# Existence of Laurent series with zeroes at $𝑒^2𝑛$ (𝑛∈ℕ0 ) and even faster coefficient decay

This is an extension of an earlier question of mine which corresponds to the case $$A = 1$$. Precisely, my question is as follows:

Given $$A > 0$$ fixed but arbitrary, is there a non-trivial sequence $$(c_n)_{n \in \Bbb{Z}}$$ with the following two properties:

1. We have $$\sum_{\ell \in \Bbb{Z}} (e^{A \ell^2} \, |c_\ell|)^2 < \infty$$, and

2. the Laurent series $$\varphi : \Bbb{C} \setminus \{0\} \to \Bbb{C}, z \mapsto \sum_{\ell \in \Bbb{Z}} c_\ell z^\ell$$ satisfies $$\varphi(e^{2n}) = 0$$ for all $$n \in \Bbb{N}_0$$.

In the earlier question, this was answered positively for $$A = 1$$ (and thus also for $$0 < A \leq 1$$) by Aleksei Kulikov, but then I realized that I need to know the answer for general $$A > 0$$.

Partial progress: For $$A > 1$$, one cannot find a "holomorphic Laurent series" (i.e., one with $$c_\ell = 0$$ for $$\ell < 0$$) satisfying the above, as a consequence of Jensen's formula. Indeed, suppose we had such a power series. The first property from above implies in particular that $$|c_\ell| \lesssim e^{-A \ell^2}$$, from which we get for $$|z| \leq R = 2 e^{2N}$$ (for $$N \in \Bbb{N}$$ fixed, but arbitrary) that \begin{align*} |\varphi(z)| & \lesssim \sum_{\ell = 0}^\infty (2 \cdot e^{2N})^\ell e^{-A \ell^2} \\ & = \sum_{\ell = 0}^\infty e^{-A \ell^2 + \ell (2N + \ln 2)} \\ & = \sum_{\ell \in \Bbb{Z}} e^{\frac{(2N + \ln 2)^2}{4 A}} e^{-A \big(\ell - \frac{2N + \ln 2}{2 A}\big)^2} \\ & \lesssim_A e^{\frac{(2N + \ln 2)^2}{4A}} \, . \end{align*} Here, the last step follows by standard arguments, essentially by writing $$\frac{2N + \ln 2}{2A} = k + r$$ with $$k \in \Bbb{Z}$$ and $$r \in [0,1)$$, and then by changing the summation variable $$t = \ell - k$$. One then arrives at $$\sum_{t \in \Bbb{Z}} e^{-A (t - r)^2}$$, which one can easily bound independently of $$r$$.

Note that $$e^{\frac{(2N + \ln 2)^2}{4 A}} \ll e^{N^2 - \theta N}$$ if $$A > 1$$ and $$\theta \geq 0$$. But (assuming $$\varphi(0) = 1$$, which one can probably remove), Jensen's formula shows $$e^{\frac{(2N + \ln2)^2}{4A}} \gtrsim \sup_{|z| \leq 2 e^{2N}} |\varphi(z)| \geq \prod_{j=1}^{N} \frac{2 e^{2N}}{e^{2j}} = 2^{N} e^{2 N^2 - N(N+1)} \geq e^{N^2 - N},$$ which is impossible.

Any help would be greatly appreciated.

As I promised, here is the proof that for $$A > 1$$ such a series does not exists.

Key idea is as follows: if $$f(z) = \sum_{n=-\infty}^\infty c_nz^n$$ has roots at $$e^{2n}$$ then $$g(z) = \sum_{n = 0}^\infty c_nz^n$$ satisfies $$g(e^{2n}) = O(e^{-2n})$$. From this we will deduce that $$g$$ has at least one root which is in some sense close to $$e^{2n}$$ which is incompatible with the possible growth of $$g$$.

Firstly, we will make some optional observations which will simplify our proof. Note that $$g$$ can not be a polynomial. Indeed, if $$g$$ is a polynomial than $$f(z) = z^Nh(\frac{1}{z})$$ for some entire function $$h$$. From our assumption $$h$$ has roots at $$e^{-2n}$$ which accumulate to $$0$$. Therefore by Identity Theorem $$h \equiv 0$$.

Also WLOG we will assume that $$g(0) = 1$$. If $$g(0) \ne 0$$ then we can just renormalize everything. Otherwise since $$f \ne 0$$ there is some nonzero coefficient $$c_m$$. Then we can consider function $$\tilde{f}(z)= z^{-m}f(z)$$ instead of $$f$$. It still has roots at $$e^{2n}$$, though our sum can accidentally diverge. But changing $$A$$ to $$A-\delta$$ for arbitrary $$\delta > 0$$ will fix this problem.

From the analysis in OP we can see that $$|g(z)| \le C\exp(\varepsilon \log^2 |z|)$$ for any $$\varepsilon > \frac{1}{4A}$$. In particular the function $$g$$ has order $$0$$. By Hadamard Factorization Theorem any entire function $$g$$ of order less than $$1$$ can be written as a Weierstrass product over its zeros (see, e.g., [1], Lecture 4, Theorem 1)

$$g(z) = Cz^m\prod_{n = 1}^\infty \left(1 - \frac{z}{a_n}\right).$$

(if the function $$g$$ is a polynomial then there will be only finitely many factors but this is not our case). Moreover, since $$g(0) = 1$$ we actually have $$C = 1, m = 0$$.

WLOG we assume that $$|a_1| \le |a_2| \le \ldots$$ We are now going to prove that $$|a_n| \ge c\exp(\frac{n}{4\varepsilon})$$ for big enough $$n$$. Suppose that it is not true. Then definitely $$|a_n| < \exp(\frac{n}{2\varepsilon}) = r$$ if $$n$$ is big enough. Consider Jensen's inequality with this $$r$$

$$C\exp(\varepsilon \log^2 r) \ge M(r) \ge \prod_{k = 1}^n \frac{r}{|a_k|} \ge \left(\frac{r}{|a_n|}\right)^n.$$

Taking logarithms $$n\log |a_n| + \log C + \varepsilon \log^2 r\ge n\log r,$$

and substituting value of $$r$$

$$\log |a_n| \ge \frac{n}{4\varepsilon} - \frac{\log C}{n} \ge \frac{n}{4\varepsilon} - \log C,$$

and we see that our estimate is satisfied with $$c = C^{-1}$$.

Rewriting this inequality via zero-counting function $$n(r)$$ we get $$n(r) \le C + 4\varepsilon \log r$$. I think it is worth mentioning that our estimate is better than the standard one $$n(r) \le \log M_g(er)$$ (see [1], Lecture 2, section 2.5) which will give us only $$n(r) = O(\log^2 r)$$ which is insufficient for our proof. I belive that the reason we can prove logarithmic bound is the "sparseness" of our problem which much more general estimate from Levin of course does not assume.

Let us now proof that function $$g$$ has at least one zero $$a_k$$ with $$\frac{e^{2n}}{2} \le|a_k| \le 2e^{2n}$$ for all big enough $$n$$. Assume the contrary that there are no such roots. Let us estimate $$g(e^{2n})$$

$$|g(e^{2n})| = \prod_{|a_k| < \frac{e^{2n}}{2}} \left|1 - \frac{e^{2n}}{a_k}\right| \prod_{|a_k| > 2e^{2n}} \left|1 - \frac{e^{2n}}{a_k}\right| = I_1 I_2.$$

Note that in $$I_1$$ all factors have absolute value at least $$1$$. Moreover, for each fixed $$K$$ we can actually say that $$I_1 \ge C_Ke^{2nK}$$ for big enough $$n$$ by considering first $$K$$ factors separately.

In $$I_2$$ we will take logarithm and estimate the corresponding sum

$$\sum_{|a_k| > 2e^{2n}} \frac{e^{2n}}{a_k} = e^{2n}\int_{2e^{2n}}^\infty \frac{1}{t}dn(t).$$

Integrating by parts we get

$$e^{2n}\left(-\frac{n(2e^{2n})}{2e^{2n}} + \int_{2e^{2n}}^\infty \frac{n(t)}{t^2}dt\right) \le e^{2n}\int_{2e^{2n}}^\infty \frac{c\log t}{t^2}dt \le Ce^{2n} \frac{\log (2e^{2n})}{2e^{2n}} \le 10Cn.$$

(everything here for big enough $$n$$)

Thus exponentiating we get $$|I_2| \ge e^{-20Cn}$$. Taking $$K > 10C$$ we get $$I_1 I_2 \ge C_K$$. On the other hand $$I_1I_2 = O(e^{-2n})$$ -- a contradiction for big enough $$n$$.

Therefore for big enough $$n$$ function $$g$$ has at least one zero $$a_k$$ with $$\frac{e^{2n}}{2} \le |a_k| \le 2e^{2n}$$ for all big enough $$n$$. Note that since $$e > 2$$ all these roots are distinct. Now we can essentially repeat application of Jensen's inequality from the OP to deduce that there are no such functions $$g$$.

As a final remark I would say that I belive everyting above can be restated as some uniquiness theorem in some radial Fock space. But I don't know enough about that and don't think it will be much shorter.

[1]Levin, B. Ya., Lectures on entire functions. In collab. with Yu. Lyubarskii, M. Sodin, V. Tkachenko. Transl. by V. Tkachenko from an original Russian manuscr, Translations of Mathematical Monographs. 150. Providence, RI: American Mathematical Society (AMS). xv, 248 p. (1996). ZBL0856.30001.