145 ■ Example 5.6.2 Let f (x)=a|x−b|+c, where a>0. Then f is a convex function and f ′ − (b)=−a, f ′+(b)=a. Thus, ∂ f (b)=[−a,a]. Since f is differentiable on(−∞,b) and(b,∞), we have ∂ f (x)= {−a}, if x <b; [−a,a], if x =b; {a}, if x >b. Definition 5.6.2 Let AandBbe two nonempty subsets of Rand let α∈R. Define A+B={a+b: a∈A,b∈B} andαA={αa: a∈A}. Figure 5.6: Set addition. Theorem 5.6.5 Let f,g: R→Rbe convex functions and let α>0. Then f +gand αf are convex functions. In addition, for anyx0 ∈Rwe have ∂(f +g)(x0)=∂ f (x0)+∂g(x0) ∂(αf )(x0)=α∂ f (x0). Proof: It is not hard to see that f +g is a convex function and (f +g)′+(x0)=f ′+(x0)+g′+(x0) (f +g)′ − (x0)=f ′ − (x0)+g′ − (x0). By Theorem 5.6.3, ∂(f +g)(x0)=[(f +g)′ − (x0),(f +g)′+(x0)] =[ f ′ − (x0)+g′ − (x0), f ′+(x0)+g′+(x0)] =[ f ′ − (x0), f ′+(x0)]+[g′ − (x0),g′+(x0)] =∂ f (x0)+∂g(x0). The proof for the second formula is similar. □
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