spectral permanence theorem

Let $𝒜$ be a unital complex Banach algebra and $ℬ\subseteq 𝒜$ a Banach subalgebra that contains the identity of $𝒜$.

For every element $x\in ℬ$ it makes sense to speak of the spectrum ${\sigma }_{ℬ}(x)$ of $x$ relative to $ℬ$ as well as the spectrum ${\sigma }_{𝒜}(x)$ of $x$ relative to $𝒜$.

We provide here three results of increasing sophistication which relate both these spectrums, ${\sigma }_{ℬ}(x)$ and ${\sigma }_{𝒜}(x)$. Any of the last two is usually refered to as the spectral permanence theorem.

- Let $ℬ\subseteq 𝒜$ be as above. For every element $x\in ℬ$ we have

$${\sigma }_{𝒜}(x)\subseteq {\sigma }_{ℬ}(x).$$

This first result is purely . It is a straightforward consequence of the fact that invertible elements in $ℬ$ are also invertible in $𝒜$.

The other inclusion, ${\sigma }_{ℬ}(x)\subseteq {\sigma }_{𝒜}(x)$, is not necessarily true. It is true, however, if one considers the boundary $\partial {\sigma }_{ℬ}(x)$ instead.

Theorem - Let $ℬ\subseteq 𝒜$ be as above. For every element $x\in ℬ$ we have

$$\partial {\sigma }_{ℬ}(x)\subseteq {\sigma }_{𝒜}(x).$$

Since the spectrum is a non-empty compact set in $ℂ$, one can decompose $ℂ-{\sigma }_{𝒜}(x)$ into its connected components, obtaining an unbounded component ${\mathrm{\Omega }}_{\mathrm{\infty }}$ together with a sequence of bounded components ${\mathrm{\Omega }}_1,{\mathrm{\Omega }}_2,\mathrm{\dots }$,

$$ℂ-{\sigma }_{𝒜}(x)={\mathrm{\Omega }}_{\mathrm{\infty }}\cup {\mathrm{\Omega }}_1\cup {\mathrm{\Omega }}_2\cup \mathrm{\cdots }$$

Of course there may be only a finite number of bounded components or none.

Theorem - Let $x\in ℬ\subseteq 𝒜$ be as above. Then ${\sigma }_{ℬ}(x)$ is obtained from ${\sigma }_{𝒜}(x)$ by adjoining to it some (possibly none) bounded components of $ℂ-{\sigma }_{𝒜}(x)$.

As an example, if ${\sigma }_{𝒜}(x)$ is the unit circle, then ${\sigma }_{ℬ}(x)$ can only possibly be the unit circle or the closed unit disk.