Linked existence of the weak integral.

src/fa/norm/linear.tex

@@ -1,4 +1,4 @@
-\section{Linear Maps}
+\section{Linear Maps Between Normed Spaces}
 \label{section:normed-linear-maps}
 
 \begin{proposition}

src/fa/norm/multilinear.tex

@@ -1,4 +1,4 @@
-\section{Multilinear Maps}
+\section{Multilinear Maps Between Normed Spaces}
 \label{section:normed-multilinear}
 
 \begin{proposition}

src/fa/tvs/complete-metric.tex

@@ -133,3 +133,20 @@
 
     Thus by \autoref{proposition:successive-approximation-all}, $B_F(0, t) \subset T(B_E(0, r)) \in \cn_F(0)$ for all $r > r_0$. As $r_0 > 0$ is arbitrary, $T(U) \in \cn_F(0)$ for all $U \in \cn_E(0)$. Therefore $T$ is open by translation-invariance of the topology on $E$.
 \end{proof}
+
+\begin{theorem}[Closed Graph Theorem]
+\label{theorem:closed-graph}
+    Let $E, F$ be complete metric TVSs over $K \in \RC$ and $T \in \hom(E; F)$. If its graph $\Gamma(T) \subset E \times F$ is closed, then $T \in L(E; F)$. 
+\end{theorem}
+\begin{proof}
+    Given that $E$ and $F$ are both complete metric TVSs, $E \times F$ is a complete metric TVS by \autoref{proposition:product-complete}. Since $\Gamma(T) \subset E \times F$ is a closed subspace of $E \times F$, it is also a complete metric TVS over $K$ by \autoref{proposition:complete-closed}. 
+    
+    Let $\pi_1: E \times F \to E$ and $\pi_2: E \times F \to F$ be the projection maps of $E \times F$. As $\Gamma(T)$ is the graph of a function, $\pi_1|_{\Gamma(T)}: \Gamma(T) \to E$ is a continuous bijection. By the \hyperref[Open Mapping Theorem]{theorem:open-mapping}, it is an isomorphism. Therefore $T$ may be expressed as the following composition of continuous linear maps
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_1|_{\Gamma(T)}^{-1}} & \Gamma(T) \ar@{->}[r]^{\pi_2|_{\Gamma(T)}} & F
+    }
+    \]
+
+\end{proof}
+

src/measure/index.tex

@@ -8,5 +8,6 @@
 \input{./measurable-maps/index.tex}
 \input{./lebesgue-integral/index.tex}
 \input{./bochner-integral/index.tex}
+\input{./weak-integral/index.tex}
 \input{./lcg/index.tex}
 \input{./notation.tex}

src/measure/weak-integral/definition.tex

@@ -9,7 +9,7 @@
 As I know so little about weak integrals, I will Dunning-Kruger myself right now, give an opinion, and laugh about it later. My gripe with seeing the definition comes from the need to test against \textit{every} continuous linear functional. To me, this seems quite inflexible: consider integrating a distribution-valued function. \textit{Surely} it is wiser to only test this function against test functions rather than \textit{the dual of $\mathcal{D}'$ (dual with respect to $\mathcal{D}'$ with the bounded convergence topology)}. As such, it may be more productive to consider a more flexible form of testing, such as using duality. 
 
 
-\begin{definition}[Weakly Integrable*]
+\begin{definition}[Weak Integrability*]
 \label{definition:weakly-integrable}
     Let $(X, \cm, \mu)$ be a measure space, $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, and $f: X \to E$, then $f$ is \textbf{Dunford $\lambda$-integrable*} if:
     \begin{enumerate}[label=(I\arabic*)]
@@ -29,7 +29,7 @@ As I know so little about weak integrals, I will Dunning-Kruger myself right now
     \begin{enumerate}
         \item[(I3+)] For each $A \in \cm$, the mapping
         \[
-        F \to K \quad \phi \mapsto \int_X \dpn{f(x), \phi}{\lambda} d\mu 
+        F \to K \quad \phi \mapsto \int_A \dpn{f(x), \phi}{\lambda} d\mu 
         \]
 
         is a $\sigma(F, E)$-continuous linear functional on $F$. 
@@ -41,4 +41,36 @@ As I know so little about weak integrals, I will Dunning-Kruger myself right now
 It is at this point that I start to understand why the bidual setup is useful: existence. 
 
 
+\begin{proposition}
+\label{proposition:dunford-existence}
+    Let $(X, \cm, \mu)$ be a measure space, $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, and $f: X \to E$. If
+    \begin{enumerate}[label=(I\arabic*)]
+        \item $f$ is weakly measurable. 
+        \item For each $\phi \in F$, $\phi \circ f \in L^1(X; K)$.
+        \item[(C)] $F$ is a Fréchet space. 
+    \end{enumerate}
+
+    then $f$ is \hyperref[Dunford $\lambda$-integrable]{definition:weakly-integrable}. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Section 3.3]{RyanTensor}}}. ]
+    Let $T: F \to L^1(X; K)$ be defined by $T\phi = \phi \circ f$. To see that $T$ is continuous, it is sufficient to apply the \hyperref[Closed Graph Theorem]{theorem:closed-graph}. 
+    
+    Let $\seq{\phi_n} \subset F$, $\phi \in F$, $\seq{g_n} \subset L^1(X; K)$, and $g \in L^1(X; K)$ such that
+    \begin{enumerate}[label=(\roman*)]
+        \item $g_n = \phi_n \circ f$ almost everywhere for all $n \in \natp$. 
+        \item $\phi_n \to \phi$ and $g_n \to g$ as $n \to \infty$.
+    \end{enumerate}
+    
+    
+    By passing through a subsequence using \autoref{theorem:cauchy-in-measure-limit}, assume further without loss of generality that $g_n \to g$ almost everywhere. In which case, $\phi_n \circ f \to g$ almost everywhere as well, and $\phi \circ f = g$ almost everywhere. Thus $T$ is continuous. 
+
+    Now, let $T^*: L^1(X; K)^* \to F^*$ be the adjoint of $T$. For each $A \in \cm$, the mapping $\Phi_A: L^1(X; K) \to K$ defined by $g \mapsto \int_A g d\mu$ is a continuous linear functional on $L^1(X; K)$. As such, for each $\phi \in F$, 
+    \[
+        \int_A \dpn{f, \phi}{\lambda} d\mu = \int_A T\phi d\mu = \dpn{T\phi, \Phi_A}{L^1(X; K)} = \dpn{\phi, T^*\Phi_A}{F}
+    \]
+
+    Therefore $T^*\Phi_A = \int_A^* f d\mu$ is the desired Dunford $\lambda$-integral of $f$ over $A$. 
+\end{proof}
+
+