Spectrum of unilateral shift and weighted operators

Question

Let $H$ be a Hilbert space with a countable orthonormal basis, and $S$ be a unilateral shift, that is $Se_n := e_{n + 1}$ operator of $H$. I am having trouble understanding what the spectrum of $S$ would be i.e. $$\sigma(S):= \{\lambda \in \mathbb{C}: \lambda I - S \text{ not invertible}\}$$I know that the spectrum is a union of other specific types of properties. I was also wondering what about(spectrum) if we are considering a weighted shift operator, essentially we have a bounded function $\omega_n:\mathbb{N} \rightarrow \mathbb{C}$ with $Te_n := \omega_{n + 1}e_{n + 1}$.(I am not thinking about this in sequence spaces)

Answer 1

It is more convenient to enumerate the basis starting at $0,$ i.e. $\{e_n\}_{n=0}^\infty.$
The adjoint operator $S^*$ satisfies $$S^*e_n=\begin{cases}e_{n-1}& n\ge 1\\ 0 & n=0\end{cases}$$ For $|z|<1$ let $$v_z=\sum_{n=0}^\infty z^{n}e_n $$ Then $v_z\in \mathcal{H}$ and $$S^*v_z=\sum_{n=0}^\infty z^nS^*e_n=\sum_{n=1}^\infty z^ne_{n-1}=zv_z$$ Therefore $\{z\in \mathbb{C}\,:\,|z|\le 1\}\subset \sigma(S^*).$ On the other hand $\|S^*\|=1,$ hence $$\{z\in \mathbb{C}\,:\,|z|\le 1\}= \sigma(S^*)$$ Thus $$\{z\in \mathbb{C}\,:\,|z|\le 1\}= \sigma(S)$$

Consider $S_\omega$ acting by $S_\omega e_n=\omega_{n+1}e_{n+1},$ $\omega_n\neq 0.$ We will show that if $|\omega_n|$ is convergent then $$\sigma(S_\omega^*)=\{z\,:\, |z|\le r\},\quad \lim_n|\omega_n|=r\quad (*)$$ In this case $$r=\lim_k\{|\omega_1|\ldots |\omega_k|\}^{1/k}$$ The adjoint operator is given by $$S^*_\omega e_n=\begin{cases}\overline{\omega_n}e_{n-1}& n\ge 1\\ 0 & n=0\end{cases}$$ We are going to determine the spectral radius of $S^*_\omega.$ We have $$\|(S^*_\omega)^k\|^2=\|(S^*_\omega)^k(S_\omega)^k\|$$ and $$(S^*_\omega)^k(S_\omega)^ke_n=|\omega_{n+1}|^2\ldots |\omega_{n+k}|^2 e_n$$ Hence $$\|(S^*_\omega)^k\|=\sup_{n\ge 0}\{|\omega_{n+1}|\ldots |\omega_{n+k}|\}$$ Thus $$\lim_{k\to\infty}\|(S^*_\omega)^k\|^{1/k}=\lim_{k\to\infty}\sup_{n\ge 0}\{|\omega_{n+1}|\ldots |\omega_{n+k}|\}^{1/k}=r$$ This gives $"\subset"$ in $(*).$ For $|z|<r$ let $$v_z=e_0+\sum_{n=1}^\infty {z^n\over \overline{\omega_1}\ldots \overline{\omega_n}}e_n $$ Then $$\|v_z\|^2=1+\sum_{n=1}^\infty{|z|^{2n}\over \{|\omega_1|\ldots |\omega_n|\}^2} $$ The series is convergent by the Cauchy test $$\lim_n{|z|^2\over \{|\omega_1|\ldots |\omega_n|\}^{2/n}}={|z|^2\over r^2}<1$$ Moreover $S^*_\omega v_z=zv_z.$ Hence $$\{z\,:\,|z|<r\}\subset \sigma(S_\omega^*)$$ This gives $"\supset"$ in $(*).$ The formula $(*)$ implies $\sigma(S_\omega)=\{z\,:\,|z|\le r\}.$