From 42eeae1679907ae23a3c315c9405aa117b1c4aa8 Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Wed, 15 Jul 2026 17:16:40 -0400
Subject: [PATCH] Every product of nuclear spaces is nuclear.

---
 src/fa/lc/nuclear-space.tex | 28 ++++++++++++++++++++++++++++
 1 file changed, 28 insertions(+)

diff --git a/src/fa/lc/nuclear-space.tex b/src/fa/lc/nuclear-space.tex
index 63dcc2b..939ed09 100644
--- a/src/fa/lc/nuclear-space.tex
+++ b/src/fa/lc/nuclear-space.tex
@@ -86,6 +86,15 @@
     Let $E$ be a complete nuclear space over $K \in \RC$, then $E$ is a projective limit of Hilbert spaces over $K$. For any Fréchet space $F$, $F$ is nuclear if and only if it is the projective limit of a sequence $\seq{H_n}$ of Hilbert spaces such that the mapping $H_m \to H_n$ is nuclear for all $1 \le m < n < \infty$. 
 \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.
@@ -139,6 +148,8 @@
 \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. 
@@ -211,3 +222,20 @@
 
     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}
+