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Results 1 - 2 of 2 |
1. CMB Online first
| The Essential Spectrum of the Essentially Isometric Operator Let $T$ be a contraction on a complex, separable, infinite dimensional
Hilbert space and let $\sigma \left( T\right) $ (resp. $\sigma _{e}\left(
T\right) )$ be its spectrum (resp. essential spectrum). We assume that $T$
is an essentially isometric operator, that is $I_{H}-T^{\ast }T$ is compact.
We show that if $D\diagdown \sigma \left( T\right) \neq \emptyset ,$ then
for every $f$ from the disc-algebra,
\begin{equation*}
\sigma _{e}\left( f\left( T\right) \right) =f\left( \sigma _{e}\left(
T\right) \right) ,
\end{equation*}
where $D$ is the open unit disc. In addition, if $T$ lies in the class
$ C_{0\cdot }\cup C_{\cdot 0},$ then
\begin{equation*}
\sigma _{e}\left( f\left( T\right) \right) =f\left( \sigma \left( T\right)
\cap \Gamma \right) ,
\end{equation*}
where $\Gamma $ is the unit circle. Some related problems are also discussed.
Keywords:Hilbert space, contraction, essentially isometric operator, (essential) spectrum, functional calculus Categories:47A10, 47A53, 47A60, 47B07 |
2. CMB 1997 (vol 40 pp. 193)
| Finite rank operators and functional calculus on Hilbert modules over abelian $C^{\ast}$-algebras We consider the problem: If $K$ is a compact normal operator on a Hilbert
module $E$, and $f\in C_0(\Sp K)$ is a function which is zero in a
neighbourhood of the origin, is $f(K)$ of finite rank? We show that
this is the case if the underlying $C^{\ast}$-algebra is abelian, and that
the range of $f(K)$ is contained in a finitely generated projective
submodule of $E$.
Categories:55R50, 47A60, 47B38 |

