Continuous ball-mean-value functions are harmonic

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.