146 5.6 Nondifferentiable Convex Functions and Subdifferentials ■ Example 5.6.3 Let a1 <a2 <· · · <an and let µi >0 for i =1, . . . ,n. Define f (x) = n ∑ i=1 µi|x−ai|. Then f is a convex function. By Theorem 5.6.5, we get ∂ f (x0) =(∑ai<x0 µi −∑ai>x0 µi, if x0 / ∈ {a1, . . . ,an} ∑ai<x0 µi −∑ai>x0 µi + [−µi0,µi0], if x0 =ai0. Theorem 5.6.6 Let fi : R→R, i =1, . . . ,n, be convex functions. Define f (x) =max{fi(x) : i =1, . . . ,n}andI(u) ={i =1, . . . ,n: fi(u) =f (u)}. Then f is a convex function. Moreover, ∂ f (x0) = [m,M], where m= min i∈I(x0) f ′ i −(x0) andM= max i∈I(x0) f ′ i+(x0). Proof: Fixu,v ∈Randλ ∈(0,1). For any i =1, . . . ,n, we have fi(λu+(1−λ)v) ≤λfi(u)+(1−λ)fi(v) ≤λf (u)+(1−λ)f (v). This implies f (λu+(1−λ)v) = max 1≤i≤n fi(λu+(1−λ)v) ≤λf (u)+(1−λ)f (v). Thus, f is a convex function. Similarly we verify that f ′ +(x0) =Mand f −′ (x0) =m. By Theorem 5.6.3, ∂ f (x0) = [m,M]. The proof is now complete. □ Remark 5.6.1 The product of two convex functions is not a convex function in general. For instance, f (x) =x andg(x) =x2 are convex functions, but h(x) =x3 is not a convex function. The following result may be considered as a version of the first derivative test for extrema in the case of non differentiable functions. Theorem 5.6.7 Let f : R→Rbe a convex function. Then f has an absolute minimum at x0 if and only if 0∈∂ f (x0) = [ f ′ −(x0), f ′ +(x0)].
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