Assignment-10. (Due 11/21) Solution: Any continuous function on a compact set is uniformly continuous.

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1 Assignment-1 (Due 11/21) 1. Consider the sequence of functions f n (x) = x n on [, 1]. (a) Show that each function f n is uniformly continuous on [, 1]. Solution: Any continuous function on a compact set is uniformly continuous. (b) For a sequence of functions f n on [, 1] write what it means for them to NOT be an equicontinuous sequence. Solution: There exists ε > such that for each k, there exists points x k, y k and functions f nk such that x k y k < 1 k but f n k (x k ) f nk (y k ) > ε. (c) Using only part (b) (and not Ascoli-Arzela theorem) show that the sequence of functions f n is not equicontinuous. Solution: Let x n = 1 1/n and y n = 1, then ( f n (y n ) f n (x n ) = ) n n 1 e 1 >. n So if we take ε > such that ε < 1 e 1 (for instance we can take ε = (1 e 1 )/2), then there exists N such that for all n > N, f n (y n ) f n (x n ) > ε, and so by part(b), {f n } is not an equicontinuous family. (d) Now prove the same statement using Ascoli-Arzela. Solution: The family {f n } is clearly pointwise bounded by 1. So if the family was equicontinuous, then by Ascoli-Arzela, there will exist a uniformly convergent subsequence. BUt the pointwise limit of the sequence (and hence also the sub-sequence) is {, x 1 f(x) = 1, x = 1, 1

2 which is not continuous. This is a contradiction since uniform limits of continuous functions have to be continuous. 2. Suppose {f n } is an equicontinuous sequence of functions on a compact set K R. If {f n } converges pointwise on K, show that it converges uniformly on K. Solution: We show that the sequence {f n } is uniformly Cauchy. So let ε >. By equicontinuity, there exists a δ > such that for all n, x y < δ = f n (x) f n (y) < ε 3. Since K is compact, there exist finitely many points {p 1,, p m } such that K m k=1 B δ(p k ). By the Cauhy criteria for pointwise convergence, for each k = 1, 2,, m, there exists N k such that whenever n, m > N k, f n (p k ) f m (p k ) < ε 3. Let N = max(n 1,, N m ). Now for any x K, there exists p k such that x p k < δ. Then for n, m > N, f n (x) f m (x) f n (x) f n (p k ) + f n (p k ) f m (p k ) + f m (p k ) f m (x) < 3 ε 3 = ε. This completes the proof of the claim, and hence it follows that the sequence {f n } converges uniformly. 3. Recall that C [, 1] denotes the set of continuous functions on [, 1]. We endow it with the usual metric d(f, g) = sup f(t) g(t). t [,1] Define a function T : C [, 1] C [, 1] by Let K C [, 1] be a bounded set. T [f](x) = f(t) dt. (a) Show that T is a continuous function. Is it injective? Solution: For f, g C [, 1], and any t [, 1], by definition, f(t) g(t) d(f, g). 2

3 So for any x [, 1], T [f](x) T [g](x) = d(f, g) d(f, g). [f(t) g(t)] dt f(t) g(t) dt dt Taking supremum over all x [, 1], we see that So given an ε >, let δ = ε. Then d(t [f], T [g]) d(f, g). d(f, g) < δ = d(t [f], T [g]) < ε. This shows that T is a continuous map. We claim that the function is injective. It is enough to show that if T [f](x) = for all x [, 1], then f(t) = for all t [, 1]. To see this, note that if T [f](x) = for all x, then for any x, y [, 1], x < y, y x f(t) dt =. Now, suppose f(t ) > at some point t [, 1]. Since f is continuous, there exists δ > such that for any t (t δ, t + δ), f(t) >. But then t +δ t δ f(t) dt >, which is a contradiction. So there is not point where f is strictly positive. Similarly there is no point where f is strictly negative. Hence f has to be identically zero. (b) Show that the set T (K) is a compact subset of C [, 1]. Hint. Use the Version-2 of Ascoli-Arzela. Solution: Since K is a bounded set, there exists an M such that d(f, ) < M for all f K. Here denotes the zero function, that is the function that vanished on all of [, 1]. By the definition of d, this means that f(t) < M, t [, 1], f K. Closed. Trivially since it is the closure of a set. 3

4 Bounded. For any f K, since f(t) < M for all t [, 1], we see that and so T [f](x) = < M, f(t) dt f(t) dt d(t [f], ) < M. This shows that T (K) is bounded, and hence T (K) is also bounded. Equicontinuous. Let ε >, and δ = δ(ε) to be picked later. For any f C [, 1], by the fundamental theorem of calculus, T [f] is differentiable on [, 1], and moreover, T [f] (x) = f(x). So if f K, we have that T [f] (x) < M for all x [, 1]. Then by the mean value theorem, for any x, y [, 1], T [f](x) T [f](y) < M x y. Another way to see this is to directly estimate the integral. That is if x, y [, 1] with say x > y, then T [f](x) T [f](y) = In any case, if we take δ = ε/m, then y y f(t) dt f(t) dt < M x y. x y < δ = T [f](x) T [f](y) < ε. This shows that T (K) is equicontinuous. To see that the closure is also equicontinuous, we use the ε/3 trick. So let ε >, and δ = ε/3m. Then by the abvove argument for any f K, x y < δ = T [f](x) T [f](y) < ε 3. Now if g T (K), there exists an f K such that d(t [f], g) < ε/3. That is for any x [, 1], T [f](x) g(x) < ε 3. 4

5 Then if x y < δ, g(x) g(y) g(x) T [f](x) + T [f(x) T [f](y) + T [f](y) g(y) ε 3 + ε 3 + ε 3 = ε. So given an ε >, there exists a δ = δ(ε) (in this case δ = ε/3m works) such that for any g T (K), and hence T (K) is equicontinuous. x y < δ = g(x) g(y) < ε, Then by the version-2 of Arzela-Ascoli, the set T (K) is compact. 4. Suppose f : [, 1] R is a continuous function such that f(t)t n dt = for all n =, 1, 2,. Show that f(t) = for all t [, 1]. Hint. theorem to show that f 2 (t) dt =. Use Stone-Weierstrass Solution: Note that by the hypothesis, if P is any polynomial, f(t)p (t) dt =. By Stone-Wieirstrass theorem, there exists a sequence of polynomials P n f uniformly on [,1]. Since f is bounded on [, 1], this also implies that P n (t)f(t) f(t) 2 uniformly. (Note that this would not have been true if f were not bounded). By the theorem on uniform convergence and integration, f 2 (t) dt = lim n P n (t)f(t) dt =. We then claim that f(t) = for all t [, 1]. If not, then there exists a point p [, 1] such that f(p), and so f 2 (p) >. By continuity, there is a δ > such that f 2 (t) > for all t (p δ, p + δ). Since f 2 (t) is at least non-negative at other points, by comparison principle for integrals, a contradiction. < p+δ p δ f 2 (t) dt f 2 (t) dt =, 5