resolvent function is analytic
Moreover, for each it has the power series
where the series converges absolutely for each in the open disk centered in given by
Proof : Analyticity is defined for functions whose domain is open.
Thus, we start by proving that is an open set in . To do so it is enough to prove that for every the open disk defined by (2) above is contained in .
Let and be such that
Then and by the Neumann series (http://planetmath.org/NeumannSeriesInBanachAlgebras) is invertible.
Since it follows that is invertible.
Hence, from the equality
we conclude that is also invertible, i.e. . Thus is open.
The above proof also pointed out that for every , is defined in the open disk of radius centered in .
We now prove the analyticity of the .
Taking inverses on the equality (3) above one obtains
Again, by the Neumann series (http://planetmath.org/NeumannSeriesInBanachAlgebras), one obtains
|Title||resolvent function is analytic|
|Date of creation||2013-03-22 17:29:36|
|Last modified on||2013-03-22 17:29:36|
|Last modified by||asteroid (17536)|