Im learning on PDE theory lectured in Evans's book, specifically Im trying to understand the proof, that if a function $v \in C^2$ is subharmonic, then $v(x) \leq \rlap{-}\!\!\int_{B_r(x)} v \, dy$ for any sphere $B_r(x) \subset U$ holds.
Within the proof there, is one equation that follows from Green's first identity:
\begin{align}
\rlap{-}\!\!\int_{\partial B_r(x)} \frac{d v(y)}{d \nu} \, dS(y) = \frac{r}{n} \rlap{-}\!\!\int_{B_r(x)} \triangle v(y) \, dy
\end{align}
for a certain setting of parameters. If the volume of $B_r(x)$ corresponds to $\alpha(n)r^n$ and the surface of $\partial B_r(x)$ corresponds to $\alpha(n)r^{n-1}$, then obviously $\frac{r}{n}$ is derived from $ \frac{B_r(x)}{\partial B_r(x)}$; but where does this even occur in Green's first identity?
So my question is: why does this equation hold?
Cheers
Best Answer
The statement in Evans, if I'm looking at the right place, is $$\rlap{-}\!\!\int_{\partial B_{r}(x)}\frac{\partial u}{\partial v}\,dS(y) = \frac{r}{n}\rlap{-}\!\!\int_{B_{r}(x)}\Delta u(y)\,dy.$$
This follows from the first of Green's formulas, i.e. if $u,v \in C^{2}(\bar{U})$, then $$\int_{U} \Delta u\,dx = \int_{\partial U}\frac{\partial u}{\partial v}\,dS.$$
Going back to the statement in question, then, we have \begin{align} \rlap{-}\!\!\int_{\partial B_{r}(x)}\frac{\partial u}{\partial v}\,dS(y) &= \frac{1}{\left|\partial B_{r}(x)\right|}\int_{\partial B_{r}(x)}\frac{\partial u}{\partial v}\,dS(y)\\ &=\frac{1}{\left|\partial B_{r}(x)\right|}\int_{B_{r}(x)}\Delta u(y)\,dy\\ &=\frac{\left|B_{r}(x)\right|}{\left|\partial B_{r}(x)\right|}\rlap{-}\!\!\int_{B_{r}(x)}\Delta u(y)\,dy\\ &=\frac{r}{n}\rlap{-}\!\!\int_{B_{r}(x)}\Delta u(y)\,dy. \end{align}
So, at the end of the day we're basically multiplying by one for the sake of a nicer equation in terms of averages.