Is the residual spectrum of every power bounded operator contained in the open unit disk?

Question

$\newcommand{\cH}{\mathcal{H}} \newcommand{\CC}{\mathbb{C}}$ 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.

Answer 1

Let us prove that $\sigma_R(T)\cap \partial D(0,1)$ is empty.

Suppose $\lambda\in \sigma_R(T)\cap \partial D(0,1) \subseteq \sigma_{ap}(T).$ We have $\overline \lambda \in \sigma_p(T^*)\cap \partial D(0,1).$

Let us write $T^* \phi=\overline\lambda \phi$ with $ \Vert \phi\Vert =1.$

As T is mean ergodic, we consider now $h\in H$ defined by: $$(y,h)= (P_{Ker (\overline\lambda- T^*)}y,\phi)=\lim_n {1\over n}\sum_{k=1}^n {{(({{T^*}\over{\overline \lambda}})^k y,\phi)}},\ \ \ y\in H$$ As $(\phi,h)=1,$ we get $h\neq 0.$ The following holds $$(y,Th)=(T^*y,h)= \lim_n {1\over n}\sum_{k=1}^n {{(({{T^*}\over{\overline \lambda}})^k T^*y,\phi)}} =(y,\lambda h), \ \ y \in H \ \Rightarrow \ \ T h =\lambda h, \ h\neq 0 $$ which is a contradiction because we assume $\lambda - T$ is injective. A similar reasoning works when H is a reflexive space.

Answer 2

Not a full answer, but some observations: First of all, if $\lambda$ is on the boundary of the disk and in the residual spectrum, then $T-\lambda$ cannot be bounded from below.

It is easy to make counterexamples of this kind in Banach spaces rather than Hilbert spaces, for instance consider multiplication by $\exp(ix)$ in $C[0,2\pi]$.