# Spectrum of shift-operator

Hoi, consider the Hilbertspace $l^2$ and the Left and Right-shift operator

\begin{align*} L(x_1,x_2,\cdots) &= (x_2,x_3,\cdots)\\ R(x_1,x_2,\cdots) &= (0,x_1,x_2,\cdots ) \end{align*}

I know that $L^*=R$ so these operators are Hilbert-space adjoints. The spectrum consists of 3 disjoint parts $\sigma(T) = \sigma_p(T)\cup \sigma_c(T)\cup \sigma_r(T)$. Assuming you are familiar with these notions: $\sigma_p(T)$ is point-spectrum, $\sigma_c(T)$ is continuous spectrum and $\sigma_r(T)$ the residual spectrum.

I want to show that $$\sigma_p(L) = \sigma_r(R) = \{\lambda :|\lambda|<1\}$$ $$\sigma_c(L)=\sigma_c(R) = \{\lambda : |\lambda|=1\}$$ $$\sigma_r(L)=\sigma_p(R) =\emptyset.$$

I stumbled upon a few problems. I can see that $\rho(L),\rho(R)<1$ so that $\{\lambda: |\lambda|>1\}$ is contained in the resolvent-sets of both $L$, and $R$. I can calculate the point-spectrum for $L$, and $R$.

So for $L$ i can calculate $\sigma_p(L)=\{\lambda : |\lambda|<1\}$ and since $\sigma(L)$ is closed, and $\{\lambda: |\lambda|>1\}$ is contained in the resolvent-set of $L$ we find that $\sigma(L) = \{\lambda: |\lambda| \leq 1\}$. Thus $$\sigma_c(L)\cup \sigma_r(L)= \{\lambda: |\lambda | =1\}.$$

Apparantly I can use the fact that $L$, and $R$ are eachothers adjoints, and reading the internet I found that $\sigma(T) = \sigma(T^*)$, or something like $\lambda \in \sigma(T)$implies $\overline{\lambda}\in \sigma(T^*)$ which is something i can't prove. I hoped to be able to use this fact by some Theorem in Rudin. (this excercise is also from Rudin CH. 12 excercise 18.c)

Apparantly the fact that $\lambda \in \sigma_r(L)$ implies that $\overline{\lambda}\in \sigma_p(L^*) = \sigma_p(R) = \emptyset$, so that we can conclude that $\sigma_r(L)=\emptyset$. I dont understand this at all.

Can someone explain this a little bit? How to go on from here? Thanks in advance.

## 2 Answers

We have \begin{align} \lambda\in\sigma(T)&\iff \lambda I-T\ \text{not invertible }\\ \ \\ &\iff (\lambda I-T)^*=\bar\lambda I-T^*\text{not invertible }\\ \ \\ &\iff \bar\lambda\in\sigma(T^*). \end{align}

And $$\bar\lambda\in\sigma_p(L)\iff \exists \text{ nonzero }v\in\ker(\bar\lambda I-L)=\text{ran}\, (\lambda I-R)^\perp\iff\lambda\in\sigma_r(R).$$ Note that the last "if and only if" requires the fact that $\sigma_p(R)=\emptyset$, since $$\sigma_r(T)=\{\lambda:\ \text{ran}\,(\lambda I-T)^\perp\ne0\}\setminus\sigma_p(T).$$

• Could you please expand your last statement? Why doesn't the double implication hold if $\sigma_p(R) \neq \emptyset$? – glS Jun 8 '15 at 7:42
• More properly, it requires that $\lambda$ is not an eigenvalue of $R$; the residual spectrum consists of those $\lambda$ that are not eigenvalues and satisfy $\text {ran }\,(\lambda\,I-R)^\perp\ne0$. – Martin Argerami Jun 8 '15 at 10:51

See article of Helein https://www.imj-prg.fr/~frederic.helein/polyspec.pdf here the spectrum of the shift operator is worked out into detail.

• Welcome to MSE. :) This more looks like a comment than an answer. – Hosein Rahnama Dec 23 '15 at 9:07
• It is better to add the main points in an answer here than leave a link only as sometimes links break (the pdf may get moved elsewhere in which case the link is useless). – Michael Albanese Dec 23 '15 at 9:18