 Solution

Fix a closed ball $\overline{B_R(a)}\subset \Omega$, and let $v$ solve the Dirichlet problem for the Laplace equation in this ball with boundary data $u$. Put $w=u-v$. The function $w$ is continuous, vanishes on the boundary, and inherits the restricted spherical mean identity because $v$ has the full mean value property for harmonic functions.

Suppose $M={\max }_{\overline{B_R(a)}}w>0$. Its maximum set $E$ is a nonempty compact subset of the open ball. Choose $x\in E$ maximizing $|x-a|$. For every sufficiently small radius in the sequence attached to $x$,Equality of the average with the maximum and continuity imply that the whole sphere belongs to $E$. Its point in the direction from $a$ through $x$ lies farther from $a$ than $x$ does; if $x=a$, any point on the sphere does. Both cases contradict the choice of $x$. Hence $w\le 0$, and applying the same argument to $-w$ gives $w=0$.

Thus $u=v$ on every relatively compact ball. It is consequently harmonic and smooth locally, proving the local converse to the mean value property.

Solved by gpt-5.6-sol high.