Let $B_r$ a ball of radius $r$ in $\mathbb{R}^n$ and $u \in H^{1}(B_r)$ with $\Delta u =0$ in the weak sense. I am reading a paper and the author says that : "since $|\nabla u |^2$ is subharmonic, then
$$ \frac{1}{\alpha_n s^n} \int_{B_s} |\nabla u|^2 dx \leq \frac{1}{\alpha_n r^n} \int_{B_r} |\nabla u|^2 dx \ " \tag{*}$$
I am not seeing how to prove this. Using the proof of Evans's PDE book of the classical mean value formula for harmonic funtions I can get that
$$ \frac{1}{\alpha_n s^{n-1}} \int_{ \partial B_s} |\nabla u|^2 \leq \frac{1}{ \alpha_n r^{n-1}} \int_{ \partial B_r} |\nabla u|^2 $$
Here $\alpha_n$ denotes the measure of the unit ball in $\mathbb{R}^n$. the part of subhamonicity I understood.
Please someone could help me to obtain formula (*)?
If spherical averages go up, so do the "solid" averages.
Let's start with an intuitive fact: if $f:[0,b]\to \mathbb{R}$ is a nondecreasing function, then the running averages $F(x)=\frac{1}{x}\int_0^x f(t)\,dt$ are also nondecreasing. Indeed, a change of variable yields $$F(x)=\int_0^1 f(xt)\,dt$$ where the integrand increases with respect to $x$. (Proof suggested by David C. Ullrich).
More generally, the weighted averages $F_p(x)=\frac{p}{x^p}\int_0^x f(t)t^{p-1}\,dt$ are nondecreasing; this is reduced to the above by substitution $u=t^p$.
As a special case, let $p=n$ and let $f(t)$ be the average of subharmonic function on the sphere of radius $t$. Then $F_n(r)$ is its average on the ball of radius $r$.