Harmonic Functions and Mean Values in Rn

The parametrization $\theta \mapsto r\theta$ gives $\mathrm{\mid }\mathrm{\partial }{B}_r\mathrm{\mid }={\omega }_{n-1}{r}^{n-1}$ [given].

Under Countable Choice, applying Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma to $1_{B_r}$ gives $\mathrm{\mid }{B}_r\mathrm{\mid }={\omega }_{n-1}{\int }_0^r{t}^{n-1}dt={\omega }_{n-1}{r}^n\mathrm{/}n$ [given, algebra]. ∎

Compact containment gives a slightly larger ball inside $\mathrm{\Omega }$. Since $u$ is $C^2$ there and $\sigma$ is finite, differentiation under the spherical integral gives $m^{\mathrm{\prime }}(r)={\omega }_{n-1}^{-1}{\int }_{S^{n-1}}\mathrm{\nabla }u(x+r\theta )\cdot \theta d\sigma (\theta )\mathrm{.}$

If $n=1$, [F4] gives $m(t)=(u(x+t)+u(x-t))\mathrm{/}2$ and $\mathrm{\mid }{B}_r(x)\mathrm{\mid }=2r$. Thus $m^{\mathrm{\prime }}(r)=(u^{\mathrm{\prime }}(x+r)-u^{\mathrm{\prime }}(x-r))\mathrm{/}2$. The one-dimensional fundamental theorem of calculus gives ${\int }_{x-r}^{x+r}{u}^{\mathrm{\prime }\mathrm{\prime }}(y)dy=u^{\mathrm{\prime }}(x+r)-u^{\mathrm{\prime }}(x-r)$, exactly the displayed formula.

Suppose $n\ge 2$. The ball is a bounded smooth domain, and $F=\mathrm{\nabla }u$ is $C^1$ on its closure. Its outward normal at $x+r\theta$ is $\theta$. Applying [F1] and using [F2] to identify and scale its surface measure gives ${\int }_{B_r(x)}\mathrm{\Delta }u(y)dy=r^{n-1}{\int }_{S^{n-1}}\mathrm{\nabla }u(x+r\theta )\cdot \theta d\sigma (\theta )={\omega }_{n-1}{r}^{n-1}{m}^{\mathrm{\prime }}(r)\mathrm{.}$ By [F3], ${\omega }_{n-1}{r}^{n-1}=n\mathrm{\mid }{B}_r(x)\mathrm{\mid }\mathrm{/}r$, so division gives the claimed identity.

Radial derivative of a spherical average gives $\frac{d}{dt}{M}_u(x,t)=0$ for $0<t\le r$ [given].

Hence $M_u(x,t)$ is constant, and continuity gives ${\lim }_{t\downarrow 0}{M}_u(x,t)=u(x)$ [given].

Its value at $r$ is therefore $u(x)$ [step 1.2]. ∎

Under Countable Choice, polar coordinates express ${\int }_{B_r(x)}u={\omega }_{n-1}{\int }_0^r{t}^{n-1}{M}_u(x,t)dt$ [given].

The spherical identity makes this ${\omega }_{n-1}u(x)r^n\mathrm{/}n=u(x)\mathrm{\mid }{B}_r\mathrm{\mid }$ [given, algebra].

Divide by the positive ball volume from Sphere and ball measures scale in Rn [step 1.2]. ∎

Under Countable Choice, Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma gives $(u\ast {\rho }_{\epsilon })(x)={\omega }_{n-1}{\int }_0^{\epsilon }{q}_{\epsilon }(t)t^{n-1}{M}_u(x,t)dt$ [given].

Substitute $M_u(x,t)=u(x)$ and use $\int {\rho }_{\epsilon }=1$ to obtain $(u\ast {\rho }_{\epsilon })(x)=u(x)$ [given]. ∎

Fix $x\in \mathrm{\Omega }$ and $R>0$ with $\overline{B_R(x)}\subseteq \mathrm{\Omega }$. Polar coordinates and the ball-volume formula give, for $0<r\le R$, ${\int }_0^r{t}^{n-1}{M}_u(x,t)dt=\frac{r^n}{n}{A}_u(x,r)=\frac{r^n}{n}u(x)$. Here $M_u(x,t)$ is continuous in $t$: $u$ is uniformly continuous on the compact closed ball, and the sphere measure is finite. Differentiating the two sides for $0<r<R$ gives $r^{n-1}{M}_u(x,r)=r^{n-1}u(x)$, hence the spherical mean-value property.

To justify smoothness locally, choose $x_0\in \mathrm{\Omega }$ and $R>0$ with $\overline{B_{3R}(x_0)}\subseteq \mathrm{\Omega }$. Define $f$ on ${\mathbb{R}}^n$ to equal $u$ on $B_{2R}(x_0)$ and zero outside it. This is locally integrable, since $u$ is bounded on the compact closed $2R$-ball. Choose a radial mollifier ${\rho }_{\epsilon }$ with $0<\epsilon <R$. For every $x\in B_R(x_0)$ its convolution with $f$ uses only values inside $B_{2R}(x_0)$, so step 1.1 and [L3] give $(f\ast {\rho }_{\epsilon })(x)=u(x)$. The convolution is smooth by [L4]. Since $x_0$ was arbitrary, $u\in C^{\mathrm{\infty }}(\mathrm{\Omega })$.

Taylor's formula at any $x$ gives $u(x+h)=u(x)+\mathrm{\nabla }u(x)\cdot h+\frac{1}{2}{h}^T{H}_u(x)h+o(\mathrm{\mid }h{\mathrm{\mid }}^2)$ uniformly as $\mathrm{\mid }h\mathrm{\mid }\le r\to 0$. Averaging over $B_r(0)$ kills the linear and off-diagonal quadratic terms by reflection symmetry. Rotation symmetry makes every diagonal second moment equal; their sum is $\mathrm{\mid }{B}_r{\mathrm{\mid }}^{-1}{\int }_{B_r}\mathrm{\mid }h{\mathrm{\mid }}^{2}dh=n{r}^2\mathrm{/}(n+2)$ by the polar formula and the ball-volume formula. Thus each diagonal moment is $r^2\mathrm{/}(n+2)$, and $A_u(x,r)=u(x)+\frac{r^2}{2(n+2)}\mathrm{\Delta }u(x)+o(r^2)$.

The ball mean-value identity makes the left side equal to $u(x)$. Divide step 3.1 by $r^2$ and let $r\downarrow 0$ to obtain $\mathrm{\Delta }u(x)=0$. The point $x$ was arbitrary.

Each $B_{r_x}(x)$ satisfies the full ball mean-value property for balls compactly contained in it, so Continuous ball-mean-value functions are harmonic applies there under the stated Countable Choice hypothesis [given].

Hence $\mathrm{\Delta }u=0$ on these balls, which cover $\mathrm{\Omega }$ [given]. ∎

The support margin makes $\tilde{\varphi }(y):=\int {\rho }_{\epsilon }(x-y)\varphi (x)dx$ a test function in $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ [given].

Differentiating the kernel and using Distributional harmonicity and Poisson's equation on an open subset of Rn gives $⟨\mathrm{\Delta }(T\ast {\rho }_{\epsilon }),\varphi ⟩=⟨\mathrm{\Delta }T,\tilde{\varphi }⟩$ [step 1.1].

This is $⟨(\mathrm{\Delta }T)\ast {\rho }_{\epsilon },\varphi ⟩$ for all $\varphi$ [step 2.1]. ∎

The distributional Laplacian commutes with local mollification makes $h_{\epsilon }=T\ast {\rho }_{\epsilon }$ smooth and harmonic on ${\mathrm{\Omega }}_{\epsilon }$ [given].

Each $h_{\epsilon }$ is harmonic by step 1.1, so the ball and spherical mean-value properties apply under Countable Choice. Radial mollification therefore fixes it: $h_{\epsilon }\ast {\rho }_{\delta }=h_{\epsilon }$ on ${\mathrm{\Omega }}_{\epsilon +\delta }$, and likewise $h_{\delta }\ast {\rho }_{\epsilon }=h_{\delta }$. Associativity of distributional convolution with these compactly supported test kernels identifies the two double convolutions on that common shrunken domain. Hence $h_{\epsilon }=h_{\delta }$ there.

On each compact subset of $\mathrm{\Omega }$, all sufficiently small regularizations agree by step 2.1; their common value defines a smooth harmonic $h$. For a test function $\varphi$ supported in that compact subset, $⟨T_{h_{\epsilon }},\varphi ⟩=⟨T,{\check{\rho }}_{\epsilon }\ast \varphi ⟩\to ⟨T,\varphi ⟩$: the mollified tests converge to $\varphi$ with all derivatives on one fixed compact support, so distributional continuity applies. Since $h_{\epsilon }=h$ on the support for small $\epsilon$, this gives $T_h=T$.

If $T_h=T_k$, then $h=k$ almost everywhere; continuity makes $h=k$ everywhere [given]. ∎

Weyl's lemma for the Laplacian supplies a smooth harmonic $h$ with $T_f=T_h$ [given].

Equality of regular distributions implies $f=h$ almost everywhere, and uniqueness is inherited [step 1.1]. ∎

Constant-coefficient derivatives commute, so $\mathrm{\Delta }({\mathrm{\partial }}^{\alpha }h)={\mathrm{\partial }}^{\alpha }\mathrm{\Delta }h=0$ [given].

Under Countable Choice, Weyl's lemma for the Laplacian supplies $T=T_h$. The derivative definition gives ${\mathrm{\partial }}^{\alpha }T=T_{{\mathrm{\partial }}^{\alpha }h}$ [step 1.1]. ∎

On every $B_r(x)⋐\mathrm{\Omega }$, Ball mean-value property for harmonic functions under Countable Choice says $u_j(x)=A_{u_j}(x,r)$ under Countable Choice [given].

Uniform convergence on $\overline{B_r(x)}$ permits passage to the integral, giving $u(x)=A_u(x,r)$ [given].

The locally uniform limit $u$ is continuous. Continuous ball-mean-value functions are harmonic therefore applies under the same Countable Choice hypothesis and proves $u$ harmonic [step 1.2]. ∎