On a variant of Wiener ' s lemma

theorem Let $\mu$ is a finite Borel measure on $R $, then

$$\lim\limits_{t\to \infty}\frac{1}{2t}\int_{-t}^t|\widehat{\mu} (\XI) |^2d\xi=\sum_{x\in R}\mu (\{x\}) ^2,$$

Where $\widehat{\mu} (\XI) $ is the Fourier transformation of the measure $\mu,$ i.e., $\widehat{\mu} (\xi) =\int e^{-2\pi i\x I x}d\mu (x). $

Proof. Observe that $|\widehat{\mu} (\XI) |^2=\widehat{\mu} (\XI) \cdot \overline{\widehat{\mu} (\xi)}=\int_{r^2}e^{-2\pi i \xi (X-y)} D\MU (x) d\mu (y). $

Let $K _t (x, y) =\frac{1}{2t}\int_{-t}^te^{-2\pi i \xi (x-y)}d\xi,$ Then

$$ k_t (x, y) = \begin{cases}
\frac{e^{2\pi it (x-y)}-e^{-2\pi it (x-y)}}{4ti (x-y)},& x\not=y\\
1, & X=y
\end{cases} $$

Moreover, $\lim\limits_{t\to \infty}k_t (x, y) =\begin{cases}
0,& x\not=y\\
1, & X=y
\end{cases}$

Since $| k_t (x, y) |\le 1$, by the dominated convergence theorem we have this for all $x $,

$$\lim\limits_{t\to \infty}\int_rk_t (x, y) d\mu (y) =\mu (\{x\}). $$

Clearly,

$$\FRAC{1}{2T}\INT_{-T}^T|\WIDEHAT{\MU} (\XI) |^2d\xi=\int_{r^2}k_t (x, y) d\mu (×) d\mu (y). $$

It follows from Fubini ' s theorem,

$$\int_{r^2}k_t (x, y) d\mu (×) d\mu (y) =\int_r\mu (\{x\}) d\mu (x) =\sum_{x\in r}\mu (\{x\}) ^2.$$

On a variant of Wiener ' s lemma