In this problem, we will consider subharmonic functions. If $U$ is a bounded, open set, we say that $v\in C^2(U)\cap C(\overline{U})$ is subharmonic if
$$-\Delta v \leq 0 \ \ \ \text{in} \ U$$
a.) Prove that if $v$ is subharmonic and $B(x,r)\subseteq U$, then
$$v(x) \leq ⨍_{\partial B(x,r)}v(y)dS_y$$
Attempted proof – Suppose $v$ is subharmonic and $B(x,r)\subseteq U$ where $U$ is a bounded, open set. Then by definition $v\in C^2(U)\cap C(\overline{U})$ and $-\Delta v \leq 0$ in $U$ or equivalently $\Delta v \geq 0$ in $U$. Then we have
$$v(x) = ⨍_{B(x,r)}v(y)S_y$$
I am having some problem with this proof, I don't see where to go. Any suggestions are greatly appreciated.
Note:
I will now try another attempt at proving this. (for consistency with comments: $\oint$ was used as a substitute for $⨍$.)
Attempted proof 1 – Let
$$f(r) = ⨍_{\partial B(x,r)}v(y)d S_y = ⨍_{|y-x| = r}v(y)d S_y = \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}v(y) dS_y = \frac{1}{n\alpha(n)}\int_{|z| = 1}u(x+rz) d S_z$$
where $y = x + rz$ and $\alpha(n) = \int_{|x|\leq 1}dx = $ volume of unit ball in $n$ dimensions. The outer unit normal at $y\in\partial B(x,r)$ is $\nu(y) = \frac{y-z}{r}$, thus
\begin{align*}
f'(r) = \frac{1}{n\alpha(n)}\int_{|z|=1}z \nabla v(x+rz) dS_z &= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y) \frac{y-x}{r}d S_y\\
&= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y)\nu(y)d S_y
\end{align*}
Since $-\Delta v\leq 0$ in $U$, we have
$$0\leq \int_{B(x,r)}\Delta v(y) dy = \int_{\partial B(x,r)}v_{\nu}d S_y$$
thus $f'(r)\geq 0$ for $r > 0$. Then we get for $r > 0$
$$v(x) = f(0) \leq f(r) = ⨍_{\partial B(x,r)}v(y)d S_y$$
Therefore we have the result.
I am sure this proof is quite messy and contains some errors. Please give me some comments or suggestions on the latter.
Best Answer
Your proof is correct, with just minor errors. Here is revised version of it.
Let $$ f(r) = ⨍_{\partial B(x,r)}v(y)d S_y = ⨍_{|y-x| = r}v(y)d S_y = \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}v(y) dS_y =\\= \frac{1}{n\alpha(n)}\int_{|z| = 1}v(x+rz) d S_z$$ where $y = x + rz$ and $\alpha(n) = \int_{|x|\leq 1}dx = $ volume of unit ball in $n$ dimensions. The outer unit normal at $y\in\partial B(x,r)$ is $\nu(y) = \frac{y-x}{r}$, thus \begin{align*} f'(r) = \frac{1}{n\alpha(n)}\int_{|z|=1}z. \nabla v(x+rz) dS_z &= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y). \frac{y-x}{r}d S_y\\ &= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y).\nu(y)d S_y \end{align*} By Green-Stokes Theorem we know that $$ \int_{|y-x| = r}\nabla v(y).\nu(y)d S_y =\int_{\partial B(x,r)}\nabla v(y).\nu(y)d S_y = \int_{B(x,r)}\Delta v(y) dy $$ Since $-\Delta v\leq 0$ in $U$, we have $$0\leq \int_{B(x,r)}\Delta v(y) dy = \int_{|y-x| = r}\nabla v(y).\nu(y)d S_y $$ thus $f'(r)\geq 0$ for $r > 0$. Then we get for $r > 0$ $$v(x) = f(0) \leq f(r) = ⨍_{\partial B(x,r)}v(y)d S_y$$ Therefore we have the result.