138 5.5 Convex Functions and Derivatives This implies that for a sufficient large n, we have 1 n f (x0) ≤ 1 n f (x) 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 inRand let f : I →Rbe 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 andt ∈(0,1), we have f (x0 +t(x−x0))−f (x0) t = f (tx+(1−t)x0)−f (x0) t ≤ t f (x)+(1−t)f (x0)−f (x0) t =f (x)−f (x0). Since f is differentiable at x0, f ′(x0)(x−x0) = lim t→0+ f (x0 +t(x−x0))−f (x0) t ≤ f (x)−f (x0), which completes the proof. □ Corollary 5.5.4 Let I be an open interval in Rand let f : I →Rbe 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 inRand suppose f : I →Ris a convex function. Fixa,b,x ∈I witha<x <b. Then f (x)−f (a) x−a ≤ f (b)−f (a) b−a ≤ f (b)−f (x) b−x . Proof: Let t = x−a b−a . Thent ∈(0,1) and f (x) =f (a+(x−a)) =f a+ x−a b−a (b−a) =f (a+t(b−a)) =f (tb+(1−t)a).
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