How to compute the spectrum of shift operators in $\ell^2(\mathbb N)$?

Question

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.

Answer 1

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 $$ \sigma_r(T)=\{\lambda:\ \text{ran}\,(\lambda I-T)^\perp\ne0\}\setminus\sigma_p(T). $$

Answer 2

You can observe directly that $\sigma_p(L)=\{\lambda:\,|\lambda|<1\}$ building the corresponding eigenvectors explicitly: $$v_\lambda = \sum_{k=0}^\infty \lambda^k e_k.$$ These satisfy $L v_\lambda= \lambda v_\lambda$, and are in $\ell^2(\mathbb{N})$ for $|\lambda|<1$. Similarly, $\sigma_p(R)=\emptyset$ because $Rx=\lambda x$ would imply $\lambda x_1=0$, $x_k=\lambda x_{k+1}$, and thus $x=0$.

On the other hand, $\|L\|=\|R\|=1$ tells you that $\sigma(L)=\sigma(R)=\{\lambda\in\mathbb{C}:\, |\lambda|\le1\}$.

Use now the general identity $\sigma_{\rm cp}(T)=\overline{\sigma_p(T^*)}$, with $\sigma_{\rm cp}$ the compression spectrum, i.e. the set of $\lambda$ such that $\operatorname{im}(T-\lambda I)$ is not dense. Using this for $T=L$, we get $\sigma_{\rm cp}(L)=\sigma_p(R)=\emptyset$. But $\sigma_r(T)=\sigma_{\rm cp}(T)\setminus\sigma_p(T)$, thus also $\sigma_r(L)=\emptyset$. We conclude that the remainder of the spectrum of $L$ must be the continuous spectrum: $\sigma_c(L)=S^1$.

Finally, $$\sigma_r(R) = \sigma_{\rm cp}(R) = \overline{\sigma_p(L) } = \{\lambda:\,|\lambda|<1\},$$ and $\sigma_c(R)=S^1$.