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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.

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    $\begingroup$ 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 $|x|$). $\endgroup$
    – Christian Remling
    Commented 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). $\endgroup$
    – Christian Remling
    Commented 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
    Commented Oct 27, 2016 at 20:16

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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.

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  • $\begingroup$ I think you overlooked that $B$ is the unit ball (not sphere) here. $\endgroup$
    – Christian Remling
    Commented Oct 27, 2016 at 19:51
  • $\begingroup$ @ChristianRemling You're right. I will update my answer. $\endgroup$
    – Liviu Nicolaescu
    Commented 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. $\endgroup$
    – gregarki khayal
    Commented Oct 28, 2016 at 4:42

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