Introduction to Mathematical Analysis I - Second Edition

130 4.6 CONVEX FUNCTIONS AND DERIVATIVES Equivalently, f ( x ) − f ( a ) x − a ≤ f ( b ) − f ( a ) b − a . Similarly, f ( x ) − f ( b ) ≤ t f ( b )+( 1 − t ) f ( a ) − f ( b ) = ( 1 − t )[ f ( a ) − f ( b )] = x − b b − a [ f ( b ) − f ( a )] . It follows that f ( b ) − f ( a ) b − a ≤ f ( b ) − f ( x ) b − x . The proof is now complete. Theorem 4.6.6 Let I be an open interval and let f : I → R be a differentiable function. Then f is convex if and only if f 0 is increasing on I . Proof: Suppose f is convex. Fix a < b with a , b ∈ I . By Lemma 4.6.5 , for any x ∈ ( a , b ) , we have f ( x ) − f ( a ) x − a ≤ f ( b ) − f ( a ) b − a . This implies, taking limits, that f 0 ( a ) ≤ f ( b ) − f ( a ) b − a . Similarly, f ( b ) − f ( a ) b − a ≤ f 0 ( b ) . Therefore, f 0 ( a ) ≤ f 0 ( b ) , and f 0 is an increasing function. Let us prove the converse. Suppose f 0 is increasing. Fix x 1 < x 2 and t ∈ ( 0 , 1 ) . Then x 1 < x t < x 2 , where x t = tx 1 +( 1 − t ) x 2 . By the Mean Value Theorem (Theorem 4.2.3 ) , there exist c 1 and c 2 such that x 1 < c 1 < x t < c 2 < x 2 with f ( x t ) − f ( x 1 ) = f 0 ( c 1 )( x t − x 1 ) = f 0 ( c 1 )( 1 − t )( x 2 − x 1 ) ; f ( x t ) − f ( x 2 ) = f 0 ( c 2 )( x t − x 2 ) = f 0 ( c 2 ) t ( x 1 − x 2 ) . This implies t f ( x t ) − t f ( x 1 ) = f 0 ( c 1 ) t ( 1 − t )( x 2 − x 1 ) ; ( 1 − t ) f ( x t ) − ( 1 − t ) f ( x 2 ) = f 0 ( c 2 ) t ( 1 − t )( x 1 − x 2 ) .