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

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.

• 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 $f(0,0) = 0$.
– usul
Mar 31, 2022 at 15:42
• 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 $\mathbb{R}^2$.
– iolo
Mar 31, 2022 at 15:56
• Indeed, see e.g this question. Any counterexample should have to be in the infinite-dimensional setting.
– iolo
Mar 31, 2022 at 16:05
• 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 $+\infty$). Apr 1, 2022 at 9:16
• (Shameless plug: Section 3.3 in arxiv.org/abs/2001.00216) Apr 1, 2022 at 10:34