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 +gis 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. □
RkJQdWJsaXNoZXIy NTc4NTAz