Let $u$ be a smooth function on the sphere $S^2$. Suppose there exists $C>0$ such that for all $R \in SO(3)$, the area of every nonempty connected component of $\{x\in S^2: u(x)> u(Rx)\}$ is at least $C$ ($u(Rx)$ is a rotation of $u$ on $S^2$). Such function can't be "too" non-symmetric. I wonder what kind of symmetry results the above condition implies.
If $u$ is a monotone axially symmetric function, then nonempty components of $\{x\in S^2: u(x)> u(Rx)\}$ are open hemispheres. So axially symmetric functions can satisfy the condition. Does the above condition necessarily imply that $u$ is axially symmetric? If not, can we deduce a weaker notion of symmetry?