Introduction to Mathematical Analysis I - Second Edition

140 4.7 NONDIFFERENTIABLE CONVEX FUNCTIONS AND SUBDIFFERENTIALS By Theorem 4.7.8 and the subdifferential sum rule, 0 ∈ ∂ g ( c ) = ∂ f ( c ) − f ( b ) − f ( a ) b − a . This implies ( 4.19 ) . The proof is now complete. Corollary 4.7.10 Let f : R → R be a convex function. Then f is Lipschitz continuous if and only if there exists ` ≥ 0 such that ∂ f ( x ) ⊂ [ − `, ` ] for all x ∈ R . Proof: Suppose f is Lipschitz continuous on R . Then there exists ` ≥ 0 such that | f ( u ) − f ( v ) | ≤ ` | u − v | for all u , v ∈ R . Then for any x ∈ R , f 0 + ( x ) = lim h → 0 + f ( x + h ) − f ( x ) h ≤ lim h → 0 + ` | h | h = `. Similarly, f 0− ( x ) ≥ − ` . Thus, ∂ f ( x ) = [ f 0 − ( x ) , f 0 + ( x )] ⊂ [ − `, ` ] . Conversely, fix any u , v ∈ R with u 6 = v . Applying Theorem 4.7.9 , we get f ( v ) − f ( u ) v − u ∈ ∂ f ( c ) ⊂ [ − `, ` ] , for some c in between u and v . This implies | f ( u ) − f ( v ) | ≤ ` | u − v | . This inequality obviously holds for u = v . Therefore, f is Lipschitz continuous. Exercises 4.7.1 B Find subdifferentials of the following functions: (a) f ( x ) = a | x | , a > 0. (b) f ( x ) = | x − 1 | + | x + 1 | . 4.7.2 Find the subdifferential of the function f ( x ) = max {− 2 x + 1 , x , 2 x − 1 } . 4.7.3 I Let f ( x ) = ∑ n k = 1 | x − k | . Find all absolute minimizers of the function.