closed epigraphs equivalence

Solution 1:

Let $E(f) = \{(x,y) : f(x) \leqslant y \}$ denote the epigraph of $f$. The $E(f)$ is closed if and only if its complement is open.

Suppose $E(f)$ is closed, let $x \in X$ and $y < f(x)$ (so $(x,y) \notin E(f)$). Since $E(f)$ is closed, there is a neighbourhood $U$ of $x$ and an $\varepsilon > 0$ such that $U\times(y-\varepsilon,y+\varepsilon)\cap E(f) =\varnothing$. By the structure of an epigraph, that entails $U\times(-\infty,y+\varepsilon)\cap E(f) = \varnothing$, hence $f(z) \geqslant y+\varepsilon$ for all $z\in U$, and that means $f$ is lower semicontinuous at $x$, since $x$ was arbitrary, $f$ is lower semicontinuous.

Conversely, suppose $f$ is lower semicontinuous. Let$(x,y) \notin E(f)$, i.e. $y < f(x)$. Define $\mu := \frac{y+f(x)}{2}$. By the lower semicontinuity of $f$, $U = \{z \in X : f(z) > \mu\}$ is open. Since$\mu < f(x)$, we have $x \in U$. Then $U\times (-\infty,\mu)$ is an open neighbourhood of $(x,y)$ that doesn't intersect $E(f)$, hence the complement of $E(f)$ is open.