# spherically symmetric function

• June 4th 2013, 04:07 PM
tommyjbaby
spherically symmetric function
Suppose f:R^3\(Ball of radius 1)--->R is smooth and satisfies f(S^2)=0, ie the unit sphere is a level set of f. does it neccessarily follow that f is a spherically symmetric function?
• June 5th 2013, 04:45 AM
nimon
Re: spherically symmetric function
I don't think so. I have an idea for a counterexample but you will have to verify it for me because I'm not 100%.

The idea is that the modulus function $\mathbb{R}^{3} \rightarrow \mathbb{R}; x \mapsto \|x\|$ is smooth, so if we smoothly stretch $\mathbb{R}^{3}$ and apply a modulus function to the stretched space we should still have a smooth function, but not a spherically symmetric one (provided that our stretching doesn't preserve spheres!).

So here's what I propose. Let $s \in S^{2}$ be any point on the unit sphere, and let $s+ts$ for $t\geq 0, t\in\mathbb{R}$ be the ray eminating from the origin through $s.$ Then $\|ts\|$ is the distance along the ray we have travelled from $s.$ Any point can be expressed in the form $s+ts$ and for any such point the function $x\mapsto t\|s\|$ is a smooth, spherically symmetrical function $\mathbb{R}^{3}/ \{ \text{Ball of radius 1} \} \rightarrow \mathbb{R}$. If we dilate that ray by a factor proportional to the altitude angle and consider the new distance from $s,$ while leaving $S^{2}$ unchanged, then that distance would be a smooth function where $S^{2} \rightarrow 0$ but which is not spherically symmetric.

I guess what I'm saying is, for any $x$ in our domain, let $x = s+ts$ for $s\in S^{2},t\in\mathbb{R}.$ Expressing $s$ as $(1,\theta,\phi)$ in spherical coordinates, then the function $x \rightarrow (|\phi|+1)t \|s\|$ has the desired properties but is not spherically symmetric.

This is more or less a sketch but if you can formalise it a bit, and if you can't find any technical problems with it, then hopefully it will work!