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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?

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    $\begingroup$ I feel like I've seen this question before. Does it appear somewhere else? $\endgroup$ Sep 25, 2014 at 20:49
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    $\begingroup$ The prequel to this question, bearing the same title, is here mathoverflow.net/questions/161600/symmetry-on-a-sphere $\endgroup$ Sep 25, 2014 at 21:27
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    $\begingroup$ Thanks Yoav. For the OP, could you let us know what you mean by a "symmetry results", and perhaps why you're interested in this question? Some context would maybe get people motivated. $\endgroup$ Sep 25, 2014 at 23:59
  • $\begingroup$ @Ricardo Andrade: substantiation for removal [real-analysis] [spherical-geometry] [symmetry] [orthogonal-groups]? Is here a damn Wikipedia to rollback good-faith edits indiscriminately? [curves-and-surfaces] [surfaces] obviously havn’t any specificity with respect to this problem. $\endgroup$ Nov 1, 2014 at 9:15
  • $\begingroup$ Dear @Incnis Mrsi, perhaps my roll-back was not the best course of action. This question is hard to tag by nature, and the current tags are not that good. Nevertheless, I think this question is not quite about real analysis, and it certainly is not about spherical geometry, even if it involves spheres and some sort of geometry. Moreover, I am not very fond of the tag 'symmetry', which is often improperly used as a meta-tag of sorts. In that light, I decided to simply roll-back your change to this month-old question. (to be continued...) $\endgroup$ Nov 1, 2014 at 11:57

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