# Prove a convex function is continuous

• Sep 9th 2011, 05:59 AM
davismj
Prove a convex function is continuous
I need to prove that a real valued convex function on R is continuous. I'm thinking this is only a special case and may not be the right way to attack the problem:

Assume the contrary. That is, there is at least one point $\displaystyle x_0$ such that for any $\displaystyle \epsilon > 0$ there is no $\displaystyle \delta > 0$ such that $\displaystyle d(x,x_0)<\delta$ implies |f(x_0) - f(x)| < \epsilon.

Now, if we further assume that there exists $\displaystyle x,y$ such that $\displaystyle x_0\in (x,y)$ and d(f(x),f(y))<\epsilon, then it follows that since x_0 = \lambda x + (1-\lambda) y for some $\displaystyle \lambda \in (0,1)$ we have

|f(x_0) - f(x)| = |f(\lambda x + (1-\lambda) y) - f(x)| \le (1-\lambda)|f(y)-f(x)| \le \epsilon

This seems like an elegant application of convexity, but it only applies to simple discontinuities, and I'm not sure I can extend it to the more general case.

Thanks!! edit: not sure whats wrong with my latex :/
• Sep 9th 2011, 01:47 PM
FernandoRevilla
Re: Prove a convex function is continuous
See theorem 3.2 here W.rudin, Real and Complex Analysis . Ask your doubts.