Introduction to Mathematical Analysis I - Second Edition

128 4.6 CONVEX FUNCTIONS AND DERIVATIVES If λ k + 1 = 1, then λ i = 0 for all i = 1 , . . . , k , and ( 4.13 ) holds. Suppose 0 ≤ λ k + 1 < 1. Then, for each i = 1 , . . . , k , λ i / ( 1 − λ k + 1 ) ≥ 0 and k ∑ i = 1 λ i 1 − λ k + 1 = 1 . It follows that f k + 1 ∑ i = 1 λ i x i ! = f ( 1 − λ k + 1 ) ∑ k i = 1 λ i x i 1 − λ k + 1 + λ k + 1 x k + 1 ≤ ( 1 − λ k + 1 ) f ∑ k i = 1 λ i x i 1 − λ k + 1 + λ k + 1 f ( x k + 1 ) = ( 1 − λ k + 1 ) f k ∑ i = 1 λ i 1 − λ k + 1 x i ! + λ k + 1 f ( x k + 1 ) ≤ ( 1 − λ k + 1 ) k ∑ i = 1 λ i 1 − λ k + 1 f ( x i )+ λ k + 1 f ( x k + 1 ) = k + 1 ∑ i = 1 λ i f ( x i ) , where the first inequality follows from the definition of convexity (or is trivial if λ k + 1 = 0) and the last inequality follows from the inductive assumption. The proof is now complete. Theorem 4.6.2 Let I be an interval and let f : I → R be a convex function. Then f has a local minimum at ¯ x if and only if f has an absolute minimum at ¯ x . Proof: Clearly if f has a global minimum at ¯ x , then it also has a local minimum at ¯ x . Conversely, suppose that f has a local minimum at ¯ x . Then there exists δ > 0 such that f ( u ) ≥ f ( ¯ x ) for all u ∈ B ( ¯ x ; δ ) ∩ I . For any x ∈ I , we have x n = ( 1 − 1 n ) ¯ x + 1 n x → ¯ x . Thus, x n ∈ B ( ¯ x ; δ ) ∩ I when n is sufficiently large. Thus, for such n , f ( ¯ x ) ≤ f ( x n ) ≤ ( 1 − 1 n ) f ( ¯ x )+ 1 n f ( x ) . This implies that for a sufficient large n , we have 1 n f ( ¯ x ) ≤ 1 n f ( x ) and, hence, f ( ¯ x ) ≤ f ( x ) . Since x was arbitrary, this shows f has an absolute minimum at ¯ x . Theorem 4.6.3 Let I be an open interval and let f : I → R be a convex function. Suppose f is differentiable at ¯ x . Then f 0 ( ¯ x )( x − ¯ x ) ≤ f ( x ) − f ( ¯ x ) for all x ∈ I . (4.14)