Math

Math 320, Real Analysis I

Solutions to Homework 7 Problems

Exercise 4.4.13. (Continuous Extension Theorem). (a) Show that a uniformly continuous function preserves Cauchy sequences; that is, if f : A → R is uniformly continuous and (xn ) ⊆ A is a Cauchy sequence, then show that f (xn ) is a Cauchy sequence. Proof. Suppose we have a Cauchy sequence (xn ). Let’s examine f (xn ). Let ε > 0 be given. Then we can find a δ > 0 such that whenever |x − y| < δ in A we have |f (x) − f (y)| < ε, because f is uniformly continuous on A. On the other hand, we know that for our δ > 0 we can find an N ∈ N such that if n, m > N we have |xn − xm | < δ because (xn ) is a Cauchy sequence. But now it’s clear that if n, m > N we’ll also have |f (xn ) − f (xm )| < ε, so f (xn ) is Cauchy. (b) Let g be a continuous function on the open interval (a, b). Prove that g is uniformly continuous on (a, b) if and only if it is possible to define values g(a) and g(b) at the endpoints so that the extended function g is continuous on [a, b]. Proof. (=⇒) Suppose g is uniformly continuous on the open interval (a, b). Pick any Cauchy sequence (xn ) in (a, b) converging to a. Then g(xn ) is Cauchy, by part (a), so it converges by the Cauchy Criterion. Define g(a) = lim g(xn ). We now show that g is continuous at a for this value of g(a). Let ε > 0 be given. ε Then there exists δ > 0 such that |x − y| < δ in (a, b) implies |g(x) − g(y)| < 2 by the uniform continuity of g on (a, b). Since (xn ) → a and δ > 0, there exists K ∈ N δ such that |xn − a| < 2 whenever n ≥ K. Also, since g(xn ) → g(a), there is a natural ε number M ∈ N such that |g(xn ) − g(a)| < 2 for all n ≥ M . Set N = max{K, M } ∈ N. δ ε Then |xN − a| < 2 and |g(xN ) − g(a)| < 2 from the above. Now suppose x ∈ [a, b] and δ |x − a| < 2 . If x = a, then |g(x) − g(a)| = 0 < ε and we are done. Otherwise, suppose x > a. Then |x − xN | = |x − a + a − xN | ≤ |x − a| + |a − xN | = |x − a| + |xN − a| < δ δ + = δ, 2 2

ε so we know from the uniform...