Density of zeros of holomorphic function It seems to me, if I remember well, that I read somewhere a theorem that says something similar:
If $f$ is holomorphic on the right complex half-plane ($\text{Re }z>0)$ and it has the real roots $0<x_1<x_2<\ldots$, maybe $\sum_{k=1}^{\infty}\frac{1}{x_k}=\infty$, and $|f(z)|\leq c_1 e^{c_2 |z|^{c_3}}$, where $c_1,c_2>0$, $0<c_3<1$ then $f\equiv 0$. I need some references, and -if not "long"- the proof also. I tried to find it using search engine without success. Any (correct) version of this theorem I would appreciate very much.
 A: Sketch of the result  (here one can allow complex zeroes under the condition that $\sum \frac{\Re z_n}{|z_n|^2} = \infty$ and still get same result)
If $g$ is holomorphic on the unit disc with zeroes $a_n$ (in increasing order of the modulus), it is not hard to show using Jensen's theorem that $\sup_{0<r<1}\int_0^{2\pi}\log |g(re^{i\theta})|d\theta < \infty$ iff $\sum (1-|a_n|) < \infty$
Also it is a standard result that $\sup_{0<r<1}\int_0^{2\pi}|\frac{1+re^{i\theta}}{1-re^{i\theta}}|^qd\theta < \infty$ for $0 <q<1$ but not for $q \ge 1$
(sometimes this is presented as the fact that the Poisson kernel which is the real part of the above analytic $g(re^{i\theta})=\frac{1+re^{i\theta}}{1-re^{i\theta}}$ is in $h^1$ being positive, but its conjugate which is the imaginary part of $g$ is only in $h^q, q<1$ where $h^p$ are the Hardy spaces for harmonic functions on the unit disc)
But now considering $g(w)=f(\frac{1+w}{1-w}), |w|<1$ we notice that its zeroes are $a_n=\frac{x_n-1}{x_n+1}$ and $1-|a_n|=\frac{2}{x_n+1}$ for $n$ large enough so $x_n >1$ and the hypothesis gives that $\sum \frac{2}{x_n+1} = \infty$, hence $\sum (1-|a_n|) =\infty$, hence
$\sup_{0<r<1}\int_0^{2\pi}\log |g(re^{i\theta})|d\theta =\infty$
On the other hand the inequality $|f(z)|\leq c_1 e^{c_2 |z|^{c_3}}$ implies $\log |g(re^{i\theta})| \le \log c_1 + c_2|\frac{1+re^{i\theta}}{1-re^{i\theta}}|^{c_3}$ and the second paragrpah above shows that $\sup_{0<r<1}\int_0^{2\pi}\log |g(re^{i\theta})|d\theta \le A+B\sup_{0<r<1}\int_0^{2\pi}|\frac{1+re^{i\theta}}{1-re^{i\theta}}|^{c_3}d\theta < \infty$ since $0<c_3<1$ and that is the required contradiction!
