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!
Best Answer
You can use the examples from Extension of bounded convex function to boundary by redefining them on the boundary to achieve lower semicontinuity.
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 $(0,0)$.