Added another draft of the Borel functional calculus.

src/op/vn/fc.tex

@@ -23,7 +23,17 @@
     Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, and $E: \cb_X \to B(H)$ be a spectral measure relative to $H$, then:
     \begin{enumerate}
         \item For each $x, y \in H$, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$. 
-        \item Let $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then for any $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{E}$ as well. 
+    \end{enumerate}
+
+    Let $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then
+    \begin{enumerate}[start=1]
+        \item For any $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{E}$ as well. 
+        \item Let 
+        \[
+        J: B^\infty(X; \complex) \to \mathscr{E}^* \quad \dpn{\mu, J(f)}{\mathscr{E}} = \int_X f d\mu
+        \]
+
+        then $\mathscr{E}^*$ admits a unique weak*-continuous involution and a unique separately weak*-continuous product, such that $\mathscr{E}^*$ is a commutative unital $C^*$-algebra, and $J$ is a unital *-homomorphism. 
     \end{enumerate}
 \end{lemma}
 \begin{proof}
@@ -45,6 +55,15 @@
     By linearity, $fdE_{x, y} \in \mathscr{E}$ for all $f \in \Sigma(X; \complex)$. For each $f \in \Sigma(X; \complex)$, the mapping $\mu \mapsto f d\mu$ is continuous in the total variation norm, so $fd\mu \in \mathscr{E}$ for all $\mu \in \mathscr{E}$ and $f \in \Sigma(X; \complex)$. By \autoref{proposition:lp-simple-dense}, $\Sigma(X; \complex)$ is dense in $L^1(\mu; \complex)$ for all $\mu \in \mathscr{E}$. Therefore $fd\mu \in \mathscr{E}$ for all $f \in L^1(\mu; \complex)$ and $\mu \in \mathscr{E}$. 
     
     Finally, let $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, then by the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(\mu; \complex)$ such that $d\nu = f d\mu \in \mathscr{E}$. 
+
+    (3): By (2), for any $\mu \in \mathscr{E}$ and $f \in L^1(\mu; \complex)$, $fd\mu \in \mathscr{E}$ as well. By \autoref{proposition:measures-dual-algebra}, there exists a unique weak*-continuous involution and separately weak*-continuous product on $\mathscr{E}^*$ making $\mathscr{E}^*$ a commutative unital $C^*$-algebra, and $J|_{C(X; \complex)}$ a unital *-homomorphism. Since
+    \begin{enumerate}[label=(\roman*)]
+        \item $J$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-$\sigma(\mathscr{E}^*, \mathscr{E})$ continuous. 
+        \item Conjugation on $B^\infty(X; \complex)$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
+        \item Multiplication on $B^\infty(X; \complex)$ is separately $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
+    \end{enumerate}
+    
+    the mapping $J$ is a unital *-homomorphism.
 \end{proof}
 
 
@@ -73,16 +92,7 @@
     For any $\phi \in \mathscr{E}^*$, $I_E(\phi) = \int_X \phi dE$ is the \textbf{integral} of $\phi$ with respect to $E$. 
 \end{definition}
 \begin{proof}
-    ($C^*$ Structure of $\mathscr{E}^*$): By \autoref{lemma:spectral-measure-properties}, for any $\mu \in \mathscr{E}$ and $f \in L^1(\mu; \complex)$, $fd\mu \in \mathscr{E}$ as well. By \autoref{proposition:measures-dual-algebra}, there exists a unique weak*-continuous involution and separately weak*-continuous product on $\mathscr{E}^*$ making $\mathscr{E}^*$ a commutative unital $C^*$-algebra, and $J|_{C(X; \complex)}$ a unital *-homomorphism. Since
-    \begin{enumerate}[label=(\roman*)]
-        \item $J$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-$\sigma(\mathscr{E}^*, \mathscr{E})$ continuous. 
-        \item Conjugation on $B^\infty(X; \complex)$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
-        \item Multiplication on $B^\infty(X; \complex)$ is separately $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
-    \end{enumerate}
-    
-    the mapping $J$ is a unital *-homomorphism.
-
-    (1): Let $\phi \in \mathscr{E}^*$ and $x, y \in H$, then 
+    (1): Let $\phi \in \mathscr{E}^*$ and $x, y \in H$, then by \autoref{lemma:spectral-measure-properties}, 
     \begin{align*}
     |\dpn{I_E(\phi) \cdot x, y}{H}| &= |\dpn{E_{x, y}, \phi}{\mathscr{E}}| \le \norm{E_{x, y}}_{\mathscr{E}} \cdot \norm{\phi}_{\mathscr{E}^{*}} \\
     &\le \norm{\phi}_{\mathscr{E}^{*}} \cdot \norm{x}_H \cdot \norm{y}_H
@@ -119,20 +129,22 @@
 \label{theorem:spectral-theorem-vn-1}
     Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then:
     \begin{enumerate}
-        \item There exists a unique spectral measure $E: \cb_{\Omega(A)} \to B(H)$ such that
+        \item There exists a unique spectral measure $E: \cb_{\Omega(A)} \to B(H)$ such that\footnote{Omitting the natural map $C(\Omega(A); \complex) \to \mathscr{E}^*$. }
         \[
         T = \int_{\Omega(A)} \Gamma_A T dE \quad \forall T \in A
         \]
-        \item Let $B \subset B(H)$ be the strong-operator closure of $A$ in $B(H)$, then
+        \item Let $B \subset B(H)$ be the strong-operator closure of $A$ in $B(H)$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then
         \[
-        B = \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}}
+        I_E: \mathscr{E}^* \to B \quad \phi \mapsto \int_{\Omega(A)}\phi dE
         \]
+
+        is a *-isomorphism. 
     \end{enumerate}
     
-    The mapping $C(\Omega(A); \complex)^{**} \to B$ defined by $\phi \mapsto \int \phi dE$ is the \textbf{extended inverse Gelfand transform} of $A$. 
+    The mapping $\mathscr{E}^* \to B$ defined by $\phi \mapsto \int \phi dE$ is the \textbf{extended inverse Gelfand transform} of $A$. 
 \end{theorem}
 \begin{proof}[Proof, {{\cite[Theorem 20.2]{Zhu}}}. ]
-    (1): By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, $\Gamma_A: A \to C(\Omega(A); \complex)$ is a unital *-isomorphism. For each $x, y \in H$, $\Gamma_A^{-1}$ induces a mapping
+    (1): By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, $\Gamma_A: A \to C(\Omega(A); \complex)$ is a *-isomorphism. For each $x, y \in H$, $\Gamma_A^{-1}$ induces a mapping
     \[
     E_{x, y}: C(\Omega(A); \complex) \to \complex \quad \dpn{f, E_{x, y}}{C(\Omega(A); \complex)} = \dpn{\Gamma_A^{-1}f \cdot x, y}{H}
     \]
@@ -141,84 +153,77 @@
 
     Since $\norm{E_{x, y}}_{C(\Omega(A); \complex)^*} \le \norm{x}_H\norm{y}_H$ for all $x, y \in H$, $\bracsn{E_{x, y}|x, y \in H}$ induces a bounded linear map
     \[
-    I_E: C(\Omega(A); \complex)^{**} \to B(H) \quad \dpn{I_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*}
+    J_E: B^\infty(\Omega(A); \complex) \to B(H) \quad \dpn{J_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*}
     \]
 
-    with $I_E(f) = \Gamma_A^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$. 
+    with $J_E(f) = \Gamma_A^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$. 
 
     For any $C \in \cb_{\Omega(A)}$, $\one_C$ is a projection in $B^\infty(\Omega(A); \complex)$. So to see that
     \[
     E: \cb_{\Omega(A)} \to B(H) \quad \dpn{E(C)x, y}{H} = E_{x, y}(C)
     \]
 
-    defines a spectral measure, it is sufficient to show that $I_E|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism. 
+    defines a spectral measure, it is sufficient to show that $J_E$ is a *-homomorphism. 
 
     Let $x, y \in H$, then as $\Gamma_A$ is a *-isomorphism, for any $f, g \in C(\Omega(A); \complex)$, 
     \begin{align*}
         \dpn{fg, E_{x, y}}{C(\Omega(A); \complex)} &= \dpn{\Gamma_A^{-1}f \cdot \Gamma_A^{-1}g \cdot x, y}{H} \\
-        &= \dpn{\Gamma_A^{-1}g \cdot x, (\Gamma_A^{-1}f)^* y}{H} = \dpn{g, E_{x, I_E(f)^*y}}{C(\Omega(A); \complex)}
+        &= \dpn{\Gamma_A^{-1}g \cdot x, (\Gamma_A^{-1}f)^* y}{H} = \dpn{g, E_{x, J_E(f)^*y}}{C(\Omega(A); \complex)}
     \end{align*}
 
-    As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y} = E_{x, I_E(f)^*y}$. Now, fix $\phi \in B^\infty(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$, 
+    As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y} = E_{x, J_E(f)^*y}$. Now, fix $\phi \in B^\infty(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$, 
     \begin{align*}
-        \dpn{E_{x, y}, \phi f}{C(\Omega(A); \complex)^*} &= \dpn{E_{x, I_E(f)^*y}, \phi}{C(\Omega(A); \complex)^*} = \dpn{I_E(\phi)x, I_E(f)^*y}{H} \\
-        &= \dpn{I_E(f)I_E(\phi)x, y}{H} = \dpn{f, E_{I_E(\phi)x, y}}{C(\Omega(A); \complex)}
+        \dpn{E_{x, y}, \phi f}{C(\Omega(A); \complex)^*} &= \dpn{E_{x, J_E(f)^*y}, \phi}{C(\Omega(A); \complex)^*} = \dpn{J_E(\phi)x, J_E(f)^*y}{H} \\
+        &= \dpn{J_E(f)J_E(\phi)x, y}{H} = \dpn{f, E_{J_E(\phi)x, y}}{C(\Omega(A); \complex)}
     \end{align*}
 
-    so $\phi E_{x, y} = E_{I_E(\phi)x, y}$ for all $\phi \in B^\infty(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^\infty(\Omega(A); \complex)$, 
+    so $\phi E_{x, y} = E_{J_E(\phi)x, y}$ for all $\phi \in B^\infty(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^\infty(\Omega(A); \complex)$, 
     \begin{align*}
-    \dpn{I_E(\phi \psi)x, y}{H} &= \dpn{E_{x, y}, \phi \psi}{C(\Omega(A); \complex)^*} = \dpn{E_{I_E(\psi) x, y}, \phi}{C(\Omega(A); \complex)^*} \\
-    &= \dpn{I_E(\phi)I_E(\psi)x, y}{H}
+    \dpn{J_E(\phi \psi)x, y}{H} &= \dpn{E_{x, y}, \phi \psi}{C(\Omega(A); \complex)^*} = \dpn{E_{J_E(\psi) x, y}, \phi}{C(\Omega(A); \complex)^*} \\
+    &= \dpn{J_E(\phi)J_E(\psi)x, y}{H}
     \end{align*}
 
-    and $I_E|_{B^\infty(\Omega(A); \complex)}$ is a homomorphism\footnote{With the same amount of writing and considerably more mental gymnastics, it can be shown that $I_E$ is a *-homomorphism on the full space $C(\Omega(A); \complex)^{**}$. However, it is not needed to show that $E$ is a spectral measure, and the homomorphism property falls out at the end anyways.}. 
+    and $J_E$ is a homomorphism. 
 
-    Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_A$ is a *-isomorphism, $I_E(f) = \Gamma_A^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)} = \dpn{I_E(f)x, x}{H} \in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^\infty(\Omega(A); \real)$ and $x \in H$, $\dpn{I_E(\phi)x, x}{H} = \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*} \in \real$ as well. Therefore $I_E(\phi)$ is self-adjoint, and $I_E|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism. 
+    Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_A$ is a *-isomorphism, $J_E(f) = \Gamma_A^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)} = \dpn{J_E(f)x, x}{H} \in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^\infty(\Omega(A); \real)$ and $x \in H$, $\dpn{J_E(\phi)x, x}{H} = \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*} \in \real$ as well. Therefore $J_E(\phi)$ is self-adjoint, and $J_E$ is a *-homomorphism. 
 
-    (2): By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$. Since the mapping $\phi \mapsto \int_{\Omega(A)}\phi dE$ is continuous from the weak* topology on $C(\Omega(A); \complex)^{**}$ to the weak operator topology on $B(H)$, 
-    \[
-    B \supset \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}}
-    \]
+    (2): By \autoref{definition:spectral-measure-integral}, $I_E$ is an injective unital *-homomorphism, so it is sufficient to show that $I_E(\mathscr{E}^*) = B$.
     
-    by \autoref{proposition:closure-of-image}. 
+    Let $J: C(\Omega(A); \complex) \to \mathscr{E}^*$ be defined by $\dpn{\mu, J(f)}{\mathscr{E}} = \int_{\Omega(A)}f d\mu$ for each $\mu \in \mathscr{E}$ and $f \in C(\Omega(A); \complex)$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$, so $J(C(\Omega(A); \complex))$ is weak*-dense in $\mathscr{E}^*$. Since $I_E$ is continuous from the weak* topology on $\mathscr{E}^*$ to the weak operator topology on $B(H)$, $I_E(\mathscr{E}^*) \subset B$ by \autoref{proposition:closure-of-image}. 
     
-    On the other hand, by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{C(\Omega(A); \complex)^{**}}(0, 1)}$ is weak*-compact, so
-    \[
-    S := \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}, \norm{\phi}_{C(\Omega(A); \complex)^{**}} \le 1}
-    \]
-
-    is weak-operator compact. As $\Gamma_A: A \to C(\Omega(A); \complex)$ is an isometric isomorphism, $S \supset \ol{B_A(0, 1)}$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $S \supset B_B(0, 1)$, and
-    \[
-    B = \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}}
-    \]
+    On the other hand, by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{\mathscr{E}^*}(0, 1)}$ is weak*-compact, so $I_E(\ol{B_{\mathscr{E}^*}(0, 1)})$ is weak-operator compact. As $\Gamma_A: A \to C(\Omega(A); \complex)$ is an isometric isomorphism, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_A(0, 1)}$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset B_B(0, 1)$, and $I_E(\mathscr{E}^*) = B$.
 \end{proof}
 
 \begin{definition}[Borel Functional Calculus]
 \label{definition:borel-functional-calculus}
-    Let $H$ be a complex Hilbert space, $T \in B(H)$ be normal, and $A \subset B(H)$ be the smallest von Neumann algebra acting on $H$ containing $T$ and $I$, then there exists a unique unital *-homomorphism
-    \[
-    C(\sigma_{B(H)}(T); \complex)^{**} \to A \quad \phi \mapsto \phi(T)
-    \]
-
-    such that:
+    Let $H$ be a complex Hilbert space, $T \in B(H)$ be normal, and $A \subset B(H)$ be the smallest von Neumann algebra acting on $H$ containing $T$ and $I$, then
     \begin{enumerate}
-        \item $\one(T) = I$, $\text{Id}(T) = T$, $\ol{\text{Id}}(T) = T^*$. 
-        \item The mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $C(\sigma_{B(H)}(T); \complex)^{**}$ to the weak operator topology on $A$. 
+        \item There exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)} \to A$ such that
+        \[
+        T = \int_{\sigma_{B(H)}(T)}\lambda E(d\lambda) \quad T^* = \int_{\sigma_{B(H)}(T)}\ol \lambda E(d\lambda)
+        \]
+        \item Let $\mathscr{E} \subset M_R(\sigma_{B(H)}(T); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $A \subset B(H)$ be the von Neumann algebra generated by $I$ and $T$, then
+        \[
+        I_E: \mathscr{E}^* \to A \quad \phi \mapsto \phi(T) := \int_{\sigma_{B(H)}(T)}\phi dE
+        \]
+
+        is a *-isomorphism. 
+        \item $I_E: \mathscr{E}^* \to A$ is the unique weak* to weak-operator continuous unital *-homomorphism such that $I_E(\text{Id}) = T$. 
     \end{enumerate}
     
-    Moreover, there exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)} \to A$ such that $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$ for all $\phi \in C(\sigma_{B(H)}(T); \complex)^{**}$. 
+    The mapping $f \mapsto f(T)$ on $\mathscr{E}^*$ is the \textbf{$L^\infty$-functional calculus} of $T$. 
 \end{definition}
 \begin{proof}
-    By the \hyperref[Spectral Theorem]{theorem:spectral-theorem-vn-1} applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping
+    (1), (2): By the \hyperref[Spectral Theorem]{theorem:spectral-theorem-vn-1} applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping
     \[
     I_E: C(\sigma_{B(H)}(T); \complex)^{**} \to A \quad \phi \mapsto \int_{\sigma_{B(H)}(T)} \phi dE
     \]
 
-    extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$. 
+    is a *-isomorphism that extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$. 
     
     For each $\phi \in C(\sigma_{B(H)}(T); \complex)^{**}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $C(\sigma_{B(H)}(T); \complex)^{**}$ to the weak operator topology on $A$ by \autoref{definition:spectral-measure-integral}.
 
-    Finally, by uniqueness of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, and (2), the mapping $\phi \mapsto \phi(T)$ is unique. 
+    (3): By uniqueness of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, and (2), the mapping $\phi \mapsto \phi(T)$ is unique. 
 \end{proof}