83 3.5 Uniform Continuity We discuss here a stronger notion of continuity. Defnition 3.5.1 Let D be a nonempty subset of R. A function f : D → R is called uniformly continuous on D if for any ε > 0, there exists δ > 0 such that | f (u) − f (v)| < ε for all u,v ∈ D with |u− v| < δ . ■ Example 3.5.1 Any constant function f : D → R, is uniformly continuous on its domain. Indeed, given ε > 0, | f (u) − f (v)| = 0 < ε for all u,v ∈ D regardless of the choice of δ . The following result is straightforward from the defnition. Theorem 3.5.1 If f : D → R is uniformly continuous on D, then f is continuous at every point x0 ∈ D. ■ Example 3.5.2 Let f : R → R be given by f (x)= 7x − 2. We will show that f is uniformly continuous on R. Let ε > 0 and choose δ = ε/7. Then, if u,v ∈ R and |u− v| < δ , we have | f (u) − f (v)| = |(7u − 2) − (7v− 2)| = |7(u− v)| = 7|u− v| < 7δ = ε. ■ 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|≤ 3+ 3 = 6. Now set δ = ε/6. Then, for u,v ∈ [−3,2] satisfying |u − v| < δ , we have | f (u) − f (v)| = |u2 − v2| = |u − v||u + v|≤ 6|u− v| < 6δ = ε. 2x ■ Example 3.5.4 Let f : R → R be given by f (x)= . We will show that f is uniformly x2 + 1 continuous on R. Let ε > 0. We observe frst that 2 2 2 2 u v u2(v + 1) − v2(u + 1) |u− v||u+ v| − = = u2 + 1 v2 + 1 (u2 + 1)(v2 + 1) (u2 + 1)(v2 + 1) 2 2 |u − v|(|u| + |v|) |u − v|((u + 1)+(v + 1)) ≤ (u2 + 1)(v2 + 1) ≤ (u2 + 1)(v2 + 1) 1 1 ≤|u − v| + ≤ 2|u− v|, v2 + 1 u2 + 1 2 (where we used the fact that |x|≤ x + 1 for all x ∈ R, which can be proved 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 2 2 u v | f (u) − f (v)| = − ≤ 2|u − v| < 2δ = ε. u2 + 1 v2 + 1
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