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. ^{[1] [2]}

Definition

Consider the space ${\displaystyle M(\mathbb {T} )}$ of all (finite) complex Borel measures on the unit circle ${\displaystyle \mathbb {T} }$ and the space ${\displaystyle C(\mathbb {T} )}$ of continuous functions on ${\displaystyle \mathbb {T} }$ as its dual space. Then ${\displaystyle C(\mathbb {T} )\subset L^{p}(\mathbb {T} )}$ for all ${\displaystyle 1\leq p<\infty }$ and ${\displaystyle L^{1}(\mathbb {T} )\subset M(\mathbb {T} )}$. ^[3]

Given ${\displaystyle \mu \in M(\mathbb {T} )}$, let $${\displaystyle \mu _{pp}=\sum _{j}c_{j}\delta _{z_{j}},}$$ 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-Stieltjes coefficient of ${\displaystyle \mu }$. ^{[4] [5]}

Similarly, given a real or complex Borel measure ${\displaystyle \mu }$ on the real line ${\displaystyle \mathbb {R} }$ and $${\displaystyle \mu _{pp}=\sum _{j}c_{j}\delta _{x_{j}},}$$ 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-Stieltjes transform of ${\displaystyle \mu }$. ^[6]

Consequences

If ${\displaystyle \mu \in M(\mathbb {T} )}$ is continuous, then $${\displaystyle \lim _{N\to \infty }{\frac {1}{2N+1}}\sum _{n=-N}^{N}|{\widehat {\mu }}(n)|^{2}=0.}$$ Furthermore, ${\displaystyle {\hat {\mu }}}$ tends to zero if ${\displaystyle \mu }$ is absolutely continuous. ^[7] Equivalently, ${\displaystyle \mu }$ is absolutely continuous if its Fourier-Stieltjes sequence belongs to the sequence space ${\displaystyle \ell ^{2}}$. ^[7] That is, if ${\displaystyle \mu }$ places no mass on the sets of Lebesgue measure zero (i.e. ${\displaystyle \mu _{pp}=0}$), then ${\displaystyle {\hat {\mu }}\to 0}$ as ${\displaystyle |N|\to \infty }$. Conversely, if ${\displaystyle {\hat {\mu }}\to 0}$ as ${\displaystyle |N|\to \infty }$, then ${\displaystyle \mu }$ places no mass on the countable sets. ^[8]

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 ${\displaystyle 1}$.

Proof

• 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

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

See also

• Lebesgue's decomposition theorem
• Trigonometric moment problem

Notes

1. Furstenberg's Conjecture on 2-3-invariant continuous probability measures on the circle (MathOverflow)
2. A complex borel measure, whose Fourier transform goes to zero (MathOverflow)
3. Helson 2010, pp. 15, 19.
4. Katznelson 1976, p. 45.
5. Helson 2010, pp. 22–24.
6. Katznelson 1976, pp. 153–154.
7. Helson 2010, p. 24.
8. Lyons 1985, pp. 155–156.

References

• Helson, Henry (2010). Harmonic Analysis. Vol. 7. Gurgaon: Hindustan Book Agency. doi:10.1007/978-93-86279-47-7. ISBN   978-93-80250-05-2.
• Katznelson, Yitzhak (1976). An introduction to harmonic analysis. New York: Dover Publications. ISBN   978-0-486-63331-2.
• Lyons, Russell (1985). "Fourier-Stieltjes Coefficients and Asymptotic Distribution Modulo 1". The Annals of Mathematics. 122 (1). doi:10.2307/1971372.