Introduction to Mathematical Analysis I - Second Edition

136 4.7 NONDIFFERENTIABLE CONVEX FUNCTIONS AND SUBDIFFERENTIALS Example 4.7.2 Let f ( x ) = a | x − b | + c , where a > 0. Then f is a convex function and f 0 − ( b ) = − a , f 0 + ( 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 4.7.2 Let A and B be two nonempty subsets of R and let α ∈ R . Define A + B = { a + b : a ∈ A , b ∈ B } and α A = { α a : a ∈ A } . Figure 4.9: Set addition. Theorem 4.7.5 Let f , g : R → R be convex functions and let α > 0. Then f + g and α f are convex functions and ∂ ( f + g )( ¯ x ) = ∂ f ( ¯ x )+ ∂ g ( ¯ x ) ∂ ( α f )( ¯ x ) = α∂ f ( ¯ x ) . Proof: It is not hard to see that f + g is a convex function and ( f + g ) 0 + ( ¯ x ) = f 0 + ( ¯ x )+ g 0 + ( ¯ x ) ( f + g ) 0 − ( ¯ x ) = f 0 − ( ¯ x )+ g 0 − ( ¯ x ) . By Theorem 4.7.3 , ∂ ( f + g )( ¯ x ) = [( f + g ) 0 − ( ¯ x ) , ( f + g ) 0 + ( ¯ x )] = [ f 0 − ( ¯ x )+ g 0 − ( ¯ x ) , f 0 + ( ¯ x )+ g 0 + ( ¯ x )] = [ f 0 − ( ¯ x ) , f 0 + ( ¯ x )] + [ g 0 − ( ¯ x ) , g 0 + ( ¯ x )] = ∂ f ( ¯ x )+ ∂ g ( ¯ x ) . The proof for the second formula is similar.