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 connected component of $\{x\in S^2: u(x)> u(Rx)\}$ is at least $C$. Is $u$ a constant function? ($u(Rx)$ is a rotation of $u$ on $S^2$)



Symmetry on a sphere
The answer is no. One may consider the function $u(x,y,z)=x$ on the sphere $x^2+y^2+z^2=1$. If $R(1,0,0)=(1,0,0)$ then your set is empty, otherwise it is an open hemisphere.


One may replace $u=x$ by any strictly monotonous function in $x$.


