Introduction to Mathematical Analysis I - 3rd Edition

137 Theorem 5.5.1 Let I be an interval in R. A function f : I →Ris convex if and only if for every λi ≥0, i =1, . . . ,n, with∑n i=1λi =1 (n≥2) and for every xi ∈I, i =1, . . . ,n, f n ∑ i=1 λixi!≤ n ∑ i=1 λi f (xi). (5.5) 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 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 fix every xi ∈I, i =1, . . . ,k+1. Then k ∑ i=1 λi =1−λk+1. If λk+1 =1, thenλi =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 ∑ i=1 λi 1−λk+1 =1. It follows that f k+1 ∑ i=1 λixi!=f (1−λk+1) ∑k i=1λixi 1−λk+1 +λk+1xk+1 ≤(1−λk+1)f ∑k i=1λixi 1−λk+1 +λk+1 f (xk+1) = (1−λk+1)f k ∑ i=1 λi 1−λk+1 xi!+λk+1 f (xk+1) ≤(1−λk+1) k ∑ i=1 λi 1−λk+1 f (xi)+λk+1 f (xk+1) = k+1 ∑ i=1 λi f (xi), where the first inequality follows from the definition 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 Rand let f : I →Rbe 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. For anyx∈I, we have xn = (1− 1 n)x0+ 1 nx→x0. Thus, xn ∈B(x0;δ)∩I whennis sufficiently large. Thus, for such n, f (x0) ≤f (xn) ≤(1− 1 n )f (x0)+ 1 n f (x).

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