto [, ] such tht F' = f, the f(t) dt = F() - F(). Proof. Let > 0 e gve, sce F' = f R[, ] therefore there exsts prtto P = {x 0,..., x } of [, ] wth P < such tht yelds tht F j '(t ) x - F j '(t)dt < where t [x -1, x ] for ech j. F'(t ) x - F'(t)dt < where t [x -1, x ] () Now y Lgrge s me vlue theorem for pot t [x -1, x ] such tht F j (x ) - F j (x -1 ) = F j '(t )x for = 1,..., d ech j yelds tht F(x ) - F(x -1 ) = F'(t )x for = 1,...,. Therefore F '(t )x = [F(x ) - F(x -1 )] = F() - F(). =1 =1 Therefore y equto (), we hve F() - F() - F '(t)dt <. Whch mples tht F() - F() = F'(t)dt = f(t)dt or f(t)dt = F() - F() Theorem If f mps [, ] to R k d f f R() for some mootoclly cresg fucto o [, ], the f R(), d f dα f dα. Proof. If f 1,, f k re the compoets of f, the f = (f f f 2 k) 1/2. Sce f R(), mples tht f j R() for ech j so ther sum s lso elogs to R(). Sce x 2 s cotuous fucto of x, the the squre-root fucto s cotuous lso cotuous t dom. Hece y theorem f R(). Let we put y = (y 1,..., y k ), where y j = f j d, the we hve y = f d d y 2 2 = y j = y j f j d = (y j f j )d. By the Schwrz equlty y j f j (t) y f(t) ( t ). Hece, we hve y 2 y f d or y f d or f dα f dα Rectfle Curves.

18 Defto. A cotuous mppg of tervl [, ] to R k s sd to e curve R k. We my lso sy tht s curve o [, ]. If (1) s oe-to-oe the t clled rc. (2) If () = () the t s sd to e closed curve. (3) Here we ote tht wth ech curve R k there s ssocted suset of R k, clled the rge of, ut dfferet curves my hve the sme rge. Now we ssocte to ech prtto P = {x 0,..., x } of [, ] d to ech curve o [, ] the pots (x 0 ), (x 1 ), (x 2 ),, (x ) re the vertces of scred polygo. We deote the legth of the polygo y (P, ) d defe y (P, ) = (x ) - (x -1 ), where the th =1 term ths sum s the dstce ( R k ) etwee the. pots (x -1 ) d (x ). As our prtto ecomes fer d fer, ths polygo pproches the rge of more d more closely d the polygol legth teds to the legth of whch s deoted y () d defed y () = sup (P, ), where the supremum s tke over ll prttos of [, ]. If () < the s sd to e rectfle. Here () s gve y Rem tegrl cert cses. Our ext theorem devote to ths for cotuously dfferetle curves Theorem. If ' s cotuous o [, ], the s rectfle, d () = '(t) dt. Proof. If x - 1 < x, the (x ) - (x -1 ) = '(t) dt '(t) dt. Hece (P, ) '(t) dt for every prtto P of [, ]. Cosequetly (P, ) '(t) dt. () x x -1 x x -1 Now for rtrry > 0, sce ' s uformly cotuous o [, ], there exsts δ > 0 such tht ' (s) - ' (t) < for s t < δ, s, t [, ]. () Let P = {x 0,, x } e y prtto of [, ], wth x < δ, for ll, the for x -1 t x, we hve ' (t) ' (x ) +. Hece x ' (t) dt '(x )x + x. x -1 x = [' (t) + ' (x ) - ' (t)] dt + x. x -1 x x ' (t)dt + ['(x ) - ' (t)] dt + x.

19 19 x -1 x -1 (x ) - (x -1 ) + 2 x. O ddg these equltes, we ot ' (t) dt (P, ) + 2 ( - ) () + 2 ( - ). Sce ws rtrry therefore, ' (t) dt (). () From () d (), we hve ' (t) dt = () Rerrgemets of terms of Seres, Rem theorem. As we hve studed the cocept of covergece d solute covergece of seres d other relted propertes erler clsses. I ths secto we study the ture of coverget seres, whe ts terms re rerrged dfferet mer. For ths frst we kow wht s the rerrgemet of terms of gve seres Defto. Let {k }, = 1, 2, 3, e sequece whch every postve teger s orepetg (I other words {k } s jectve mp o the set of postve teger to tself). We put ' = k, = 1, 2, 3,, the ' s clled the rerrgemet of the gve. If we suppose tht {s } d {s' } re the prtl sums of, ' respectvely, the t c e esly see tht the terms of thee sequeces re etrely dfferet umers d {s } s coverget wheever the seres s coverget. The there rse turl questo out the ture of the sequece {s' } (s well s the seres ' lso),.e., t s coverget or dverget. Furthermore, thus led to the prolem of determg uder wht codto ll rerrgemets of coverget seres wll coverget d where the sums re ecessrly the sme. We c uderstd ths y the followg exmple: Exmple. Cosder the coverget seres 1 ½ + 1 / 3 ¼ + d ts rerrgemet / 3 ½ + 1 / / 7 - ¼ + 1 / / 11-1 / 6 d f {s } d {s' } re ther prtl sums respectvely, the s = lms < 1- ½ + 1 / 3 = 5/6. Now sce fore k 1, we se tht {1/(4k 3)} +{1/(4k 1)} (1/2k) > 0, we se tht s' 3 < s' 6 < s' 9 <, mples tht lm sup s' > s' 3 = 5/6. Hece s' does ot coverge to s. Ths exmple llustrte the ext theorem due to Rem Theorem.(Rem theorem) Let e seres of rel umers whch s coverges, ut ot solutely. Suppose there re umers, such tht -. The there exsts rerrgemet ' of wth prtl sum s' such tht

20 20 lm f s' = d lm sup s' =. () I other words the terms of codtolly coverget seres c e rerrge such tht the ew seres my e coverget, dverget or osclltory. Proof. Cosder p = ( + )/2 d q = ( )/2 ( = 1, 2, 3, ). The p - q = d p + q = d sce p 0, q 0 lso f the seres p d q re coverget the the seres (p + q ) = s coverget cotrdcts the fct tht the seres coverget codtolly. Hece the seres p d q must e dverge. Ag sce = (p - q ) = p - q the the dvergece of p d covergece of q (or vce vers) mples dvergece of g cotrdcts the fct tht the seres s coverget. Now suppose tht P 1, P 2, P 3, deote the oegtve terms of the order whch they occur d Q 1, Q 2, Q 3, deote the solute vlues of egtve terms of d they re lso ther orgl order. I such wy we see tht the seres P, Q re dffer from the seres p, q y zero terms d hece dverget lso. Now we costruct sequeces {m } d {k } such tht the seres P P m1 - Q Q k1 + P m P m2 Q k Q k2 +, () s rerrgemet of the seres d stsfes lm f s' = d lm sup s' =. Further choose sequeces { } d { } such tht d, <, 1 > 0. I such wy tht, let m 1 d k 1 e the smllest tegers for whch P P m1 > 1, P P m1 - Q Q k1 < 1 ; let m 1 d k 1 e the smllest tegers for whch P P m1 - Q Q k1 + P m P m2 > 2, P P m1 - Q Q k1 + P m P m2 Q k Q k2 < 2 ; d cotue ths wy, sce t s possle ecuse of P, Q re dverge. If s d t re the prtl sums of () whose lst terms re P m d - Q k, the s - P m d t - Q k. Sce s coverget the P, Q 0 s, we hve s d t. Flly we see tht o umer less th or greter th c e su-sequetl lmt of the prtl sum of (). Hece lm f s' = d lm sup s' = Theorem. Let e seres of complex umers whch s solutely coverget the every rerrgemets of re coverges d ll coverges to sme sum.. Or, The sum of every rerrgemets of solutely coverget seres s rems uchged. Proof. Let e y rerrgemet of d s, s e the prtl sums of, respectvely. Sce covergece solutely therefore y Cuchy s crtero of coverget of seres for gve > 0 there exsts postve teger N such tht m

21 21 r <, m N. () r= Now we choose p such tht the tegers 1, 2, 3,, N les the set {k 1, k 2,, k p }. The for > p the umers 1, 2,, N wll ccel the dfferece s s, therefore s s < y (). Cosequetly {s } coverges to the sme sum s {s } Uty Summry Deftos : Let [, ] e gve tervl. The set P = {x 0, x 1,, x -1, x } of [, ] such tht = x 0 x 1 x -1 x = s sd to e Prtto of [, ]. The set of ll prttos of [, ] s deoted y P([, ]). The tervls [x 0, x 1 ], [x 0, x 1 ], [x 1, x 2 ],, [x -1, x ] re clled the sutervls of [, ]. Wrte x = x x -1 s clled the legth of the tervl [x -1, x ] ( = 1,, ) d mx x s clled the orm of the prtto P d s deoted y P or (P). A prtto Q of [, ] such tht P Q s clled the refemet or fer of the prtto P. Let f : [, ] R s ouded fucto d e mootoclly cresg fucto o [, ] d P e y Prtto of [, ]. We wrte =(x ) - (x -1 ), = 1, 2, 3,,. Let M = sup f(x), m = f f(x) (x -1 x x ) for ech P P( [, ]). We defe U(P, f, ) = M, d L(P, f, ) = m, =1 =1 re clled the Upper d Lower Rem Steltjes sums respectvely. Further we defe - f d = f U(P, f, ) d f d = sup L(P, f, ), - where the f d the sup re tke over ll prttos P of [, ], re clled the Upper d Lower Rem Steltjes tegrls of f over [, ], respectvely. If the upper d lower Rem Steltjes tegrls re equl, we sy tht f s Rem Steltjes tegrle o [, ] () If f : [, ] R s ouded fucto d e mootoclly cresg fucto o [, ]. Let P e y Prtto of [, ].The the upper d lower Rem-Stetjes tegrls of f wth respect to re ouded. () If P* s refemet of the prtto P of [, ], the L(P, f, ) L(P*, f, ) d U(P*, f, ) U(P, f, ). () If P 1, P 2 re y two prttos of [, ] the L(P 1, f, ) U(P 2, f, ) d L(P 2, f, ) U(P 1, f, ). - (v) f dα f dα - (v) f R(α) o [, ] f d oly f for every ε > 0 there exsts prtto P such tht U(P, f, α) - L(P, f, α ) < ε (v) If for every ε > 0 there exsts prtto P = {x o, x 1,..., x } such tht U(P,f, α) - L(P, f, α) < ε holds d f s, t, re rtrry pots [x -1, x ], the

22 22 f(s ) f(t ) I. =1 Further f f R(), the f(t ) f d. =1 (v) If f s cotuous o [, ] d s mootoc cresg o [, ] the f R() o [, ]. (v) If f s mootoc o [, ] d f s cotuous o [, ], the f R(). (x) Suppose f s ouded o [, ], f hs oly ftely my pots of dscotuty o [, ], d s cotuous t every pot t whch f s dscotuous. The f R(). (x) Suppose f R() o [, ], m f M, s cotuous o [m, M], d h(x) = (f(x)) o [, ]. The h R() o [, ] Propertes of Rem-Stetjes () lf f 1, f 2 R() o [, ], the f 1 + f 2 R(), cf R() for every costt c, d () If f 1 (x) f 2 (x) o [,], the (f 1 +f 2 )dα = f 1 dα + f 2 dα ; cf dα = c f dα. f 1 dα f 2 dα. (c) If f R() o [, ] d f < c <, the f R() o [, c] d o [c, ], d c f dα + f dα = f dα. c (d) If f R() o [, ] d f f(x) M o [, ], the f dα M[() - ()]. (e) If f R( 1 ) d f R( 2 ), the f R( ) d f d( ) = f dα 1 + f dα 2 ; If f R() d c s postve costt, the f R(c) d f d(cα) = c f dα. (f) If f R() d g R() o [, ] the fg R(), f R() d f dα f dα.

23 23 (g) If < s <, f s ouded o [, ], f s cotuous t s d (x) = I(x - s), the f dα = f(s). (h) Suppose c 0 for = 1, 2, 3,..., c coverges, {s } s sequece of dstct pots (, ), d α(x) = c I(x - s ). Let f e cotuous o [, ]. The =1 f dα = c f(s ). =1 () Assume α creses mootoclly d α R[, ]. Let f e ouded rel fucto o [, ]. The f R(α) f d oly f f α R[, ]. Moreover f dα = f (x) α(x)dx (j) Suppose s strctly cresg cotuous fucto tht mps tervl [A, B] oto [, ]. Suppose α s mootoclly cresg o [, ] d f R(α) o [, ]. Defe d g o [A, B] y (y) = α ((y)), g(y) = f((y)). B The g R() d g d = f d. A x (k) Let f R [, ]. For x, put F(x) = f(t) dt. The F s cotuous o [, ]. Furthermore, f f s cotuous t pot x 0 of [, ], the F s dfferetle t x 0, d F'(x 0 ) = f(x 0 ). I other word The tegrl of Rem tegrle fucto s cotuous d s dfferetle f f s cotuous. (l) If f R[, ] d f there s dfferette fucto F o [, ] such tht F ' = f, the f(x) dx = F() - F(). (m) Suppose F d G re dfferetle fuctos o [, ], F ' = f R[, ] d G' = g R[, ]. The F(x)g(x)dx = F()G() - F()G() - f(x)g(x)dx Itegrto of vector- vlued fuctos. Let f 1,...,f k e rel fuctos o [, ], d let f = (f,...,f k ) e the correspodg mppg of [, ] to R k. If creses mootoclly o [,], we sy tht f R() f f j R() for j = 1,..., k. I ths cse, we hve f dα = ( f 1 dα,, f k dα)..e., f dα s the pot R k whose j th co-ordte s f j dα.

24 Rectfle Curves. A cotuous mppg of tervl [, ] to R k s sd to e curve R k. We my lso sy tht s curve o [, ]. If (1) s oe-to-oe the t clled rc. (2) If () = () the t s sd to e closed curve. (3) The legth of whch s deoted y () d defed y d f f () < the s sd to e rectfle. (4) If ' s cotuous o [, ], the s rectfle, d () = '(t) dt Rerrgemets of terms of Seres : Let {k }, = 1, 2, 3, e sequece whch every postve teger s o-repetg (I other words {k } s jectve mp o the set of postve teger to tself). We put ' = k, = 1, 2, 3,, the ' s clled the rerrgemet of the gve. () The terms of codtolly coverget seres c e rerrge such tht the ew seres my e coverget, dverget or osclltory. () The sum of every rerrgemets of solutely coverget seres s rems uchged Assgmets/Check your Progress. 1. Suppose creses o [, ], f(x) = k x [, ]. The prove tht f R() d fd = () - () 2. If f(x) = x d (x) = x 2 o [0, 1]. Is f R(), f yes the fd ts vlue of 0 1 fd. As 2/3. 3. If f(x) = (x) = x 2 o [0, 1]. Is f R(), f yes the fd ts vlue of 0 1 fd. As 1/2. 4. Suppose creses o [, ] d s cotuous t x 0, x 0, f(x 0 ) = 1, d f(x) = 0 f x x 0. Prove tht f R() d tht f d = Suppose f 0,f s cotuous o [, ], d f(x)dx = 0. Prove tht f(x) = 0 for ll x [, ]. 6. () If f, g R(), f, g 0 d f p d = 1 = g q d, the f g d 1. where 1/p + 1/q = 1. () If f d g re complex fuctos R(), the f g d { f p d} 1/p { g q d} 1/q. 7. Suppose creses mootoclly o [, ], g s cotuous, d g(x) = G (x) for x. Prove tht (x) g(x)dx = G() () - G() () - Gd 8. Let 1, 2, 3 e curves the complex ple, defed o [0, 2] y 1 (t) = e t, 2 (t) = e 2t, 3 (t) = e 2ts(1/t) Show tht these three curves hve the sme rge, tht 1 d 2 re rectfle, tht the legth of 1 s 2, tht the legth of 2 s 4, d tht 3 s ot rectfle.

25 25 9. Let e cotuous fucto of ouded vrto o [, ], ssume g R() o [, ] d defe (x) = x gd f x [, ] the prove tht f f s cresg o [, ] there exts pot x 0 [, ] such tht fd = f() x 0 gd + f() x0 gd. Furthermore f f s cotuous o [, ] the fg d = f() x 0 gd + f() x0 gd Pots for dscusso/ Clrfcto Suggested Study mterl 1. H. L. Royde, Rel Alyss, Mcmll Pu. Co. 4 th Edto, New York Apostl T.M. Mthemtcl Alyss, Nros Pulshg House, New Delh, P.K. J d V.P. Gupt Leesgu Mesure d Itegrto,New ge Itertol (P) Lmted pulcto, New Delh Wlter Rud, Prcples of Mthemtcl lyss McGrw Hll, Kogkush, 1976, Itertol studet edto. 5. H.K. Pthk - Rel Alyss 6. Lecture Notes o Rel lyss Rchrd F. Bss

26 26 Sequeces d seres of Fuctos Itroducto: The sequeces {f } whose terms re rel or complex-vlued fuctos hvg commo dom o the rel le R or the complex ple C. For ech x the dom we c form other sequece {f (x)} whose terms re the correspodg fucto vlues. Let S deote the set of x for whch {f (x)} coverges. The fucto f defed y the equto f ( x) lm f ( x), f x S s clled the lmt fucto of the sequece {f }, d we sy tht {f } coverges pomwse to f o the set S. If ech fucto of sequece {f } hs cert property, such s cotuty, dfferetlty, or tegrlty, to wht extet s ths property trsferred to the lmt fucto? For exmple, f ech fucto f s cotuous t c, s the lmt fucto f lso cotuous t c? We shll see tht, geerl, t s ot. I fct, we shll fd tht potwse covergece s usully ot strog eough to trsfer y of the propertes metoed ove from the dvdul terms f to the lmt fucto f. Therefore we re led to study stroger methods of covergece tht do preserve these propertes. The most mportt of these s the oto of uform covergece. Ojectve: As prevous clsses we hve ee studed the cocept d propertes of sequeces. I ths ut we del wth sequeces {f } of rel or complex vlued fuctos hvg dom R or C. For every x the dom we c form sequece {f (x)}. For exmple f f : [0, 1] R defed y f (x) = x, N, the we c form dfferet sequeces s {0,0,0 }, {1/2, 1/2 2, 1/2 3, }, etc., d the propertes lke covergece of these sequeces deped o s well s x the dom. Here we shll dscuss the pot wse d uform covergece d other relted cocepts of sequeces d seres of fuctos Sequece d Seres of Fuctos: Defto2.1.1: Let E e y o-empty set of R d N d f : E R e y fucto the {f } s sd to e sequece of fuctos from E to R d f s sd to e seres of fuctos from E to R. Clerly, for ech x E, we hve sequece {f (x)} d seres f (x). For some vlues of x ths sequece (or seres) s coverget d dverget o the other vlues of x, for exmple f f : N R defed y f (x) = x, N, the {f (x)} s coverget oly whe x = 1 d dverget for other vlues of x N. However the seres x s dverget for ll vlues of x N Potwse d Uform Covergece: Potwse covergece: Defto2.2.1: Let {f } e sequece of fuctos defed o set E d suppose tht for ech x E the sequece of umers {f (x)} coverges to fucto f(x),.e., f(x) = lm f (x), for ech x E... () The we sy tht {f } coverges o E d f s the lmt or lmt fucto of {f } d the covergece of {f } s clled potwse covergece of {f } o E. I other words the sequece {f } s sd to e potwse coverges to f f for > 0 d ech x E, 0 = 0 (, x) N

27 27 such tht f (x) f(x) <, 0. Rememer tht here 0 s deped o x d oth. Smlrly f the seres f (x) coverges for ech x E d = 1f (x) = f(x) the the fucto f s clled the sum (or potwse sum) fucto of the seres f. For exmple f f : [0, 1] R defed y f (x) = x, N d f f(x) s the lmt fucto of f (x) the f(x) = 0 f x (0, 1) d f(x) = 1 f x = 1. Therefore {f } coverges potwse. Smlrly for ll rel x cosder the geometrc seres 0 {x 2 /(1 + x 2 ) } d f f(x) s the sum fucto of the seres the f(x) = 0 f x = 0 d f(x) = 1 + x 2 f x 0, therefore the seres s potwse coverges to f(x). Here questo rses f ech fucto of the sequece {f } hs cert property lke cotuty, dfferetlty or tegrlty etc; s the sme property s true for the lmt (or sum) fucto?. Cosder f ech fucto of the sequece {f } s cotuous t pot c,.e., lm f (x) = f (c), the lm f(x) = f(c) or lm xc lm f (x) = lm lm xc f (x) s true? () Therefore t ths stge the questo rses out cotuty s tht c we terchge the order of lmts ()? We shll see tht, geerl tht we c ot.e. ether the lmt does ot exsts or f t dose exst, t eed ot e equl to f(c). We shll dscuss ths stuto y mes of some exmples tht lmt processes c ot e terchged geerl wthout ffectg the result. Exmple 2.2.1: Cosder the doule sequece s m, = m/(m + ); m, = 1, 2, 3,. Clerly t s the potwse coverget sequece. Keepg fxed d lettg m, s m, 1 mples tht lm lm m s m, = 1. O the other hd, keepg m fxed d lettg, s m, 0 mples tht lm m lm s m, = 0. Hece lm lm m s m, lm m lm s m,. Exmple 2.2.2: Cosder the sequece f m (x) = lm (cos m!x) 2 ; m = 1, 2, 3,. Clerly t s the potwse coverget sequece of cotuous fuctos. Whe m!x s teger the f m (x) = 1 d for ll other vlues of x f m (x) = 0. If we suppose f(x) = lm m f m (x), the f(x) = 0 f x s rrtol d f(x) = 1 f x s rtol. Thus we hve oted dscotuous lmt fucto. Exmple 2.2.3: Cosder the sequece f (x) = s x/ (x s rel d = 1, 2, 3, ). Clerly t s the potwse coverget sequece. Sce f(x) = lm f (x) = 0 for ll x. The lm f (0) = lm cos x = lm f (0) = 0. Exmple 2.2.4: Cosder the sequece f (x) = 2 x(1 x 2 ) (0 x 1, = 1, 2, 3, ). Clerly t s the potwse coverget sequece. If f(x) = lm f (x), the f(x) = 0 for ll x [0, 1]. Now y smple clculto we hve x(1 x 2 ) dx = 2 /(2 + 2) s, wheres 0 1 [lm f (x)]dx = 0. Hece lm 0 1 f (x)dx 0 1 [lm f (x)]dx. From the ove exmples we coclude tht the potwse covergece s ot suffcet to llow us to terchge the order of lmts, lmt d dfferetto, lmt d tegrto. Therefore potwse covergece s usully ot strog eough to trsfer y of the propertes metoed ove from the dvdul term f to the lmt fucto f. Hece we re led to the study stroger

28 28 method of covergece tht do preserve these propertes. A cocept of gret mportce ths regrd s the oto of uform covergece. Uform Covergece: Defto2.2.2: A sequece {f } of fuctos defed o set E s sd to e coverges uformly o E to fucto f f for > 0 0 = 0 () N such tht f (x) f(x) <, 0, for ll x E. Symolclly t s wrtte s f f uformly o E or lm f (x) = f(x) uformly o E or f f or f (x) f(x), x E. Clerly uform covergece mples potwse covergece ut ot coversely. The dfferece etwee these cocepts s ths: If {f } coverges potwse o E, the there exsts fucto f such tht, for every > 0 d ech x E 0 = 0 (, x) N depedg o d x such tht f (x) f(x) < 0. But f {f } coverges uformly o E t s possle, for every > 0 0 = 0 () N depedg oly o such tht f (x) f(x) < 0. Smlrly seres f (x) coverges uformly o E f the sequece {s } of prtl sums defed y = 1f (x) = s (x) coverges uformly o E. Suppose s s uformly o E, the we wrte = 1f (x) = s(x) uformly o E. I other words the seres f (x) s sd to e uformly coverget to s(x) f for > 0 0 = 0 () N depedet of x, such tht s (x) s(x) < 0 for ll x E. Let ech term of the uformly coverget sequece {f } s rel-vlued the y f (x) f(x) < 0 we me tht f(x) - < f (x) < f(x) +, 0 mples tht the etre grph of f (.e. the set [{(x, y) : y = f (x), x E}] les wth d of heght 2 stuted symmetrclly out the grph of f s show fg (). Ths s geometrcl terpretto of uform covergece. y =f(x) + y = f(x) A sequece {f } s sd to e uformly ouded o the set E f there exsts costt M > 0 such tht f (x) M, x E d ll d the umer M s clled the uform oud for {f }. Clerly f ech dvdul fucto s ouded d f f uformly o E, the the sequece {f } s uformly ouded o E. If there s pot x E s such tht sequece {f } s ot coverget uformly y eghorhood of x the x s clled the pot of o-uform covergece of the sequece {f }. Now we llustrte the oto of uform covergece through some exmples: Exmple 2.2.6: Show tht the sequece {f } where f : R R defed y f (x) = x/ x R, N, coverget potwse ut ot uformly.

29 29 Soluto: Let for rtrry > 0 there exsts m N such tht f (x) f(x) < m. Sce here f (x) = x/ 0 = f(x) x R, therefore x/ 0 < m, I prtculr x /m < or, m > x /. Clerly m depeds o x d oth. Hece the sequece {f } coverget potwse ut ot uformly. But t coverget uformly o [0, 1], sce here m > x / > 1/, clerly m depeds oly o. Exmple 2.2.7: Show tht the sequece {f } where f : [0, 1] R defed y f (x) = x x [0, 1], N, coverget potwse ut ot uformly. Soluto: Clerly ths sequece s potwse coverget, sce f(x) = lm f (x) = 0 f x [0, 1) 1 f x =1. Let for rtrry > 0 there exsts m N such tht f (x) f(x) < m. Or, x 0 < m d x [0, 1). I prtculr x m < or 1/ x m > 1/ m log(1/x) > log(1/) m > [log(1/) / log(1/x)]. Clerly m depeds o x d oth d t s ot possle to determe m N such tht f (x) f(x) < m x [0, 1]. Hece the sequece {f } coverget potwse ut ot uformly. Exmple 2.2.8: Show tht the sequece {f } where f : (0, ) R defed y f (x) = 1 + {x / (1 + x)} x (0, ), N, coverges uformly. Soluto: Sce f(x) = lm f (x) = lm [1 + {x / (1 + x)}] = 1 x (0, ) d N. Let for rtrry > 0 there exsts m N such tht f (x) f(x) < m. Or, [1 + {x / (1 + x)}] - 1 < m. I prtculr {x / (1 + mx)} < Or, (1 + mx) / x > 1/ or, 1 > 1/x > (1/) m mples tht m > (1/) 1. Here m s deped oly o. Hece the sequece {f } coverget uformly. Exmple 2.2.9: Show tht the sequece {f } where f :XR defed y f (x)= 1/(1+ x 2 ), N, ot coverget uformly f X = R ut coverget uformly f X = (0, ). Soluto: Sce f(x) = lm f (x) = lm {1 / (1 + x 2 )} = 1 f x = 0 0 f x 0. Hece s exmples d the sequece {f } ot coverget uformly f X = R ut coverget uformly f X = (0, ) Cuchy Crtero for uform covergece: Followg s the Cuchy crtero for uform covergece Theorem (Cuchy Crtero for uform covergece for sequece) Ay sequece {f } of fuctos defed o set E s coverges uformly o E to fucto f ff the followg codto (clled the Cuchy codto) hold: For every > 0, p N such tht f (x) f m (x) <, m p d x E () Proof. Suppose tht f f uformly o E, the for every > 0, p = p() N such tht f (x) f(x) < /2 p d x E. Therefore,

30 30 f (x) f m (x) f (x) f(x) + f(x) f m (x) < /2 + /2 =, m, p d x E. Coversely, suppose tht the codto (1) stsfed for ll x E. The y Cuchy crtero of covergece of sequeces for fxed x E the sequece {f (x)} s coverget d sy coverges to fucto f(x). Now for > 0 usg the codto (), we hve f (x) f +k (x) < p, k > 0 d x E () Lettg k, y (), we hve f (x) f(x) < p x E, yelds f f uformly o E. Note: Sometme s the mmedte cosequece of defto of uform covergece f lm k f (x) = f(x) (x E ) d put M = sup xe f (x) f(x). The f f uformly o E ff M 0 s. Theorem (Cuchy Crtero for uform covergece for seres) A seres f of fuctos defed o set E s coverges uformly o E to fucto f ff the followg codto (clled the Cuchy codto) hold: +p For every > 0, m N such tht f k (x) < m, p = 1, 2, 3, d x E k=+1 Proof. Cosder the sequece {s (x) of prtl sum of the f, where s (x) = f 1 (x) + f 2 (x) + f 3 (x) + + f (x). +p The s +p (x) - s (x) = f +1 (x) + f +2 (x) + f +3 (x) + + f +p (x) = f k (x) k=+1 Suppose tht the seres f coverges uformly to f o E ff s (x) f(x) uformly. Now y Cuchy crtero for uformly covergece of sequece of fuctos sequece {s (x)} coverges uformly to f(x) ff for every > 0, m N such tht s +p (x) s (x) < m d x E +p ff f k (x) < m d x E. k= Test for uform covergece: Here we gve some test for uform covergece. Theorem (M -test for sequece): A sequece {f } of fuctos defed o set E d f lm k f (x) = f(x) (x E ) d M = sup xe f (x) f(x). The f f uformly o E ff M 0 s. Proof. Suppose tht f f uformly o E, the for every > 0, p = p() N such tht f (x) f(x) <, p d x E. M = sup xe f (x) f(x) <, p. M 0 s. Coversely, suppose tht M 0 s, the for > 0, p N such tht M < p d x E M <. p d x E, sup xe f (x) f(x) <, p f (x) f(x) <, p d x E, yelds f f uformly o E.

31 31 Theorem (Werstrss M-test for seres): A sequece {f } of fuctos defed o set E d suppose tht f (x) M (x E, = 1, 2, 3, ). The seres f coverges uformly (d solutely) o E f M coverges. Proof. Suppose tht f M coverges, the y Cuchy s crtero for covergece of seres for every > 0, m N such tht +p M k <, m, p = 1, 2, 3,. k= +1 Sce f (x) M (x E, = 1, 2, 3, ). Therefore, +p f k (x) = f +1 (x) + f +2 (x) + f +3 (x) + + f +p (x) k=+1 f +1 (x) + f +2 (x) + f +3 (x) + + f +p (x) +p M +1 + M +2 + M M +p = M k < m, p = 1, 2, 3, x E, k=+1 whch mples tht f coverges uformly o E. Exmple Show tht the sequece {f } where f (x) = x(1 x) does ot coverge uformly o [0, 1]. Soluto. Here f(x) = lm f (x) = lm [ x /(1 x) - ] [/ form] = - lm [x /(1 x) - log(1 x)] [y L Hosptl Lw] = - lm [x (1 x) / log(1 x)] 0, sce (1 x) 0 s. Hece f(x) = 0, x [0, 1]. Now M = sup{ f (x) f(x) : x [0, 1]} = sup{ x(1 x) : x [0, 1]} (1/)[1 (1/)] [ for x = 1/ [0, 1]] = [1 (1/)] e 0 s. Hece y theorem the gve sequece s ot coverges uformly. Exmple Show tht the seres f where f (x) = cos x / p, = 1, 2, 3, coverge uformly o R for p > 1. Soluto. Here sce f (x) = cos x / p 1/ p, x R. Now the seres o rght hd sde coverget for p > 1. Hece y Weerstrss s M-test the gve seres covergece uformly o R, for p > 1. Note : Weerstrss s M-test s oly suffcet ut ot ecessry.e., o-covergece of M does ot mply the ture of f. Although Weestrss s M- Test s pplcle to restrcted clss of seres,.e., seres whch re solutely coverget s well, yet t s of gret prctcl mportce s ths test c e ppled to erly ll seres whch re frequetly used the lterture. I ll other cses we hve to mke use of more delcte tests, whch we costruct y logy wth those for seres of rtrry terms- Ael s d Drchlet s test. Theorem (Ael s test for seres): A seres u (x)v (x) coverges uformly o [,] f () u (x) s uformly coverget [, ] () the sequece {v (x)} s mootoc for every x [, ]