146 5.6 Nondifferentiable Convex Functions and Subdifferentials ■ Example 5.6.3 Let a1 < a2 < ··· < an and let µi > 0 for i = 1,...,n. Defne n f (x)= ∑ µi|x − ai|. i=1 Then f is a convex function. By Theorem 5.6.5, we get (∑ai<x0 ∈{a1,...,an} µi − ∑ai>x0 µi, if x0 / ∂ f (x0)= µi +[−µi0 , µi0 ], if x0 = ai0 . ∑ai<x0 µi − ∑ai>x0 Theorem 5.6.6 Let fi : R → R, i = 1,...,n, be convex functions. Defne f (x)= max{ fi(x) : i = 1,...,n} and I(u)= {i = 1,...,n : fi(u)= f (u)}. Then f is a convex function. Moreover, ∂ f (x0)=[m,M], where ′ ′ m = min fi − (x0) and M = max f (x0). i+ i∈I(x0) i∈I(x0) Proof: Fix u,v ∈ R and λ ∈ (0,1). For any i = 1,...,n, we have fi(λ u+(1 − λ )v) ≤ λ fi(u)+(1− λ ) fi(v) ≤ λ f (u)+(1− λ ) f (v). This implies f (λ u +(1 − λ )v)= max fi(λ u+(1 − λ )v) ≤ λ f (u)+(1 − λ ) f (v). 1≤i≤n ′ ′ Thus, f is a convex function. Similarly we verify that f (x0)= M and 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, 2 f (x)= x and g(x)= x are convex functions, but h(x)= x3 is not a convex function. The following result may be considered as a version of the frst derivative test for extrema in the case of non differentiable functions. Theorem 5.6.7 Let f : R → R be 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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