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Added some linfty.
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Bokuan Li
2026-08-18 21:47:50 -04:00
parent e1fd4219a5
commit f65e49fccf
4 changed files with 117 additions and 66 deletions
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30
src/fa/lp/duality.tex
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@@ -178,7 +178,7 @@ The typical argument for $L^p$ duality requires using the Radon-Nikodym theorem
Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $H$ be a Hilbert space over $K$, $p, q \in [1, \infty]$ be Hölder conjugates such that one of the following holds:
\begin{enumerate}[label=(\alph*)]
\item $p \in (1, \infty)$ and $q \in (1, \infty)$.
\item $p = 1$, $q = \infty$, and $\mu$ is $\sigma$-finite\footnote{This should become localisable, under the additional hypothesis that $H$ is separable. }.
\item $p = 1$, $q = \infty$, $H$ is separable, and $\mu$ is localisable.
\end{enumerate}
For each $g \in L^q(X, \cm, \mu; H)$, let
@@ -203,25 +203,27 @@ The typical argument for $L^p$ duality requires using the Radon-Nikodym theorem
By \autoref{theorem:lp-dual-function}, $g \in L^q(X; H)$.
(Arbitrary): In the case of (a), by \autoref{lemma:lp-functional-support}, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^p(X; H)$, $\dpn{f, \phi}{L^p(X; H)} = \dpn{\one_A \cdot f, \phi}{L^p(X; H)}$. In the case of (b), $A = X$ is a $\sigma$-finite set satisfying the same restriction condition.
(Arbitrary): In the case of (a), by \autoref{lemma:lp-functional-support}, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^p(X; H)$, $\dpn{f, \phi}{L^p(X; H)} = \dpn{\one_A \cdot f, \phi}{L^p(X; H)}$. In the case of (b), $A = X$ is a localisable set satisfying the same restriction condition.
Let $\seq{A_n} \subset \cm$ such that $\mu(A_n) < \infty$ for all $n \in \natp$, and $A = \bigsqcup_{n \in \natp}A_n$. By the finite case, there exists $\seq{g_n} \subset L^q(X; H)$ such that for each $n \in \natp$ and $f \in L^p(X; H)$,
Let $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$. By the finite case, there exists $g_F \in L^q(F; H)$ such that for every $f \in L^p(X; H)$,
\[
\int \dpn{f, g_n}{H} d\mu = \dpn{\one_{A_n} \cdot f, \phi}{L^p(X; H)}
\]
Let $g = \sum_{n = 1}^\infty g_n$. If $q < \infty$, then $g \in L^q(X; H)$ by the \hyperref[Monotone Convergence Theorem]{theorem:mct}. Otherwise,
\[
\norm{g}_{L^\infty(X; H)} \le \sup_{n \in \natp}\norm{g_n}_{L^\infty(X; H)} \le \norm{\phi}_{L^1(X; H)^*}
\int_F \dpn{f, g_F}{H} d\mu = \dpn{\one_{F} \cdot f, \phi}{L^p(X; H)}
\]
In the case of (a), there exists a countable exhaustion $\seq{F_n} \subset \cm$ of $A$ with sets of finite measure. For each $n \in \natp$, a representative of $g_{F_n}$ may be taken to have separable range. In the case of (b), such a representative may be chosen for every $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$. Thus by the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists a measurable function $g: X \to H$ such that $g|_{F} = g_F$ almost everywhere for all $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$.
If $q < \infty$, then $g \in L^q(X; H)$ by the \hyperref[Monotone Convergence Theorem]{theorem:mct}. Otherwise,
\[
\norm{g}_{L^\infty(X; H)} \le \sup_{\substack{F \in \cm \\ F \subset A \\ \mu(F) < \infty}}\norm{g_F}_{L^\infty(F; H)} \le \norm{\phi}_{L^1(X; H)^*}
\]
Hence $g \in L^q(X; H)$ with $\norm{g}_{L^q(X; H)} \le \norm{\phi}_{L^1(X; H)^*}$.
For every $f \in L^p(X; H)$,
Finally, let $f \in L^p(X; H)$, then there exists $\seq{F_n} \subset \cm$ such that $F_n \upto \bracsn{f \ne 0} \cap A$ and $\mu(F_n) < \infty$ for all $n \in \natp$. In which case, by the \hyperref[Dominated Convergence Theorem]{theorem:dct},
\begin{align*}
\int \dpn{f, g}{H} d\mu &= \sum_{n = 1}^\infty \int \dpn{f, g_n}{H} d\mu = \sum_{n = 1}^\infty \dpn{\one_{A_n} \cdot f, \phi}{L^p(X; H)} \\
&= \dpn{f, \phi}{L^p(X; H)}
\int \dpn{f, g}{H} d\mu &= \limv{n}\int_{F_n} \dpn{f, g}{H} d\mu = \limv{n}\int_{F_n} \dpn{f, g_{F_n}}{H} d\mu \\
&= \limv{n}\dpn{\one_{F_n} \cdot f, \phi}{L^p(X; H)} = \limv{n}\dpn{f, \phi}{L^p(X; H)}
\end{align*}
by the \hyperref[Dominated Convergence Theorem]{theorem:dct}.
Therefore the mapping is surjective, and hence an isomorphism.
\end{proof}

40
src/measure/measure/localisable.tex
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@@ -198,43 +198,3 @@
\item[(U)] For all $A \in \cf$, $f|_A = g|_A$ almost everywhere. Since $\cf$ is a scaffold for $\mu$, $f = g$ almost everywhere.
\end{enumerate}
\end{proof}
\begin{corollary}
\label{corollary:l-infty-dedekind-complete}
Let $(X, \cm, \mu)$ be a localisable measure space, then $L^\infty(X; \real)$ is order complete.
\end{corollary}
\begin{proof}
Let $\seqi{f} \subset L^\infty(X; \real)$ and $M \in \real$ such that $f_i \le M$ almost everywhere for all $i \in I$.
Fix $A \in \cm$ with $\mu(A) < \infty$, and let
\[
\mathcal{S}_A = \bracs{g \in L^\infty(A; \real)| f_i|_A \le g \text{ almost everywhere }\forall i \in I}
\]
then since $f_i \le M$ almost everywhere for all $i \in I$, $\mathcal{S}_A \ne \emptyset$, and $m_A = \inf_{g \in \mathcal{S}_A}\int g d\mu \in \real$.
Let $\seq{g_{A, n}} \subset \mathcal{S}_A$ such that $\seq{g_{A, n}}$ is decreasing pointwise and $\limv{n}\int_A g_{A, n} d\mu \downto m_A$. Take $g_A = \limv{n}g_{A, n}$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $\int g_A d\mu = m_A$.
For each $i \in I$, since $g_{A, n} \ge f_i|_A$ almost everywhere for all $n \in \natp$, $g_A \ge f_i|_A$ almost everywhere as well. Thus $g_A \in \mathcal{S}_A$. For any $h \in \mathcal{S}_A$, $g_A \wedge h \in \mathcal{S}_A$ with
\[
m_A \le \int_A g_A \wedge h d\mu \le \int_A g_A d\mu = m_A
\]
Thus $g_A \wedge h = g_A$ almost everywhere, so $g_A \le h$ almost everywhere, and $g_A$ is an essential supremum of $\bracsn{f_i|_A}_{i \in I}$.
Now, let $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$, then $\one_{A \cap B} g_B + \one_{A \setminus B}M \in \mathcal{S}_A$, and
\[
m_A \le \int_A g_A \wedge (\one_{A \cap B} g_B + \one_{A \setminus B}M)d\mu \le \int_A g_A d\mu = m_A
\]
Thus $g_A \wedge (\one_{A \cap B} g_B + \one_{A \setminus B}M) = g_A$ almost everywhere, so $g_A|_{A \cap B} \le g_B|_{A \cap B}$ almost everywhere. As the argument is symmetric, $g_A|_{A \cap B} = g_B|_{A \cap B}$ almost everywhere.
By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists a measurable function $g:X \to \real$ such that $g|_A = g_A$ for all $A \in \cm$ with $\mu(A) < \infty$.
Let $h \in L^\infty(X; \real)$ with $h \ge f_i$ almost everywhere for all $i \in I$, then for any $A \in \cm$ with $\mu(A) < \infty$,
\[
\mu(\bracs{h < g} \cap A) \le \mu(\bracs{h|_A < g_A} \cup \bracs{g|_A \ne g_A}) = 0
\]
As $\mu$ is semifinite, $\mu(\bracs{h < g}) = 0$. Finally, since $g_A \le M$ almost everywhere for all $A \in \cm$ with $\mu(A) < \infty$, $g \le M$ almost everywhere. Therefore $g \in L^\infty(X; \real)$ is indeed the essential supremum of $\seqi{f}$.
\end{proof}

8
src/op/c-star/gelfand.tex
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@@ -65,13 +65,5 @@
By \autoref{theorem:gelfand-naimark}, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^*$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the \hyperref[Stone-Nakano Theorem]{theorem:stone-nakano-extremely-disconnected} implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.
\end{proof}
\begin{corollary}
\label{corollary:linfinity-extremely-disconnected}
Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected.
\end{corollary}
\begin{proof}
By \autoref{corollary:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremely disconnected.
\end{proof}

105
src/op/example/bb.tex
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@@ -19,9 +19,9 @@
\item $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^*$.
\item There exists a unique weak*-continuous involution on $\mathscr{M}^*$ such that $J(f^*) = J(f)^*$ for all $f \in C(X; \complex)$, given by
\[
\dpn{\mu, \phi^*}{C(X; \complex)^*} = \ol{\dpn{\mu, \phi}{C(X; \complex)^*}}
\dpn{\mu, \phi^*}{\mathscr{M}^*} = \ol{\dpn{\mu, \phi}{\mathscr{M}^*}}
\]
\item There exists a unique seperately weak*-continuous bilinear map on $\mathscr{M}^*$ such that $J(fg) = J(f)J(g)$ for all $f, g \in C(X; \complex)$.
\item There exists a unique separately weak*-continuous bilinear map on $\mathscr{M}^*$ such that $J(fg) = J(f)J(g)$ for all $f, g \in C(X; \complex)$.
\item $\mathscr{M}^{*}$ equipped with the above involution and product is a commutative unital $C^*$-algebra.
\end{enumerate}
\end{proposition}
@@ -40,7 +40,7 @@
(2): For each $g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$, let $g^* = \ol g$. For any $\mu \in [l^1(I); L^1(\mu_i; \complex)]$,
\[
\dpn{\mu, g^*}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{f, \ol g}{[l^1(I); L^1(\mu_i; \complex)]} = \ol{\dpn{f, g}{[l^1(I); L^1(\mu_i; \complex)]}}
\dpn{\mu, g^*}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{\mu, \ol g}{[l^1(I); L^1(\mu_i; \complex)]} = \ol{\dpn{\mu, g}{[l^1(I); L^1(\mu_i; \complex)]}}
\]
so the conjugation map is weak*-continuous.
@@ -53,4 +53,101 @@
so the composition map is separately weak*-continuous.
(4): $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is a commutative unital $C^*$-algebra.
\end{proof}
\end{proof}
\begin{proposition}
\label{proposition:linfty-von-neumann-algebra}
Let $(X, \cm, \mu)$ be a localisable measure space, and let $L^\infty(X; \complex)$ act on $L^2(X; \complex)$ by multiplication, then:
\begin{enumerate}
\item The weak* topology on $L^\infty(X; \complex)$ is equal to the weak operator topology of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$.
\item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra.
\end{enumerate}
\end{proposition}
\begin{proof}
(2): Let $A \subset B(L^2(X; \complex))$ be the von Neumann algebra generated by $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$. By \autoref{theorem:lp-duality}, $L^\infty(X; \complex)$ is the dual of $L^1(X; \complex)$, so (1) and the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu} imply that the closed unit ball of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is weak-operator closed. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_{L^\infty(X; \complex)}(0, 1)} = \ol{B_A(0, 1)}$. Therefore $L^\infty(X; \complex) = A$.
\end{proof}
\begin{theorem}[Order Structure of $L^\infty$]
\label{theorem:l-infty-dedekind-complete}
Let $(X, \cm, \mu)$ be a localisable measure space, then
\begin{enumerate}
\item $L^\infty(X; \real)$ is order complete.
\item For each $\phi \in L^1(X; \real)$ with $\phi \ge 0$ and bounded directed subset $S \subset L^\infty(X; \real)$,
\[
\sup_{f \in S} \dpn{\phi, f}{L^1(X; \real)} = \bigg\langle\phi, \sup_{f \in S}f\bigg\rangle_{L^1(X; \real)}
\]
\end{enumerate}
\end{theorem}
\begin{proof}
By \autoref{proposition:linfty-von-neumann-algebra}, $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra.
(1): Since $L^\infty(X; \real)$ is a lattice, it is order complete by \autoref{theorem:existence-of-projections-vna}.
(2): By \autoref{theorem:existence-of-projections-vna}, $\sup(S) = \sotlim_{f \in S}f = \wotlim_{f \in S}f$. By (1) of \autoref{proposition:linfty-von-neumann-algebra}, $\sup_{f \in S}\dpn{\phi, f}{L^1(X; \real)} = \dpn{\phi, \sup_{f \in S}f}{L^1(X; \real)}$.
\end{proof}
\begin{corollary}
\label{corollary:linfinity-extremely-disconnected}
Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected.
\end{corollary}
\begin{proof}
By \autoref{theorem:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremely disconnected.
\end{proof}
\begin{lemma}
\label{lemma:l-infty-order-isotone}
Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be localisable measure spaces, and $T: L^\infty(X; \complex)\to L^\infty(Y; \complex)$ such that:
\begin{enumerate}[label=(\alph*)]
\item $T$ is an isometric isomorphism.
\item For each $f, g \in L^\infty(X; \complex)$, $f \ge g$ if and only if $Tf \ge Tg$.
\end{enumerate}
then:
\begin{enumerate}
\item $T^*(L^1(Y; \complex)) \subset L^1(X; \complex)$.
\item $T$ is weak*-continuous.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $\phi \in L^1(Y; \complex)$ with $\phi \ge 0$, then $T^*\phi \in L^\infty(X; \complex)^*$. For each $\seq{B_n} \subset \cm$ and $B \in \cm$ with $B_n \upto B$, $\one_B = \sup_{n \in \natp}\one_{B_n}$ as an element of $L^\infty(X; \complex)$. Thus (b) and \autoref{theorem:l-infty-dedekind-complete} imply that
\begin{align*}
\sup_{n \in \natp}\dpn{\one_{B_n}, T^*\phi}{L^\infty(X; \complex)} &= \sup_{n \in \natp} \dpn{\phi, T\one_{B_n}}{L^1(Y; \complex)} = \dpn{\phi, T\one_B}{L^1(Y; \complex)} \\
&= \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}
\end{align*}
Hence the mapping $B \mapsto \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}$ is a finite positive measure on $(X, \cm)$, which is absolutely continuous with respect to $\mu$. By the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(X; \complex)$ with $f \ge 0$ such that $\int_B f d\mu = \dpn{\one_{B}, T^*\phi}{L^\infty(X; \complex)}$ for all $B \in \cm$.
Let $g \in L^\infty(X; [0, 1])$. By \autoref{lemma:separable-metric-space-approx-identity}, there exists simple functions $\seq{g_n} \subset \Sigma(X; [0, 1])$ such that $g_n \upto g$ pointwise. In which case, $g = \sup_{n \in \natp}g_n$ as an element of $L^\infty(X; \real)$, so (b) and \autoref{theorem:l-infty-dedekind-complete} imply that,
\begin{align*}
\dpn{\phi, Tg}{L^1(Y; \complex)} &= \sup_{n \in \natp}\dpn{\phi, Tg_n}{L^1(Y; \complex)} = \sup_{n \in \natp}\dpn{f, g_n}{L^1(X; \complex)} \\
&= \dpn{f, g}{L^1(X; \complex)}
\end{align*}
By linearity, $\dpn{\phi, Tg}{L^1(Y; \complex)} = \dpn{f, g}{L^1(X; \complex)}$ for all $g \in L^\infty(X; \complex)$. Therefore $T^*(L^1(Y; \complex)) \subset L^1(X; \complex)$, and $T$ is weak*-continuous.
\end{proof}
\begin{theorem}[Uniqueness of $L^\infty$]
\label{theorem:linfty-uniqueness}
Let $X$ be a compact Hausdorff space, $\mu, \nu: \cb_X \to [0, \infty)$ be Radon measures on $X$, and $\Phi: L^\infty(\mu; \complex) \to L^\infty(\nu; \complex)$ such that:
\begin{enumerate}[label=(\alph*)]
\item $\Phi$ is a *-isomorphism.
\item $\Phi|_{C(X; \complex)}$ is the identity.
\end{enumerate}
then:
\begin{enumerate}
\item $\mu$ and $\nu$ are equivalent.
\item $L^\infty(\mu; \complex) = L^\infty(\nu; \complex)$.
\item $\Phi$ is the identity map.
\end{enumerate}
\end{theorem}
\begin{proof}[Proof, {{\cite[Theorem 21.4]{Zhu}}}. ]
As $\Phi$ is a *-isomorphism, $\Phi f \ge \Phi g$ if and only if $f \ge g$ for any $f, g \in L^\infty(\mu; \complex)$.
(1): By \autoref{lemma:l-infty-order-isotone}, there exists $f \in L^1(\mu; \complex)$ such that $\int f g d\mu = \int g d\nu$ for all $g \in C(X; \complex)$. Thus the uniqueness of the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0} implies that $\nu = f d\mu$. As the argument is symmetric, the two measures are equivalent.
(3): By \autoref{lemma:l-infty-order-isotone}, $\Phi$ is also weak*-continuous. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex)$ is weak*-dense in $L^\infty(\mu; \complex)$ and $L^\infty(\nu; \complex)$. As $\Phi$ is the identity on $C(X; \complex)$, $\Phi$ is the identity on $L^\infty(\mu; \complex)$.
\end{proof}