# Hyponormal operator and approximate spectrum

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.

## 1 Answer

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

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