3
$\begingroup$

Let $d$ be a large positive integer and let $S_{d-1}$ be the unit-sphere in $\mathbb R^d$ and let $K_\gamma:S_{d-1} \times S_{d-1} \to \mathbb R$ be defined by $K_\gamma(x,x') = e^{-\|x-x'\|_2^\gamma}$, for $\gamma > 0$. Note that $K_\gamma$ is positive-semidefinite in the sense that $\sum_{i=1}^n \sum_{j=1}^n a_ia_jK_\gamma(x_i,x_j) \ge 0$ for all $x_1,\ldots,x_n \in S_{d-1}$ and $a_1,\ldots,a_n \in \mathbb R$. Let $L^2(S_{d-1})$ be the (Hilbert space) of square-integrable functions $S_{d-1} \to \mathbb R$ w.r.t the uniform probability measure $\tau_d$ on $S_{d-1}$. Consider the induced kernel integral operator $T_{K_\gamma}:L^2(S_{d-1}) \to L^2(S_{d-1})$ defined by

$$ f \mapsto T_{K_\gamma}f: x \mapsto \int K_\gamma(x,x')f(x')d\tau_d(x'). $$

It is clear that $T_{K_\gamma}$ is a compact positive operator. Let $\lambda_1 \ge \lambda_2 \ge \ldots$ be its eigenvalues.

Question. What are good estimates (upper-bounds and lower-bounds) for the $\lambda_j$'s valid for large $d$ ?

I'm particularly interested in the cases $\gamma = 1$ and $\gamma = 2$.

Related: General strategy for studying the decay of eigenvalues of kernel integral operators

$\endgroup$
1
  • $\begingroup$ Were you able to find the answer to it? $\endgroup$
    – newbie
    Aug 18, 2021 at 4:42

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.