Suppose $U$ is a convex open set in $\mathbb R^n$, and $f$ is a bounded convex function on $U$. $f$ is then automatically continuous on $U$.
The question is: Can we always extend $f$ continuously to $\bar{U}$?
I know boundedness of $f$ is necessary, otherwise we can construct $g=1/[x-x^2]$ on unit open interval. Boundedness of $f$ is also sufficient in one dimensional case. If the extension in general exists, the only reasonable way to construct it is via:
find a fixed point $x_0$ in $U$; for each point x on $\partial U$, consider the line $L$ joining $x_0$ and $x$; $f$ restricting on $L \cap U$ is convex; we do the extension in this one dimensional case.
But I can not prove the continuity of the extended function, nor can I prove the extension is irrelevant to choice of $x_0$. I know $f$ is locally Lipschitz; perhaps it can help in some way.