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. $f$ is lower semicontinuous on $S$.
  2. The epigraph of $f$ is a closed set in $S \times \R$ with the product topology.
  3. 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

Assume that $f:S\to\R\cup\left\{-\infty,\infty\right\}$ is lower semicontinuous.

Take a sequence $\left\langle \left(x_n,a_n \right) \right\rangle_{n\in\N}\in\operatorname{epi}f$ so that $\left(x_n,a_n \right)\to\left(x,a\right)$.

This implies that $x_n\to x$ and $a_n\to a$ while $f(x_n)\leq a_n$.

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

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

Let $\left\langle f\left(x_{n_k}\right)\right\rangle_{k\in\N}$ be a subsequence of $\left\langle f\left(x_n\right)\right\rangle_{n\in\N}$ so that $f\left(x_{n_k}\right)\to \beta$. Then $\beta \leq a$ and since:

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

it follows that $\liminf_{t\to x} f\left(t\right)\leq\beta$.

Therefore $f\left(x\right)=\liminf_{t\to x} f\left(t\right)\leq a$ thus $\left(x,a\right)\in\operatorname{epi}f$.

$\Box$


Closed Epigraph implies Closed Level Sets

Let us assume that $\operatorname{epi}f$ is a closed set in $S\times\R$ and $a\in\R$. Then the set

$\operatorname{lev}_{\leq a}=\operatorname{epi}f\cap S\times\left\{a\right\}$

is closed in $S$ because closeness is preserved under 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

Assume that all level sets of $f$ are closed. Let $x\in S$ and set $a=\liminf_{t\to x}f\left( t \right)$.

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

We know already that $a\leq f\left( x \right)$; thus it suffices to show that $f\left( x \right) \leq a$.

The proof is trivial if $a=\infty$ since $\infty\leq \infty$. Assume that $a<\infty$. Then we can find a

sequence $\left\langle x_n \right\rangle_{n\in\N}$ so that $x_n \to x$ and $f\left( x_n \right) \to a$.

For any $b > a$ we see that $f\left( x_n \right) \leq b$, or equivalently $x_n \in \operatorname{lev}_{\leq b}f$ which is closed,

therefore it contains the limit point $x$. We have that for all $b > a$, $f\left( x \right) \leq b$, therefore $f\left( x \right) \leq a$.

$\Box$

This completes the proof since we have shown that $1.\implies 2. \implies 3.$

$\blacksquare$