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Thread: spherically symmetric function

  1. #1
    Dec 2011

    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?
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  2. #2
    Junior Member nimon's Avatar
    Sep 2009
    Edinburgh, UK

    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!
    Last edited by nimon; Jun 5th 2013 at 04:50 AM.
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