# Wiener Corollary in “An introduction to harmonic analysis” by Yitzhak Katznelson

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)

• it may help if you give the source more precisely (which corollary of which book? Katznelson's?) – Carlo Beenakker Feb 28 at 10:48
• 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. – Matthew Daws Feb 28 at 11:44
• I'm really sorry, It's a typo, the correct is: $\mu(\{\tau\})=\lim\limits_{N\rightarrow\infty}\sum\limits_{-N}^N\hat \mu(n)e^{in\tau}$. – Christoff_ferland Feb 28 at 12:09
• 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... – Asaf Feb 28 at 20:54

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.
• well, $\mu(n)=\int y^{-n} d\mu(y)$, $\bar{\mu}(n)=\int x^{n}d\mu(x)$, multiply to get $|\mu(n)|^2=\int \int (x/y)^{n} d\mu(x)d\mu(y)$ – Fedor Petrov Feb 28 at 13:00