0

Let $\{e_n\}_{n=-\infty}^{\infty}$ be an orthonormal basis of Hilbert space $l^2(\mathbb{Z})$. Let $U$ be an unitary operator such that $U e_n = e_{n+1}$, then what is the spectrum of $U$?

First I know $\sigma(U)\subset S^1$...

Thanks in advance!

Edit:

  1. if $B$ is any (infinite) set, then one can form a Hilbert space of sequences with index set $B$, defined by $$ \ell^{2}(B)=\left\{x:B \stackrel{x}{\rightarrow} \mathbb{C}: \sum_{b \in B}\left| x(b)\right|^{2}<\infty\right\}. $$ The summation over $B$ is here defined by $$ \sum_{b \in B}|x(b)|^{2}=\sup \sum_{n=1}^{N}\left|x\left(b_{n}\right)\right|^{2} $$ the supremum being taken over all finite subsets of B. (From wikipedia).

  2. $S^1=\{z\in\mathbb{C}: |z|=1\}$.

Ryze
  • 1,136
  • I’m a little new to this, can you define what $l^2(Z)$ and $S^1$ are? – paulinho Jun 27 '20 at 13:50
  • @paulinho I've edited. – Ryze Jun 27 '20 at 14:08
  • 1
    The question has been anwered here already. In particular consider Robert Israel's comment: Use the Fourier transform $\mathcal{F}:L^2(S^1)\rightarrow \ell^2(\mathbb{Z})$ to turn $U$ into a multiplication operator $f\mapsto zf$ on $L^2(S^1)$ (with same spectrum). It is then easy to check that $\sigma(U)=S^1$. – Jan Bohr Jun 27 '20 at 14:33
  • @JanBohr Thank you very much! Could you please think about this problem? https://math.stackexchange.com/q/3736344/509160 – Ryze Jun 27 '20 at 15:46

0 Answers0