86 3.5 Uniform Continuity Figure 3.4: Continuous but not uniformly continuous on(0,∞). Proof: Suppose first 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 ∈Dand |u−v| <δ. Let N∈Nbe 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 anyδ >0, there exist u,v ∈Dwith |u−v| <δ and| f (u)−f (v)|≥ε0. Thus, for everyn∈N, there exist un,vn ∈Dwith |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 lim n→∞ (f (un)−f (vn))=lim n→∞ 1 1/(n+1) − 1 1/n =lim n→∞ (n+1−n)=1̸=0. 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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