Added nuclear spaces.

src/fa/lc/index.tex

@@ -14,3 +14,4 @@
 \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/norm/hilbert.tex

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

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}
+
+