Added nuclear operators.

src/fa/lc/index.tex

@@ -13,3 +13,4 @@
 \input{./hahn-banach.tex}
 \input{./spaces-of-linear.tex}
 \input{./tensor.tex}
+\input{./nuclear.tex}

src/fa/lc/nuclear.tex

@@ -38,7 +38,7 @@
         \begin{enumerate}[label=(\alph*)]
             \item The auxiliary space $F_B$ is a Banach space.
             \item $T(U) \subset B$. 
-            \item The induced map $\tilde E_U \to F_B$ is nuclear. 
+            \item The induced map $\wh E_U \to F_B$ is nuclear. 
         \end{enumerate}
         \item There exists an equicontinuous sequence $\seq{\phi_n} \subset E^*$, a convex, circled, and bounded subset $B \subset F$, $\seq{y_n} \subset B$, and $\seq{\lambda_n} \subset K$ such that
         \begin{enumerate}[label=(\alph*)]
@@ -48,7 +48,7 @@
         \end{enumerate}
     \end{enumerate}
     
-    If the above holds, then $T$ is \textbf{nuclear}. 
+    If the above holds, then $T$ is \textbf{nuclear}. The set $N(E; F)$ is the \textbf{space of nuclear operators} from $E$ to $F$. 
 \end{definition}
 \begin{proof}[Proof, {{\cite[Theorem III.7.1]{SchaeferWolff}}}. ]
     (1) $\Rightarrow$ (2): Let $\pi: E \to E_U$ be the canonical projection map associated with $E_U$ and $\iota: F_B \to F$ be the canonical inclusion map associated with $F_B$. By assumption (1b), there exists an induced map $\hat T: E_U \to F_B$ such that the following diagram commutes:
@@ -59,7 +59,7 @@
     }
     \]
 
-    By the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-normed}, $E_U^* = (\tilde E_U)^*$. Assume without loss of generality that $E_U$ is a Banach space, then (1c) implies that $\hat T \in L(E_U; F_B)$ is a nuclear operator. By \autoref{lemma:nuclear-operator-normed-tensor} and \autoref{theorem:metrisable-tensor-product}, there exists $\seq{\phi_n} \subset E_U^*$, $\seq{y_n} \subset F_B$, and $\seq{\lambda_n} \subset K$ such that:
+    By the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-normed}, $E_U^* = (\wh E_U)^*$. Assume without loss of generality that $E_U$ is a Banach space, then (1c) implies that $\hat T \in L(E_U; F_B)$ is a nuclear operator. By \autoref{lemma:nuclear-operator-normed-tensor} and \autoref{theorem:metrisable-tensor-product}, there exists $\seq{\phi_n} \subset E_U^*$, $\seq{y_n} \subset F_B$, and $\seq{\lambda_n} \subset K$ such that:
     \begin{enumerate}[label=(\roman*)]
         \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
         \item $\limv{n}\phi_n = 0$ and $\limv{n}y_n = 0$. 
@@ -93,15 +93,15 @@
 
     Thus for each $x \in E$, 
     \[
-    Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E} = \sum_{n = 1}^\infty y_n \dpn{\pi(x), \lambda_n \hat \phi_n}{E}
+    Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E} = \sum_{n = 1}^\infty y_n \dpn{\pi(x), \lambda_n \hat \phi_n}{E_U}
     \]
     
-    and the induced map $\hat T: \tilde E_U \to F_B$ takes the form $\hat Tx = \sum_{n = 1}^\infty y_n \dpn{x, \lambda_n \hat \phi_n}{\tilde E_U}$. Finally, for each $n \in \natp$, $U \subset \phi_n^{-1}(B_K(0, 1))$, and $\normn{\hat \phi_n}_{E_U^*} \le 1$. Similarly, since $y_n \in B$, $\norm{y_n}_{F_B} \le 1$ as well. Therefore
+    and the induced map $\hat T: \wh E_U \to F_B$ takes the form $\hat Tx = \sum_{n = 1}^\infty y_n \dpn{x, \lambda_n \hat \phi_n}{\wh E_U}$. Finally, for each $n \in \natp$, $U \subset \phi_n^{-1}(B_K(0, 1))$, and $\normn{\hat \phi_n}_{E_U^*} \le 1$. Similarly, since $y_n \in B$, $\norm{y_n}_{F_B} \le 1$ as well. Therefore
     \[
-        \norm{\hat T}_{N(\tilde E_U; F_B)} \le \sum_{n \in \natp}|\lambda_n| \cdot \norm{y_n}_{F_B} \cdot \normn{\hat \phi_n}_{E_U^*} \le \sum_{n \in \natp}|\lambda_n| < \infty
+        \normn{\hat T}_{N(\wh E_U; F_B)} \le \sum_{n \in \natp}|\lambda_n| \cdot \norm{y_n}_{F_B} \cdot \normn{\hat \phi_n}_{E_U^*} \le \sum_{n \in \natp}|\lambda_n| < \infty
     \]
 
-    and $\hat T: \tilde E_U \to F_B$ is nuclear. 
+    and $\hat T: \wh E_U \to F_B$ is nuclear. 
     
     
 \end{proof}
@@ -110,17 +110,17 @@
 \label{proposition:nuclear-gymnastics}
     Let $E, F, G, H$ be separated locally convex spaces and $S \in N(F; G)$, then:
     \begin{enumerate}
-        \item $T$ is compact.
+        \item $S$ is compact.
         \item For any $T \in L(E; F)$, $S \circ T \in N(E; G)$. 
         \item For any $R \in L(G; H)$, $R \circ S \in N(F; H)$. 
-        \item There exists a unique $\tilde S \in L(\tilde F; G)$ such that $\tilde S|_{F} = S$. Moreover, $\tilde S \in N(\tilde F; G)$. 
+        \item There exists a unique $\wh S \in L(\wh F; G)$ such that $\wh S|_{F} = S$. Moreover, $\wh S \in N(\wh F; G)$. 
     \end{enumerate}
 \end{proposition}
 \begin{proof}[Proof, {{\cite[Corollary III.7.1.1-III.7.1.3]{SchaeferWolff}}}. ]
     Let $\seq{\phi_n} \subset F^*$ be an equicontinuous sequence, $B \in B(G)$ be convex and circled, $\seq{y_n} \subset B$, and $\seq{\lambda_n} \subset K$ such that
     \begin{enumerate}[label=(\alph*)]
         \item The auxiliary space $G_B$ is a Banach space. 
-        \item For each $x \in F$, $Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{F}$. 
+        \item For each $x \in F$, $Sx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{F}$. 
         \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
     \end{enumerate}
 
@@ -131,28 +131,28 @@
     \end{CD}
     \]
 
-    By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, $\overline{B_K(0,1)}^{\natp}$ is compact. Since $T(U)$ is contained in its image in the above diagram, $T(U)$ is relatively compact. 
+    By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, $\overline{B_K(0,1)}^{\natp}$ is compact. Since $S(U)$ is contained in its image in the above diagram, $S(U)$ is relatively compact. 
 
-    (2): Since $T \in L(E; F)$, for each $\seq{\phi_n \circ T} \subset E^*$ is equicontinuous. Thus for any $x \in E$, 
+    (2): Since $T \in L(E; F)$, $\seq{\phi_n \circ T} \subset E^*$ is equicontinuous. Thus for any $x \in E$, 
     \[
     (S \circ T)x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n \circ T}{E}
     \]
 
     and $S \circ T \in N(E; G)$. 
 
-    (3): Using (1), assume without loss of generality that $B$ is also compact. In which case, $R(C)$ is a convex, circled, and compact set in $H$ containing $0$. Thus $H_{R(C)}$ is a Banach space. For each $x \in F$, 
+    (3): Using (1), assume without loss of generality that $B$ is also compact. In which case, $R(B)$ is a convex, circled, and compact set in $H$ containing $0$. Thus $H_{R(B)}$ is a Banach space. For each $x \in F$, 
     \[
     (R \circ S)x = \sum_{n = 1}^\infty \lambda_n R(y_n) \dpn{x, \phi_n}{F}
     \]
 
     and $R \circ S \in N(F; H)$. 
 
-    (4): By the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-tvs}, $\seq{\phi_n} \subset F^*$ extend into an equicontinuous family $\bracsn{\tilde \phi_n}_1^\infty \subset \tilde F^*$. In which case, since $G_B$ is complete, the extension $\tilde S \in L(\tilde F; G)$ takes the form
+    (4): By the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-tvs}, such an extension exists and is unique. Moreover, $\seq{\phi_n} \subset F^*$ extend into an equicontinuous family $\bracsn{\wh \phi_n}_1^\infty \subset \wh F^*$. Since $G_B$ is complete, the extension $\wh S \in L(\wh F; G)$ takes the form
     \[
-    \tilde S x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \tilde \phi_n}{F}
+    \wh S x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \wh \phi_n}{\wh F}
     \]
 
-    Therefore $\tilde S \in N(\tilde F; G)$.
+    Therefore $\wh S \in N(\wh F; G)$.
 \end{proof}