From e1c2d65b16253eeba5120e08697eca31aefca664 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 13 Oct 2026 15:07:35 -0400 Subject: [PATCH] First draft of 1.10. --- src/op/example/disk.tex | 125 ++++++++++++++++++++++++++++++++++++++++ 1 file changed, 125 insertions(+) diff --git a/src/op/example/disk.tex b/src/op/example/disk.tex index 1db6809..230c479 100644 --- a/src/op/example/disk.tex +++ b/src/op/example/disk.tex @@ -26,4 +26,129 @@ 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 + \[ + \varphi: \complex \setminus \bracsn{b/|b|^2} \quad \varphi(z) = \frac{z - b}{1 - \ol b z} + \] + + 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(a)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), + \[ + \varphi(T) = \begin{bmatrix} + \varphi(a) & c\frac{\varphi(a)}{a - b} \\ + 0 & 0 + \end{bmatrix} \quad \norm{\varphi(T)}_{M_2(\complex)} = |\varphi(a)|^2 + \abs{c\frac{\varphi(a)}{a - b}}^2 + \] + + 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}, $\varphi \in A(D)$ 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 + \[ + \norm{T}_{M_2(\complex)} = \norm{\psi(\varphi(T))}_{M_2(\complex)} \le \norm{\psi}_{A(D)} = 1 + \] + + (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\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} +