# 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, the left-hand limit equals $f(y)$ and for $y, the right-hand limit also equals $f(y)$, hence the limit is $f(y)$.

Title continuity of convex functions, alternate proof ContinuityOfConvexFunctionsAlternateProof 2013-03-22 18:25:28 2013-03-22 18:25:28 yesitis (13730) yesitis (13730) 4 yesitis (13730) Proof msc 26B25 msc 26A51