# 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.

• I don't get what you're asking here. What is $g_\infty$ ? You call it an analytical continuation, but a continuation of what function? Your notation suggests that you are not asking about analytic continuation, but about sequences of analytic function that converge (in what sense?) to an analytic function which is something very different. Note for instance, that in your example $g_N$ converges to $g_\infty$ uniformly on compact subsets of $\{(-i)(z-0.5) \mid \mathfrak{Re}(z)>1\}=\{w \mid \mathfrak{Im}(w)<-0.5\}$. – Johannes Hahn Feb 4 '18 at 19:33
• So for a better example of what I'm talking about, consider the function $F\{G(t) \cdot \zeta_N(s+it)\}(\omega)$, where $F$ is the Fourier transform, $G$ is a Gaussian function with mean 0 and std. 1, and $\zeta_N$ is the "incomplete zeta function" which is only summed up to $N$. For any finite $N$, this is a well-defined Fourier transform, and since we are multiplying by the Gaussian we don't even need to worry about tempered distributions. (continued) – Mike Battaglia Feb 4 '18 at 19:41
• For $N=\infty$, this series only converges if $s>1$ - otherwise it does not, and hence the Fourier transform is not defined. However, we can still use the analytic continuation to get a function for $\zeta(0.5+it)$, and hence for $F\{G(t) \cdot \zeta_N(s+it)\}(\omega)$. – Mike Battaglia Feb 4 '18 at 19:41
• That doesn't make your question any clearer. What exactly are you asking here? What exactly are the assumptions you're making? For example: What is the domain of all these functions? Your post as well as your comments suggest that you work with $t\in\mathbb{R}$ which of course if not sufficient (at least not without some further clarification) to talk about holomorphic functions. You need some open subset $\Omega\subseteq\mathbb{C}$ or something similar to talk about holomorphic functions. – Johannes Hahn Feb 4 '18 at 19:48
• Here is an attempt at a precise question which you may have intended to ask: Assume $g_N, g_\infty: \Omega\to\mathbb{C}$ are holomorphic functions s.t. $g_N$ converges pointwise to $g_\infty$ on some open subset $\emptyset\neq\Omega_0\subseteq\Omega$. Given an entire function $f$, 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 $\Omega$? – Johannes Hahn Feb 4 '18 at 20:01

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.

• Thanks - I think I worded it wrong, but this was helpful nonetheless. I'll tighten up my original question and ask it differently, but I've accepted this one. – Mike Battaglia Feb 14 '18 at 4:57
• Actually, to clarify, I didn't want that $f$ be an entire function, just that it be meromorphic on the complex plane. Does that change this answer at all? I'm thinking specifically about $f(z) = 1/z$ here. The issue is, there might be open sets for which the $g_N$ blow up to infinity as $N$ increases, meaning that on those open sets, $f(g_N(z))$ converges to 0. – Mike Battaglia Feb 14 '18 at 5:01
• A good example is, $g_N(z) = \sum_{n=0}^N z^n$, $g_\infty(z) = \frac{1}{1-z}$, $f(z) = \frac{1}{z}$. It is clear that $g_N(z) \to g_\infty(z)$ for a small disk of radius 1 at the origin. Within that disk, $(f \circ g_N)(z) \to (f \circ g_\infty)(z)$, and we can holomorphically extend this to a larger portion of z. However, outside of that disk, $(f \circ g_N)(z) \to 0$, and the holomorphic extension of this is the zero function! Does this mean that we might get non-unique analytic continuations when meromorphic functions get involved? – Mike Battaglia Feb 14 '18 at 5:25
• Ah, I think I see the problem - it only converges to 0 on the positive real line > 1, which is not an open set in the complex plane. – Mike Battaglia Feb 14 '18 at 5:44
• No, meromorphic functions don't change anything, because you can consider them as holomorphic functions $\mathbb{C}\to\widehat{\mathbb{C}}$ which makes the proof go through just as well. – Johannes Hahn Feb 14 '18 at 12:33