proof of f(t)-f(s)t-sf(u)-f(s)u-sf(u)-f(t)u-t for convex f

We will prove

f(t)-f(s)t-sf(u)-f(s)u-s. (1)

The proof of the right-most inequalityMathworldPlanetmath is similar.

Suppose (1) does not hold. Then for some s,t,u,

f(t)-f(s)t-s>f(u)-f(s)u-s. (2)

This inequality is just the statement of the slope of the line segment AB¯,A=(t,f(t)),B=(s,f(s)), being larger than the slope of the segment CB¯,C=(u,f(u)). Since t is between s and u, and f is continuousMathworldPlanetmathPlanetmath, this implies

f(t)>h(x)=f(u)-f(s)u-s(x-s)+f(s), (3)

s<x<u. This contradicts convexity of f on (a,b). Hence, (2) is false and (1) follows.

Note that we have tacitly use the fact that x=λu+(1-λ)s and h(x)=λf(u)+(1-λ)f(s) for some λ.

Title proof of f(t)-f(s)t-sf(u)-f(s)u-sf(u)-f(t)u-t for convex f
Canonical name ProofOffracftfstsleqfracfufsusleqfracfuftutForConvexF
Date of creation 2013-03-22 18:25:34
Last modified on 2013-03-22 18:25:34
Owner yesitis (13730)
Last modified by yesitis (13730)
Numerical id 6
Author yesitis (13730)
Entry type Proof
Classification msc 26A51