Let $H$ be a complex Hilbert space.

It is well known that if $T:H \to H$ is a normal operator, then $$\sigma(T)=\sigma_{ap}(T),$$

where $\sigma_{ap}(T)$ is defined as: $\lambda \in\sigma_{ap}(T)$ iff there exists a sequence $(x_n)_{n \in \mathbb N}$ with $\Vert x_n \Vert = 1$ for all $n \in \mathbb N$ such that $$\lim_{n \to \infty} \Vert Tx_n - \lambda x_n \Vert = 0,$$

If $T$ is hyponormal i.e. $T^*T\geq TT^*$, is $$\sigma(T)=\sigma_{ap}(T)?$$

Thanks for your help.


No. As usual, the counterexample is the unilateral shift $S$. We have $\sigma_{ap}(S)=\mathbb T$, while $\sigma(S)=\overline{\mathbb D}$.

  • $\begingroup$ Please what is $\mathbb T$? $\endgroup$ – Schüler Apr 27 '19 at 16:30
  • $\begingroup$ The unit circle (the "one-dimensional torus"). $\endgroup$ – Martin Argerami Apr 27 '19 at 16:47

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