spectrum is a non-empty compact set

Theorem - Let $𝒜$ be a complex Banach algebra with identity element. The spectrum of each $a\in 𝒜$ is a non-empty compact set in $ℂ$.

Remark : For Banach algebras over $ℝ$ 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 $𝒜$. Let $\sigma (a)$ denote the spectrum of the element $a\in 𝒜$.

• - For each $\lambda \in ℂ$ 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 $\varphi :ℂ⟶𝒜$ be the function defined by

  $$\varphi (\lambda )=a-\lambda e$$

  It is known that the set $𝒢$ of the invertible elements of $𝒜$ is open (see this entry (http://planetmath.org/InvertibleElementsInABanachAlgebraFormAnOpenSet)).

  Since ${\varphi }^{-1}(𝒢)=ℂ-\sigma (a)$ and $\varphi$ is a continuous function we see that that $\sigma (a)$ is a closed set in $ℂ$.

  As $\sigma (a)$ is a bounded closed subset of $ℂ$, it is compact.

• Non-emptiness - Suppose that $\sigma (a)$ was empty. Then the resolvent $R_a$ is defined in $ℂ$.

  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$${\displaystyle \frac{{|\lambda |}^{-1}}{1-{|\lambda |}^{-1}\parallel a\parallel }}$

  $=$${\displaystyle \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 $ℂ$, 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 non-empty.$\mathrm{\square }$