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it is all in the title : what is $\overline{\partial} \frac{1}{z^2}$ in the sense of distributions ? I remember that $\overline{\partial} \frac{1}{z}$ is a dirac at 0, but I can't seem to find a way to deduce the result for $1/z^2$ from that. I also know that the support of $\overline{\partial} \frac{1}{z^2}$ is included in $\{0 \}$ so it's a dirac or a derivative of dirac. I was unable to find the proof.

EDIT : as pointed out, the question doesn't make sense since $1/z^2$ is not a distribution.

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$1/z^2$ is not locally integrable, so is not instantly a distribution near $0$. That is, some sort of "regularization" is needed, or needs to be prescribed, in order for the question to have a precise sense. Is there a preference? – paul garrett Sep 26 '12 at 13:40
you're absolutely right. the question doesn't make sense. I don't think it can be saved through a regularization. Thanks ! – Glougloubarbaki Sep 26 '12 at 13:47
But we could take the position that $z^{-2}=-\partial (z^{-1})$ in the distributional sense: since $z^{-1}$ is a distribution, so is its derivative. – user31373 Sep 26 '12 at 13:49
Yes, as LVK suggests, choosing to take $z^{-2}=\partial z^{-1}$ would be compatible with my own choice of regularization. This is also consistent with construing the thing as meromorphic continuation of $|z|^s\cdot z^{-2}$. – paul garrett Sep 26 '12 at 13:52
OK, so then what would $\overline{\partial}\partial (z^{-1})$ be ? – Glougloubarbaki Sep 26 '12 at 14:06

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