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!

  • $\begingroup$ You shouldn't have to do distributions on $\mathbb{C}$. Essentially all the arguments relating complex analysis with vector analysis use the interpretation of $\mathbb{C}$ as $\mathbb{R}^2$, so you are dealing with (in fact harmonic) functions from $\mathbb{R}^2$ to $\mathbb{R}^2$. $\endgroup$ – mlk Jun 13 '14 at 19:00
  • $\begingroup$ @CameronWilliams: obviously my comment is not considered to be offensive. I hope you gladly take my advice. apart from that "when one restricts to holomorphic functions" doesn't make sense, and what you are asking for is the fundamental solution of the C-R operator, which is well known and can be found in any PDE book (E.g. treves)...... $\endgroup$ – user55315 Jun 21 '14 at 21:51
  • $\begingroup$ @Karl Your comment came off very rude and unkind (even if that was not your intention) but I see your point now. My phrasing was not the best since I was a bit frustrated and frantic but I am by no means a novice at complex analysis. I did extensive searches on this topic before asking here (probably spent a few hours). I was not using the right terms by which to search it seems. I apologize for getting upset but your first comment was horribly presumptuous. I hope you recognize that. $\endgroup$ – Cameron Williams Jun 21 '14 at 22:13
  • $\begingroup$ I can assure you that it wasn't my intention. you can read about this in e.g. Taylor, PDE I, chap. 3.4; or Treves, Basic linear PDE, chapter 1.5. (fundamental solutions of the cauchy-riemann operator). there some cool things are pointed out, such every distributional solution of $\bar{\partial}u=0$ is in fact an analytic function, derives cauchy's formula (from a distributional point of view), and also has many nice exercises (in case you want to prove some related results). Another is Krantz, Geometric function theory, chap. 7.2. I think it is exactly what you are looking for! $\endgroup$ – user55315 Jun 21 '14 at 23:23
  • $\begingroup$ Thanks @Karl. Sorry for the misunderstanding. $\endgroup$ – Cameron Williams Jun 21 '14 at 23:27

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

$$\frac{1}{\pi}\frac{\partial}{\partial\overline{z}} \frac{1}{z-w} = \delta_w.$$

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  • $\begingroup$ Very nice Daniel! The problem I was having was exactly Stokes' theorem. I recognized that I would have $dxdy$ but did not think to write it in the language of differential forms. This is exactly what I was looking for. Thanks a bunch. $\endgroup$ – Cameron Williams Jun 13 '14 at 20:20
  • $\begingroup$ One concern that I have is that we say that $\varphi$ has compact support but then make use of the fact that it is holomorphic at the very end. Are these two ideas not at odds with each other? If $\varphi$ is holomorphic and has compact support, it must be identically zero. $\endgroup$ – Cameron Williams Jun 13 '14 at 20:26
  • 3
    $\begingroup$ No, we do not assume that $\varphi$ is holomorphic. (There is a certain scarcity of holomorphic test functions ;-) But since it is smooth, we can write $$\varphi(z) = \varphi(w) + \frac{\partial\varphi}{\partial z}(w)(z-w) + \frac{\partial\varphi}{\partial\overline{z}}(w)\overline{z-w} + O(\lvert z-w\rvert^2).$$ The constant term gives the $\varphi(w)$ (times $\pi$) independent of $\varepsilon$. The other terms are all bounded when divided by $z-w$, and so the integral of these terms is $\leqslant C\cdot 2\pi \varepsilon$ in absolute value. In the limit, only the constant term survives. $\endgroup$ – Daniel Fischer Jun 13 '14 at 20:34
  • $\begingroup$ Oh of course. That's very clever. Thanks again. $\endgroup$ – Cameron Williams Jun 13 '14 at 20:38

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