Summer Jump-Start Program for Analysis, 2012 Song-Ying Li. 1 Lecture 7: Equicontinuity and Series of functions
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1 Summer Jump-Start Program for Analysis, 0 Song-Ying Li Lecture 7: Equicontinuity and Series of functions. Equicontinuity Definition. Let (X, d) be a metric space, K X and K is a compact subset of X. C(K) denotes the set of continuous functions on f on K. Let F (K) be the subset of C(K). Then (a) F (K) is bounded pointwise on K if for any given x K there is M x > 0 such that f(x) M x for all f F (K); (b) F (K) is bounded uniformly on K if there is M > 0 such that f(x) M for all f F (K) and x K; (c) F (K) is equicontinuous on K if for any ɛ > 0, there is δ > 0 such that if d X (x, y) < δ, x, y K, then f(x) f(y) < ɛ for all f F (K). EXAMPLE () F = { nx : n =,,...} is pointwise bounded on K = (0, ], but not bounded on K; () F = {nx : n =,,...} is not equicontinuous on [0, ]; (3) F = {f α : α } with f α(x) x K = [a, b], α Λ then F is equicontinuous on [a, b]. Proof. () and () are straight forward. Now we prove (3). For any ɛ > 0, let δ = ɛ. When x, y [a, b] and x y < δ, we have f(x) f(y) = f (ξ) x y x y δ = ɛ By deifnition, F is equicontinuous on [a, b]. THEOREM. (Arzelà - Ascoli Theorem) Let F be a family of continuous functions on [a, b] IR. Then the following two statements are equivalent: (a) F is equi-continuous on [a, b] and F is bounded pointwise on [a, b]. (b) F is bounded on [a, b] and (b) every sequence {f n } F has a uniformly convergent subsequence {f nk } on [a, b]. Proof. b) a) (i) By (b), F is bounded pointwise on [a, b]. Suppose that F is not equicontinuous on [a, b], then there exists ɛ 0 > 0 such that for any δ = k (k IN), there are two points x k, y k [a, b] with x k y k < δ = k, and there is f k F such that f k (x k ) f k (y k ) ɛ 0.
2 By (b), {f k } has a subsequence {f kl } and f on [a, b] such that f kl f uniformly on [a, b] as l. Therefore, f C([a, b]). We may choose a convergent subsequence {x klj } of {x kl }, without loss of generality, we may assume x kl x [a, b] for some x. Since x kl y kl < k l 0 as l. So, y kl x as l. Therefore, when l, one has ɛ 0 f kl (x kl ) f kl (y kl ) f kl (x kl ) f(x kl ) + f(x kl ) f(y kl ) + f(y kl ) f kl (y kl ) 0. This is a contradiction. So, F is equicontinuous. Next, we prove a) b.) ( i) It is easily to prove that (a) implies (b). (ii) Next we prove (a) implies (b). For any sequence {f n } F, we need to choose a subsequence {f nk } and f such that f nk f uniformly on [a, b]. Step : Choose a subsequence {f nk } k= which converges pointwise on Q [a, b]. Since Q is dense in IR, Q is countable. Thus, Q [a, b] = {x, x,..., x n,...}. Since {f n (x )} is bounded sequence in IR. There is a convergent subsequence: {f,n (x )} with f,n (x ) f(x ) (some number in IR, we call it f(x )). Chose a subsequence {f,n (x)} from {f,n (x)} such that f,n (x ) f(x ) IR,. In particular, f,n (x ) f(x ). Continuing in this fashion, choose {f k,n (x)} from {f (k ),n } such that f k,n (x j ) f(x j ) j k as n for all k =, 3,. Let f nk (x) = f k,k (x). Then {f nk } is a subsequence of {f n } with f nk (x j ) f(x) as k for each x Q [a, b]. Step : Since {f nk } is a family of equicontinuous on [a, b], one limit function f(x) is uniformly continuous on Q [a, b]. For any ɛ > 0, since {f nk } is equicontinuous on [a, b], there is δ > 0 such that for any x, y [a, b] with x y < δ, one has f nk (x) f nk (y) < ɛ, k IN. For any x, y [a, b] Q with x y < δ, there is k >> such that f nk (x) f(x), f nk (y) f(y) < ɛ. Thus, f(x) f(y) f nk (x) f(x) + f nk (x) f nk (y) + f nk (y) f(y) 3ɛ. Therefore, f is uniformly continuous on [a, b] Q. So, we can extend f as a continuous function on [a, b]. For the ɛ > 0, there is δ > 0 such that if x y < δ, we have f(x) f(y) < ɛ. Choose rational numbers {r,, r m } [a, b] such that for any x [a, b], there is r k {r,, r m } such that x r k < δ. Thus for any x [a, b], we choose
3 r j {r,, r m } such that r j x < δ. There is k 0 such that k k 0, we have f nk (r j ) f(r j ) < ɛ for all j =,,, m. Therefore, f nk (x) f(x) f nk (x) f nk (x j ) + f(r j ) f nk (x j ) + f(x j ) f nk (x) < 3ɛ, So, f nk f uniformly on [a, b] as k.. Series of functions Let (X, d) be a metric space or IR n and let K X. We consider a sequence of functions {f n (x)} x= on K. Let s n (x) = n f k (x), x K, n =,, 3,. k= Definition.3 (a) f n(x) converges on K if lim n s n (x) exists for each x K. (b) f n(x) converges absolutely on K if f n(x) converges on K. (c) f n(x) converges uniformly on K if {s n } uniform Cauchy sequence on K. Question: How to test if a series converges uniformly on a set K? THEOREM.4 (Weirstrass M-test) Let {f n (x)} be a sequence of functions on K such that f n (x) M n, x K, n IN. If M n < +, then f n(x) converges absolutely and uniformly on K. Proof. For the ɛ > 0, we need to fin N such that if m > n N, then m k=n f k(x) < ɛ, x K. Since k= M k < +, for ɛ > 0, there is N s.t. if m > n N, then m k=n M k < ɛ. Therefore, when m > n N, m k=n f k(x) m k=n M k < ɛ, x K. Thus, k= f k(x) converges absolutely and uniformly on K. EXAMPLE Determine if (, ). sin(nx) n converges uniformly for x IR = Solution. We claim: sin(nx) n converges uniformly on IR. Since sin(nx) n, x IR, for all n =,,... n Notice that n < + (p-series with p = ). By Weierstrass M-test, we have converges absolutely and uniformly on IR. sin(nx) n 3
4 EXAMPLE 3 Find x (+x ), x (, ). n Solution Since For any x 0, since Therefore x ( + x ) n = x f(0) = 0 ( + 0) n = 0. ( ) n + x = x +x = x + x + x +x x =. { 0 x = 0, x 0. EXAMPLE 4 Discuss : f n(x) = +n x by answering the following questions. (a) Where does the series converge? i.e., where is f(x) well-defined? (b) Where does the series converge uniformly? (c) Where does the series not converge uniformly? (d) Where is f(x) continuous? Solution Since f n (x) is well-defined on IR\{ /n } and it is continuous there. Let K = (, )\{ /n : n =,,...}. (a) when x = 0, f n (0) = 0, so series does not converge at x = 0, and f(0) = +. Let K 0 = K\{0}. We claim f n(x) converges on K 0. For any x 0 K 0, we divide it into two cases: (i) x 0 > 0, f n (x 0 ) = +n x 0 x 0 n. By the comparison test, f n(x 0 ) converges. (ii) x 0 < 0. There is N such that + n x 0 < 0 if n N. Thus, f n (x 0 ) = + n x 0 n x 0 n x 0 = x 0 n, n N. Since n < by comparison test, f n(x 0 ) converges absolutely. (b) +n x converges uniformly on {x IR : x δ} K 0 for any fixed δ > 0. Proof. For any δ > 0, there is N such that if n N then n δ n δ +. Thus, for any n N, x δ, x K 0, we have (i) if x δ, then f n (x) = +n x n x n δ ; (ii) If x < δ and n > /δ, then f n (x) = + n x n x n δ + n δ. 4
5 So f n(x) converges uniformly on { x δ}. The proves the claim (b). (c) For any δ > 0, f n(x) does not converge uniformly on ( δ, δ) K 0. Proof. Let ɛ 0 = /. For N >, m = n +, n = N; x N = (+N) ( δ, δ). m f k (x N ) = f N+(x N ) = + (N + ) = = ɛ 0 (N+) k=n+ So, f n(x) does not converge uniformly on ( δ, δ) K 0. (d) We know by part (a) that f(x) is well-defined on K 0. We know f n (x) is continuous on K 0 for n =,, 3..., we also know by (b) that f n(x) converges uniformly on K 0 {x IR : x δ} for any given δ > 0. This implies that f n(x) is continuous on K 0 { x δ}, δ > 0. Therefore, f(x) is continuous on K 0 δ>0 { x δ} = K 0 IR \ {0} = K 0. EXAMPLE 5 Let : (nx) n, where (x) is fractional part of x. i.e. (x) = x [x], [x] = integer part of x. Discuss where f(x) is continuous. Solution Since (i) f n (x) = (nx) n IR \ Q for all n. = nx [nx] n is continuous on K =: IR\{x = k n, k Z} = (ii) (nx) n n, for all n =,,... We know that n converges. By the Weierstrass M-Test, we know that f(x) is well-defined on IR, and f(x) is continuous on K. THEOREM.5 (Stone-Weierstrass Theorem) Let f(x) be continuous on [a, b]. Then there is a sequence of polynomials p n (x) such that p n (x) f(x) uniformly on [a, b]..3 Exercise. Prove that every uniformly convergent sequence of bounded functions is uniformly bounded.. If sequences of functions {f n } and {g n } converge uniformly on a set E. (a) Prove that {f n + g n } converges uniformly on E. (b) Does {f n g n } converge uniformly on E? 3. Consider + n x For what values of x does the series converges absolutely? On what intervals does it converges uniformly? On what interval does it fail to converge uniformly? Is f continuous wherever the series converges? Is f bounded? 5
6 4. Let 0, if x < n+, sin ( π x ), n+ x n 0, if n < x Show that {f n } converges to a continuous function, but not uniformly. Use the series f n to show that absolute convergence, even for all x, does not implies uniform convergence. 5. Prove that the series ( ) n x + n converges uniformly in every bounded interval, but does not converge absolutely for any value of x. 6. If I(x) = 0 if x 0 and I(x) = if x > 0. Let {x n } be a sequence of distinct points of (a, b), and if c n converges, prove that the series n c n I(x x n ) (a x b) converges uniformly on [a, b], and that f is continuous for every x x n. 7. Letting (x) denote the fractional part of the real number x, consider the function (nx) n, x IR Find all discontinuities of f, and show that they form a countable dense set. 8. Suppose {f}, {g n } are two sequences of functions defined on set E, and (a) f n has uniformly bounded partial sums; (b) g n 0 uniformly on E; (c) g (x) g (x) g 3 (x) for every x E. Prove that f ng n converges uniformly on E. 9. Prove or disprove {f n (x) = x n : n IN} is equicontinuous on [0, ). 6
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