바나흐-알라오글루 정리

$X{\displaystyle \mathbb{}}$ ${\displaystyle X}$의 기본 필드${\displaystyle (}$K $,${\$displaystyle \mathb$ {$K}),$ 즉$$ 실제 $숫자{\displaystyle \mathbb{}}$ R$${\$displaystyle \mathb$ {$R}$ 또는 복잡한 $숫자{\displaystyle \mathbb{}}$ C${\displaystyle .}${\$displaystyle \mathb {C}).$ 이 증명은$$ 문서에 나열된 기본 속성 중 일부를 사용한다. 극성 세트, 이중 시스템m 및 연속 선형 연산자.

증명을 시작하기 위해 일부 정의와 쉽게 검증된 결과를 회수한다.When ${\displaystyle X^{\#}}$ is endowed with the weak-* topology ${\displaystyle \sigma \left(X^{\#},X\right),}$ then this Hausdorff locally convex topological vector space is denoted by ${\displaystyle \left(X^{\#},\sigma \left(X^{\#},X\right)\right).$$}$ The space ${\displaystyle \left(X^{\#},\sigma \left(X^{\#},X\right)\right)}$ is always a complete TVS; however, ${\displaystyle \left(X^{\prime },\sigma \left(X^{\prime },X\right)\right)}$ may fail to be a complete space, which is the reason why this proof는 공간$({\displaystyle X{}^{\mathrm{}}}$#,${\displaystyle {}^{}\sigma }$ ${\displaystyle (X{\mathrm{}}^{}}$ X${\displaystyle }$)을 포함한다$.$ {\$displaystyle \left($X^{\#},\$sigma \left($X^{\#},$X\$right)\$right).$$}$ 구체적으로$$ 이 증거는 완전한 하우스도르프 공간의 부분집합이 닫히고 완전히 경계된 경우에만 압축된다는 사실을 사용할 것이다.Importantly, the subspace topology that ${\displaystyle X^{\prime }}$ inherits from ${\displaystyle \left(X^{\#},\sigma \left(X^{\#},X\right)\right)}$ is equal to ${\displaystyle \sigma \left(X^{\prime },X\right).$$}}$ 이는$$ $주어진{\displaystyle }$ f ${\displaystyle \in }$ ${\displaystyle {}^{}}$ ${\displaystyle {\prime \mathrm{}}^{}}$${\displaystyle {}^{}}$, ${\displaystyle f\in$ X$^{\prime$ $}}}$에 있는$$ net이 다른 위상의$$ $f{\displaystyle }$ ${\displaystyle$ X$^{\prime$}에 수렴되는 경우에만 위상 중 하나에서$$ $f{\displaystyle }$ ${\displaystystyle$ $f}$에 수렴된다는$$ 것을 쉽게 확인할 수 있다(결론).두 가지 토폴로지는 정확히 동일한 수렴 그물을 가지고 있는 경우에만 동일하다.

트리플 $⟨{\displaystyle }$ ${\displaystyle {}^{}}$, X ${\displaystyle {}^{}}${ {\$displaystyle \left\langle$ X$,X^{\prime$$prime \langle X,X^{\#}\right\angle }}$은$({\displaystyle }$는) ${\displaystyle 일반적}$으로 이중$$ 시스템임을 보장하지는 않지만 이중 페어링이다${\displaystyle .}$전체적으로, 달리 명시되지 않는 한, 모든 극세트는 표준 쌍인${\displaystyle pairing}$ X${\displaystyle ,}$ ${\displaystyle {}^{}}$ ${\displaystyle {\prime \mathrm{}}^{}{}^{}}$ ${\displaystyle \left\langle X,$ X$^{\prime }\rigle.}$에 대해 취해진다.

$U{\displaystyle }$ ${\displaystyle U}$을(를) $X{\displaystyle }$ ${\displaystyle X}$의 원점 부근으로 설정하고$$ 다음을$$ 수행하십시오.

• ${\displaystyle U^{\circ }=\left\{f\in X^{\prime }~:~\sup _{u\in U} f(u) \leq 1\right\}}$ be the polar of ${\displaystyle U}$ with respect to the canonical pairing ${\displaystyle \left\langle X,X^{\prime }\right\rangle }$;
• ${\displaystyle U^{\circ \circ }=\left\{x\in X~:~\sup _{f\in U^{\circ }} f(x) \leq 1\right\}}$ be the bipolar of ${\displaystyle U}$ with respect to ${\displaystyle \left\langle X,X^{\prime }\right\rangle }$;
• ${\displaystyle U^{\#}=\left\{f\in X^{\#}~:~\sup _{u\in U} f(u) \leq 1\right\}}$ be the polar of ${\displaystyle U}$ with respect to the canonical dual system ${\displaystyle \left\langle X,X^{\#}\right\rangle .}$ Note that ${\displaystyle U^{\circ }=}$$displaystyle$

극지방 집합에 대해 잘 알려진 사실은 $U{\displaystyle {}^{}}$ ${\displaystyle {\circ }^{}}$ ${\displaystyle {\circ }^{}}$u ${\displaystyle {}^{}}$ U ${\displaystyle {}^{}}$ ${\displaystyle U^{\circle \circle$\circle \$circle \}\subseteq$ U$^{\circircle }}}}$이다.

1. Show that ${\displaystyle U^{\#}}$ is a ${\displaystyle \sigma \left(X^{\#},X\right)}$-closed subset of ${\displaystyle X^{\#}:}$ Let ${\displaystyle f\in X^{\#}}$ and suppose that ${\displaystyle f_{\bullet }=\left(f_{i}\right)_{i\in$$I}}$은$({\displaystyle }$는) $U{\displaystyle {}^{\mathrm{}}}$# ${\$ U$^{\#}$의${\displaystyle }$$f{\displaystyle }$ ${\displaystyle f$${\displaystyle \}}$에 수렴되는$$ ${\displaystyle 그물}$로서 (${\displaystyle X{}^{}{}^{}}$ $(X^{\#},$)의$${\$displaystystyle \}에 수렴한다.$$}$ To conclude that ${\displaystyle f\in U^{\#},}$ it is sufficient (and necessary) to show that ${\displaystyle f(u) \leq 1}$ for every ${\displaystyle u\in U.}$ Because ${\displaystyle f_{i}(u)\to f(u)}$ in the scalar field ${\displaystyle \mathbb$${K} }$ and every value ${\displaystyle f_{i}(u)}$ belongs to the closed (in ${\displaystyle \mathbb {K} }$) subset ${\displaystyle \left\{s\in \mathbb {K} : s \leq 1\right\},}$ so too must this net's limit ${\displaystyle f(u)}$ belong to this set.따라서 ${\displaystyle f}$ (${\displaystyle u}$ ) ${\displaystyle \mathrm{\le }}$1. ${\displaystyle f(u) \leq 1.}$
2. $U{\displaystyle {}^{\mathrm{}}}$#${\displaystyle {}^{}}$= ${\displaystyle {}^{}}$ ${\displaystyle {\circ }^{}}$ ${\$ U$^{\#}=$ 표시$U^{\circ }}$ and then conclude that ${\displaystyle U^{\circ }}$ is a closed subset of both ${\displaystyle \left(X^{\#},\sigma \left(X^{\#},X\right)\right)}$ and ${\displaystyle \left(X^{\prime },\sigma \left(X^{\prime },X\right)\right):}$포함된 $U{\displaystyle {}^{}}$ ${\displaystyle {}^{}}$ ${\displaystyle {}^{\mathrm{}}}$# ${\$ U$^{\#}$ 모든 연속 선형 기능은 (특히) 선형 기능이기 때문에 유지된다$$.For the reverse inclusion ${\displaystyle \,U^{\#}\subseteq U^{\circ },\,}$ let ${\displaystyle f\in U^{\#}}$ so that ${\displaystyle \;\sup _{u\in U} f(u) \leq 1,\,}$ which states exactly that the linear functional ${\displaystyle f}$ is bounded on t그는 이웃 $U{\displaystyle }$ ${\displaystyle$ U$\};$ 따라서 $f{\displaystyle }$ ${\displaystyle f}$은($즉{\displaystyle }$, f${\displaystyle {}^{}}$ X ${\displaystyle {\prime \mathrm{}}^{}}$ ${\$ X$^{\prime$ $\})$ 연속적인 선형 기능이며, ${\displaystyle {}^{}}$ f${\displaystyle {}^{}}$ U ${\displaystyle {\circ }^{}}$ ,$${\$displaystystysty$ f$\in$ U$^{\circir}}}}.$Using (1) and the fact that the intersection ${\displaystyle U^{\#}\cap X^{\prime }=U^{\circ }\cap X^{\prime }=U^{\circ }}$ is closed in the subspace topology on ${\displaystyle X^{\prime },}$ the claim about ${\displaystyle U^{\circ }}$ being closed follows.
3. Show that ${\displaystyle U^{\circ }}$ is a ${\displaystyle \sigma \left(X^{\prime },X\right)}$-totally bounded subset of ${\displaystyle X^{\prime }:}$ By the bipolar theorem, ${\displaystyle U\subseteq U^{\circ \circ }}$ where because the neighborhood ${\displays$$tyle U}$ is an absorbing subset of ${\displaystyle X,}$ the same must be true of the set ${\displaystyle U^{\circ \circ };}$ it is possible to prove that this implies that ${\displaystyle U^{\circ }}$ is a ${\displaystyle \sigma \left(X^{\prime },X\right)}$-bounded subset of $$${\displaystyle X^{\prime }.}$ Because ${\displaystyle X}$ distinguishes points of ${\displaystyle X^{\prime },}$ a subset of ${\displaystyle X^{\prime }}$ is ${\displaystyle \sigma \left(X^{\prime },X\right)}$-bounded if and only if it is ${\d$$isplaystyle \sigma \left(X^{\premy },X\right)}$ -totally$$ 경계.그래서 특히 $∘{\$ U$^{\circle }}$도 $\sigma {\displaystyle }$(${\displaystyle X{}^{}}$ ${\displaystyle {\prime \mathrm{}}^{}}$,${\displaystyle X}$ )$${\$displaystyle \sigma \left(X^{\prime }}}X\right)}$이 완전 경계로 되어 있다$$.
4. Conclude that ${\displaystyle U^{\circ }}$ is also a ${\displaystyle \sigma \left(X^{\#},X\right)}$-totally bounded subset of ${\displaystyle X^{\#}:}$ Recall that the ${\displaystyle \sigma \left(X^{\prime },X\right)}$ topology on ${\displaystyle X^{$$\prime$ $}}}{\$ ${\displaystyle \}\}\}}$이(가) $상속$하는 서브스페이스 토폴로지${\displaystyle (X}$#, { ($X^{\#,$ ${\displaystyle X}$) .${\displaystyle \reft;\sigma \left(X^{#, 오른쪽)\$rig)\rig)\rig)\right$)$와 동일하다.$}}$ 이 사실은$$ (3)과 "완전 경계"의 정의와 함께 $U{\displaystyle }$ ${\displaystyle {\circ }^{}}$ ${\displaystyle {\{\backslash \mathrm{}}^{}}$$displaystyle$ U${\displaystyle {}^{}\mathrm{circle}}$$$$}}{\$ -완전$$ 경계된 $X{\displaystyle {}^{\mathrm{}}}$#${\displaystyle {}^{}}$. ${\displaystysty$ X$^{\}$의 하위 집합임을 암시한다.$}$
5. Finally, deduce that ${\displaystyle U^{\circ }}$ is a ${\displaystyle \sigma \left(X^{\prime },X\right)}$-compact subset of ${\displaystyle X^{\prime }:}$ Because ${\displaystyle \left(X^{\#},\sigma \left(X^{\#},X\right)\right)}$ is a complete TVS and ${\displaystyle U^{\circ }}$ is a closed (by (2)) and totally bounded (by (4)) subset of ${\displaystyle \left(X^{\#},\sigma \left(X^{\#},X\right)\right),}$ it follows that ${\displaystyle U^{\circ }}$ is compact.${\displaystyle \blacksquare }$

Assume that ${\displaystyle X}$ is a topological vector space with continuous dual space ${\displaystyle X^{\prime }}$ and that ${\displaystyle U}$ is a neighborhood of the origin. Because ${\displaystyle U}$ is a neighborhood of the origin in ${\displaystyle X,}$ it is also an absorbing subset of ${\displaystyle X,}$ so for every ${\displaystyle x\in X,}$ there exists a real number ${\displaystyle r_{x}>0}$ such that ${\displaystyle x\in r_{x}U.}$ Thus the hypotheses of the above proposition are satisfied, and so the set ${\displaystyle U^{\#}}$ is therefore compact in the weak-* topology. The proof of the Banach–Alaoglu theorem will be complete once it is shown that ${\displaystyle U^{\#}=U^{\circ },}$^{[note 2]} where recall that ${\displaystyle U^{\circ }}$ was defined as

$${\displaystyle U^{\circ }~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\Big \{}f\in X^{\prime }~:~\sup _{u\in U} f(u) \leq 1{\Big \}}~=~U^{\#}\cap X^{\prime }.}$$

Proof that ${\displaystyle U^{\circ }=U^{\#}:}$ Because ${\displaystyle U^{\circ }=U^{\#}\cap X^{\prime },}$ the conclusion is equivalent to ${\displaystyle U^{\#}\subseteq X^{\prime }.}$ If ${\displaystyle f\in U^{\#}}$ then ${\displaystyle \;\sup _{u\in U} f(u) \leq 1,\,}$ which states exactly that the linear functional ${\displaystyle f}$ is bounded on the neighborhood ${\displaystyle U;}$ thus ${\displaystyle f}$ is a continuous linear functional (that is, ${\displaystyle f\in X^{\prime }}$), as desired. ${\displaystyle \blacksquare }$