$\bar{\partial}$-Poincaré lemma

This is $\bar{\partial}$-Poincaré lemma: Given a holomorphic funtion $f:U\subset \mathbb{C} \to \mathbb{C}$ ,locally on $U$ there is a holomorphic function $g$ such that : $$\frac{\partial g}{\partial \bar z}=f$$

The author says that this is a local statement so we may assume $f$ with compact support and defined on the whole plane $\mathbb{C}$, my question is why she says that... thanks.

*Added*

$f,g$ are suppose to be $C^k$ not holomorphic, by definition $$\frac{\partial g}{\partial \bar z}=0$$ if $g$ were holomorphic...

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What book is this from? It would be helpful if you gave the title and page number so we could see the precise statement and surrounding discussion. –  Potato Jun 9 '12 at 1:45
It is Voisin's book Hodge theory and complex algebraic geometry, p.35, theorem 1.28. –  Jr. Jun 9 '12 at 1:49
Just as a general note, you should be aware the author is a "she," not a "he." –  Potato Jun 9 '12 at 1:52
I fixed it , thanks. –  Jr. Jun 9 '12 at 2:05
Dear Jr., There is something strange in your statement: if $g$ were truly holomorphic, then $\partial g/\partial \bar{z}$ would equal $0$ (this is the Cauchy--Riemann equations). So the $g$ you are looking for should probably not be holomorphic. And it seems likely to me that $f$ should not be required to be holomorphic either, since a compactly supported holomorphic function also necessarily vanishes. Regards, –  Matt E Jun 9 '12 at 3:11
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2 Answers

I don't have the book, and thus I can't check the statement. However, I believe that the statement holds for smooth $f$.

Basically we want to construct/find $g$ as the following integral:

$$g(z) = \frac{1}{2 \pi i}\int_{w\in \mathbb{C}} \frac{f(w)}{z-w} d\overline{w}\wedge dw$$

In order to do this, $f$ must be defined over the whole complex plane.

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The statement is defined on a local subset $U$... so we can make $f$ have compact support which vanishes outside $U$, thus trivially extending to the whole complex plane (defined to be zero outside the support).

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I don't understand why we get no loss of generality if we suppose $f$ with compact support. –  Jr. Jun 9 '12 at 3:01
So $f: U\to \mathbb{C}$ and the statement applies to a local subset $V\subset U$... i.e. we don't care about outside $V$ (nor $U$ in $\mathbb{C}$)... so taking a compact set $N\subset U$ containing $V$, we haven't lost any information, and $f$ vanishes outside $N$ (hence can be extended to $\mathbb{C}$ trivially). In other words, the original function $f$ and the "new" function with compact support do not look any different in the local region that you care about. –  Chris Gerig Jun 9 '12 at 4:59
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