# Is a convex and lower semicontinuous function defined on a closed and convex subset of $\mathbb{R}^n$ continuous?

Let $A\subset \mathbb{R}^n$ be a nonempty convex and closed set, and let $f:A\to\mathbb{R}$ be a lower semicontinuous and convex function.

Does this imply that $f$ is continuous on $A$?

I know that in the (relative) interior of $A$, $f$ is locally lipchitz so my question is about the points on the boundary. Of course, I consider the continuity of $f$ with respect to $A$ and not the continuity of the extended function $\tilde{f}:\mathbb{R}^{n}\to\mathbb{R}\cup\left\{+\infty\right\}$ defined by $\tilde{f}(x):=f(x)$ for $x\in A$ and $\tilde f(x):=+\infty$ for $x\notin A$, which is obviously not upper semicontinuous on the boundary of $A$.

If the answer to my question is negative a counterexample would be welcome. Thanks!

For example, you can do the following. We consider the set $$A:= \{(x,y) \in \mathbb R^2 \mid x^2 \le y \}$$ and the function $$f(x,y) = \frac{x^2}y \qquad\forall (x,y) \in A \setminus \{(0,0)\}$$ and $f(0,0) = 0$. If I did not miss something, this should satisfy your assumptions while being discontinuous in $(0,0)$.
• Your function is not lower semicontinuous at $(0,0)$. it is indeed upper semicontinuous ! – Red shoes Oct 24 '17 at 18:14
• Indeed. Would it help to set it to $0$ in $0$? – gerw Oct 24 '17 at 18:26