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?



Mean value principle reversed
*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 $r = R$.




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*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](https://doi.org/10.1112/blms/4.3.311).


