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I can't understand a lemma in "An introduction to harmonic analysis" by Yitzhak Katznelson which is stated as follows:

Corollary. Let $\mu\in M(\mathbb T)$. Then $$\sum\limits_{\tau\in\mathbb T}|\mu(\{\tau\})|^2=\lim\limits_{N\rightarrow\infty }\frac{1}{2N+1}\sum\limits_{-N}^N|\hat\mu(n)|^2.$$

It looks bizzare to me since the left hand sum up all variable $\tau$, while the other side not. What I only know is that $\mu(\{\tau\})=\frac{1}{2N+1}\sum\limits_{-N}^N|\hat\mu(n)|,$ but it seems impossible to derive the corollary from this relation.

(note added by YC: the result is stated without proof at the end of Chapter I Section 7.11 in the 1976 Dover edition)

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    $\begingroup$ it may help if you give the source more precisely (which corollary of which book? Katznelson's?) $\endgroup$
    – Carlo Beenakker
    Commented Feb 28, 2021 at 10:48
  • $\begingroup$ Notice that both sides of the stated equation are computing something like an $L^2$-norm. This is not surprising, because it's presumably some form of the Plancherel theorem. Though I do not quite understand it as stated. However, your "hoped for" formula looks very different: a point evaluation one side, and something like an $L^2$-norm on the other. That looks very unlikely to be true. $\endgroup$
    – Matthew Daws
    Commented Feb 28, 2021 at 11:44
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    $\begingroup$ I'm really sorry, It's a typo, the correct is: $\mu(\{\tau\})=\lim\limits_{N\rightarrow\infty}\sum\limits_{-N}^N\hat \mu(n)e^{in\tau}$. $\endgroup$
    – Christoff_ferland
    Commented Feb 28, 2021 at 12:09
  • $\begingroup$ A finite measure has at-most countably many atoms, that should make the definition of the LHS clear as it is the sum of at-most countably many non-zero values... $\endgroup$
    – Asaf
    Commented Feb 28, 2021 at 20:54

1 Answer 1

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As far as I understand, $\mu$ is a Borel probability measure on the unit circle $\mathbb{T}$. Then $$ \frac{1}{2N+1}\sum\limits_{n=-N}^N|\hat\mu(n)|^2=\int\int\frac1{2N+1}\sum_{n=-N}^N (x/y)^nd\mu(x)d\mu(y). $$ The integrand $$ \frac1{2N+1}\sum_{n=-N}^N (x/y)^n $$ has absolute value at most 1 and converges pointwise to 1 for $x=y$ and to $0$ for $x\ne y$ (the sum is a geometric progression; you may sum it up to see that it is bounded when $x\ne y$). Thus, by the Dominated Convergence Theorem, the limit is the $\mu\times \mu$-measure of the diagonal $\{x=y\}$, which is exactly the LHS.

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  • $\begingroup$ Why is the first equality right? I'm not quite sure of this, please let me know. $\endgroup$
    – Christoff_ferland
    Commented Feb 28, 2021 at 12:44
  • $\begingroup$ well, $\mu(n)=\int y^{-n} d\mu(y)$, $\bar{\mu}(n)=\int x^{n}d\mu(x)$, multiply to get $|\mu(n)|^2=\int \int (x/y)^{n} d\mu(x)d\mu(y)$ $\endgroup$
    – Fedor Petrov
    Commented Feb 28, 2021 at 13:00
  • $\begingroup$ I got your idea. Thanks a lot. $\endgroup$
    – Christoff_ferland
    Commented Feb 28, 2021 at 13:22

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