Matthew Straughn Math 402 Homework 6 Homework 6 (p. 452) 14.3.3, 14.3.4, 14.3.5, 14.3.8 (p. 455) 14.4.3* (p. 458) 14.5.3 (p. 460) 14.6.1 (p. 472) 14.7.2* Lemma 1. If (f (n) ) converges uniformly to some function f, then the sequence (of f (n) s) is uniformly Cauchy. Proof. Since f (n) f uniformly, we know that given ɛ 1 > 0 there exists N 1 > 0 such that d Y (f (n) (x), f(x)) < ɛ 1 for all n > N 1 and x X. So we also have d Y (f (m) (x), f(x)) < ɛ 1 for all m > N 1 and x X. So given ɛ > 0 take ɛ 1 /2 and N = N 1. Then we get that for all n, m > N and x X. d Y (f (n) (x), f (m) (x)) d Y (f (n) (x), f(x)) + d Y (f(x), f (m) (x)) < ɛ Hence (f (n) ) is uniformly Cauchy. Exercise 14.3.1. Prove Proposition 14.3.1. Proposition 14.3.1 Suppose (f (n) ) is a sequence of functions from one metric space (X, d X ) to another (Y, d Y ), and suppose that this sequence converges uniformly to another function f : X Y. Let x 0 be a point in X. If the functions f (n) are continuous at x 0 for each n, then the limiting function f is also continuous at x 0. Proof. Since f (n) converges uniformly to f, we know that given ɛ > 0 there exists N > 0 such that d Y (f (n) (x), f(x)) < ɛ for all n > N and x X. Since f (n) is continuous at x 0 X we know that given ɛ > 0 there exists δ > 0 such that if d X (x, x 0 ) < δ then d Y (f (n) (x), f (n) (x 0 )) < ɛ So given ɛ > 0 take ɛ /3, ɛ /3 and δ = δ. If d X (x, x 0 ) < δ = δ, then d Y (f(x), f(x 0 )) d Y (f(x), f (n) (x)) + d Y (f (n) (x), f (n) (x 0 )) + d Y (f (n) (x 0 ), f(x 0 )) < ɛ/3 + ɛ/3 + ɛ/3 (by uniform convergence and continuity) Thus f is continuous at x 0. 1
Exercise 14.3.2. Prove Proposition 14.3.3. Proposition 14.3.3. Let (X, d X ) and (Y, d Y ) be metric spaces, with Y complete, and let E be a subset of X. Let (f (n) ) be a sequence of functions from E to Y, and suppose that this sequence converges uniformly in E to some function f : E Y. Let x 0 X be an adherent point of E, and suppose that for each n the limit lim x x0 ;x E f (n) (x) exists. Then the limit lim x x0 ;x E f(x) also exists, and is equal to the limit of the sequence (lim x x0 ;x E f (n) (x)) ; in other words we have the interchange of limits lim n lim f (n) (x) = lim lim f (n) (x). x x 0 ;x E x x 0 ;x E n Proof. Suppose f (n) converges uniformly to f. That is, given ɛ 1 > 0 there exists N 1 > 0 such that d Y (f (n) (x), f(x)) < ɛ 1 for all n > N 1 and x X. Let x n x 0. And so suppose for each n that lim k f (n) (x k ) = L n. So we also have given ɛ 2 > 0 there exists N 2 > 0 such that d Y (f (n) (x k ), L n ) < ɛ 2 for all n > N 2 and x X. Now let us show that (L n ) is a Cauchy sequence. Given ɛ > 0 take ɛ 1 2 /4 and N = min(n 1, N 2 ). Then d Y (L n, L m ) d Y (L n, f (n) (x k )) + d Y (f (n) (x k ), f(x k )) + d Y (f(x k ), f (m) (x k )) + d Y (f (m) (x k ), L m ) < ɛ/4 + ɛ/4 + ɛ/4 + ɛ/4 So (L n ) is a Cauchy sequence, and since Y is complete we know that this sequence converges to, say, L 0. That is, given ɛ 3 > 0 there exists N 3 > 0 such that d Y (L n, L 0 ) < ɛ for all n > N 3. Now we will show that lim f(x k) = L 0 = lim lim f (n) (x k ) k n k Given ɛ > 0 take ɛ 1 2 3 /3 and N = min(n 1, N 2, N 3 ). Then we get that As desired. d Y (f(x k ), L 0 ) d Y (f(x k ), f (n) (x k )) + d Y (f (n) (x k ), L n ) + d Y (L n, L 0 ) < ɛ/3 + ɛ/3 + ɛ/3 Exercise 14.3.3. Compare Proposition 14.3.3 with Example 1.2.8. Can you now explain why the interchange of limits in Example 1.2.8 led to a false statement, whereas the interchange of limits in Proposition 14.3.3 is justified? Solution. You cannot change the limits in Example 1.2.8 because the sequence of functions f (n) = x n does not converge uniformly on [0, 1], and hence does not contradict Proposition 14.3.3. 2
Exercise 14.3.4. Prove Proposition 14.3.4. Proposition 14.3.4: Let (f (n) ) be a sequence of continuous functions from one metric space (X, d X ) to another (Y, d Y ), and suppose that this sequence converges uniformly to another function f : X Y. Let x (n) be a sequence of points in X which converge to some limit x. Then f (n) (x (n) ) converges to f(x). Proof. Let x (n) be a sequence that converges to x. By the continuity of the f (n) s we have that f (n) (x (m) ) f (n) (x) as m. That is, given ɛ 1 > 0 there exists N 1 > 0 such that d Y (f (n) (x (m) ), f (n) (x)) < ɛ 1 for all m > N 1. But since f (n) f uniformly we know given ɛ 2 > 0 there exists N 2 > 0 such that d Y (f (n) (x), f(x)) < ɛ 2 for all n > N 2 and x X. Given ɛ > 0 take ɛ 1 2 /3 and N = max{n 1, N 2 }. Then we have that d Y (f (n) (x (m) ), f (n) (x)) < ɛ/3 and d Y (f (n) (x), f(x)) < ɛ/3 for all m, n > N and x X. In particular we have that d Y (f (n) (x (m) ), f (n) (x (n) )) < ɛ/3. But by the triangle inequality we have: d Y (f (n) (x (n) ), f(x)) d Y (f (n) (x (n) ), f (n) (x (m) )) + d Y (f (n) (x (m) ), f (n) (x)) Thus f (n) (x (n) ) converges to f(x). + d Y (f (n) (x), f(x)) < ɛ for all n > N and x X. Exercise 14.3.5. Give an example to show that Proposition 14.3.4 fails if the phrase converges uniformly is replaced by converges pointwise. Solution. Take our good friend f (n) = x n where f (n) : ( 1, 1) R and converges pointwise to the zero function (f(x) = 0 for all x). And take a sequence in ( 1, 1) that converges to 1. Then the theorem states that f (n) (x (n) ) converges to f(1) = 0, but if we take the sequence x (n) = 1 1 n we get that f (n) (x (n) ) = (1 1 n )n e 1 0. Exercise 14.3.6. Prove Proposition 14.3.6. Proposition 14.3.6. Let (f (n) ) be a sequence of functions from one metric space (X, d X ) to another (Y, d Y ), and suppose that this sequence converges uniformly to another function f : X Y. If the functions f (n) are bounded on X for each n, then the limiting function f is also bounded on X. Proof. Suppose that (f (n) ) is a Convergent sequence, to say f. Then from Lemma 1 we have that (f (n) ) is also uniformly Cauchy. That is, given ɛ > 0 there exists N > 0 such that d Y (f (n) (x), f(x)) < ɛ for all n > N and x X. And also that d Y (f (n) (x), f (m) (x)) < ɛ for all n, m > N and x X. Also suppose that for each n that f (n) is bounded. That is, there is some ball B (Y,dY )(y n, R n ) such that f (n) (x) B (Y,dY )(y n, R n ) for all x X. That is, d Y (f (n) (x), y n ) < R n for all x X. 3
But, d Y (f (m) (x), y n ) d Y (f (m) (x), f (n) (x)) + d Y (f (n) (x), y n ) < ɛ + R n Thus, f (m) (x) B (Y,dY )(y n, R n + ɛ) for all m > N and x X. Hence there exists N > 0 (same N as above) such that f (n) (x) B (Y,dY )(y n, R n + ɛ) for all n > N and x X. But, d Y (f(x), y n ) d Y (f(x), f (n) (x)) + d Y (f (n) (x), y n ) < ɛ + R n + ɛ = R n + 2ɛ So f(x) B (Y,dY )(y n, R n + 2ɛ) for all x X. Therefore f(x) is bounded on X. Exercise 14.3.7. Give an example to show that Proposition 14.3.6 fails if the phrase converges uniformly is replaced by converges pointwise. Solution. Consider the same example as in Exercise 14.2.2(c). That is, g (N) = N xn. Each of these is a finite sum and hence bounded. However, this sequence converges pointwise to g(x) = x 1 x which is not bounded on ( 1, 1). Exercise 14.3.8. Let (X, d) be a metric space, and for every positive integer n, let f (n) : X R and g (n) : X R be functions. Suppose that (f (n) ) converges uniformly to another function f : X R, and that (g (n) ) converges uniformly to another function g : X R. Suppose also that the functions (f (n) ) and (g (n) ) are uniformly bounded. Prove that the functions f (n) g (n) : X R converge uniformly to fg : X R. Proof. Suppose that (f (n) ) converges uniformly to another function f : X R, and that (g (n) ) converges uniformly to another function g : X R That is, given ɛ > 0 there exists N > 0 such that f (n) (x) f(x) < ɛ and g (n) (x) g(x) < ɛ for all n > N and x X. Also suppose that the functions (f (n) ) and (g (n) ) are uniformly bounded. That is, there exists M > 0 such that f (n) (x) M and g (n) (x) M for all n 1 and x X. And hence from the previous exercise we know that f(x), g(x) M for all x X. (Note that I am being lazy and using the same M as above. This is not necessarily true, but we can take the max between the two M s and just call that M). Given ɛ > 0 take ɛ /M and choose N = N. Thus we have that f (n) (x) f(x) < ɛ = ɛ/m and g (n) (x) g(x) < ɛ /M for all n > N and x X. But, f (n) g (n) (x) fg(x) = f (n) g (n) (x) fg(x) + f (n) g(x) f (n) g(x) Thus f (n) g (n) fg uniformly. = f (n) (x)(g (n) (x) g(x)) + g(x)(f (n) (x) f(x)) f (n) (x) g (n) (x) g(x) + g(x) f (n) (x) f(x) M ɛ M + M ɛ M 4
Exercise 14.5.3. Prove Theorem 14.5.7. Theorem 14.5.7. Let (X, d) be a metric space, and let (f (n) ) be a sequence of bounded real-valued continuous functions on X such that the series f (n) is convergent. Then the series f (n) converges uniformly to some function f on X, and that function f is also continuous. Proof. Let us show that N i=1 f (i) is a Cauchy sequence in C(X R). Since f (n) is convergent, we know that lim n f (n) = 0. That is, given ɛ > 0 there exists N > 0 such that f (n) < ɛ for all n > N. So, f (n) = sup{f (n) (x) : x X} < ɛ for all n > N. Given ɛ > 0 take ɛ = ɛ and N = N. Then n m n f (i) i=1 m f (i) = i=1 n i=m+1 f (i) n m sup{f (n) (x) : x X} ɛ < n m n m Note that we assumed n > m, but the same argument gives us the case n < m, and n = m is obvious. Thus, N i=1 f (i) is a Cauchy sequence in C(X R) (I did not state this earlier, but a finite sum of continuous and bounded functions is also continuous and bounded, so each term is in C(X R)). And now using Theorem 14.4.5. we have that f (n) converges uniformly to some function f on X, and that function f is also continuous. Exercise 14.6.1. Use Theorem 14.6.1 to prove Corollary 14.6.2. Theorem 14.6.1. Let [a, b] be an interval, and for each integer n 1, let f (n) : [a, b] R be a Riemann-integrable function. Suppose f (n) converges uniformly on [a, b] to a function f : [a, b] R. Then f is also Riemann integrable, and lim n f (n) = Corollary 14.6.2. Let [a, b] be an interval, and let (f (n) ) be a sequence of Riemann integrable functions on [a, b] such that the series f (n) is uniformly convergent. Then we have f (n) = f (n). f. 5
Proof. Let g (N) = N f (n). So we know that g (N) is Riemann integrable since each f (n) is Riemann integrable. Also g (N) f (n) = g, when N, and convergence is uniformy by assumption. Then from Theorem 14.6.1 we have that So we have But, Thus, g = g = g = lim g (N) N N = lim f (n) N [ = lim f (1) + f (2) + + N = f (1) + f (2) + = f (n). f (n). f (n) f (n) = f (n). f (N) ] 6