# spectrum is a non-empty compact set

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

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 $a\in\mathcal{A}$.

• - For each $\lambda\in\mathbb{C}$ such that $|\lambda|>\|a\|$ one has $\|\lambda^{-1}a\|<1$, 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 $\|a\|$, and therefore it is bounded.

Let $\phi:\mathbb{C}\longrightarrow\mathcal{A}$ be the function defined by

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

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

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

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

We can see that $R_{a}$ is bounded since it is continuous in the closed disk $|\lambda|<\|a\|$ and, for $\lambda>\|a\|$, we have (again, by the Neumann series (http://planetmath.org/NeumannSeriesInBanachAlgebras))

 $\displaystyle\|R_{a}(\lambda)\|$ $\displaystyle=$ $\displaystyle\|(a-\lambda e)^{-1}\|$ $\displaystyle=$ $\displaystyle\|\lambda^{-1}(e-\lambda^{-1}a)^{-1}\|$ $\displaystyle\leq$ $\displaystyle\frac{|\lambda|^{-1}}{1-|\lambda|^{-1}\|a\|}$ $\displaystyle=$ $\displaystyle\frac{1}{|\lambda|-\|a\|}$

and therefore $\displaystyle\lim_{|\lambda|\rightarrow\infty}R_{a}(\lambda)=0$, 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 $\mathbb{C}$, 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 $|\lambda|\rightarrow\infty$ 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.$\square$

Title spectrum is a non-empty compact set SpectrumIsANonemptyCompactSet 2013-03-22 17:25:05 2013-03-22 17:25:05 asteroid (17536) asteroid (17536) 10 asteroid (17536) Theorem msc 46H05