1
$\begingroup$

Let $u \in C(\Omega),$ where $\Omega \subset \mathbb{R}^{n}$ is open. If for all $x \in \Omega$ there exists a sequence of radius (positive) $\left\{r_{k}(x)\right\}_{k \in \mathbb{N}}$ such that $\lim _{k \rightarrow \infty} r_{k}(x)=0$ and $$ u(x)=\dfrac{1}{n\omega_{n}r_k^{n-1}}\int_{\partial B(x,r_k(x))} u\quad\mathrm{dS} \quad \forall k \in \mathbb{N} $$ Then $u$ is harmonic.

My attempt: Let $y \in \Omega$ and $R > 0$ such that $\overline{B(y,R)} \subset \Omega.$ Let $v$ a harmonic function such that \begin{array}{r} \Delta v=0 \text{ when } x \in B(y,R) \\ v=u \text{ when } x \in \partial B(y,R) \end{array} Suppose by contradiction that $v \neq u$ in $\overline{B(y,R)}$ then there exists $x \in \overline{B(y,R)}$ such that $$(v-u)(x)>0$$ Recall that $v=u$ in $\partial B(y,R)$ then $x \in B(y,R).$ So I am stuck. Any help? I was thinking about to use something like maximum principle but I am not sure, indeed I have not used the hypothesis about the radius or mean value property.

$\endgroup$
2
  • $\begingroup$ Does there exist $v$ such that $\Delta v = 0$ for $x \in \Omega$ and $v = u$ for $x \in \partial \Omega$? $\endgroup$ – mathworker21 Oct 10 '20 at 14:51
  • 1
    $\begingroup$ One can use exactly the proof as described in this answer, with some changes from $r\to 0^+$ to $r_k \to 0$. $\endgroup$ – Arctic Char Oct 10 '20 at 22:01
2
$\begingroup$

We can use the following corollary Corollary 10 (Comparison principle). Let $\Omega$ be a bounded open set, and let $u$ and $v$ be elements of $C^{2}(\Omega) \cap C(\bar{\Omega}) .$ Assume that $\Delta u \geq \Delta v$ in $\Omega$ and that $u \leq v$ on $\partial \Omega$. Then $u \leq v$ in $\Omega$.

But we need $u$ be an element of $C^{2}(\Omega).$

For that we use Theorem: Let $\Omega \subset \mathbb{R}^N$, $u \in C(\Omega)$ be such that $$\frac{1}{|B(x_0,R)|}\int_{B(x_0,R)}u(y)\ dy = u(x_0) = \frac{1}{|\partial B(x_0,R)|}\int_{\partial B(x_0,R)}u\ dS$$ for every ball $\overline{B(x_0,R)} \subset \Omega$. Then $u \in C^{\infty}(\Omega)$ and it is harmonic

Proof: Consider the standard mollifier: $$\rho(x) := \begin{cases}Ce^{-\frac{1}{1 - \|x\|^2}} & \text{if $\|x\|$ < 1} \\0 & \text{otherwise.} \end{cases}$$ Here $C$ is a constant such that $\|\rho\|_{L^1} = 1.$ Let $\epsilon > 0$ and consider $$\rho_{\epsilon}(x) = \epsilon^{-N}\rho(x\epsilon^{-N}).$$ Set $\Omega_{\epsilon} = \{x \in \Omega : \text{dist}(x,\partial \Omega) > \epsilon\}$ and define for $x \in \Omega_{\epsilon}$ $$u_{\epsilon}(x) = \rho_{\epsilon} * u(x) = \int_{\Omega}\rho_{\epsilon}(x - y)u(y)\ dy.$$ The following is a well know theorem in analysis, if it is new to you you can look for a proof Analysis by Lieb and Loss or anywhere else.

**Theorem:**If $u \in C(\Omega)$, then $u_{\epsilon} \to u$ uniformly on compact subsets of $\Omega$, $u_{\epsilon} \in C^{\infty}(\Omega_{\epsilon})$ and for any multindex $\alpha$ we have $$\frac{\partial^{\alpha}u_{\epsilon}}{\partial x^{\alpha}}(x) = \int_{\Omega}\frac{\partial^{\alpha}\rho_{\epsilon}}{\partial x^{\alpha}}(x - y)u(y)\ dy.$$

Finally we can proceed with the proof!

Fix $x_0 \in \Omega_{\epsilon}$. $$u_{\epsilon}(x_0) = \int_{B(x_0,\epsilon)}\rho_{\epsilon}(x - y)u(y)\ dy = \int_{B(0,\epsilon)}\rho_{\epsilon}(z)u(x_0 - z)\ dz = $$ $$ = \int_0^{\epsilon}r^{N - 1}\int_{\partial B(0,1)}\rho_{\epsilon}(rw)u(x_0 - rw)\ dS(w)dr = $$ $$ \int_0^{\epsilon}r^{N - 1}\rho(r)\int_{\partial B(0,1)}u(x_0 - rw)\ dS(w)dr = \int_0^{\epsilon}r^{N-1}\rho_{\epsilon}(r)\frac{\alpha_N N}{|\partial B(x_0,r)|}\int_{\partial B(x_0,r)}u(y)\ dS(y)dr $$ $$ = u(x_0)\|\rho\|_{L^1} = u(x_0).$$

This proves that $u = u_{\epsilon}$ and hence $u \in C^{\infty}(\Omega_{\epsilon})$, for every $\epsilon$.

Therefore $u$ is harmonic.

$\endgroup$
1
  • $\begingroup$ Julioprofe, note that the definition of harmonicity you are investigating has been studied long ago by Ivan Privalov (“Sur les fonctions harmoniques” (in French), Matematicheskii Sbornik (Recueil Mathématique), 32:3 (1925), 464–471, JFM 51.0363.02. The operator defined by taking the limiting value of the integral term in the mean value is called, as a matter of fact, Privalov's Laplacian and is de facto one of the first ever proposed generalized formulations of a partial differential operator. $\endgroup$ – Daniele Tampieri Oct 12 '20 at 9:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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