Are closed convex functions continuous? A function is called closed if its epigraph is closed -- see here for a little more detail. Suppose $D \subset \mathbb{R}^n$ is a convex set and $f: D \rightarrow \mathbb{R}$ is a closed convex function. Does it follow that $f$ is continuous on $D$? 
If $D$ is an interval, the answer is yes, and this is Proposition 1.3.12 in the textbook Convex Optimization Theory by Bertsekas. The fact that the proposition is stated for an interval suggests that the answer to my question might be negative, but I'm having trouble thinking of a counterexample.
 A: A convex function is closed if and only if it is lower semi-continuous, but a closed convex function is not neccesarily continuous. Here is a celebrated example:
$$D=\{(x,y) \mid y\geq x^2\}$$
$$f(x,y) = \begin{cases} \frac{x^2}{y} & \text{if } y\geq x^2 \\ 0 & \text{if } x = 0 \text{ and } y=0. \end{cases} $$
This function is discontinuous in $(0,0)$ since you could walk over the curve $y=x^2$ or over $y=2x^2$ and end up with different function values. However, it is lower-semicontinuous and convex (see next paragraph), and hence closed.
Convexity for $y>0$ is trivial (quadratic over linear is known to be convex). For any $(x_1,y_1) \in D$ with $y_1=0$ (and hence $x_1=0$) and for any $(x_2,y_2) \in D$ with $y_2>0$ and for any $\lambda \in [0,1]$ we have:
$$\lambda f(x_1,y_1) + (1-\lambda)f(x_2,y_2) = \frac{(1-\lambda)x_2^2}{y_2}$$
$$f(\lambda x_1 + (1-\lambda) x_2,\lambda y_1 + (1-\lambda) y_2) = \frac{((1-\lambda) x_2)^2}{(1-\lambda)y_2} = \frac{(1-\lambda) x_2^2}{y_2}$$
from which convexity follows.
