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A distribution $u$ on a plane open set with ${\partial }_{\bar{z}}u=0$ is represented by a holomorphic function. Indeed $\Delta u=4{\partial }_z{\partial }_{\bar{z}}u=0$, and elliptic regularity makes $u$ smooth. For a direct proof, mollify locally with a radial mollifier ${\eta }_{\epsilon }$. The resulting smooth functions are harmonic. Their mean value property gives $(u\ast {\eta }_{\epsilon })\ast {\eta }_r=u\ast {\eta }_{\epsilon }$ on a smaller ball for fixed small $r$. Passing to the weak convergence of distributions limit gives $u\ast {\eta }_r=u$ there. This proves smoothness without assuming local integrability beforehand. The ordinary Cauchy-Riemann equations now prove holomorphy.

The equation ${\partial }_{z}u=0$ similarly characterizes antiholomorphic functions. On the whole plane, the homogeneous solutions are respectively arbitrary entire functions $h(z)$ and the functions $h(\bar{z})$ with $h$ entire. Consequently any two fundamental solutions for one of these Wirtinger derivatives differ by the corresponding homogeneous solution.

Interchange the two Wirtinger derivatives in part (i). The same real fundamental solution of the Laplace equation $L$ givesThe calculation is valid for distributions by the same vanishing boundary term estimate. Any other fundamental solution differs by a distribution $H$ with ${\partial }_{z}H=0$. Then $\Delta H=0$, so the regularity argument of part (ii) makes $H$ smooth, and its Cauchy-Riemann equations make it an antiholomorphic function. On the whole plane it can be written $H(z)=h(\bar{z})$ with $h$ an entire function. Conversely every such $H$ has ${\partial }_{z}H=0$. Therefore