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\section{Nuclear Operators}
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\label{section:nuclear-operator}
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\begin{definition}[Nuclear Operator Between Banach Spaces]
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\label{definition:nuclear-operator-normed}
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Let $E, F$ be Banach spaces, $E^*$ be the dual of $E$, equipped with the uniform topology, and $T \in L(E; F)$, then $T$ is \textbf{nuclear} if there exists $\seq{\phi_n} \subset E^*$ and $\seq{y_n} \subset F$ such that:
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\begin{enumerate}
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\item For each $x \in E$, $Tx = \sum_{n = 1}^\infty y_n \dpn{x, \phi_n}{E}$.
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\item $\sum_{n \in \natp}\norm{y_n}_F\norm{\phi_n}_{E^*} < \infty$.
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\end{enumerate}
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The set $N(E; F)$ is the \textbf{space of nuclear operators} from $E$ to $F$. For each $T \in N(E; F)$, let
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\[
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\norm{T}_{N(E; F)} = \inf\bracs{\sum_{n \in \natp}\norm{y_n}_F\norm{\phi_n}_{E^*} \bigg | Tx = \sum_{n = 1}^\infty y_n \dpn{x, \phi_n}{E} \forall x \in E}
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\]
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then $\norm{\cdot}_{N(E; F)}$ is a norm on $N(E; F)$, and $N(E; F)$ is a Banach space.
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\end{definition}
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\begin{lemma}
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\label{lemma:nuclear-operator-normed-tensor}
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Let $E, F$ be Banach spaces, $E^*$ be the dual of $E$, equipped with the uniform topology, then the mapping
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\[
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E^* \otimes F \to N(E; F) \quad \sum_{j = 1}^n \phi_j \otimes y_j \mapsto \sum_{j = 1}^n y_j\dpn{\cdot, \phi_j}{E}
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\]
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extends continuously into a surjective linear map $E^* \tilde \otimes_\pi F \to N(E; F)$.
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\end{lemma}
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\begin{definition}[Nuclear Operator]
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\label{definition:nuclear-operator}
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Let $E, F$ be separated locally convex spaces over $K \in \RC$ and $T \in L(E; F)$, then the following are equivalent:
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\begin{enumerate}
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\item There exists convex and circled sets $U \in \cn_E(0)$ and $B \in B(F)$ such that:
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\begin{enumerate}[label=(\alph*)]
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\item The auxiliary space $F_B$ is a Banach space.
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\item $T(U) \subset B$.
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\item The induced map $\wh E_U \to F_B$ is nuclear.
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\end{enumerate}
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\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
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\begin{enumerate}[label=(\alph*)]
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\item The auxiliary space $F_B$ is a Banach space.
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\item For each $x \in E$, $Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}$.
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\item $\sum_{n \in \natp}|\lambda_n| < \infty$.
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\end{enumerate}
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\end{enumerate}
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If the above holds, then $T$ is \textbf{nuclear}. The set $N(E; F)$ is the \textbf{space of nuclear operators} from $E$ to $F$.
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\end{definition}
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\begin{proof}[Proof, {{\cite[Theorem III.7.1]{SchaeferWolff}}}. ]
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(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:
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\[
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\xymatrix{
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E \ar@{->}[r]^{T} \ar@{->}[d]_{\pi} & F \\
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E_U \ar@{->}[r]_{\hat T} & F_B \ar@{->}[u]_{\iota}
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}
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\]
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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:
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\begin{enumerate}[label=(\roman*)]
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\item $\sum_{n \in \natp}|\lambda_n| < \infty$.
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\item $\limv{n}\phi_n = 0$ and $\limv{n}y_n = 0$.
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\item For each $x \in E_U$, $\hat Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E_U}$.
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\end{enumerate}
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By (ii), $\sup_{n \in \natp}\norm{\phi_n}_{E_U^*} < \infty$ and $\sup_{n \in \natp}\norm{y_n}_{F_B} < \infty$, so $\seq{\phi_n}$ is equicontinuous, and there exists $R > 0$ such that $\seq{y_n} \subset RB$. After rescaling, assume without loss of generality that $\seq{y_n} \subset B$. By unraveling the factorisation, (iii) shows that for each $x \in E$,
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\[
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Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n \circ \pi}{E}
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\]
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Therefore the decomposition using $\seq{\phi_n \circ \pi} \subset E^*$, $B \subset F$, $\seq{y_n} \subset B$, and $\seq{\lambda_n} \subset K$ given above satisfies (2).
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(2) $\Rightarrow$ (1): Since $\seq{\phi_n}$ is equicontinuous, $U = \bigcap_{n \in \natp}\phi_n^{-1}(B_K(0, 1))$ is a convex and circled neighbourhood of $0$ in $E$.
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(1b): Using assumption (2c) and rescaling, assume without loss of generality that $\sum_{n = 1}^\infty |\lambda_n| < 1$. Let $\rho: F_B \to [0, \infty)$ be the gauge of $B$, then for any $x \in U$,
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\begin{align*}
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\rho(Tx) &= \rho\braks{\sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}} \le \sum_{n \in \natp} |\lambda_n| \cdot \underbrace{|\dpn{x, \phi_n}{E}|}_{\le 1} \cdot \underbrace{\rho(y_n)}_{\le 1} \\
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&\le \sum_{n \in \natp}|\lambda_n| < 1
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\end{align*}
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so $\rho(Tx) < 1$ and $Tx \in B$. Therefore $T(U) \subset B$.
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(1c): Let $\pi: E \to E_U$ be the canonical projection map associated with $E_U$ and $n \in \natp$. By construction, $U \subset \phi_n^{-1}(B_K(0, 1))$, so there exists $\hat \phi_n \in E_U^*$ such that the following diagram commutes:
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\[
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\xymatrix{
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E \ar@{->}[d]_{\pi} \ar@{->}[rd]^{\phi_n} & \\
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E_U \ar@{->}[r]_{\hat \phi_n} & K
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}
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\]
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Thus for each $x \in E$,
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\[
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Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E} = \sum_{n = 1}^\infty y_n \dpn{\pi(x), \lambda_n \hat \phi_n}{E_U}
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\]
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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
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\[
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\normn{\hat T}_{N(\wh E_U; F_B)} \le \sum_{n \in \natp}|\lambda_n| \cdot \norm{y_n}_{F_B} \cdot \normn{\hat \phi_n}_{E_U^*} \le \sum_{n \in \natp}|\lambda_n| < \infty
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\]
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and $\hat T: \wh E_U \to F_B$ is nuclear.
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\end{proof}
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\begin{proposition}
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\label{proposition:nuclear-gymnastics}
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Let $E, F, G, H$ be separated locally convex spaces and $S \in N(F; G)$, then:
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\begin{enumerate}
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\item $S$ is compact.
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\item For any $T \in L(E; F)$, $S \circ T \in N(E; G)$.
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\item For any $R \in L(G; H)$, $R \circ S \in N(F; H)$.
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\item There exists a unique $\wh S \in L(\wh F; G)$ such that $\wh S|_{F} = S$. Moreover, $\wh S \in N(\wh F; G)$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Corollary III.7.1.1-III.7.1.3]{SchaeferWolff}}}. ]
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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
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\begin{enumerate}[label=(\alph*)]
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\item The auxiliary space $G_B$ is a Banach space.
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\item For each $x \in F$, $Sx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{F}$.
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\item $\sum_{n \in \natp}|\lambda_n| < \infty$.
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\end{enumerate}
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(1): Let $U = \bigcap_{n \in \natp}\phi_n^{-1}(B_K(0, 1))$, then since $\seq{\phi_n}$ is equicontinuous, $U$ is a convex and circled neighbourhood of $0$ in $F$. Given that $G_B$ is complete, $S$ is the following composition of continuous maps:
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\[
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\begin{CD}
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U @>{\prod_{n \in \natp} \phi_n}>> \overline{B_K(0,1)}^{\natp} @>{x \mapsto \sum_{n=1}^\infty \lambda_n x_n y_n}>> G_B @>>> G
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\end{CD}
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\]
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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.
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(2): Since $T \in L(E; F)$, $\seq{\phi_n \circ T} \subset E^*$ is equicontinuous. Thus for any $x \in E$,
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\[
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(S \circ T)x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n \circ T}{E}
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\]
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and $S \circ T \in N(E; G)$.
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(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 $x \in F$,
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\[
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(R \circ S)x = \sum_{n = 1}^\infty \lambda_n R(y_n) \dpn{x, \phi_n}{F}
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\]
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and $R \circ S \in N(F; H)$.
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(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
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\[
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\wh S x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \wh \phi_n}{\wh F}
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\]
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Therefore $\wh S \in N(\wh F; G)$.
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\end{proof}
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