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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