Is there an injective, self-adjoint operator on $\ell_2(\mathbb{Z})$ with spectrum $[0,1]$?

Question

Is there an injective, self-adjoint operator on $\ell_2(\mathbb{Z})$ with spectrum $[0,1]$ ?

I basically need to find two self adjoint non unitarily equivalent operators on that Hilbert space which have spectrum $[0,1]$ and empty point spectrum.

Answer 1

Since $\ell^2(\mathbb Z)$ is a separable Hilbert space, it is isomorphic to $L^2[0,1]$. On $L^2[0,1]$, consider the multiplication operator $T$ given by $$ (Tf)(t)=tf(t). $$ Then $T$ is injective, $\sigma(t)=[0,1]$ and $\sigma_p(T)=\varnothing$.

The isomorphism $\Gamma:\ell^2(\mathbb Z)\to L^2[0,1]$ can be made explicit, as all we need to do is map one orthonormal basis onto another. So for instance we can take the canonical basis $\{e_n\}$ on $\ell^2(\mathbb Z)$ and have $$ \Gamma(e_n)=g_n,\qquad\text{ where }\quad g_n(t)=e^{2\pi i n t}. $$ Calculating the Fourier coefficients for $T$, we get that $$ T_{nm}=\langle Tg_m,g_n\rangle=\int_0^1t\,e^{2\pi i (m-n) t}\,dt=-\frac1{2\pi i (m-n)} $$ when $n\ne m$, and $T_{nn}=\frac12$ for all $n$. So the operator $T'\in B(\ell^2(\mathbb Z))$ with entries $$ T'_{nm}=\langle Te_m,e_n\rangle=\begin{cases}\displaystyle-\frac1{2\pi i (m-n)},&\ n\ne m\\ \ \\ \displaystyle\frac12,&\ n=m\end{cases} $$ satisfies that it is selfadjoint, injective, $\sigma(T')=[0,1]$, $\sigma_p(T')=\varnothing$.