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.

  • 1
    $\begingroup$ This is of course not correct in this form, as $K$ need not have any eigenfunctions at all (for example, take $K$ to be multiplication by $|x|$). $\endgroup$ Oct 27, 2016 at 15:06
  • $\begingroup$ What is true is that the spaces $\{ h(|x|)Y(x/|x|)\}$ are reducing subspaces for $K$. I don't think that's very easy to prove, but this is fairly standard material (try a search for "commuting self-adjoint operators" or some such keywords perhaps). $\endgroup$ Oct 27, 2016 at 15:16
  • $\begingroup$ As far as I've seen the paper does not mention an argument like you stated. You probably refer to the (weaker) statement "Since $K$ commutes with all rotations, we may restrict our search for eigenfunctions $g$ of $K$ to functions of the form…" which sound more plausible. $\endgroup$
    – Dirk
    Oct 27, 2016 at 20:16

1 Answer 1


In the paper you mention at page 345 it is shown that the operator $K$ is selfadjoint and, for every $N\geq 0$, the operator $K$ preserves the space $\newcommand{\eP}{\mathscr{P}}$ $\eP_N$ of polynomials of degree $\leq N$. This space has an $SO(3)$-invariant orthogonal decomposition

$$\eP_N= \bigoplus_{k+2\ell\leq N} r^{2\ell} H_\ell $$

where $H_k$ denotes the space of degree $k$ harmonic homogeneous polynomial. The spaces $H_k$ are irreducible $SO(3)$-representations and Schur Lemma will imply that, for every $k$ and $m$ the spaces

$$\bigoplus_{k=0}^m r^{2\ell} H_k $$

are $K$-invariant since $K$ is $SO(3)$-equivariant.

  • $\begingroup$ I think you overlooked that $B$ is the unit ball (not sphere) here. $\endgroup$ Oct 27, 2016 at 19:51
  • $\begingroup$ @ChristianRemling You're right. I will update my answer. $\endgroup$ Oct 27, 2016 at 20:47
  • $\begingroup$ It seems that I made a lot of mistakes in my question... I will try to edit it and accept the answer. Thank you very much. $\endgroup$ Oct 28, 2016 at 4:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.