TITLE: Eigenvalues of Left Shift + Right Shift in $l_2([0,\infty))$
QUESTION [2 upvotes]: This question appeared on an old final exam and I am having difficulty completing it for practice.
Let $S_r$ and $S_l$ be defined on the hilbert space $l_2[0,\infty)\to l_2[0,\infty)$ as the following: $S_r (x_0, x_1, x_2, \dots) \mapsto (0,x_0, x_1, \dots)$ and $S_l (x_0, x_1, x_2,\dots) \mapsto (x_1, x_2, x_3,\dots)$.  That is to say, $S_r$ is the right shift operator and $S_l$ is the left shift operator.
I fully understand how to describe the spectrum of each individually, however things become clouded once they are combined.
The question asks to:

Describe the spectrum of $S_r + S_l$

That is, $(S_r + S_l)(x_0, x_1, x_2,\dots) \mapsto (x_1, x_0+x_2, x_1+x_3,\dots)$
To begin I noted first that $S_r + S_l$ is self-adjoint which implies two things.  First, the spectrum will necessarily be a subset of $\mathbb{R}$.  Second, it implies that $R_\sigma(S_r+S_l)$ is empty.
Suppose that $\mu\in P_\sigma(S_r+S_l)$, then $\mu x = (S_r+S_l)x$ for some $x=(x_0,x_1,x_2,\dots)$.
Then $\mu(x_0,x_1,\dots) = (x_1,x_0+x_2, x_1+x_3, \dots, x_{n-1}+x_{n+1},\dots)$ and $x_1 = \mu x_0$, $x_{n+1} = \mu x_n - x_{n-1}$
So, for $x$ to be an eigenvector of $\mu$, we must have $x=x_0(1, \mu, \mu^2-1, \mu^3-2\mu, \mu^4-3\mu^2+1,\dots,\mu^n - (n-1)\mu^{n-2} + O(\mu^{n-4}+1),\dots) ~~~(\star)$
It follows that as $n\to\infty$ if $|\mu|>1$, then $|((S_l+S_r)x)_n|\to\infty$ and so no eigenvalues of $S_l+S_r$ are greater than 1.
For $\mu=1$ we must have $x_1 = x_0$, $x_2 = 0$, $x_3 = -x_0$, $x_{n+1} = x_n - x_{n-1}$
And we get that $x = x_0(1,1,0,-1,0,1,0,-1,\dots)$ which is only in $l_2[0,\infty)$ if $x_0 = 0$, so $1$ is not an eigenvalue of $S_l+S_r$.
In the case that $\mu=0$, by $(\star)$
$x = x_0(1,0,-1,0,1,\dots)$, and since we are in $l_2[0,\infty)$, $x_0=0$ and $x=0$, so $0$ is not an eigenvalue of $S_l+S_r$.

With this work so far, I was able to show as well that the operator norm is $2$ and the operator is definitely not compact (else $\pm 2$ would have been an eigenvalue)
It is at this point that I lose track of where to continue from here.  With what argument can I show that for $|\mu|<1$ that they are or aren't eigenvalues?  I expect but cannot find the right wording to show that there are in fact no eigenvalues.

REPLY [5 votes]: Even though this question is almost three years old: for anyone that stumbles across this problem like me, I want to give a simple argument on why the operator given by the sum of the right shift $S_r$ and left shift $S_l$ on $\ell_2(\mathbb N)$ does not have any eigenvalues. I will need a few well known results which can for example be looked up in Chapter VII.§1 of the book "Introduction to Hilbert Space" by Berberian (1976).

All eigenvalues of a bounded operator $T$ are bound by $\Vert T\Vert$.
All eigenvalues of a self-adjoint operator are real.
$\mu\in\mathbb C$ is eigenvalue of $S_l$ if and only if $|\mu|<1$.

As stated in the original post, $S_l+S_r$ is self-adjoint and $\Vert S_l+S_r\Vert\leq \Vert S_l\Vert+\Vert S_r\Vert=2$ so if that operator has eigenvalues, those have to be elements of the real interval $[-2,2]$. Let $\lambda\in[-2,2]$ be given and assume there exists $x\in\ell_2(\mathbb N)$ such that $(S_l+S_r)x=\lambda x$ (our goal will be to show that $x$ has to be the zero sequence). Applying $S_l$ to this equation from the left gives
$$
0=(S_l^2+S_lS_r-\lambda S_l)x=(S_l^2-\lambda S_l+\operatorname{id})x
$$
since $S_lS_r$ obviously is the identity $\operatorname{id}$ on $\ell_2(\mathbb N)$. Looking at the related equation $t^2-\lambda t+1=0$ yields the solutions
$$
t_\pm=\frac{\lambda}2\pm i\sqrt{1-\frac{\lambda^2}4}
$$
and thus we have
$$
(  S_l-t_+\operatorname{id} )(  S_l-t_-\operatorname{id} )  x=0.
$$
Note that $|t_\pm|=1$ since $\lambda\in[-2,2]$ so $t_\pm$ can not be an eigenvalue of $S_l$ as stated at the beginning. This implies $(  S_l-t_-\operatorname{id} )  x=0$ and further $ x=0$ so $\lambda$ is not an eigenvalue of $S_l+S_r$. In total we showed that $S_l+S_r$ does not have any eigenvalues.
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