40.5 The Disk Algebra
Definition 40.5.1 (Disk Algebra).label Let $D = B_{\complex}(0, 1)$, then the disk algebra $A(D) = H(D; \complex) \cap C(\ol D; \complex)$ is the space of holomorphic functions on $D$ that extend to $\ol D$, equipped with the uniform norm.
Proposition 40.5.2.label Let $f \in A(D)$, then $\sigma(f) = f(\ol D)$.
Proposition 40.5.3.label Let $D = B_{\complex}(0, 1)$ and $A(D)$ be the disk algebra, then the mapping
is a homeomorphism.
Proof. Let $\phi \in \Omega(D)$. By Proposition 37.7.2, $\norm{\phi}_{A(D)^*}= 1$. 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 Proposition 5.16.5.$\square$
Proposition 40.5.4.label Let $a, b, c \in \complex$ with $a \ne b$, and
then:
- (1)
For each $p \in \complex[z]$,
\[p(T) = \begin{bmatrix}p(a) &c \frac{p(a) - p(b)}{a - b} \\ 0 & p(b)\end{bmatrix}\] - (2)
If $|a|, |b| \le 1$, then for each $f \in A(D)$,
\[f(T) = \begin{bmatrix}f(a) &c \frac{f(a) - f(b)}{a - b} \\ 0 & f(b)\end{bmatrix}\] - (3)
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
\[|\varphi(a)|^{2}(|a - b|^{2} + |c|^{2}) \le |a - b|^{2}\] - (4)
If $|a|, |b| \le 1$, then $\norm{T}_{M_2(\complex)}\le 1$ if and only if
\[|\varphi(a)|^{2}(|a - b|^{2} + |c|^{2}) \le |a - b|^{2}\]
Proof. (1): Suppose that (1) holds for $p, q \in \complex[z]$, then
As (1) already holds for $1, z \in \complex[z]$, by induction, (1) holds for all $p \in \complex[z]$.
(3): For each $z \in \partial B_{\complex}(0, 1)$,
By the Maximum Modulus Theorem, $\varphi \in A(D)$ with $\norm{\varphi}_{A(D)}= 1$. By (2),
so $\norm{\varphi(T)}_{M_2(\complex)}\le 1$ if and only if
(3, $\Rightarrow$): Suppose that $\norm{T}_{M_2(\complex)}\le 1$. By continuity of the holomorphic functional calculus and von Neumann’s inequality, $\norm{\varphi(T)}_{M_2(\complex)}\le 1$, 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 $z \in \partial B_{\complex}(0, 1)$,
By the Maximum Modulus Theorem, $\psi \in A(D)$ with $\norm{\psi}_{A(D)}= 1$. Since $\norm{\psi(T)}_{M_2(\complex)}\le 1$, the holomorphic functional calculus and von Neumann’s inequality imply that
(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
On the other hand, if $|\varphi(a)|^{2}(|a - b|^{2} + |c|^{2}) \le |a - b|^{2}$, then
so
and $\norm{tT}_{M_2(\complex)}\le 1$ for all $t \in (0, 1)$. Therefore $\norm{T}_{M_2(\complex)}\le 1$.$\square$
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