When reading a paper I saw an argument like the following.
Given an operator $K: L^2(B) \rightarrow L^2(B)$, where $B$ is the unit ball in $\mathbb R^3$ equipped with the Lebesgue measure. We know that $K$ is bounded, self-adjoint. We also know that $K$ commutes with all rotations in $\mathbb R^3$: if $R$ is a rotation on $\mathbb R^3$, then $$K(g \circ R) = (Kg) \circ R. $$ Then the author gives the conclusion that to study the spectrum of $K$, we only need to consider the eigenfunctions $g$ of $K$ with the following form: $$g(v) = h(|v|)Y_l^m(v/|v|), $$ where $Y_l^m$ is some spherical harmonic function and $h$ is some function on $\mathbb R_+$.
I guess here he means that for every eigenvalue $\lambda$, we at least have one eigenfunction $g$ in the above form such that $Kg = \lambda g$.
It seems to be some well-known argument about the spherical harmonics, but still confused me a lot. Could anyone give me a proof or some reference? Thanks in advance.
This is a very known statement, but you need some Theorem: if the operators are commuting, then they have simultaneous eigenfunctions.
But in my opinion this is not too important here. What the Author really did was to use separation of variables to define the eigenvectors $g$ as a product of two functions, one of them depending only on the radial coordinate and another depending on the angular ones (that is, the spherical harmonics which must be eigenfunctions of the angular part of your operator).
In this kind of situation it is natural to take an ordinary differential equation for the radial variables (only!). Please, see the resolution of the Schrödinger equation for the hydrogen atom in any book on mathematical physics.
Best regards.