Let $X$ be a vector space, $\Omega$ a convex subset thereof and $f:\Omega \to \mathbb R$ a convex function. Then $f$ need not be bounded from below - not even if it is strictly convex, as the example of $\Omega=X=\mathbb R$ and $f:x\mapsto x+e^x$ shows.

Things change, however, if additionally $X$ is a normed space and $\Omega$ is compact: then indeed I am able to prove that $f$ cannot be unbounded from below (unless $f\equiv -\infty$, which is excluded by assumption). However, the only proof I can come up with is based on a geometrical intuition (basically, separation of the epigraph of $f$ and a point $x_0$ below it by means of a hyperplane) that can be elementarily justified only if $X$ is a separable Hilbert space (and in particular if $\dim X <\infty$) and relies instead upon Hahn-Banach in the case of general $X$. Is there a simpler way to prove this assumption?


I think it is not true as stated, you need some further assumption on $f$, like lower semicontinuity.

Without further assumptions, I think that it is possible to to construct a discontinuous linear function, which is unbounded on a convex set.

Your argument also uses the closedness of the epigraph, hence the lower semicontinuity of $f$. And in this case, it can be proven quite simple by using the compactness:

  • Assume $f$ is unbounded, then there exists a sequence $\{x_n\}\subset \Omega$ with $f(x_n) \to -\infty$.
  • By compactness, you choose a convergent subsequence of $\{x_n\}$ (without relabeling) with limit $x$.
  • By lower semicontinuity you have $$f(x) \le \liminf_{n \to \infty} f(x_n) = -\infty,$$ which is a contradiction to $f(x) \in \mathbb{R}$.
  • $\begingroup$ I think I don't understand. I would say it is not necessary to impose closedness of the epigraph: if it is not close, then it suffices to apply my proof to the closed convex hull thereof. And how could a discontinuous linear function that blows up (or rather: blows down) within a compact set look like? $\endgroup$ – Delio Mugnolo Dec 15 '15 at 14:24
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    $\begingroup$ The Hilbert cube $H = \{x \in \ell^2 : |x_n| \le 1/n\}$ is compact in $\ell^2$. Now, you do almost the usual construction of an unbounded linear functional. Extend the unit vectors $\{e_1,\ldots\}$ to an algebraic basis, and define the linear functional $f$ by $f(e_n) = n^2$, $f(b) = 0$ for the remaining basis elements and extend by linearity. Then, $e_n/n$ lies in the compact set $H$, but $f(e_n/n) = n$. $\endgroup$ – gerw Dec 15 '15 at 14:31
  • $\begingroup$ And to comment on your proof: How do you guarantee that there is actually a point below the closed convex hull of the epigraph? $\endgroup$ – gerw Dec 15 '15 at 14:32
  • $\begingroup$ Your proof seems to work (and in Hilbert space!). On the other hand, your remark about my own proof ("proof"?) puzzles me. The closed convex hull of the epigraph is the simply the closure $\overline{Epi(f)}$ of the epigraph, so if we can show that there is point outside it, then Hahn-Banach can be immediately applied (the point being compact and convex, the epigraph's closure being closed and convex). So, you are suggesting that it may well happen that $\overline{Epi(f)}=\Omega\times \mathbb R$ although the complement of $Epi(f)$ contains a bunch of halflines with origin in $x_0$. $\endgroup$ – Delio Mugnolo Dec 15 '15 at 22:39
  • $\begingroup$ By the way: by this post (math.stackexchange.com/questions/992663/…) $\overline{Epi(f)}$ is the epigraph of the convex conjugate $f^*$ of $f$, so the question is whether the convex conjugate of a convex (but not lower semicontinuous) function can be identically $-\infty$. $\endgroup$ – Delio Mugnolo Dec 16 '15 at 0:04

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