Added nuclear spaces.

src/fa/lc/convex.tex

@@ -159,7 +159,7 @@
         \item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
         \item If $A$ is convex, then for any $x, y \in E$, $[x + y]_A \le [x]_A + [y]_A$.
         \item If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_A = \abs{\lambda}[x]_A$.
-        \item If $A$ is circled, then $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$.
+        \item If $A$ is circled, then $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$, with respect to any vector space topology on $E$. 
     \end{enumerate}
 
     In particular,

src/fa/lc/hahn-banach.tex

@@ -40,7 +40,7 @@
     Let $E$ be a vector space over $K \in \RC$, $\rho: E \to \real$ be a sublinear functional, and $F \subsetneq E$ be a subspace, then
     \begin{enumerate}
         \item For any $\phi \in \hom(F; \real)$ with $\phi \le \rho|_F$, there exists $\Phi \in \hom(E; \real)$ such that $\Phi \le \rho$ and $\Phi|_F = \phi$.
-        \item If $\rho$ is a seminorm, then for any $\phi \in \hom(F; \complex)$ with $\abs{\phi} \le \rho|_F$, there exists $\Phi \in \hom(E; \complex)$ such that $\abs{\Phi} \le \rho$ and $\Phi|_F = \phi$.
+        \item If $\rho$ is a seminorm, then for any $\phi \in \hom(F; K)$ with $\abs{\phi} \le \rho|_F$, there exists $\Phi \in \hom(E; K)$ such that $\abs{\Phi} \le \rho$ and $\Phi|_F = \phi$.
     \end{enumerate}
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 5.6, 5.7]{Folland}}}. ]
@@ -60,7 +60,7 @@
 
     By Zorn's lemma, $\mathbf{F}$ admits a maximal element $\Phi$. If $\cd(\Phi) \subsetneq E$, then $\Phi$ is not maximal by the preceding discussion. Therefore $\cd(\Phi) = E$ and $\Phi$ is a desired extension.
 
-    (2): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$.
+    (2): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$. Assume without loss of generality that $K = \complex$. 
 
     Let $u = \re{\phi}$, then $u \in \hom(E; \real)$ by \autoref{proposition:polarisation-linear}. By (1), there exists $U \in \hom(E; \real)$ such that $\abs{U} \le \rho$ and $U|_F = u$. For each $x \in E$, let $\Phi(x) = U(x) - iU(ix)$, then $\Phi \in \hom(E; \complex)$ and $\Phi|_F = \phi$ by \autoref{proposition:polarisation-linear}. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then
     \[

src/fa/lc/index.tex

@@ -13,3 +13,5 @@
 \input{./hahn-banach.tex}
 \input{./spaces-of-linear.tex}
 \input{./tensor.tex}
+\input{./nuclear.tex}
+\input{./nuclear-space.tex}

src/fa/lc/nuclear-space.tex

@@ -0,0 +1,85 @@
+\section{Nuclear Spaces}
+\label{section:nuclear-space}
+
+\begin{definition}[Nuclear Space]
+\label{definition:nuclear-space}
+    Let $E$ be a separated locally convex space over $K \in \RC$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists a fundamental system of convex and circled neighbourhoods $\fB \subset \cn_E(0)$ such that for each $U \in \fB$, the canonical projection $\pi_U: E \to \wh E_U$ is nuclear. 
+        \item For each Banach space $F$ and $T \in L(E; F)$, $T$ is nuclear. 
+        \item For each convex and circled neighbourhood $U \in \cn_E(0)$, there exists $V \in \cn_E(0)$ with $V \subset U$ such that the induced map $\wh E_V \to \wh E_U$ is nuclear. 
+    \end{enumerate}
+    
+    If the above holds, then $E$ is a \textbf{nuclear space}. 
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $U = T^{-1}(B_F(0, 1))$, then there exists $V \in \fB$ with $V \subset U$. In which case, there exists $\wh T \in L(\wh E_V; F)$ such that the following diagram commutes: 
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{T} \ar@{->}[d]_{\pi_V} & F \\
+    \wh E_V \ar@{->}[ru]_{\wh T} & 
+    }
+    \]
+
+    Since $\pi_V \in N(E; \wh E_V)$, $T = \wh T \circ \pi_V$ is nuclear by \autoref{proposition:nuclear-gymnastics}. 
+
+    (2) $\Rightarrow$ (3): Let $U \in \cn_E(0)$, then the canonical map $\pi_U: E \to \wh E_U$ is nuclear. Thus there exists an equicontinuous sequence $\seq{\phi_n} \subset E^*$, $\seq{y_n} \subset B_{\wh E_U}(0, 1)$, and $\seq{\lambda_n} \subset K$ such that
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $x \in E$, $\pi_U x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}$. 
+        \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+    \end{enumerate}
+
+    Let $V = U \cap \bigcap_{n \in \natp}\phi_n^{-1}(B_K(0, 1))$, then by equicontinuity of $\seq{\phi_n}$, $V \in \cn_E(0)$. Moreover, for each $n \in \natp$, there exists $\wh \phi_n \in \wh E_V^*$ such that the following diagram commutes:
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\phi_n} \ar@{->}[d]_{\pi_V} & K \\
+    \wh E_{V} \ar@{->}[ru]_{\widehat \phi_n} & 
+    }
+    \]
+    
+    As $V \subset \phi_n^{-1}(B_K(0, 1))$, $\normn{\widehat \phi_n}_{\wh E_V^*} \le 1$. Thus the induced map $\widehat \pi_U: \wh E_{V} \to \wh E_U$ takes the form
+    \[
+    \wh \pi_U x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \wh \phi_n}{\wh E_{V}}
+    \]
+
+    with
+    \[
+    \normn{\wh \pi_U}_{N(\wh E_{V}; \wh E_U)} \le \sum_{n \in \natp}|\lambda_n| \cdot \underbrace{\norm{y_n}_{\wh E_U}}_{\le 1} \cdot \underbrace{\normn{\wh \phi_n}_{\wh E_V^*}}_{\le 1} \le \sum_{n \in \natp}|\lambda_n| < \infty
+    \]
+
+    Therefore $\wh \pi_U$ is nuclear. 
+
+    (3) $\Rightarrow$ (1): Let $U \in \cn_E(0)$ be convex and circled, then there exists a convex circled neighbourhood $V \in \cn_E(0)$ such that the induced map $\wh \pi_U: \wh E_V \to \wh E_U$ is nuclear. In which case, the canonical map $\pi_U: E \to \wh E_U$ is the composition of $\pi_V$ and $\wh \pi_U$. Thus $\pi_U: E \to \wh E_U$ is nuclear by \autoref{proposition:nuclear-gymnastics}. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:nuclear-lp}
+    Let $E$ be a nuclear space over $K \in \RC$, $U \in \cn_E(0)$, and $p \in [1, \infty]$, then there exists $V \in \cn_E(0)$ with $V \subset U$ such that $\wh E_V$ is isomorphic to a subspace of $l^p(\natp; K)$ with equal norms.  
+\end{theorem}
+\begin{proof}[Proof, {{\cite[III.7.3]{SchaeferWolff}}}. ]
+    Assume without loss of generality that $U$ is convex and circled, and the canonical projection $\pi_U: E \to \wh E_U$ is nuclear. In which case, there exists an equicontinuous sequence $\seq{\phi_n} \subset E^*$, $\seq{y_n} \subset B_{\wh E_U}(0, 1)$, and $\seq{\lambda_n} \subset K$ such that
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $x \in E$, $\pi_U x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}$. 
+        \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+    \end{enumerate}
+
+    By rescaling, further assume without loss of generality that $\sum_{n \in \natp}|\lambda_n| = 1$ and $\lambda_n > 0$ for all $n \in \natp$. Under the convention that $1/\infty = 0$, define
+    \[
+    T: E \to l^p(\natp; K) \quad (Tx)_n = \lambda_n^{1/p}\dpn{x, \phi_n}{E}
+    \]
+
+    then for each $x \in E$, $\norm{Tx}_{l^p(\natp; K)} \le \norm{\pi_U x}_{\wh E_U}$, so $T$ is continuous. 
+    
+
+    On the other hand, 
+    \[
+        \normn{\pi_U x}_{\wh E_U} = \norm{\sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}}_{\wh E_U} \le \sum_{n = 1}^\infty \lambda_n |\dpn{x, \phi_n}{E}|
+    \]
+
+    Let $q \in [1, \infty]$ be the Hölder conjugate of $p$. By \hyperref[Hölder's inequality]{theorem:holder} applied to $\bracsn{\lambda_n^{1/p}\dpn{x, \phi_n}{E}}_1^\infty$ and $\bracsn{\lambda_n^{1/q}}_1^\infty$, $\normn{\pi_U x}_{\wh E_U} \le \norm{Tx}_{l^p(\natp; K)}$. 
+
+    Finally, let $V = T^{-1}(B_{l^p(\natp; K)})$, then $V \subset U$, and $\wh E_V$ is isomorphic to $\ol{T(E)}$, with equal norms. 
+\end{proof}
+
+
+

src/fa/lc/nuclear.tex

@@ -0,0 +1,158 @@
+\section{Nuclear Operators}
+\label{section:nuclear-operator}
+
+\begin{definition}[Nuclear Operator Between Banach Spaces]
+\label{definition:nuclear-operator-normed}
+    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:
+    \begin{enumerate}
+        \item For each $x \in E$, $Tx = \sum_{n = 1}^\infty y_n \dpn{x, \phi_n}{E}$. 
+        \item $\sum_{n \in \natp}\norm{y_n}_F\norm{\phi_n}_{E^*} < \infty$. 
+    \end{enumerate}
+
+    The set $N(E; F)$ is the \textbf{space of nuclear operators} from $E$ to $F$. For each $T \in N(E; F)$, let
+    \[
+    \norm{T}_{N(E; F)} = \inf\bracs{\sum_{n \in \natp}\norm{y_n}_F\norm{\phi_n}_{E^*} \bigg | Tx = \sum_{n = 1}^\infty y_n \dpn{x, \phi_n}{E} \forall x \in E}
+    \]
+
+    then $\norm{\cdot}_{N(E; F)}$ is a norm on $N(E; F)$, and $N(E; F)$ is a Banach space. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:nuclear-operator-normed-tensor}
+    Let $E, F$ be Banach spaces, $E^*$ be the dual of $E$, equipped with the uniform topology, then the mapping
+    \[
+    E^* \otimes F \to N(E; F) \quad \sum_{j = 1}^n \phi_j \otimes y_j \mapsto \sum_{j = 1}^n y_j\dpn{\cdot, \phi_j}{E}
+    \]
+
+    extends continuously into a surjective linear map $E^* \tilde \otimes_\pi F \to N(E; F)$. 
+\end{lemma}
+
+
+
+
+\begin{definition}[Nuclear Operator]
+\label{definition:nuclear-operator}
+    Let $E, F$ be separated locally convex spaces over $K \in \RC$ and $T \in L(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists convex and circled sets $U \in \cn_E(0)$ and $B \in B(F)$ such that:
+        \begin{enumerate}[label=(\alph*)]
+            \item The auxiliary space $F_B$ is a Banach space.
+            \item $T(U) \subset B$. 
+            \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*)]
+            \item The auxiliary space $F_B$ is a Banach space. 
+            \item For each $x \in E$, $Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}$. 
+            \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+        \end{enumerate}
+    \end{enumerate}
+    
+    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:
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{T} \ar@{->}[d]_{\pi} & F \\
+    E_U \ar@{->}[r]_{\hat T} & F_B \ar@{->}[u]_{\iota}
+    }
+    \]
+
+    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 $\limv{n}y_n = 0$. 
+        \item For each $x \in E_U$, $\hat Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E_U}$. 
+    \end{enumerate}
+
+    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 $x \in E$, 
+    \[
+    Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n \circ \pi}{E}
+    \]
+
+    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).
+
+    (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 $E$. 
+    
+    (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 $x \in U$, 
+    \begin{align*}
+        \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} \\
+        &\le \sum_{n \in \natp}|\lambda_n| < 1
+    \end{align*}
+
+    so $\rho(Tx) < 1$ and $Tx \in B$. Therefore $T(U) \subset B$. 
+
+    (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: 
+    \[
+    \xymatrix{
+    E \ar@{->}[d]_{\pi} \ar@{->}[rd]^{\phi_n} &  \\
+    E_U \ar@{->}[r]_{\hat \phi_n} & K
+    }
+    \]
+
+    Thus for each $x \in E$, 
+    \[
+    Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E} = \sum_{n = 1}^\infty y_n \dpn{\pi(x), \lambda_n \hat \phi_n}{E_U}
+    \]
+    
+    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
+    \[
+        \normn{\hat T}_{N(\wh E_U; F_B)} \le \sum_{n \in \natp}|\lambda_n| \cdot \norm{y_n}_{F_B} \cdot \normn{\hat \phi_n}_{E_U^*} \le \sum_{n \in \natp}|\lambda_n| < \infty
+    \]
+
+    and $\hat T: \wh E_U \to F_B$ is nuclear. 
+    
+    
+\end{proof}
+
+\begin{proposition}
+\label{proposition:nuclear-gymnastics}
+    Let $E, F, G, H$ be separated locally convex spaces and $S \in N(F; G)$, then:
+    \begin{enumerate}
+        \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 $\wh S \in L(\wh F; G)$ 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 $x \in F$, $Sx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{F}$. 
+        \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+    \end{enumerate}
+
+    (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:
+    \[
+    \begin{CD}
+    U @>{\prod_{n \in \natp} \phi_n}>> \overline{B_K(0,1)}^{\natp} @>{x \mapsto \sum_{n=1}^\infty \lambda_n x_n y_n}>> G_B @>>> G
+    \end{CD}
+    \]
+
+    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 $T \in L(E; F)$, $\seq{\phi_n \circ T} \subset E^*$ 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, $R(B)$ 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}, 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
+    \[
+    \wh S x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \wh \phi_n}{\wh F}
+    \]
+
+    Therefore $\wh S \in N(\wh F; G)$.
+\end{proof}
+
+

src/fa/lc/tensor.tex

@@ -135,7 +135,7 @@
     (5): By (6) of \autoref{definition:projective-tensor-product}. 
 \end{proof}
 
-\begin{theorem}[{{\cite[III.6.4]{SchaeferWolff}}}]
+\begin{theorem}
 \label{theorem:metrisable-tensor-product}
     Let $E, F$ be metrisable locally convex spaces over $K \in \RC$, then for any $z \in E \td{\otimes}_\pi F$, there exists $\seq{\lambda_n} \subset K$ and $\seq{(x_j, y_j)} \subset E \times F$ such that:
     \begin{enumerate}
@@ -146,7 +146,7 @@
 
     
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[III.6.4]{SchaeferWolff}}}.]
     Let $\seq{p_n}$ and $\seq{q_n}$ be increasing sequences of continuous seminorms that induce the topology on $E$ and $F$, respectively. For each $n \in \natp$, let $r_n = p_n \otimes q_n$, and $\td r_n$ be the continuous extension of $r_n$ to $E \td{\otimes}_\pi F$. 
 
     Let $u \in E \td{\otimes}_\pi F$, then there exists $\seq{u_n} \subset E \otimes_\pi F$ such that $\td r_n(u - u_n) < 2^{-n}/n^2$ for all $n \in \natp$. For each $N \in \natp$, let $v_N = u_{N+1} - u_N$, then

src/fa/norm/hilbert.tex

@@ -369,3 +369,4 @@ A significant property of Hilbert spaces is that every closed subspace is comple
 \end{proof}
 
 
+

src/fa/notation.tex

@@ -26,6 +26,7 @@
     $E \otimes_\pi F$ & Projective tensor product of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
     $E \,\widetilde{\otimes}_\pi F$ & Projective completion of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
     $p \otimes q$ & Cross seminorm of $p$ and $q$. & \autoref{definition:cross-seminorm} \\
+    $N(E; F)$ & Nuclear mappings from $E$ to $F$. & \autoref{definition:nuclear-operator-normed} \\ 
     % ---- Order Structures ----
     $x \vee y$, $x \wedge y$ & $\sup$ and $\inf$ in vector lattice. & \autoref{definition:vector-lattice} \\
     $|x|$ & Absolute value $x \vee (-x)$ in a vector lattice. & \autoref{definition:order-absolute-value} \\

src/op/c-star/gns.tex

@@ -0,0 +1,176 @@
+\section{The GNS Construction}
+\label{section:gns}
+
+\begin{definition}[Cyclic Representation]
+\label{definition:cyclic-representation}
+    Let $A$ be a $C^*$-algebra, $(H, \pi)$ be a representation of $A$, and $\xi \in H$, then $\xi$ is a \textbf{cyclic vector} for $(H, \pi)$ if $\bracsn{\pi(x)(\xi)|x \in A}$ is dense in $H$. The representation $(H, \pi)$ is \textbf{cyclic} if it admits a cyclic vector. 
+\end{definition}
+
+
+\begin{lemma}
+\label{lemma:cstar-state-kernel}
+    Let $A$ be a $C^*$-algebra, $\phi \in S(A)$, and
+    \[
+    N_\phi = \bracsn{x \in A| \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = 0}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item For any $x, y \in A$ with $x \in N_\phi$ or $y \in N_\phi$, $\dpn{x, y}{A} = 0$. 
+        \item $N_\phi$ is a closed left ideal of $A$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, for any $x, y \in A$, 
+    \[
+    |\dpn{x, y}{\phi}|^2 \le \dpn{x, x}{\phi} \cdot \dpn{y, y}{\phi}
+    \]
+
+    If $x \in N_\phi$ or $y \in N_\phi$, then the above inequality shows that $\dpn{x, y}{\phi} = 0$. 
+
+    (2): As the zero set of a continuous function on $A$, $N_\phi$ is closed.
+
+    For any $x, y \in N_\phi$, 
+    \begin{align*}
+        \dpn{x + y, x + y}{\phi} &= \dpn{x, x}{\phi} + \dpn{x, y}{\phi} + \dpn{y, x}{\phi} + \dpn{y, y}{\phi} \\
+        &= \dpn{x, y}{\phi} + \dpn{y, x}{\phi} 
+    \end{align*}
+
+    By (1), $\dpn{x, y}{\phi} = \dpn{y, x}{\phi} = 0$. Therefore $x + y \in N_\phi$. 
+
+    Finally, for each $x \in N_\phi$ and $y \in A$, 
+    \[
+        \dpn{yx, yx}{\phi} = \dpn{x^*y^*yx, \phi}{A} = \dpn{x^*(y^*yx), \phi}{A} = \dpn{y^*yx, x}{\phi} =0 
+    \]
+
+    by (1). 
+\end{proof}
+
+
+\begin{definition}[GNS Triple]
+\label{definition:gns-triple}
+    Let $A$ be a unital $C^*$-algebra, $\phi \in S(A)$, and
+    \[
+    N_\phi = \bracsn{x \in A| \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = 0}
+    \]
+
+    Let $H_\phi^0 = A/N_\phi$, $H_\phi$ be its completion with respect to $\dpn{\cdot, \cdot}{\phi}$, and
+    \[
+    \pi_\phi^0: A \to B(H_\phi^0) \quad \pi_\phi^0(x)(y + N_\phi) = xy + N_\phi
+    \]
+
+    For each $x \in A$, let $\pi_\phi(x)$ be the continuous extension of $\pi_\phi^0(x)$ to an element of $B(H_\phi)$, then:
+    \begin{enumerate}
+        \item $(H_\phi, \dpn{\cdot, \cdot}{\phi})$ is a Hibert space.
+        \item $(H_\phi, \pi_\phi)$ is a well-defined representation of $A$. 
+        \item $\xi_\phi = 1_A + N_\phi$ is a unit vector in $H_\phi$, and $\bracsn{\pi_\phi(x)\xi_\phi| x \in A}$ is dense in $H_\phi$. Moreover, for each $x, y \in A$,
+        \[
+        \dpn{x, y}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}
+        \]
+    \end{enumerate}
+    
+    The representation $(H_\phi, \pi_\phi)$ is the \textbf{cyclic representation of $A$ induced by $\phi$}, and the triple $(H_\phi, \pi_\phi, \xi_\phi)$ is the \textbf{Gelfand-Naimark-Segal (GNS) triple associated with $\phi$}. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition 14.2]{Zhu}}}. ]
+    (2): Fix $x \in A$, then for each $y_1, y_2 \in A$ with $y_1 - y_2 \in N_\phi$, $x(y_1 - y_2) \in N_\phi$ by \autoref{lemma:cstar-state-kernel}, so $\pi_\phi^0(x)$ is well-defined on $A/N_\phi$. 
+
+    By rescaling, assume without loss of generality that $\norm{x}_A \le 1$. In which case, for each $y \in A$, 
+    \[
+        \dpn{y, y}{\phi} - \dpn{xy, xy}{\phi} = \dpn{y^*y, \phi}{A} - \dpn{y^*x^*xy, \phi}{A} = \dpn{y^*(1 - x^*x)y, \phi}{A}
+    \]
+
+    Since $\sigma_A(x^*x) \subset [0, 1]$, $\sigma_A(1 - x^*x) \subset [0, 1]$ and is positive by \autoref{corollary:spectrum-characterisation-iff}. Thus there exists $z \in A$ positive such that $(1 - x^*x) = z^*z$, so
+    \[
+        \dpn{y, y}{\phi} - \dpn{xy, xy}{\phi} = \dpn{y^*z^*zy, \phi}{A} = \dpn{zy, zy}{\phi} \ge 0
+    \]
+
+    and $\dpn{y, y}{\phi} \ge \dpn{xy, xy}{\phi}$. Therefore $\pi_\phi^0(x)$ extends continuously into an element of $B(H_\phi)$ by the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-normed}. 
+    
+
+    Now, let $x, y, z \in A$, then
+    \[
+    \pi_\phi^0(x)[\pi_\phi^0(y)(z + N_\phi)] = \pi_\phi^0(x)(yz + N_\phi) = xyz + N_\phi = \pi_\phi^0(xy)(z + N_\phi)
+    \]
+
+    and by uniqueness of continuous extensions, $\pi_\phi(x)\pi_\phi(y) = \pi_\phi(xy)$, so $\pi_\phi$ is a homomorphism. 
+
+    Finally, 
+    \[
+        \dpn{\pi_\phi^0(x^*)y, z}{\phi} = \dpn{z^*x^*y, \phi}{A} = \dpn{y, xz}{\phi} = \dpn{y, \pi_\phi^0(x)z}{\phi}
+    \]
+
+    By uniqueness of continuous extensions, $\pi_\phi(x^*) = \pi_\phi(x)^*$. Therefore $\pi_\phi$ is a *-homomorphism, and $(H_\phi, \pi_\phi)$ is a representation of $A$. 
+
+    (3): Since $\phi$ is a state, $\dpn{1_A, 1_A}{\phi} = 1$, and $1_A$ is a unit vector. As $H_\phi$ is the completion of $A/N_\phi$ and $A/N_\phi = \bracsn{\pi_\phi(x)(1_A + N_\phi)| x \in A}$, $\bracsn{\pi_\phi(x)(1_A + N_\phi)| x \in A}$ is dense in $H_\phi$. 
+
+    For each $x, y \in A$, $\dpn{x, y}{\phi} = \dpn{\pi_\phi(x)(1_A + N_\phi), \pi_\phi(y)(1_A + N_\phi)}{H_\phi}$ by well-definedness of the inner product on $H_\phi$.
+\end{proof}
+
+\begin{theorem}[Gelfand-Naimark-Segal]
+\label{theorem:gns}
+    Let $A$ be a unital $C^*$-algebra, then:
+    \begin{enumerate}
+        \item For each $\phi \in S(A)$, there exists a triple $(H_\phi, \pi_\phi, \xi_\phi)$ where $(H_\phi, \pi_\phi)$ is a representation of $A$, $\xi_\phi$ is a cyclic unit vector of $(H_\phi, \pi_\phi)$, and
+        \[
+        \dpn{x, y}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}
+        \]
+        \item For each representation $(H, \pi)$ of $A$ with cyclic unit vector $\xi$, the mapping
+        \[
+        \phi: A \to \complex \quad x \mapsto \dpn{\pi(x)\xi, \xi}{H}
+        \]
+
+        is a state on $A$. Moreover, if $(H_\phi, \pi_\phi, \xi_\phi)$ is the GNS triple associated with $\phi$, then there exists a unitary equivalence $U: H \to H_\phi$ such that $U\xi = \xi_\phi$. 
+        \item For each $\mathcal{S} \subset S(A)$, the mapping
+        \[
+        \pi_{\mathcal{S}}: A \to B([l^2(\mathcal{S}); H_\phi]) \quad \pi_{\mathcal{S}}(x)(\eta)_\phi = \pi_{\phi}(x)(\eta_\phi)
+        \]
+
+        is a representation of $A$, which is injective if for every $x \in A$, there exists $\phi \in \mathcal{S}$ with $\dpn{x^*x, \phi}{A} \ne 0$. 
+
+        In particular, $A$ is isomorphic to a closed subalgebra of $B([l^2(P(A)); H_\phi])$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): By the \hyperref[GNS construction]{definition:gns-triple}. 
+
+    (2): For each $x \in A$, if $x$ is positive, then so is $\pi(x)$, so $\dpn{\pi(x)\xi, \xi}{H} \ge 0$. Since $\xi$ is a unit vector, $\dpn{\pi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H} = 1$, and $\phi$ is a state. 
+
+    Let $H^0 = \bracsn{\pi(x)\xi|x \in A}$ and $H_\phi^0 = \bracsn{\pi_\phi(x)\xi_\phi|x \in A}$. Define
+    \[
+    U: H^0 \to H_\phi^0 \quad \pi(x)\xi \mapsto \pi_\phi(x)\xi_\phi
+    \]
+
+    then for each $x, y \in A$ with $\pi(x - y)\xi = 0$,
+    \begin{align*}
+    0 &= \dpn{\pi(x - y)\xi, \pi(x - y)\xi}{H} = \dpn{(x - y)^*(x - y), \phi}{A} \\
+    &= \dpn{x - y, x- y}{\phi} = \dpn{\pi_\phi(x - y)\xi_\phi, \pi_\phi(x - y)\xi_\phi}{H_\phi}
+    \end{align*}
+    
+
+    and $\pi_\phi(x - y)\xi_\phi = 0$ as well. Thus $U$ is well-defined. Moreover, for each $x \in A$,
+    \[
+        \dpn{\pi(x)\xi, \pi(x)\xi}{H} = \dpn{x^*x, \phi}{A} = \dpn{x^*x, 1_A}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(x) \xi_\phi}{H_\phi}
+    \]
+
+    so $U$ is an isometry. For each $x, y \in A$,
+    \begin{align*}
+    U(\pi(x)[\pi(y)\xi]) &= U(\pi(xy)\xi) = \pi_\phi(xy)\xi_\phi \\
+    &= \pi_\phi(x)[\pi_\phi(y)\xi_\phi] = \pi_\phi(x)[U(\pi(y)\xi)]
+    \end{align*}
+    
+
+    so $U$ \hyperref[extends continuously]{theorem:linear-extension-theorem-normed} to a unitary equivalence between $(H, \pi)$ and $(H_\phi, \pi_\phi)$, with $U(\xi) = \xi_\phi$. 
+
+    (3): Suppose that for each $x \in A$, there exists $\phi \in \mathcal{S}$ such that $\dpn{x^*x, \phi}{A} \ne 0$. In which case,
+    \[
+    0 \ne \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(x)\xi_\phi}{H_\phi}
+    \]
+
+    so $\pi_\phi(x) \ne 0$, and $\pi_{\mathcal{S}}(x) \ne 0$ as well. 
+
+    By \autoref{corollary:cstar-positive-weakstar-dense}, for each $x \in A$, there exists $\phi \in P(A)$ with $\dpn{x^*x, \phi}{A} \ne 0$, so $\pi_{P(A)}$ is injective. By \autoref{theorem:continuity-of-homomorphism-c-star}, $\pi_{P(A)}(A)$ is closed in $B([l^2(P(A)); H_\phi])$. 
+    
+\end{proof}
+
+
+

src/op/c-star/index.tex

@@ -11,3 +11,4 @@
 \input{./order.tex}
 \input{./positive.tex}
 \input{./state.tex}
+\input{./gns.tex}

src/op/c-star/state.tex

@@ -9,8 +9,8 @@
     The set of states $S(A) \subset A^*$ of $A$ equipped with the weak* topology is the \textbf{state space} of $A$.
 \end{definition}
 
-\begin{lemma}
-\label{lemma:cstar-state-cauchy-schwarz}
+\begin{definition}
+\label{definition:cstar-state-pseudo-inner-product}
     Let $A$ be a unital $C^*$-algebra and $\phi \in A^*$ be a positive linear functional, then the mapping
     \[
     A \times A \to \complex \quad (x, y) \mapsto \dpn{x, y}{\phi} := \dpn{y^*x, \phi}{A}
@@ -20,7 +20,9 @@
     \[
     |\dpn{y^*x, \phi}{A}|^2 = |\dpn{x, y}{\phi}|^2 \le \dpn{x, x}{\phi} \cdot \dpn{y, y}{\phi}
     \]
-\end{lemma}
+
+    The pairing $\dpn{\cdot, \cdot}{\phi}$ is the \textbf{pseudo inner product associated with $\phi$}. 
+\end{definition}
 \begin{proof}
     By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}. 
 \end{proof}
@@ -60,7 +62,7 @@
 \begin{proof}
     (1): Let $\phi \in \Omega(A)$. By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A^*} = \dpn{1, \phi}{A} = 1$. Thus $\phi$ is a state by \autoref{theorem:cstar-positive-algebraic}, and $\Omega(A) \subset S(A)$. 
 
-    Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\phi = (1 - t)\psi + t\rho$, then for each $x \in \ker(\phi)$, $x^*x \in \ker(\phi)$ as well. As $t \ne 0$, $x^*x \in \ker(\psi)$ and $x^*x \in \ker(\rho)$. By the \hyperref[Cauchy-Schwarz inequality]{lemma:cstar-state-cauchy-schwarz}, 
+    Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\phi = (1 - t)\psi + t\rho$, then for each $x \in \ker(\phi)$, $x^*x \in \ker(\phi)$ as well. As $t \ne 0$, $x^*x \in \ker(\psi)$ and $x^*x \in \ker(\rho)$. By the \hyperref[Cauchy-Schwarz inequality]{definition:cstar-state-pseudo-inner-product}, 
     \[
     |\dpn{x, \psi}{A}|^2 = |\dpn{1^*x, \psi}{A}|^2 \le \dpn{1, \psi}{A} \cdot \dpn{x^*x, \psi}{A} = 0
     \]

src/op/example/bounded.tex

@@ -3,9 +3,60 @@
 
 \begin{definition}[$B(H)$]
 \label{definition:hilbert-endomorphism}
-    Let $H$ be a Hilbert space, then $B(H) = L(H; H)$ is the algebra of all bounded linear operators on $H$. 
+    Let $H$ be a complex Hilbert space, then $B(H) = L(H; H)$ is the algebra of all bounded linear operators on $H$. 
 \end{definition}
 
 
 % 1. Every non-trivial ideal of B(H) contains the finite-rank operators.
 % 2. If H is separable, then the only non-trivial closed idela of B(H) are the compact operators. 
+
+
+\begin{definition}[Partial Isometry]
+\label{definition:partial-isometry}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a \textbf{partial isometry} if $T|_{\ker(T)^\perp}$ is an isometry. In which case, $\ker(T)^\perp$ is the \textbf{initial space} of $T$, and $T(H)$ is the \textbf{final space} of $T$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:partial-isometry-characterisation}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^*T$ is a projection. 
+\end{proposition}
+\begin{proof}
+    ($\Rightarrow$): Suppose that $T$ is a partial isometry. Let $x \in \ker(T)^\perp$, then $\dpn{Tx, Tx}{H} = \norm{x}_H^2$ and $\dpn{T^*Tx, x}{H} = \norm{x}_H^2$. By \hyperref[polarisation]{proposition:polarisation-complex}, for each $x, y \in \ker(T)^\perp$, 
+    \begin{align*}
+        \dpn{T^*Tx, y}{H} &= \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{T^*T(x + i^ky), x + i^ky}{H} \\
+        &= \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{x + i^ky, x + i^ky}{H} = \dpn{x, y}{H} 
+    \end{align*}
+
+    Therefore $T^*T$ is idempotent. As $T^*T$ is self-adjoint, it is a projection. 
+
+    ($\Leftarrow$): Suppose that $T^*T$ is a projection, then for each $x \in \ker(T)^\perp$, $\dpn{Tx, Tx}{H} = \dpn{T^*Tx, x}{H} = \norm{x}_H^2$. 
+\end{proof}
+
+\begin{theorem}[Polar Decomposition]
+\label{theorem:hilbert-polar-decomposition}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then there exists a unique pair $(P, V) \in B(H)^2$ such that:
+    \begin{enumerate}
+        \item $P$ is positive.
+        \item $V$ is a partial isometry.
+        \item $T = VP$. 
+        \item $\ker P = \ker V$.
+    \end{enumerate}
+
+    The pair $(P, V)$ is the \textbf{polar decomposition} of $T$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 12.8]{Zhu}}}. ]
+    Let $P = |T| = \sqrt{T^*T}$, then $P$ is positive (1). For each $x \in H$, 
+    \[
+        \norm{Px}_H^2 = \dpn{Px, Px}{H} = \dpn{P^*Px, x}{H} = \dpn{T^*Tx, x}{H} = \norm{Tx}_H^2
+    \]
+
+    Let $V_0: P(H) \to H$ be defined by $V(Px) = Tx$, then $V_0$ extends to a well-defined isometry $\ol{P(H)} \to H$. Further extend $V_0$ to $V$ by setting its value to $0$ on $P(H)^\perp$, then $V$ is a partial isometry (2). Moreover, for any $x \in H$, $Tx = V_0Px = VPx$ (3).
+
+    Finally, since the initial space of $V$ is $\ol{P(H)}$, $\ker(V) = P(H)^\perp = \ker(P)$ (4). 
+    
+    It remains to show uniqueness. Let $T = WQ$ be a polar decomposition of $T$ satisfying (1)-(4). By \autoref{proposition:partial-isometry-characterisation}, $W^*W$ is a projection onto $\ker(W)^\perp = \ker(Q)^\perp = \ol{Q(H)}$. Thus $P^2 = T^*T = QW^*WQ = Q^2$, and $P = Q$ by uniqueness of the positive square root. 
+
+    Now, since $VP = WP$ and $\ker(V) = \ker(W) = P(H)^\perp$, $V = W$ on $H$, and the polar decomposition is unique. 
+\end{proof}
+
+

src/op/notation.tex

@@ -17,6 +17,9 @@
     $A[S]$ & $C^*$-subalgebra of $A$ generated by $S \subset A$. & \autoref{definition:generated-subalgebra} \\ 
     $S(A)$ & State space of a $C^*$-algebra $A$. & \autoref{definition:cstar-state} \\ 
     $P(A)$ & Pure state space of a $C^*$-algebra $A$. & \autoref{definition:pure-state} \\ 
+    $\dpn{x, y}{\phi}$ & Defined as $\dpn{y^*x, \phi}{A}$, the pseudo inner product associated to a positive linear functional. & \autoref{definition:cstar-state-pseudo-inner-product} \\ 
+    $(H_\phi, \pi_\phi, \xi_\phi)$ & GNS triple associated with $\phi \in S(A)$. & \autoref{definition:gns-triple} \\ 
+
     $M_n(\complex)$ & Algebra of $n \times n$ matrices over $\complex$.  & \autoref{definition:matrix-algebra} \\
     $B(H)$ & Algebra of bounded operators on a Hilbert space. & \autoref{definition:hilbert-endomorphism} \\
     $A(D)$ & The disk algebra. & \autoref{definition:disk-algebra} \\

src/topology/main/compact.tex

@@ -19,9 +19,7 @@
     U_J = \bigcup_{j \in J}E_j^c
     \]
 
-    then $U_J \subset X$ is open. For any $J, J' \subset I$, $U_J \cup U_{J'} = U_{J \cup J'}$.
-
-    Suppose for contradiction that $\bigcap_{i \in I}E_i = \emptyset$, then
+    then $U_J \subset X$ is open. Suppose for contradiction that $\bigcap_{i \in I}E_i = \emptyset$, then
     \[
     \mathbf{U} = \bracs{U_J|J \subset I \text{ finite}}
     \]