139 By convexity of f , we obtain f (x) ≤ t f (b)+(1−t) f (a). Thus, x − a f (x) − f (a) ≤ t f (b)+(1−t) f (a) − f (a)= t[ f (b) − f (a)] = ( f (b) − f (a)). b− a Equivalently, f (x) − f (a) f (b) − f (a) x− a ≤ b − a . Similarly, x − b f (x) − f (b) ≤ t f (b)+(1 − t) f (a) − f (b)=(1 −t)[ f (a) − f (b)] = [ f (b) − f (a)]. b − a It follows that f (b) − f (a) f (b) − f (x) ≤ . b − a b − x The proof is now complete. □ Theorem 5.5.6 Let I be an open interval in R and let f : I → R be a differentiable function. Then f ′ is convex if and only if f is increasing on I. Proof: Suppose f is convex. Fix a < b with a,b ∈ I. By Lemma 5.5.5, for any x ∈ (a,b), we have f (x) − f (a) f (b) − f (a) ≤ . x − a b− a This implies, taking limits, that f (b) − f (a) f ′ (a) ≤ . b − a Similarly, f (b) − f (a) ≤ f ′ (b). b − a ′ Therefore, f ′ (a) ≤ f ′ (b), and f is an increasing function. ′ Let us prove the converse. Suppose f is increasing. Fix x1 < x2 and t ∈ (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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