A professor I talked to showed me a proof of the mean value property. (He actually showed it for functions solving the heat equation instead of Laplace's equation, but it seems like the argument is the same.) The proof involves distributions, which I am not very familiar with, so there is a step that I am confused about.
Here is how the proof goes:
Let $\Phi(\vec x)$ be the fundamental solution of Laplace's equation.
For simplicity, suppose we are in two dimensions. Then $\Phi(\vec x) = - \frac{1}{2\pi} \log |\vec x|$, so $-\Delta \Phi(\vec x) = \delta_0(\vec x)$, where $\delta_0$ is the delta distribution.
Fix $k \in \mathbb R$ and let $B = \{ \vec x \in \mathbb R^n : \Phi(\vec x) > k \} = \{ \vec x \in \mathbb R^n : |\vec x| < e^{-2\pi k} \}$. Let $\chi_B(\vec x)$ be the characteristic function of $B$, and consider the function $\Theta(\vec x) = \chi_B(\vec x)(\Phi(\vec x) - k)$. Note in particular that $\Theta$ is continuous at the boundary $\partial B$.
If we take the Laplacian of $\Theta$, we will have a $\delta_0$ due to $\Phi$, but also, since the first derivative of $\Theta$ has a jump discontinuity at $\partial B$, there will be another distribution that only involves $\partial B$. Thus, $-\Delta\Theta(\vec x) = \delta_0(\vec x) + f(\vec x)$, where $f$ is a distribution that only involves $\partial B$. (Here is where I start to get confused.)
If $u$ is a harmonic function, then $(\Delta\Theta, u) = (\Theta, \Delta u) = (\Theta, 0) = 0$. But $-\Delta\Theta = \delta_0 + f$, so $(\delta_0, u) = - (f, u)$. Since $(\delta_0, u) = u(0)$, we need to show that $(f, u) = \frac{1}{2\pi e^{-2\pi k}} \int_{\partial_B} u $ to complete the proof.
My question is, how do we show that $f$ satisfies $(f, u) = \frac{1}{2\pi e^{-2\pi k}} \int_{\partial_B} u$? I know this comes from differentiating $\Theta$ at the boundary, but I am not that familiar with distributions. (I can see why in one dimension, the distributional derivative of the Heaviside function is the delta function, but I am not sure how to proceed in higher dimensions.)