Spectrum of Element of Banach Algebra is Bounded

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 $\map {\sigma_A} x$ is bounded, and in particular:
 * $\cmod \lambda \le \norm x$ for all $\lambda \in \map {\sigma_A} x$

Proof
Suppose first that $\struct {A, \norm {\, \cdot \,} }$ is unital.

Let $\map G A$ be the group of units.

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

Then from, we have:
 * $\ds \norm {\frac x \lambda} < 1$

From Element of Unital Banach Algebra Close to Identity is Invertible:
 * $\ds {\mathbf 1}_A - \frac x \lambda \in \map G A$

Hence:
 * $\lambda {\mathbf 1}_A - x \in \map G A$

So, we have:
 * $\set {\lambda \in \C : \cmod \lambda > \norm x} \subseteq \set {\lambda \in \C : \lambda {\mathbf 1}_A - x \in \map G A}$

so that:
 * $\set {\lambda \in \C : \lambda {\mathbf 1}_A - x \not \in \map G A} \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm x}$

So we have:
 * $\map {\sigma_A} x \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm x}$

in the case that $\struct {A, \norm {\, \cdot \,} }$ is unital.

Now suppose that $\struct {A, \norm {\, \cdot \,} }$ is not unital.

Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.

From the definition of the spectrum in a non-unital algebra, we have:
 * $\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$

From the unital case, we have that $\map {\sigma_{A_+} } {\tuple {x, 0} }$ is bounded with:
 * $\cmod \lambda \le \norm {\tuple {x, 0} }_{A_+}$ for all $\lambda \in \map {\sigma_A} x$.

We have:
 * $\norm {\tuple {x, 0} }_{A_+} = \norm x$

and so:
 * $\cmod \lambda \le \norm x$ for all $\lambda \in \map {\sigma_A} x$.