Introduction to Mathematical Analysis I - Second Edition

127 Figure 4.6: A Convex Function. Definition 4.6.1 Let I be an interval of R and let f : I → R . We say that f is convex on I if f ( λ u +( 1 − λ ) v ) ≤ λ f ( u )+( 1 − λ ) f ( v ) for all u , v ∈ I and for all λ ∈ ( 0 , 1 ) . Example 4.6.1 The following functions are convex. (a) f : R → R , f ( x ) = x . This is straightforward. (b) f : R → R , f ( x ) = x 2 . Here note first that 2 xy ≤ x 2 + y 2 for all real numbers x , y . Then, if 0 < λ < 1 and x , y ∈ R , we get f ( λ x +( 1 − λ ) y ) = ( λ x +( 1 − λ ) y ) 2 = λ 2 x 2 + 2 λ ( 1 − λ ) xy +( 1 − λ ) 2 y 2 ≤ λ 2 x 2 + λ ( 1 − λ )( x 2 + y 2 )+( 1 − λ ) 2 y 2 = λ ( λ x 2 +( 1 − λ ) x 2 )+( 1 − λ )( λ y 2 +( 1 − λ ) y 2 ) = λ x 2 +( 1 − λ ) y 2 = λ f ( x )+( 1 − λ ) f ( y ) . (c) f : R → R , f ( x ) = | x | . This follows from the triangle inequality and other basic properties of absolute value. Theorem 4.6.1 Let I be an interval of R . A function f : I → R is convex if and only if for every λ i ≥ 0, i = 1 , . . . , n , with ∑ n i = 1 λ i = 1 ( n ≥ 2) and for every x i ∈ I , i = 1 , . . . , n , f n ∑ i = 1 λ i x i ! ≤ n ∑ i = 1 λ i f ( x i ) . (4.13) Proof: Since the converse holds trivially, we only need to prove the implication by induction. The conclusion holds for n = 2 by the definition of convexity. Let k be such that the conclusion holds for any n with 2 ≤ n ≤ k . We will show that it also holds for n = k + 1. Fix λ i ≥ 0, i = 1 , . . . , k + 1 , with ∑ k + 1 i = 1 λ i = 1 and fix every x i ∈ I , i = 1 , . . . , k + 1. Then k ∑ i = 1 λ i = 1 − λ k + 1 .