7
$\begingroup$

I am stuck on a representation of the Dirac Delta function that is used in several books I am reading. They state: $\begin{equation} \delta^{(2)} = \frac{1}{2\pi} \partial_{\bar{z}} \frac{1}{z} = \frac{1}{2\pi} \partial_{z} \frac{1}{\bar{z}}. \end{equation} $

They all either do not give a proof or I am not understanding the short motivations they give for this. Does anyone know how to understand this? Do you have some references where I can read on this? Help is greatly appreciated.

$\endgroup$

1 Answer 1

8
$\begingroup$

According to the definition of distributional derivative, the statement $\delta = \frac{1}{2\pi} \partial_{z} (1/ \bar{z})$ means: for every compactly supported $C^\infty$ function $\phi$, $$\phi(0)= - \frac{1}{2\pi} \int_{\mathbb C} \frac{1}{\bar z} \frac{\partial \phi}{\partial z}\tag{1}$$ The formula (1) is known as the Cauchy-Pompeiu formula, the generalization of the Cauchy integral formula to smooth (not necessarily holomorphic) functions. There is no boundary term in (1) since $\phi$ is compactly supported.

One can express (1) by saying that convolution with $-\frac{1}{2\pi}\frac{1}{\bar z}$ (the Cauchy kernel) is an operator inverting $z$-derivative. This operator is known as the Cauchy transform. You can read about it in The Cauchy Transform, Potential Theory and Conformal Mapping by Bell.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .