Context: During a mathematical discussion with a good friend, I was brought to think about the functional equation $q(t)q(1-t) = 1$ for $q$ with some regularity. (To be honest the exact context doesn't really matter, since I am not trying to solve the original problem that this stemmed from). Upon brainstorming a few solutions ($exp(t-1/2)$, $\Gamma(t)\sqrt{\sin \pi t}$ up to a constant)), I came to realize that the solutions are plentiful. By reparameterizing the functions and choosing the regularity to be meromorphic (motivated by the fact that all the pretty solutions were meromorphic), I set myself to the following challenge:
Characterise/Classify all the functions $f \in \mathcal{M}(\mathbb{D})$ which satisfy the reflection $f(z)f(-z)=1$
My attempt goes as follows:
Denote by $G$ the set of such functions. After some verifications, it becomes clear that $G$ is actually a group with respect to the operation of pointwise multiplication* of functions (*one has to take care of removable singularities). Moreover, define the function \begin{align} K : \mathcal{M}(\mathbb{D}) &\to \mathcal{M}(\mathbb{D}) \\\\ f(z) &\to K(f)(z) = f(z)/f(-z) \end{align}
Easy computations show the following three facts:
- The image of $K$ is inside $G$, ie $Im(K) \subset G$
- K is in fact a morphism of groups with respect to multiplication
- The kernel of $K$ is the subgroup of even meromorphic functions on the unit disk, let's call it $\mathcal{E}(\mathbb{D})$
Now using the first isomorphism theorem, one obtains the following isomorphism of groups: $\mathcal{M}(\mathbb{D})/\mathcal{E}(\mathbb{D}) \cong Im(K)$
Obviously for my purpose, I'd like $K$ to be onto $G$, in other words, I want to be able to express every $g \in G$ as $g(z) = f(z) / f(-z)$ for some meromorphic $f$, obtaining the isomorphism of groups: $\mathcal{M}(\mathbb{D})/\mathcal{E}(\mathbb{D}) \cong G$
There is some non formal evidence for that being the case, but I'd like to formalize it. The path I took was the following: if $g \in G$ is without any zeroes, then it is also without poles, so there exists a holomorphic branch of its square root on $\mathbb{D}$. Then $f(z) = \sqrt{g(z)}$ satisfies the following:
$f(z) / f(-z) = \sqrt{g(z)}/\sqrt{g(-z)} = g(z) / (\sqrt{g(z)}\sqrt{g(-z)})$, where the last denominators square is equal to one (since $g \in G$), so that it itself must be equal to one (if one takes the correct branch?). Thus any $g\in G$ without zeroes can be expressed as $f(z) / f(-z)$ for some meromorphic $f$.
To tackle the case where $g$ is allowed to be zero, we'll show a little lemma:
lemma: $g\in G$ admits the desired decomposition iff there exists an even meromorphic $e$ such that there exists a meromorphic branch of $\sqrt{e(z)g(z)}$
proof: if such an $e$ exists then one can set $f(z) = \sqrt{e(z)g(z)}$ and redo the previous computations. For the converse, consider $f^2(z) / g(z) \blacksquare$
In our case, the poles and zeros of $g$ come in pairs: if $z$ is a zero of order $n$ then $-z$ is a pole of order $n$ and vice versa. By multiplying by our magic $e$, we want to make all the orders of the zeroes/poles even, then $eg$ will admit a square root. An easy example is if $g$ has a finite number of zeroes of odd order, say $\alpha_1, ..., \alpha_n$: one considers: $e(z) = \prod_k (z^2 - \alpha_k^2)$. This finite product is an even function that has simple zeroes at all the points where $g$ has an odd ordered zero, thus $eg$ will only have even ordered zeroes and hence admits a square root. (and obviously the problem of the poles is also taken care of, since for any odd ordered pole $-\alpha_k$, the product $eg$ will have the pole's order decreased by one, giving an even ordered pole)
This is where the problem comes: what if $g$ has an infinite number of odd ordered zeroes? In that case the aforementioned product has no reason to converge (that I'm aware of). Is there a workaround with more well chosen functions than $(z^2 - \alpha^2)$ to make the infinite product converge? Or is this approach just not good enough to obtain the full result?