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Let f be an anti-symmetric function in the Bergman space of the bi-disc. That is, f is square integrable, holomorphic on the bi-disc and $f(z_1,z_2)=-f(z_2,z_1).$ So, $f(z_1,z_2)/(z_1-z_2)$ extends as a holomorphic function on the bi-disc, by the Riemann's Extension Theorem. I ask, is it true that $f(z_1,z_2)/(z_1-z_2)$ is square integrable? In other words, is $f(z_1,z_2)/(z_1-z_2)$ an element of the Bergman space on the bi-disc?

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No. Try $f(z_{1},z_{2})=\frac{1}{\sqrt{z_{1}+1}}-\frac{1}{\sqrt{z_{2}+1}}$.

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