Let $Z \subseteq \mathbb{C}$ without limit point. By the Weierstrass factorization theorem there is an entire function $h$ those zero set is $Z$. Let $a_n > 0$ be a sequence where $\lim_n \sqrt[n]{a_n}=0$.

Question: Can $h(z) = \sum_{n \ge 0}b_nz^n$ as above be chosen such that $|b_n| \le a_n$ for all $n$ ?

Informally I wonder if the functions coming from the Weierstrass factorization theorem have any particular growth condition on their coefficients.

  • $\begingroup$ The zero set is related to the growth of your function, which is related to the Taylor coefficients. The standard books on the subject (Boas, for example) will discuss this in great detail. $\endgroup$ – Christian Remling Feb 23 '19 at 19:42

The answer is in general no because the zero set's convergence exponent imposes a lower bound on the order of growth of the function, which in turn imposes a growth condition on a subsequence of the Taylor coefficients.

In detail:

Assume $Z$ infinite of course and order all its non-zero elements (obviously $Z$ has finitely many zeroes and for each $r > 0$ there are only finitely many elements in $Z$ bounded by $r$ in absolute value) by the increasing absolute value of its elements $0 < |z_1| \leq |z_2| ... \leq |z_n| ...$, then the convergence exponent $\rho$ of $Z$ is the greatest lower bound of all $\lambda$ s.t. $\Sigma{\frac{1}{|z_n|^{\lambda}}}$ converges, where $\rho$ can be zero or infinity and we can find $Z$ to give us any such $\rho$ (for finite non-zero $\rho$ just pick $z_n = {n^{\frac{1}{\rho}}}$, for zero, $z_n = {2^n}$, for infinity, $z_n = n^{\frac{1}{loglogn}}$)

Assume now $f$ with zero set $Z$ has order $\rho$ as an entire function (finite or infinite, the order being the superior limit of $\frac{\log{\log{M_f(r)}}}{\log{r}}$, when $r$ goes to infinity, where $M_f(r)$ is the maximum of $|f|$ on the disc of radius $r$ centered at the origin), then a basic theorem (Hadamard) about entire functions says that the convergence exponent of its zero set defined above is less or equal than its order (obviously by multiplying $f$ with something like $e^{e^z}$ or $e^{z^q}$, $q$ high integer, we can always increase the order, even make it infinite, and keep same zero set, though in this case the order becomes integral or infinite as for finite non-integral orders there is a converse of the above).

But now there is a formula that gives the order of $f$ in terms of the Taylor coefficients and it essentially says that if $f(z) = \Sigma{b_nz^n}$ has order $\rho$, there is a subsequence of Taylor coefficients $c_n$ s.t $\frac{1}{|c_n}|$ is about $n^{\frac{n}{\rho})}$ - here we need $\rho$ non-zero, while for $\rho$ infinity we just interpret it as $\frac{1}{|c_n}|$ is smaller than $n^{\frac{n}{\rho})}$ for any finite $\rho$. Hence the sequence of Taylor coefficients of an entire function of non-zero order has a subsequence with a definite rate of growth ($\sqrt[n]{|c_n|}$ being about $n^{-\frac{1}{\rho}}$, where again we interpret that as usual for $\rho$ infinite), so in particular we have proved the negative answer to the original question as long as the zero set $Z$ has non-zero convergence exponent.

  • $\begingroup$ References for the two main properties used in the answer are: (1) $\rho \le \text{ord}(f)$: Lemma 4.8 uvm.edu/~tdupuy/complexspring2017/hadamard.pdf (2) $\text{ord}(f) = \limsup n\log(n)/\log(1/|b_n|)$: math.tamu.edu/~boas/courses/618-2006a/feb08.pdf $\endgroup$ – tj_ Feb 23 '19 at 8:01
  • $\begingroup$ Thanks for your answer. That is what I was looking for. A minor question: You say $\sqrt[n]{|c_n|}$ is about $\rho/n$. According to the 2nd ref. $\rho = \limsup n\log(n)/\log(1/|b_n|)$. So isn't $\frac{1}{|c_n|}$ about $n^{n/\rho}$ i.e. $\sqrt[n]{|c_n|}$ is about $\frac{1}{n^{1/\rho}}$? $\endgroup$ – tj_ Feb 23 '19 at 8:12
  • $\begingroup$ yes, corrected - misplaced the exponential $\endgroup$ – Conrad Feb 23 '19 at 13:34
  • $\begingroup$ And it is actually easy to check with simple functions like $e^z$ and $e^{z^2}$ where $\sqrt[n]{|c_n|}$ is about $\frac{e}{n}$ and $(\frac{2e}{n})^{\frac{1}{2}}$ by Stirling, while orders are 1 and 2 respectively $\endgroup$ – Conrad Feb 23 '19 at 13:54

The main obstacle is called Jensen's inequality. Let $n(r)$ be the counting function of zeros (=sum of multiplicities of zeros in $|z|\leq r$). Assume for simplicity that $h(0)\neq 0$. Then Jensen's formula says that $$\int_0^r\frac{n(t)}{t}dt\leq \log M(r)-\log|h(0)|.$$ where $$M(r)=\max_{|z|=r}|f(z)|\leq\sum_{0}^\infty|b_n||z|^n.$$ So, if there are many zeros in disks $|z|\leq r$ then there is an estimate for $|b_n|$ from below. This estimate is essentially best possible, and everything that Conrad wrote can be derived from it.


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