Mean-value theorem for subharmonic functions:

Question

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.

Answer 1

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.

Answer 2

Do you know the proof for the corresponding property for harmonic functions (the "mean value property")? If so, try to look at all the places where you use the harmonicity and use the inequality instead of the equality there. You should see that it works perfectly.

Or, to be a bit more specific, in that problem, one applies Green's identity to find that (using the subharmonicity first) $$ 0 \geq\int_{B(x_{0},r)} \Delta u (y) dy = \int_{\partial B(x_{0},r)} \langle \nabla u(y), \nu(y) \rangle d S(y). $$ Then you make a substitution to get an integral over $B(0,1)$ (because you would like to get rid of the $r$ in the boundaries of the integral) and note that by the chain rule you can write the integral as $r^{d-1}$ times a total derivative w.r.t. $r$ and that this derivative equals zero. Then you can integrate the resulting inequality w.r.t. $r$ which yields the desired result.

Hope that helps!

Andre