Wiener's lemma

In mathematics, Wiener's lemma is a well-known identity which relates the asymptotic behaviour of the Fourier coefficients of a Borel measure on the circle to its atomic part. This result admits an analogous statement for measures on the real line. It was first discovered by Norbert Wiener.

Statement

 * Given a real or complex Borel measure $$\mu$$ on the unit circle $$\mathbb T$$, let $$\mu_a=\sum_j c_j\delta_{z_j}$$ be its atomic part (meaning that $$\mu(\{z_j\})=c_j\neq 0$$ and $$\mu(\{z\})=0$$ for $$z\not\in\{z_j\}$$. Then
 * $$\lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^N|\widehat\mu(n)|^2=\sum_j|c_j|^2,$$

where $$\widehat{\mu}(n)=\int_{\mathbb T}z^{-n}\,d\mu(z)$$ is the $$n$$-th Fourier coefficient of $$\mu$$.


 * Similarly, given a real or complex Borel measure $$\mu$$ on the real line $$\mathbb R$$ and called $$\mu_a=\sum_j c_j\delta_{x_j}$$ its atomic part, we have
 * $$\lim_{R\to\infty}\frac{1}{2R}\int_{-R}^R|\widehat\mu(\xi)|^2\,d\xi=\sum_j|c_j|^2,$$

where $$\widehat{\mu}(\xi)=\int_{\mathbb R}e^{-2\pi i\xi x}\,d\mu(x)$$ is the Fourier transform of $$\mu$$.

Proof

 * First of all, we observe that if $$\nu$$ is a complex measure on the circle then
 * $$\frac{1}{2N+1}\sum_{n=-N}^N\widehat{\nu}(n)=\int_{\mathbb T}f_N(z)\,d\nu(z),$$

with $$f_N(z)=\frac{1}{2N+1}\sum_{n=-N}^N z^{-n}$$. The function $$f_N$$ is bounded by $$1$$ in absolute value and has $$f_N(1)=1$$, while $$f_N(z)=\frac{z^{N+1}-z^{-N}}{(2N+1)(z-1)}$$ for $$z\in\mathbb{T}\setminus\{1\}$$, which converges to $$0$$ as $$N\to\infty$$. Hence, by the dominated convergence theorem,
 * $$\lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^N\widehat{\nu}(n)=\int_{\mathbb T}1_{\{1\}}(z)\,d\nu(z)=\nu(\{1\}).$$

We now take $$\mu'$$ to be the pushforward of $$\overline\mu$$ under the inverse map on $$\mathbb T$$, namely $$\mu'(B)=\overline{\mu(B^{-1})}$$ for any Borel set $$B\subseteq\mathbb T$$. This complex measure has Fourier coefficients $$\widehat{\mu'}(n)=\overline{\widehat{\mu}(n)}$$. We are going to apply the above to the convolution between $$\mu$$ and $$\mu'$$, namely we choose $$\nu=\mu*\mu'$$, meaning that $$\nu$$ is the pushforward of the measure $$\mu\times\mu'$$ (on $$\mathbb T\times\mathbb T$$) under the product map $$\cdot:\mathbb{T}\times\mathbb{T}\to\mathbb{T}$$. By Fubini's theorem
 * $$\widehat{\nu}(n)=\int_{\mathbb{T}\times\mathbb{T}}(zw)^{-n}\,d(\mu\times\mu')(z,w)

=\int_{\mathbb T}\int_{\mathbb T}z^{-n}w^{-n}\,d\mu'(w)\,d\mu(z)=\widehat{\mu}(n)\widehat{\mu'}(n)=|\widehat{\mu}(n)|^2.$$ So, by the identity derived earlier, $$\lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^N|\widehat{\mu}(n)|^2=\nu(\{1\})=\int_{\mathbb T\times\mathbb T}1_{\{zw=1\}}\,d(\mu\times\mu')(z,w).$$ By Fubini's theorem again, the right-hand side equals
 * $$\int_{\mathbb T}\mu'(\{z^{-1}\})\,d\mu(z)=\int_{\mathbb T}\overline{\mu(\{z\})}\,d\mu(z)=\sum_j|\mu(\{z_j\})|^2=\sum_j|c_j|^2.$$


 * The proof of the analogous statement for the real line is identical, except that we use the identity
 * $$\frac{1}{2R}\int_{-R}^R\widehat\nu(\xi)\,d\xi=\int_{\mathbb R}f_R(x)\,d\nu(x)$$

(which follows from Fubini's theorem), where $$f_R(x)=\frac{1}{2R}\int_{-R}^R e^{-2\pi i\xi x}\,d\xi$$. We observe that $$|f_R|\le 1$$, $$f_R(0)=1$$ and $$f_R(x)=\frac{e^{2\pi iRx}-e^{-2\pi iRx}}{4\pi iRx}$$ for $$x\neq 0$$, which converges to $$0$$ as $$R\to\infty$$. So, by dominated convergence, we have the analogous identity
 * $$\lim_{R\to\infty}\frac{1}{2R}\int_{-R}^R\widehat\nu(\xi)\,d\xi=\nu(\{0\}).$$

Consequences

 * A real or complex Borel measure $$\mu$$ on the circle is diffuse (i.e. $$\mu_a=0$$) if and only if $$\lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^N|\widehat\mu(n)|^2=0$$.
 * A probability measure $$\mu$$ on the circle is a Dirac mass if and only if $$\lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^N|\widehat\mu(n)|^2=1$$. (Here, the nontrivial implication follows from the fact that the weights $$c_j$$ are positive and satisfy $$1=\sum_j c_j^2\le\sum_j c_j\le 1$$, which forces $$c_j^2=c_j$$ and thus $$c_j=1$$, so that there must be a single atom with mass $$1$$.)