Spectrum of Element in Maximal Commutative Subalgebra of Unital Banach Algebra

Theorem

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

Let $C$ be a commutative subalgebra of $A$ that is maximal with respect to set inclusion.

Let $x \in C$.

Let $\map {\sigma_A} x$ and $\map {\sigma_C} x$ be the spectra of $x$ in $A$ and $C$ respectively.

Then:

$\map {\sigma_C} x = \map {\sigma_A} x$

Proof

Let $\map G A$ and $\map G C$ be the groups of units of $A$ and $C$ respectively.

From Maximal Commutative Subalgebra of Unital Algebra is Unital, $C$ is unital.

From Spectrum of Element in Unital Subalgebra, we have $\map {\sigma_A} x \subseteq \map {\sigma_C} x$.

Now let $\lambda \in \C \setminus \map {\sigma_A} x$.

Then $\lambda {\mathbf 1}_A - x \in \map G A$.

Let $z = \paren {\lambda {\mathbf 1}_A - x}^{-1}$.

We want to show that $z \in C$.

Let $y \in C$.

Since $y$ commutes with $\lambda {\mathbf 1}_A$ and $z$, we have:

$z y \paren {\lambda {\mathbf 1}_A - x} z = z \paren {\lambda {\mathbf 1}_A - x} y z$

Since:

$\paren {\lambda {\mathbf 1}_A - x} z = {\mathbf 1}_A$

and:

$z \paren {\lambda {\mathbf 1}_A - x} = {\mathbf 1}_A$

we have $z y = y z$.

So $C \cup \set z$ is a set of commuting elements.

Let $\widetilde C$ be the subalgebra generated by $C$.

From Subalgebra Generated by Commuting Elements is Commutative, $\widetilde C$ is commutative.

We also have $C \subseteq \widetilde C$.

Since $C$ maximal with respect to set inclusion, we have:

$\widetilde C = C$

Hence $z \in C$.

So, we have $\lambda {\mathbf 1}_A x \in \map G C$.

Hence, we have $\lambda \in \C \setminus \map {\sigma_C} x$.

So, we have:

$\C \setminus \map {\sigma_A} x \subseteq \C \setminus \map {\sigma_C} x$

From Set Complement inverts Subsets, we have:

$\map {\sigma_C} x \subseteq \map {\sigma_A} x$

Hence, we have:

$\map {\sigma_C} x = \map {\sigma_A} x$

$\blacksquare$