83 3.5 Uniform Continuity We discuss here a stronger notion of continuity. Definition 3.5.1 Let Dbe a nonempty subset of R. A function f : D→Ris called uniformly continuous on Dif for anyε >0, there exists δ >0 such that | f (u)−f (v)| <ε for all u,v ∈Dwith|u−v| <δ. ■ Example 3.5.1 Any constant function f : D→R, is uniformly continuous on its domain. Indeed, givenε >0, | f (u)−f (v)| =0<ε for all u,v ∈Dregardless of the choice of δ. The following result is straightforward from the definition. Theorem 3.5.1 If f : D→Ris uniformly continuous on D, then f is continuous at every point x0 ∈D. ■ Example 3.5.2 Let f : R→Rbe given by f (x)=7x−2. We will show that f is uniformly continuous onR. Let ε >0 and choose δ =ε/7. Then, if u,v ∈Rand|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] →Rbe 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. Nowset 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. ■ Example 3.5.4 Let f : R→Rbe given by f (x)= x2 x2 +1 . We will show that f is uniformly continuous onR. Let ε >0. We observe first that u2 u2 +1− v2 v2 +1 = u2(v2 +1)−v2(u2 +1) (u2 +1)(v2 +1) = | u−v||u+v| (u2 +1)(v2 +1) ≤ | u−v|(|u|+|v|) (u2 +1)(v2 +1) ≤ | u−v|((u2 +1)+(v2 +1)) (u2 +1)(v2 +1) ≤|u−v| 1 v2 +1 + 1 u2 +1 ≤ 2|u−v|, (where we used the fact that |x|≤x2 +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, givenu,v ∈Rsatisfying|u−v| <δ we have | f (u)−f (v)| = u2 u2 +1− v2 v2 +1 ≤ 2|u−v| <2δ =ε.
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