Pointwise convergence need not behave well with respect to standard properties such as continuity.

Transcription

1 Chpter 3 Uniform Convergence Lecture 9 Sequences of functions re of gret importnce in mny res of pure nd pplied mthemtics, nd their properties cn often be studied in the context of metric spces, s in Exmples nd.5. Unless otherwise stted, the functions under considertion tke the form f : D R, where D R denotes generic domin in the Eucliden line. Usully, D is n intervl. Definition 3.1. A sequence (f n : n 1) of such functions converges pointwise to f on D whenever the sequence of rel numbers (f n (x)) converges to f(x) in R for every x D. Pointwise convergence need not behve well with respect to stndrd properties such s continuity. Exmple 3.2. Let f n (x) = x n for ll n 1 nd x [, 1]. Then f n (x) is the function for x < 1 1 for x = 1. So f is discontinuous on [, 1], even though f n is continuous for every n. Definition 3.3. A sequence (f n : n 1) of functions converges uniformly to f on D whenever for every x D. ɛ >, N(ɛ) N such tht n N(ɛ) f n (x) f(x) < ɛ The force of this definition is tht the sme N = N(ɛ) must work for every x D simultneously. This requirement fils in Exmple 3.2. For tke ɛ = 1/2; 21

2 then for ny N, however lrge, x N 1/2 for ll 1 > x 2 1/N (tht is, for ll x sufficiently close to 1). On the other hnd, (3.4) f n f uniformly on D f n f pointwise on D. lwys holds. Exmples The sequence (x n ) converges to the constnt function on the subintervl [, 1). However, the rgument bove remins vlid, nd shows tht this convergence lso fils to be uniform on [, 1). 2. Let n in R s n. Then the sequence (f n ) of constnt functions f n (x) = n converges both pointwise nd uniformly to the constnt function on ny domin D. 3. If f n f uniformly on the domin D, then f n f uniformly on ny subdomin D D. It is now possible to link uniform convergence with the sup metric described in Lemm Proposition 3.6. Let f nd f n : D R be functions on the domin D; then f n f uniformly on D iff sup x D f n (x) f(x) exists for sufficiently lrge n, nd tends to s n. Proof. Assume tht M n = sup x D f n (x) f(x) for sufficiently lrge n, nd tht M n. So for ny ɛ >, there exists n integer N such tht n N implies M n < ɛ; hence f n (x) f(x) < ɛ for every x D, nd f n f uniformly on D. Conversely, ssume tht f n f uniformly on D. So for ny ɛ >, there exists n integer N such tht f n (x) f(x) < ɛ/2 for ll n N nd every x D. The set f n (x) f(x) : x D} is therefore bounded bove by ɛ/2, nd its supremum M n stisfies M n ɛ/2 < ɛ for ll n N. The sup metric rises on spces of bounded or continuous functions, nd ws tenttively relted to uniform convergence in Exmples In tht context, Proposition 3.6 confirms tht ordinry convergence in the function spces C[, b] B[, b] is the sme s uniform convergence of the corresponding sequence of functions on the domin [, b]. Exmple 3.7. Let A sup C[, 1] denote the subspce h : h() }, with respect to the sup metric. For ech h A sup, let ɛ = h() > nd consider the open bll B ɛ (h) C[, 1]. Then f B ɛ (h) implies tht sup x [,1] f(x) h(x) < ɛ, nd therefore tht f() h() < ɛ = h(). So f(), nd f must lso lie in A sup. Hence A sup is open in C[, 1]. 22

3 Exmple 3.7 shows tht the complement h : h() = } of A sup is closed. Such properties depend on the metric, s is pprent from Exmples Exmple 3.8. Let A 1 L 1 [, 1] denote the subspce h : h() }, with respect to the L 1 metric of Lemm Assume h A 1 hs h() >, nd for ny ɛ > define g L 1 [, 1] to be liner on [, ɛ/h()], with h() if x = g(x) = if x [ɛ/h(), 1]. Hence g := h g hs g () =, so g / A 1. But d 1 (g, h) = g(t)dt = ɛ/2 < ɛ, so g lies in B ɛ (h). Thus h / A 1. A similr rgument pplies to the cse h() <, so A 1 is empty; in prticulr, A 1 cnnot be open. Lecture 1 Uniformly convergent sequences of functions often interct well with nlytic properties such s continuity, nd procedures such s integrtion. By wy of motivtion, recll tht Exmple 3.2 exhibits non-uniformly convergent sequence of continuous functions whose pointwise it is discontinuous. Exmple 3.9. On the intervl [, 1] R, let (g n : n 1) denote the sequence of functions n if x (, 1/n) g n (x) = otherwise. Pointwise, g n (x) s n, but the convergence is not uniform, becuse sup x [,1] g n (x) = n does not tend to. Ech g n is discontinuous t points x = (from the right) nd 1/n; nevertheless, it is bounded nd integrble, with 1 g n(t)dt = 1 for ll n. Thus g n (t)dt = 1 = g n(t)dt. So the processes of integrtion nd tking pointwise its cnnot necessrily be interchnged, t lest in situtions where convergence is not uniform. Neither of these nomlies cn occur for uniformly convergent sequence of functions. Theorem 3.1. If f n : [, b] R is continuous for every n N, nd f n f uniformly on [, b], then f is continuous on [, b]. 23

4 Proof. Since the convergence is uniform, for ny ɛ > there exists n integer N such tht n N implies f n (x) f(x) < ɛ/3 for every x [, b]. Moreover, f N is continuous t ech point x, so there exists δ > for which x x < δ implies f N (x) f N (x ) < ɛ/3. For ny such x, it follows tht f(x) f(x ) f(x) f N (x) + f N (x) f N (x ) + f N (x ) f(x ) < ɛ from the tringle inequlity. Thus f is continuous t x. Cre is needed to interpret certin steps of this proof t x = or b. Corollry Suppose tht f n : [, b] R is continuous for every n N nd tht the pointwise it of the sequence (f n ) is discontinuous on [, b]; then the convergence cnnot be uniform. Exmples In Lecture 9, the proof tht the sequence (x n ) fils to converge uniformly on [, 1] involves the ɛ nd N of Definition 3.3. But Exmple 3.2 notes tht its pointwise it is discontinuous t x = 1; so Corollry 3.1 shows immeditely tht convergence cnnot be uniform. 2. The function f n (x) = 1/(1 + nx) is continuous on [, 1], for ll n 1. Moreover, (f n (x)) converges pointwise to the function 1 if x = if x (, 1], which is discontinuous t x =. So Corollry 3.1 confirms tht the convergence is not uniform. 3. The function f n (x) = cos n x is continuous on [ π/2, π/2], for ll n 1. Moreover, (f n (x)) converges pointwise to the function 1 if x = otherwise, which is discontinuous t x =. So Corollry 3.1 confirms tht the convergence is not uniform. Of course, Exmples 3.12 my lso be pproched directly vi Proposition 3.6. In ech cse, sup x D f n (x) f(x) is esily computed to be 1 for ll n, nd therefore ; so the convergence of (f n ) is not uniform. It is instructive to sketch the grphs of f n (x) for severl vlues of n. 24

5 Theorem If f n : [, b] R is integrble on [, b] for every n N, nd (f n ) converges uniformly on [, b], then f n (t)dt = f n(t)dt. Proof. Let f n f uniformly. For ny ɛ > there exists n integer N such tht n N implies sup x [,b] f n (x) f(x) < ɛ, by Proposition 3.6. Hence ( fn (t) f(t) ) dt f n (t) f(t) dt < ɛdt = ɛ(b ) for ll n N. Thus ( f n(t)dt f(t)dt) tends to s n. Exmple Let f n (x) = xe nx2 on [, 1], for n 1. Since < f n (x) < 1/nx for ech < x 1, the pointwise it of (f n ) is the zero function. To check for uniform convergence, consider sup x [,1] f n (x) ; this requires identifying the turning points of f n (x) on [, 1]. But df n (x)/dx = e nx2 + ( 2nx)xe nx2, which is iff x = 1/(2n) 1/2 ; furthermore, the derivtive is > to the left of the turning point, nd < to the right. Thus f n (1/(2n) 1/2 ) = 1/(2ne) 1/2 is the mximum vlue of f n (x). Hence sup x [,1] f n (x) tends to n (1/(2ne) 1/2 ) =, nd the convergence is uniform. Then Theorem 3.13 confirms the well-known fct tht te nt2 dt =. 25