149 is differentiable at c and, hence, ∂h(c)={h′(c)}= − f (b)−f (a) b−a . By Theorem 5.6.7 and the subdifferential sum rule, 0∈∂g(c)=∂ f (c)− f (b)−f (a) b−a . This implies (5.11). The proof is now complete. □ Corollary 5.6.9 Let f : R→Rbe a convex function. Then f is Lipschitz continuous if and only if there exists ℓ ≥0 such that ∂ f (x) ⊂[−ℓ,ℓ] for all x ∈R. Proof: Suppose f is Lipschitz continuous onR. Then there exists ℓ ≥0 such that | f (u)−f (v)|≤ℓ|u−v| for all u,v ∈R. Then for anyx ∈R, f ′+(x)= lim h→0+ f (x+h)−f (x) h ≤ lim h→0+ ℓ|h| h =ℓ. Similarly, f ′ − (x) ≥−ℓ. Thus, ∂ f (x)=[ f ′ − (x), f ′+(x)] ⊂[−ℓ,ℓ]. Conversely, fix anyu,v ∈Rwithu̸=v. Applying Theorem 5.6.8, we get f (v)−f (u) v−u ∈ ∂ f (c) ⊂[−ℓ,ℓ], for some c in betweenu andv. This implies | f (u)−f (v)|≤ℓ|u−v|. This inequality obviously holds for u=v. Therefore, f is Lipschitz continuous. □ Exercises 5.6.1 ▷Find subdifferentials of the following functions: (a) f (x)=a|x|, a>0. (b) f (x)=|x−1|+|x+1|. 5.6.2 Find the subdifferential of the function f (x)=max{−2x+1,x,2x−1}.
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