149 is differentiable at c and, hence, f (b) − f (a) ∂ h(c)= {h ′ (c)} = − . b − a By Theorem 5.6.7 and the subdifferential sum rule, f (b) − f (a) 0 ∈ ∂ g(c)= ∂ f (c) − . b− a This implies (5.11). The proof is now complete. □ Corollary 5.6.9 Let f : R → R be 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 on R. Then there exists ℓ ≥ 0 such that | f (u) − f (v)|≤ ℓ|u − v| for all u,v ∈ R. Then for any x ∈ R, f (x + h) − f (x) ℓ|h| ′ f (x)= lim ≤ lim = ℓ. + h→0+ h h→0+ h ′ Similarly, f − (x) ≥−ℓ. Thus, ′ ′ ∂ f (x)=[ f − (x), f (x)] ⊂ [−ℓ,ℓ]. + Conversely, fx any u,v ∈ R with u ̸ = v. Applying Theorem 5.6.8, we get f (v) − f (u) ∈ ∂ f (c) ⊂ [−ℓ,ℓ], v − u for some c in between u and v. 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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