Spectrum of Element of Banach Algebra is Non-Empty

Theorem

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

Let $x \in A$.

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

Then $\map {\sigma_A} x \ne \O$.

Proof

Suppose first that $\struct {A, \norm {\, \cdot \,} }$ is a unital banach algebra.

Let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$.

Aiming for a contradiction, suppose that $\map {\sigma_A} x = \O$.

Then $\map {\rho_A} x = \C$.

Define $R : \C \to A$ by:

$\map R \lambda = \paren {\lambda {\mathbf 1}_A - x}^{-1}$

for each $\lambda \in \C$.

From Resolvent Mapping is Analytic: Banach Algebra, $R$ is analytic.

With view to apply Liouville's Theorem (Complex Analysis): Banach Space, we show that $R$ is bounded.

Let $\lambda \in \C$ be such that $\cmod \lambda > \norm x + 1$.

Then $\cmod \lambda > \norm x$, and so:

$\ds \norm {\frac x \lambda} < 1$

from Norm Axiom $\text N 2$: Positive Homogeneity.

From Element of Unital Banach Algebra Close to Identity is Invertible, we have:

$\ds \norm {\paren { {\mathbf 1}_A - \frac x \lambda}^{-1} } \le \frac 1 {1 - \norm {\frac x \lambda} } = \frac {\cmod \lambda} {\cmod \lambda - \norm x}$

by Norm Axiom $\text N 2$: Positive Homogeneity.

Hence, we have:

$\ds \norm {\paren {\lambda {\mathbf 1}_A - x}^{-1} } = \cmod \lambda^{-1} \norm {\paren { {\mathbf 1}_A - \frac x \lambda}^{-1} } \le \frac 1 {\cmod \lambda - \norm x} < 1$

From Resolvent Mapping is Continuous: Banach Algebra, $R$ is continuous.

From Continuous Function on Compact Space is Bounded, there exists $M > 0$ such that:

$\norm {\paren {\lambda {\mathbf 1}_A - x}^{-1} } \le M$ for $\lambda \in \C$ with $\cmod \lambda \le \norm x + 1$.

Then:

$\norm {\paren {\lambda {\mathbf 1}_A - x}^{-1} } \le \max \set {1, M}$ for all $\lambda \in \C$.

From Liouville's Theorem (Complex Analysis): Banach Space, $R$ is constant.

Then:

$\paren {2 {\mathbf 1}_A - x}^{-1} = \paren { {\mathbf 1}_A - x}^{-1}$.

Note that for $y, z \in A$ we have:

$y^{-1} = z^{-1}$ if and only if $y = z$.

So, we have:

$2 {\mathbf 1}_A - x = {\mathbf 1}_A - x$

giving:

${\mathbf 1}_A = {\mathbf 0}_A$

This is impossible since $\struct {A, \norm {\, \cdot \,} }$ is unital.

Hence we have reached a contradiction and we have $\map {\sigma_A} x \ne \O$.

Suppose now that $\struct {A, \norm {\, \cdot \,} }$ is not unital.

Then, we have:

$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$

where $A_+$ is the normed unitization of $A$.

From the previous case, we have $\map {\sigma_{A_+} } {\tuple {x, 0} } \ne \O$ and hence $\map {\sigma_A} x \ne \O$.

$\blacksquare$