If $u$ is a subharmonic function, $$\mu_n = \frac{1}{2\pi} \Delta u$$ is defined as the Riesz measure of $u$. Let $u(z)=\log |z|$, prove that $$ \mu_u = \frac{1}{2\pi} \Delta \log|u| = \delta_0 = \text{Dirac Mass at 0} $$
When I tried to compute the laplacian directly, the results did not seem to be anything close to the definition of Dirac Mass at 0. I am not that familiar with the distribution theory, how do I approach this question?
Let $\phi \in \mathbb{S}$ and $u=\log(|\vec \rho|)$. Then using polar coordinates, we have
$$\begin{align} \langle \Delta u, \phi\rangle &=\langle u,\Delta \phi\rangle\\\\ &=\int_0^{2\pi}\int_0^\infty\log(\rho)\Delta \phi(\vec \rho)\,\rho\,d\rho\,d\theta\\\\ &=\int_0^{2\pi}\int_0^\infty \nabla \cdot \left(\log(\rho)\nabla u(\vec \rho)\right)-\nabla \left(\log(\rho)\right)\cdot \nabla \left(u(\vec \rho)\right)\,\rho\,d\rho\,d\theta\\\\ &=-\int_0^{2\pi}\int_0^\infty \frac{\partial u(\vec \rho)}{\partial \rho}\,d\rho\,d\theta\\\\ &=2\pi u(0) \end{align}$$
Hence, we see that in distribution $\frac1{2\pi}\Delta \log(\vec \rho)=\delta(\vec \rho)$. And we are done!