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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$\Box$
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 $1.7$