$\newcommand{\cH}{\mathcal{H}}
\newcommand{\CC}{\mathbb{C}}$
Let $\cH$ be a Hilbert space. A linear operator $T: \cH \to \cH$ is said to be *power bounded* if $\sup_{n \geq 0} \|T^n\| < \infty$.

If $T$ is a power bounded operator and $r(T)$ is the spectral radius of $T$, then clearly $r(T) = \lim_{n \to \infty} \|T^n\|^{1/n} \leq 1$, so that the spectrum of $T$ is contained in the closed unit disk. In fact, the spectrum may even be the whole closed disk, as is the case for the left and right shifts in $\ell_2(\mathbb{N})$. Now, in the case of the right shift $S(x_0, x_1, \ldots) = (0, x_0, x_1, \ldots)$, the residual spectrum is the open disk $\sigma_R(S) = \{|\lambda| < 1\}$, whereas the continuous spectrum is the circle $\sigma_C(S) = \{|\lambda| = 1\}$.

In his book An Introduction to Models and Decompositions in Operator Theory, Kubrusly asks: does every power bounded operator satisfy $\sigma_R(T) \subset \{|\lambda| < 1\}$? (I can't give the exact page as I don't have the book with me right now).

Now I couldn't find anything online about this. Does anyone know the status of this problem?

My initial feeling was that it should be false, but some attempts at a counterexample via a modification of the shift were not successful.