The statements of the explicit formula for L-functions that I am aware of require the test function to be a Schwartz function (see, e.g., equation (4.11) in Section 4 of Low lying zeros of families of L-functions by Iwaniec, Luo, and Sarnak). However, test functions of the form $$ \phi(x)=\left(\frac{\sin(\pi v x)}{\pi v x}\right)^2 $$ are often used (e.g., equation (1.42) of Iwaniec-Luo-Sarnak). However, the Fourier transform of such a function, \begin{align*} \widehat{\phi}(y)=\begin{cases} \frac{1}{v}\left(1-\frac{|y|}{v}\right) & \text{ if } |y|<v \newline 0 & \text{ if } |y|\geq v, \end{cases} \end{align*} is not smooth, and therefore $\phi$ is not rapidly decreasing. In particular, $\phi$ is not a Schwartz function.


  1. Why does the explicit formula hold for these non-Schwartz test functions?

  2. Is it only modified versions of the explicit formula (such as (10.16) in Iwaniec-Luo-Sarnak) in which there is an error term, that allow a more flexible choice of test functions?

  3. In the case that more general test functions are permitted, can one choose the test function to be any smooth, even, absolutely integrable function with compactly supported Fourier transform?


1 Answer 1


The class of "weights" $\phi(t)$ for which the explicit formula holds is pretty flexible. The particular choice of weight depends on the problem one studies. On one hand, sums over primes (possibly also prime powers) need to converge, and on the other hand so do sums over zeros. By the uncertainty principle, $\phi$ and its Mellin/Fourier transform cannot simultaneously have compact support unless $\phi\equiv 0$. Therefore, one decide whether it is more important for the sum over zeros or the sum over primes to be weighted by a test function with compact support, but one cannot have both.

For simplicity, let's only consider the zeros of $\zeta(s)$ and assume RH (not absolutely necessary, but it's ok for here). In your example, the choice of $\phi(\gamma)$ decays like $|\gamma|^{-2}$ as $|\gamma|\to\infty$, which is enough to ensure absolute convergence (since there are $O(\log(|t|+2))$ nontrivial zeros $1/2+i\gamma$ such that $|\gamma-t|\leq 1$). On the other hand, the sum over primes will be weighted by $\widehat{\phi}(y)$ (or some dilation of it), which has compact support. So all of the sums converge nicely, and it is more convenient in the context of Iwaniec-Luo-Sarnak to place the compact support on the sum over primes.

To see another example of a non-Schwartz weight that ends up being crucial to prove the best result available, see the work of Chandee and Soundararajan on bounding $|\zeta(1/2+it)|$.

If you want to see a fully general explicit formula, look at Theorem 5.12 in Iwaniec and Kowalski. It is stated for Schwartz functions, but instead, think about the proof and about the properties a function $g$ must possess in order to ensure that all sums converge absolutely. That is the collection of weights for which the explicit formula works. There exist smooth, even, absolutely integrable weights with compactly supported Fourier transform that are not in the Schwartz class for which that theorem is true.

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    $\begingroup$ What's an example of a situation where it's useful to make the test function have compact support on the spectral/zeroes side? $\endgroup$ Nov 8, 2023 at 22:59
  • $\begingroup$ Thank you! This is an insightful description of the role played by the test function in the explicit formula. I think Lemma 1 of A NOTE ON S(t) AND THE ZEROS OF THE RIEMANN ZETA-FUNCTION by Goldston and Gonek was the sort of result I was looking for (this is what is used in Chandee-Soundararajan). $\endgroup$ Nov 14, 2023 at 14:03
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    $\begingroup$ One can often derive formulas for non-Schwartz functions from the Schwartz case, instead of from its proof, by approximating by a sequence of Schwartz function and calculating the limit of sums on both sides. $\endgroup$
    – Will Sawin
    Nov 14, 2023 at 14:21
  • $\begingroup$ @WillSawin This is what I originally tried to do, but had some difficulty. I was trying to get this to work for a general test function $\phi:\mathbb{R}\to\mathbb{R}_{\geq 0}$ that is smooth, even, absolutely integrable, and whose Fourier transform is compactly supported. I was able to show that $\phi$ is a limit of Schwartz functions, but was not able to switch the limit and sum over the zeros (I am concerned that mass can escape horizontally). Perhaps one needs some growth condition on $\phi$, or to make a careful choice of the Schwartz functions that converge to $\phi$? In any case, ... $\endgroup$ Nov 29, 2023 at 17:59
  • $\begingroup$ ... I would be interested in any reference that carries out the strategy you mention, especially if it is carried out in some generality. $\endgroup$ Nov 29, 2023 at 17:59

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