Compute the spectral measure of an operator $(T\mathbf{a})_{n}=a_{n+1}-a_{n}$ on the space of square summable sequences Consider the space $\mathcal{H}$ of square summable sequences $\mathbf{a}=\{a_{n}\}_{n=-\infty}^{\infty}$. The operator is defined by $$(T\mathbf{a})_{n}=a_{n+1}-a_{n}.$$
I want to compute the spectral resolution of this operator. From this website, https://encyclopediaofmath.org/wiki/Spectral_resolution, the spectrum measure $\mu$ can define resolution by $\mu((-\infty, t))$. So I guess the key is to get the spectral measure of this operator.

Edit 1: I think I got at least the spectrum
Given $\Psi\in\ell^{2}(\mathbb{Z})$, we can define $\hat{\Psi}\in L^{2}([0,2\pi))$ by $\hat{\Psi}(\xi):=\sum_{n\in\mathbb{Z}}e^{in\xi}\Psi_{n},$ with $$\Psi_{n}=\dfrac{1}{2\pi}\int_{0}^{2\pi}e^{in\xi}d\xi.$$
Then, the map $$U:\ell^{2}(\mathbb{Z})\longrightarrow L^{2}([0,2\pi),\dfrac{d\xi}{2\pi}),\ \ \Psi\longrightarrow_{U}\hat{\Psi},$$ is unitary.
We want to know what does $UTU^{-1}$ give us. Take $f(\xi)\in L^{2}([0,2\pi))$, then $$(TU^{-1}f)(n)=\dfrac{1}{2\pi}\int_{0}^{2\pi}(e^{-i(n+1)\xi}-e^{-in\xi})f(\xi)d\xi=\dfrac{1}{2\pi}\int_{0}^{2\pi}e^{-in\xi}(e^{-i\xi}-1)f(\xi)d\xi,$$ which implies that $$(UTU^{-1})(\xi)=(e^{-i\xi}-1)f(\xi).$$ Now we have the form of our operator in Fourier space, we are equipped to determine its spectrum.
Look at the resolvent:
\begin{align*}
R(T):=\{\lambda:(\lambda-T)\ \text{is invertible}\}&=\{\lambda:\lambda-e^{-i\xi}+1\neq 0\ \ \text{for all}\ \ \xi\in [0,2\pi)\}\\
&=\{\lambda:\lambda\neq e^{-i\xi}-1\ \ \ \text{for all}\ \ \xi\in [0,2\pi)\}\\
&=\mathbb{C}\setminus \{|\lambda+1|=1\}.
\end{align*}
This implies that the spectrum is $$\sigma(T)= \{|\lambda+1|=1\}.$$ We denote this set to be $\mathbb{S}_{1}^{*}(-1,1)$.
Am I correct?

Edit 2: I possibly get the spectral measure. But it confuses me
Now, for $f\in C(\mathbb{S}^{*}(-1,1))$ in Fourier space $f(T)$ is just multiplication by $f(e^{-i\xi}-1)$. Fix $\Psi\in\ell^{2}(\mathbb{Z})$, and let $\hat{\Psi}$ be the corresponding function in $L^{2}([0,2\pi))$. Then
\begin{align*}
\int f(T)d\mu_{\Psi}=\langle \Psi|f(T)\psi\rangle_{\ell^{2}}&=\langle U^{-1}U\Psi|f(T)U^{-1}U\Psi\rangle_{\ell{^2}}\\
&=\Big\langle U\Psi\Bigg|\Big(Uf(T)U^{-1}\Big)U\Psi\rangle_{L^{2}}\\
&=\langle \hat{\Psi}|f(e^{-i\xi}-1)\hat{\Psi}\rangle_{L^{2}}\\
&=\dfrac{1}{2\pi}\int_{0}^{2\pi}\overline{\hat{\Psi}}f(e^{-i\xi}-1)\hat{\Psi}d\xi\\
&=\dfrac{1}{2\pi}\int_{0}^{2\pi}f(e^{-i\xi}-1)|\hat{\Psi}(\xi)|^{2}d\xi.
\end{align*}
Write $\lambda:=e^{-i\xi}-1$, then $d\xi=\frac{d\lambda}{-e^{-i\xi}}$ and $\xi=i\log(\lambda+1)$ where $\log$ is taking over the principal branch (it's okay if just a general $\log$ since $\xi\in [0,2\pi]$). Then,
\begin{align*}
\int fd\mu_{\Psi}&=\dfrac{1}{2\pi}\int_{\mathbb{S}_{1}^{*}(-1,0)}f(\lambda)|\hat{\Psi}(i\log(\lambda+1))|^{2}\dfrac{d\lambda}{-e^{-i\xi}},
\end{align*}
therefore, $$d\mu_{\Psi}=|\hat{\Psi}(i\log(\lambda+1))|^{2}\dfrac{d\lambda}{-e^{-i\xi}}\Bigg|_{\mathbb{S}_{1}^{*}(-1,0)}.$$
According to this website, https://encyclopediaofmath.org/wiki/Spectral_resolution, the spectrum measure $\mu$ can define resolution by $\mu((-\infty, t))$. But you see my spectrum density is restricted to the unit circle. I am not sure how to go back to the real line.

Are the above things correct? How can I proceed?
 A: By definition a spectral resolution  on a Hilbert space $H$,
defined on a measurable space $(X, \mathscr B)$, is a function
$$
  P:\mathscr B \mapsto  B(H),
  $$
such that

*

*$P(E)$ is a projection for each $E$ in $\mathscr B$,


*$P(\emptyset) = 0$, and $P(X) = I$,


*whenever $\{E_n\}_{n\in \mathbb N}$ is a pairwise disjoint collection of measurable sets, one has that
$$
  P\left(\bigcup_{n\in \mathbb N}E_n\right) = \sum_{n\in \mathbb N}P(E_n),
  $$
the sum converging in the weak operator topology.

A well known form of the spectral theorem for bounded normal operators
asserts that, given any such operator $T$, there exists a spectral resolution $P$ defined on the Borel subsets of $\sigma (T)$ such
that
$$
  T=\int_{\sigma (T)} \lambda  \, dP(\lambda ).
  \tag 1
  $$
This integral is supposed to be interpreted in the weak sense, meaning that for every $\xi $ and $\eta $ in $H$, one
has that
$$
  \langle T\xi , \eta \rangle =\int_{\sigma (T)} \lambda  \, d P_{\xi , \eta }(\lambda ),
  $$
where $P_{\xi , \eta }$ is the actual (complex) measure on $\sigma (T)$, defined for every Borel set $E$ by
$$
  P_{\xi , \eta }(E)=  \langle P(E)\xi , \eta \rangle.
  $$

When $T$ is a self-adjoint operator, it is well known that $\sigma (T)\subseteq \mathbb R$, so its spectral resolution $P$ gives
rise to a so-called cummulative distributiuon, namely
$$
  \nu :t\in \mathbb R \mapsto  P(\sigma(T)\cap (-\infty ,t)) \in  B(H).
  $$
The importance of $\nu $ is that the integral in (1) can be expressed as a sort of Stieltjes integral
$$
  T=\int_{-\infty }^\infty  \lambda  \, d\nu (\lambda ).
  $$
For that reason, some authors prefer to emphasize $\nu $ over $P$, perhaps also because students have not yet learned any measure
theory at this point, but have already seen the Riemann-Stieltjes integral.  However I think  this is
not such a good idea because it has no counterpart for normal operators.

Finally,  if  $H=L^2(X, \mathscr B, \mu )$,  and $T$ is the so called multiplication operator defined on $H$ by
$$
  T(\xi )|_x = \varphi (x)\xi (x),  \quad \forall \xi \in  H,  \quad \forall x\in  X,
  $$
where $\varphi $ is a fixed bounded measurable complex valued function on $X$,  one may prove that the spectrum of $T$ is the essential range of
$\varphi $, and the spectral resolution provided by the spectral theorem is such that, for every $E\subseteq \sigma (T)$,
$P(E)$
is the orthogonal projection on the subspace $L_2(\varphi^{-1}(E))$.

Regarding the case in point,  we have that  $X=[0, 2π)$  and $φ(t) = e^{-it}+1$,  so the spectral resolution may be  obtained
as above.
A: HINT:
Enough to get the spectrum of $T(e_n)= e_{n-1}$, where $(e_n)_{n\in \mathbb{Z}}$ is the orthnormal basis of $\mathcal{H}$.  Now, $\mathcal{H}$ is isometric with $L^2(S^1)$ with the norm
$$f\mapsto \frac{1}{2\pi} \int_{0}^{2\pi} |f(e^{it}|^2 dt $$
with
$$e_n \mapsto \epsilon_n$$
with $\epsilon_n(t) = e^{i n t}$.
(this is the inverse Fourier transform)
Now, on the $L^2(S^1)$, $T$ becomes the map $\epsilon_{n}\mapsto \epsilon_{n-1}$, that is, the multiplication by the function $\epsilon_{-1}(t)= e^{-i t}$
