# Characterization of Lower Semicontinuity

## Contents

## 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

The proof is carried out in the following three steps:

### LSC implies Closed Epigraph

Let $f: S \to \overline \R$ be LSC on $S$.

Let $\operatorname{epi} \left({f}\right)$ denote the epigraph of $f$.

Take a sequence $\left\langle{\left({x_n, a_n}\right)}\right\rangle_{n \mathop \in \N} \in \operatorname{epi} \left({f}\right)$ such that:

- $\left({x_n, a_n}\right) \to \left({x, a}\right)$

as $n \to \infty$.

This implies that $x_n \to x$ and $a_n \to a$ as $f \left({x_n}\right) \le a_n$.

Since $f$ is LSC, the sequence $\left\langle{f \left({x_n}\right)}\right\rangle_{n \mathop \in \N}$ has at least one limit point.

Equivalently $\left\langle{f \left({x_n}\right)}\right\rangle_{n \mathop \in \N}$ has a convergent subsequence.

Let $\left\langle{f \left({x_{n_k} }\right)}\right\rangle_{k \mathop \in \N}$ be a subsequence of $\left\langle{f \left({x_n}\right)}\right\rangle_{n \mathop \in \N}$ so that:

- $f \left({x_{n_k} }\right) \to \beta$

Then $\beta \le a$.

We have that:

- $\displaystyle \liminf_{t \mathop \to x} f \left({t}\right) = \min \left\{{a \in \overline \R, \exists \left\langle{x_n}\right\rangle_{n \mathop \in \N} \text{ such that } x_n \to x}\right\}$

Hence follows that:

- $\displaystyle \liminf_{t \mathop \to x} f \left({t}\right) \le \beta$

Therefore:

- $\displaystyle f \left({x}\right) = \liminf_{t \mathop \to x} f \left({t}\right) \le a$

and thus:

- $\left({x, a}\right) \in \operatorname{epi} \left({f}\right)$

$\Box$

### Closed Epigraph implies Closed Level Sets

Let $\operatorname{epi} \left({f}\right)$ be a closed set in $S \times \R$.

Let $a \in\R$.

Then the $\alpha$-lower level set:

- $\displaystyle \operatorname{lev} \limits_{\mathop \le a} = \operatorname{epi} f \cap S \times \left\{{a}\right\}$

is closed in $S$ because:

- closedness is preserved under Intersection|intersection

and:

- $S \times \left\{{a}\right\}$ is a closed set with respect with the product topology of $S \times \R$.

$\Box$

### Closed Level Sets implies LSC

Let all lower level sets of $f$ be closed.

Let $x \in S$.

Let $\displaystyle a: = \liminf_{t \mathop \to x} f \left({t}\right)$.

In order to prove that $f$ is LSC it suffices to show that $a = f \left({x}\right)$.

We know already that $a \le f \left({x}\right)$.

Thus it suffices to show that $f \left({x}\right) \le a$.

Let $a = \infty$.

As $\infty \le \infty$ the proof is complete.

Let $a < \infty$.

Then we can find a sequence $\left\langle{x_n}\right\rangle_{n \mathop \in \N}$ such that:

- $x_n \to x$ and $f \left({x_n}\right) \to a$

as $n \to \infty$.

For any $b > a$:

- $f \left({x_n}\right) \le b$

or equivalently:

- $x_n \in \displaystyle \operatorname{lev} \limits_{\mathop \le b} f$

which by hypothesis is closed.

Therefore $\displaystyle \operatorname{lev} \limits_{\mathop \le b} f$ contains the limit point $x$.

We have that for all $b > a$:

- $f \left({x}\right) \le b$

Therefore:

- $f \left({x}\right) \le a$

Hence the result.

$\Box$

So:

- $(1) \implies (2) \implies (3)$

and the proof is complete.

$\blacksquare$