proof of criterion for convexity


Theorem 1.

Suppose f:(a,b)R is continuousMathworldPlanetmathPlanetmath and that, for all x,y(a,b),

f(x+y2)f(x)+f(y)2.

Then f is convex.

Proof.

We begin by showing that, for any natural numbersMathworldPlanetmath n and m2n,

f(mx+(2n-m)y2n)mf(x)+(2n-m)f(y)2n

by inductionMathworldPlanetmath. When n=1, there are three possibilities: m=1, m=0, and m=2. The first possibility is a hypothesisMathworldPlanetmathPlanetmath of the theoremMathworldPlanetmath being proven and the other two possibilities are trivial.

Assume that

f(mx+(2n-m)y2n)mf(x)+(2n-m)f(y)2n

for some n and all m2n. Let m be a number less than or equal to 2n+1. Then either m2n or 2n+1-m2n. In the former case we have

f(12mx+(2n-m)y2n+y2) 12(f(mx+(2n-m)y2n)+f(y))
12mf(x)+(2n-m)f(y)2n+f(y)2
=mf(x)+(2n+1-m)f(y)2n+1.

In the other case, we can reverse the roles of x and y.

Now, every real number s has a binary expansion; in other words, there exists a sequence {mn} of integers such that limnmn/2n=s. If 0s1, then we also have 0mn2n so, by what we proved above,

f(mnx+(2n-mn)y2n)mnf(x)+(2n-mn)f(y)2n.

Since f is assumed to be continuous, we may take the limit of both sides and conclude

f(sx+(1-s)y)sf(x)+(1-s)f(y),

which implies that f is convex. ∎

Title proof of criterion for convexity
Canonical name ProofOfCriterionForConvexity
Date of creation 2013-03-22 17:00:23
Last modified on 2013-03-22 17:00:23
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 5
Author rspuzio (6075)
Entry type Proof
Classification msc 52A41
Classification msc 26A51
Classification msc 26B25