A problem on the equation $\bar{\partial} g=f$ in complex analysis I'm reading Voisin's famous book Hodge theory and Complex algebraic geometry, page 30. And in this section Voisin proved the following fact: given a smooth function $f$, we can solve the equation $\bar{\partial} g=f$ locally. To be more precise, we can suppose that $f$ is of compact support and write down the explicit formula for $u$ as:
$$
u(z)=\frac{1}{2 i \pi} \int_{\mathbb{C}} \frac{f(\zeta)}{\zeta-z} d \zeta \wedge d \bar{\zeta}.
$$
Of course this looks very reasonable. But I also read Hormander's famous book Introduction to complex analysis in several variables. In page 30, theorem 2.3.1 (1990 edition), he wrote a remark, which said $\bar{\partial} g=f$ needn't have a solution even when $f$ is of compact support! He said [take an arbitrary $f$ with nonzero Lebesgue integral on $\mathbb{C}$].
I am very much confused for the conclusions on two masters' books look like quite different. Can anyone explain why Hormander said we can [take an arbitrary $f$ with nonzero Lebesgue integral on $\mathbb{C}$] as a counterexample? Or have I misunderstood anything? Thanks in advance!
 A: A comment that got too long but answering the three questions in the last two comments of the OP:
For the first question (here since $ n \ge 2$ we have a form of type $(0,1)$ with compact support $f=(f_1,..f_k)$ which satisfies the condition $\bar \partial f=0$ and we construct $u$ with compact support st $\bar \partial u=f$ - in one dimension there is no condition of course since automatically $\bar \partial (fd\bar z)=0$ as $d\bar z \wedge d\bar z=0$))
we construct $u$ like in the OP in the first variable only: $u(z_1,z_2,..z_n)=\frac{1}{2 i \pi} \int_{\mathbb{C}} \frac{f_1(\zeta, z_2,..z_n)}{\zeta-z_1} d \zeta \wedge d \bar{\zeta}$; then obviously $u=0$ when $|z_2+..z_n| \to \infty$ as the integrand is zero (and then by analytic continuation since $u$ is analytic outside the compact support of $f$ we have that $u=0$ there, hence it has indeed compact support).
It is easy to see that $u$ works and gives $\bar \partial_j u=f_j$ as by defintion we have this for $j=1$ and then we have $\bar \partial f=0$ which is equivalent to $\bar \partial_j f_k=\bar \partial_k f_j$, so we differentiate under the integral sign and switch $\bar \partial_j f_1$ with $\bar \partial_1 f_k$ and again apply the one dimensional case.
In general, we can add to $u$ any analytic function $g$ and we still have $\bar \partial (u+g)=f$, but the point is that we can construct one such $u$ with compact support, not that every solution has compact support (which is of course not true)
For the second question, we need $f$ with compact support so the integral exists, we can differentiate at will under the integral sign, we can switch series with the integral etc (obviously weaker conditions work and part of the theory is to understand that).
For the third question - notice that if $z \to \infty$, $zu(z) \to \frac{1}{\pi}\int_{\mathbb C} fdA$ which is then non-zero, so $u$ cannot be zero for large $|z|$ when the Lebesgue integral of $f$ is not zero, hence $u$ doesn't have compact support in general. The difference between $n=1$ and $n \ge 2$ is then clear and the two statements in the OP are not contradictory as they refer to two different situations.
($u(z)=\frac{1}{2 i \pi} \int_{\mathbb{C}} \frac{f(\zeta)}{\zeta-z} d \zeta \wedge d \bar{\zeta}=-\frac{1}{\pi}\int_{\mathbb C}\frac{f(\zeta)}{\zeta-z}dA=\frac{1}{\pi}\sum_{k \ge 0}\int_{\mathbb C}f(\zeta)\zeta^k/z^{k+1}dA$ hence $zu(z) \to \frac{1}{\pi}\int_{\mathbb C} fdA$ as noted)
