According to Lindelöf's theorem, given points $z_i\in \mathbb C\setminus \{0\}$ ordered by increasing modulus with possible repetitions, we can define a function $$ f(z)=\prod_{n=1}^\infty (1-z/z_n)e^{z/z_n}$$ which is of exponential type iff the number of $z_i$ in the ball $B(0,R)$ is bounded by $cR$ for some constant $C>0$ and $$\sup_{R>0}\sum_i z_i^{-1}<\infty,$$ (see the book of Koosis, The Logarithmic Integral). My question is whether, under the same conditions, by adapting the definition of $f$, we can also prescribe the type, i.e. the infimum of the $a>0$ such that $$\sup_z |f(z)|\exp(-a|z|)<\infty.$$

As a step in this direction, I believe I managed to do it under the condition $\sum_i |z_i|^{-1}<\infty$, by defining $$f(z)=\prod_n \frac{s\left(\frac{z_n-z}{\lambda z_n}\right)}{s(\lambda^{-1})}$$ with the right choice of the function $s$ and the parameter $\lambda>0$ chosen sufficiently large (one can show that the spectrum of $f$ is bounded by $a=K\lambda^{-1}\sum_i |z_i|^{-1}$ for some constant $K$ depending on $s$, which means the type is smaller than $a$ by the Schwartz–Paley–Wiener Theorem).

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    $\begingroup$ If $\sum \frac{1}{|z_k|}$ is convergent, then there is such a function of minimal type (or, more precisely, the function that you wrote has a type $|\sum \frac{1}{z_k}|$ coming from the exponential factors, as the bracket factors give a function of minimal type). In general, there are two-sided estimates for the type, see Lectures on entire functions by Boris Yakovlevich Levin, Lecture 5 Theorem 4, but I'm not aware of the precise formula for the type. $\endgroup$ Commented Jun 24 at 8:32
  • $\begingroup$ Thanks you for this answer. I will go see these lectures, but if I understand correctly what you are saying, my construction to have a type as small as I want (in the second part of the question) while still keeping the zeros is doomed to fail, right? $\endgroup$ Commented Jun 24 at 8:57
  • $\begingroup$ Yes, if the sequence is dense enough (say, if there are arbitrarily big $R$ such that in $B(0, R)$ there are at least $R$ points), then there is an absolute lower bound on the type. This can be established by a simple application of the Jensen's formula on the ball $B(0, 2R)$. $\endgroup$ Commented Jun 24 at 12:39
  • $\begingroup$ Yes ok I understand that here there is a bound on the type, but in this case the inverse moduli are not summable, so I still have some hope. $\endgroup$ Commented Jun 24 at 14:17
  • $\begingroup$ I am confused. Can you state your question absolutely precisely, about which expression exactly, under which extra conditions and what do you ask? $\endgroup$ Commented Jun 24 at 20:12

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One restriction on the type follows from Jensen's formula which can be written as $$\frac{1}{2\pi}\int_{-\pi}^{\pi}\log|f(re^{i\theta})|d\theta= \int_0^r\frac{n(t)}{t}dt+\log|f(0)|,$$ assuming that $f(0)\neq 0$, where $n(t)$ is the number of zeros in $|z|\leq t.$ So, for example the exponential type $\sigma$ implies that $$ \limsup\frac{n(r)}{r}\leq e\sigma.$$ This estimate is exact for some distributions of zeros, but in general the type depends in a very subtle way not only on $n(t)$ but also on the distribution of the arguments of zeros.

So the exact relation between $n(t)$ and type of the function is complicated. But there is a formula for it in terms angular density of zeros:

MR0164036 Golʹdberg, A. A. The integral over a semi-additive measure and its application to the theory of entire functions. I-IV (Russian). Mat. Sb. (N.S.) 58(100) (1962), 289–334; 61 (1962) 334–349. III. 65(107) (1964), 414–453; 66(108):3 (1965), 411–457.

Russian originals are available here (free), but there are also English translations (not free).

To write these explicit formulas, Goldberg had to generalize the notion of the integral (he introduced the integral over a semi-additive measure), so the answer cannot be called very explicit.

He considers the general case of functions of finite order, but I do not think that his formulas can be substantially simplified in the case of exponential type.

See also my paper An extremal problem for a class of entire functions with Peter Yuditskii, which addresses a subset of functions of exponential type arising in harmonic analysis.

  • $\begingroup$ Thanks, so apparently the right assumption to make is not that the partial sum is bounded but rather that the counting measure grows linearly (or not) $\endgroup$ Commented Jun 25 at 13:48
  • $\begingroup$ @kaleidoscop: "right assumption" for what conclusion? $\endgroup$ Commented Jun 26 at 13:49
  • $\begingroup$ My initial question was wether the condition $\sup_R \sum_{i:|z_i|\leq R}z_i^{-1}<\infty$ is enough to guarantee minimal type, but there was no answer to that. $\endgroup$ Commented Jun 28 at 8:38
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    $\begingroup$ @kaleidoscope: certainly it is not enough: an example is $\cos z$ (or any even placing of zeros) for which all these sums are 0. But there is no function of zero type with these zeros. $\endgroup$ Commented Jun 28 at 17:00

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