Problem 2
Let: $\Delta u \ge 0$ in $B\left(y,r\right)$, $\omega_n$ be the volume of the $n$-dimensional unit ball, and $\sigma_n = n \omega_n$. Prove that:
Part a. $u\left(y\right)$ does not exceed the mean value of $u$ over the sphere $S\left(y,r\right)$ bounding the ball $B\left(y,r\right)$:
\begin{equation}
u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \label{2.a.0}
\end{equation}
Answer:
In $\Delta u \ge 0$, $u$ is twice differentiable, so the spherical mean is continuous and we have:
\begin{equation}
\lim_{r \rightarrow 0^+} \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \rightarrow u\left(y\right) \label{2.a.1}
\end{equation}
Thus, to prove $ u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S $, it is sufficient to show that $\frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$ increases with $r > 0$. By the divergence theorem we have for a $C^1$ function $F$ on a compact set $\Omega$ with boundary $\partial\Omega = \Sigma$:
$$ \int_\Omega \left(\nabla \cdot F\right) \,dV = \int_\Sigma \left(F \cdot n\right) \, dS \text{, namely: } $$
$$\int_{B\left(y,r\right)} \left(\nabla \cdot \nabla \right)\left(x\right) \,dV = \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = \int_{S\left(y,r\right)} \left(\nabla u \cdot n\right)\left(x\right) \, dS = \int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS $$
As $\Delta u \ge 0$, we then have:
$$ 0 \le \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = \int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS $$
Parametrizing by the normal $\nu = \frac{x-y}{|x-y|}$ on the boundary where $|x-y| = r$, we have: $\frac{\partial u}{\partial n} \left(x\right) = \partial_r u\left(y+ r \nu\right)$ on $ T : \{ \nu : | \nu | = 1 \} $, and $ dS = r^{n-1} d \nu $
What? See note below V.I..
Then:
$$ \int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS = \int_{T} \partial_r\left( u\left(y+ r \nu\right)\right) r^{n-1} \, d \nu $$
By differentiating under the sign with $\partial_r$ and pulling the $r^{n-1}$ term out of the integral, we have:
$$ = r^{n-1} \partial_r \int_{T} u\left(y+ r \nu\right) \, d \nu = r^{n-1} \partial_r \left(r^{1-n} \int_{S\left(y,r\right)} u\left(x\right) \, d S\right) $$
$$ = r^{n-1} \partial_r\left(\sigma_n \frac{r^{1-n}}{\sigma_n } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right) = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right) $$
As $r>0$ and $\sigma_n>0$ by \eqref{2.a.1} we have:
$$ 0 \le \int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right) \implies $$
\begin{equation}
0 \le \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right) \label{2.a.2}
\end{equation}
Then by \eqref{2.a.1} as $r \rightarrow 0$ we have equality in \eqref{2.a.0}, and as $r$ increases the spherical mean is non-decreasing by \eqref{2.a.2}, $u\left(y\right)$ constant. Thus we have \eqref{2.a.0} for $r \ge 0$:
$$ u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \phantom{O} \blacksquare $$
Part b. $u\left(y\right)$ does not exceed the mean value of $u$ over the ball $B\left(y,r\right)$:
\begin{equation}
u\left(y\right) \le \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V \label{2.b.0}
\end{equation}
Answer:
$$u\left(y\right) = \frac{\sigma_n \left(n r^n\right)}{\left(n \sigma_n\right) r^n} u\left(y\right) = \frac{\sigma_n \int_{0}^{r} t^{n-1} \, dt}{\omega_n r^n} u\left(y\right) $$
By \eqref{2.a.0} we have that:
$$ u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S $$
So because $ \{r, \omega_n, \sigma_n\} \ge 0 $ we have that:
$$\implies \frac{\sigma_n \int_{0}^{r} t^{n-1} \, dt}{\omega_n r^n} u\left(y\right) \le \frac{\sigma_n \int_{0}^{r} t^{n-1}\, dt}{\omega_n r^n} \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S $$
$$ = \frac{1}{\omega_n r^n} \sigma_n\int_{0}^{r}t^{n-1} \,d t \frac{1}{\sigma_n r^{n-1}} \int_T u\left(y +r \nu\right) r^{n-1} \, d \nu $$
Where we parameterized as in part a. with $\nu = \frac{x-y}{|x-y|}$ on the boundary $|x-y| = r$, and $ \left(x\right) = u\left(y+ r \nu\right)$ on $ T : \{ \nu : | \nu | = 1 \} $, $ dS = r^{n-1} d \nu $. Canceling terms gives:
$$ = \frac{1}{\omega_n r^n} \int_{0}^{r}t^{n-1} \,d t \int_T u\left(y +r \nu\right) \, d \nu $$
Which allows us to change the order of integration to yield our result:
$$ = \frac{1}{\omega_n r^n} \int_{0}^{r} t^{n-1} \int_T u\left(y +r \nu\right) \, d \nu d t = \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V $$
$$ \implies u\left(y\right) \le \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V \phantom{O} \blacksquare$$
Part c. Formulate similar statements for functions satisfying $ \Delta u \le 0$.
Answer:
In part a. we used the fact that $ \Delta u \ge 0 $, and $ \{ \sigma_n, r \} \ge 0$ to show that the spherical mean is non-decreasing in $r$:
$$ 0 \le \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
$$ \implies 0 \le \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right) $$
Because we have equality at $r \rightarrow 0$:
$$ \lim_{r \rightarrow 0^+} \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \rightarrow u\left(y\right) $$
and $u\left(y\right)$ is constant with respect to $r$, we showed that for all $r > 0$
$$ u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S $$
If instead $ \Delta u \le 0$, we would have the inequalities:
$$ 0 \ge \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
$$ \implies 0 \ge \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right) $$
And $ \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S$ is non-increasing with $r$, which gives us by the same argument that for all $r>0$
$$ u\left(y\right) \ge \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \phantom{O} \square$$
In part b. we merely integrated this result radially to extend the inequality for the interior of the sphere. Doing the same in the case of $ \Delta u \le 0$ yields us:
$$ u\left(y\right) \ge \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V \phantom{O} \blacksquare$$
