I am trying to compute the spectral measure of a stationary sequence and would like to know whether my thoughts are correct. I can use the result for a more involved question, thus just seeking confirmation for now :-)
The setup is :
$\mathcal{H} = L^2(\mathbb{R})$ and let $I : L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})$ be the identity operator on $L^2(\mathbb{R})$, i.e. $I : f \rightarrow f$. I choose an arbitrary $s \in L^2(\mathbb{R})$ and define the sequence $\mathbf{s} = \{s_n = I^ns\}_{n \in \mathbb{Z}}$. I want to determine the spectral measure $\mu_s$ of $\mathbf{s}$ which is defined via the auto-correlation
$R_s(n) = \langle s_n, s_0 \rangle = \frac{1}{2 \pi} \cdot \int_{- \pi}^{\pi} e^{- i n \theta} d \mu_s(\theta)$ with $d\mu_s(\theta) = \Phi_s(\theta) d \theta + d\mu_s^{singular}(\theta)$
My thoughts :
Since the identity operator is unitary, the sequence $\mathbf{s}$ is indeed. stationary to begin with.
The spectral measure can be further broken down into the absolutely integrable part and the singular part via : $R_s(n) = \langle s_n, s_0 \rangle = \frac{1}{2 \pi} \cdot \int_{-\pi}^{\pi} [\Phi_s(\theta) + \sum_{k} c_k \delta(\theta - \theta_k)] d \theta = \frac{1}{2 \pi} \cdot \int_{- \pi}^{\pi} \Phi_s(\theta) e^{-i n \theta} d \theta + \frac{1}{2 \pi} \sum_k c_k e^{-i n \theta_k}$
For the identity operator, we obviously have that $s_n(t) = (I^n s_0)(t) = (I^n s)(t) = s(t)$ for all $n$.
This gives the auto-correlation to be $R_s(n) = \langle s_n, s_0 \rangle = \int_{- \infty}^{\infty} s(t) \overline{s(t)} dt = \left\| s \right\|^2$
The $\Phi_s$ of the absolutely integrable part must be 0 because if it is not, $R_s(n)$ would contain a component (which is this absolutely integrable part) that converges to 0 for $n \rightarrow \infty$. The absolutely integrable part is thus the $n$-th Fourier coefficient of $\Phi_s$.
With this, the autocorrelation becomes simply $R_s(n) = \left\| s \right\|^2 = \frac{1}{2 \pi} \sum_k c_k e^{- i n \theta_k}$. It remains to find out the parameters $c_k$ and $\theta_k$.
I would argue that this expression is "1-periodic". Am I thus correct in assuming $\theta_k = 0$ for all $k$ ? And if so, how to obtain $c_k$?
The result should then look like this : $\mu_s(d\theta) = \sum_k c_k \delta(\theta - \theta_k) d \theta$
I would appreciate somebody giving me the last tipp. I need this proof to show something different in my thesis. Would appreciate any help!
Ok, so I think you are asking for the measure $\mu_s$ on $[-\pi,\pi]$ whose Fourier coefficients are given by $\hat\mu_s(n)=\|s\|^2$, right? In that case, the answer is simply $$ \mu_s=2\pi\|s^2\|\delta_0,$$ where $\delta_0$ is Dirac's measure supported on $\{0\}$. Thus the absolutely continuous part vanishes since $\mu_s$ is singular.