Let $\Delta$ be the discrete Laplacian on $\ell^2(\mathbb{Z})$, that is, $$\Delta = 1- \frac{S+S^*}{2},$$ where $S$ is the right shift operator. I know that $\Delta$ is a bounded self-adjoint operator. Do you know if there is some kind of explicit expression for the spectral resolution of $\Delta$?
Thanks in advance.
On
$S^*$ is the left shift. Consider the functions on $\psi_\theta(n) = e^{i n \theta}$ for $n\in \mathbb{Z}, \theta\in[0,2\pi)$. Then $$S \psi_\theta(n) = e^{i (n-1) \theta} = e^{-i\theta} \psi_\theta(n)$$ and $$S^* \psi_\theta(n) = e^{i (n+1) \theta} = e^{i\theta} \psi_\theta(n)$$ So the functions $\psi_\theta$ are generalized eigenvectors (yes, I'm aware they're not in $l^2(\mathbb{Z})$) for $S, S^*$ - unitary operators whose spectrum is the entire unit circle. For your Laplacian $\Delta$ you get $$\Delta \psi_\theta(n) = \psi_\theta(n) - \frac{(e^{-i\theta} + e^{i\theta})\psi_\theta(n)}{2} = (1 - \cos(\theta)) \psi_\theta(n)$$ So $\Delta$ has spectrum the interval $[0, 2]$ with the same generalized eigenvectors.
This is not intended as a rigorous proof.
Consider the unitary mapping $U:\ell^2(\mathbb{Z})\to L^2(-\pi,\pi)$ $$U\{a_n\}=(2\pi)^{-1/2}\sum_{-\infty}^\infty a_ne^{int}$$ Then $(USU^{-1}f)(t)=e^{it}f(t),$ hence $$(U\Delta U^{-1}f)(t)=(1-\cos t )f(t),\quad f\in L^2(-\pi,\pi)$$ The operator $U\Delta U^{-1}$ is unitarily equivalent to $\Delta$ and is represented in generalized diagonal form. The operator $U\Delta U^{-1}$ is unitarily equivalent to the direct sum of the operator $M$ with itself, where $M:L^2(0,\pi)\to L^2(0\,\pi)$ and $$(Mf)(t)=(1-\cos t)f(t)$$ The spectral resolution of $M$ can be easily read off, by substituting $x=1-\cos t.$ Indeed, consider the mapping $$V:L^2((0,2),\mu)\to L^2(0,\pi)$$ given by $$(Vf)(t)=f(1-\cos t),\quad d\mu(x)=x^{-1/2}(2-x)^{-1/2}\,dx$$ Then $V$ is unitary and $$(V^{-1}MV)f(x)=xf(x),\quad f\in L^2((0,2),\mu)$$