A function $f:X\to\mathbb{R}$, with $X$ being a topological space, is termed as lower semicontinuous (lsc) at $x_0\in X$ if: $$\forall\epsilon>0\,\,\exists V\text{ an open neighborhood of }x_0:x\in V\implies f(x)>f(x_0)-\epsilon.$$ Equivalently, for any net $x_n\to x_0$, $\limsup_{n\to\infty}f(x_n)\geq f(x_0)$ (sequences are sufficient in the case of metric spaces $(X,d)$). Suppose $X$ is now a normed space.
- Is linearity sufficient for a function $f$ to be lsc at all points of $X$?
- Do we need $X$ to be Banach (i.e. metrically complete) for that to hold?
- Do we need something more to be satisfied by the function?
- How do I prove anything that emerged from the previous points?
For the moment, I managed to show that for linear functions, lsc-ity in the origin is equivalent to lsc-ity everywhere. I tried then contradicting lsc-ity in the origin, i.e. supposing that: $$\exists\epsilon>0:\forall\delta>0\,\,\exists x\in X:\|x\|<\delta,f(x)\leq-\epsilon,$$ but did not get anywhere with that.
It seems the following.
Proposition. Let $X$ be a linear topological space and $f:X\to\Bbb R$ be a lower semicontinuous linear function. Then $f$ is continuous.
Proof. Since the function $f$is linear, it suffices to show that $f$ is continuous at the origin $0$. Let $\varepsilon>0$ be an arbitrary number. Since the function $f$ is lower semicontinuous at the origin, there exists an open neighborhood $V$ of the origin such that $f(x)>f(0)-\varepsilon=-\varepsilon$ for each point $x\in V$. Then $V\cap (-V)$ is a neighborhood of the origin and $|f(x)|<\varepsilon $ for each point $x\in V\cap (-V)$. $\square$