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}