Gelfand's Spectral Radius Formula/Banach Algebra

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Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.

Let $x \in A$.

Let $\map {r_A} x$ be the spectral radius of $x$ in $A$.


Then, we have:

$\ds \map {r_A} x = \inf_{n \mathop \in \N_{> 0} } \norm {x^n}^{1/n} = \lim_{n \mathop \to \infty} \norm {x^n}^{1/n}$


Proof

First suppose that $A$ is unital.

Let $\map {\sigma_A} x$ be the spectrum of $x$ in $A$.

Let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$.

Let $\lambda \in \map {\sigma_A} x$.

Let $n \in \N_{> 0}$.

From Spectral Mapping Theorem for Polynomials, we have $\lambda^n \in \map {\sigma_A} {x^n}$.

From Spectrum of Element of Banach Algebra is Bounded, we then have $\cmod \lambda^n \le \norm {x^n}$.

So, we have $\cmod \lambda \le \norm {x^n}^{1/n}$ for each $\lambda \in \map {\sigma_A} x$.

Taking the supremum over $\lambda \in \map {\sigma_A} x$, we have:

$\map {r_A} x \le \norm {x^n}^{1/n}$ for each $n \in \N_{> 0}$.

Taking the infimum over $n \in \N_{> 0}$, we have:

$\ds \map {r_A} x \le \inf_{n \mathop \in \N_{> 0} } \norm {x^n}^{1/n}$


Let $\phi \in A^\ast$.

Define $g : \map {\rho_A} x \to \C$ by:

$\map g \lambda = \map \phi {\paren {\lambda {\mathbf 1}_A - x}^{-1} }$

for each $\lambda \in \map {\rho_A} x$.

Let $\lambda \in \C$ be such that $\cmod \lambda > \norm x$.

From Spectrum of Element of Banach Algebra is Bounded, we have:

$\map {\sigma_A} x \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm x}$

Hence from Set Complement inverts Subsets:

$\set {\lambda \in \C : \cmod \lambda > \norm x} \subseteq \map {\rho_A} x$

So $\lambda$ is in the domain of $g$.

Then, we have:

$\ds \map g \lambda = \frac 1 \lambda \map \phi {\paren { {\mathbf 1}_A - \frac x \lambda} }^{-1}$

As shown in Element of Unital Banach Algebra Close to Identity is Invertible, we have:

$\ds \paren { {\mathbf 1}_A - \frac x \lambda}^{-1} = \sum_{n \mathop = 0}^\infty \paren {\frac x \lambda}^n$

Then, we have:

\(\ds \map g \lambda\) \(=\) \(\ds \frac 1 \lambda \map \phi {\sum_{n \mathop = 0}^\infty \paren {\frac x \lambda}^n}\)
\(\ds \) \(=\) \(\ds \frac 1 \lambda \map \phi {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {\frac x \lambda}^n}\)
\(\ds \) \(=\) \(\ds \frac 1 \lambda \lim_{N \mathop \to \infty} \map \phi {\sum_{n \mathop = 0}^N \paren {\frac x \lambda}^n}\) $\phi$ is continuous
\(\ds \) \(=\) \(\ds \frac 1 \lambda \lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \map \phi {\paren {\frac x \lambda}^n}\) $\phi$ is linear
\(\ds \) \(=\) \(\ds \frac 1 \lambda \sum_{n \mathop = 0}^\infty \map \phi {\paren {\frac x \lambda}^n}\)

This gives a Laurent series expansion for $g$ in $\set {\lambda \in \C : \cmod \lambda > \norm x}$.

By the definition of the spectral radius, we have:

$\set {\lambda \in \C : \cmod \lambda > \map {r_A} x} \subseteq \map {\rho_A} x$

From Uniqueness of Laurent Series, the Laurent series expansion we have determined for $g$ also holds on $\set {\lambda \in \C : \cmod \lambda > \map {r_A} x}$.

That is:

$\ds \map g \lambda = \frac 1 \lambda \sum_{n \mathop = 0}^\infty \map \phi {\paren {\frac x \lambda}^n}$

for $\cmod \lambda > \map {r_A} x$.

From Terms in Convergent Series Converge to Zero, we have:

$\ds \map \phi {\paren {\frac x \lambda}^n} \to 0$ for $\phi \in A^\ast$ and $\cmod \lambda > \map {r_A} x$.

Hence:

$\ds \frac {x^n} {\lambda^n} \weakconv 0$ for $\cmod \lambda > \map {r_A} x$.

From Weakly Convergent Sequence in Normed Vector Space is Bounded:

for each $\cmod \lambda > \map {r_A} x$ there exists $M_\lambda \in \openint 0 \infty$ such that:
$\ds \norm {\frac {x^n} {\lambda^n} } \le M_\lambda$ for each $n \in \N_{> 0}$.

From Norm Axiom $\text N 2$: Positive Homogeneity, we have:

$\norm {x^n} \le M_\lambda \size \lambda^n$

That is:

$\norm {x^n}^{1/n} \le M_\lambda^{1/n} \size \lambda$

for each $\cmod \lambda > \map {r_A} x$.

So, we have:

$\ds \limsup_{n \mathop \to \infty} \norm {x^n}^{1/n} \le \size \lambda$

Taking $\size \lambda \to \map {r_A} x$, we obtain:

$\ds \limsup_{n \mathop \to \infty} \norm {x^n}^{1/n} \le \map {r_A} x$

We now have:

\(\ds \map {r_A} x\) \(\le\) \(\ds \inf_{n \mathop \in \N_{> 0} } \norm {x^n}^{1/n}\)
\(\ds \) \(\le\) \(\ds \liminf_{n \mathop \to \infty} \norm {x^n}^{1/n}\)
\(\ds \) \(\le\) \(\ds \limsup_{n \mathop \to \infty} \norm {x^n}^{1/n}\)
\(\ds \) \(\le\) \(\ds \map {r_A} x\)

Then, we have:

$\ds \map {r_A} x = \inf_{n \mathop \in \N_{> 0} } \norm {x^n}^{1/n} = \limsup_{n \mathop \to \infty} \norm {x^n}^{1/n} = \inf_{n \mathop \to \infty} \norm {x^n}^{1/n}$

From Convergence of Limsup and Liminf:

$\ds \lim_{n \mathop \to \infty} \norm {x^n}^{1/n}$ exists and:
$\ds \map {r_A} x = \inf_{n \mathop \in \N} \norm {x^n}^{1/n} = \lim_{n \mathop \to \infty} \norm {x^n}^{1/n}$


Now, suppose that $\struct {A, \norm {\, \cdot \,} }$ is not unital.

Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.

We have:

\(\ds \map {r_A} x\) \(=\) \(\ds \sup_{\lambda \in \map {\sigma_A} x} \cmod \lambda\) Definition of Spectral Radius in Banach Algebra
\(\ds \) \(=\) \(\ds \sup_{\lambda \in \map {\sigma_{A_+} } {\tuple {x, 0} } } \cmod \lambda\) Definition of Spectrum in Non-Unital Algebra
\(\ds \) \(=\) \(\ds \map {r_{A_+} } {\tuple {x, 0} }\) Definition of Spectral Radius in Banach Algebra

Applying the formula for unital Banach algebras, we obtain:

$\ds \map {r_{A_+} } {\tuple {x, 0} } = \inf_{n \mathop \in \N_{> 0} } \norm {\tuple {x, 0}^n}_{A_+}^{1/n} = \lim_{n \mathop \to \infty} \norm {\tuple {x, 0}^n}_{A_+}^{1/n}$

We have:

$\norm {\tuple {x, 0}^n}_{A_+} = \norm {\tuple {x^n, 0} }_{A_+} = \norm {x^n}$

So, we have:

$\ds \map {r_A} x = \map {r_{A_+} } {\tuple {x, 0} } = \inf_{n \mathop \in \N_{> 0} } \norm {x^n}^{1/n} = \lim_{n \mathop \to \infty} \norm {x^n}^{1/n}$

$\blacksquare$

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