Introduction to Mathematical Analysis I - Second Edition

84 3.5 UNIFORM CONTINUITY Example 3.5.3 Let f : [ − 3 , 2 ] → R be given by f ( x ) = x 2 . This function is uniformly continuous on [ − 3 , 2 ] . Let ε > 0. First observe that for u , v ∈ [ − 3 , 2 ] we have | u + v | ≤ | u | + | v | ≤ 6. Now set δ = ε / 6. Then, for u , v ∈ [ − 3 , 2 ] satisfying | u − v | < δ , we have | f ( u ) − f ( v ) | = | u 2 − v 2 | = | u − v || u + v | ≤ 6 | u − v | < 6 δ = ε . Example 3.5.4 Let f : R → R be given by f ( x ) = x 2 x 2 + 1 . We will show that f is uniformly continuous on R . Let ε > 0. We observe first that u 2 u 2 + 1 − v 2 v 2 + 1 = u 2 ( v 2 + 1 ) − v 2 ( u 2 + 1 ) ( u 2 + 1 )( v 2 + 1 ) = | u − v || u + v | ( u 2 + 1 )( v 2 + 1 ) ≤ | u − v | ( | u | + | v | ) ( u 2 + 1 )( v 2 + 1 ) ≤ | u − v | (( u 2 + 1 )+( v 2 + 1 )) ( u 2 + 1 )( v 2 + 1 ) ≤ | u − v | 1 v 2 + 1 + 1 u 2 + 1 ≤ 2 | u − v | , (where we used that | x | ≤ x 2 + 1 for all x ∈ R , which can be easily seen by considering separately the cases | x | < 1 and | x | ≥ 1). Now, set δ = ε / 2. In view of the previous calculation, given u , v ∈ R satisfying | u − v | < δ we have | f ( u ) − f ( v ) | = u 2 u 2 + 1 − v 2 v 2 + 1 ≤ 2 | u − v | < 2 δ = ε . Definition 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 ` ≥ 0 and α > 0 such that | f ( u ) − f ( v ) | ≤ ` | 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 ` ≥ 0 and α > 0 such that | f ( u ) − f ( v ) | ≤ ` | u − v | α for every u , v ∈ D . If ` = 0, then f is constant and, thus, uniformly continuous. Suppose next that ` > 0. For any ε > 0, let δ = ε ` 1 / α . Then, whenever u , v ∈ D , with | u − v | < δ we have | f ( u ) − f ( v ) | ≤ ` | u − v | α < ` δ α = ε . The proof is now complete. Example 3.5.5 (1) Let D = [ a , ∞ ) , where a > 0. Then the function f ( x ) = √ x is Lipschitz continuous on D and, hence, uniformly continuous on this set. Indeed, for any u , v ∈ D , we have | f ( u ) − f ( v ) | = | √ u − √ v | = | u − v | √ u + √ v ≤ 1 2 √ a | u − v | , which shows f is Lipschitz with ` = 1 / ( 2 √ a ) .