87 Proof: Suppose by contradiction that f is not uniformly continuous on [a,b]. Then there exists ε0 >0 such that for anyδ >0, there exist u,v ∈[a,b] with |u−v| <δ and| f (u)−f (v)|≥ε0. Thus, for everyn∈N, there exist un,vn ∈[a,b] with |un −vn|≤1/n and| f (un)−f (vn)|≥ε0. Applying the Bolzano-Weierstrass theorem (Theorem 2.4.1) there exists a subsequence {unk} of {un} andu0 ∈[a,b] such that unk →u0 as k →∞. Then |unk −vnk|≤ 1 nk , for all k and, hence, we also have vnk →u0 as k →∞. By the continuity of f , f (unk) →f (u0) and f (vnk) →f (u0). Therefore, {f (unk)−f (vnk)} converges to zero, which is a contradiction. The proof is now complete. □ We now prove a result that characterizes uniform continuity on open bounded intervals. We first make the observation that if f : D→Ris uniformly continuous on Dand A⊂D, then f is uniformly continuous on A. More precisely, the restriction f |A : A→Ris uniformly continuous on A (see Section 1.2 for the notation). This follows by noting that if | f (u)−f (v)| <ε whenever u,v∈D with|u−v| <δ, then we also have | f (u)−f (v)| <ε when we restrict u,v tobe inA. Theorem 3.5.5 Let a,b∈Rand a<b. A function f : (a,b) →Ris uniformly continuous if and only if f can be extended to a continuous function ˜f : [a,b] →R(that is, there is a continuous function ˜f : [a,b] →Rsuch that f = ˜f |(a,b) ). Proof: Suppose first that there exists a continuous function ˜f : [a,b] →Rsuch that f = ˜f |(a,b) . By Theorem 3.5.4, the function ˜f is uniformly continuous on [a,b]. Therefore, it follows from our early observation that f is uniformly continuous on(a,b). For the converse, suppose f : (a,b) →R is uniformly continuous. We will show first that limx →a+ f (x) exists. Note that the one sided limit corresponds to the limit in Theorem 3.2.2. We will check that the ε-δ condition of Theorem 3.2.2 holds. Let ε >0. Choose δ0 >0 so that | f (u)−f (v)| <ε whenever u,v∈(a,b) and|u−v| <δ0. Set δ =δ0/2. Then, if u,v ∈(a,b), |u−a| <δ, and|v−a| <δ we have |u−v|≤|u−a|+|a−v| <δ +δ =δ0 and, hence, | f (u)−f (v)| <ε. We can now invoke Theorem 3.2.2 to conclude limx →a+ f (x) exists. In a similar way we can show that limx →b− f (x) exists. Now define, ˜f : [a,b] →Rby ˜f (x)= f (x), if x ∈(a,b); limx →a+ f (x), if x =a; limx →b− f (x), if x =b. By its definition ˜f |(a,b) =f and, so, ˜f is continuous at every x ∈(a,b). Moreover, limx →a+ ˜f (x)= limx →a+ f (x)= ˜f (a) and limx →b− ˜f (x)=limx →b− f (x)= ˜f (b), so ˜f is also continuous at aandbby Theorem 3.3.1. Thus ˜f is the desired continuous extension of f . □
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