TITLE: The ground state is signed and symmetric
QUESTION [12 upvotes]: Background
In Berestycki and Lions it is asserted that (on page 316), if I am not misreading, that the "ground state", i.e. action minimizer among nontrivial solutions, corresponding to the action 
$$ S[u] = \int_{\mathbb{R}^d} |\nabla u|^2 + G(u) dx $$
where $G$ satisfies certain conditions must be signed and spherically symmetric. In the paper they showed that under certain conditions on $G$ that such a ground state (that is, signed and spherically symmetric) exists; but they do not show that all grounds states must have these symmetries (by which I mean that they don't show the "action minimizing solutions" must be both single-signed and spherically symmetric). 
The only reference I can find on this claim is the paper of Coleman, Glaser, Martin; but I am not entirely convinced that they have established the necessity. Their proof uses the fact that a ground state must be a minimizer of $\int |\nabla u|^2$ under the constraint that $\int G(u)$ is fixed. They then use the Polya-Szego principle: under spherical rearrangements the latter integral is unchanged, while the former can only decrease. 
But the decrease is not necessarily strict. Brothers and Ziemer gave a counterexample in the case that the distribution function of $u$ is not absolutely continuous. 
My question
I know how to complete the proof and get that ground states must be spherically symmetric, and monotonic radially, provided one assume that $G$ is $C^{1,1}$. This one can do by unique continuation principles for elliptic PDEs or equivalently by uniqueness theorems for ODEs. But in Berestycki-Lions or in Coleman-Glaser-Martin, $G$ is only assumed to be $C^1$, for which non-uniqueness, at least in the case of ODEs, is well known as a possibility. 
So, is the "uniqueness" statement true for $G$ merely $C^1$? Are there known counterexamples?

REPLY [3 votes]: A little bit more digging turned up this paper of Mihai Mariş. 
He shows that under two technical conditions:

Minimizers are $C^1$  (which we have from elliptic regularity)
If $u$ is an admissible function ($H^1$ in our case) and $v$ is a unit vector in $\mathbb{R}^d$, the function $\tilde{u}(x) = u(x)$ when $x\cdot v \geq 0$ and $\tilde{u}(x) = u(x - 2 x\cdot v v)$ otherwise is also admissible. 

The solution to a constrained minimisation problem with $k$ constraints must be radially symmetric about an axis of dimension $k-1$ if the functionals are translation and rotationally invariant. 
A very bad description of the process is follows:
Roughly speaking, the main idea is the following: by the Borsuk-Ulam Lemma we can find inductively $\sim d-k$ mutually orthogonal hyperplanes each of which split all the constraints exactly in half. The reflection symmetry inherent in the problem implies that from each orthant we can extend to a solution that is symmetric across each of the planes, which implies that it is symmetric under $x\to -x$ inside a subspace. This however implies that any linearly dependent hyperplane of the given ones will also split all the constraints in half. 
The last step is to show that using multiple reflections and a bit of geometry, one can show that the original solution much also be reflectively symmetric across each of the planes. This shows radial symmetry.