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?) 



Factorization of antisymmetric bounded holomorphic functions
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 $+\infty$. 


