4
$\begingroup$

$u$ satisfies "mean value property locally " on $\Omega$ if for every $x\in \Omega \exists \delta=\delta (x)>0 $ such that

$u(x) \le \frac {1}{|\mu(B(x,r)|}\int_{\partial B(x,r)} u(y) dS_y$

for all $r\le \delta(x)$

Does this imply that if $u\in C^2(\Omega)$ and satisfies mean value property locally in $\Omega$ then $u$ is subharmonic ?? Any hints will be nice.

$\endgroup$
  • $\begingroup$ Yes, I am sure this is covered in standard texts on potential theory, such as Helms ("Potential Theory"). In most of potential theory, we don't require $C^2$, just upper or lower semi-continuity. $\endgroup$ – Old John Jul 17 '12 at 12:00
  • $\begingroup$ @Old John : What is the basic argument to extend local property to the global property ? $\endgroup$ – Theorem Jul 17 '12 at 12:03
  • $\begingroup$ Hint: use the 2nd order Taylor expansion of u at x. The first derivatives cancel out in integration, the second derivatives contribute the Laplacian, and nothing else matters. $\endgroup$ – user31373 Jul 17 '12 at 12:06
  • $\begingroup$ From memory, I think it is a connectedness argument, but I think you might have to adjust the statement of the problem above, as I think it would fail if $\Omega$ were not connected. $\endgroup$ – Old John Jul 17 '12 at 12:06
  • $\begingroup$ @LeonidKovalev : I will try and will try to get back with an answer . $\endgroup$ – Theorem Jul 17 '12 at 12:09
3
$\begingroup$

We first claim that if a $C^0(\bar{\Omega})$ function $u$ satisfies the following local mean value inequality

$$u(y)\leq\frac{1}{n\omega_nR^{n-1}}\int_{\partial B_R(y)}u ds \text{ for all } R\leq \delta$$

then maximal principle applies, i.e. $u$ attains its maximum at $\partial \Omega$.

Assume our claim now. Let $\tilde{u}$ be the harmonic function defined by the boundary value $u|_{\partial B}$ on the boundary of an arbitrary ball $B\subset\subset \Omega$. Then $\tilde{u}$ satisfies mean value equality in $B$ since it is harmonic in $B$. Then $u-\tilde{u}$ satisfies the above mean value inequality. Then by our claim (maximal principle), we have $\sup_{x\in B}(u-\tilde{u})=\sup_{x\in\partial B}(u-\tilde{u})$. But $u-\tilde{u}=0$ on $\partial B$. Hence we have $u-\tilde{u}\leq 0$. $u\leq \tilde{u}$ in $B$.

Now we show that $u$ is subharmonic, given an harmonic function $h$ in $B\subset \subset \Omega$, such that $u\leq h$ on $\partial B$. We need to show that $u\leq h$ in B. Since we know $u\leq\tilde{u}$ in $B$. It suffices to show that $\tilde{u}\leq h$ in $B$. To see this one uses the Possion kernel $$\tilde{u}(x)=\int_{\partial B}K(x,y)u(y)ds_y$$ $$h(x)=\int_{\partial B}K(x,y)h(y)ds_y$$
Since $u(y)\leq h(y)$ on $\partial B$ and $K(x,y)\geq 0$. We have $\tilde{u}(x)\leq h(x)$.

Now we show our claim that the local mean value inequality $u(y)\leq\frac{1}{n\omega_nR^{n-1}}\int_{\partial B_R(y)}u ds $

implies maximal principle.

We use the standard topological argument. Define a set $S:=\{q\in\bar{\Omega}| u(q)=\sup_{x\in\bar{\Omega}}u(x)\}$. If $u$ attains maximum at a point $p\in\Omega$ then $S$ is nonempty since $p\in S$. Note that $S$ is closed since every sequence in $S$ converges to a point in $S$ since $u$ is continuous. Next if $p\in S\cap\Omega$, then $u(p)\leq\frac{1}{n\omega_nR^{n-1}}\int_{\partial B_R(p)}u ds $ and $u(p)\geq u(x)$ implies $u(x)=u(p)$ for all $x\in B_R(p)$ since $u$ is continuous. Hence $B_R(p)\in S$ Hence $S$ is open. Now $S$ is open and closed and nonempty, hence we have $S=\bar{\Omega}$. This proves our claim and finished our proof.

$\endgroup$
  • $\begingroup$ Thanks for your editing, Byron. $\endgroup$ – Dianbin Bao Feb 2 '17 at 19:43
0
$\begingroup$

No, a function satisfying a mean-value inequality is not necessarily $C^2$ (think of the norm function $\mathbb R^d\supseteq\Omega\ni\vec x\mapsto|\vec x|\in\mathbb R$) but if it satisfies a mean-value equality it is harmonic and therefore $C^\infty$ (in fact, analytic).

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.