Let $f:\mathbb{H}\to \mathbb{C}$ be a holomorphic function on the upper half plane. Even it is not defined on the real line, we will define $\mathrm{ord}_{z=z_0}f(z)$ to be the unique value $\xi\in\mathbb{R}$ such that $\lim_{z\to z_0}f(z)/(z-z_0)^{\xi}$ exists and is non-zero; if there is no such number, then we will say that $\mathrm{ord}_{z=z_0}f(z)$ is just not defined. Residue is defined analagously.
We now assume the following conditions on $f$:
- $\mathrm{ord}_{z=z_0}f(z)$ is a well defined integer for every $z_0\in\mathbb{R}$
- $\mathrm{ord}_{z=n}f(z)=\mathrm{ord}_{z=1}f(z)<0$ and $\mathrm{res}_{z=n}f(z)=\mathrm{res}_{z=1}f(z)$ for all $n\in\mathbb{N}$
- $\lim_{|z|\to\infty, \Im(z)\geq1}f(z)/z^C=0$ for some $C>0$
Do conditions 1-3 imply that $\mathrm{ord}_{z=n}f(z)=\mathrm{ord}_{z=1}f(z)$ and $\mathrm{res}_{z=n}f(z)=\mathrm{res}_{z=1}f(z)$ for negative integers $n$ as well?
This is very similar to Carlson's theorem, except that we are looking at poles with a given residue instead of zeros, and we have the extremely large added difficulty of working on the boundary of where our function is defined, so we cannot make contours around it; to make up for this added difficulty we have the much stronger growth condition of polynomial assumed instead of the more general "of exponential type<$\pi$" the regular Carlson's theorem allows.
