# Mean-value theorem for subharmonic functions:

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.

• I have replaced the instances of $\oint$ with $⨍$s, which was accomplished with the unicode character U+2A0D FINITE PART INTEGRAL. I got the idea from tex.stackexchange.com/questions/759/average-integral-symbol ; if it doesn't work for you, feel free to revert the edit. Commented Jul 17, 2017 at 22:32
• The solution does not feel perfect yet but thank you, that's a big improvement! :) Commented Jul 18, 2017 at 8:43

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.

• Do you know how to prove this:math.stackexchange.com/questions/2362075/…? Commented Jul 19, 2017 at 1:21
• @Wolfgang-1 I have posted an answer there, with a detailed proof. Please, let me know if you have any questions. Commented Jul 19, 2017 at 3:41

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

• I tried to do a proof if you can see my edit. Commented Jul 16, 2017 at 22:05
• This is pretty much the proof that I suggested and I can't find any obvious mistakes. :) Good job, I'd say :) Commented Jul 16, 2017 at 22:13
• If I had to know prove $$v(x)\leq \oint_{B(x,r)}v(y)dy$$ I feel like I could use what I have already have done, any ideas? Commented Jul 17, 2017 at 22:12
• I think the simplest idea is to integrate your "spherical" mean value property from a) from $0$ to $r$ on both sides. The left-hand side yields $v(x) r^{n-1}$ and the right-hand side yields $\frac{1}{\alpha(n)} ⨍_{B(x,r)} v(y) dy$. Division by $r^{n-1}$ then yields the result. Commented Jul 18, 2017 at 8:24