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 $\mu (\{z_j\})=c_j\ne 0$ and $\mu (\{z\})=0$ for $z\in ̸\{z_j\}$. Then

$\underset{N\to \mathrm{\infty }}{\lim }\frac{1}{2N+1}\sum _{n=-N}^N|\hat{\mu }(n){|}^2=\sum _j|c_j{|}^2,$

where $\hat{\mu }(n)=\underset{\mathbb{𝕋}}{\int }{z}^{-n}d\mu (z)$ 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

$\underset{R\to \mathrm{\infty }}{\lim }\frac{1}{2R}\underset{-R}{\overset{R}{\int }}|\hat{\mu }(\xi ){|}^{2}d\xi =\sum _j|c_j{|}^2,$

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

Proof

• First of all, we observe that if $\nu$ is a complex measure on the circle then

$\frac{1}{2N+1}\sum _{n=-N}^N\hat{\nu }(n)=\underset{\mathbb{𝕋}}{\int }{f}_N(z)d\nu (z),$

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 $N\to \mathrm{\infty }$. Hence, by the dominated convergence theorem,

$\underset{N\to \mathrm{\infty }}{\lim }\frac{1}{2N+1}\sum _{n=-N}^N\hat{\nu }(n)=\underset{\mathbb{𝕋}}{\int }{1}_{\{1\}}(z)d\nu (z)=\nu (\{1\}).$

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

$\hat{\nu }(n)=\underset{\mathbb{𝕋}\times \mathbb{𝕋}}{\int }(zw{)}^{-n}d(\mu \times {\mu }^{\prime })(z,w)=\underset{\mathbb{𝕋}}{\int }\underset{\mathbb{𝕋}}{\int }{z}^{-n}{w}^{-n}d{\mu }^{\prime }(w)d\mu (z)=\hat{\mu }(n)\widehat{{\mu }^{\prime }}(n)=|\hat{\mu }(n){|}^2.$

So, by the identity derived earlier, $\underset{N\to \mathrm{\infty }}{\lim }\frac{1}{2N+1}\sum _{n=-N}^N|\hat{\mu }(n){|}^2=\nu (\{1\})=\underset{\mathbb{𝕋}\times \mathbb{𝕋}}{\int }{1}_{\{zw=1\}}d(\mu \times {\mu }^{\prime })(z,w).$ By Fubini's theorem again, the right-hand side equals

$\underset{\mathbb{𝕋}}{\int }{\mu }^{\prime }(\{z^{-1}\})d\mu (z)=\underset{\mathbb{𝕋}}{\int }\overline{\mu (\{z\})}d\mu (z)=\sum _j|\mu (\{z_j\}){|}^2=\sum _j|c_j{|}^2.$

• The proof of the analogous statement for the real line is identical, except that we use the identity

$\frac{1}{2R}\underset{-R}{\overset{R}{\int }}\hat{\nu }(\xi )d\xi =\underset{\mathbb{ℝ}}{\int }{f}_R(x)d\nu (x)$

(which follows from Fubini's theorem), where $f_R(x)=\frac{1}{2R}\underset{-R}{\overset{R}{\int }}{e}^{-2\pi i\xi x}d\xi$. 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 $x\ne 0$, which converges to $0$ as $R\to \mathrm{\infty }$. So, by dominated convergence, we have the analogous identity

$\underset{R\to \mathrm{\infty }}{\lim }\frac{1}{2R}\underset{-R}{\overset{R}{\int }}\hat{\nu }(\xi )d\xi =\nu (\{0\}).$

Consequences

• A real or complex Borel measure $\mu$ on the circle is diffuse (i.e. ${\mu }_a=0$) if and only if $\underset{N\to \mathrm{\infty }}{\lim }\frac{1}{2N+1}\sum _{n=-N}^N|\hat{\mu }(n){|}^2=0$.
• A probability measure $\mu$ on the circle is a Dirac mass if and only if $\underset{N\to \mathrm{\infty }}{\lim }\frac{1}{2N+1}\sum _{n=-N}^N|\hat{\mu }(n){|}^2=1$. (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 $1$.)