Mean value principle reversed Suppose that $\Omega \subset \mathbb R^3$ is a domain with smooth boundary $\partial \Omega$ and suppose that $0\in \Omega$. Given any $f \in C^{\infty}(\partial \Omega)$ let $u^f$ denote the unique harmonic function on $\Omega$ with Dirichlet data $f$. Finally, suppose that given any smooth $f$ there holds:
$$ u^f(0)= \frac{1}{|\partial \Omega|}\int_{\partial \Omega} f(y) \,d\sigma_y.$$
Does it follow that $\Omega$ is a sphere?
 A: Edit: I misread the question. New answer:
The question asks whether the Poisson kernel $P_\Omega(0, \cdot)$ is constant only when the domain $\Omega$ is a ball centred at $0$.
This is indeed true: let $r$ be the radius of the largest ball $B(0, r)$ contained in $\Omega$, and $R$ the radius of the smallest ball $B(0, R)$ containing $\Omega$. If $y \in \partial \Omega$ and $|y| = r$, then $P_\Omega(0, y) \geqslant P_{B(0,r)}(0, y)$. Similarly, if $z \in \partial \Omega$ and $|z| = R$, then $P_\Omega(0, z) \leqslant P_{B(0,R)}(0, z)$. If $P_\Omega(0, \cdot)$ is constant, it follows that $P_{B(0,r)}(0, y) = P_{B(0,R)}(0, z)$, and therefore $r = R$.

Old answer, to a different question, whether $u_f(0) = \frac{1}{|\Omega|} \int_\Omega u_f(x) dx$ implies that $\Omega$ necessarily a ball?
Yes, it does. This is sometimes called Kuran's theorem, proved by Ülkü Kuran in 1972, see [1], after preliminary works by a number of other researchers.
Reference:
[1] Ülkü Kuran, On the mean-value property of harmonic functions. Bull. London Math. Soc. 4 (1972): 311–312, DOI:10.1112/blms/4.3.311.
