spectrum is a nonempty compact set
Theorem  Let $\mathcal{A}$ be a complex Banach algebra^{} with identity element. The spectrum of each $a\in \mathcal{A}$ is a nonempty compact set in $\u2102$.
Remark : For Banach algebras over $\mathbb{R}$ the spectrum of an element is also a compact set, although it can be empty. To assure that it is not the empty set^{}, proofs usually involve Liouville’s theorem (http://planetmath.org/LiouvillesTheorem2) for of a complex with values in a Banach algebra.
Proof : Let $e$ be the identity element of $\mathcal{A}$. Let $\sigma (a)$ denote the spectrum of the element $a\in \mathcal{A}$.

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 For each $\lambda \in \u2102$ such that $\lambda >\parallel a\parallel $ one has $$, and so, by the Neumann series (http://planetmath.org/NeumannSeriesInBanachAlgebras), $e{\lambda}^{1}a$ is invertible. Since
$$a\lambda e=\lambda (e{\lambda}^{1}a)$$ we see that $a\lambda e$ is also invertible.
We conclude that $\sigma (a)$ is contained in a disk of radius $\parallel a\parallel $, and therefore it is bounded^{}.
Let $\varphi :\u2102\u27f6\mathcal{A}$ be the function defined by
$$\varphi (\lambda )=a\lambda e$$ It is known that the set $\mathcal{G}$ of the invertible elements of $\mathcal{A}$ is open (see this entry (http://planetmath.org/InvertibleElementsInABanachAlgebraFormAnOpenSet)).
Since ${\varphi}^{1}(\mathcal{G})=\u2102\sigma (a)$ and $\varphi $ is a continuous function^{} we see that that $\sigma (a)$ is a closed set^{} in $\u2102$.
As $\sigma (a)$ is a bounded closed subset of $\u2102$, it is compact.

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Nonemptiness  Suppose that $\sigma (a)$ was empty. Then the resolvent ${R}_{a}$ is defined in $\u2102$.
We can see that ${R}_{a}$ is bounded since it is continuous in the closed disk $$ and, for $\lambda >\parallel a\parallel $, we have (again, by the Neumann series (http://planetmath.org/NeumannSeriesInBanachAlgebras))
$\parallel {R}_{a}(\lambda )\parallel $ $=$ $\parallel {(a\lambda e)}^{1}\parallel $ $=$ $\parallel {\lambda}^{1}{(e{\lambda}^{1}a)}^{1}\parallel $ $\le $ $\frac{{\lambda }^{1}}{1{\lambda }^{1}\parallel a\parallel}$ $=$ $\frac{1}{\lambda \parallel a\parallel}$ and therefore $\underset{\lambda \to \mathrm{\infty}}{lim}{R}_{a}(\lambda )=0$, which shows that ${R}_{a}$ is bounded.
The resolvent function, ${R}_{a}$, is analytic^{} (http://planetmath.org/BanachSpaceValuedAnalyticFunctions) (see this entry (http://planetmath.org/ResolventFunctionIsAnalytic)). As it is defined in $\u2102$, it is a bounded entire function. Applying Liouville’s theorem (http://planetmath.org/LiouvillesTheorem2) we conclude that it must be constant (see this this entry (http://planetmath.org/BanachSpaceValuedAnalyticFunctions) for an idea of how holds for Banach space^{} valued functions).
Since ${R}_{a}(\lambda )$ converges^{} to $0$ as $\lambda \to \mathrm{\infty}$ we see that ${R}_{a}$ must be identically zero.
Thus, we have arrived to a contradiction^{} since $0$ is not invertible.
Therefore $\sigma (a)$ is nonempty.$\mathrm{\square}$
Title  spectrum is a nonempty compact set 

Canonical name  SpectrumIsANonemptyCompactSet 
Date of creation  20130322 17:25:05 
Last modified on  20130322 17:25:05 
Owner  asteroid (17536) 
Last modified by  asteroid (17536) 
Numerical id  10 
Author  asteroid (17536) 
Entry type  Theorem 
Classification  msc 46H05 