continuity of convex functions, alternate proof

Let $f$ be convex and $y\in(a,b)$ be arbitrary but fixed. Then

	$\displaystyle f(\lambda x+(1-\lambda)y)$	$\displaystyle\leq$	$\displaystyle\lambda f(x)+(1-\lambda)f(y)$		(1)
	$\displaystyle f(\lambda x+(1-\lambda)y)-f(y)$	$\displaystyle\leq$	$\displaystyle\lambda(f(x)-f(y))\leq\lambda\|f(x)-f(y)\|.$		(2)

Fix a number $c>\sup\{|f(u)-f(v)|:u,v\in(a,b)\}$ . Then

$|f(\lambda x+(1-\lambda)y)-f(y)|\leq\lambda|f(x)-f(y)|<\lambda c.$

(3)

Given $\epsilon>0$ , let $\lambda$ range over $(0,\epsilon/c)$ if $\epsilon/c<1$ , or $\lambda=1$ otherwise. Then it is easy to see that $f(\lambda x+(1-\lambda)y)$ and $f(y)$ lie within $\epsilon$ distance of each other when $\lambda$ varies as specified.

Continuity of $f$ now follows–for $x<y$ , the left-hand limit equals $f(y)$ and for $y<x$ , the right-hand limit also equals $f(y)$ , hence the limit is $f(y)$ .

Title	continuity of convex functions, alternate proof
Canonical name	ContinuityOfConvexFunctionsAlternateProof
Date of creation	2013-03-22 18:25:28
Last modified on	2013-03-22 18:25:28
Owner	yesitis (13730)
Last modified by	yesitis (13730)
Numerical id	4
Author	yesitis (13730)
Entry type	Proof
Classification	msc 26B25
Classification	msc 26A51