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While thinking about convex functions, I managed to put together the following proof which I find a bit too good to be true. $X$ is a topological vector space that is also a Baire space.

Lemma: Let $f : X \to \mathbb{R}$ be convex and locally bounded. Then $f$ is continuous.

proof: Let $x \in X$ and $U \subseteq X$ a balanced neighbourhood of zero such that $\sup_{y \in U} \vert f( x + y ) \vert \le C$ for some $C > 0$. Then for all $t > 0$, $y \in t U$, \begin{equation} \begin{aligned} f \left( x \right) &= f \left( \frac{1}{1 + t} \left[ x + y \right] + \frac{t}{1+t} \left[ x - \frac{y}{t} \right] \right) \\ &\le \frac{1}{1 + t} f \left( x + y \right) + \frac{t}{1 + t} f \left( x - \frac{y}{t} \right) \\ \implies f \left( x \right) - f \left( x + y \right) &\le t \left[ f \left( x - \frac{y}{t} \right) - f \left( x \right) \right] \le 2 C t \, . \end{aligned} \end{equation} Likewise, for all $t \in (0,1)$, \begin{equation} \begin{aligned} f \left( x + y \right) &= f \left( t \left[ x + \frac{y}{t} \right] + \left( 1 - t \right) x \right) \\ &\le t f \left( x + \frac{y}{t} \right) + \left( 1 - t \right) f \left( x \right) \\ \implies f \left( x + y \right) - f \left( x \right) &\le t \left[ f \left( x + \frac{y}{t} \right) - f \left( x \right) \right] \le 2 C t \end{aligned} \end{equation} whenever $y \in t U$.

Theorem: Let $f : X \to \mathbb{R}$ be convex, lower semicontinuous and bounded from below. Then $f$ is continuous.

proof: By the lemma it suffices to show that $f$ is locally bounded. Let $m \in \mathbb{R}$ be lower bound of $f$ and define $A_K = f^{-1}( [m,K]) = f^{-1}( (-\infty,K])$ for all $K \in \mathbb{N}$. These sets are closed by the lower semicontinuity of $f$ and $\cup_{K \in \mathbb{N}} A_K = X$.

Hence, by the Baire category theorem some $A_K$ has nonempty interior, i.e there are $K \in \mathbb{N}$, $x \in X$ and an open neighbourhood $U \subseteq X$ of zero such that \begin{equation} \sup_{y \in U} f \left( x + y \right) \le K \, . \end{equation}

Now, for any $z \in X$ and $y \in U/2$, \begin{equation} \begin{aligned} m \le f \left( z + y \right) &= f \left( \frac{1}{2} \left[ 2 z - x \right] + \frac{1}{2} \left[ x + 2 y \right] \right) \\ &\le \frac{1}{2} f \left( 2 z - x \right) + \frac{1}{2} f \left( x + 2 y \right) \\ &\le \frac{1}{2} f \left( 2 z - x \right) + \frac{K}{2} \, . \end{aligned} \end{equation} Thus, $f$ is locally bounded since $z$ was arbitrary.

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  • $\begingroup$ Did you look at examples of lower semicontinuous convex functions that are not continuous? [This answer]math.stackexchange.com/a/2487999/9759) seems to give one that is bounded below: $f(x,y) = x^2/y$ defined on the set $\{(x,y) : y \geq x^2\}$ where we define $f(0,0) = 0$. $\endgroup$
    – usul
    Commented Mar 31, 2022 at 15:42
  • $\begingroup$ I tried to find some and the closest I got was seminorms that are measurable with respect to some Gaussian measure. But these seem overly abstract and are usually not lower semicontinuous. Your example is not a contradiction, since it is not defined on all of $\mathbb{R}^2$. $\endgroup$
    – iolo
    Commented Mar 31, 2022 at 15:56
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    $\begingroup$ Indeed, see e.g this question. Any counterexample should have to be in the infinite-dimensional setting. $\endgroup$
    – iolo
    Commented Mar 31, 2022 at 16:05
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    $\begingroup$ Yes. This is a standard result in convex analysis that can be found in most textbooks. Note that lower semicontinuity (which you assume explicitly) is an essential requirement in infinite dimensions; you also have to assume that you are in the interior of the effective domain (which you assume implicitly since your $f$ is real-valued and cannot attain the value $+\infty$). $\endgroup$ Commented Apr 1, 2022 at 9:16
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    $\begingroup$ (Shameless plug: Section 3.3 in arxiv.org/abs/2001.00216) $\endgroup$ Commented Apr 1, 2022 at 10:34

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Though not entirely in the same setting, as can be seen from these lecture notes my reasoning seems to hold. In the lecture notes, one considers barrelled spaces but the local boundedness at some point can easily be obtained in either the barrelled or Baire setting, by noticing that a closed, balanced, absorbing subset then has non-empty interior.

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it is true that a convex function is continuous on the relative interior of it's domain, but not necessarily on the relative boundary, consider the function $f(x)=0$ if $0<x<1$ and $f(x)=1$ if $x=0,1$, this is a convex function, but not continuous at zero or one.

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    $\begingroup$ Note that the full statement in the question body is about functions defined on (the whole of) a topological vector space, so this is not a counterexample. $\endgroup$ Commented Jun 10 at 18:37

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