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 R and let f : I → R be 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 on I. 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)= 1/(x2 + 1)3/2 x/ x2 + 1 and f . 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 → R be a convex function. Then it is locally Lipschitz continuous in the sense that for any x0 ∈ 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 any x0 ∈ I. Choose four numbers a,b,c,d satisfying a < b < x0 < c < d with a,d ∈ I. Choose δ > 0 such that B(x0;δ ) ⊂ (b,c). Let u,v ∈ B(x0;δ ) with v < u. Then by Lemma 5.5.5, we see that f (b) − f (a) f (u) − f (a) f (u) − f (v) f (d) − f (v) f (d) − f (c) ≤ ≤ ≤ ≤ . b− a u− a u − v d − v d − c Using a similar approach for the case u < v, we get f (b) − f (a) f (u) − f (v) f (d) − f (c) ≤ ≤ for all u,v ∈ B(x0;δ ). b − a u− v d − c Choose ℓ ≥ 0 suffciently large so that f (b) − f (a) f (u) − f (v) f (d) − f (c) −ℓ ≤ ≤ ≤ ≤ ℓ for all u,v ∈ B(x0;δ ). b− a u − v d − c Then (5.7) holds. The proof is now complete. □
RkJQdWJsaXNoZXIy NTc4NTAz