I am working on a problem in harmonic analysis, which I converted into the following existence problem concerning Laurent series. I am a bit at a loss concerning this problem, since my knowledge of complex analysis is limited; essentially I know a subset of what is in Rudin's "Real and Complex Analysis".



> 
> Precisely, I would like to know if there is a coefficient sequence $(c\_\ell)\_{\ell \in \Bbb{Z}} \subset \Bbb{C}$ satisfying the following:
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> 
> 1) $\sum\_{\ell \in \Bbb{Z}} (e^{\ell^2} \cdot |c\_\ell|)^2 < \infty$;
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> 2) The Laurent series $\varphi(z) := \sum\_{\ell \in \Bbb{Z}} c\_\ell z^\ell$ satisfies $\varphi(e^{2n}) = 0$ for all $n \in \Bbb{N}\_0 = \{0,1,2,\dots\}$.
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> 
> 


If the Laurent series was actually a power series, then I know Jensen's formula (see for instance [Rudin, Real and Complex Analysis, Theorem 15.20]) which shows that if $f : \Bbb{C} \to \Bbb{C}$ is holomorphic, then
$$
M(2r) \geq \prod\_{n=1}^{n(2r)} \frac{2r}{|\alpha\_n|},
$$
where the $\alpha\_i$ are the zeroes of $f$, and where $|\alpha\_1| \leq |\alpha\_2| \leq \dots$, and where $n(2r)$ is the number of zeroes of $f$ satisfying $|z| < 2r$.


But even this estimate does not seem to rule out the existence of such a function: For $r = e^{2N}$, the RHS of Jensen's inequality is $2^N e^{N^2 - N}$. Conversely, given my assumptions on the decay of the coefficients, I can estimate
$M(2r) \lesssim \sum\_{\ell \in \Bbb{Z}} e^{-\ell^2} (2 e^{2N})^\ell$, where I think the estimate is quite sharp, and where the $N$-th term is precisely $(2 e^{N})^N$, so that I don't seem to get a contradiction to Jensen's estimate.


Finally, I also know that there are results about the existence of holomorphic functions with prescribed zeros, but I don't know of any result that gives decent control over the coefficients of the power (or Laurent) series.


Any help would be greatly apprechiated.



Existence of Laurent series with zeroes at $e^{2n}$ ($n \in \Bbb{N}_0$) and extremely fast coefficient decay
I believe that there is such a function and even more it is (almost) entire.


Indeed, the most standard way to construct function with prescribed zeros is to consider Weierstrass product.


Put $f(z) = \prod\_{n = 0}^\infty \left(1 - \frac{z}{e^{2n}}\right)$. We will use the well-known estimate $|c\_n| \le \frac{M(r)}{r^n}$ so let us calculate $M(r)$. 


let $|z| = r$. We have $$|f(z)| \le \prod\_{n = 0}^\infty \left(1 + \frac{r}{e^{2n}}\right) = \prod\_{n = 0}^{[\frac{\log r}{2}]}\frac{r}{e^{2n}} \prod\_{n = 0}^{[\frac{\log r}{2}]} \left(1 + \frac{e^{2n}}{r}\right) \prod\_{n = [\frac{\log r}{2}]+1}^\infty \left(1 + \frac{r}{e^{2n}}\right).$$


Note that second and third products are $O(1)$ because we can take logarithm and estimate sum of geometric progression. Thus, putting $[\frac{\log r}{2}] = m$


$$|f(z)| \le C\prod\_{n = 0}^{m} \frac{r}{e^{2n}} = cr^{m+1}e^{-m(m+1)} \le Cr^{\frac{\log r}{2} + 1}e^{-(\frac{\log r}{2} - 1)\frac{\log r}{2}} = Ce^{\frac{\log^2 r}{4} + \log r}.$$


Therefore choosing $r = e^{2n}$ we get


$$|c\_n| \le \frac{M(r)}{r^n} = Ce^{n^2 + 2n - 2n^2} = Ce^{2n - n^2}.$$


So this bound is just barely not enough to make $\sum\_{n\in \mathbb{Z}} (|c\_n|e^{n^2})^2$ convergent. This is why I said in the beginning that our function will be almost entire. It is not clear to me whether we can push our estimates far enough so that there is entire function satisfying your conditions so here is a dirty trick which gives us Laurent series with the required properties:


consider $g(z) = \frac{f(z)}{z^2}$. For this function coefficients satisfies


$$|c\_n| \le Ce^{2(n+2)-(n+2)^2} = Ce^{2n + 4 - n^2 - 4n - 4} = Ce^{-n^2 - 2n}$$


and now series $\sum\_{n\in \mathbb{Z}} (|c\_n|e^{n^2})^2$ converges.


Being a bit more careful I believe we can prove that $\frac{f(z)}{z}$ suffices but it is not clear to me whether $f$ suffices on its own.


