Dilation situation.

src/op/example/bounded.tex

@@ -81,3 +81,123 @@
 \end{proof}
 
 
+\begin{definition}[Unitary Dilation]
+\label{definition:unitary-dilation}
+    Let $H$ be a complex Hilbert space, $V \in B(H)$ be isometric, $P = I_H - VV^*$ be the projection onto $V(H)^\perp$, and
+    \[
+    U = \begin{bmatrix}
+        V & P \\
+        0 & V^*
+    \end{bmatrix}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $U^*U = UU^* = I_{H^2}$. In other words, $U$ is an isometry on $H^2$. 
+        \item Under the identification that $H = H \times \bracsn{0} \subset H^2$, let $Q \in B(H^2)$ be the projection onto $H$, then $V^n = QU^n|_H$ for all $n \ge 0$. 
+    \end{enumerate}
+    
+    The operator $U \in B(H^2)$ is a \textbf{unitary dilation} of $V$. 
+\end{definition}
+\begin{proof}
+    (1): By direct computation, 
+    \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*}
+    
+
+    (2): For each $n \in \natp$, 
+    \[
+    U^n = \begin{bmatrix}
+            V & P \\
+            0 & V^*
+        \end{bmatrix}^n = \begin{bmatrix}
+            V^n & \cdots \\
+            0 & (V^*)^n
+        \end{bmatrix}
+    \]
+
+    so $QU^n|_{H} = V^n$.
+\end{proof}
+
+\begin{definition}[Isometric Dilation]
+\label{definition:isometric-dilation}
+    Let $H$ be a complex Hilbert space, $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, $D_T = (1 - T^*T)^{1/2}$, 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:
+    \begin{enumerate}
+        \item $V$ is isometric on $l^2(\natp; H)$. 
+        \item Under the identification that $H = H \oplus 0 \oplus \cdots$, let $Q$ be the orthogonal projection onto $H$, then $T^n = QV^n|_H$ for all $n \ge 0$. 
+    \end{enumerate}
+
+    The operator $V$ is an \textbf{isometric dilation} of $T$. 
+\end{definition}
+\begin{proof}
+    (1): Since $\norm{T}_H \le 1$, $1 - T^*T \ge 0$, and $(1 - T^*T)^{1/2}$ is well-defined. 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}_H^2 + \sum_{n = 2}^\infty \norm{x_{n}}_H^2 = \norm{x}_{l^2(\natp; H)}^2
+    \end{align*}
+\end{proof}
+
+\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 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 $H$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ]
+    Let $K = l^2(\natp; H)^2$ and identify $H = (H \oplus \bracsn{0} \oplus \cdot) \times \bracsn{0}$. Let $V \in B(l^2(\natp; H))$ be the isometric dilation of $T$, and $U \in B(l^2(\natp; H)^2)$ be the unitary dilation of $V$. Identify $l^2(\natp; H)$ as a subspace of $K$ as $l^2(\natp; H) \times \bracsn{0}$, and let $P_{l^2(\natp; H)}$ be the orthogonal projection onto $l^2(\natp; H)$, then for each $n \ge 0$, 
+    \begin{align*}
+        V^n &= P_{l^2(\natp; H)}U^n|_{l^2(\natp; H)} \\
+        T^n = P_{H}V^n|_{H} &= P_HU^n|_{H}
+    \end{align*}
+\end{proof}
+
+\begin{corollary}[Von Neumann's Inequality]
+\label{corollary: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{corollary}
+\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}
+
+

src/op/example/matrix.tex

@@ -82,5 +82,3 @@
     \]
 \end{example}
 
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