Characterization of Lower Semicontinuity

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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:

$\left({x_n, a_n}\right) \to \left({\bar x, \bar a}\right) \in S \times \R$

as $n \to \infty$.

This implies that $x_n \to \bar x$ and $a_n \to \bar a$ by definition of the product topology on $S \times \R$ and because a Continuous Mapping is Sequentially Continuous.

Consequently

\(\displaystyle f \left({\bar x}\right)\) \(=\) \(\displaystyle \liminf_{x \mathop \to \bar x} f \left({x}\right)\)          Definition of LSC          
\(\displaystyle \) \(\le\) \(\displaystyle \liminf_{n \mathop \to \infty} f \left({x_n}\right)\)          Relationship between Limit Inferior and Lower Limit.          
\(\displaystyle \) \(\le\) \(\displaystyle \liminf_{n \mathop \to \infty} a_n\)          because $\left({x_n, a_n}\right) \in \operatorname{epi} \left({f}\right)$          
\(\displaystyle \) \(=\) \(\displaystyle \bar a\)          because $a_n \to \bar a$          

Thus:

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

$\Box$


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.

$\blacksquare$


Sources