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I am looking a proof of Parseval's identity for Dirichlet series, if it is possible a free source or well if you know and it is possible get an idea of the proof a such theorem.

I am saying the quantitative version of Parseval's theorem for Dirichlet series, I mean the Lema 2.3.5 in page 48 of this (in spanish but is a concise formula, in the last paragraph) Granados, La distribución de los ceros de la función zeta de Riemann, (2014), available from this web of the Universidad Autónoma de Madrid.

Question. I am interested in details of a proof of the quantitative version of Parseval's identity for Dirichlet series. Do you know some reference where can I read it (if there is a free access it is the best)? Alternatively, if it is possible, provide us an outline or some idea to get such proof. Thanks in advance.

I recommend the above reference because has very high quality. I believe that there are no reference of such Lema 2.3.5 in the above text.

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  • $\begingroup$ Yes many thanks @JackD'Aurizio $\endgroup$ – user243301 Apr 18 '17 at 20:12
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We want to prove that $$ \int_{0}^{T}\left|\sum_{n}a_n n^{-it}\right|\,dt \leq \sum_{n\geq 1}\left|a_n\right|^2(T+O(n))$$ where the hidden constant in the $O(n)$ terms are uniform.


We have $$\begin{eqnarray*}\left|\sum_{n\geq 1}a_n n^{-it}\right|^2 &=& \left(\sum_{n\geq 1}a_n e^{-it\log n}\right)\left(\sum_{m\geq 1}\overline{a_m} e^{it\log m}\right)\\&=&\sum_{n\geq 1}a_n\overline{a_n}+\sum_{n\neq m}a_n \overline{a_m} e^{it(\log m-\log n)}\end{eqnarray*}$$ but if $m>n$ $$ \left|\int_{0}^{T}e^{it(\log m-\log n)}\,dt\right|\leq \min\left(2,\frac{2}{\log\frac{m}{n}}\right)$$ that is enough to prove the wanted inequality through Cauchy-Schwarz.


Here Besicovitch uses a slightly different notation and a partitioning argument, but the key idea is indeed the same, i.e. to prove that the dominant part of a quadratic form is concentrated in a suitable neighbourhood of the diagonal, since some exponential sums/integrals are almost vanishing far from it.

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  • $\begingroup$ Many thanks for your attention, I will check the calculations. $\endgroup$ – user243301 Apr 18 '17 at 20:45
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    $\begingroup$ @user243301: Besicovitch uses a partitioning argument and a slightly different notation, but essentially the same idea, here: onlinelibrary.wiley.com/doi/10.1112/plms/s2-26.1.25/epdf $\endgroup$ – Jack D'Aurizio Apr 18 '17 at 20:58

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