From d1ddf9f64ba3dad95190410b44c95230ad4134a6 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 13 Jul 2026 15:36:06 -0400 Subject: [PATCH] Added nuclear operators. --- src/fa/lc/index.tex | 1 + src/fa/lc/nuclear.tex | 32 ++++++++++++++++---------------- 2 files changed, 17 insertions(+), 16 deletions(-) diff --git a/src/fa/lc/index.tex b/src/fa/lc/index.tex index b605d8a..204bed6 100644 --- a/src/fa/lc/index.tex +++ b/src/fa/lc/index.tex @@ -13,3 +13,4 @@ \input{./hahn-banach.tex} \input{./spaces-of-linear.tex} \input{./tensor.tex} +\input{./nuclear.tex} diff --git a/src/fa/lc/nuclear.tex b/src/fa/lc/nuclear.tex index d9c4f49..3f07cd1 100644 --- a/src/fa/lc/nuclear.tex +++ b/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}