137 Theorem 5.5.1 Let I be an interval in R. A function f : I → R is convex if and only if for every λi ≥ 0, i = 1,...,n, with ∑n i=1 λi = 1 (n ≥ 2) and for every xi ∈ I, i = 1,...,n, ! n n f ∑ λixi ≤ ∑ λi f (xi). (5.5) i=1 i=1 Proof: Since the converse holds trivially, we only need to prove the implication by induction. The conclusion holds for n = 2 by the defnition of convexity. Let k be such that the conclusion holds for n = k with 2 ≤ k. We will show that it also holds for n = k + 1. Fix λi ≥ 0, i = 1,...,k + 1, with ∑k+1 i=1 λi = 1 and fx every xi ∈ I, i = 1,...,k + 1. Then k ∑ λi = 1− λk+1. i=1 If λk+1 = 1, then λi = 0 for all i = 1,...,k, and (5.5) holds. Suppose 0 ≤ λk+1 < 1. Then, for each i = 1,...,k, λi/(1 − λk+1) ≥ 0 and k λi ∑ = 1. i=1 1 − λk+1 It follows that ! k+1 ∑k i=1 λixi f ∑ λixi = f (1 − λk+1) + λk+1xk+1 i=1 1− λk+1 ∑k i=1 λixi ≤ (1 − λk+1) f + λk+1 f (xk+1) 1− λk+1 ! k λi =(1 − λk+1) f ∑ xi + λk+1 f (xk+1) i=1 1− λk+1 k λi ≤ (1 − λk+1) ∑ f (xi)+ λk+1 f (xk+1) i=1 1− λk+1 k+1 = ∑ λi f (xi), i=1 where the frst inequality follows from the defnition of convexity (or is trivial if λk+1 = 0) and the last inequality follows from the inductive assumption. The proof is now complete. □ Theorem 5.5.2 Let I be an interval in R and let f : I → R be a convex function. Then f has a local minimum at x0 if and only if f has an absolute minimum at x0. Proof: Clearly if f has a global minimum at x0, then it also has a local minimum at x0. Conversely, suppose that f has a local minimum at x0. Then there exists δ > 0 such that f (u) ≥ f (x0) for all u ∈ B(x0;δ ) ∩ I. 1 For any x ∈ I, we have x n =(1− 1 )x 0 + x → x0. Thus, xn ∈ B(x0;δ ) ∩ I when n is suffciently large. n n Thus, for such n, 1 1 f (x0) ≤ f (xn) ≤ (1− ) f (x0)+ f (x). n n
RkJQdWJsaXNoZXIy NTc4NTAz