# 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 discrete part. This result admits an analogous statement for measures on the real line. It was first discovered by Norbert Wiener.

## Definition

Consider the space $M(\mathbb {T} )$ of all (finite) complex Borel measures on the unit circle $\mathbb {T}$ and the space $C(\mathbb {T} )$ of continuous functions on $\mathbb {T}$ as its dual space. Then $C(\mathbb {T} )\subset L^{p}(\mathbb {T} )$ for all $1\leq p<\infty$ and $L^{1}(\mathbb {T} )\subset M(\mathbb {T} )$.
Given $\mu \in M(\mathbb {T} )$, let $\mu _{pp}=\sum _{j}c_{j}\delta _{z_{j}},$ be its discrete 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-Stieltjes coefficient of $\mu$.
Similarly, on the real line $\mathbb {R}$, the space $C_{0}(\mathbb {R} )$ of continuous functions which vanish at infinity is the dual space of $M(\mathbb {R} )$ and $C_{0}(\mathbb {R} )\subset L^{p}(\mathbb {R} )$ for all $1\leq p\leq \infty$.
Given $\mu \in M(\mathbb {R} )$, let $\mu _{pp}=\sum _{j}c_{j}\delta _{x_{j}},$ its discrete part. Then $\lim _{R\to \infty }{\frac {1}{2R}}\int _{-R}^{R}|{\widehat {\mu }}(\xi )|^{2}\,d\xi =\sum _{j}|c_{j}|^{2},$ where ${\widehat {\mu }}(\xi )=\int _{\mathbb {R} }e^{-2\pi i\xi x}\,d\mu (x)$ is the Fourier-Stieltjes transform of $\mu$.

### Consequences

If $\mu \in M(\mathbb {T} )$ is continuous, then $\lim _{N\to \infty }{\frac {1}{2N+1}}\sum _{n=-N}^{N}|{\widehat {\mu }}(n)|^{2}=0.$ Furthermore, ${\hat {\mu }}$ tends to zero if $\mu$ is absolutely continuous. Equivalently, $\mu$ is absolutely continuous if its Fourier-Stieltjes sequence belongs to the sequence space $\ell ^{2}$. That is, if $\mu$ places no mass on the sets of Lebesgue measure zero (i.e. $\mu _{pp}=0$), then ${\hat {\mu }}\to 0$ as $|N|\to \infty$. Conversely, if ${\hat {\mu }}\to 0$ as $|N|\to \infty$, then $\mu$ places no mass on the countable sets.
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 $1=\sum _{j}c_{j}^{2}\leq \sum _{j}c_{j}\leq 1,$ which forces $c_{j}^{2}=c_{j}$ and thus $c_{j}=1$, so that there must be a single atom with mass $1$.

## 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 $|f_{R}|\leq 1$, $f_{R}(0)=1$ 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
    $\lim _{R\to \infty }{\frac {1}{2R}}\int _{-R}^{R}{\widehat {\nu }}(\xi )\,d\xi =\nu (\{0\}).$