29.4 The Spectrum
Definition 29.4.1 (Spectrum).label Let $A$ be a unital Banach algebra and $x \in A$, then
is the spectrum of $x$ in $A$, and its complement is the resolvent set of $x$.
Definition 29.4.2 (Spectral Radius).label Let $A$ be a unital Banach algebra and $x \in A$, then
is the spectral radius of $x$.
Definition 29.4.3 (Resolvent).label Let $A$ be a unital Banach algebra and $x \in A$, then
is the resolvent function of $x$, which is holomorphic on $\complex \setminus \sigma_{A}(x)$.
Proof. By Proposition 29.2.3, the mapping $y \mapsto y^{-1}$ is smooth on $G(A)$. Hence $R_{x}: \complex \setminus \sigma_{A}(x) \to A$ is holomorphic.$\square$
Lemma 29.4.4 (Resolvent Equation).label Let $A$ be a unital Banach algebra, $x \in A$, and $\lambda, \mu \in \complex \setminus \sigma_{A}(x)$, then
Proof.
$\square$
Proposition 29.4.5.label Let $A$ be a unital Banach algebra and $x \in A$, then $\sigma_{A}(x) \ne \emptyset$.
Proof. Assume for contradiction that $\sigma_{A}(x) = \emptyset$, then by Definition 29.4.3, 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.$\square$
Theorem 29.4.6 (Gelfand-Naimark).label Let $A$ be a unital Banach algebra. If every non-zero element of $A$ is invertible, then $A$ is isometrically isomorphic to $\complex$.
Proof. Let $x \in A$. By Proposition 29.4.5, 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.$\square$
Proposition 29.4.7.label Let $A$ be a unital Banach algebra and $x \in A$, then $[\cdot]_{sp}= \limsup_{n \to \infty}\normn{x^n}_{A}^{1/n}$.
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 $\limsup_{n \to \infty}\norm{x^n}_{A}^{1/n}\le [x]_{sp}$.$\square$
Proposition 29.4.8.label Let $A$ be a unital Banach algebra and $x \in A$, then $\sigma_{A}(x)$ is a compact subset of $B_{\complex}(0, \norm{x}_{A})$.
Proof. By Proposition 29.2.3, $G(A)$ and hence the resolvent set of $x$ is open, so $\sigma_{A}(x)$ is closed. By Proposition 29.4.7, $[x]_{sp}\le \norm{x}_{A}$.$\square$
Proposition 29.4.9.label Let $A$ be a unital Banach algebra and $x, y \in A$, then:
- (1)
$\sigma(xy) \cup \bracs{0}= \sigma(yx) \cup \bracs{0}$.
- (2)
$[xy]_{sp}= [yx]_{sp}$.
Proof. (1): Let $\lambda \in \complex \setminus \bracs{0}$, then by Proposition 29.2.4, $\lambda - xy \in G(A)$ if and only if $\lambda - yx \in G(A)$.$\square$
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