1

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!

Anton
  • 230

1 Answers1

5

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 $(0,0)$.

gerw
  • 31,359
  • One question: if we take square from this function, we come up to function which is homogeneous of degree 2 while keep all other properties . Am I right? – Red shoes Mar 20 '19 at 17:11
  • @Redshoes: Yes, I think so. – gerw Mar 20 '19 at 19:15
  • This is a very nice example. I've been trying to construct a discontinuous, lower semicontinuous convex function for a few days, probably a week. I was trying to produce one where the discontinuity would "come from the boundary" rather than from the inside. Do you know if the function defined by $e^{ix}\mapsto x$ for $x\in[0,2\pi)$ has a convex extension to the closed unit disk in $\mathbb{C}$ ? – Olivier Bégassat Aug 14 '21 at 16:26
  • @OlivierBégassat: Yes, I think so. You can just take the largest convex function which lies below your function on the boundary. Its graph is the union of the segments from $(1,0,0)$ to $(\cos(\phi),\sin(\phi),\phi)$ for $\phi \in (0,2\pi)$. – gerw Aug 16 '21 at 06:07