85 Therefore, √ n ≤ ℓ or n ≤ ℓ2 for every n ∈ N. This is a contradiction. Hence, f is not Lipschitz continuous on D. Now we prove that f is Hölder continuous on D by showing that | f (u) − f (v)|≤|u− v|1/2 for every u,v ∈ D. (3.7) The inequality in (3.7) holds obviously for u = v = 0. For u > 0 or v > 0, we have √ √ | f (u) − f (v)| = | u − v| u − v = √ √ u + v p|u− v| = p|u − v| √ √ u + v √ u + v p ≤√ √ |u− v| u + v p≤ |u− v|. √ u+v Note that one can justify the inequality √ √ ≤ 1 by squaring both sides since they are both u+ v positive. Thus, (3.7) holds and it follows that f (x) is Hölder continuous on [0,∞). 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 1 f (x)= . x We already know that this function is continuous at every x0 ∈ (0,1). We will show that f is not uniformly continuous on (0,1). Let ε0 = 2 and δ > 0. Set δ0 = min{δ /2,1/4}, x = δ0, and y = 2δ0. Then x,y ∈ (0,1) and |x − y| = δ0 < δ , but 1 1 y− x δ0 1 | f (x) − f (y)| = − = = = ≥ 2 = ε0. 2 x y xy 2δ0 2δ0 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 3.3.2. Theorem 3.5.3 Let D be a nonempty subset of R and f : D → R. Then the following are equivalent: (i) f is uniformly continuous on D. (ii) For every two sequences {un}, {vn} in D such that limn →∞(un − vn)= 0, it follows that limn →∞( f (un) − f (vn)) = 0.
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