Spectrum of Element of Unital Commutative Banach Algebra

Theorem

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

Let $\Phi_A$ be the spectrum of $A$.

Let $x \in A$.

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

Then:

$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A}$

Corollary 1

Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative Banach algebra over $\C$ that is not unital as an algebra.

Then:

$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A} \cup \set 0$

Corollary 2

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

Let $\Phi_A$ be the spectrum of $A$.

Then $\Phi_A \ne \O$.

Proof

We have $\lambda \in \map {\sigma_A} x$ if and only if:

$\lambda {\mathbf 1}_A - x$ is not invertible.

From Element of Unital Commutative Banach Algebra is Invertible iff not in Kernel of Character, this is the case if and only if there exists a character $\phi$ such that:

$\map \phi {\lambda {\mathbf 1}_A - x} = 0$

From linearity, we have:

$\map \phi x = \lambda \map \phi { {\mathbf 1}_A}$

in this case.

From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have $\map \phi { {\mathbf 1}_A} = 1$.

Hence we have:

$\lambda \in \map {\sigma_A} x$ if and only if there exists a character $\phi$ with $\lambda = \map \phi x$.

This is precisely the statement:

$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A}$

$\blacksquare$