Added the Arens extension.

refs.bib

@@ -268,4 +268,18 @@
   year      = {2002},
   isbn      = {978-1-85233-437-6},
   doi       = {10.1007/978-1-4471-3903-4}
-}
+}
+
+@article{ArensBilinear,
+ ISSN = {00029939, 10886826},
+ URL = {http://www.jstor.org/stable/2031695},
+ author = {Richard Arens},
+ journal = {Proceedings of the American Mathematical Society},
+ number = {6},
+ pages = {839--848},
+ publisher = {American Mathematical Society},
+ title = {The Adjoint of a Bilinear Operation},
+ urldate = {2026-08-13},
+ volume = {2},
+ year = {1951}
+}

src/fa/norm/arens.tex

@@ -0,0 +1,117 @@
+\section{The Arens Product}
+\label{section:arens-product}
+
+\begin{definition}[Arens Extension]
+\label{definition:arens-product}
+    Let $E, F, G$ be normed vector spaces over $K \in \RC$ and $\lambda \in L^2(E, F; G)$ be a continuous bilinear mapping, then there exists a unique bilinear mapping $\Lambda_1: E^{**} \times F^{**} \to G^{**}$ such that:
+    \begin{enumerate}
+        \item For each $(x, y) \in E \times F$, $\Lambda_1(x, y) = \lambda(x, y)$. 
+        \item For each $x \in E$, $\Lambda_1(x, \cdot)$ is weak*-continuous. 
+        \item For each $y \in F^{**}$, $\Lambda_1(\cdot, y)$ is weak*-continuous. 
+    \end{enumerate}
+
+    Similarly, there exists a unique bilinear mapping $\Lambda_2: E^{**} \times F^{**} \to G^{**}$ such that:
+    \begin{enumerate}
+        \item For each $(x, y) \in E \times F$, $\Lambda_2(x, y) = \lambda(x, y)$. 
+        \item[(2')] For each $x \in E^{**}$, $\Lambda_2(x, \cdot)$ is weak*-continuous. 
+        \item[(3')] For each $y \in F$, $\Lambda_2(\cdot, y)$ is weak*-continuous. 
+    \end{enumerate}
+
+    The mappings $\Lambda_1, \Lambda_2: E^{**} \times F^{**} \to G^{**}$ are the \textbf{first} and \textbf{second} \textbf{Arens extensions} of $\lambda$, respectively. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Section 1, Theorem 3.2]{ArensBilinear}}}. ]
+    For each $x \in E$, the mapping $\lambda(x, \cdot) \in L(F; G)$ admits an adjoint $\lambda^*(x, \cdot) \in L(G^*; F^*)$, which induces an adjoint of the bilinear map as follows
+    \[
+    \lambda^*: E \times G^* \to F^* \quad \dpn{y, \lambda^*(x, \phi)}{F} = \dpn{\lambda(x, y), \phi}{G}
+    \]
+
+    Applying the above operation again yields a second adjoint
+    \[
+    \lambda^{**}: F^{**} \times G^* \to E^* \quad \dpn{x, \lambda^{**}(y, \phi)}{E} = \dpn{\lambda^*(x, \phi), y}{F^*}
+    \]
+
+    and finally, applying the adjoint operation a third time gives
+    \[
+    \Lambda_1 = \lambda^{***}: E^{**} \times F^{**} \to G^{**} \quad \dpn{\phi, \lambda^{***}(x, y)}{G^*} = \dpn{\lambda^{**}(y, \phi), x}{E^*}
+    \]
+
+    (1): Let $(x, y) \in E \times F$, then for each $\phi \in G^*$, 
+    \[
+        \dpn{\phi, \Lambda_1(x, y)}{G^*} = \dpn{\lambda^{**}(y, \phi), x}{E^*} = \dpn{\lambda^*(x, \phi), y}{F^*} = \dpn{\lambda(x, y), \phi}{G}
+    \]
+
+    By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $\Lambda_1$ is an extension of $\lambda$. 
+
+    (2): Fix $x \in E$ and $\phi \in G^*$, then for each $y \in F^{**}$, 
+    \begin{align*}
+        \dpn{\phi, \Lambda_1(x, y)}{G^*} &= \dpn{\lambda^{**}(y, \phi), x}{E^*} = \dpn{x, \lambda^{**}(y, \phi)}{E} \\
+        &= \dpn{\lambda^*(x, \phi), y}{F^*}
+    \end{align*}
+
+    Since $\lambda^*(x, \phi) \in F^*$, $\Lambda_1(x, \cdot)$ is weak*-continuous. 
+
+    (3): Fix $y \in F^{**}$ and $\phi \in G^*$, then for each $x \in E^{**}$, $\dpn{\phi, \Lambda_1(x, y)}{G^*} = \dpn{\lambda^{**}(y, \phi), x}{E^*}$. Since $\lambda^{**}(y, \phi) \in E^*$, $\Lambda_1(\cdot, y)$ is weak*-continuous. 
+
+    (Uniqueness): By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $E$ is weak*-dense in $E^{**}$, and $F$ is weak*-dense in $F^{**}$, so the extension is uniquely determined.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:arens-extension-continuous}
+    Let $E, F, G$ be normed vector spaces over $K \in \RC$, $\lambda \in L^2(E, F; G)$ be a continuous bilinear mapping, then $\Lambda_1, \Lambda_2 \in L^2(E^{**}, F^{**}; G^{**})$, where
+    \[
+    \norm{\lambda}_{L^2(E, F; G)} = \norm{\Lambda_1}_{L^2(E^{**}, F^{**}; G^{**})} = \norm{\Lambda_2}_{L^2(E^{**}, F^{**}; G^{**})}
+    \]
+\end{proposition}
+\begin{proof}
+    Assume without loss of generality that $\norm{\lambda}_{L^2(E, F; G)} = 1$. Fix $x \in \ol{B_{E}(0, 1)}$, then
+    \[
+    \Lambda_1(x, \ol{B_F(0, 1)}) = \lambda(x, \ol{B_F(0, 1)}) \subset \ol{B_{G^{**}}(0, 1)}
+    \]
+    
+    
+    Since for any $z \in G^{**}$, $\norm{z}_{G^{**}} = \sup_{\phi \in G^*, \norm{\phi}_{G^*} \le 1}\dpn{z, \phi}{G^*}$, the norm on $G^{**}$ is weak*-lower semicontinuous, and $\ol{B_{G^{**}}(0, 1)}$ is weak*-closed. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $\ol{B_{F}(0, 1)}$ is weak*-dense in $\ol{B_{F^{**}}(0, 1)}$. As such, weak*-continuity of $\Lambda_1(x, \cdot)$ and \autoref{proposition:closure-of-image} implies that $\Lambda_1(x, \ol{B_{F^{**}}(0, 1)}) \subset \ol{B_{G^{**}}(0, 1)}$ as well. 
+
+    Now, fix $y \in \ol{B_{F^{**}}(0, 1)}$, then $\Lambda_1(\ol{B_E(0, 1)}, y) \subset \ol{B_{G^{**}}(0, 1)}$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $\ol{B_{E}(0, 1)}$ is weak*-dense in $\ol{B_{E^{**}}(0, 1)}$. Thus the weak*-continuity of $\Lambda_1(\cdot, y)$ and \autoref{proposition:closure-of-image} implies that $\Lambda_1(\ol{B_{E^{**}}(0, 1)}, y) \subset \ol{B_{G^{**}}(0, 1)}$. Therefore
+    \[
+    \Lambda_1(\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}) \subset \ol{B_{G^{**}}(0, 1)}
+    \]
+
+    and $\norm{\lambda}_{L^2(E, F; G)} = \norm{\Lambda_1}_{L^2(E^{**}, F^{**}; G^{**})}$. 
+\end{proof}
+
+
+\begin{definition}[Arens Regularity]
+\label{definition:arens}
+    Let $E, F, G$ be normed vector spaces over $K \in \RC$, $\lambda \in L^2(E, F; G)$ be a continuous bilinear mapping, and $\Lambda_1, \Lambda_2: E^{**} \times F^{**} \to G^{**}$ be its first and second Arens extensions, respectively, then the following are equivalent:  
+    \begin{enumerate}
+        \item $\Lambda_1 = \Lambda_2$. 
+        \item There exists an extension $\Lambda: E^{**} \times F^{**} \to G^{**}$ of $\lambda$ that is separately weak*-continuous. 
+        \item There exists an extension $\Lambda: E^{**} \times F^{**} \to G^{**}$ of $\lambda$ that is separately weak*-continuous when restricted to $\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}$.
+    \end{enumerate}
+    
+    If the above holds, then $\lambda$ is an \textbf{Arens regular} bilinear map, and $\Lambda = \Lambda_1 = \Lambda_2$ is \textit{the} \textbf{Arens extension} of $\lambda$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem 3.3]{ArensBilinear}}}. ]
+    (3) $\Rightarrow$ (1): By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $\ol{B_{E}(0, 1)}$ is weak*-dense in $\ol{B_{E^{**}}(0, 1)}$, and $\ol{B_{F}(0, 1)}$ is weak*-dense in $\ol{B_{F^{**}}(0, 1)}$. Thus the restrictions of $\Lambda_1$ and $\Lambda_2$ to $\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}$ are uniquely determined by the value of $\lambda$ on $\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}$, in the following sense:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\Lambda_1|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}$ is the unique extension of $\lambda|_{\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}}$ such that
+        \begin{enumerate}[label=(\alph*)]
+            \item For each $x \in \ol{B_E(0, 1)}$, $\Lambda_1(x, \cdot)$ is weak*-continuous. 
+            \item For each $y \in \ol{B_{F^{**}}(0, 1)}$, $\Lambda_1(\cdot, y)$ is weak*-continuous. 
+        \end{enumerate}
+        \item $\Lambda_2|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}$ is the unique extension of $\lambda|_{\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}}$ such that
+        \begin{enumerate}[label=(\alph*)]
+            \item For each $x \in \ol{B_{E^{**}}(0, 1)}$, $\Lambda_2(x, \cdot)$ is weak*-continuous. 
+            \item For each $y \in \ol{B_{F}(0, 1)}$, $\Lambda_2(\cdot, y)$ is weak*-continuous. 
+        \end{enumerate}
+    \end{enumerate}
+
+    Since the given extension $\Lambda$ satisfies (i.a), (i.b), (ii.a), and (ii.b), 
+    \[
+    \Lambda_1|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}} = \Lambda = \Lambda_2|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}
+    \]
+
+    As $\Lambda_1, \Lambda_2$ are bilinear, the above implies that $\Lambda_1 = \Lambda_2$. 
+\end{proof}
+
+

src/fa/norm/index.tex

@@ -6,6 +6,7 @@
 \input{./linear.tex}
 \input{./separable.tex}
 \input{./multilinear.tex}
+\input{./arens.tex}
 \input{./hilbert.tex}
 \input{./compact.tex}
 \input{./ap.tex}

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:

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}