84 3.5 Uniform Continuity Defnition 3.5.2 (Hölder continuity). Let D be a nonempty subset of R. A function f : D → R is said to be Hölder continuous if there are constants ℓ ≥ 0 and α > 0 such that | f (u) − f (v)|≤ ℓ|u − v| α for every u,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 → R is Hölder continuous, then it is uniformly continuous. Proof: Since f is Hölder continuous, there are constants ℓ ≥ 0 and α > 0 such that | f (u) − f (v)|≤ ℓ|u − v| α for every u,v ∈ D. If ℓ = 0, then f is constant and, thus, uniformly continuous. Suppose next that ℓ> 0. For any �ε 1/α ε > 0, let δ = . Then, whenever u,v ∈ D, with |u− v| < δ we have ℓ | 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 on D and, hence, uniformly continuous on this set. Given u,v ∈ D, we have √ √ |u− v| 1 | f (u) − f (v)| = | u − v| = √ √ ≤ √ |u − v|, u+ v 2 a √ which shows f is Lipschitz continuous with ℓ = 1/(2 a). √ (b) Let D =[0,∞). We will show that f (x)= x is not Lipschitz continuous on D, but it is Hölder continuous on D and, 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 every u,v ∈ D. Considering v = 0 and un = 1/n for every n ∈ N,we have 1 √ − 0 ≤ ℓ 1 − 0 . n n
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