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 seethat $f(\\lambda x+(1-\\lambda)y)$ 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 $f(y)$ , hencethe limit is $f(y)$ .