Holomorphic almost periodic function: how do the number of zeros in a box of height $T$ scale with $T$?

Question

Define

$$f(z) = \sum_{n} {x_n}^z - 1,$$

where $0< x_n$ and $\displaystyle\sum_n x_n = 1$ (the sum is allowed to be infinite).

You can see that this defines a holomorphic almost periodic function on a half plane, that $f(1) = 0$. and $Re(z)>1 \implies f(z) \neq 0$.

Further assume that $f$ converges absolutely on some strip $\{Re (z)\geq \sigma\}$ for some $\sigma < 1$.

Question. Is it known whether there is a definitive growth rate for the number of zeros of $f$ in a box $[\sigma, 1] + i[-T,T]$, as $T \to \infty$?

Here is a picture: the yellow region denotes the box in question, and crosses denote zeros of $f$.

In particular, does the following limit exist? $$\lim_{T\to \infty}\left(\frac1T\#\bigg\{z \in [\sigma, 1] + i[-T,T]\,\bigg|\,f(z) = 0\bigg\}\right) $$

If not, are there any examples where the size of this set grows faster or slower than linear? Thanks!

Answer 1

Let $f(s)=\sum_{n} a_n b_n^{-s}$ converge absolutely for $\Re(s)\ge 1-2c$ and such that $f(1)=0$.

Assume for simplicity that $f$ has no zero on two tiny strips containing the vertical lines $\Re(s)=1-c,\Re(s)=1+c$.

(if no such vertical lines exist we need to consider some zigzagging curves, not sure if it changes the result)

Let $f_N(s)=\sum_{n=1}^N a_n b_n^{-s}$. There is some $r> 0$ such that for $N$ large enough, $|f_N(s)|\ge r$ on $\Re(s)= 1\pm c$. Consider a branch of $\log f_N$ which is continuous on the two vertical lines.

Then $\log(f_N)\to \log(f)$ uniformly on $\Re(s)=1\pm c$.

Let $Z(f,c,T)$ be the number of zeros of $f$ on the rectangle $ [1-c,1+c]+i [0,T]$.

$f(1)=0$ plus the almost-periodicity implies $\lim_{T\to \infty} Z(f,c,T)= \infty$.

On the other hand $$Z(f,c,T)=\frac1{2i\pi}\int_{\partial\ [1-c,1+c]+i[0,T]}\frac{f'(s)}{f(s)}ds\qquad\qquad\qquad$$ $$\qquad\qquad\qquad= \frac{1}{2i\pi}(\log f(1+c+iT)-\log f(1-c+iT)+O(1))$$ which means that $$\lim_{T\to \infty} \frac{Z(f_N,c,T)}{ Z(f,c,T)}=1$$

$\lim_{T\to \infty} Z(f_N,\infty,T)/T$ exists because for $\Re(s)\to \pm \infty$ we can exploit that $f_N(s)\to a_1 b_1^{-s}$, $f_N(s)\to a_N b_N^{-s}$.

The zeros of $f_N$ are located on finitely many vertical strips, it seems hard to imagine how the density of zeros would oscillate from one strip to the other, suggesting that $\lim_{T\to \infty} Z(f_N,c,T)/T$ exists and hence $$Z(f,c,T)\sim A T$$