# Factorization of antisymmetric bounded holomorphic functions

A basic principle in complex function theory is that one can split off zeros of holomorphic functions in a similar way as for polynomials: If $f$ is holomorphic near $0$ and $f(0) = 0$, then $f(z) = zg(z)$ for some holomorphic function $g$. A crucial property of this type of factorization is that it preserves boundedness: If, say, $f \in H^\infty(\mathbb D)$, then also $g \in H^\infty(\mathbb D)$. I am interested in generalizations to several complex variables:

Thus, let $f$ be a holomorphic function on the $2$-disc $\mathbb D \times \mathbb D$ with the following properties: $f$ is antisymmetric (i.e. $f(z,w) = -f(w,z)$), bounded and $f(z,0) = f(0,z) = 0$ for all $z \in \mathbb D$. In particular, $f$ vanishes along the diagonal $\{w=z\}$. By an easy reduction to the one-dimensional case (using a linear change of variables), we see that $f$ can be factorized as $f(z,w) = (z-w)g(z,w)$ for some holomorphic function $g$ on the $2$-disc.

My first question is, whether $g$ is necessarily bounded. I suspect that the answer is negative, but a concrete counterexample would be nice. In this case, my second question would be: What can be said about the boundary behaviour of $g$?

(If I am not mistaken, then the radial limit function $\partial g$ of $g$ on the $2$-torus exists almost everywhere and is bounded on the complement of any neighbourhood of the diagonal circle. The question is, what happens near the diagonal. If $g$ is not bounded, then $\partial g$ will blow up near the diagonal - the question is, how badly?)

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## 2 Answers

No, $g$ is not necessarily bounded. Let $h(z,w)$ be (a branch of) $log (z+w-2)$ and let $f(z,w)$ be $(z-w)g(z,w)=(z-w)zwh(z,w)$.

Then on the one hand $zwh(z,w)$ is unbounded as $z$ and $w$ (in $\mathbb D$) both approach $1$, but on the other hand $f(z,w)$ remains bounded: the inequality $|z-w|^2+|z+w|^2=2|z|^2+2|w|^2\le 4$ shows that $|z-w|$ is going to $0$ faster than $|\log (z+w-2)\ |$ is going to $+\infty$.

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Ah, yes that is what I was looking for. I was playing around with logarithms myself, but I had the problem of finding a branch defined on all of $\mathbb D^2$. Of course, since the image of $\mathbb D^2$ under $z+w-2$ avoids $0$ and is simply-connected there is no problem here. Still, this answers my first question (so I will accept it), but not my second one. In this example we have that $(z-w)^\epsilon\cdot h(z,w)$ is bounded, so the unboundedness is rather mild. Can we modify the example to get something worse? –  Tobias Hartnick Jan 8 '13 at 17:01
You can use a power of $z+w-2$ instead of log. –  Tom Goodwillie Jan 9 '13 at 2:30
Sure, but it seems I still get a rate, i.e. I can find $\epsilon$ such that $f(z,w)/|z-w|^\epsilon$ is bounded. –  Tobias Hartnick Jan 9 '13 at 9:20
Well, you could choose something smaller than $(z+w-2)^{-1}$ but bigger than $(z+w-2)^{-1+\epsilon}$. But that's only a little worse. –  Tom Goodwillie Jan 9 '13 at 13:37

Under you assumptions, $f$ vanishes on $(\lbrace 0\rbrace\times\mathbb C)\cup (\mathbb C\times \lbrace 0\rbrace)\cup \lbrace (z,z): z\in \mathbb C\rbrace$ intersected with the domain. Thus $f(z,w)=z.w.(z-w).h(z,w)$ and $h$ is symmetric. Since you just removed zero divisors, any polar divisor that was there in $f$ is still there in $h$.

So if $f$ was bounded on an open domain, $h$ is still bounded there.

Edit: Tom's answer woke me up. The last conclusion is wrong. If the polar divisor touches $D\times D$ in the boundary, you get a counterexample. Take $h(z,w)=1/(z+w-2)$.

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It seems that I am missing a point here: I understand that by splitting of the zeros one does not introduce any new poles, i.e. h is holomorphic in the domain. However, I don't understand how that affects boundedness, which concerns only the behaviour of f and h near the boundary. I guess there is a basic fact about holomorphic functions here that I miss? –  Tobias Hartnick Jan 8 '13 at 13:24