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Suppose $f_j(z)$, $j=1,2,..$ is a sequence of meromorphic functions on the complex plane $\mathbb{C}$. With a common set of all poles given by $S = \{-i,-2i, -3i,..\}$.

Assume that each of them is holomorphic in the upper half plane $\mathbb{H}$ with their restrictions converging (in some appropriate sense) to a holomorphic function $f(z)$ on $\mathbb{H}$.

Is there a precise sense for the convergence on $\mathbb{H}$ that guarantees that the limit function $f$ also has a meromorphic extension to the entire plane $\mathbb{C}$?

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    $\begingroup$ I don't think so. Consider the sum $\sum_{j=1}^\infty c_j\frac1{z-(i+1/j)}$. Choose $c_j$ to converge to zero as fast as you like, the limit will have accumulating poles and hence not be meromorphic. $\endgroup$
    – user473423
    Commented Sep 2, 2023 at 5:51
  • $\begingroup$ For any holomorphic function on the half plane, there is a sequence of polynomials that converges to it locally uniformly. $\endgroup$
    – Christian Remling
    Commented Sep 2, 2023 at 14:48
  • $\begingroup$ Actually, I also happen to know the set of poles for the functions $f_j$ is the same. I have edited the question accordingly. I think in Echo's example the set of poles is changing. $\endgroup$
    – 2inftyandBeyond
    Commented Sep 2, 2023 at 22:06

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