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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 $\overline U$?

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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
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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.

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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
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