diff --git a/src/measure/vector/fin.tex b/src/measure/vector/fin.tex index faf5086..3e836b0 100644 --- a/src/measure/vector/fin.tex +++ b/src/measure/vector/fin.tex @@ -117,6 +117,51 @@ Despite not covering the full dual space, the bounded Borel functions still form (2) $\Rightarrow$ (1): By the \hyperref[Dominated Convergence Theorem]{theorem:dct-bochner-vector}. \end{proof} + +The following is an explicit construction of what should be the Arens product on $C(X; \complex)^{**}$ thanks to the above description of $M_R(X; \complex)$ and its dual. + +\begin{proposition} +\label{proposition:measures-dual-algebra} + Let $X$ be a compact Hausdorff space, $J: C(X; \complex) \to C(X; \complex)^{**}$ be the natural embedding, and $B$ be the closed unit ball of $C(X; \complex)^{**}$, then + \begin{enumerate} + \item There exists a unique weak*-continuous involution $C(X; \complex)^{**}$ such that $J(f^*) = J(f)^*$ for all $f \in C(X; \complex)$, given by + \[ + C(X; \complex)^{**} \to C(X; \complex)^{**} \quad \dpn{\mu, \phi^*}{C(X; \complex)^*} = \ol{\dpn{\mu, \phi}{C(X; \complex)^*}} + \] + \item There exists a unique seperately weak*-continuous bilinear map on $C(X; \complex)^{**}$ such that $J(fg) = J(f)J(g)$ for all $f, g \in C(X; \complex)$. + \item $C(X; \complex)^{**}$ equipped with the above involution and product is a commutative unital $C^*$-algebra. + \end{enumerate} +\end{proposition} +\begin{proof} + It is sufficient to construct the maps in (1) and (2). Let $\seqi{\mu}$ be a maximal mutually singular family of Radon measures on $X$. Using \autoref{theorem:hilbert-measures-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, identify + \[ + C(X; \complex)^{*} = M_R(X; \complex) = [l^1(I); L^1(\mu_i; \complex)] \quad C(X; \complex)^{**} = [l^\infty(I); L^\infty(\mu_i; \complex)] + \] + + Under the above, $C(X; \complex)$ may be identified as the diagonal + \[ + \bracsn{f \in C(X; \complex)^I|f_i = f_j \forall i, j \in I} \subset [l^\infty(I); L^\infty(\mu_i; \complex)] + \] + + which is weak*-dense in $[l^\infty(I); L^\infty(\mu_i; \complex)]$ by \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}. + + (1): For each $g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$, let $g^* = \ol g$. For any $\mu \in [l^1(I); L^1(\mu_i; \complex)]$, + \[ + \dpn{\mu, g^*}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{f, \ol g}{[l^1(I); L^1(\mu_i; \complex)]} = \ol{\dpn{f, g}{[l^1(I); L^1(\mu_i; \complex)]}} + \] + + so the conjugation map is weak*-continuous. + + (2): Let $f, g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$ and $\mu \in [l^1(I); L^1(\mu_i; \complex)]$, + \[ + \dpn{\mu, fg}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{f\mu, g}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{g\mu, f}{[l^1(I); L^1(\mu_i; \complex)]} + \] + + so the composition map is separately weak*-continuous. + + (3): $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is a commutative unital $C^*$-algebra. +\end{proof} + \begin{proposition} \label{proposition:space-of-measures-extreme-points} Let $X$ be an LCH space and $\cm \subset \overline{B_{M_R(X; \complex)}(0, 1)}$ be a compact convex set such that: diff --git a/src/op/vn/fc.tex b/src/op/vn/fc.tex new file mode 100644 index 0000000..7b97a2b --- /dev/null +++ b/src/op/vn/fc.tex @@ -0,0 +1,19 @@ +\section{The Borel Functional Calculus} +\label{section:borel-functional-calculus} + +\begin{definition}[Borel Functional Calculus] +\label{definition:borel-functional-calculus} + Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $x \in A$ be normal, then there exists a unique continuous unital *-homomorphism + \[ + C(\sigma_A(x); \complex)^{**} \to A[x] \quad f \mapsto f(x) + \] + + such that: + \begin{enumerate} + \item $\one(x) = 1_A$, $\text{Id}(x) = x$, and $\overline{\text{Id}}(x) = x^*$. + \item The mapping $f \mapsto f(x)$ is continuous from the weak* topology on $C(\sigma_A(x); \complex)^{**}$ to the strong operator topology on $B(H)$. + \end{enumerate} +\end{definition} +\begin{proof} + Since the \autoref{definition:continuous-functional-calculus} +\end{proof}