Convergence in Metric Spaces Functional Analysis Lecture 3: Convergence and Continuity in Metric Spaces Bengt Ove Turesson September 4, 2016 Suppose that (X, d) is a metric space. A sequence (x n ) X is said to be convergent with the limit x X if d(x, x n ) 0 as n. We denote this circumstance by writing x n x or lim n x n = x. The limit of a convergent sequence in a metric space is unique. Suppose that (X, d) is a metric space and that x n x and x n x. Then 0 d(x, x ) d(x, x n ) + d(x n, x ) 0 as n, so d(x, x ) = 0 and hence x = x. 1 / 19 2 / 19 Convergence in Metric Spaces Convergence in Metric Spaces In R d or C d, x n = (x (n) 1,..., x (n) d )t x = (x 1,..., x d ) t as n if and only if ( d ) 1/2 d 2 (x, x n ) = x 2 0 as n. =1 This is in turn is equivalent to x (n) x as n for = 1, 2,..., d. Conclusion: Convergence in R d and C d is equivalent to convergence in every coordinate. In l, take and x n = (1, 1,..., 1, 0, 0...) for n = 1, 2,..., }{{} n 1:s x = (1, 1,...). Then x (n) x as n for = 1, 2,..., but d (x, x n ) = 1 for every n, so x n x. Conclusion: In the space l, convergence in every coordinate does not imply convergence. 3 / 19 4 / 19
Convergence in Metric Spaces In l (M), f n f if d(f, f n ) = sup f (x) f n (x) 0 as n. x M This type of convergence is known as uniform convergence. Uniform convergence implies pointwise convergence: f (x) f n (x) sup f (y) f n (y) 0 for every x X. y M The previous example shows that the converse is not true. Characterisation of Accumulation Points Suppose that (X, d) is a metric space and let A be a subset of X. Then x A if and only if there exists a sequence (x n ) A such that x n x for every n and x n x. Recall that A denotes the set of accumulation points of A. Necessity: If x A, then, for n = 1, 2,..., there exists an element x n A such that x n x and x n B 1/n (x). But then d(x, x n ) < 1 n, so x n x. Sufficiency: Suppose that x X and x n x, where x n A and x n x for every n. Let r > 0 be arbitrary. Since x n x, x n B r (x) if n is large enough. This proves that x A. 5 / 19 6 / 19 Characterisation of Closed Sets Suppose that (X, d) is a metric space and let F be a subset of X. Then F is closed if and only if the limit of any convergent sequence of elements in F belongs to F. Necessity: Suppose that F is closed and that (x n ) F satisfies x n x X. If x n = x for some n, then x F. Otherwise, it follows from the previous proposition that x F and therefore that x F. But F = F since F is closed, so x F. Sufficiency: Suppose that x F and choose a sequence (x n ) F such that x n x. Then, by the assumption, x F. Thus, F F, which implies that F = F, so F is closed. Characterisation of Closed Sets Let us show that if M R d, then C b (M) is a closed subset of l (M). Suppose that (f n ) C b (X ) and that f n f l (M). Let ε > 0 be given and choose n so large that d (f, f n ) < ε. Let x M be arbitrary. Since f n is continuous at x, there exists a δ > 0 such that f n (x) f n (y) < ε for every y X such that d 2 (x, y) < δ. It then follows that f (x) f (y) f (x) f n (x) + f n (x) f n (y) + f n (y) f (y) 2d (f, f n ) + ε < 3ε if d 2 (x, y) < δ. This shows that f is continuous at x. By assumption, f is bounded. Hence, f belongs to C b (M). This example shows that continuity is preserved under uniform convergence. 7 / 19 8 / 19
Cauchy Sequences Cauchy Sequences Suppose that (X, d) is a metric space. A sequence (x n ) X is a called a Cauchy sequence if d(x m, x n ) 0 as m, n. The sequence ( 1 n ) n=1 (0, 1] is a Cauchy sequence since 1 m 1 n 1 m + 1 0 as m, n. n Every convergent sequence in a metric space is a Cauchy sequence. Suppose that (X, d) is a metric space and that x n x. Then 0 d(x m, x n ) d(x m, x) + d(x, x n ) 0 as m, n, so d(x m, x n ) 0 as m, n. According to the previous example, the converse to this proposition is false: ( 1 n ) n=1 is a Cauchy sequence in (0, 1], but does not converge in (0, 1]. 9 / 19 10 / 19 A metric space (X, d) is said to be complete if every Cauchy sequence in X is convergent. According to the previous example, (0, 1] is not a complete metric space. The metric space Q (with the same metric as in R) is not complete: Take for instance x n = (1 + 1 n )n for n = 1, 2,.... This is a Cauchy sequence in Q since it converges to e in R. But since e / Q, the sequence is not convergent in Q. Theorem (The Cauchy Convergence Principle) Every Cauchy sequence in R is convergent, i.e., R is a complete metric space. See, e.g., the proof in the appendix to Kreysig s book. We shall give another proof of this fact later on. The space C is complete. The spaces R d and C d are complete. 11 / 19 12 / 19
The space l p is complete for 1 p <. Beginning of proof Let (x n ) n=1 be a Cauchy sequence in lp, where x n = (x (n) ) =1 lp. Given an arbitrary ε > 0, choose N so large that ( ) 1/p d p (x m, x n ) = x (m) p < ε if m, n N. =1 This choice implies that x (m) < ε if m, n N for every fixed index 1, so (x (n) ) n=1 is a Cauchy sequence for every index 1. The completeness of C now implies that there exist numbers x C such that x (n) x as n. End of proof Let x = (x ) =1. It follows from the above that k =1 x (m) p < ε p if m, n N for every k 1. If we first let m and then k in this inequality, we obtain that =1 x p ε p if n N. This shows that x x n l p. But l p is a vector space, so x = (x x n ) + x n l p. The last inequality also shows that d(x, x n ) ε if n N. Hence, x n x. 13 / 19 14 / 19 The space l (M) is complete for any nonempty set M. Exercise The space l is complete. Suppose that (X, d) is a complete metric space and F X. Then (F, d) is complete if and only F is closed. Necessity: Suppose that (F, d) is complete. If F x n x X, then (x n ) is a Cauchy sequence, so x F since (F, d) is complete. This shows that F is closed. Sufficiency: Suppose that F is closed. If (x n ) F is a Cauchy sequence, then x n x X since X is complete. It follows that x F since F is closed. This shows that (F, d) is complete. The space C b (M) is complete for any nonempty subset M of R d. Notice that C b (M) is a closed subset of the complete space l (M). 15 / 19 16 / 19
Continuity in Metric Spaces Recall that a function f from R d to R is said to be continuous at a point x R d if there for every number ε > 0 exists a number δ > 0 such that f (x) f (x ) < ε if x x = d 2 (x, x ) < δ. This definition carries over directly to metric spaces. Suppose that (X, d X ) and (Y, d Y ) are two metric spaces. A function f from X to Y is continuous at x X if there for every number ε > 0 exists a number δ > 0 such that d Y (f (x), f (x )) < ε if d X (x, x ) < δ. A function f from X to Y is continuous if it is continuous at every point in X. Continuity in Metric Spaces Suppose that (X, d X ) and (Y, d Y ) are two metric spaces. Then a function f : X Y is continuous at x X if and only if f (x n ) f (x) for every sequence (x n ) X such that x n x. The equivalent condition to continuity in the proposition is sometimes called sequential continuity. In a metric space, continuity is therefore equivalent to sequential continuity. Necessity: Suppose that f is continuous at x and that x n x. Given ε > 0, let δ > 0 be as in the definition of continuity. Then d X (x, x n ) < δ if n is sufficiently large, so it follows that d Y (f (x), f (x n )) < ε for those n. Thus, f (x n ) f (x). Sufficiency: Suppose that f is not continuous at x. Then there exists a number ε > 0 and a sequence (x n ) X such that d X (x, x n ) < 1 n for every n, but d Y (f (x), f (x n )) ε. This contradicts the assumption. 17 / 19 18 / 19 Continuity in Metric Spaces Given a metric space (X, d), C b (X ) denotes the space of bounded, continuous functions from X to C. The metric space C b (X ) is complete. Same proof as for the case X = R d. 19 / 19