spectrum is a non-empty compact set
Remark : For Banach algebras over 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 be the identity element of . Let denote the spectrum of the element .
- For each such that one has , and so, by the Neumann series (http://planetmath.org/NeumannSeriesInBanachAlgebras), is invertible. Since
we see that is also invertible.
Let be the function defined by
It is known that the set of the invertible elements of is open (see this entry (http://planetmath.org/InvertibleElementsInABanachAlgebraFormAnOpenSet)).
As is a bounded closed subset of , it is compact.
Non-emptiness - Suppose that was empty. Then the resolvent is defined in .
We can see that is bounded since it is continuous in the closed disk and, for , we have (again, by the Neumann series (http://planetmath.org/NeumannSeriesInBanachAlgebras))
and therefore , which shows that is bounded.
The resolvent function, , is analytic (http://planetmath.org/BanachSpaceValuedAnalyticFunctions) (see this entry (http://planetmath.org/ResolventFunctionIsAnalytic)). As it is defined in , 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 converges to as we see that must be identically zero.
Thus, we have arrived to a contradiction since is not invertible.
Therefore is non-empty.
|Title||spectrum is a non-empty compact set|
|Date of creation||2013-03-22 17:25:05|
|Last modified on||2013-03-22 17:25:05|
|Last modified by||asteroid (17536)|