User:S.anzengruber/Sandbox/LSC

Theorem
Let $f: S \to \overline \R$ be an extended real valued function.

Let $S$ be endowed with a topology $\tau$.

The following are equivalent:


 * $(1): \quad$ $f$ is lower semicontinuous (LSC) on $S$.
 * $(2): \quad$ The epigraph $\operatorname{epi} \left({f}\right)$ of $f$ is a closed set in $S \times \R$ with the product topology.
 * $(3): \quad$ All lower level sets of $f$ are closed in $S$.

Proof
Throughout the proof we use that the lower level set $\operatorname{lev} \limits_{\mathop \le a} \left({f}\right)$ is closed if and only if


 * $\ds \operatorname{lev} \limits_{\mathop > a} \left({f}\right) := \left\{ s \in S : ~ f \left({s}\right) > a \right\}$

is open.

The proof is carried out in the following steps:

LSC implies Closed Level Sets
Let $a \in \R$.

We prove that $\operatorname{lev} \limits_{\mathop > a} \left({f}\right)$ is open.

Let $s_0 \in \operatorname{lev} \limits_{\mathop > a} \left({f}\right)$.

Because $f$ is LSC this implies


 * $\displaystyle f\left({s_0}\right) = \liminf_{s \mathop \to s_0} f \left({s}\right) > a$

Thus, by definition of the lower limit, there exists an open neighborhood $V_0$ of $s_0$ such that


 * $\displaystyle \inf_{s \mathop \in V_0} f \left({s}\right) > a$

This implies


 * $\displaystyle V_0 \subseteq \operatorname{lev} \limits_{\mathop > a} \left({f}\right)$

and hence proves that $\operatorname{lev} \limits_{\mathop > a} \left({f}\right)$ is open.

Closed Level Sets implies LSC
Let $\operatorname{lev} \limits_{\mathop \le a} \left({f}\right)$ be closed for all $a \in \R$.

Then $\operatorname{lev} \limits_{\mathop > a} \left({f}\right)$ is open for all $a \in \R$.

Let $s_0 \in S$.

The definition of the lower limit implies


 * $\displaystyle \liminf_{s \mathop \to s_0} f \left({s}\right) \le f \left({s_0}\right)$

It remains to show that

$(4) \quad \displaystyle f \left({s_0}\right) \le \liminf_{s \mathop \to s_0} f \left({s}\right)$


 * Case 1: $f \left({s_0}\right) = - \infty$. Then (4) is trivial.


 * Case 2: $f \left({s_0}\right) = + \infty$.


 * Let $n \in \N$.


 * Then $V_n := \operatorname{lev} \limits_{\mathop > n} \left({f}\right)$ is an open neighborhood of $s_0$


 * Thus, by definition of the lower limit:


 * $\displaystyle \liminf_{s \mathop \to s_0} f \left({s}\right) \geq \inf_{s \mathop \in V_n} \ f \left({s}\right) = n$


 * Since $n$ was arbitrary it follows that $\displaystyle \liminf_{s \mathop \to s_0} f \left({s}\right) = +\infty$.


 * Hence (4).


 * Case 3: $f \left({s_0}\right) \in \R$.


 * Let $\varepsilon > 0$.


 * Then $V_\varepsilon := \operatorname{lev} \limits_{\mathop > f \left({s_0}\right)- \varepsilon} \left({f}\right)$ is an open neighborhood of $s_0$


 * Thus, by definition of the lower limit:


 * $\displaystyle \liminf_{s \mathop \to s_0} f \left({s}\right) \geq \inf_{s \mathop \in V_\varepsilon} \ f \left({s}\right) \geq f \left({s_0}\right)- \varepsilon$


 * Since $\varepsilon > 0$ was arbitrary, (4) follows.

Closed Epigraph implies Closed Level Sets
So:
 * $(1) \iff (2)$ and $(1) \iff (3)$

and the proof is complete.