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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