Introduction to Mathematical Analysis I - Second Edition

85 Figure 3.4: The square root function. (2) Let D = [ 0 , ∞ ) . Then f is not Lipschitz continuous on D , but it is Hölder continuous on D and, hence, f is also uniformly continuous on this set. Indeed, suppose by contradiction that f is Lipschitz continuous on D . Then there exists a constant ` > 0 such that | f ( u ) − f ( v ) | = | √ u − √ v | ≤ ` | u − v | for every u , v ∈ D . Thus, for every n ∈ N , we have 1 √ n − 0 ≤ ` 1 n − 0 . This implies √ n ≤ ` or n ≤ ` 2 for every n ∈ N . This is a contradiction. Therefore, f is not Lipschitz continuous on D . Let us show that f is Hölder continuous on D . We are going to prove that | f ( u ) − f ( v ) | ≤ | u − v | 1 / 2 for every u , v ∈ D . (3.9) The inequality in ( 3.9 ) 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 ≤ p | u | + | v | √ u + √ v p | u − v | = p | u − v | . Note that one can justify the inequality p | u | + | v | √ u + √ v ≤ 1 by squaring both sides since they are both positive. Thus, ( 3.9 ) is satisfied.