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 Dby 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 √u+√vp| u−v| ≤p|u−v|. Note that one can justify the inequality √u+v √u+√v ≤1 by squaring both sides since they are both positive. Thus, (3.7) holds and it follows that f (x) is Hölder continuous on [0,∞). While every uniformly continuous function on a set Dis 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) →Rbe given by f (x) = 1 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 andδ >0. Set δ0 =min{δ/2,1/4}, x=δ0, andy=2δ0. Then x,y ∈(0,1) and|x−y| =δ0 <δ, but | f (x)−f (y)| = 1 x − 1 y = y−x xy = δ0 2δ2 0 = 1 2δ0 ≥ 2=ε0. 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 Dbe a nonempty subset of Rand f : D→R. Then the following are equivalent: (i) f is uniformly continuous onD. (ii) For every two sequences {un}, {vn} in Dsuch that limn→∞(un −vn) =0, it follows that limn→∞(f (un)−f (vn)) =0.
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