6 Sequece d Series of Fuctios 6. Sequece of Fuctios 6.. Poitwise Covergece d Uiform Covergece Let J be itervl i R. Defiitio 6. For ech N, suppose fuctio f : J R is give. The we sy tht sequece (f ) of fuctios o J is give. More precisely, sequece of fuctios o J is mp F : N F(J), where F(J) is the set of ll rel vlued fuctios defied o J. If f := F () for N, the we deote F by (f ), d cll (f ) s sequece of fuctios. Defiitio 6.2 Let (f ) be sequece of fuctios o itervl J. () We sy tht (f ) coverges t poit x 0 J if the sequece (f (x 0 )) of rel umbers coverges. (b) We sy tht (f ) coverges poitwise o J if (f ) coverges t every poit i J, i.e., for ech x J, the sequece (f (x)) of rel umbers coverges. Defiitio 6.3 Let (f ) be sequece of fuctios o itervl J. If (f ) coverges poitwise o J, d if f : J R is defied by f(x) = lim f (x), x J, the we sy tht (f ) coverges poitwise to f o J, d f is the poitwise limit of (f ), d i tht cse we write f f poitwise o J. Thus, (f ) coverges to f poitwise o J if d oly if for every ε > 0 d for ech x J, there exists N N (depedig, i geerl, o both ε d x) such tht f (x) f(x) < ε for ll N. Exercise 6. Poitwise limit of sequece of fuctios is uique. 63
64 Sequece d Series of Fuctios M.T. Nir Exmple 6. Cosider f : R R defied by f (x) = si(x), x R d for N. The we see tht for ech x R, f (x) N. Thus, (f ) coverges poitwise to f o R, where f is the zero fuctio o R, i.e., f(x) = 0 for every x R. Suppose (f ) coverges to f potwise o J. As we hve metioed, it c hppe tht for ε > 0, d for ech x J, the umber N N stisfyig f (x) f(x) < ε N depeds ot oly o ε but lso o the poit x. For istce, cosider the followig exmple. Exmple 6.2 Let f (x) = x for x [0, ] d N. The we see tht for 0 x <, f (x) 0, d f () s. Thus, (f ) coverges poitwise to fuctio f defied by { 0, x, f(x) =, x =. I prticulr, (f ) coverges poitwise to the zero fuctio o [0, ). Note tht if there exists N N such tht x < ε for ll N d for ll x [0, ), the, lettig x, we would get < ε, which is ot possible, hd we chose ε <. For ε > 0, if we re ble to fid N N which does ot vry s x vries over J such tht f (x) f(x) < ε for ll N, the we sy tht (f ) coverges uiformly to f o J. Followig is the precise defiitio of uiform covergece of (f ) to f o J. Defiitio 6.4 Suppose (f ) is sequece of fuctios defied o itervl J. We sy tht (f ) coverges to fuctio f uiformly o J if for every ε > 0 there exists N N (depedig oly o ε) such tht d i tht cse we write We observe the followig: f (x) f(x) < ε N d x J, f f uiformly o J. If (f ) coverges uiformly to f, the it coverges to f poitwise s well. Thus, if sequece does ot coverge poitwise to y fuctio, the it c ot coverge uiformly.
Sequece of Fuctios 65 If (f ) coverges uiformly to f o J, the (f ) coverges uiformly to f o every subitervl J 0 J. I Exmple 6.2 we obtied sequece of fuctios which coverges poitwise but ot uiformly. Here is other exmple of sequece of fuctios which coverges poitwise but ot uiformly. Exmple 6.3 For ech N, let f (x) = x + 2, x [0, ]. x2 Note tht f (0) = 0, d for x 0, f (x) 0 s. Hece, (f ) coverges poitwise to the zero fuctio. We do ot hve uiform covergece, s f (/) = /2 for ll. Ideed, if (f ) coverges uiformly, the there exists N N such tht I prticulr, we must hve f N (x) < ε x [0, ]. 2 = f N(/N) < ε x [0, ]. This is ot possible if we hd chose ε < /2. Exmple 6.4 Cosider the sequece (f ) defied by f (x) = t (x), x R. Note tht f (0) = 0, d for x 0, f (x) π/2 s. Hece, the give sequece (f ) coverges poitwise to the fuctio f defied by f(x) = { 0, x = 0, π/2, x 0. However, it does ot coverge uiformly to f o y itervl cotiig 0. To see this, let J be itervl cotiig 0 d ε > 0. Let N N be such tht f (x) f(x) < ε for ll N d for ll x J. I prticulr, we hve f N (x) π/2 < ε x J \ {0}. Lettig x 0, we hve π/2 = f N (0) π/2 < ε which is ot possible if we hd chooses ε < π/2. Now, we give theorem which would help us to show o-uiform covergece of certi sequece of fuctios. Theorem 6. Suppose f d f re fuctios defied o itervl J. If there exists sequece (x ) i J such tht f (x ) f(x ) 0, the (f ) does ot coverge uiformly to f o J.
66 Sequece d Series of Fuctios M.T. Nir Proof. Suppose (f ) coverges uiformly to f o J. The, for every ε > 0, there exists N N such tht I prticulr, f (x) f(x) < ε N, x J. f (x ) f(x ) < ε N. Hece, f (x ) f(x ) 0 s. This is cotrdictio to the hypothesis tht f (x ) f(x ) 0. Hece our ssumptio tht (f ) coverges uiformly to f o J is wrog. I the cse of Exmple 6.2, tkig x = /( + ), we see tht f (x ) = ( ) + e. Hece, by Theorem 6., (f ) does ot coverge to f 0 uiformly o [0, ). I Exmple 6.3, we my tke x = /, d i the cse of Exmple 6.4, we my tke x = π/, d pply Theorem 6.. Exercise 6.2 Suppose f d f re fuctios defied o itervl J. If there exists sequece (x ) i J such tht [f (x ) f(x )] 0, the (f ) does ot coverge uiformly to f o J. Why? [Suppose := [f (x ) f(x )] 0. The there exists δ > 0 such tht δ for ifiitely my. Now, if f f uiformly, there exists N N such tht f (x) f(x) < δ/2 for ll N. I prticulr, < δ/2 for ll N. Thus, we rrive t cotrdictio.] Here is sufficiet coditio for uiform covergece. exercise. Its proof is left s Theorem 6.2 Suppose f for N d f re fuctios o J. If there exists sequece (α ) of positive rels stisfyig α 0 s d the (f ) coverges uiformly to f. f (x) f(x) α N, x J, Exercise 6.3 Supply detiled proof for Theorem 6.2. Here re few exmples to illustrte the bove theorem. Exmple 6.5 For ech N, let f (x) = 2x + 4, x [0, ]. x2
Sequece of Fuctios 67 Sice + 4 x 2 2 2 x (usig the reltio 2 + b 2 2b), we hve 0 f (x) 2x 2 2 x =. Thus, by Theorem 6.2, (f ) coverges uiformly to the zero fuctio. Exmple 6.6 For ech N, let f (x) = 3 log( + 4 x 2 ), x [0, ]. The we hve 0 f (x) 3 log( + 4 ) =: α N. Tkig g(t) := t 3 log( + t 4 ) for t > 0, we see, usig L Hospitl s rule tht I prticulr, 4t 3 lim g(t) = lim t t 3t 2 ( + t 4 ) = 0. lim 3 log( + 4 ) = 0. Thus, by Theorem 6.2, (f ) coverges uiformly to the zero fuctio. We my observe tht i Exmples 6.2 d 6.4, the limit fuctio f is ot cotiuous, lthough every f is cotiuous. This mkes us to sk the followig: Suppose ech f is cotiuous fuctio o J d (f ) coverges to f poitwise. If f is Riem itegrble, the do we hve for every [, b] J? f(x)dx = lim f (x)dx Suppose ech f is cotiuously differetible o J. The, is the fuctio f differetible o J? If f is differetible o J, the do we hve the reltio d d f(x) = lim dx dx f (x)dx? The swers to the bove questios eed ot be ffirmtive s the followig exmples show. Exmple 6.7 For ech N, let f (x) = x( x 2 ), 0 x.
68 Sequece d Series of Fuctios M.T. Nir The we see tht lim f (x) = 0 x [0, ]. Ideed, for ech x (0, ), ( ) f + (x) + = x( x 2 ) x( x 2 ) f (x) s. Sice x( x 2 ) < for x (0, ), we obti lim f (x) = 0 for every x [0, ]. But, 0 f (x)dx = 2 + 2 2 s. Thus, limit of the itegrls is ot the itegrl of the limit. Exmple 6.8 For ech N, let f (x) = si(x), x R. The we see tht lim f (x) = 0 x [0, ]. But, f (x) = cos(x) for ll N, so tht f (0) = s. Thus, limit of the derivtives is ot the derivtive of the limit. 6..2 Cotiuity d uiform covergece Theorem 6.3 Suppose (f ) is sequece of cotiuous fuctios defied o itervl J which coverges uiformly to fuctio f. The f is cotiuous o J. Proof. Suppose x 0 J. The for y x J d for y N, f(x) f(x 0 ) f(x) f (x) + f (x) f (x 0 ) + f (x 0 ) f(x 0 ). ( ) Let ε > 0 be give. Sice (f ) coverges to f uiformly, there exists N N such tht f (x) f(x) < ε/3 N, x J. Sice f N is cotiuous, there exists δ > 0 such tht Hece from ( ), we hve f N (x) f N (x 0 ) < ε/3 wheever x x 0 < δ. f(x) f(x 0 ) f(x) f N (x) + f N (x) f N (x 0 ) + f N (x 0 ) f(x 0 ) < ε wheever x x 0 < δ. Thus, f is cotiuous t x 0. This is true for ll x 0 J. Hece, f is cotiuous fuctio o J.
Sequece of Fuctios 69 6..3 Itegrtio-Differetitio d uiform covergece Theorem 6.4 Suppose (f ) is sequece of cotiuous fuctios defied o itervl [, b] which coverges uiformly to fuctio f o [, b]. The f is cotiuous d lim f (x)dx = f(x)dx. Proof. We lredy kow by Theorem 6.3 tht f is cotiuous fuctio. Next we ote tht f (x)dx f(x)dx f (x) f(x) dx. Let ε > 0 be give. By uiform covergece of (f ) to f, there exists N N such tht f (x) f(x) < ε/(b ) N, x [, b]. Hece, for ll N, f (x)dx This completes the proof. f(x)dx f (x) f(x) dx < ε. Theorem 6.5 Suppose (f ) is sequece of cotiuously differetible fuctios defied o itervl J such tht (i) (f ) coverges uiformly to fuctio, d (ii) (f ()) coverges for some J. The (f ) coverges to cotiuously differetible fuctio f d lim f (x) = f (x) x J. Proof. Let g(x) := lim f (x) for x J, d α := lim f (). Sice the covergece of (f ) to g is uiform, by Theorem 6.4, the fuctio g is cotiuous d lim x f (t)dt = x g(t)dt. Let ϕ(x) := x g(t)dt, x J. The ϕ is differetible d ϕ (x) = g(x) for x J. But, x f (t)dt = f (x) f (). Hece, we hve lim [f (x) f ()] = ϕ(x). Thus, (f ) coverges poitwise to differetible fuctio f defied by f(x) = ϕ(x) + α, x J, d (f ) coverges to f. Remrk 6. I Theorem 6.5, it be show tht the covergece of the sequece (f ) is uiform.
70 Sequece d Series of Fuctios M.T. Nir 6.2 Series of Fuctios Defiitio 6.5 By series of fuctios o itervl J, we me expressio of the form or f (x), = f = where (f ) is sequece of fuctios defied o J. Defiitio 6.6 Give series = f (x) of fuctios o itervl J, let s (x) := f i (x), x J. i= The s is clled the -th prtil sum of the series = f. Defiitio 6.7 Cosider series = f (x) of fuctios o itervl J, d let s (x) be its -th prtil sum. The we sy tht the series = f (x) () coverges t poit x 0 J if (s ) coverges t x 0, (b) coverges poitwise o J if (s ) coverges poitwise o J, d (c) coverges uiformly o J if (s ) coverges uiformly o J. The proof of the followig two theorems re obvious from the sttemets of Theorems 6.4 d 6.5 respectively. Theorem 6.6 Suppose (f ) is sequece of cotiuous fuctios o J. If = f (x) coverges uiformly o J, sy to f(x), the f is cotiuous o J, d for [, b] J, f(x)dx = = f (x)dx. Theorem 6.7 Suppose (f ) is sequece of cotiuously differetible fuctios o J. If = f (x) coverges uiformly o J, d if = f (x) coverges t some poit x 0 J, the = f (x) coverges to differetible fuctio o J, d ( ) d f (x) = f dx (x). = Next we cosider useful sufficiet coditio to check uiform covergece. First defiitio. Defiitio 6.8 We sy tht = f is domited series if there exists sequece (α ) of positive rel umbers such tht f (x) α for ll x J d for ll N, d the series = α coverges. =
Series of Fuctios 7 Theorem 6.8 A domited series coverges uiformly. Proof. Let = f be domited series defied o itervl J, d let (α ) be sequece of positive rels such tht (i) f (x) α for ll N d for ll x J, d (ii) = α coverges. Let s (x) = i= f i(x), N. The for > m, s (x) s m (x) = f i (x) f i (x) i=m+ i=m+ i=m+ α i = σ σ m, where σ = k= α k. Sice = α coverges, the sequece (σ ) is Cuchy sequece. Now, let ε > 0 be give, d let N N be such tht σ σ m < ε, m N. Hece, from the reltio: s (x) s m (x) σ σ m, we hve s (x) s m (x) < ε, m N, x J. This, i prticulr implies tht {s (x)} is lso Cuchy sequece t ech x J. Hece, {s (x)} coverges for ech x J. Let f(x) = lim s (x), x J. The, we hve f(x) s m (x) = lim s (x) s m (x) < ε m N, x J. Thus, the series = f coverges uiformly to f o J. Exmple 6.9 The series cos x = 2 cos x 2 2, si x 2 d = si x 2 2 re domited series, sice N d = 2 is coverget. Exmple 6.0 The series =0 x is domited series o [ ρ, ρ] for 0 < ρ <, sice x ρ for ll N d =0 ρ is coverget. Thus, the give series is domited series, d hece, it is uiformly coverget. Exmple 6. Cosider the series = d 3/2 = coverges uiformly o R. x (+x 2 ) x ( + x 2 ) ( 2 o R. Note tht coverges. Thus, the give series is domited series, d hece it ),
72 Sequece d Series of Fuctios M.T. Nir Exmple 6.2 Cosider the series x = for x [c, ), c > 0. Note tht + 2 x 2 d = x + 2 x 2 x 2 x 2 = 2 x 2 c coverges. Thus, the give series is domited series, d hece it 2 coverges uiformly o [c, ). Exmple 6.3 The series = tht d the series =! ( ) xe x is domited o [0, ): To see this, ote (xe x) x = e x coverges. x (x) /! =! It c lso be see tht xe x /2 for ll x [0, ). Exmple 6.4 The series = x is ot uiformly coverget o (0, ); i prticulr, ot domited o (0, ). This is see s follows: Note tht s (x) := k= x k = x x f(x) := x s. Hece, for ε > 0, f(x) s (x) < ε x x < ε. Hece, if there exists N N such tht f(x) s (x) < ε for ll N for ll x (0, ), the we would get x N < ε x (0, ). x This is ot possible, s x N / x s x. However, we hve see tht the bove series is domited o [, ] for 0 < <. Exmple 6.5 The series = ( x)x is ot uiformly coverget o [0, ]; i prticulr, ot domited o [0, ]. This is see s follows: Note tht s (x) := ( x)x k = k= { x if x 0 if x =. I prticulr, s (x) = x for ll x [0, ) d N. By Exmple 6.2, we kow tht (s (x)) coverges to f(x) poitwise, but ot uiformly.
Series of Fuctios 73 Remrk 6.2 Note tht if series = f coverges uiformly to fuctio f o itervl J, the we must hve β := sup s (x) f(x) 0 s. x J Here, s is the -th prtil sum of the series. Coversely, if β 0, the the series is uiformly coverget. Thus, if = f coverges to fuctio f o J, d if sup x J s (x) f(x) 0 s, the we c ifer tht the covergece is ot uiform. As illustrtio, cosider the Exmple 6.5. There we hve { x if x s (x) f(x) = 0 if x =. Hece, sup x s (x) f(x) =. Moreover, the limit fuctio f is ot cotiuous. Hece, the o-uiform covergece lso follows from Theorem 6.6. Exercise 6.4 Cosider series = f d := sup x J f (x). Show tht this series is domited series if d oly if = coverges. Next exmple shows tht i Theorem 6.7, the coditio tht the derived series coverges uiformly is ot ecessry coditio for the the coclusio. Exmple 6.6 Cosider the series =0 x. We kow tht it coverges to /( x) for x <. It c be see tht the derived series = x coverges uiformly for x ρ for y ρ (0, ). This follows sice = ρ coverges. Hece, ( x) 2 = d dx x = x for x ρ. = The bove reltio is true for x i y ope itervl J (, ); becuse we c choose ρ sufficietly close to such tht J [ ρ, ρ]. Hece, we hve ( x) 2 = x for x <. = We kow tht the give series is ot uiformly coverget (see, Exmple 6.4). Remrk 6.3 We hve see tht if = f (x) is domited series o itervl J, the it coverges uiformly d bsolutely, d tht bsolutely coverget series eed ot be domited series. Are there series which coverge uiformly but ot domited. The swer is i ffirmtive. Look t the followig series: + x ( ), x [0, ]. = Sice is diverget, the give series is ot bsolutely coverget t x = d = hece it is ot domited series. However, the give series coverges uiformly o [0, ].
74 Sequece d Series of Fuctios M.T. Nir 6.3 Additiol Exercises. Let f (x) = coverge uiformly. x 2 ( + x 2 ) for x 0. Show tht the series = f (x) does ot x 2. Let f (x) = + x 2, x R. Show tht (f ) coverge uiformly, wheres (f ) ( ) does ot coverge uiformly. Is the reltio lim f (x) = lim f (x) true for ll x R? 3. Let f (x) = log( + 3 x 2 ) 2, d g (x) = 2x + 3 for x [0, ]. Show tht x2 the sequece (g ) coverges uiformly to g where g(x) = 0 for ll x [0, ]. Usig this fct, show tht (f ) lso coverges uiformly to the zero fuctio o [0, ]. 2 x, 0 x /, 4. Let f (x) = 2 x + 2, / x 2/, 0, 2/ x. Show tht (f ) does ot coverge uiformly of [0, ]. [Hit: Use termwise itegrtio.] 5. Suppose ( ) is such tht = is bsolutely coverget. Show tht is domited series o R. 6. Show tht for ech p >, the series = x p = x 2 + x 2 is coverget o [, ] d the limit fuctio is cotiuous. 7. Show tht the series {(+) 2 x + 2 x }( x) coverges to cotiuous = fuctio o [0, ], but it is ot domited. 8. Show tht the series ot domited, d 0 = 9. Show tht = [ + (k + )x ] is coverget o [0, ], but + kx [ + (k + )x ] dx = + kx 0 = x ( + x 2 ) 2 dx = 2. = 0 [ + (k + )x ] dx. + kx