# The spectrum of a self-adjoint operator on $\mathcal l^2$

Let $S$ be the unilateral shift operator on $\mathcal l^2$ (which shifts one place to the right) and $S^*$ its adjoint, the backward shift (which shifts one place to the left). I've been trying to find the spectrum of $T=S+S^*$.

Since $T$ is self-adjoint, any eigenvalues would be real, and I've shown that no $|\lambda|\ge 2$ can be an eigenvalue. The case $|\lambda|< 2$ corresponds to the roots of $t^2=\lambda t-1$ being complex (this is the characteristic polynomial corresponding to the recurrence relation $a_n=\lambda a_{n-1}+a_{n-2}$). These complex roots have absolute value 1, and this implies that $a_n$ does not converge to zero.

So I've shown that $T$ has no eigenvalues. But I'm still looking for an elementary way to find that the spectrum is $[-2,2]$, as answered by Joel below. Although I appreciate Joel's answer below and I'm sure it'll be valuable for many in the community, unfortunately I only know basic Hilbert space theory. Any hints would be very helpful!

• Note that, for example, $(1,1,1,\dotsc)$ is an eigenvector with eigenvalue $2$. – Chappers Apr 29 '15 at 18:32
• No, it's not. It's not even in $\ell^2$. – Robert Israel Apr 29 '15 at 18:33

You know that $\|S\| \le 1$ which gives $\|S^{\star}\|=\|S\| \le 1$ and, hence, $\|S+S^{\star}\| \le 2$. However $S+S^{\star}$ is selfadjoint and, so, its norm and spectral radius are the same, and its spectrum is real. That means $\sigma(S+S^{\star})\subseteq[-2,2]$. So that part is fairly straightforward.

Suppose $\lambda \in [-2,2]$. You want to show that $\lambda$ is definitely in the spectrum. So, try to solve $$(S+S^{\star}-\lambda I)x = y$$ and see if something goes wrong for some value of $y=\{ y_0,y_1,\cdots\}$. The algebra is a mess, unless you use a power series. $$x(z) = \sum_{n=0}^{\infty}x_{n}z^{n},\;\;\; y(z)=\sum_{n=0}^{\infty}y_{n}z^{n}.$$ These power series converge in $|z| < 1$ because the coefficients are square integrable. What's nice, though, is that $$(Sx)(z) = zx(z),\;\;\;\; (S^{\star}x)(z) = \frac{x(z)-x(0)}{z}.$$ Now, the equation you want to solve is $$zx(z)+\frac{x(z)-x(0)}{z}-\lambda x(z)=y(z).$$ So you don't have to use full-blown function theory, but the power series is really helpful. Here you've got $$(z+\frac{1}{z}-\lambda)x(z)=y(z)+\frac{x(0)}{z} \\ (z^{2}+1-\lambda z)x(z) = zy(z)+x(0).$$ The roots of $z^{2}-\lambda z+1$ are $$\frac{\lambda}{2}\pm i\sqrt{1-\frac{\lambda^{2}}{4}}$$ For $-2 \le \lambda \le 2$, the modulus of the above is $$\frac{\lambda^{2}}{4}+\left(1-\frac{\lambda^{2}}{4}\right)=1.$$ Here's where it's best to take a specific $y$ and show that you can't get a solution. The obvious choice is $y(z)\equiv 1$. Then the power series equation gives you $$x(z) = \frac{z+x(0)}{z^{2}-\lambda z+1}.$$ You've got two distinct roots of the denominator on the unit circle for $-2 < \lambda < 2$, and there's no way for the numerator to dampen the effect of both roots. So $|x(re^{i\theta})| \sim \frac{1}{1-r}$ if $e^{i\theta}$ is an undamped root of the denominator. But look at the growth rate of $x(z)$ just by using the Cauchy-Schwarz inequality: \begin{align} |x(z)| & \le \sum_{n=0}^{\infty}|x_n||z|^{n} \\ & \le \left(\sum_{n=0}^{\infty}|x_n|^{2}\right)^{1/2}\left(\sum_{n=0}^{\infty}|z|^{2n}\right)^{1/2} \\ & = \|x\|\frac{1}{\sqrt{1-|z|^{2}}} \\ & \le \|x\|\frac{1}{\sqrt{1-|z|}}. \end{align} That's a contradiction for any $-2 < \lambda < 2$. So $(-2,2)\subseteq\sigma(S+S^{\star})$. Because spectrum is closed, then $[-2,2]\subseteq \sigma(S+S^{\star})$.

The spectrum is not empty. We will change our view from $l^2$ to $H^2$ (a common trick) to re-express the problem.

If we look at the Hardy space, $H^2$, given by those functions analytic in the disc whose Taylor series at the origin are square summable, we can represent $S$ by multiplication by $z$.

$$H^2= \left\{ f(z) = \sum_{n=0}^\infty a_n z^n : \sum |a_n|^2 < \infty \right\}.$$

Every member of this space has radial limits almost everywhere, i.e. $\hat f(\theta) = \lim_{r\to 1^-} f(re^{i\theta})$ exists almost everywhere for all $f\in H^2$. Moreover, these radial limits are in $L^2(\mathbb{T})$. Viewing $H^2$ in this way, $H^2$ becomes those elements in $L^2(\mathbb{T})$, where all of the negative frequency Fourier coefficients vanish.

Every element in $l^2$ has a natural identification with $H^2$ by $$\{ a_n \} \mapsto \sum_{n=0}^\infty a_n z^n.$$ Therefore, we can use $S=M_z = T_z$ and $S^* = T_{\overline{z}}= P_{H^2} M_{\overline{z}}$. $T_\phi = P_{H^2} M_\phi$ is called a Toeplitz operator with symbol $\phi \in L^\infty(\mathbb{T})$.

This yields $S+S^* = T_{z+\bar z} = T_{2\cos(\phi)}$. This operator has no eigenvalues, but it's essential spectrum is $[-2,2]$. In general the spectrum is given by $[essinf \phi, esssup \phi]$. Details can be found in the book by Ron Douglas, "Banach Algebra Techniques in Operator Theory". An essential work for anyone studying operator theory.

Also this can be found in Rosenblum's work: https://projecteuclid.org/download/pdf_1/euclid.pjm/1103038245

The identification with $l^2$ means that $S+S^*$ has the same spectrum, $[-2,2]$.

In Douglas' book, this boils down to a handful of ideas.

Proposition 7.6 If $\phi \in L^\infty(\mathbb{T})$ is such that $T_\phi$ is invertible, then $\phi$ is invertible in $L^\infty(\mathbb{T})$.

From here we have:

Corollary 7.7 (Hartman-Wintner) If $\phi \in L^\infty(\mathbb{T})$ then $\sigma(M_\phi) \subset \sigma(T_\phi)$.

Note that $T_{\phi} -\lambda = T_{\phi - \lambda}$, and if $T_{\phi-\lambda}$ is invertible, then $M_{\phi-\lambda} = M_{\phi} - \lambda$ is invertible. Thus $\rho(T_\phi) \subset \rho(M_{\phi})$ which yields $\sigma(M_{\phi}) \subset \sigma(T_\phi)$.

Finally, the spectrum of a multiplication operator with symbol $\phi$ is the essential range of $\phi$. In the case for $2\cos(\theta)$, this is $[-2,2]$.

Here is an attempt without using function theory. For any $|\lambda| \le 2$ we want to show that $S+S^* - \lambda I$ is not invertible. One way to do this is to show that $$\inf_{x \in l^2} \frac{\|(S+S^* - \lambda)x\|}{\|x\|} = 0.$$

I will demonstrate this with $\lambda=0$. Taking $x_n \in l^2$ to be $$x_n =(1,1,-1,-1,1,1,-1,-1,...,1,1,-1,-1,0,0,0,0,...)$$ or in otherwords $$x_n(i) = \left\{ \begin{array}{cc} 1 & \text{ if } i \equiv 0,1 \mod 4\\ -1 & \text{ if } i \equiv 2,3 \mod 4 \end{array}\right.$$ for $i \le 4n$ and $x_n(i) = 0$ for $i > 4n$. Thus $\|x_n\| = 2\sqrt{n}$.

Note that $(S+S^*)x_n = (1,0,0,0,...,0,0,-1,0,0,...)$ and $\|(S+S^*)x_n\| = \sqrt{2}$.

Therefore, $$\inf_{x \in l^2} \frac{\|(S+S^*)x\|}{\|x\|} \le \inf_{n} \frac{\|(S+S^*)x_n\|}{\|x_n\|} = \lim_{n\to\infty} \frac{\sqrt{2}}{2\sqrt{n}} = 0,$$ and $S+S^*$ is not invertible. Therefore, $0$ is in the spectrum of $S+S^*$.

This edit is to reply to your last comment as to why the operator cannot have any $\lambda$ with $|\lambda| > 2$ in its spectrum.

Note that $(I-\lambda T)^{-1} = \sum_{n=0}^\infty \lambda^n T^n$ whenever $\|\lambda T\| < 1$.

The norm of the operator $S+S^*$ is bounded by $2$ by the triangle inequality. Therefore $(S+S^*)/\lambda$ has norm less than $1$. This means $$\left( \frac{S+S^*}{\lambda} - I\right)^{-1} = - \sum_{n=0}^\infty \left(\frac{S+S^*}{\lambda}\right)^n$$ and so $\left( \frac{S+S^*}{\lambda} - I\right)$ is invertible. Of course $$\left( \frac{S+S^*}{\lambda} - I\right) = \lambda^{-1} \left( S+S^* - \lambda I\right)$$ so this quantity is invertible as well. Therefore, $\lambda$ is not in the spectrum of $S+S^*$ for any $\lambda$ with $|\lambda|>2$.

• Thanks a lot for the answer, but unfortunately it's too advanced for my background. Perhaps there's a more elementary way to obtain the result, using only basic Hilbert space methods and without using Toeplitz operators? – Aubrey Apr 29 '15 at 19:27
• I hope that the example at the end is helpful. It should be possible to do with the other elements of the spectrum, I wasn't motivated enough to work it out myself. – Joel Apr 29 '15 at 20:18
• Thank you, that's very helpful. It'd be great if you could please explain why the spectrum does not contain any $\lambda$ with $\lambda>2$? – Aubrey Apr 29 '15 at 20:48
• Thanks again, I appreciate your answer to my comment. – Aubrey Apr 29 '15 at 21:05