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Proceedings of the American Mathematical Society
Proceedings of the American Mathematical Society
ISSN 1088-6826(online) ISSN 0002-9939(print)

 

On asymptotic properties of several classes of operators


Authors: Stephen L. Campbell and Ralph Gellar
Journal: Proc. Amer. Math. Soc. 66 (1977), 79-84
MSC: Primary 47B15; Secondary 47A50
MathSciNet review: 0461187
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Abstract: Let $ p(T,{T^ \ast })$ be a polynomial in T and $ {T^ \ast }$ where T is a bounded linear operator on a separable Hilbert space. Let $ \Delta = \{ T\vert p(T,{T^ \ast }) = 0\} $. Then $ \Delta $ is said to be asymptotic for p if for every $ K > 0$, there exists an $ {\varepsilon _0} > 0$ and function $ \delta (\varepsilon ,K),{\lim _{\varepsilon \to 0}}\delta (\varepsilon ,K) = 0$, such that if $ \varepsilon < {\varepsilon _0},\left\Vert T \right\Vert \leqslant K$ , and $ \left\Vert {p(T,{T^ \ast })} \right\Vert < \varepsilon $, then there exists $ \hat T \in \Delta $ such that $ \left\Vert {T - \hat T} \right\Vert < \delta (\varepsilon ,K)$. It is observed that the hermitian, unitary, and isometric operators are asymptotic for the obvious polynomials. It is known that the normals are not asymptotic for $ p(T,{T^ \ast }) = {T^ \ast }T - T{T^ \ast }$. An example gives several negative results including one that says the quasinormals are not asymptotic for $ p(T,{T^\ast}) = T{T^\ast}T - {T^\ast}{T^2}$. It is shown that if p is any polynomial in just one of T or $ {T^ \ast }$, then $ \Delta $ is asymptotic for p.


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Additional Information

DOI: http://dx.doi.org/10.1090/S0002-9939-1977-0461187-5
PII: S 0002-9939(1977)0461187-5
Keywords: Normal, quasinormal, hyponormal, approximation, asymptotic, algebraic
Article copyright: © Copyright 1977 American Mathematical Society