Wiener's lemma

• Given a real or complex Borel measure ${\displaystyle \mu }$ on the unit circle ${\displaystyle \mathbb {T} }$, let ${\displaystyle \mu _{a}=\sum _{j}c_{j}\delta _{z_{j}}}$ be its atomic part (meaning that ${\displaystyle \mu (\{z_{j}\})=c_{j}\neq 0}$ and ${\displaystyle \mu (\{z\})=0}$ for ${\displaystyle z\not \in \{z_{j}\}}$. Then

${\displaystyle \lim _{N\to \infty }{\frac {1}{2N+1}}\sum _{n=-N}^{N}|{\widehat {\mu }}(n)|^{2}=\sum _{j}|c_{j}|^{2},}$

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

• Similarly, given a real or complex Borel measure ${\displaystyle \mu }$ on the real line ${\displaystyle \mathbb {R} }$ and called ${\displaystyle \mu _{a}=\sum _{j}c_{j}\delta _{x_{j}}}$ its atomic part, we have

${\displaystyle \lim _{R\to \infty }{\frac {1}{2R}}\int _{-R}^{R}|{\widehat {\mu }}(\xi )|^{2}\,d\xi =\sum _{j}|c_{j}|^{2},}$

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

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

${\displaystyle {\frac {1}{2N+1}}\sum _{n=-N}^{N}{\widehat {\nu }}(n)=\int _{\mathbb {T} }f_{N}(z)\,d\nu (z),}$

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

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

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

${\displaystyle {\widehat {\nu }}(n)=\int _{\mathbb {T} \times \mathbb {T} }(zw)^{-n}\,d(\mu \times \mu ')(z,w)=\int _{\mathbb {T} }\int _{\mathbb {T} }z^{-n}w^{-n}\,d\mu '(w)\,d\mu (z)={\widehat {\mu }}(n){\widehat {\mu '}}(n)=|{\widehat {\mu }}(n)|^{2}.}$

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

${\displaystyle \int _{\mathbb {T} }\mu '(\{z^{-1}\})\,d\mu (z)=\int _{\mathbb {T} }{\overline {\mu (\{z\})}}\,d\mu (z)=\sum _{j}|\mu (\{z_{j}\})|^{2}=\sum _{j}|c_{j}|^{2}.}$

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

${\displaystyle {\frac {1}{2R}}\int _{-R}^{R}{\widehat {\nu }}(\xi )\,d\xi =\int _{\mathbb {R} }f_{R}(x)\,d\nu (x)}$

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

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

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