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) $.

1$\begingroup$ 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 $xy$. $\endgroup$– Francesco PolizziCommented Aug 13, 2021 at 12:57

$\begingroup$ 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 $. $\endgroup$– Luis Yanka AnnaliscCommented Aug 13, 2021 at 14:10

2$\begingroup$ The group generated by reflections in the planes (passing through the origin) contains the group of all rotations. $\endgroup$– Alexandre EremenkoCommented Aug 13, 2021 at 14:46

$\begingroup$ A theorem of Cartan and Dieudonne says that every orthogonal transformation is a product of reflections, hence Alexandre Eremenko's remark. $\endgroup$– Ben McKayCommented Aug 13, 2021 at 15:46

2$\begingroup$ 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 $\{(x  x_0)\cdot v = 0\}$. CartanDieudonne (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. $\endgroup$– Willie WongCommented Aug 13, 2021 at 15:58
1 Answer
Notations:
 given $P$ a hyperplane, let $r_P$ denote the reflection operation $\mathbb{R}^d\to\mathbb{R}^d$ about the plane $P$.
 given $R$ a codimension 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 nontrivial 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 nonparallel 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 nongeneric), 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 $d(d1)/2$ codimension2 "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.

$\begingroup$ 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). $\endgroup$ Commented Aug 13, 2021 at 16:11