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 $M(𝕋)$ of all (finite) complex Borel measures on the unit circle $𝕋$ and the space $C(𝕋)$ of continuous functions on $𝕋$ as its dual space. Then $C(𝕋)\subset L^p(𝕋)$ for all $1\le p<\mathrm{\infty }$ and $L^1(𝕋)\subset M(𝕋)$.Template:Sfn

Given $\mu \in M(𝕋)$, let $${\mu }_{pp}=\sum _j{c}_j{\delta }_{z_j},$$ be its discrete 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 $\hat{\mu }(n)=\underset{𝕋}{\int }{z}^{-n}d\mu (z)$ is the $n$-th Fourier-Stieltjes coefficient of $\mu$.Template:SfnTemplate:Sfn

Similarly, on the real line $ℝ$, the space $C_0(ℝ)$ of continuous functions which vanish at infinity is the dual space of $M(ℝ)$ and $C_0(ℝ)\subset L^p(ℝ)$ for all $1\le p\le \mathrm{\infty }$.Template:Sfn

Given $\mu \in M(ℝ)$, let $${\mu }_{pp}=\sum _j{c}_j{\delta }_{x_j},$$ its discrete part. Then $$\underset{R\to \mathrm{\infty }}{\lim }\frac{1}{2R}\underset{-R}{\overset{R}{\int }}|\hat{\mu }(\xi ){|}^{2}d\xi =\sum _j|c_j{|}^2,$$ where $\hat{\mu }(\xi )=\underset{ℝ}{\int }{e}^{-2\pi i\xi x}d\mu (x)$ is the Fourier-Stieltjes transform of $\mu$.Template:Sfn

Consequences

If $\mu \in M(𝕋)$ is continuous, then $$\underset{N\to \mathrm{\infty }}{\lim }\frac{1}{2N+1}\sum _{n=-N}^N|\hat{\mu }(n){|}^2=0.$$ Furthermore, $\hat{\mu }$ tends to zero if $\mu$ is absolutely continuous.Template:Sfn Equivalently, $\mu$ is absolutely continuous if its Fourier-Stieltjes sequence belongs to the sequence space ${\ell }^2$.Template:Sfn 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 $|N|\to \mathrm{\infty }$. Conversely, if $\hat{\mu }\to 0$ as $|N|\to \mathrm{\infty }$, then $\mu$ places no mass on the countable sets. Template:Sfn

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 $1$.

Proof

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

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

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 $z\in 𝕋\setminus \{1\}$, which converges to $0$ as $N\to \mathrm{\infty }$. Hence, by the dominated convergence theorem,

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

We now take ${\mu }^{\prime }$ to be the pushforward of $\bar{\mu }$ under the inverse map on $𝕋$, namely ${\mu }^{\prime }(B)=\overline{\mu (B^{-1})}$ for any Borel set $B\subseteq 𝕋$. 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 $𝕋\times 𝕋$) under the product map $\cdot :𝕋\times 𝕋\to 𝕋$. By Fubini's theorem

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

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

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

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

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

(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

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

See also

• Lebesgue's decomposition theorem
• Trigonometric moment problem

References

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