33.5 The Spectrum

Proof. By Proposition 33.2.3, the mapping $y \mapsto y^{-1}$ is smooth on $G(A)$. Hence $R_{x}: \complex \setminus \sigma_{A}(x) \to A$ is holomorphic.$\square$

Proof.

$\square$

Proof. Assume for contradiction that $\sigma_{A}(x) = \emptyset$, then by Definition 33.5.3, the resolvent

is an entire function. Since

which tends to $0$ as $|\lambda| \to \infty$, $R_{x} \in H(\complex; A) \cap C_{0}(\complex; A)$. By Liouville’s Theorem, $R_{x} = 0$, which is impossible.$\square$

Proof. Let $x \in A$. By Proposition 33.5.5, there exists $\lambda \in \sigma_{A}(x)$. Since $\lambda \cdot 1 - x$ is not invertible and every non-zero element of $A$ is invertible, $x = \lambda \cdot 1$. Therefore the mapping $\complex \to A$ defined by $\lambda \mapsto \lambda \cdot 1$ is an isometric isomorphism.$\square$

Proof, [Theorem 1.8, Fol16]. Let $r = \limsup_{n \to \infty}\normn{x^n}_{A}^{1/n}$, then for any $\lambda \in \complex$ with $|\lambda| > r$, the series $\sum_{n = 0}^{\infty} \lambda^{-n-1}x^{n}$ converges absolutely, and to the inverse of $(x - \lambda)$. Therefore $r \ge [\cdot]_{sp}$.

Let $D = \bracs{\lambda \in \complex|\ |\lambda| > [x]_{sp}}$, then since the series $\sum_{n = 0}^{\infty} \lambda^{-n-1}x^{n}$ is the expansion of $R_{x}$ at infinity, which is defined on $D$, it must converge on $D$. Let $\phi \in A^{*}$, then by the preceding discussion, the series $\sum_{n = 0}^{\infty} \lambda^{-n-1}\dpn{x^n, \phi}{A}$ converges on $D$. Thus for any $\lambda \in D$,

By the Uniform Boundedness Principle,

Therefore $\limsup_{n \to \infty}\norm{x^n}_{A}^{1/n}\le [x]_{sp}$.$\square$

Proof. By Proposition 33.2.3, $G(A)$ and hence the resolvent set of $x$ is open, so $\sigma_{A}(x)$ is closed. By Proposition 33.5.7, $[x]_{sp}\le \norm{x}_{A}$.$\square$

Proof. (1): Let $\lambda \in \complex \setminus \bracs{0}$, then by Proposition 33.2.5, $\lambda - xy \in G(A)$ if and only if $\lambda - yx \in G(A)$.$\square$

Proof, [Proposition I.2.9, Tak01]. Let $x \in A$ with $\sigma_{A}(x) \subset U$, and $\lambda \in U^{c}$. By Proposition 33.2.3, for any $y \in A$ with $\norm{y}_{A} \le \normn{(\lambda - x)^{-1}}_{A}^{-1}$, $\lambda - x - y \in G(A)$ as well. Since the mapping $\lambda \mapsto \normn{(\lambda - x)^{-1}}_{A}$ vanishes at infinity and $U^{c}$ is closed, $\delta = \inf_{\lambda \in U^c}\normn{(\lambda - x)^{-1}}_{A}^{-1}> 0$. Therefore for every $z \in B_{A}(x,\delta)$, $\sigma_{A}(x) \subset U$.$\square$

Proof. For any $z \in A$, denote $R(z) = \bracs{(\lambda - z)^{-1}|\lambda \in \sigma_A(z)}$. Let $B \subset A$ be the closed subalgebra generated by $1$, $x$, $y$, $R(x)$, $R(y)$, $R(xy)$, $R(x + y)$, then $B$ is a commutative algebra with $\sigma_{B}(x) = \sigma_{A}(x)$, $\sigma_{B}(y) = \sigma_{A}(y)$, $\sigma_{B}(xy) = \sigma_{A}(xy)$, and $\sigma_{B}(x + y) = \sigma_{A}(x + y)$.

By Proposition 33.8.2, for each $z \in B$, $(\Gamma_{B}z)(\Omega(B)) = \sigma_{B}(z)$. Since for any $u, v \in C(\Omega(B); \complex)$,

1. (1)

   $\sigma_{C(\Omega(B); \complex)}(u + v) \subset \sigma_{C(\Omega(B); \complex)}(u) + \sigma_{C(\Omega(B); \complex)}(v)$.

2. (2)

   $\sigma_{C(\Omega(B); \complex)}(uv) \subset \sigma_{C(\Omega(B); \complex)}(u)\sigma_{C(\Omega(B); \complex)}(v)$.

The above holds for $x$ and $y$ with respect to $\sigma_{A}$.$\square$

Proof. (1): $G(B) \subset G(A)$.

(2): Let $\lambda \in \partial \sigma_{B}(x)$, then there exists $\seq{\lambda_n}\subset \complex \setminus \sigma_{B}(x)$ such that $\lambda_{n} - x \in G(B)$ for all $n \in \natp$, and $\lambda_{n} \to \lambda$ as $n \to \infty$. By Corollary 33.2.4, $\norm{(\lambda_n - x)^{-1}}_{A} \to \infty$ as $n \to \infty$. If $\lambda - x \in G(A)$, then $(\lambda_{n} - x)^{-1}\to (\lambda - x)^{-1}$ as $n \to \infty$. Thus $\norm{(\lambda - x)^{-1}}_{A} = \infty$, which is impossible. Therefore $\lambda - x \not\in G(A)$, and $\lambda \in \sigma_{A}(x)$.$\square$

Proof, [Theorem 4.9, Mar21]. By construction, $P \subset \complex \setminus \sigma_{B}(x)$. In addition, for any polynomial $p \in \complex[z]$, $p(x) \in B$. Thus for every rational function $f \in \complex(z) \cap H(\complex_{\infty} \setminus (P \cup \bracs{\infty}); \complex)$, $f(x) \in B$.

By Runge’s Theorem, $H(\complex_{\infty} \setminus (P \cup \bracs{\infty}); \complex)$ is dense in $H(\sigma_{A}(x); \complex)$. The continuity of the holomorphic functional calculus then implies that $f(x) \in B$ for all $f \in H(\sigma_{A}(x); \complex)$. In particular, $(\lambda - x)^{-1}\in B$ for all $\lambda \in \complex \setminus \sigma_{A}(x)$. Therefore $\sigma_{B}(x) \subset \sigma_{A}(x)$, and $\sigma_{B}(x) = \sigma_{A}(x)$ by Proposition 33.5.12.$\square$