Disc algebra corrections.

src/op/example/disk.tex

@@ -21,9 +21,134 @@
     is a homeomorphism. 
 \end{proposition}
 \begin{proof}
-    Let $\phi \in \Omega(D)$. By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A(D)^*} = 1$. Let $p$ be the identity polynomial, and $z_0 = \phi(p)$, then for every $q \in \complex[x]$, $\phi(q) = q(x_0)$. By density of polynomials in $A(D)$, $\phi(f) = f(z_0)$ for all $f \in A(D)$. Therefore $\ol D$ is in bijection with $\Omega(D)$. 
+    Let $\phi \in \Omega(D)$. By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A(D)^*} = 1$. Let $p$ be the identity polynomial, and $z_0 = \phi(p)$, then for every $q \in \complex[x]$, $\phi(q) = q(z_0)$. By density of polynomials in $A(D)$, $\phi(f) = f(z_0)$ for all $f \in A(D)$. Therefore $\ol D$ is in bijection with $\Omega(D)$. 
 
     Since $A(D) \subset C(\ol D)$, $E$ is continuous. As $\ol D$ is compact and $\Omega(D)$ is Hausdorff, $E$ is a homeomorphism by \autoref{proposition:compact-hausdorff-homeomorphism}. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:disk-algebra-calculus-upper-triangular}
+    Let $a, b, c \in \complex$ with $a \ne b$, and
+    \[
+    T = \begin{bmatrix}
+        a &c \\
+        0 &b
+    \end{bmatrix}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item For each $p \in \complex[z]$, 
+        \[
+        p(T) = \begin{bmatrix}
+            p(a) &c \frac{p(a) - p(b)}{a - b} \\
+            0 & p(b)
+        \end{bmatrix}
+        \]
+
+        \item If $|a|, |b| \le 1$, then for each $f \in A(D)$, 
+        \[
+        f(T) = \begin{bmatrix}
+            f(a) &c \frac{f(a) - f(b)}{a - b} \\
+            0 & f(b)
+        \end{bmatrix}
+        \]
+
+        \item Suppose that $|a|, |b| < 1$ and let
+        \[
+        \varphi: \complex \setminus \bracsn{b/|b|^2} \to \complex \quad \varphi(z) = \frac{z - b}{1 - \ol b z}
+        \]
+
+        then $\norm{T}_{M_2(\complex)} \le 1$ if and only if
+        \[
+        |\varphi(a)|^2(|a - b|^2 + |c|^2) \le |a - b|^2
+        \]
+        \item If $|a|, |b| \le 1$, then $\norm{T}_{M_2(\complex)} \le 1$ if and only if
+        \[
+        |\varphi(a)|^2(|a - b|^2 + |c|^2) \le |a - b|^2
+        \]
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}
+    (1): Suppose that (1) holds for $p, q \in \complex[z]$, then 
+    \begin{align*}
+        p(T)q(T) &= \begin{bmatrix}
+            p(a) &c \frac{p(a) - p(b)}{a - b} \\
+            0 & p(b)
+        \end{bmatrix}\begin{bmatrix}
+            q(a) &c \frac{q(a) - q(b)}{a - b} \\
+            0 & q(b)
+        \end{bmatrix} \\
+        &= \begin{bmatrix}
+            p(a)q(a) &c p(a)\frac{q(a) - q(b)}{a - b} + cq(b)\frac{p(a) - p(b)}{a - b} \\
+            0 & p(b)q(b)
+        \end{bmatrix}\\
+        &= \begin{bmatrix}
+            p(a)q(a) &c \frac{p(a)q(a) + p(b)q(b)}{a - b} \\
+            0 & p(b)q(b)
+        \end{bmatrix}
+    \end{align*}
+
+    As (1) already holds for $1, z \in \complex[z]$, by induction, (1) holds for all $p \in \complex[z]$. 
+
+    (3): For each $z \in \partial B_\complex(0, 1)$, 
+    \[
+    |\varphi(z)| = \abs{\frac{z - b}{1 - \ol{b}z}} = \abs{\frac{z - b}{\ol{z} - \ol{b}}} = 1
+    \]
+
+    By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, $\varphi \in A(D)$ with $\norm{\varphi}_{A(D)} = 1$. By (2), 
+    \[
+    \varphi(T) = \begin{bmatrix}
+        \varphi(a) & c\frac{\varphi(a)}{a - b} \\
+        0 & 0
+    \end{bmatrix} \quad \norm{\varphi(T)}_{M_2(\complex)}^2 = |\varphi(a)|^2  + \abs{c\frac{\varphi(a)}{a - b}}^2
+    \]
+
+    so $\norm{\varphi(T)}_{M_2(\complex)} \le 1$ if and only if
+    \begin{align*}
+        1 &\ge |\varphi(a)|^2 \paren{1 + \frac{|c|^2}{|a - b|^2}} \\
+        |a - b|^2 &\ge |\varphi(a)|^2 (|a - b|^2 + |c|^2)
+    \end{align*}
+    
+
+    (3, $\Rightarrow$): Suppose that $\norm{T}_{M_2(\complex)} \le 1$. By continuity of the \hyperref[holomorphic functional calculus]{definition:holomorphic-functional-calculus} and \hyperref[von Neumann's inequality]{theorem:von-neumann-inequality-polynomial}, $\norm{\varphi(T)}_{M_2(\complex)} \le 1$, and $|a - b|^2 \ge |\varphi(a)|^2 (|a - b|^2 + |c|^2)$.
+    
+    (3, $\Leftarrow$): Suppose that $|a - b|^2 \ge |\varphi(a)|^2 (|a - b|^2 + |c|^2)$, or equivalently that $\norm{\varphi(T)}_{M_2(\complex)} \le 1$. 
+    
+    Let $\psi(z) = \frac{z + b}{1 + \ol b z}$, then $\psi \circ \varphi = \text{Id}$. For each $z \in \partial B_\complex(0, 1)$, 
+    \[
+    |\psi(z)| = \abs{\frac{z + b}{1 + \ol b z}} = \abs{\frac{z + b}{\ol z + \ol b}} = 1
+    \]
+
+    By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, $\psi \in A(D)$ with $\norm{\psi}_{A(D)} = 1$. Since $\norm{\psi(T)}_{M_2(\complex)} \le 1$, the \hyperref[holomorphic functional calculus]{definition:holomorphic-functional-calculus} and \hyperref[von Neumann's inequality]{theorem:von-neumann-inequality-polynomial} imply that 
+    \[
+    \norm{T}_{M_2(\complex)} = \norm{\psi(\varphi(T))}_{M_2(\complex)} \le \norm{\psi}_{A(D)} = 1
+    \]
+    
+    (4): For each $t \in (0, 1)$, let $\varphi_t(z) = (z - tb)/(1 - t\ol b z)$. 
+    
+    If $\norm{T}_{M_2(\complex)} \le 1$, then $\norm{tT}_{M_2(\complex)} \le 1$ for all $t \in (0, 1)$, so
+    \begin{align*}
+        |\varphi(a)|^2(|a - b|^2 + |c|^2) &= \lim_{t \upto 1}|\varphi_t(ta)|^2(|ta - tb|^2 + |tc|^2) \\
+        &\le \lim_{t \upto 1}|ta - tb|^2 = |a - b|^2
+    \end{align*}
+    
+    On the other hand, if $|\varphi(a)|^2(|a - b|^2 + |c|^2) \le |a - b|^2$, then
+    \begin{align*}
+        |1 - t^2\ol b a|^2 &= 1 - 2t^2 \text{Re}(\ol b a) + t^4|a|^2|b|^2 \\
+        &= t^2(1 - 2\text{Re}(\ol b a) + |a|^2|b|^2) + (1 - t^2)(1 - t^2|a|^2|b|^2) \\
+        &\ge t^2(1 - 2\text{Re}(\ol b a) + |a|^2|b|^2) = t^2|1 - \ol ba|^2
+    \end{align*}
+    so
+
+    \begin{align*}
+        |\varphi_t(ta)|^2(|ta - tb|^2 + |tc|^2) &= t^2\abs{\frac{ta - tb}{1 - t^2\ol b a}}^2(|a - b|^2 + |c|^2) \\
+        &\le t^2 \abs{\frac{a - b}{1 - \ol b a}}^2(|a - b|^2 + |c|^2) \\
+        &\le |ta - tb|^2
+    \end{align*}
+
+    and $\norm{tT}_{M_2(\complex)} \le 1$ for all $t \in (0, 1)$. Therefore $\norm{T}_{M_2(\complex)} \le 1$. 
+\end{proof}
+
 

src/op/system/os.tex

@@ -181,7 +181,7 @@
     
     (3): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $\dpn{T\xi, \xi}{H} \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_H$. The sharp case is given by $I_H$. 
 
-    On the other hand, if $w(T) \le 1$, then $2 + T + T^* = 2 + 2\text{Re}(T) \ge 0$ implies that $\norm{\text{Re}(T)}_{B(H)} \le 1$. Similarly, $2 + iT - iT^* = 2 + 2\text{Im}(T) \ge 0$ implies that $\norm{\text{Im}(T)}_{B(H)} \le 1$, so $\norm{T}_{B(H)} \le \norm{\text{Re}(T)}_{B(H)} + \norm{\text{Im}(T)}_{B(H)} \le 2$. Therefore $\norm{T}_{B(H)} \le 2w(T)$. 
+    On the other hand, if $w(T) \le 1$, then $2 + T + T^* = 2 + 2\text{Re}(T) \ge 0$ and $2 - T - T^* = 2 - 2\text{Re}(T) \ge 0$ imply that $\norm{\text{Re}(T)}_{B(H)} \le 1$. Similarly, $\norm{\text{Im}(T)}_{B(H)} \le 1$ as well, so $\norm{T}_{B(H)} \le \norm{\text{Re}(T)}_{B(H)} + \norm{\text{Im}(T)}_{B(H)} \le 2$. Therefore $\norm{T}_{B(H)} \le 2w(T)$. 
 
     For any $\xi, \eta \in \complex$ with $\norm{(\xi, \eta)}_{\complex^2} = 1$, by \hyperref[Young's inequality]{lemma:young-inequality},
     \[