Theorem 3.1.7.A60BA3B2DBFC A continuous function on $\Omega $ is harmonic if and only if it has the sphere mean-value property. Every harmonic function is smooth. If $h$ is harmonic on $\Omega $ and $\overline{B(x,r)}\subset \Omega $, then for every multi-index $\alpha $,
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3.1.8
\begin{equation} |\partial ^\alpha h(x)|\leq C_{d,\alpha }r^{-d-|\alpha |}\int _{B(x,r)}|h(y)|\, \mathrm d y, \end{equation}
where the constant $C_{d,\alpha }$ depends only on $d$ and $\alpha $. Moreover, every harmonic function $h$ on $\Omega $ satisfies $h*\rho _\varepsilon =h$ on $\Omega _\varepsilon $ for every $\varepsilon >0$, where $\rho _\varepsilon $ and $\Omega _\varepsilon $ are as in section 2.5.
Proof.
As in section 2.5, we write $u_\varepsilon =u*\rho _\varepsilon $. Let $u$ be continuous with the sphere mean-value property, and let $\overline{B(x,\varepsilon )}\subset \Omega $. The compact set $\overline{B(x,\varepsilon )}$ has positive distance from $\mathbb R^ d\setminus \Omega $, so $x\in \Omega _\varepsilon $, and $M_ u(x,r)=u(x)$ for $0<r\leq \varepsilon $. Since $\rho $ is radial, we may write $\rho _\varepsilon (y)=\phi _\varepsilon (|y|)$; as $\rho _\varepsilon $ is even and vanishes outside $B(0,\varepsilon )$, (3.1.2) gives
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3.1.9
\begin{equation} \begin{split} u_\varepsilon (x)& =\int _{B(0,\varepsilon )}u(x+y)\rho _\varepsilon (y)\, \mathrm dy=s_{d-1}\int _0^\varepsilon r^{d-1}\phi _\varepsilon (r)M_ u(x,r)\, \mathrm dr\\ & =u(x)\, s_{d-1}\int _0^\varepsilon r^{d-1}\phi _\varepsilon (r)\, \mathrm dr=u(x)\int _{\mathbb R^ d}\rho _\varepsilon \, \mathrm dy=u(x). \end{split} \end{equation}
Every point of $\Omega _\varepsilon $ is the center of such a closed ball, so $u=u_\varepsilon $ on the open set $\Omega _\varepsilon $, and $u_\varepsilon $ is smooth there (section 2.5). As $\Omega $ is the union of the sets $\Omega _\varepsilon $, the function $u$ is smooth.
Next we show that $\Delta u=0$. Fix $x\in \Omega $. Taylor's formula gives
\[ u(x+r\omega )=u(x)+r\sum _ j\partial _ ju(x)\, \omega _ j+\frac{r^2}{2}\sum _{j,k}\partial _ j\partial _ ku(x)\, \omega _ j\omega _ k+o(r^2), \]
uniformly in $\omega \in S^{d-1}$. We average over the sphere. The reflections $\omega \mapsto -\omega $ and $\omega _ j\mapsto -\omega _ j$ preserve $S$, so $\int \omega _ j\, \mathrm dS=0$ and $\int \omega _ j\omega _ k\, \mathrm dS=0$ for $j\neq k$. Permutations of the coordinates also preserve $S$, so the integrals $\int \omega _ j^2\, \mathrm dS$ are all equal, and their sum is $s_{d-1}$ because $\sum _ j\omega _ j^2=1$. Hence
\[ M_ u(x,r)=u(x)+\frac{r^2}{2d}\Delta u(x)+o(r^2). \]
Since $M_ u(x,r)=u(x)$ for all small $r$, it follows that $\Delta u(x)=0$. Thus $u$ is harmonic. Conversely, harmonic functions have the sphere mean-value property by proposition 3.1.5; in particular they are smooth, and they satisfy $h=h_\varepsilon =h*\rho _\varepsilon $ on $\Omega _\varepsilon $ by the first part of the proof.
Now let $h$ be harmonic and $\overline{B(x,r)}\subset \Omega $. Then $x$ lies in the open set $\Omega _ r$, and $h=h_ r$ on $\Omega _ r$ by the first part of the proof. By the differentiation formula $\partial ^\alpha h_ r=h*\partial ^\alpha \rho _ r$ of section 2.5, and since $\partial ^\alpha \rho _ r$ vanishes outside $B(0,r)$, we get
\[ \partial ^\alpha h(x)=\int _{B(x,r)} h(y)\, (\partial ^\alpha \rho _ r)(x-y)\, \mathrm d y. \]
Since $\partial ^\alpha \rho _ r(z)=r^{-d-|\alpha |}(\partial ^\alpha \rho )(z/r)$, this proves (3.1.8) with $C_{d,\alpha }=\sup |\partial ^\alpha \rho |$, where $\rho $ is the mollifier on $\mathbb R^ d$ fixed in section 2.5.
$\square$
Pome
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