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:

1) $\sum_{\ell \in \Bbb{Z}} (e^{\ell^2} \cdot |c_\ell|)^2 < \infty$;

2) The Laurent series $\varphi(z) := \sum_{\ell \in \Bbb{Z}} c_\ell z^\ell$ satisfies $\varphi(e^{2n}) = 0$ for all $n \in \Bbb{N}_0 = \{0,1,2,\dots\}$.

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.


1 Answer 1


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.

  • $\begingroup$ Thank you very much for your answer! It made me rediscover a lot of complex analysis (Weierstrass products and Cauchy's estimate). Somehow, I expected that there is no such Laurent series. While translating your construction back to my setting, I realized that I actually need the following: For each fixed $A > 0$, we can find a Laurent series as above, but such that $\sum_{\ell \in \Bbb{Z}} (e^{A \ell^2} |c_\ell|)^2 < \infty$. I don't see how to generalize your answer to this setting (the shifting trick with dividing by $z^n$ does not give enough decay). Do you have an idea concerning this? $\endgroup$
    – PhoemueX
    Feb 9, 2019 at 20:19
  • $\begingroup$ One observation: For $A > 1$, Jensen's inequality rules out the existence of an entire function with the desired property. Indeed, we have $|c_\ell| \lesssim e^{-A \ell^2}$, and this implies for $r = e^{2N}$ that $M(2r) \lesssim \sum_{\ell \in \Bbb{Z}} e^{-A \ell^2} (2 e^{2N})^\ell = e^{(2N + \ln 2)^2 / (4A)} \sum_{\ell \in \Bbb{Z}} e^{-A (\ell + \frac{2N + \ln 2}{2A})^2}$, where I completed the square in the exponent. It is not hard to see that the last series is bounded by $C_A > 0$, independently of $N$. Since $e^{(2N + \ln 2)^2 / (4A)} \ll e^{N^2 - N}$, we get a contradiction to Jensen. $\endgroup$
    – PhoemueX
    Feb 9, 2019 at 21:03
  • $\begingroup$ @PhoemueX Wow, what a fascinating question! It seems I can prove that there are no such series for big enough $A$ (and probably even $A > 1$), but argument is more involved than the construction above. I can add it to my answer if you are interested or perhaps you should just create another question and I answer there. $\endgroup$ Feb 10, 2019 at 12:23
  • $\begingroup$ Thanks! I will open a new question when I am back home :) $\endgroup$
    – PhoemueX
    Feb 10, 2019 at 14:25
  • $\begingroup$ I just posted the new question (mathoverflow.net/questions/323064/…). Thanks again for your help; I am looking forward to your ideas. $\endgroup$
    – PhoemueX
    Feb 12, 2019 at 15:24

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