140 5.5 Convex Functions and Derivatives with f (xt)−f (x1) =f ′(c1)(xt −x1) =f ′(c1)(1−t)(x2 −x1); f (xt)−f (x2) =f ′(c2)(xt −x2) =f ′(c2)t(x1 −x2). This implies t f (xt)−t f (x1) =f ′(c1)t(1−t)(x2 −x1); (1−t)f (xt)−(1−t)f (x2) =f ′(c2)t(1−t)(x1 −x2). Since f ′(c1) ≤f ′(c2), we have t f (xt)−t f (x1) =f ′(c1)t(1−t)(x2−x1) ≤f ′(c2)t(1−t)(x2−x1) = (1−t)f (x2)−(1−t)f (xt). Rearranging terms, we get f (xt) ≤t f (x1)+(1−t)f (x2). Therefore, f is convex. The proof is now complete. □ Corollary 5.5.7 Let I be an open interval in Rand let f : I →Rbe a function. Suppose f is twice differentiable on I. Then f is convex if and only if f ′′(x) ≥0 for all x ∈I. Proof: It follows from Proposition 4.3.2 that f ′′(x) ≥0 for all x ∈I if and only if the derivative function f ′ is increasing onI. The conclusion then follows directly from Theorem 5.5.6. □ ■ Example 5.5.2 Consider the function f : R→R given by f (x) =√x2 +1. Now, f ′(x) = x/√x2 +1 and f ′′(x) =1/(x2 +1)3/2. Since f ′′(x) ≥0 for all x, it follows from the corollary that f is convex. Theorem 5.5.8 Let I be an open interval and let f : I →Rbe a convex function. Then it is locally Lipschitz continuous in the sense that for anyx0 ∈I, there exist ℓ ≥0 and δ >0 such that | f (u)−f (v)| ≤ℓ|u−v| for all u,v ∈B(x0;δ). (5.7) In particular, f is continuous. Proof: Fix anyx0 ∈I. Choose four numbers a,b,c,d satisfying a<b<x0 <c <d witha,d ∈I. Choose δ >0 such that B(x0;δ) ⊂(b,c). Let u,v ∈B(x0;δ) withv <u. Then by Lemma 5.5.5, we see that f (b)−f (a) b−a ≤ f (u)−f (a) u−a ≤ f (u)−f (v) u−v ≤ f (d)−f (v) d−v ≤ f (d)−f (c) d−c . Using a similar approach for the case u<v, we get f (b)−f (a) b−a ≤ f (u)−f (v) u−v ≤ f (d)−f (c) d−c for all u,v ∈B(x0;δ). Choose ℓ ≥0 sufficiently large so that −ℓ ≤ f (b)−f (a) b−a ≤ f (u)−f (v) u−v ≤ f (d)−f (c) d−c ≤ ℓ for all u,v ∈B(x0;δ). Then (5.7) holds. The proof is now complete. □
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