Characterization of Lower Semicontinuity

From ProofWiki
Jump to navigation Jump to search

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

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 $\sequence{\tuple{x_n, a_n}}_{n \mathop \in \N} \in \map {\operatorname{epi}} f$ such that:

$\tuple{x_n, a_n} \to \tuple{\bar x, \bar a} \in S \times \R$

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:

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

Thus:

$\tuple{\bar x, \bar a} \in \map {\operatorname{epi}} f$

$\Box$


Closed Epigraph implies Closed Level Sets

Let $\map {\operatorname{epi}} f$ be a closed set in $S \times \R$.

Let $a \in \R$.

Then the $\alpha$-lower level set:

$\displaystyle \operatorname{lev} \limits_{\mathop \le a} = \map {\operatorname{epi}} f \cap S \times \set{a}$

is closed in $S$ because:

closedness is preserved under Intersection|intersection

and:

$S \times \set{a}$ 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 $\displaystyle a: = \liminf_{t \mathop \to x} \map f t$.

In order to prove that $f$ is LSC it suffices to show that $a = \map f x$.

We know already that $a \le \map f x$.

Thus it suffices to show that $\map f x \le a$.


Let $a = \infty$.

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


Let $a < \infty$.

Then we can find a sequence $\sequence{x_n}_{n \mathop \in \N}$ such that:

$x_n \to x$ and $\map f {x_n} \to a$

as $n \to \infty$.

For any $b > a$:

$\map f {x_n} \le b$

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 $b > a$:

$\map f x \le b$

Therefore:

$\map f x \le a$

Hence the result.

$\Box$


So:

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

and the proof is complete.

$\blacksquare$


Sources