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Semicontinuous extreme value theorem on compact Euclidean sets

Statement

Let $n \geq 1$ and let $K \subseteq R^{n}$ be nonempty and compact. Every lower semicontinuous function $f : K \to R$ is bounded below and attains a minimum. Dually, every upper semicontinuous function $f : K \to R$ is bounded above and attains a maximum.

Facts & Assumptions

Given: A nonempty compact Euclidean set $K$ and a lower semicontinuous function $f : K \to R$ , with infima as in Greatest lower bound (infimum).

[L1]

Lower semicontinuity is equivalent to relative openness of every strict superlevel set and to relative closedness of every weak sublevel set (Semicontinuity on $R^{n}$ is characterized by strict open level sets and weak closed level sets).

[L2]

In a compact space, every family of closed sets with the finite intersection property has nonempty intersection (A metric space is compact if and only if every family of closed subsets with the finite intersection property has nonempty intersection).

[L3]

Every nonempty real set bounded below has a greatest lower bound in $R$ (Every nonempty set bounded below has an infimum).

[L4]

If $m$ is the infimum of a nonempty real set, then for every $ε > 0$ the set has an element $s < m + ε$ (Epsilon characterisation of the infimum).

[L5]

For every real $x$ there is a natural number $n \geq 1$ with $x < n$ (Every complete ordered field is Archimedean).

[L6]

For every real $ε > 0$ there is a natural number $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L7]

A metric space is compact when every open cover has a finite subcover (Open cover, subcover, compact metric space, and compact subset of a metric space).

[L8]

Every nonempty finite set of real numbers has a maximum and a minimum (Every nonempty finite set of reals has a maximum and a minimum).

Proof

technique · direct

1.1

For positive naturals $N$ , the relatively open sets $U_{N} = {x \in K : f (x) > - N}$ cover $K$ : apply [L5] to $- f (x)$ . By [L7], finitely many $U_{N}$ cover, and [L8] gives a largest index $M$ among them. Since the sets are nested, $U_{M} = K$ , so $f > - M$ on $K$ and $f [K]$ is bounded below.

L1L5L7L8givenalgebra

2.1

By [L3], $m = \inf f [K]$ exists. For every positive natural $N$ , the set $F_{N} = K \cap {f \leq m + 1 / N}$ is closed by [L1] and nonempty by [L4]. The family is nested, so every finite subfamily has nonempty intersection.

step 1.1L1L3L4algebra

3.1

By [L2], choose $x_{*} \in ⋂_{N} F_{N}$ . If $f (x_{*}) > m$ , [L6] gives $N \geq 1$ with $1 / N < f (x_{*}) - m$ , contradicting $x_{*} \in F_{N}$ . Thus $f (x_{*}) = m$ . Applying the same argument to $- f$ proves that an upper semicontinuous function attains its maximum.

step 2.1L2L6choosealgebra∎

Depends on

Used by

A lower semicontinuous function on [0,1] with no maximum Counterexample

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

D. Bertsekas, MIT 6.253 Convex Analysis and Optimization, Lectures 5 and 6 (standard reference, not scraped)