If we take as definition that if $\Omega \subseteq \Bbb R^n$ is open and $u \in \mathscr{C}^2(\Omega)$ is subharmonic if $\triangle u \geq 0$, then for every $x \in \Omega$ and small enough $r>0$ we have the inequality $$u(x) \leq \frac{1}{|B_r(x)|}\int_{B_r(x)} u(y)\,{\rm d}y,$$where $|B_r(x)|$ denotes the volume of the ball. There's a proof for the harmonic case in Chapter 1 of Evans which is easy to follow and adapt, and he would first prove that $$u(x) \leq \frac{1}{|\partial B_r(x)|}\int_{\partial B_r(x)} u(y)\,{\rm d}S(y),$$to then integrate using polar coordinates and get the result. This is all fine.
I tried to give a direct proof avoiding the surface integral and got something absurd.
Fix $x \in \Omega$ and define $\varphi\colon [0,\epsilon) \to \Bbb R$ (for some maximal $\epsilon>0$ which exists due to openness of $\Omega$) by $$\varphi(r) = \frac{1}{|B_r(x)|}\int_{B_r(x)} u(y)\,{\rm d}y.$$Lebesgue's differentiation theorem says that $$ \varphi(0) = \lim_{r\to 0^+} \varphi(r) = u(x),$$ so our goal is to show that $\varphi'(r)\geq 0$ (as the desired conclusion is just $\varphi(0) \leq \varphi(r)$ for $r>0$). And this is done as follows: first we rewrite $\varphi(r)$ in a more convenient fashion, using that $|B_r(x)| = r^n|B_1(0)|$ and setting $y = x+rz$ with $z \in B_1(0)$, so that ${\rm d}{y} = r^n \,{\rm d}{z}$ and $$ \varphi(r) = \frac{1}{|B_1(0)|} \int_{B_1(0)} u(x+rz)\,{\rm d}{z}. $$So differentiating under the integral (allowed by smoothness of $u$ and compactness of $B_1(0)$) gives $$ \varphi'(r) = \frac{1}{|B_1(0)|} \int_{B_1(0)} \langle \nabla u(x+rz),z\rangle\,{\rm d}{z}. $$To apply Green's identities, let $v(z) = u(x+rz)$ and $q(z) = \langle z,z\rangle/2$, so that $\nabla v(z) = r\nabla u(x+rz)$ and $\nabla q(z) = z$. Also $\triangle v(z) = r^2\triangle u(x+rz)$. So
\begin{align*}
\varphi'(r) &= \frac{1}{r|B_1(0)|}\int_{B_1(0)} \langle \nabla v(z),\nabla q(z)\rangle\,{\rm d}{z} \\ &= \frac{1}{r|B_1(0)|}\left( -\int_{B_1(0)} q(z)\triangle v(z)\,{\rm d}{z} + \int_{\partial B_1(0)} q(z)\frac{\partial v}{\partial \nu}(z)\,{\rm d}{S}(z)\right) \\ &= -\frac{r}{2|B_1(0)|} \int_{B_1(0)} \|z\|^2 \triangle u(x+rz)\,{\rm d}{z} \color{red}{\leq 0}
\end{align*}
I cannot find the mistake. What's the screw up?
Best Answer
Why is that boundary integral in the last formula zero?
What about using $q(z) =\frac12( \|z\|^2 -1 )$? Then $\nabla q = z$ and $-q(z) \ge0$ for $\|z\|\le1$, moreover $q(z) =0$ for all $\|z\|=1$. And that last inequality becomes \begin{align*} \varphi'(r) &= \frac{1}{r|B_1(0)|}\int_{B_1(0)} \langle \nabla v(z),\nabla q(z)\rangle\,{\rm d}{z} \\ &= \frac{1}{r|B_1(0)|}\left( -\int_{B_1(0)} q(z)\triangle v(z)\,{\rm d}{z} + \int_{\partial B_1(0)} q(z)\frac{\partial v}{\partial \nu}(z)\,{\rm d}{S}(z)\right) \\ &= \frac{1}{r|B_1(0)|}\left( \int_{B_1(0)} \frac{1-\|z\|^2}2\triangle v(z)\,{\rm d}{z} + 0\right) \ge0 \end{align*}