I've read some complex analysis texts and often there is some appeals to Green's theorems when proving facts about contour integrals of holomorphic functions yet there seems to be a lack of appeals to Green's functions (at least explicitly). We know that for a holomorphic function $f$, we have
$$\frac{\partial f}{\partial\bar{z}} = 0$$
and also that
$$f(z) = \frac{1}{2\pi i}\int_{\gamma} \frac{f(\zeta)}{\zeta-z}\,d\zeta,$$
where $\gamma$ is a circular contour around $z$ (with a winding number of one). Given the duality between partial differential operators and integral operators, this has made me a bit curious as to what the Green's function for $\dfrac{\partial}{\partial \bar{z}}$ is (when one restricts to holomorphic functions). Intuitively, I would hazard to guess that the Green's function is related to $\dfrac{1}{z}$ based on the above observations but I haven't figured out a way to show this. Solving
$$\frac{\partial}{\partial\bar{z}}G = \delta(z)$$
for $G$ seems to be a bit of a tall task. Distributions seem to be a little bit complicated on $\mathbb{C}$. Ultimately, I want to use this understanding to tackle some aspects of Green's functions in several complex variables. Any help would be greatly appreciated!
Best Answer
Actually, it's not so hard to find a solution of $\frac{\partial}{\partial\overline{z}}G =\delta$. You have the right thing there already, modulo normalisation. We have, in the sense of distributions, for any $\varphi\in \mathscr{D}(\mathbb{C})$,
$$\begin{align} \left(\frac{\partial}{\partial\overline{z}}\frac{1}{z-w}\right)[\varphi] &= - \int_{\mathbb{C}} \frac{\partial\varphi}{\partial\overline{z}}(z)\frac{1}{z-w}\,d\lambda\\ &= -\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert z-w\rvert < R} \frac{\partial\varphi/\partial\overline{z}(z)}{z-w}\,d\lambda \end{align}$$
where $\lambda$ is the Lebesgue measure on $\mathbb{C}$ and $R$ is sufficiently large, so that the support of $\varphi$ is contained in $D_R(w)$. Now write the integral with differential forms, that is, replace $d\lambda$ with $dx\wedge dy$, and write the latter as $\frac{1}{2i}d\overline{z}\wedge dz$. We get
$$\begin{align} \left(\frac{\partial}{\partial\overline{z}}\frac{1}{z-w}\right)[\varphi] &= -\lim_{\varepsilon\to 0} \frac{1}{2i}\int_{\varepsilon < \lvert z-w\rvert < R} \frac{\partial\varphi/\partial\overline{z}(z)}{z-w}\,d\overline{z}\wedge dz\\ &= -\frac{1}{2i}\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert z-w\rvert < R} d\left(\frac{\varphi(z)}{z-w}\, dz\right) \tag{Stokes}\\ &= \frac{1}{2i}\lim_{\varepsilon\to 0} \int_{\lvert z-w\rvert = \varepsilon} \frac{\varphi(z)}{z-w}\,dz\\ &= \frac{1}{2i}\lim_{\varepsilon\to 0} \left(\int_{\lvert z-w\rvert = \varepsilon} \frac{\varphi(w)}{z-w}\,dz + \int_{\lvert z-w\rvert =\varepsilon} \frac{\varphi(z)-\varphi(w)}{z-w}\,dz \right)\\ &= \pi \varphi(w), \end{align}$$
since by the differentiability of $\varphi$ in $w$, the integrand of the second integral in the penultimate line remains bounded as $\varepsilon\to 0$, and so the second integral is $\leqslant C\cdot 2\pi\varepsilon$ in absolute value by the ML-inequality. That means
$$\frac{1}{\pi}\frac{\partial}{\partial\overline{z}} \frac{1}{z-w} = \delta_w.$$