Sums of nuclear spaces are nuclear.

src/fa/lc/inductive.tex

@@ -17,7 +17,7 @@
         is a fundamental system of neighbourhoods for $E$ at $0$.
         \item If $E$ is spanned by $\bigcup_{i \in I}T_i(E_i)$, then
         \[
-        \fB = \bracs{\Gamma\paren{\bigcup_{i \in I}T_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
+        \fB = \bracs{\aconv\paren{\bigcup_{i \in I}T_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
         \]
 
         is a fundamental system of neighbourhoods for $E$ at $0$. 
@@ -42,7 +42,7 @@
 
     (6): If $E$ is spanned by $\bigcup_{i \in I}T_i(E_i)$, then each set in $\fB$ is radial. Hence $\fB$ is a family of neighbourhoods of $E$ at $0$. 
 
-    Let $U \in \cn_E(0)$ be convex, circled, and radial, then for each $i \in I$, $T_i^{-1}(U) \in \cn_{E_i}(0)$, so $U \supset \bigcup_{i \in I}T_i[T_i^{-1}(U)]$. Since $U$ is convex and circled, $U \supset \Gamma\paren{\bigcup_{i \in I}T_i[T_i^{-1}(U)]} \in \fB$. Therefore $\fB$ forms a fundamental system of neighbourhoods for $E$ at $0$.
+    Let $U \in \cn_E(0)$ be convex, circled, and radial, then for each $i \in I$, $T_i^{-1}(U) \in \cn_{E_i}(0)$, so $U \supset \bigcup_{i \in I}T_i[T_i^{-1}(U)]$. Since $U$ is convex and circled, $U \supset \aconv\paren{\bigcup_{i \in I}T_i[T_i^{-1}(U)]} \in \fB$. Therefore $\fB$ forms a fundamental system of neighbourhoods for $E$ at $0$.
 \end{proof}
 
 \begin{definition}[Locally Convex Direct Sum]
@@ -61,7 +61,7 @@
         
         \item The family
         \[
-        \fB = \bracs{\Gamma\paren{\bigcup_{i \in I}\iota_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
+        \fB = \bracs{\aconv\paren{\bigcup_{i \in I}\iota_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
         \]
 
         is a fundamental system of neighbourhoods for $E$ at $0$. 

src/fa/lc/nuclear-space.tex

@@ -90,7 +90,7 @@
 \label{proposition:nuclear-subspace}
     Let $E$ be a nuclear space over $K \in \RC$ and $F \subset E$ be a subspace, then $F$ is also nuclear.
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Theorem III.7.4]{SchaeferWolff}}}. ]
     Firstly, a setup about auxiliary spaces and subspaces is required. Let $U \in \cn_E(0)$ be convex and circled, then the composition of the inclusion map $\iota: F \to E$ and the canonical projection $\pi_U: E \to E_U$ factors through $F_{U \cap F}$ as follows:  
     \[
     \xymatrix{
@@ -143,7 +143,7 @@
 \label{proposition:nuclear-quotient}
     Let $E$ be a nuclear space over $K \in \RC$, and $F$ be a closed subspace of $E$, then $E/F$ is also nuclear. 
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Theorem III.7.4]{SchaeferWolff}}}. ]
     Firstly, a setup about auxiliary spaces and quotients is required. Let $p: E \to E/F$ be the canonical projection and $U \in \cn_E(0)$ be a convex and circled neighbourhood, then the composition of maps $E \to E/F \to (E/F)_{p(U)}$ factors through $E_{U}$ as follows: 
     \[
     \xymatrix{
@@ -190,3 +190,24 @@
 
     Therefore $\widehat \pi_{p(U)}$ is nuclear, and $E/F$ is a nuclear space. 
 \end{proof}
+
+\begin{proposition}
+\label{proposition:nuclear-direct-sum}
+    Let $\seq{E_n}$ be nuclear spaces over $K \in \RC$, then $\bigoplus_{n = 1}^\infty E_n$ is also nuclear. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem III.7.4]{SchaeferWolff}}}. ]
+    For each $n \in \natp$, identify $E_n$ as a subspace of $\bigoplus_{n = 1}^\infty E_n$. Let $F$ be a Banach space and $T \in L(\bigoplus_{n = 1}^\infty E_n; F)$. For each $n \in \natp$, $E_n$ is a nuclear space, so $T|_{E_n}: E_n \to F$ is a nuclear operator, and there exists $\bracsn{\phi_{n, k}}_{k = 1}^\infty \subset E_n^*$ equicontinuous, $\bracsn{y_{n, k}}_{k = 1}^\infty \subset B_F(0, 1)$, and $\bracsn{\lambda_{n, k}}_{k = 1}^\infty \subset K$ such that $\sum_{k \in \natp}|\lambda_{n, k}| \le 2^{-n}$ and
+    \[
+    Tx = \sum_{k = 1}^\infty \lambda_{n, k}y_{n, k} \dpn{x, \phi_{n, k}}{E_n}
+    \]
+
+    for all $x \in E_n$. Thus for any $x \in \bigoplus_{n = 1}^\infty E_n$, 
+    \begin{align*}
+    Tx &= \sum_{n = 1}^\infty \sum_{k = 1}^\infty \lambda_{n, k}y_{n, k}\dpn{x_n, \phi_{n, k}}{E_n} \\
+    &= \sum_{n = 1}^\infty \sum_{k = 1}^\infty \lambda_{n, k}y_{n, k}\dpn{x, \phi_{n, k} \circ \pi_n}{\bigoplus_{n = 1}^\infty E_n}
+    \end{align*}
+    
+    where $\sum_{n \in \natp}\sum_{k \in \natp}|\lambda_{n, k}| \le \sum_{n \in \natp}2^{-n} < \infty$ and $\bracsn{y_{n, k}|n, k \in \natp} \subset B_F(0, 1)$. 
+
+    Finally, for each $n \in \natp$, let $U_n = \bigcap_{k \in \natp}\phi_{n, k}^{-1}(B_K(0, 1))$, then $U_n \in \cn_{E_n}(0)$ by equicontinuity of $\bracsn{\phi_{n, k}}_{k = 1}^\infty \subset E_n^*$. Let $U = \aconv(\bigcup_{n \in \natp}U_n)$, then $U \in \cn_{\bigoplus_{n = 1}^\infty E_n}(0)$ and $U \subset \bigcap_{n \in \natp}\bigcap_{k \in\natp}(\phi_{n, k} \circ \pi_n)^{-1}(B_K(0, 1))$. Hence $\bracsn{\phi_{n, k} \circ \pi_n|n, k \in \natp}$ is equicontinuous, and $T$ is a nuclear operator. 
+\end{proof}