Is a convex, lower semicontinuous function that is bounded from below, actually continuous?

Question

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.

Answer 1

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.

Answer 2

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.