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 $\sigma(S)=\overline{\mathbb D}$.