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 $\map {\operatorname{epi} } f$ 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 $\map {\operatorname{lev} \limits_{\mathop \le a} } f$ is closed if and only if


 * $\ds \map {\operatorname{lev} \limits_{\mathop > a} } f := \set {s \in S : ~ \map f s > a}$

is open.

The proof is carried out in the following steps:

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

We prove that $\map {\operatorname{lev} \limits_{\mathop > a} } f$ is open.

Let $s_0 \in \map {\operatorname{lev} \limits_{\mathop > a} } f$.

Because $f$ is LSC this implies


 * $\ds \map f {s_0} = \liminf_{s \mathop \to s_0} \map f s > a$

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


 * $\ds \inf_{s \mathop \in V_0} \map f s > a$

This implies


 * $\ds V_0 \subseteq \map {\operatorname{lev} \limits_{\mathop > a} } f$

and hence proves that $\map {\operatorname{lev} \limits_{\mathop > a} } f$ is open.

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

Then $\map {\operatorname{lev} \limits_{\mathop > a} } f$ is open for all $a \in \R$.

Let $s_0 \in S$.

The definition of the lower limit implies


 * $\ds \liminf_{s \mathop \to s_0} \map f s \le \map f {s_0}$

It remains to show that

$(4) \quad \ds \map f {s_0} \le \liminf_{s \mathop \to s_0} \map f s$


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


 * Case 2: $\map f {s_0} = +\infty$.


 * Let $n \in \N$.


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


 * Thus, by definition of the lower limit:


 * $\ds \liminf_{s \mathop \to s_0} \map f s \ge \inf_{s \mathop \in V_n} \map f s = n$


 * Since $n$ was arbitrary it follows that $\ds \liminf_{s \mathop \to s_0} \map f s = +\infty$.


 * Hence $(4)$.


 * Case 3: $\map f {s_0} \in \R$.


 * Let $\varepsilon > 0$.


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


 * Thus, by definition of the lower limit:


 * $\ds \liminf_{s \mathop \to s_0} \map f s \ge \inf_{s \mathop \in V_\varepsilon} \map f s \ge \map f {s_0} - \varepsilon$


 * 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.