87 Proof: Suppose by contradiction that f is not uniformly continuous on [a,b]. Then there exists ε0 > 0 such that for any δ > 0, there exist u,v ∈ [a,b] with |u− v| < δ and | f (u) − f (v)|≥ ε0. Thus, for every n ∈ 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} and u0 ∈ [a,b] such that unk → u0 as k → ∞. Then 1 |unk − vnk |≤ , 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 frst make the observation that if f : D → R is uniformly continuous on D and A ⊂ D, then f is uniformly continuous on A. More precisely, the restriction f |A : A → R is 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 to be in A. Theorem 3.5.5 Let a,b ∈ R and a < b. A function f : (a,b) → R is 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] → R such that f = f˜ |(a,b) ). Proof: Suppose frst that there exists a continuous function f˜ : [a,b] → R such 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 frst 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 defne, f˜ : [a,b] → R by f (x), if x ∈ (a,b); f˜(x)= limx →a+ f (x), if x = a; limx →b− f (x), if x = b. By its defnition 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 a and b by Theorem 3.3.1. Thus f˜ is the desired continuous extension of f . □
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