Added the start to operator systems.

src/op/system/dilation.tex

@@ -0,0 +1,122 @@
+\section{Dilations}
+\label{section:bh-dilation}
+
+\begin{definition}[Compression]
+\label{definition:bh-compression}
+    Let $H$ be a complex Hilbert space, $K \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $K$, $S \in B(K)$, and $T \in B(H)$, then $S$ is a \textbf{compression} of $T$ if $S = PT|_K$. 
+\end{definition}
+
+\begin{definition}[Dilation]
+\label{definition:bh-dilation}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then a \textbf{dilation} of $T$ is a pair $(K, S)$ where:
+    \begin{enumerate}
+        \item $K$ is a Hilbert space containing $H$. 
+        \item $S \in B(K)$. 
+        \item If $P \in B(K)$ is the orthogonal projection onto $H$, then $T = PS|_H$. 
+    \end{enumerate}
+
+    If $T^n = PS^n|_H$ for all $n \in \natz$, then $S$ is a \textbf{power dilation} of $T$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:bh-dilation-transitive}
+    Let $H$ be a complex Hilbert space, $R \in B(H)$, $(K, S)$ be a dilation of $R$, and $(L, T)$ be a dilation of $S$, then $(L, T)$ is a dilation of $R$. 
+
+    Similarly, if $(K, S)$ is a power dilation of $T$ and $(L, T)$ is a power dilation of $S$, then $(L, T)$ is a power dilation of $R$. 
+\end{lemma}
+
+\begin{theorem}[Sz.-Nagy's Dilation Theorem]
+\label{theorem:sznagy-dilation}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a unitary power dilation of $T$.
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ]
+    Using \autoref{lemma:bh-dilation-transitive} it is sufficient to show the following statements:
+    \begin{enumerate}[label=(\roman*)]
+        \item For any $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, there exists an isometric power dilation of $T$. 
+        \item For any isometry $V \in B(H)$, there exists a unitary power dilation of $V$. 
+    \end{enumerate}
+
+    (i): Since $\norm{T}_{B(H)} \le 1$, $1 - T^*T \ge 0$, and $(1 - T^*T)^{1/2}$ is well-defined. Let $D_T = (1 - T^*T)^{1/2}$, and
+    \[
+    V: l^2(\natp; H) \to l^2(\natp; H) \quad (Vx)_n = \begin{cases}
+        Tx_1 &n = 1 \\
+        D_Tx_1 &n = 2 \\
+        x_{n-1} &n \ge 3
+    \end{cases}
+    \]
+    
+    then for each $x \in H$,
+    \begin{align*}
+        \norm{Tx}_H^2 + \norm{D_Tx}_H^2 &= \dpn{T^*Tx, x}{H} + \dpn{(1 - T^*T)x, x}{H} \\
+        &= \dpn{x, x}{H} = \norm{x}_H^2
+    \end{align*}
+
+    Thus for each $x = \seq{x_n} \in l^2(H)$, 
+    \begin{align*}
+        \norm{Vx}_{l^2(\natp; H)}^2 &= \norm{Tx_1}_H^2 + \norm{D_Tx_1}_H^2 + \sum_{n = 3}^\infty \norm{x_{n-1}}_H^2 \\
+        &= \norm{x_1}_H^2 + \sum_{n = 2}^\infty \norm{x_{n}}_H^2 = \norm{x}_{l^2(\natp; H)}^2
+    \end{align*}
+
+    So $V$ is an isometry. Moreover, $[V^n(x, 0, \cdots)]_1 = T^nx$ for all $n \in \natz$. Therefore under the identification that $H = \bracsn{(x, 0, \cdots)|x \in H} \subset l^2(\natp; H)$, $V$ is an isometric power dilation of $T$. 
+
+    (ii): Let $P = I_H - VV^*$ be the projection onto $V(H)^\perp$, and $U \in B(H^2)$ be defined by
+    \[
+    U = \begin{bmatrix}
+        V & P \\
+        0 & V^*
+    \end{bmatrix}
+    \]
+
+    then
+    \begin{align*}
+        UU^* &= \begin{bmatrix}
+            V & P \\
+            0 & V^*
+        \end{bmatrix} \cdot 
+        \begin{bmatrix}
+            V^* & 0 \\
+            P & V
+        \end{bmatrix} = \begin{bmatrix}
+            VV^* + P & PV \\
+            V^*P & V^*V
+        \end{bmatrix} = I_{H^2} \\
+        U^*U &= \begin{bmatrix}
+            V^* & 0 \\
+            P & V
+        \end{bmatrix} \cdot 
+        \begin{bmatrix}
+            V & P \\
+            0 & V^*
+        \end{bmatrix}
+         = \begin{bmatrix}
+            V^*V & V^*P \\
+            PV & VV^* + P
+        \end{bmatrix} = I_{H^2} \\
+    \end{align*}
+
+    so $U$ is unitary. Moreover, for each $n \in \natp$, 
+    \[
+    U^n = \begin{bmatrix}
+            V & P \\
+            0 & V^*
+        \end{bmatrix}^n = \begin{bmatrix}
+            V^n & \cdots \\
+            0 & (V^*)^n
+        \end{bmatrix}
+    \]
+
+    Under the identification that $H = \bracsn{(x, 0)|x \in H} \subset H^2$, $QU^n|_{H} = V^n$ for all $n \in \natp$. Therefore $U$ is a unitary power dilation of $V$. 
+\end{proof}
+
+\begin{theorem}[Von Neumann's Inequality]
+\label{theorem:von-neumann-inequality-polynomial}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then for any $p \in \complex[x]$, 
+    \[
+    \norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|
+    \]    
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Corollary 1.2]{Paulsen}}}. ]
+    By \hyperref[Sz.-Nagy's Dilation Theorem]{theorem:sznagy-dilation}, there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^n = P_HU^n|_H$ for all $n \in \natp$, where $P_H$ is the orthogonal projection onto $H$. 
+
+    Thus for any $p \in \complex[x]$, $p(T) = P_Hp(U)|_H$. As $U$ is unitary, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} shows that $\norm{p(U)}_{B(K)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. Therefore $\norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. 
+\end{proof}