complex analysis - Analytic continuation "playing nice" with function composition

Suppose I have a meromorphic function $f(s)$, and a sequence of functions $g_N(t)$ that diverge to infinity, but for which the analytic continuation exists. A good example would be $g_N(t)= \sum_{n=1}^N \frac{1}{n^{0.5+it}}$, which diverges as $N \to \infty$, but which can be analytically continued to $g_\infty(t)=\zeta(0.5+it)$.

Now consider the function $f(g_N(t))$. As we increase $N$, the argument to $f$ blows up. Naively, there are two ways to do the analytic continuation:

1. 
   

   First analytically continue $g_N(t)$ to $g_\infty(t)$, and then take $f(g_\infty(t))$

   
   


2. 
   

   Analytically continue $f(g_N(t))$ all at once

   
   

In other words, for the second approach, we define a new family of functions $(f \circ g)_N(t)$, which has a different limit than the analytic continuation of $f(g_\infty(t))$.

Am I correct that this shows that analytic continuation and function composition do not play nice with one another? Is there a general theory of when the two approaches will agree?

For example, look at $f(t) = \frac{1}{t}$. Then as $N$ blows up, $f(g_N(t)) \to 0$, so the composition is the zero function. On the other hand, the analytic continuation $g_\infty(t)$ need.not be strictly positive at all, or could even be zero at points, leading to poles.

Answer

Just to make my comment an official answer:

Assuming you really intend to ask this question

Assume $g_N,g_\infty:\Omega\to\mathbb{C}$ are holomorphic functions s.t. $g_N \xrightarrow{N\to\infty} g_\infty$ pointwise on some open subset $\emptyset\neq\Omega_0\subseteq\Omega$. Given any entire function $f:\mathbb{C}\to\mathbb{C}$, is it true that $f\circ g_N$ converges pointwise to a holomorphic function $h$ on $\Omega_0$ and that $f\circ g_\infty$ is the analytic continuation of $h$ to all of $\Omega$?

In that case, the answer is "yes" for obvious reasons: $f$ is continuous, therefore $f(g_N(z)) \to f(g_\infty(z))$ for all $z\in\Omega_0$. We can therefore define $h:=(f\circ g_\infty)_{|\Omega_0}$ and have found a holomorphic function on $\Omega_0$ such that $f\circ g_N$ converges pointwise to $h$. Furthermore: By construction $f\circ g_\infty$ is a holomorphic extension of $h$ to all of $\Omega$ and by the identity theorem it is the unique such function, i.e. the analytic continuation of $h$.

Note that the two instances of pointwise convergence can be replace by a lot of other convergence modes. For example one could ask for uniform convergence, locally uniform convergence, and many more.