TITLE: Extension of bounded convex function to boundary
QUESTION [8 upvotes]: 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.

REPLY [6 votes]: By constructing a counterexample for a different problem, I encountered an explicit function which can serve as a counterexample here.
On
$$D = \{(x,y) \in \mathbb{R}^2 : x^2 < y\}$$
consider the function
$$f(x,y) = \frac{x^2}{y}.$$
On $D$, $f$ takes values in the interval $(0,1)$.
We can extend $f$ continuously to $\bar D \setminus \{(0,0)\}$ and have $f(0,y) = 0$, $y > 0$, and $f(x,x^2) = 1$, $x > 0$. Hence, it does not extend continuously to $(0,0)$.
Similarly to Normal Human's example, the epigraph of this function is the convex hull of $\{(x,x^2,1), x \in \mathbb{R}\} \cup \{(0,y,0),  y \in \mathbb{R}\}$.
Edit: Here is an image (upside down):