# Characterization of Lower Semicontinuity

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

Consequently:

\(\ds \map f {\bar x}\) | \(=\) | \(\ds \liminf_{x \mathop \to \bar x} \map f x\) | 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 $a_n \to \bar a$ |

Thus:

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

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The validity of the material on this page is questionable.In particular: This only proves sequential closedness.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Questionable}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

$\Box$

### 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 $f$.

Then:

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

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is closed in $S$ because:

- closedness is preserved under Intersection|intersection

and:

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

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 $b > a$:

- $\map f x \le b$

Therefore:

- $\map f x \le a$

Hence the result.

$\Box$

So:

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

and the proof is complete.

$\blacksquare$

## Sources

- 1986: Viorel Barbu and Th. Precupanu:
*Convexity and Optimization in Banach spaces*: $\S 2$: Proposition $1.3$ - 1990: Ioana Cioranescu:
*Geometry of Banach Spaces, Duality Mappings and Nonlinear Problems*: Proposition $1.7$