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 {−a}, if x < b; ∂ f (x)= [−a,a], if x = b; {a}, if x > b. Defnition 5.6.2 Let A and B be two nonempty subsets of R and let α ∈ R. Defne A+ B = {a + b : a ∈ A,b ∈ B} and αA = {αa : a ∈ A}. Figure 5.6: Set addition. Theorem 5.6.5 Let f ,g: R → R be convex functions and let α > 0. Then f + g and α f are convex functions. In addition, for any x0 ∈ R we have ∂ ( f + g)(x0)= ∂ f (x0)+ ∂ g(x0) ∂ (α f )(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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