# Elegant proofs of $\bar{\partial}z^{-1} = 2\pi \delta_0$

For a function $$f(x,y)$$ on $$\mathbb{R}^2,$$ defined possibly outside the origin, write

$$\int_\epsilon ' f \,dx\,dy : = \int_{\mathbb{R}^2\setminus D_\epsilon}f \, dx\,dy,$$

(the integral on the complement to the $$\epsilon$$-disk) and

$$\int' f \,dx \,dy : = \lim_{\epsilon\to 0} \int'_\epsilon f \,dx\,dy,$$ when defined.

We view $$\mathbb{R}^2$$ as the complex line with coordinate $$z = x+iy$$. Then I claim that the assignment $$\phi:f(x,y) \mapsto \int_{\mathbb{R}^2}' f\cdot z^{-1} \,dx\,dy$$ makes sense as a functional on compactly supported, smooth functions (indeed, if $$f$$ is rotationally symmetric then $$\phi(f) = 0$$ even before taking the limit, and if $$f(0) = 0$$ then $$f\cdot z^{-1}$$ is bounded, hence integrable, at $$0$$; now in the space of smooth compactly supported functions, any function can be written as a rotationally symmetric function plus a function that vanishes at the origin). We can thus formally define a distribution $$z^{-1}\in C^{-\infty}(\mathbb{R}^2)$$ as $$z^{-1}: = \frac{\phi}{dx\,dy}.$$

Now write $$\bar{\partial} = \partial_x + i \partial_y$$ .Like any vector field, this acts on distributions, and so we have a distribution $$\bar{\partial} z^{-1}.$$ Since $$z^{-1}$$ is holomorphic where it is smooth, we must have $$\bar{\partial} z^{-1}$$ be a distribution supported at the origin. In fact, it is known to mathematical physicists that the result is a delta function at the origin: $$\bar{\partial} z^{-1} = 2\pi \delta_0,$$ and this fact is useful in conformal field theory.

I would like to see a proof of this result (the physics sources I have seen do not prove this). In fact, I know how to give one using a direct calculation with polar coordinates: but this seems ad hoc and is not very satisfying to me.

I suspect there might be "nicer" proofs using one or more of the following three techniques, and I am hoping that more analytically literate MO users can provide them.

1. If one can show that $$\bar{\partial} z^{-1}$$ is determined by its values on holomorphic (near the origin) functions, the Cauchy residue formula would imply that $$\bar{\partial} z^{-1} = 2\pi \delta_0.$$
2. I suspect there should be a proof using the stationary phase approximation.
3. This is probably overkill, but it would be nice if there were a proof using pseudodifferential operators.
• I believe that this directly and easily follows from my identity: mathoverflow.net/questions/131670/… (this is as elementary as it can be). Commented Apr 9, 2022 at 19:04

The identification of $$\partial_{\bar{z}}z^{-1}$$ with a delta function follows directly from the Cauchy–Pompeiu formula $$f(\zeta) = \frac{1}{2\pi i}\int_{\partial D} \frac{f(z) \,dz}{z-\zeta} - \frac{1}{\pi}\iint_D \frac{\partial f(z)}{\partial \bar{z}} \frac{dx\wedge dy}{z-\zeta},\qquad\qquad(\ast)$$ for $$f$$ a complex-valued $$C^1$$ function on the closure of the disk $$D$$ in $$\mathbb{C}$$. (Note that $$f$$ need not be holomorphic, as in the usual Cauchy formula.) If the support of $$f$$ is within $$D$$ the boundary integral $$\int_{\partial D}$$ does not contribute, while the area integral $$\iint_D$$ states that$$^\ast$$ $$\frac{\partial }{\partial \bar{z}} \frac{1}{z-\zeta}=\pi\delta^{(2)}(z-\zeta).$$

Historical note: Wikipedia gives a 1905 paper as the source of the formula ($$\ast$$), but I could not locate it there. I did find it in a 1913 paper by Pompeiu (equation 3):

The first term $$h(z)$$ is the boundary integral. The capital $$S$$ denotes the area integral, the function $$\varphi$$ is the derivative $$\partial f/\partial \bar{\zeta}$$.

$$^\ast$$ The OP has a factor $$2\pi$$ instead of $$\pi$$, because of a different definition of the Wirtinger derivative. Here I follow the definition $$\partial/\partial\bar{z}=\tfrac{1}{2}(\partial/\partial x+i\partial/\partial y)$$, while in the OP there is no coefficient $$\tfrac{1}{2}$$.

• Thank you! This is the kind of thing I was looking for Commented Apr 10, 2022 at 19:27

You can conclude it from the well known fact that $$\frac{1}{2\pi}\log|z|$$ is a fundamental solution to the Laplace equation. The argument presented below is taken from my lecture notes: Harmonic Analysis, see Proposition 6.18 on pg. 89. Please, refer to the notes for more details.

• Not known only to mathematical ohysicists. Even Laurent Schwartz was aware of the formula $$\frac{\partial}{\partial \bar z}{\frac 1 z}=\pi \delta$$ and included it on p. 49 of his seminal text. Whether his proof is elegant is a matter of taste. Commented Apr 9, 2022 at 13:36

Following Gelfand-Graev, Grothdieck, and Schwartz: The right-most pole of $$u_s=|z|^{2s}=z^s\cdot \overline{z}^s$$ (or, properly, the distribution given by integration-against that function) is at $$s=-1$$, when the function ceases to be $$L^1_{\mathrm{loc}}$$. As usual, $$u_s(f) \;=\; u_s(f-f(0)\cdot e^{-z\overline{z}})+f(0)\cdot u_s(e^{-z\overline{z}})$$ The value $$u_s(f-f(0)\cdot e^{-z\overline{z}})$$ is computable by integration against $$|z|^{2s}$$, since (by design) that difference is a Schwartz function vanishing at $$0$$. This vanishing does not imply divisibility by $$z$$, nor by $$\overline{z}$$, nor by $$x$$, nor by $$y$$, but still does imply, by Taylor-Maclaurin expansion with remainder, that $$f(z) - f(0)\cdot e^{-z\overline{z}} \;=\; O(|z|) \hskip30pt \hbox{(as |z|\to 0)}$$

In particular, the residue of $$u_s$$ at $$s=-1$$ is a multiple of $$\delta$$ (because $$f(0)=\delta(f)$$), and the multiple can be determined by integrating against $$e^{-z\overline{z}}$$: $$\int_{\mathbb C} e^{-z\overline{z}}\,(z\overline{z})^s\;dx\,dy \;=\; 2\pi \int_0^\infty e^{-r^2}\,r^{2s}\;r\,dr \;=\; 2\pi \int_0^\infty e^{-r^2}\,r^{2s+2}\;{dr\over r} \;=\; \pi \int_0^\infty e^{-r}\,r^{s+1}\;{dr\over r} \;=\; \pi\cdot \Gamma(s+1)$$ The residue at $$s=-1$$ is $$\pi$$. Then, with interchange of evaluation and differentiation justified by the Schwartz-Grothendieck vector-valued extension of Cauchy-Goursat theory, $$\overline{\partial}\,{1\over z} \;=\; \overline{\partial}\Big(z^s \overline{z}^{s+1}\Big|_{s=-1}\Big) \;=\; \Big(\overline{\partial}(z^s\overline{z}^{s+1})\Big)\Big|_{s=-1} \;=\; \Big((s+1)z^s\overline{z}^s\Big)\Big|_{s=-1}$$ $$\;=\; \mathrm{Res}_{s=-1}(z^s\overline{z}^s) \;=\; \pi\cdot \delta$$