Introduction to Mathematical Analysis I - Second Edition

87 u , v ∈ D and | u − v | < δ . Let N ∈ N be such that | u n − v n | < δ for n ≥ N . For such n , we have | f ( u n ) − f ( v n ) | < ε . This shows lim n → ∞ ( f ( u n ) − f ( v n )) = 0. To prove the converse, assume condition (C) holds and suppose, by way of contradiction, that f is not uniformly continuous. Then there exists ε 0 > 0 such that for any δ > 0, there exist u , v ∈ D with | u − v | < δ and | f ( u ) − f ( v ) | ≥ ε 0 . Thus, for every n ∈ N , there exist u n , v n ∈ D with | u n − v n | ≤ 1 / n and | f ( u n ) − f ( v n ) | ≥ ε 0 . It follows that for such sequences, lim n → ∞ ( u n − v n ) = 0, but { f ( u n ) − f ( v n ) } 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 u n = 1 / ( n + 1 ) and v n = 1 / n for all n ≥ 2. Then clearly, lim n → ∞ ( u n − v n ) = 0, but lim n → ∞ ( f ( u n ) − f ( v n )) = lim n → ∞ 1 1 / ( n + 1 ) − 1 1 / n = lim n → ∞ ( n + 1 − n ) = 1 6 = 0 . The following theorem shows one important case in which continuity implies uniform continuity. Theorem 3.5.4 Let f : D → R be a continuous function. Suppose D is compact. Then f is uniformly continuous on D . Proof: Suppose by contradiction that f is not uniformly continuous on D . Then there exists ε 0 > 0 such that for any δ > 0, there exist u , v ∈ D with | u − v | < δ and | f ( u ) − f ( v ) | ≥ ε 0 . Thus, for every n ∈ N , there exist u n , v n ∈ D with | u n − v n | ≤ 1 / n and | f ( u n ) − f ( v n ) | ≥ ε 0 . Since D is compact, there exist u 0 ∈ D and a subsequence { u n k } of { u n } such that u n k → u 0 as k → ∞ . Then | u n k − v n k | ≤ 1 n k , for all k and, hence, we also have v n k → u 0 as k → ∞ . By the continuity of f , f ( u n k ) → f ( u 0 ) and f ( v n k ) → f ( u 0 ) . Therefore, { f ( u n k ) − f ( v n k ) } converges to zero, which is a contradiction. The proof is now com- plete.