When reading the paper
E. Carlen, J. Geronimo & M. Loss: SIAM J. MATH. ANAL., vol. 40, no. 1, 327-374
I found an argument like the following.
Given an bounded and self-adjoint linear operator $K: L^2(B) \rightarrow L^2(B)$, where $B$ is the unit ball in $\mathbb R^3$ equipped with the Lebesgue measure. Suppose 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 $K$ has a complete basis of eigenfunctions of the 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 $\mathbb R_+$.
It seems to be some well-known fact about the spherical harmonics, but still confused me a lot. I consider in the following way.
Since $K$ commutes with all rotations $R$, if $\lambda$ is an eigenvalue of $K$ and $X_\lambda$ is the corresponding eigen subspace, then it is clear that for all $g \in X_\lambda$, $$K(g \circ R) = (Kg) \circ R = \lambda g \circ R. $$ Thus $X_\lambda$ should be a subspace of $L^2(B)$, which is invariant under all rotations $R$. It is also clear that for fixed $l$, the linear span in $L^2(B)$ of the functions $$g(v) = h(|v|)Y_l^m(v/|v|), \ -l \leq m \leq l $$ is invariant under all $R$. But I am not sure how to continue.
Could anyone give me a proof or some reference? Thanks in advance.