84 3.5 Uniform Continuity Definition 3.5.2 (Hölder continuity). Let Dbe a nonempty subset of R. A function f : D→Ris said tobe Hölder continuous if there are constants ℓ ≥0 andα>0 such that | f (u)−f (v)|≤ℓ|u−v| α for everyu,v ∈D. The number α is called the Hölder exponent of the function. If α=1, then the function f is called Lipschitz continuous. Theorem 3.5.2 If a function f : D→Ris Hölder continuous, then it is uniformly continuous. Proof: Since f is Hölder continuous, there are constants ℓ ≥0 andα>0 such that | f (u)−f (v)|≤ℓ|u−v| α for everyu,v ∈D. If ℓ =0, then f is constant and, thus, uniformly continuous. Suppose next that ℓ>0. For any ε >0, let δ = ε ℓ 1/α . Then, whenever u,v ∈D, with|u−v| <δ we have | f (u)−f (v)|≤ℓ|u−v| α <ℓδ α =ε. The proof is now complete. □ Figure 3.3: The square root function. ■ Example 3.5.5 (a) Let D=[a,∞), where a >0. We will show that f (x)=√x is Lipschitz continuous onDand, hence, uniformly continuous on this set. Givenu,v ∈D, we have | f (u)−f (v)| =| √u − √v | = | u−v| √u+√v ≤ 1 2√a| u−v|, which shows f is Lipschitz continuous withℓ =1/(2√a). (b) Let D=[0,∞). We will show that f (x)=√x is not Lipschitz continuous onD, but it is Hölder continuous onDand, hence, f is also uniformly continuous on this set. Suppose by contradiction that f is Lipschitz continuous on D. Then there exists a constant ℓ>0 such that | f (u)−f (v)| =| √u − √v |≤ℓ|u−v| for everyu,v ∈D. Considering v =0 andun =1/n for everyn∈N,we have 1 √n− 0 ≤ℓ 1 n− 0 .
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