139 By convexity of f , we obtain f (x) ≤t f (b)+(1−t)f (a). Thus, f (x)−f (a) ≤t f (b)+(1−t)f (a)−f (a) =t[ f (b)−f (a)] = x−a b−a (f (b)−f (a)). Equivalently, f (x)−f (a) x−a ≤ f (b)−f (a) b−a . Similarly, f (x)−f (b) ≤t f (b)+(1−t)f (a)−f (b) = (1−t)[ f (a)−f (b)] = x−b b−a [ f (b)−f (a)]. It follows that f (b)−f (a) b−a ≤ f (b)−f (x) b−x . The proof is now complete. □ Theorem 5.5.6 Let I be an open interval in Rand let f : I →Rbe a differentiable function. Then f is convex if and only if f ′ is increasing onI. Proof: Suppose f is convex. Fix a<bwitha,b∈I. By Lemma 5.5.5, for any x ∈(a,b), we have f (x)−f (a) x−a ≤ f (b)−f (a) b−a . This implies, taking limits, that f ′(a) ≤ f (b)−f (a) b−a . Similarly, f (b)−f (a) b−a ≤ f ′(b). Therefore, f ′(a) ≤f ′(b), and f ′ is an increasing function. Let us prove the converse. Suppose f ′ is increasing. Fix x1 <x2 andt ∈(0,1). Then x1 <xt <x2, where xt =tx1 +(1−t)x2. By the Mean Value Theorem (Theorem 4.2.3), there exist c1 and c2 such that x1 <c1 <xt <c2 <x2
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