Disc algebra corrections.

src/op/example/disk.tex

@@ -21,7 +21,7 @@
     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}
@@ -56,7 +56,7 @@
 
         \item Suppose that $|a|, |b| < 1$ and let
         \[
-        \varphi: \complex \setminus \bracsn{b/|b|^2} \quad \varphi(z) = \frac{z - b}{1 - \ol b z}
+        \varphi: \complex \setminus \bracsn{b/|b|^2} \to \complex \quad \varphi(z) = \frac{z - b}{1 - \ol b z}
         \]
 
         then $\norm{T}_{M_2(\complex)} \le 1$ if and only if
@@ -85,7 +85,7 @@
             0 & p(b)q(b)
         \end{bmatrix}\\
         &= \begin{bmatrix}
-            p(a)q(a) &c \frac{p(a)q(a) + p(a)q(b)}{a - b} \\
+            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*}
@@ -102,7 +102,7 @@
     \varphi(T) = \begin{bmatrix}
         \varphi(a) & c\frac{\varphi(a)}{a - b} \\
         0 & 0
-    \end{bmatrix} \quad \norm{\varphi(T)}_{M_2(\complex)} = |\varphi(a)|^2  + \abs{c\frac{\varphi(a)}{a - b}}^2
+    \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
@@ -121,7 +121,7 @@
     |\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}, $\varphi \in A(D)$ with $\norm{\varphi}_{A(D)} = 1$. Since $\norm{\varphi(T)}_{M_2(\complex)} \le 1$, the \hyperref[holomorphic functional calculus]{definition:holomorphic-functional-calculus} and \hyperref[von Neumann's inequality]{theorem:von-neumann-inequality-polynomial} imply that 
+    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 
     \[
     \norm{T}_{M_2(\complex)} = \norm{\psi(\varphi(T))}_{M_2(\complex)} \le \norm{\psi}_{A(D)} = 1
     \]
@@ -136,9 +136,9 @@
     
     On the other hand, if $|\varphi(a)|^2(|a - b|^2 + |c|^2) \le |a - b|^2$, then
     \begin{align*}
-        |1 - t\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
+        |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