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!


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 $(0,0)$.

  • $\begingroup$ Your function is not lower semicontinuous at $(0,0)$. it is indeed upper semicontinuous ! $\endgroup$ – Red shoes Oct 24 '17 at 18:14
  • $\begingroup$ Indeed. Would it help to set it to $0$ in $0$? $\endgroup$ – gerw Oct 24 '17 at 18:26
  • $\begingroup$ Then it is fine . $\endgroup$ – Red shoes Oct 24 '17 at 18:51
  • $\begingroup$ 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? $\endgroup$ – Red shoes Mar 20 at 17:11
  • $\begingroup$ @Redshoes: Yes, I think so. $\endgroup$ – gerw Mar 20 at 19:15

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