# spectral permanence theorem

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

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

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

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

 $\sigma_{\mathcal{A}}(x)\subseteq\sigma_{\mathcal{B}}(x).$

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

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

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

 $\partial\sigma_{\mathcal{B}}(x)\subseteq\sigma_{\mathcal{A}}(x).$

Since the spectrum is a non-empty compact set in $\mathbb{C}$, one can decompose $\mathbb{C}-\sigma_{\mathcal{A}}(x)$ into its connected components, obtaining an unbounded component $\Omega_{\infty}$ together with a sequence of bounded components $\Omega_{1},\Omega_{2},\dots$,

 $\mathbb{C}-\sigma_{\mathcal{A}}(x)=\Omega_{\infty}\cup\Omega_{1}\cup\Omega_{2}\cup\cdots$

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

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

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

Title spectral permanence theorem SpectralPermanenceTheorem 2013-03-22 17:29:50 2013-03-22 17:29:50 asteroid (17536) asteroid (17536) 5 asteroid (17536) Theorem msc 46H10 msc 46H05