Spectral radius

Assume the Axiom of Choice (The Axiom of Choice). Let $A$ be a nonzero unital complex Banach algebra and let $a\in A$. By Spectrum is nonempty compact and norm bounded the spectrum ${\sigma }_A(a)$ (Spectrum and resolvent set in a Banach algebra) is a nonempty compact subset of $\mathbb{C}$ contained in the closed disc of radius $\mathrm{\parallel }a\mathrm{\parallel }$. The real-valued function $z\mapsto \mathrm{\mid }z\mathrm{\mid }$ is continuous, so its image $\mathrm{\mid }{\sigma }_A(a)\mathrm{\mid }$ is a nonempty compact subset of $[0,\mathrm{\infty })$ and has a maximum. The spectral radius of $a$ is the real number

$r(a):=\max \{\mathrm{\mid }z\mathrm{\mid }:z\in {\sigma }_A(a)\}\mathrm{.}$

It satisfies $0\le r(a)\le \mathrm{\parallel }a\mathrm{\parallel }$. When the ambient algebra must be recorded, the notation is $r_A(a)$; for a bounded operator $T$ on a nonzero complex Banach space the convention is that $r(T)=\max \{\mathrm{\mid }z\mathrm{\mid }:z\in \sigma (T)\}$ is computed in the algebra $\mathcal{B}(X)$ of bounded operators, whose spectrum convention is fixed by the spectrum definition above.

• The maximum is a maximum because the spectrum is compact and nonempty. This is the only place where the Axiom of Choice enters the definition: it is inherited from Spectrum is nonempty compact and norm bounded, whose nonemptiness proof uses Hahn–Banach separation. Selecting the maximum of the compact set of moduli uses no further choice, since a nonempty compact subset of $\mathbb{R}$ contains its supremum.

• Monotonicity under containment. If $B$ is a closed unital subalgebra of $A$ containing $a$ and the same unit, then $B$ is a unital Banach algebra in the inherited norm and ${\sigma }_A(a)\subseteq {\sigma }_B(a)$: invertibility in $B$ implies invertibility in $A$. Consequently $r_A(a)\le r_B(a)$.

• Constancy on scalar multiples. For $\lambda \in \mathbb{C}$ one has $\sigma (\lambda 1)=\{\lambda \}$ and hence $r(\lambda 1)=\mathrm{\mid }\lambda \mathrm{\mid }$: the element $\lambda 1$ is the unit rescaled, and $z1-\lambda 1=(z-\lambda )1$ is invertible exactly when $z\mathrm{\ne }\lambda$. This computation is used in the counterexample cex-norm-need-not-equal-spectral-radius.

• The radius is not the norm in general. The inequality $r(a)\le \mathrm{\parallel }a\mathrm{\parallel }$ is strict for many elements; the definitive relation $r(a)={\lim }_n\mathrm{\parallel }{a}^n{\mathrm{\parallel }}^{1\mathrm{/}n}$ is the theorem Spectral radius formula. In particular $r(a)=0$ is possible for nonzero $a$, and then ${\sigma }_A(a)=\{0\}$.

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