Why the symmetry for all directions implies the radial symmetry?

I am recently reading the pde book of Evans's. I am reading the chapter 9.5.2 which is a topic about the radial symmetry for the solution of a elliptic equation. In the book there is a method of moving planes. That is when a function is symmetric for all directions, we can have that it is radial symmetric. I do not know why it is true, can you give me some information about it ? Here if we assume that $$u:\mathbb{R}^n\rightarrow \mathbb{R}$$ is a function with $$n$$ variables, the symmetry for all directions means that for all $$v\in\mathbb{R}^n$$, here is a plane $$P_v\subset\mathbb{R}^n$$ such that $$P_v\perp v$$ and for all $$x,y\in\mathbb{R}^n$$ that are symmetric about the plane $$P_v$$, we have $$u(x)=u(y)$$.

• Assume $n=2$, the general proof is the same. If there are two points $x$, $y$ such that $||x||=||y||$ and $u(x) \neq u(y)$, then $u$ would not be symmetric with respect to the direction given by the axis of the segment $xy$. Commented Aug 13, 2021 at 12:57
• Here the plane $P_v$ may not contain $O$. So how to prove that $u$ would not be symmetric with respect to the direction given by the axis of the segment $xy$. Commented Aug 13, 2021 at 14:10
• The group generated by reflections in the planes (passing through the origin) contains the group of all rotations. Commented Aug 13, 2021 at 14:46
• A theorem of Cartan and Dieudonne says that every orthogonal transformation is a product of reflections, hence Alexandre Eremenko's remark. Commented Aug 13, 2021 at 15:46
• Note however that in the method of moving planes, one gets that for every direction $v\in \mathbb{R}^n$ there exists a base point $x_0$ such that the function is reflection symmetric across the plane $\{(x - x_0)\cdot v = 0\}$. Cartan-Dieudonne (or the version that states the set of ALL reflections generate ALL Euclidean motions) do not quite apply directly. In particular, the result is not that $u$ is radial about the origin, but that $u$ is radial about some unspecified center. Commented Aug 13, 2021 at 15:58

1. given $$P$$ a hyperplane, let $$r_P$$ denote the reflection operation $$\mathbb{R}^d\to\mathbb{R}^d$$ about the plane $$P$$.
2. given $$R$$ a co-dimension 2 affine subspace of $$\mathbb{R}^d$$, denote by $$\rho_{R,\theta}$$ the rotation by angle $$\theta$$ that fixes the "axis" $$R$$.
Basic linear algebra/affine geometry gives you that if $$P, Q$$ are hyperplanes with non-trivial intersection, then $$r_P\circ r_Q = \rho_{P\cap Q, \theta}$$ for some $$\theta$$. (Two reflections generate a rotation.) The angle $$\theta$$ depends only on the angle between the two hyperplanes.
Now let $$P, Q$$ be generic non-parallel hyperplanes. Then the corresponding $$\theta$$ is an irrational multiple of $$\pi$$. And so if $$u$$ is a continuous function satisfying $$u = u\circ r_P = u\circ r_Q$$, then we must have $$u = u \circ \rho_{P\cap Q, \phi}$$ for any angle $$\phi$$. (Because by alternating between $$r_P$$ and $$r_Q$$, and composing as many times as you want, you find that for all $$n\in \mathbb{N}$$ that $$u = u\circ \rho_{P\cap Q, n\theta}$$. And $$\{n\theta \mod 2\pi\}$$ is dense on the circle.)
Returning to your setting, start by choosing $$d$$ generic directions (in particular, you don't want them to be mutually orthogonal, as that is non-generic), and set $$P^k$$ be the corresponding planes of symmetry. Let $$x_0$$ be the common intersection. The above argument shows that about each $$P^i\cap P^j$$ (forming a set of $$d(d-1)/2$$ codimension-2 "axes") through $$x_0$$, your function $$u$$ is rotationally symmetric. These rotations however generate the full rotation group about $$x_0$$, and hence you are done.
• Note that a posteriori, if for some direction the corresponding plane of symmetry does not pass through $x_0$, then your function $u$ also has a discrete translation symmetry in addition to the rotational symmetry; this immediately implies that $u$ is constant (by drawing isosceles triangles). Commented Aug 13, 2021 at 16:11