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 anyi =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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