144 5.6 Nondifferentiable Convex Functions and Subdifferentials Thus, f (x) − f (x0) ′ u ≤ lim + = f (x0). + x − x0 x→x0 Similarly, we have u· (x − x0) ≤ f (x) − f (x0) for all x < x0. Thus, f (x) − f (x0) u ≥ for all x < x0. x − x0 ′ This implies u ≥ f − (x0). So ′ ′ ∂ f (x0) ⊂ [ f − (x0), f (x0)]. + ′ ′ To prove the reverse inclusion, take u ∈ [ f − (x0), f (x0)]. By Theorem 5.6.2 + ′ ′ sup φx0 (x)= f − (x0) ≤ u ≤ f+(x0)= inf φx0 (x). x>x0 x<x 0 ′ Using the upper estimate by f (x0) for u, one has + f (x) − f (x0) u ≤ φx0 (x)= for all x > x0. x − x0 It follows that u· (x − x0) ≤ f (x) − f (x0) for all x ≥ x0. Similarly, one also has u· (x − x0) ≤ f (x) − f (x0) for all x < x0. Thus, (5.8) holds and, hence, u ∈ ∂ f (x0). Therefore, (5.10) holds. □ Corollary 5.6.4 Let f : R → R be a convex function and let x0 ∈ R. Then f is differentiable at x0 if and only if ∂ f (x0) is a singleton. In this case, ∂ f (x0)= { f ′ (x0)}. Proof: Suppose f is differentiable at x0. Then ′ ′ f − (x0)= f (x0)= f ′ (x0). + By Theorem 5.6.3, ′ ′ ∂ f (x0)=[ f − (x0), f (x0)] = { f ′ (x0)}. + Thus, ∂ f (x0) is a singleton. ′ ′ Conversely, if ∂ f (x0) is a singleton, we must have f − (x0)= f (x0). Thus, f is differentiable at + x0. □
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