Exercise 1.8.

src/op/system/dilation.tex

@@ -18,6 +18,28 @@
     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-polynomial}
+    Let $H$ be a complex Hilbert space, $T \in B(H)$, $(K, S)$ be a power dilation of $T$, and $P \in B(K)$ be the orthogonal projection onto $H$, then
+    \begin{enumerate}
+        \item $(K, S^*)$ is a power dilation of $T^*$. 
+        \item For any $p, q \in \complex[z]$, 
+        \[
+        p(T) + q(T)^* = P[p(S) + q(S)^*]|_{H}
+        \]
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): For any $\xi, \eta \in H$ and $n \in \natp$, $P\xi = \xi$ and $P\eta = \eta$, so
+    \begin{align*}
+        \dpn{(T^*)^n\xi, \eta}{H} &= \dpn{\xi, T^n\eta}{H} = \dpn{\xi, PS^n\eta}{K} = \dpn{P\xi, S^n\eta}{K} \\
+        &= \dpn{(S^*)^nP\xi, P\eta}{K} = \dpn{P(S^*)^n\xi, \eta}{H}
+    \end{align*}
+\end{proof}
+
+
+
 \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 $R$. 
@@ -25,6 +47,92 @@
     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}[Stinespring's Dilation Theorem]
+\label{theorem:stinespring-dilation}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a completely positive map, then there exists a complex Hilbert space $K$, a unital *-homomorphism $\pi: A \to B(K)$, and $V \in L(H; K)$ such that
+    \begin{enumerate}
+        \item $\normn{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. 
+        \item For all $a \in A$, $\phi(a) = V^*\pi(a)V$. 
+    \end{enumerate}
+
+    Moreover, if $\phi$ is unital, then $V$ is an isometry, and
+    \begin{enumerate}[start=2]
+        \item Under the identification of $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$. 
+    \end{enumerate}
+
+    If $\phi$ is a state, then $(K, \pi)$ is the GNS representation of $A$ associated with $\phi$. 
+
+    The triple $(K, V, \pi)$ is a \textbf{Stinespring representation} for $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(K, V, \pi)$ is \textbf{minimal}.  
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ]
+    ($K$): For each $(a \otimes \xi), (b \otimes \eta) \in K_0$, let
+    \[
+    \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
+    \]
+
+    and extend by linearity. For each $\seqf{(a_j \otimes \xi_j)} \subset A \otimes H$, let
+    \[
+    T = \sum_{i, j = 1}^n E_{ij} \otimes a_i^*a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}} \in M_n(A)
+    \]
+
+    Then $T \ge 0$. As $\phi$ is completely positive, 
+    \[
+    \angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_{\phi} = \angles{\phi_n(T)\begin{bmatrix}
+        \xi_1 \\
+        \vdots \\
+        \xi_n
+    \end{bmatrix}, \begin{bmatrix}
+        \xi_1 \\
+        \vdots \\
+        \xi_n
+    \end{bmatrix}}_{H^n} \ge 0
+    \]
+
+    so $\dpn{u, u}{\phi}$ is a pseudo inner product. Let
+    \[
+    N = \bracsn{u \in A \otimes H| \dpn{u, u}{\phi} = 0}
+    \]
+
+    then by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \[
+    N = \bracsn{u \in A \otimes H| \dpn{u, v}{\phi} = 0 \forall v \in A \otimes H}
+    \]
+
+    Let $K_0 = (A \otimes H)/N$ and $K$ be the completion of $K_0$, then $K$ is a Hilbert space. 
+
+    ($\pi$): For each $a \in A$, let
+    \[
+    \pi_0(a): K_0 \to K_0 \quad \pi_0(a)(b \otimes \xi + N) = (ab) \otimes \xi + N
+    \]
+
+    For any $b \in A$, $b^*a^*ab \le b^*\norm{a}_A^2b = \norm{a}_A^2b^*b$. As such, for any $u = \sum_{j = 1}^n b_j \otimes \xi_j \in  A \otimes H$, 
+    \begin{align*}
+        \angles{\pi(a)u, \pi(a)u}_{\phi} &= \sum_{i, j = 1}^n \dpn{\pi(a)(b_i \otimes \xi_i), \pi(a)(b_j \otimes \xi_j)}{\phi} \\
+        &= \sum_{i, j = 1}^n \dpn{(ab_i) \otimes \xi_i, (ab_j) \otimes \xi_j}{\phi} = \sum_{i, j = 1}^n \dpn{\phi(b_j^*a^*ab_i) \otimes \xi_i,\xi_j}{H} \\
+        &\le \sum_{i, j = 1}^n \norm{a^*a}_{A}\dpn{\phi(b_j^*b_i)\xi_i, \xi_j}{H} = \norm{a}_A^2 \dpn{u, u}{\phi}
+    \end{align*}
+    
+    Thus $\pi_0(a)$ is well-defined with $\norm{\pi_0(a)}_{B(K_0)} \le \norm{a}_A$, and extends into a continuous linear operator $\pi(a) \in B(K)$. The map $\pi: A \to B(K)$ is then a *-homomorphism. 
+
+    ($V$): Let $V: H \to K$ be defined by $\xi \mapsto 1_A \otimes \xi + N$. 
+
+    (1): For each $\xi \in H$, 
+    \[
+    \dpn{V\xi, V\xi}{K} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H}
+    \]
+
+    so $\normn{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. In particular, if $\phi$ is unital, then $V$ is an isometry by \hyperref[polarisation]{proposition:polarisation-complex}. 
+    
+    (2): Finally, for any $\xi, \eta \in H$, 
+    \begin{align*}
+        \dpn{V^*\pi(a)V\xi, \eta}{H} &= \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{a \otimes \xi, 1_A \otimes \eta}{\phi} \\
+        &= \dpn{\phi(a)\xi, \eta}{H}
+    \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 unitary power dilation of $T$.
@@ -110,15 +218,21 @@
 
 \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 $p \in \complex[x]$, 
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then for any $p, q \in \complex[x]$, 
     \[
-    \norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|
+    \normn{p(T) + q(T)^*}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z) + \ol{q(z)}|
     \]    
 \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 $H$. 
+    By \hyperref[Sz.-Nagy's Dilation Theorem]{theorem:sznagy-dilation}, $T$ admits a unitary power dilation $(K, U)$. Let $P \in B(K)$ be the orthogonal projection onto $H$, then by \autoref{lemma:bh-dilation-transitive}, 
+    \[
+    p(T) + q(T)^* = P[p(U) + q(U)^*]|_{H}
+    \]
 
-    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 $\norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. 
+    As $U$ is unitary, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} shows that $\norm{p(U) + q(U)^*}_{B(K)} \le \sup_{z \in B_\complex(0, 1)}|p(z) + \ol{q(z)}|$. Therefore
+    \[
+    \normn{p(T) + q(T)^*}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z) + \ol{q(z)}|
+    \] 
 \end{proof}
 \begin{proof}[Proof, {{\cite[Exercise 2.15]{Paulsen}}}. ]
     Assume without loss of generality that $\norm{T}_{B(H)} < 1$. Let
@@ -175,11 +289,12 @@
         &= (1 - e^{-it}T)^{-1}\braks{1 - TT^*}(1 - e^{it}T^*)^{-1} \ge 0
     \end{align*}
 
-    For each $\xi, \eta \in H$, let $\dpn{\xi, \eta}{t} = \dpn{P(T; t)\xi, \eta}{H}$, then $\dpn{\cdot, \cdot}{t}$ is a pseudo inner product on $H$. Hence by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    Let $M = \sup_{z \in \partial B_\complex(0, 1)}|p(z) + \ol{q(z)}|$. For each $\xi, \eta \in H$, let $\dpn{\xi, \eta}{t} = \dpn{P(T; t)\xi, \eta}{H}$, then $\dpn{\cdot, \cdot}{t}$ is a pseudo inner product on $H$. Hence by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
     \begin{align*}
-        \dpn{p(T)\xi, \eta}{H} &= \frac{1}{2\pi}\int_0^{2\pi} p(T)\dpn{P(T; t)\xi, \eta}{H}dt = \frac{1}{2\pi}\int_0^{2\pi} p(T)\dpn{\xi, \eta}{t}dt \\ 
+        \dpn{[p(T) + q(T)^*]\xi, \eta}{H} &= \frac{1}{2\pi}\int_0^{2\pi} [p(T) + q(T)^*]\dpn{P(T; t)\xi, \eta}{H}dt \\
-        |\dpn{p(T)\xi, \eta}{H}| &\le \frac{1}{2\pi}\int_0^{2\pi} \norm{p}_{H^\infty(D)}\dpn{\xi, \xi}{t}^{1/2}\dpn{\eta, \eta}t^{1/2}dt\\
+        &= \frac{1}{2\pi}\int_0^{2\pi} [p(T) + q(T)^*]\dpn{\xi, \eta}{t}dt \\ 
-        &\le \norm{p}_{H^\infty(D)}\braks{\frac{1}{2\pi}\int_0^{2\pi}\dpn{\xi, \xi}{t} dt}^{1/2}\braks{\frac{1}{2\pi}\int_0^{2\pi}\dpn{\eta, \eta}{t} dt}^{1/2}
+        |\dpn{[p(T) + q(T)^*]\xi, \eta}{H}| &\le \frac{1}{2\pi}\int_0^{2\pi} M\dpn{\xi, \xi}{t}^{1/2}\dpn{\eta, \eta}t^{1/2}dt\\
+        &\le M\braks{\frac{1}{2\pi}\int_0^{2\pi}\dpn{\xi, \xi}{t} dt}^{1/2}\braks{\frac{1}{2\pi}\int_0^{2\pi}\dpn{\eta, \eta}{t} dt}^{1/2}
     \end{align*}
 
     Since $I_H(\xi) = \frac{1}{2\pi}\int_0^{2\pi} P(T; t)dt$, 
@@ -187,7 +302,72 @@
     \frac{1}{2\pi}\int_0^{2\pi}\dpn{\xi, \xi}{t} dt = \frac{1}{2\pi}\int_0^{2\pi} \dpn{P(T; t)\xi, \xi}{H}dt = \norm{\xi}_H^2
     \]
 
-    Therefore $\dpn{p(T)\xi, \eta}{H} \le \norm{p}_{H^\infty(D)}\norm{\xi}_H\norm{\eta}_H$, and $\norm{p(T)}_H \le \norm{p}_{H^\infty(D)}$.
+    Therefore $\dpn{[p(T) + q(T)^*]\xi, \eta}{H} \le M\norm{\xi}_H\norm{\eta}_H$, and $\norm{[p(T) + q(T)^*]}_{B(H)} \le M$.
-    
     
 \end{proof}
+
+
+\begin{theorem}[Von Neumann's Inequality for Matrices]
+\label{theorem:von-neumann-matrix-wermer}
+    Let $n \in \natp$, $T \in M_n(\complex)$ with $\norm{T}_{M_n(\complex)} \le 1$, and $U, V, S \in M_n(\complex)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $U, V$ are unitary.
+        \item $S = \text{diag}(s_1, \cdots, s_n)$, where $\seqf{s_j} \subset [0, 1]$. 
+    \end{enumerate}
+
+    Let $B = \ol{B_\complex(0, 1)}^n \subset \complex^n$ and 
+    \[
+    T: B \to M_n(\complex) \quad T(z_1, \cdots, z_n) = U\text{diag}(z_1, \cdots, z_n)V
+    \]
+
+    and $p \in \complex[z]$, then:
+    \begin{enumerate}
+        \item For each $\xi, \eta \in \complex^n$, let
+        \[
+        f_{\xi, \eta}: B \to \complex \quad (z_1, \cdots, z_n) \mapsto \dpn{p(T(z_1, \cdots, z_n))\xi, \eta}{\complex^n}
+        \]
+
+        then $f_{\xi, \eta}$ achieves its maximum modulus on $\partial B$. 
+        \item For any $(z_1, \cdots, z_n) \in B$, if
+        \[
+        \normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} = \max_{w \in \overline{B}}\normn{p(T(w_1, \cdots, w_n))}_{M_n(\complex)} 
+        \]
+
+        then $T(z_1, \cdots, z_n)$ is unitary. 
+        \item $\norm{p(T)}_{M_n(\complex)} \le \max_{W \in U(M_n(\complex))}\norm{p(W)}_{M_n(\complex)}$. 
+        \item $\norm{p(T)}_{M_n(\complex)} \le \norm{p}_{H^\infty(D)}$. 
+    \end{enumerate}
+    
+
+    For each $\seqf{z} \in \overline{B_{\complex^n}(0, 1)}$
+    
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Exercise 2.16]{Paulsen}}}, attributed to Wermer. ]
+    (1): By the Maximum Modulus Principle. 
+
+    (2): Let $(z_1, \cdots, z_n) \in \complex^n$ such that $\normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)}$ is maximal. By compactness of the unit sphere in $\complex^n$, there exists $\xi, \eta \in \partial B_{\complex^n}(0, 1)$ such that
+    \[
+    \normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} = \dpn{p(T(z_1, \cdots, z_n))\xi, \eta}{\complex^n} = f_{\xi, \eta}(z_1, \cdots, z_n)
+    \]
+
+    By maximality, $f_{\xi, \eta}$ achieves its maximum modulus at $(z_1, \cdots, z_n)$, so (1) implies that $(z_1, \cdots, z_n) \in \partial B$. At which point, $|z_j| = 1$ for all $1 \le j \le n$, and $\text{diag}(z_1, \cdots, z_n)$ is unitary. Therefore $T(z_1, \cdots, z_n) = U\text{diag}(z_1, \cdots, z_n)V$ is also unitary. 
+
+    (3): Since $(s_1, \cdots, s_n) \in B$, (2) implies that  
+    \begin{align*}
+    \norm{p(T)}_{M_n(\complex)} &\le \max_{z \in B}\norm{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} \\
+    &\le \max_{W \in U(M_n(\complex))}\norm{p(U(W))}_{M_n(\complex)}
+    \end{align*}
+    
+    (4): For any $W \in U(M_n(\complex))$, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} implies that
+    \[
+    \norm{p(W)}_{M_n(\complex)} \le \max_{\lambda \in \sigma_{M_n(\complex)}(W)}|p(\lambda)| \le \max_{\lambda \in \partial B_\complex(0, 1)}|p(\lambda)| = \norm{p}_{H^\infty(D)}
+    \]
+
+    As the above holds for all unitary matrices, 
+    \[
+    \norm{p(T)}_{M_n(\complex)} \le \max_{W \in U(M_n(\complex))}\norm{p(U(W))}_{M_n(\complex)} \le \norm{p}_{H^\infty(D)}
+    \]
+
+    by (3).
+\end{proof}
+