138 5.5 Convex Functions and Derivatives This implies that for a suffcient large n, we have 1 1 f (x0) ≤ f (x) n n and, hence, f (x0) ≤ f (x). Since x was arbitrary, this shows f has an absolute minimum at x0. □ Theorem 5.5.3 Let I be an open interval in R and let f : I → R be a convex function. Suppose f is differentiable at x0. Then f ′ (x0)(x − x0) ≤ f (x) − f (x0) for all x ∈ I. (5.6) Proof: For any x ∈ I and t ∈ (0,1), we have f (x0 +t(x − x0)) − f (x0) f (tx +(1−t)x0) − f (x0) = t t t f (x)+(1 − t) f (x0) − f (x0) ≤ t = f (x) − f (x0). Since f is differentiable at x0, f (x0 + t(x − x0)) − f (x0) f ′ (x0)(x − x0)= lim ≤ f (x) − f (x0), t→0+ t which completes the proof. □ Corollary 5.5.4 Let I be an open interval in R and let f : I → R be a convex function. Suppose f is differentiable at x0. Then f has an absolute minimum at x0 if and only if f ′ (x0)= 0. Proof: Suppose f has an absolute minimum at x0. By Theorem 4.2.1, f ′ (x0)= 0. Let us prove the converse. Suppose f ′ (x0)= 0. It follows from Theorem 5.5.3 that 0 = f ′ (x0)(x − x0) ≤ f (x) − f (x0) for all x ∈ I. This implies f (x0) ≤ f (x) for all x ∈ I. Thus, f has an absolute minimum at x0. □ Lemma 5.5.5 Let I be an interval in R and suppose f : I → R is a convex function. Fix a,b,x ∈ I with a < x < b. Then f (x) − f (a) f (b) − f (a) f (b) − f (x) ≤ ≤ . x − a b− a b− x Proof: Let x − a t = . b− a Then t ∈ (0,1) and x − a f (x)= f (a +(x − a)) = f a+ (b − a) = f (a +t(b − a)) = f (tb +(1 − t)a). b − a
RkJQdWJsaXNoZXIy NTc4NTAz