Spherically Symmetric Function

Question

Suppose $f:\mathbb{R}^3\setminus B(0,1) \to \mathbb{R}$ is smooth and satisfies $f(S^2)=0$, i.e. the unit sphere is a level set of $f$. Does it necessarily follow that $f$ is a spherically symmetric function?

Answer 1

Let $g: \mathbb{R}^3 \to \mathbb{R}$ be any smooth function (for example, $g(x) = x_1$, which is not spherically symmetric). Let $f(x) = g(x) (\sum_k x_k^2 -1)$. Then $f$ is smooth and $f(x) = 0 $ if $x \in S^2$.

If $g$ is not spherically symmetric for $\|x\| >1$, then $f$ is not either.

Answer 2

No. Consider $f(x) := \exp(-1/(1-|x|^2))$ when $|x|<1$ and $f(x):=0$ when $|x|\geq1$. This function is in $C^\infty$ and has compact support $B(0,1)$. In particular, if $g(x) := f(x+u)$ where $u \in \mathbb{R}^3$ has $|u|>2$, then $g$ is zero on $S^2$ but is clearly not spherically symmetric.