# Sequence of radius implies that a function is harmonic

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.

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

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.

• 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. – Daniele Tampieri Oct 12 '20 at 9:18