Characterization of Lower Semicontinuity

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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 $S$.


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 $\map {\operatorname {epi} } f$ 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.


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


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


Closed Epigraph implies Closed Level Sets

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

Let $\alpha \in \R$.

Let $\operatorname {lev} \limits_{\mathop \le \alpha} f$ denote the $\alpha$-lower level set of $f$.


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

is closed in $S$ because:

closedness is preserved under Intersection|intersection


$S \times \set \alpha$ is a closed set with respect with the product topology of $S \times \R$.


Closed Level Sets implies LSC

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

Let $x \in S$.

Let $\ds a: = \liminf_{t \mathop \to x} \map f t$.

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:

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

which by hypothesis is closed.

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

We have that for all $b > a$:

$\map f x \le b$


$\map f x \le a$

Hence the result.



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

and the proof is complete.