Introduction to Mathematical Analysis I - Second Edition

131 Since f 0 ( c 1 ) ≤ f 0 ( c 2 ) , we have t f ( x t ) − t f ( x 1 ) = f 0 ( c 1 ) t ( 1 − t )( x 2 − x 1 ) ≤ f 0 ( c 2 ) t ( 1 − t )( x 2 − x 1 ) = ( 1 − t ) f ( x 2 ) − ( 1 − t ) f ( x t ) . Rearranging terms, we get f ( x t ) ≤ t f ( x 1 )+( 1 − t ) f ( x 2 ) . Therefore, f is convex. The proof is now complete. Corollary 4.6.7 Let I be an open interval and let f : I → R be a function. Suppose f is twice differentiable on I . Then f is convex if and only if f 00 ( x ) ≥ 0 for all x ∈ I . Proof: It follows from Proposition 4.3.2 that f 00 ( x ) ≥ 0 for all x ∈ I if and only if the derivative function f 0 is increasing on I . The conclusion then follows directly from Theorem 4.6.6 . Example 4.6.2 Consider the function f : R → R given by f ( x ) = √ x 2 + 1. Now, f 0 ( x ) = x / √ x 2 + 1 and f 00 ( x ) = 1 / ( x 2 + 1 ) 3 / 2 . Since f 00 ( x ) ≥ 0 for all x , it follows from the corollary that f is convex. Theorem 4.6.8 Let I be an open interval and let f : I → R be a convex function. Then it is locally Lipschitz continuous in the sense that for any ¯ x ∈ I , there exist ` ≥ 0 and δ > 0 such that | f ( u ) − f ( v ) | ≤ ` | u − v | for all u , v ∈ B ( ¯ x ; δ ) . (4.15) In particular, f is continuous. Proof: Fix any ¯ x ∈ I . Choose four numbers a , b , c , d satisfying a < b < ¯ x < c < d with a , d ∈ I . Choose δ > 0 such that B ( ¯ x ; δ ) ⊂ ( b , c ) . Let u , v ∈ B ( ¯ x ; δ ) with v < u . Then by Lemma 4.6.5 , we see that f ( b ) − f ( a ) b − a ≤ f ( u ) − f ( a ) u − a ≤ f ( u ) − f ( v ) u − v ≤ f ( d ) − f ( v ) d − v ≤ f ( d ) − f ( c ) d − c . Using a similar approach for the case u < v , we get f ( b ) − f ( a ) b − a ≤ f ( u ) − f ( v ) u − v ≤ f ( d ) − f ( c ) d − c for all u , v ∈ B ( ¯ x ; δ ) . Choose ` ≥ 0 sufficiently large so that − ` ≤ f ( b ) − f ( a ) b − a ≤ f ( u ) − f ( v ) u − v ≤ f ( d ) − f ( c ) d − c ≤ ` for all u , v ∈ B ( ¯ x ; δ ) . Then ( 4.15 ) holds. The proof is now complete.