Does it follow that $F^{(1)}(z)=F^{(2)}(z)$ for all $z \in \mathbb H$?

Question

Let $\mathbb H= \{z \in \mathbb C\,:\, \textrm{Re}\,(z)\geq 0\}$ and for $j=1,2$ suppose that $F^{(j)}:\mathbb H\to \mathbb H$ is defined via $$ F^{(j)}(z) = \sum_{k=1}^{\infty} \frac{a_k^{(j)}}{z+\lambda^{(j)}_k},$$ where $|a_{k}^{(j)}|\leq k^{-2}$ and $0<\lambda^{(j)}_1<\lambda^{(j)}_2<...$ with $\lim_{k\to \infty} \lambda^{(j)}_k=\infty$. Suppose that $$ F^{(1)}(n)=F^{(2)}(n)\quad \text{for all $n \in \mathbb N$}.$$ Does it follow that $F^{(1)}(z)=F^{(2)}(z)$ for all $z \in \mathbb H$?

Answer 1

The answer is positive. Indeed, $f=F^{(1)}-F^{(2)}$ is a bounded analytic function in the right half-plane (this follows from your conditions $|a_k|\leq k^{-2}$ and $\lambda_k\to\infty$). But a bounded analytic function cannot be zero at positive integers, unless it is identically equal to zero. This follows from the Blaschke condition on zeros of bounded analytic functions.

Your assumption that $F$ maps the right half-plane into itself is redundant, and the assumption that $\lambda_k>0$ can be very much relaxed; all you need is that your functions are bounded in the right half-plane.