# Extension of Poisson Summation formula

Under the condition f continuous, integrable and:

$|f(t)| + |\hat{f}(t)| \le C (1+|t|)^{-1-a}$ (with a>0)

we have the twisted Poisson formula that holds (where $\chi(n)$ is a primitive Dirichlet character):

$\sum_{-\infty}^{\infty} \chi(n) f(\frac{nx}{\sqrt{q}}) = \frac{A}{x} \sum_{-\infty}^{\infty} \overline{\chi(n)} \widehat{f}(\frac{n}{x \sqrt{q}})$

In the latest sum the term in zero is null as the character is primitve ($\chi(0)=0$) so I think this equation also holds with some functions integrable but not defined in zero like: $|x|^{-s} e^{-|x|}$ (with $0<s<1$).

Do you have any idea of a reference where I could find such result? (I did not find reference in litterature with such extension of Poisson formula, it seems it does not exists?)

What would be the simplest way to demonstrate it? (I thought about multiplying previous function by a function $g_n$ to use classical Poisson summation formula and then make $n \to \infty$, but any idea simpler than that?)

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I suspect that your proof is correct, simply becaus changing $f$ locally changes $\hat{f}$ globally. I think such an approach will fail. The Fourier transform of your $f$ will be a distribution and formally the PS is an identity of distribution. Applying distribution to distributions may or may not be welldefined. Btw, you lost a "$\hat$" in your PS. – Marc Palm Oct 11 '13 at 13:35
I corrected and proved your formula below. – GH from MO Oct 11 '13 at 15:33
Marc, I do not think the Fourier of $|x|^{-s} e^{-|x|}$ is a distribution, it can be calculated. I am thinking about defining $g_n(x)=|1/p|^{-s} e^{-|1/p|}$ for $-1/p<x<1/p$ and $g_n(x)=|x|^{-s} e^{-|x|}$ elsewhere in order to apply classic Poisson formula then to make $n \to \infty$, it seems to work as both side of sums converge. – Bertrand Oct 12 '13 at 1:51
Not (yet?) a published reference, but see the theorem stated on page 4 in math.harvard.edu/~elkies/M259.02/lsx.pdf . – Noam D. Elkies Oct 12 '13 at 2:49
@Bertrand: Lucia's comment made me realize that the "Added" section in my response contained no additional information, because the condition there implied the original condition. So I replaced this section with a short remark. – GH from MO Oct 14 '13 at 11:26

The correct formula is, for $\chi$ primitive of conductor $q$, $$\sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) = \frac{A}{x}\sum_{n\in\mathbb{Z}}\bar\chi(n)\hat f\left(\frac{n/x}{\sqrt{q}}\right).$$ Here $A:=\sqrt{q}/\tau(\bar\chi)$ is the so-called root number, it is of modulus $1$.

This formula is essentially equivalent to the functional equation of $L(s,\chi)$, but let me provide a direct proof. We start from the well-known formula $$\chi(n) = \frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m)e\left(\frac{mn}{q}\right),$$ where $e(x)$ abbreviates $e^{2\pi i x}$. See Davenport: Multiplicative number theory, Chapter IX, equation (6). Then we can rewrite the left hand side in the formula as $$\sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) = \frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m) \sum_{n\in\mathbb{Z}} e\left(\frac{mn}{q}\right) f\left(\frac{nx}{\sqrt{q}}\right).$$ Applying the Poission summation formula for the inner sum on the right hand side, $$\sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) = \frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m) \sum_{k\in\mathbb{Z}}\int_{-\infty}^\infty e\left(\frac{mt}{q}+kt\right) f\left(\frac{tx}{\sqrt{q}}\right)\,dt.$$ Denoting $n:=m+qk$ on the right hand side, we obtain $$\sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) = \frac{1}{\tau(\bar\chi)}\sum_{n\in\mathbb{Z}}\bar\chi(n) \int_{-\infty}^\infty e\left(\frac{nt}{q}\right) f\left(\frac{tx}{\sqrt{q}}\right)\,dt.$$ By a change of variable, $$\sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) = \frac{A}{x}\sum_{n\in\mathbb{Z}}\bar\chi(n) \int_{-\infty}^\infty e\left(\frac{nt/x}{\sqrt{q}}\right) f(t)\,dt,$$ where $A$ is as above. This is the stated (corrected) formula.

Remark of 10/14/2013. The formula as proved above holds for integrable continuous functions satisfying $|f(t)|+|\hat f(t)|\ll (1+|t|)^{-1-\delta}$ for some $\delta>0$. I had an Addendum of 10/12/2013 here with a seemingly more relaxed condition for $q>1$, but this condition turned out to be equivalent to the original one by basic facts on the Fourier transform.

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Yes you are right, I corrected my question. – Bertrand Oct 12 '13 at 1:27
Thanks, yes it is a good way to obtain the result! I consider my question has been fully answered. And it seems it is not in classical litterature. – Bertrand Oct 13 '13 at 6:26
If $f$ is like $|x|^{-s}e^{-x}$ is it obvious that ${\hat f}(t)$ decays like $(1+|t|)^{-1-\delta}$? I'm not sure that is so. – Lucia Oct 13 '13 at 14:48
To GH from MO: Or simply because if ${\hat f}$ is in $L^1$ then $f$ can only have removable discontinuities. – Lucia Oct 13 '13 at 15:48
To GH from MO: I'm not sure I understand. I think you do assume that $f$ is in $L^1$. But then it follows that ${\hat f}$ is continuous and decays at infinity. If you now assume that ${\hat f}$ is bounded for large $|t|$ by $(1+|t|)^{-1-\delta}$ then it does follow that ${\hat f}$ is in $L^1$. And now you can look at ${\hat {\hat f}}$ which will be continuous. Doesn't this show that $f$ (which is essentially ${\hat {\hat f}}$ only has removable discontinuities? – Lucia Oct 13 '13 at 23:53

I wanted to simply make a comment in GH's answer, but it's getting too long.

Another very similar way to obtain the formula is the following. (EDIT: This proof is essentially the same as the one given by Noam Elkies in a comment under the question.) After having appropriately normalized, we reduce the formula to showing that $$\sum_n f(n)\chi(n) = \frac{\tau(\chi)}{q} \sum_n \hat{f}\left(\frac{n}{q}\right)\bar{\chi}(n) .$$ (Remark: I use the definition $\hat{f}(\xi) = \int_{\mathbb{R}} f(t) e(-\xi t)dt$, whereas in GH's proof the exponential is $e(\xi t)$. This explains some discrepancies.) Now, note that the arithmetic progression $n\equiv a\pmod q$ is a shifted lattice of $\mathbb{R}$. In fact, $$\sum_{n\equiv a\pmod q}f(n) = \frac{1}{q}\sum_{n} e\left(\frac{an}{q}\right) \hat{f}\left(\frac{n}{q}\right),$$ a corollary of the full Poisson summation formula. Then we have that $$\sum_{n}f(n)\chi(n)= \sum_{a=1}^q \chi(a) \sum_{n\equiv a\pmod q} f(n) = \frac{1}{q} \sum_{a=1}^q \chi(a) \sum_{n} e\left(\frac{an}{q}\right) \hat{f}\left(\frac{n}{q}\right) = \frac{\tau(\chi)}{q} \sum_n \hat{f}\left(\frac{n}{q}\right)\bar{\chi}(n) ,$$ since $$\sum_{a=1}^q \chi(a) e\left(\frac{an}{q}\right) = \bar{\chi}(n)\tau(\chi)$$ for all $n$ if $\chi$ is a primitive character mod $q$. (A fancy way to read this last formula which might be useful conceptually is "a primitive character $\chi$ is an eigenvector of the operator $\bar{F}$, where $F$ is the Fourier transform mod $q$, with eigenvalue $\overline{\tau(\chi)}$.)

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I think your formula needs some scaling of $f$ and ${\hat f}$ by appropriate factors of $q$ to match with the correct version given by GH from MO. – Lucia Oct 12 '13 at 20:35
Dear Dimitris, this proof is similar to the one by Noam Elkies, see the link in his comment to the original post. In your first and second displays, $\hat f(n)$ should be $\hat f(n/q)$, with your definition of the Fourier transform. In your third display, the leftmost $f(n)$ should be multiplied by $\chi(n)$. – GH from MO Oct 12 '13 at 23:30
@GHfromMO: Thanks, I have made the correction and added a reference to Elkies's comment. – Dimitris Koukoulopoulos Oct 13 '13 at 5:11