Every product of nuclear spaces is 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

@@ -87,3 +87,155 @@
 \end{corollary}
 
 
+\begin{summary}
+\label{summary:nuclear-extension}
+    Every subspace and separated qoutient space of a nuclear space is nuclear. The product of nuclear spaces is nuclear. The locally convex direct sum of countably many nuclear spaces is nuclear. 
+\end{summary}
+\begin{proof}
+    See \autoref{proposition:nuclear-quotient}, \autoref{proposition:nuclear-subspace}, \autoref{proposition:nuclear-direct-sum}, and \autoref{proposition:nuclear-product}. 
+\end{proof}
+
+\begin{proposition}
+\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}[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:  
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_U} & E_U \\
+    F \ar@{->}[u]^{\iota} \ar@{->}[r]_{\pi_{U \cap F}} & F_{U \cap F} \ar@{->}[u]_{\widehat \pi_U}
+    }
+    \]
+
+    where $\widehat \pi_U$ is an isometric embedding. As a result, the factored map $\widehat \pi_U: F_{U \cap F} \to E_U$ extends to an isometric embedding on the completions: 
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_U} & E_U \ar@{->}[r] & \widehat E_{U} \\
+    F \ar@{->}[u]^{\iota} \ar@{->}[r]_{\pi_{U \cap F}} & F_{U \cap F} \ar@{->}[u]_{\widehat \pi_U} \ar@{->}[r] & \widehat F_{U \cap F} \ar@{->}[u]_{\widehat \pi_U}
+    }
+    \]
+
+    which enables identifying $\widehat F_{U \cap F}$ as a closed subspace of $\widehat E_{U}$.
+
+    To start the proof, let $U \in \cn_E(0)$ be a given convex and circled neighbourhood. Since $E$ is nuclear, there exists a convex and circled neighbourhood $V \in \cn_E(0)$ with $V \subset U$ such that the induced map $\widehat \pi_U: \widehat E_V \to \widehat E_U$ is nuclear. By prior discussion, the following diagram commutes:
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_V} & \widehat E_V \ar@{->}[r]^{\widehat \pi_{U}} & \widehat E_U \\
+    F \ar@{->}[u] \ar@{->}[r] & \widehat F_{V \cap F} \ar@{->}[u] \ar@{->}[r]_{\widehat \pi_{U \cap F}} & \widehat F_{U \cap F} \ar@{->}[u]
+    }
+    \]
+
+    Thus the induced map $\widehat \pi_{U \cap F}: \widehat F_{V \cap F} \to \widehat F_{U \cap F}$ corresponds to the restriction of $\widehat \pi_U$ to $\widehat F_{V \cap F}$. Since $\widehat \pi_U$ is nuclear, there exists $\seq{\phi_n} \subset E_V^*$ and $\seq{y_n} \subset \widehat E_U$ such that
+    \[
+    \widehat \pi_U x = \sum_{n = 1}^\infty y_n\dpn{x, \phi_n}{\widehat E_V} \quad \forall x \in \widehat E_V
+    \]
+
+    and $\sum_{n \in \natp} \norm{y_n}_{\widehat E_U}\norm{\phi_n}_{E_V^*} < \infty$. 
+
+    Now, using \autoref{theorem:nuclear-lp}, further assume without loss of generality that $\widehat E_{U}$ is a Hilbert space. Let $P: \widehat E_{U} \to \widehat F_{U \cap F}$ be the orthogonal projection of $\widehat E_U$ onto $\widehat F_{U \cap F}$, then
+    \[
+    \widehat \pi_{U \cap F}x = \sum_{n = 1}^\infty Py_n \dpn{x, \phi_n}{\widehat F_{V \cap F}} \quad \forall x \in \widehat F_{V \cap F}
+    \]
+
+    with
+    \[
+    \normn{\widehat \pi_{U \cap F}}_{N(\widehat F_{V \cap F}; \widehat F_{U \cap F})}
+    \le \sum_{n \in \natp} \norm{Py_n}_{\widehat F_{U \cap F}} \norm{\phi_n}_{F_{V \cap F}^*} \le \sum_{n \in \natp} \norm{y_n}_{\widehat E_U}\norm{\phi_n}_{E_V^*} < \infty
+    \]
+
+    Therefore the induced map $\widehat \pi_{U \cap F}$ is nuclear, and $F$ is a nuclear space.
+\end{proof}
+
+
+
+
+\begin{proposition}
+\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}[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: 
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_U} \ar@{->}[d]_{p} & E_U \ar@{->}[d] \\
+    E/F \ar@{->}[r]_{\pi_{p(U)}} & (E/F)_{p(U)}
+    }
+    \]
+
+    This extends through the completion
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_U} \ar@{->}[d]_{p} & E_U \ar@{->}[d] \ar@{->}[r] & \widehat E_U \ar@{->}[d] \\
+    E/F \ar@{->}[r]_{\pi_{p(U)}} & (E/F)_{p(U)} \ar@{->}[r] & \widehat{(E/F)}_{p(U)}
+    }
+    \]
+
+    and yields that $\widehat{(E/F)}_{p(U)}$ is a quotient space of $\widehat E_{U}$. 
+
+    To begin the proof, let $U \in \cn_E(0)$ be a convex and circled neighbourhood. Since $E$ is nuclear, there exists a convex and circled neighbourhood $V \in \cn_E(0)$ with $V \subset U$ such that the induced map $\widehat \pi_{U}: \widehat E_V \to \widehat E_{U}$ is nuclear. The composition of maps $\wh E_V \to \wh E_U \to \wh{(E/F)}_{p(U)}$ then factors through $\widehat{(E/F)}_{p(V)}$ as $\widehat \pi_{p(U)}$:
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_V} \ar@{->}[d]_{p} & \widehat E_V \ar@{->}[d] \ar@{->}[r]^{\widehat \pi_U} & \widehat E_U \ar@{->}[d]^{\widehat p} \\
+    E/F \ar@{->}[r]_{\pi_{p(V)}} & \widehat{(E/F)}_{p(V)} \ar@{->}[r]_{\widehat \pi_{p(U)}} & \widehat{(E/F)}_{p(U)}
+    }
+    \]
+
+    Since $\wh \pi_U: \wh E_V \to \wh E_U$ is nuclear, there exists $\seq{\phi_n} \subset E_V^*$ and $\seq{y_n} \subset \wh E_U$ such that
+    \[
+    \wh \pi_U x = \sum_{n = 1}^\infty y_n \dpn{x, \phi_n}{\wh E_V} \quad \forall x \in \wh E_V
+    \]
+
+    and $\sum_{n \in \natp}\norm{y_n}_{\wh E_U}\norm{\phi_n}_{E_V^*} < \infty$. 
+
+    Now, using \autoref{theorem:nuclear-lp}, further assume without loss of generality that $\wh E_V$ is a Hilbert space. Identify $(\widehat{E/F})_{p(V)}$ as a closed subspace of $\widehat E_V$, and let $P: \widehat E_V \to (\widehat{E/F})_{p(V)}$ be the orthogonal projection of $\widehat E_V$ onto $(\widehat{E/F})_{p(V)}$. This allows rewriting
+    \[
+        \widehat \pi_{p(U)}x = \sum_{n = 1}^\infty \widehat p(y_n) \dpn{Px, \phi_n}{\wh E_V} = \sum_{n = 1}^\infty \widehat p(y_n) \dpn{x, P\phi_n}{(\widehat{E/F})_{p(V)}}
+    \]
+
+    where
+    \begin{align*}
+    \normn{\widehat \pi_{p(U)}}_{N((\widehat{E/F})_{p(V)}; (\widehat{E/F})_{p(U)})} &\le \sum_{n \in \natp}\normn{\widehat p(y_n)}_{(\widehat{E/F})_{p(U)}}\norm{P\phi_n}_{(\widehat{E/F})_{p(V)}} \\
+    &\le \sum_{n \in \natp}\norm{y_n}_{\wh E_U}\norm{\phi_n}_{E_V^*} < \infty
+    \end{align*}
+
+    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}
+
+\begin{proposition}
+\label{proposition:nuclear-product}
+    Let $\seqi{E}$ be nuclear spaces over $K \in \RC$, then $\prod_{i \in I}E_i$ is nuclear. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem III.7.4]{SchaeferWolff}}}. ]
+    Let $F$ be a Banach space and $T \in L(\prod_{i \in I}E_i; F)$, then there exists $J \subset I$ finite and $\wh T \in L(\prod_{j \in J}E_j; F)$ such that the following diagram commutes: 
+    \[
+    \xymatrix{
+    \prod_{i \in I} E_i \ar@{->}[r]^{T} \ar@{->}[d]_{\pi_J} & F \\
+    \prod_{j \in J}E_j \ar@{->}[ru]_{\widehat T} & 
+    }
+    \]
+
+    By \autoref{proposition:finite-lc-product}, $\prod_{j \in J}E_j = \bigoplus_{j \in J}E_j$. By \autoref{proposition:nuclear-direct-sum}, $\bigoplus_{j \in J}E_j$ is a nuclear space, so $\widehat T: \bigoplus_{j \in J}E_j \to F$ is a nuclear operator. As the composition of a continuous operator and a nuclear operator, $T$ is nuclear by \autoref{proposition:nuclear-gymnastics}. Therefore $\prod_{i \in I}E_i$ is a nuclear space. 
+\end{proof}
+