For uniform $X$ on unit-sphere, under what conditions on $f$ is the mapping $(a, b) \mapsto \mathbb E_X[f(X^\top a)f(X^\top b)]$ a dot-product kernel?

Question

Let $X$ be random vector uniform on the unit-sphere $\mathcal S^{d-1}$ in $\mathbb R^d$. Given a continuous function $f:\mathbb R \to \mathbb R$, consider the kernel function $K_f:\mathcal S^{d-1} \times \mathcal S^{d-1} \to \mathbb R$ defined by $K_f(a,b) := \mathbb E_X[f(X^\top a)f(X^\top b)]$.

Question. What is a complete characterization of the functions $f$ for which there exists a function $\varphi_f:\mathbb R \to \mathbb R$ such that $K_f(a,b) = \varphi_f(a^\top b)$ for all $a,b \in \mathcal S^{d-1}$ ?

Example

If $f$ is positive $p$-homoegenous for some $p \ge 0$, i.e if $f(tx) = t^pf(x)$ for all $t \ge 0$ and all $x \in \mathbb R$, then one can show that (see Lemma 2.2.2 of this paper) $K_f(a,b) \equiv \varphi_f(a^\top b)$, where $\varphi_f(t) = A_p \int_0^{2\pi} \varphi(\cos t)\varphi(\cos (t - \arccos(a^\top b))dt$, where $A_p \ge 0$ is a constant which only depends on $p$.

Answer 1

For any unitary matrix (i.e orthgonal projection) $U$ of size $d$, and any $a,b \in \mathbb R^d$ we have $K_f(Ua,Ub) = K_f(a,b)$, thanks to the unitary-invariance of the uniform distribution on $\mathcal S_{d-1}$. Thus, $K_f$ is unitary invariant and thus must be of the form $K_f(a,b) \equiv \varphi_f(a^\top b)$ for some $\varphi_f:\mathbb R \to \mathbb R$.