Let $f$ be a continuous function on $S^2$ and suppose there exists a constant $C>0$ such that for every $\mathcal{R} \in SO(3)$ the area of every connected component of $\{f(x)\geq f(\mathcal{R}x)\}$ is at least $C$ ($f(\mathcal{R}x)$ is a rotation of $f$ on $S^2$). Does there exist $I\neq\mathcal{R}_0 \in SO(3)$ such that $f(x)=f(\mathcal{R}_0x)$ on an open subset of $S^2$?
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asked Nov 29, 2017 at 20:42
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$\begingroup$ Yes, I revised the question. $\endgroup$– A random mathematicianCommented Nov 30, 2017 at 15:57
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Note that any eigen-function satisfies your condition for some $C>0$. It remains to find a non-symmetric one. Take for example $$f(x,y,z)=10\cdot x^3+y^3+\tfrac1{10}\cdot z^3.$$
answered Nov 30, 2017 at 21:42
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$\begingroup$ Thanks Anton. Is it easy to see why such C exists for your example? $\endgroup$ Commented Dec 1, 2017 at 17:35
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$\begingroup$ @MathStudent assume it does not, then passing to the limit, we get a point $x_0\in\mathbb{S}^2$ such that $f(x_0)=(f\circ\mathcal{R})(x_0)$, $d_{x_0}f=d_{x_0}(f\circ\mathcal{R})$ and $\mathrm{Hess}_{x_0}f\le\mathrm{Hess}_{x_0}f \circ\mathcal{R}$, which is impossible; at least for the given $f$ (I might be wrong for general eigen-function). $\endgroup$ Commented Dec 1, 2017 at 18:40