16
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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 $T$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.