MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $U$ be an open convex set in a locally convex space $X$, and let $f : U \to [0,1]$ be a log-concave function on $U$ (i.e., bounded and real-valued). Under what conditions does $f$ have a continuous extension to the closure $\overline U$?

share|cite|improve this question
My simple-minded reaction to this is to study $F = e^{-f}$, which is convex and bounded (if you assume the image of $f$ is in $[0,1]$). The properties of convex functions are rather well known. – Deane Yang Nov 24 '10 at 18:55
Thanks for the thought, Deane. Here it's that $-\log f$ is convex, not $\mathrm e^{-f}$. Originally, I had phrased the question in terms of convex functions instead. Since $-\log 0 = \infty$, though, I figured the question would be clearer if I just asked the log-concave version. You're right: this question translates to one about convex functions, which I also do not know the answer to. – Tom LaGatta Nov 24 '10 at 20:32
Define $f(0) = 0$ and $f(1) = \mathrm e^{-1}$. I don't understand the point of your example; perhaps you misunderstood my question? – Tom LaGatta Nov 24 '10 at 22:17
Tom, sorry for the dyslexic comment. – Deane Yang Nov 24 '10 at 22:49
No problem, Deane. It was a good suggestion in intent, and will perhaps help somebody answer the question. – Tom LaGatta Nov 24 '10 at 22:54

I do not see how $\log$-concavity should imply any form of continuity. For instance, if $\|\cdot\|$ is any semi-norm on the locally convex space $X$ then $f(x) = e^{-\|x\|}$ will be bounded and $\log$-concave but it will only be continuous if the semi-norm is continuous.

What you really want in order to be able to extend $f$ to the boundary of $U$ is uniform continuity and for that the fact that $f$ is $\log$-concave may help but is certainly not sufficient.

share|cite|improve this answer
Yes, certainly if $f$ is not even continuous on $U$, then it has no continuous extension to the closure... Perhaps the wording of the problem needs to be changed. – Gerald Edgar Nov 25 '10 at 14:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.