Sufficient condition on convex function such that $f(x) > -\infty$ for all $x$.

Let $f : \mathbb R^n \to [-\infty, \infty]$ convex and let $f(\overline x) > -\infty$ for $\overline x \in \mbox{int}(\mbox{dom}(f)$. Show that $f(x) > -\infty$ for all $x \in \mathbb R$.

The involved definitions could be found here. I have conceptual difficulty with this exercise, because I read somewhere that convex functions are continuous, so intuitively with a jump from a value $x \in \mathbb R$ to some value $\pm \infty$, this continuity might be violated, or could the extended real line $[-\infty, \infty]$ incorporated with some suitable topology on it, such that continuity is still valid. Or does it simply mean that continuity is just supposed on $\mbox{dom}(f) = \{ x \in \mathbb R : f(x) \ne \pm \infty \}?$ I am asking because I guess continuity issues might play a role in solving this exercise.

• Suppose $f(x) = -\infty$ if $x = 0$, and $f(x) = \infty$ otherwise. Then $f$ is convex and it's vacuously true that $f(x) > -\infty$ for all $x \in \text{int}( \text{dom}(f))$. Is this a counterexample to the statement you're asked to prove? – littleO Jun 26 '14 at 10:44
• "because I read somewhere that convex functions are continuous" That holds for convex functions defined on open sets. $f\colon [0,1]\to\mathbb{R}$ with $f(0) = f(1) = 1$ and $f(x) = 0$ for $0 < x < 1$ is convex, but not continuous in the end points. – Daniel Fischer Jun 26 '14 at 10:44
• @littleO: Yes thanks for your example, I was also skeptical about this statement, but is there any way to (by certain definitions) give it meaning and made it a valid statement? – StefanH Jun 26 '14 at 10:50
• Note that there is a problem with the definition of "convex function" in "your" book. Take any function $f$ which takes only the values $\pm \infty$. Then, $\text{dom}f$ is empty, hence, it is convex by the first definition. However, it is, in general, not convex by using the second definition via the epigraph. – gerw Jul 4 '14 at 19:58

To get a true statement, I think we need to add the assumption that $\text{dom} f$ has a non-empty interior.
Let $y \in \text{int}(\text{dom} f )$ and suppose (for a contradiction) that $f(x) = -\infty$ for some $x \in \mathbb R^n$. Convexity implies that $f(z) =- \infty$ for all points $z$ between $x$ and $y$. But some of the points between $x$ and $y$ belong to $\text{dom} f$, which is a contradiction.
Here is the last step in more detail. There exists $\epsilon > 0$ such that the closed ball of radius $\epsilon$ centered at $y$ is a subset of $\text{dom} f$. Choose $0 < t \leq \epsilon$ small enough that $z = y + t(x - y)$ is between $x$ and $y$. Then $f(z) = -\infty$, yet $z \in \text{dom} f$, which is a contradiction.