Added second proof for von Neumann.

src/op/system/dilation.tex

@@ -120,3 +120,74 @@
 
     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}
+\begin{proof}[Proof, {{\cite[Exercise 2.15]{Paulsen}}}. ]
+    Assume without loss of generality that $\norm{T}_{B(H)} < 1$. Let
+    \[
+    P(T; t) = \frac{1}{1 - e^{-it}T} + \frac{1}{1 - e^{it}T^*} - 1
+    \]
+    
+    
+    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},
+    \[
+        p(T) = \frac{1}{2\pi i}\int_\gamma \frac{p(\lambda)}{\lambda - T} d\lambda
+    \]
+
+    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}, 
+    \begin{align*}
+    \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 \\
+    &= \sum_{k = 0}^\infty T^k\int_{\gamma} \lambda^{-(n+k+1)}d\lambda = \begin{cases}
+        1 & n = 0 \\
+        0 & n > 0
+    \end{cases}
+    \end{align*}
+
+    By linearity and \hyperref[Cauchy's integral formula]{theorem:cauchy-formula}, 
+    \[
+    \frac{1}{2\pi i}\int_\gamma\frac{\ol{q(\lambda)}}{\lambda - T} d\lambda = \ol{q(0)} = \frac{1}{2\pi i}\int_\gamma \frac{\ol{q(\lambda)}}{\lambda} d\lambda
+    \]
+
+    so
+    \begin{align*}
+        p(T) + \overline{q(0)} &= \frac{1}{2\pi i} \int_\gamma \frac{p(\lambda) + \ol{q(\lambda)}}{\lambda - T} d\lambda \\
+        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 \\
+        &= \frac{1}{2\pi}\int_0^{2\pi}\frac{p(e^{it}) + \ol{q(e^{it})}}{1 - e^{-it}T}dt
+    \end{align*}
+    
+    Similarly, 
+    \[
+    q(T)^* + p(0) = \frac{1}{2\pi}\int_0^{2\pi}\frac{p(e^{it}) + \ol{q(e^{it})}}{1 - e^{it}T^*}dt
+    \]
+
+    Therefore
+    \begin{align*}
+        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 \\
+        p(T) + q(T)^* &= \frac{1}{2\pi}\int_0^{2\pi}[p(e^{it}) + \ol{q(e^{it})}]P(T; t)dt
+    \end{align*}
+    
+    Now, for each $t \in [0, 2\pi]$, 
+    \[
+    (1 - e^{-it}T)(1 - e^{it}T^*) = 1 - e^{-it}T - e^{it}T^* + TT^*
+    \]
+    
+    so by the resolvent identity, 
+    \begin{align*}
+        P(T; t) &= \frac{1}{1 - e^{-it}T} + \frac{1}{1 - e^{it}T^*} - 1 \\
+        &= (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}, 
+    \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)\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\\
+        &\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}
+    \end{align*}
+
+    Since $I_H(\xi) = \frac{1}{2\pi}\int_0^{2\pi} P(T; t)dt$, 
+    \[
+    \frac{1}{2\pi}\int_0^{2\pi}\dpn{\xi, \xi}{t} dt = \frac{1}{2\pi}\int_0^{2\pi} \dpn{P(T; t)\xi, \xi}{H}dt = \norm{\xi}_H^2
+    \]
+
+    Therefore $\dpn{p(T)\xi, \eta}{H} \le \norm{p}_{H^\infty(D)}\norm{\xi}_H\norm{\eta}_H$, and $\norm{p(T)}_H \le \norm{p}_{H^\infty(D)}$.
+    
+    
+\end{proof}