Harmonic Functions and Mean Values in Rn

Let $n\ge 1$ be an integer and let $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$ be open. Write $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ for smooth real functions with compact support in $\mathrm{\Omega }$. A distribution is a continuous linear functional $T:C_c^{\mathrm{\infty }}(\mathrm{\Omega })\to \mathbb{R}$. Here continuity means that $T({\varphi }_j)\to T(\varphi )$ whenever the supports of ${\varphi }_j$ and $\varphi$ lie in one compact subset of $\mathrm{\Omega }$ and every partial derivative of ${\varphi }_j$ converges uniformly to the corresponding derivative of $\varphi$. Set $({\mathrm{\partial }}_{i}T)(\varphi ):=-T({\mathrm{\partial }}_i\varphi ),\mathrm{\Delta }T:={\sum }_{i=1}^n{\mathrm{\partial }}_i^{2}T\mathrm{.}$ For $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathrm{\Omega })$, $T_f(\varphi ):={\int }_{\mathrm{\Omega }}f\varphi$ is its regular distribution. We say $T$ is distributionally harmonic when $\mathrm{\Delta }T=0$. Given another distribution $F:C_c^{\mathrm{\infty }}(\mathrm{\Omega })\to \mathbb{R}$, we say that $T$ solves the distributional Poisson equation $-\mathrm{\Delta }T=F$ when that equality holds as distributions. These are distinct from the classical conditions on a $C^2$ function in The Laplacian of a $C^2$ function and of a $C^2$ vector field.

Let $n\ge 1$, let $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$ be open, and let $u:\mathrm{\Omega }\to \mathbb{R}$ be locally Lebesgue integrable. Assume also that $\theta \mapsto u(x+r\theta )$ is $\sigma$-integrable on $S^{n-1}$ whenever $\overline{B_r(x)}\subseteq \mathrm{\Omega }$. Put ${\omega }_{n-1}:=\sigma (S^{n-1})$. For $x\in \mathrm{\Omega }$ and $r>0$ with $\overline{B_r(x)}\subseteq \mathrm{\Omega }$, define $M_u(x,r):=\frac{1}{{\omega }_{n-1}}{\int }_{S^{n-1}}u(x+r\theta )d\sigma (\theta ),A_u(x,r):=\frac{1}{\mathrm{\mid }{B}_r(x)\mathrm{\mid }}{\int }_{B_r(x)}u(y)dy\mathrm{.}$ The spherical (respectively ball) mean-value property says $u(x)=M_u(x,r)$ (respectively $u(x)=A_u(x,r)$) for every such ball. The polar formula Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma makes the normalizations meaningful.

For $n\ge 1$ and $r>0$, $\mathrm{\mid }\mathrm{\partial }{B}_r\mathrm{\mid }={\omega }_{n-1}{r}^{n-1}$ and $\mathrm{\mid }{B}_r\mathrm{\mid }={\omega }_{n-1}{r}^n\mathrm{/}n$; both factors are finite and positive.

If $u\in C^2(\mathrm{\Omega })$, $B_r(x)⋐\mathrm{\Omega }$, and $m(t)=M_u(x,t)$, then $m^{\mathrm{\prime }}(r)=\frac{r}{n}\frac{1}{\mathrm{\mid }{B}_r(x)\mathrm{\mid }}{\int }_{B_r(x)}\mathrm{\Delta }u(y)dy$.

If $u\in C^2(\mathrm{\Omega })$ and $\mathrm{\Delta }u=0$, then $u(x)=M_u(x,r)$ whenever $B_r(x)⋐\mathrm{\Omega }$.

Under Spherical mean-value property for harmonic functions, $u(x)=A_u(x,r)$ for every $B_r(x)⋐\mathrm{\Omega }$.

A radial mollifier family is ${\rho }_{\epsilon }(x)={\epsilon }^{-n}\rho (x\mathrm{/}\epsilon )$ as in The mollifier family generated by a unit-mass smooth bump, where $\rho \ge 0$, $\int \rho =1$, $\rho \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$, $\rho (x)=q(\mathrm{\mid }x\mathrm{\mid })$, and $\mathrm{supp}\rho \subset B_1(0)$. Thus $\mathrm{supp}{\rho }_{\epsilon }\subset B_{\epsilon }(0)$.

Let $u\in C(\mathrm{\Omega })$ have the spherical mean-value property, and let $({\rho }_{\epsilon })$ be a radial mollifier family as in A radial mollifier family in Rn. Then $(u\ast {\rho }_{\epsilon })(x)=u(x)$ whenever $B_{\epsilon }(x)⋐\mathrm{\Omega }$.

If $u\in C(\mathrm{\Omega })$ has the ball mean-value property, then $u\in C^{\mathrm{\infty }}(\mathrm{\Omega })$ and $\mathrm{\Delta }u=0$.

If $u\in C(\mathrm{\Omega })$ and every $x\in \mathrm{\Omega }$ has $r_x>0$ with $B_{r_x}(x)⋐\mathrm{\Omega }$ such that $u(y)=A_u(y,r)$ whenever $y\in B_{r_x}(x)$ and $0<r<r_x-\mathrm{\mid }y-x\mathrm{\mid }$, then $u$ is harmonic.

For $T\in {\mathcal{D}}^{\mathrm{\prime }}(\mathrm{\Omega })$, put ${\mathrm{\Omega }}_{\epsilon }:=\{x\in \mathrm{\Omega }:B_{\epsilon }(x)⋐\mathrm{\Omega }\}$ and define $(T\ast {\rho }_{\epsilon })(x)=T({\rho }_{\epsilon }(x-\cdot ))$ on ${\mathrm{\Omega }}_{\epsilon }$. Then $\mathrm{\Delta }(T\ast {\rho }_{\epsilon })=(\mathrm{\Delta }T)\ast {\rho }_{\epsilon }$ there.

If $T\in {\mathcal{D}}^{\mathrm{\prime }}(\mathrm{\Omega })$ and $\mathrm{\Delta }T=0$, there is a unique smooth harmonic $h$ with $T=T_h$.

If $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathrm{\Omega })$ and $\mathrm{\Delta }{T}_f=0$, then $f=h$ almost everywhere for a unique smooth harmonic $h$.

Every partial derivative of a smooth harmonic function is smooth harmonic. If $\mathrm{\Delta }T=0$, every distributional derivative of $T$ is induced by the corresponding smooth harmonic derivative.

If harmonic $u_j$ converge locally uniformly on $\mathrm{\Omega }$ to $u$, then $u$ is harmonic.

The real-variable results on this page include $n=2$. The holomorphic disc Poisson kernel, Perron theory, and conformal invariance remain in the complex-analysis track; no plane-specific result is reproved here.