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Proceedings of the American Mathematical Society

Published by the American Mathematical Society since 1950, Proceedings of the American Mathematical Society is devoted to shorter research articles in all areas of pure and applied mathematics.

ISSN 1088-6826 (online) ISSN 0002-9939 (print)

The 2020 MCQ for Proceedings of the American Mathematical Society is 0.85.

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On asymptotic properties of several classes of operators
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by Stephen L. Campbell and Ralph Gellar PDF
Proc. Amer. Math. Soc. 66 (1977), 79-84 Request permission

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 $\Delta = \{ T|p(T,{T^ \ast }) = 0\}$. 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 $\varepsilon < {\varepsilon _0},\left \| T \right \| \leqslant K$ , and $\left \| {p(T,{T^ \ast })} \right \| < \varepsilon$, then there exists $\hat T \in \Delta$ such that $\left \| {T - \hat T} \right \| < \delta (\varepsilon ,K)$. 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
  • © Copyright 1977 American Mathematical Society
  • Journal: Proc. Amer. Math. Soc. 66 (1977), 79-84
  • MSC: Primary 47B15; Secondary 47A50
  • DOI: https://doi.org/10.1090/S0002-9939-1977-0461187-5
  • MathSciNet review: 0461187