Let $\ell^2 =\ell^2(\mathbb{Z})$. Choose $\theta \in ]0,1[$ and set:

$$Tx=(\theta x_{n-1} +(1-\theta)x_{n+1})_{n\in \mathbb{Z}}$$

for each $x=(x_n)_{n\in \mathbb{Z}}\in \ell^2$ (thus $T$ is a convex combination of the right and left shift operators).

It is easy to prove that, for every $\theta$, $T$ is a bounded linear operator of $\ell^2$ into itself, that $\lVert T\rVert =1$ and that $T$ is selfadjoint iff $\theta =\ frac{1}{2}$. Moreover $T$ is not compact: in fact, if $e^m:=(\delta_n^m)$ (so $e^m$ is a vector of the canonical base of $\ell^2$), one has:

$$|Te^m -Te^p|^2=\begin{cases} 0 &\text{, if } p=m \\ \theta^2 +(1-\theta)^2+1 &\text{, if } m=p+2 \text{ or } p=m+2 \\ 2\theta^2+2(1-\theta)^2 &\text{, otherwise} \end{cases} \; ,$$

thus $|Te^m-Te^p|^2> \theta^2+(1-\theta)^2>0$ for $m\neq p$; therefore the sequence $\{ Te^m\}_{m\in \mathbb{N}}$ does not contain any Cauchy's subsequence.

The problem is:

I am not able to find the spectrum of $T$.

About the eigenvalues, the only thing I know for sure is that $1$ is not in the point spectrum of $T$ for any value of $\theta$: in fact if $1$ were in the point spectrum $\sigma_P(T)$, then the eigenvectors would satisfy the linear recurrence:

$$x_n=\theta x_{n-1}+(1-\theta) x_{n+1} \; ,$$

hence they have to be sequences of the type:

$$x_n=A \left( \frac{\theta}{1-\theta}\right)^n +B$$

($A,B$ suitable constants); but a sequence like this doesn't belong to $\ell^2$ except in the trivial case $A=B=0$, which however doesn't give a valid eigenvector. Therefore $1\notin \sigma_P(T)$.

But now, what about other eigenvalues? And what about the residue and continuous spectra of $T$?

Any hint is welcome.

  • $\begingroup$ You can use your argument to show that the point spectrum is empty. If $\lambda$ is an eigenvalue, the eigenvectors satisfy $\lambda x_n=\theta x_{n-1}+(1-theta)x_{n+1}$, and heus are of the form $x_n=Ar_1^n+Br_2^n$ where $r_1,r_2$ are the solutions of $\lambda r=\theta+(1-\theta)r^2$. No such sequence is in $\ell^2(\mathbb{Z})$ unless $A=B=0$. $\endgroup$ – Julián Aguirre Mar 27 '11 at 9:38
  • $\begingroup$ question on notation, does your statement $\theta \in ]0,1[$ mean that the set is non-inclusive? $\endgroup$ – rcollyer Apr 18 '11 at 17:43

To determine the spectrum of $T$, let us first determine the one of the right shift $\tau$. Since $||\tau||=1$, $\mathrm{Sp}(\tau) \subset \bar{B}(0,1)$. But the same goes for the left shift $\tau^{-1}$, so $\mathrm{Sp}(\tau) \subset C(0,1)$. It is actually equal to $C(0,1)$: $(\ldots,0,1,\lambda,\lambda^2,\ldots,\lambda^n,0,\ldots)$ is an "almost eigenvector".

For every $c \in \mathbb{C}^{\times}$, $(c \theta \mathrm{Id} - \tau)(c (1-\theta) \mathrm{Id} - \tau^{-1}) = (1+c^2 \theta (1-\theta)) \mathrm{Id} - cT$, so $f(c)=\frac{1+c^2 \theta (1-\theta)}{c} \in \mathrm{Sp} (T)$ iff $c \theta \in \mathrm{Sp}(\tau)$ or $c (1- \theta) \in \mathrm{Sp}(\tau^{-1})$ iff $|c|=\theta^{-1}$ or $(1-\theta)^{-1}$.

Now $f(\mathbb{C}^{\times})=\mathbb{C}$, and $f(\theta^{-1} (1-\theta)^{-1} c^{-1})=f(c)$, so $\mathrm{Sp}(T)= \left\{ f(\theta e^{i \alpha}) \right\} = \left\{ \cos \alpha + i (1-2 \theta) \sin \alpha \right\}$ which is an ellipsis (flat when $\theta=1/2$).

EDIT: It is easy to check that for all $\lambda$, $\lambda \mathrm{Id} - \tau$ is injective and has dense range, hence the same is true for $T$.

  • $\begingroup$ I do really thank you, but I'm not too much into your notations: can you explain $C(0,1)$ and $\mathbb{C}^\times$? I guess $C(0,1)=\{ z|\ |z|=1\}$ and $\mathbb{C}^\times =\mathbb{C}\setminus \{ 0\}$, is it correct? $\endgroup$ – Pacciu Mar 27 '11 at 18:58
  • $\begingroup$ Yes, and $\bar{B}(0,1) = \left\{ z \ |\ |z| \leq 1 \right\}$. $\endgroup$ – Plop Mar 28 '11 at 19:28

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