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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.
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3 Answers 3

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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 $\tfrac{1}{2}$.

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

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    $\begingroup$ 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. $\endgroup$ Commented Apr 9, 2022 at 13:36
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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, $$ \overline{\partial}\,{1\over z} \;=\; \overline{\partial}\Big(z^s \overline{z}^{s+1}\Big|_{s=-1}\Big) \;=\; \Big(\overline{\partial}(z^s\overline{z}^{s+1})\Big)\Big|_{s=-1} \;=\; \Big((s+1)z^s\overline{z}^s\Big)\Big|_{s=-1} $$ $$ \;=\; \mathrm{Res}_{s=-1}(z^s\overline{z}^s) \;=\; \pi\cdot \delta $$

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