374 lines
16 KiB
TeX
374 lines
16 KiB
TeX
\section{Dilations}
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\label{section:bh-dilation}
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\begin{definition}[Compression]
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\label{definition:bh-compression}
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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 $S = PT|_K$.
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\end{definition}
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\begin{definition}[Dilation]
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\label{definition:bh-dilation}
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Let $H$ be a complex Hilbert space and $T \in B(H)$, then a \textbf{dilation} of $T$ is a pair $(K, S)$ where:
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\begin{enumerate}
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\item $K$ is a Hilbert space containing $H$.
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\item $S \in B(K)$.
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\item If $P \in B(K)$ is the orthogonal projection onto $H$, then $T = PS|_H$.
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\end{enumerate}
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If $T^n = PS^n|_H$ for all $n \in \natz$, then $S$ is a \textbf{power dilation} of $T$.
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\end{definition}
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\begin{lemma}
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\label{lemma:bh-dilation-polynomial}
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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
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\begin{enumerate}
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\item $(K, S^*)$ is a power dilation of $T^*$.
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\item For any $p, q \in \complex[z]$,
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\[
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p(T) + q(T)^* = P[p(S) + q(S)^*]|_{H}
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\]
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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(1): For any $\xi, \eta \in H$ and $n \in \natp$, $P\xi = \xi$ and $P\eta = \eta$, so
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\begin{align*}
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\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} \\
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&= \dpn{(S^*)^nP\xi, P\eta}{K} = \dpn{P(S^*)^n\xi, \eta}{H}
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\end{align*}
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\end{proof}
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\begin{lemma}
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\label{lemma:bh-dilation-transitive}
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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 $R$.
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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 $R$.
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\end{lemma}
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\begin{theorem}[Stinespring's Dilation Theorem]
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\label{theorem:stinespring-dilation}
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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
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\begin{enumerate}
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\item $\normn{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$.
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\item For all $a \in A$, $\phi(a) = V^*\pi(a)V$.
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\end{enumerate}
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Moreover, if $\phi$ is unital, then $V$ is an isometry, and
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\begin{enumerate}[start=2]
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\item Under the identification of $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$.
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\end{enumerate}
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If $\phi$ is a state, then $(K, \pi)$ is the GNS representation of $A$ associated with $\phi$.
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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}.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ]
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($K$): For each $(a \otimes \xi), (b \otimes \eta) \in K_0$, let
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\[
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\dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
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\]
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and extend by linearity. For each $\seqf{(a_j \otimes \xi_j)} \subset A \otimes H$, let
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\[
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T = \sum_{i, j = 1}^n E_{ij} \otimes a_i^*a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}} \in M_n(A)
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\]
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Then $T \ge 0$. As $\phi$ is completely positive,
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\[
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\angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_{\phi} = \angles{\phi_n(T)\begin{bmatrix}
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\xi_1 \\
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\vdots \\
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\xi_n
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\end{bmatrix}, \begin{bmatrix}
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\xi_1 \\
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\vdots \\
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\xi_n
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\end{bmatrix}}_{H^n} \ge 0
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\]
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so $\dpn{u, u}{\phi}$ is a pseudo inner product. Let
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\[
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N = \bracsn{u \in A \otimes H| \dpn{u, u}{\phi} = 0}
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\]
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then by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz},
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\[
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N = \bracsn{u \in A \otimes H| \dpn{u, v}{\phi} = 0 \forall v \in A \otimes H}
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\]
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Let $K_0 = (A \otimes H)/N$ and $K$ be the completion of $K_0$, then $K$ is a Hilbert space.
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($\pi$): For each $a \in A$, let
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\[
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\pi_0(a): K_0 \to K_0 \quad \pi_0(a)(b \otimes \xi + N) = (ab) \otimes \xi + N
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\]
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For any $b \in A$, $b^*a^*ab \le b^*\norm{a}_A^2b = \norm{a}_A^2b^*b$. As such, for any $u = \sum_{j = 1}^n b_j \otimes \xi_j \in A \otimes H$,
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\begin{align*}
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\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} \\
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&= \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} \\
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&\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}
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\end{align*}
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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.
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($V$): Let $V: H \to K$ be defined by $\xi \mapsto 1_A \otimes \xi + N$.
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(1): For each $\xi \in H$,
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\[
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\dpn{V\xi, V\xi}{K} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H}
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\]
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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}.
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(2): Finally, for any $\xi, \eta \in H$,
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\begin{align*}
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\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} \\
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&= \dpn{\phi(a)\xi, \eta}{H}
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\end{align*}
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\end{proof}
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\begin{theorem}[Sz.-Nagy's Dilation Theorem]
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\label{theorem:sznagy-dilation}
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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 $T$.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ]
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Using \autoref{lemma:bh-dilation-transitive} it is sufficient to show the following statements:
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\begin{enumerate}[label=(\roman*)]
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\item For any $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, there exists an isometric power dilation of $T$.
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\item For any isometry $V \in B(H)$, there exists a unitary power dilation of $V$.
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\end{enumerate}
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(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
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\[
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V: l^2(\natp; H) \to l^2(\natp; H) \quad (Vx)_n = \begin{cases}
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Tx_1 &n = 1 \\
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D_Tx_1 &n = 2 \\
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x_{n-1} &n \ge 3
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\end{cases}
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\]
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then for each $x \in H$,
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\begin{align*}
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\norm{Tx}_H^2 + \norm{D_Tx}_H^2 &= \dpn{T^*Tx, x}{H} + \dpn{(1 - T^*T)x, x}{H} \\
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&= \dpn{x, x}{H} = \norm{x}_H^2
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\end{align*}
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Thus for each $x = \seq{x_n} \in l^2(H)$,
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\begin{align*}
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\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 \\
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&= \norm{x_1}_H^2 + \sum_{n = 2}^\infty \norm{x_{n}}_H^2 = \norm{x}_{l^2(\natp; H)}^2
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\end{align*}
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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 $T$.
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(ii): Let $P = I_H - VV^*$ be the projection onto $V(H)^\perp$, and $U \in B(H^2)$ be defined by
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\[
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U = \begin{bmatrix}
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V & P \\
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0 & V^*
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\end{bmatrix}
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\]
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then
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\begin{align*}
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UU^* &= \begin{bmatrix}
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V & P \\
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0 & V^*
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\end{bmatrix} \cdot
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\begin{bmatrix}
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V^* & 0 \\
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P & V
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\end{bmatrix} = \begin{bmatrix}
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VV^* + P & PV \\
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V^*P & V^*V
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\end{bmatrix} = I_{H^2} \\
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U^*U &= \begin{bmatrix}
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V^* & 0 \\
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P & V
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\end{bmatrix} \cdot
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\begin{bmatrix}
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V & P \\
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0 & V^*
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\end{bmatrix}
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= \begin{bmatrix}
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V^*V & V^*P \\
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PV & VV^* + P
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\end{bmatrix} = I_{H^2} \\
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\end{align*}
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so $U$ is unitary. Moreover, for each $n \in \natp$,
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\[
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U^n = \begin{bmatrix}
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V & P \\
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0 & V^*
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\end{bmatrix}^n = \begin{bmatrix}
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V^n & \cdots \\
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0 & (V^*)^n
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\end{bmatrix}
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\]
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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 $V$.
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\end{proof}
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\begin{theorem}[Von Neumann's Inequality]
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\label{theorem:von-neumann-inequality-polynomial}
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Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then for any $p, q \in \complex[x]$,
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\[
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\normn{p(T) + q(T)^*}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z) + \ol{q(z)}|
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\]
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Corollary 1.2]{Paulsen}}}. ]
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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},
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\[
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p(T) + q(T)^* = P[p(U) + q(U)^*]|_{H}
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\]
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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
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\[
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\normn{p(T) + q(T)^*}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z) + \ol{q(z)}|
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\]
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\end{proof}
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\begin{proof}[Proof, {{\cite[Exercise 2.15]{Paulsen}}}. ]
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Assume without loss of generality that $\norm{T}_{B(H)} < 1$. Let
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\[
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P(T; t) = \frac{1}{1 - e^{-it}T} + \frac{1}{1 - e^{it}T^*} - 1
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\]
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Let $p, q \in \complex[z]$ and $\gamma: [0, 2\pi] \to \complex$ be defined by $\gamma(t) = e^{it}$, then by the \hyperref[holomorphic functional calculus]{definition:holomorphic-functional-calculus},
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\[
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p(T) = \frac{1}{2\pi i}\int_\gamma \frac{p(\lambda)}{\lambda - T} d\lambda
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\]
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For each $n \in \natz$, since $\gamma([0, 2\pi]) \subset \partial B_\complex(0, 1)$, $\ol{\lambda^n} = \lambda^{-n}$. After passing through a \hyperref[Neumann series]{lemma:neumann-series},
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\begin{align*}
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\int_\gamma \frac{\ol{\lambda^n}}{\lambda - T} d\lambda &= \int_\gamma \frac{\lambda^{-n}}{\lambda - T} d\lambda = \int_\gamma \frac{\lambda^{-{(n+1)}}}{1 - T/\lambda} d\lambda \\
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&= \sum_{k = 0}^\infty T^k\int_{\gamma} \lambda^{-(n+k+1)}d\lambda = \begin{cases}
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1 & n = 0 \\
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0 & n > 0
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\end{cases}
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\end{align*}
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By linearity and \hyperref[Cauchy's integral formula]{theorem:cauchy-formula},
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\[
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\frac{1}{2\pi i}\int_\gamma\frac{\ol{q(\lambda)}}{\lambda - T} d\lambda = \ol{q(0)} = \frac{1}{2\pi i}\int_\gamma \frac{\ol{q(\lambda)}}{\lambda} d\lambda
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\]
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so
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\begin{align*}
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p(T) + \overline{q(0)} &= \frac{1}{2\pi i} \int_\gamma \frac{p(\lambda) + \ol{q(\lambda)}}{\lambda - T} d\lambda \\
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p(T) + \overline{q(0)} &= \frac{1}{2\pi i} \int_0^{2\pi}\frac{ie^{it}[p(e^{it}) + \ol{q(e^{it})}]}{e^{it} - T} dt \\
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&= \frac{1}{2\pi}\int_0^{2\pi}\frac{p(e^{it}) + \ol{q(e^{it})}}{1 - e^{-it}T}dt
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\end{align*}
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Similarly,
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\[
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q(T)^* + p(0) = \frac{1}{2\pi}\int_0^{2\pi}\frac{p(e^{it}) + \ol{q(e^{it})}}{1 - e^{it}T^*}dt
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\]
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Therefore
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\begin{align*}
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p(T) + q(T)^* + p(0) + \ol{q(0)} &= \frac{1}{2\pi}\int_0^{2\pi}[p(e^{it}) + \ol{q(e^{it})}](P(T; t) - 1)dt \\
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p(T) + q(T)^* &= \frac{1}{2\pi}\int_0^{2\pi}[p(e^{it}) + \ol{q(e^{it})}]P(T; t)dt
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\end{align*}
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Now, for each $t \in [0, 2\pi]$,
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\[
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(1 - e^{-it}T)(1 - e^{it}T^*) = 1 - e^{-it}T - e^{it}T^* + TT^*
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\]
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so by the resolvent identity,
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\begin{align*}
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P(T; t) &= \frac{1}{1 - e^{-it}T} + \frac{1}{1 - e^{it}T^*} - 1 \\
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&= (1 - e^{-it}T)^{-1}\braks{1 - TT^*}(1 - e^{it}T^*)^{-1} \ge 0
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\end{align*}
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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},
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\begin{align*}
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\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 \\
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&= \frac{1}{2\pi}\int_0^{2\pi} [p(T) + q(T)^*]\dpn{\xi, \eta}{t}dt \\
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|\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\\
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&\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}
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\end{align*}
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Since $I_H(\xi) = \frac{1}{2\pi}\int_0^{2\pi} P(T; t)dt$,
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\[
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\frac{1}{2\pi}\int_0^{2\pi}\dpn{\xi, \xi}{t} dt = \frac{1}{2\pi}\int_0^{2\pi} \dpn{P(T; t)\xi, \xi}{H}dt = \norm{\xi}_H^2
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\]
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Therefore $\dpn{[p(T) + q(T)^*]\xi, \eta}{H} \le M\norm{\xi}_H\norm{\eta}_H$, and $\norm{[p(T) + q(T)^*]}_{B(H)} \le M$.
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\end{proof}
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\begin{theorem}[Von Neumann's Inequality for Matrices]
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\label{theorem:von-neumann-matrix-wermer}
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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:
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\begin{enumerate}[label=(\alph*)]
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\item $U, V$ are unitary.
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\item $S = \text{diag}(s_1, \cdots, s_n)$, where $\seqf{s_j} \subset [0, 1]$.
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\end{enumerate}
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Let $B = \ol{B_\complex(0, 1)}^n \subset \complex^n$ and
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\[
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T: B \to M_n(\complex) \quad T(z_1, \cdots, z_n) = U\text{diag}(z_1, \cdots, z_n)V
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\]
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and $p \in \complex[z]$, then:
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\begin{enumerate}
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\item For each $\xi, \eta \in \complex^n$, let
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\[
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f_{\xi, \eta}: B \to \complex \quad (z_1, \cdots, z_n) \mapsto \dpn{p(T(z_1, \cdots, z_n))\xi, \eta}{\complex^n}
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\]
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then $f_{\xi, \eta}$ achieves its maximum modulus on $\partial B$.
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\item For any $(z_1, \cdots, z_n) \in B$, if
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\[
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\normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} = \max_{w \in \overline{B}}\normn{p(T(w_1, \cdots, w_n))}_{M_n(\complex)}
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\]
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then $T(z_1, \cdots, z_n)$ is unitary.
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\item $\norm{p(T)}_{M_n(\complex)} \le \max_{W \in U(M_n(\complex))}\norm{p(W)}_{M_n(\complex)}$.
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\item $\norm{p(T)}_{M_n(\complex)} \le \norm{p}_{H^\infty(D)}$.
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\end{enumerate}
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For each $\seqf{z} \in \overline{B_{\complex^n}(0, 1)}$
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Exercise 2.16]{Paulsen}}}, attributed to Wermer. ]
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(1): By the Maximum Modulus Principle.
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(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
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\[
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\normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} = \dpn{p(T(z_1, \cdots, z_n))\xi, \eta}{\complex^n} = f_{\xi, \eta}(z_1, \cdots, z_n)
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\]
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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.
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(3): Since $(s_1, \cdots, s_n) \in B$, (2) implies that
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\begin{align*}
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\norm{p(T)}_{M_n(\complex)} &\le \max_{z \in B}\norm{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} \\
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&\le \max_{W \in U(M_n(\complex))}\norm{p(U(W))}_{M_n(\complex)}
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\end{align*}
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(4): For any $W \in U(M_n(\complex))$, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} implies that
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\[
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\norm{p(W)}_{M_n(\complex)} \le \max_{\lambda \in \sigma_{M_n(\complex)}(W)}|p(\lambda)| \le \max_{\lambda \in \partial B_\complex(0, 1)}|p(\lambda)| = \norm{p}_{H^\infty(D)}
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\]
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As the above holds for all unitary matrices,
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\[
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\norm{p(T)}_{M_n(\complex)} \le \max_{W \in U(M_n(\complex))}\norm{p(U(W))}_{M_n(\complex)} \le \norm{p}_{H^\infty(D)}
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\]
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by (3).
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\end{proof}
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