Why the symmetry for all directions implies the radial symmetry?

Question

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) $.

Answer 1

Notations:

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.