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 $S^2$)
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1$\begingroup$ How do you get nonempry components for the constant function? $\endgroup$– Alex DegtyarevCommented Mar 27, 2014 at 12:48
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1$\begingroup$ @AlexDegtyarev: If $u$ is constant, then for any $R$, $\{x\in S^2:u(x)>u(Rx) \} = \emptyset$, so all of its connected components (all 0 of them) have area $> C$. I don't see a problem here :-) $\endgroup$– EdgarTheWiseCommented Mar 27, 2014 at 13:10
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$\begingroup$ OK, this is an interesting way to interpret the area of the empty set, but let it be :) $\endgroup$– Alex DegtyarevCommented Mar 27, 2014 at 13:12
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3$\begingroup$ @AlexDegtyarev Edgar's comment was not about the area of the empty set. It was about the area of every one of the components, and the empty set is not a component. Since there are no components when $u$ is constant, what he wrote about all their areas is vacuously true. $\endgroup$– Andreas BlassCommented Mar 27, 2014 at 13:45
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$\begingroup$ Yes, I got it. The ":)" was supposed to explain that. $\endgroup$– Alex DegtyarevCommented Mar 27, 2014 at 13:55
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1 Answer
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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 $x$.
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1$\begingroup$ Thanks Ilya. The example you provided is axially symmetric. Do you think the above property can imply some sort of symmetry result on u? $\endgroup$ Commented Mar 28, 2014 at 0:02