86 3.5 Uniform Continuity Figure 3.4: Continuous but not uniformly continuous on (0,∞). Proof: Suppose frst that f is uniformly continuous and let {un}, {vn} be sequences in D such that limn →∞(un − vn)= 0. Let ε > 0. Choose δ > 0 such that | f (u) − f (v)| < ε whenever u,v ∈ D and |u − v| < δ . Let N ∈ N be such that |un − vn| < δ for n ≥ N. For such n, we have | f (un) − f (vn)| < ε. This shows limn →∞( f (un) − f (vn)) = 0. To prove the converse, assume (ii) holds and suppose, by way of contradiction, that f is not uniformly continuous. Then there exists ε0 > 0 such that for any δ > 0, there exist u,v ∈ D with |u− v| < δ and | f (u) − f (v)|≥ ε0. Thus, for every n ∈ N, there exist un,vn ∈ D with |un − vn|≤ 1/n and | f (un) − f (vn)|≥ ε0. It follows that for such sequences, limn →∞(un − vn)= 0, but { f (un) − f (vn)} does not converge to zero, which contradicts the assumption. □ ■ Example 3.5.7 Using this theorem, we can give an easier proof that the function in Example 3.5.6 is not uniformly continuous. Consider the two sequences un = 1/(n + 1) and vn = 1/n for all n ≥ 2. Then clearly, limn →∞(un − vn)= 0, but 1 1 lim ( f (un) − f (vn)) = lim − = lim (n+ 1 − n)= 1 ̸ = 0. n→∞ n→∞ 1/(n + 1) 1/n n→∞ The following theorem shows one important case in which continuity implies uniform continuity. Theorem 3.5.4 Let f : [a,b] → R. If f is continuous on [a,b], then f is uniformly continuous on [a.b].
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