Introduction to Mathematical Analysis I - Second Edition

86 3.5 UNIFORM CONTINUITY While every uniformly continuous function on a set D is also continuous at each point of D , the converse is not true in general. The following example illustrates this point. Example 3.5.6 Let f : ( 0 , 1 ) → R be given by f ( x ) = 1 x . Figure 3.5: Continuous but not uniformly continuous on ( 0 , ∞ ) . We already know that this function is continuous at every ¯ x ∈ ( 0 , 1 ) . We will show that f is not uniformly continuous on ( 0 , 1 ) . Let ε = 2 and δ > 0. Set δ 0 = min { δ / 2 , 1 / 4 } , x = δ 0 , and y = 2 δ 0 . Then x , y ∈ ( 0 , 1 ) and | x − y | = δ 0 < δ , but | f ( x ) − f ( y ) | = 1 x − 1 y = y − x xy = δ 0 2 δ 2 0 = 1 2 δ 0 ≥ 2 = ε . This shows f is not uniformly continuous on ( 0 , 1 ) . The following theorem offers a sequential characterization of uniform continuity analogous to that in Theorem 3.3.3 . Theorem 3.5.3 Let D be a nonempty subset of R and f : D → R . Then f is uniformly continuous on D if and only if the following condition holds (C) for every two sequences { u n } , { v n } in D such that lim n → ∞ ( u n − v n ) = 0, it follows that lim n → ∞ ( f ( u n ) − f ( v n )) = 0. Proof: Suppose first that f is uniformly continuous and let { u n } , { v n } be sequences in D such that lim n → ∞ ( u n − v n ) = 0. Let ε > 0. Choose δ > 0 such that | f ( u ) − f ( v ) | < ε whenever