# 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.

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