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Let $X$ be the unit sphere in $\mathbb R^n$ and let $\sigma$ be the uniform measure on $X$. Consider a positive definite kernel $K:X \times X \rightarrow \mathbb R$, which is $L^2$ w.r.t the measure $\sigma \otimes \sigma$ on $X \times X$. Let $L^2(X,\sigma)$ be the Hilbert space of all $\sigma$-measurable functions which are square-integrable w.r.t the measure $\sigma$, and consider the kernel integral operator induced by $K$, namely $I_K: L^2(X,\sigma) \to L^2(X,\sigma)$ be the associated kernel integral operator, defined by

$$ (I_K f)(x) := \int_X K(x,y)f(y)d\sigma(y). $$

Note that $I_K$ is compact and self-adjoint operator (to check!) on the nontrivial Hilbert space $L^2(X,\sigma)$. Consequently, it has only countably many eigenvalues (including multiplicities) $\lambda_1 \ge \lambda_2 \ge \ldots \lambda_n \ge \ldots$. Moreover, $\lambda_n \to 0$ as $n \to \infty$.

Question. Is there a general strategy (tools, theorems, etc.) for studying the rate of decay of $\lambda_n$ as a function of the regularity properties (Lipschitz, Hoelder class, etc.) of the input kernel $K$ ?

Note. I'm particularly interested in kernels of the form $K(x,y) \equiv k(x^Ty)$, for some $k:[-1,1] \to \mathbb R$.

  • 1
    $\begingroup$ If by makes sense you mean $I_K$ is bounded, then all the eigenvalues are contained in a disk it is not clear to me what you mean by their decay. There are strategies but they are not general in the sense that they require additional assumptions on $K$ so they do not apply to general $K$. By definite kernel do you mean positive definite? $\endgroup$ Oct 31, 2020 at 14:06
  • $\begingroup$ @LiviuNicolaescu Thanks for the comment. Fixed some issues with the question as I stated it. I meant general strategies which only use regularity properties of $K$, say. I'm particularly interested in kernels of the form $K(x,y) = k(x^Ty)$, for some $k:\mathbb R \to \mathbb R$. $\endgroup$
    – dohmatob
    Oct 31, 2020 at 16:23
  • 2
    $\begingroup$ Since $I_K$ is a Hilbert-Schmidt operator, the eigenvalue sequence is in $\ell_2$. -- A detailed analysis of the eigenvalue distribution of compact (and other) operators on Banach spaces can be found in the monographs by H. K\"onig (Eigenvalue distribution of Compact Operators) and A. Pietsch ($s$-Numbers and Eigenvalues). $\endgroup$ Oct 31, 2020 at 20:35
  • $\begingroup$ @DirkWerner Thanks for generic $\ell_2$ result and the refs. I was hoping for a result / tool which would incomporate information about smoothness of $k$ (e.g Hoelder class, Lipschitz, etc., with explicit dependence on exponents, etc.) $\endgroup$
    – dohmatob
    Oct 31, 2020 at 20:50
  • 1
    $\begingroup$ @dohmatob You will find results like this in the books that I have mentioned! $\endgroup$ Oct 31, 2020 at 20:52

1 Answer 1


Birman and Solomyak have studied this question quite intensivly.

The paper may not be the easiest to understand, but it does cover in a very general setup, what regularity conditions on the kernel imply in terms of singular number estimates. As you are working with a positive kernel, those are equivalent to the eigenvalues.

If you identify the sphere with a cube $[0,1]^{m-1}$ and transform the measure in a suitable way, you can apply Propositions 2.1-2.3. The technical conditions are a bit more complicated then Hölder continuity, and are defined in section 1.

doi link to the paper

Let $W^{\alpha,p}(X)$ be the fractional Sobolev space with $\alpha$ degrees of differentiability and $p$-integrable functions. Let $W^{\alpha,p}_{hom}(X)$ be the associated homogenuous space (we only care for the highest differential and the function does not need to be $p$-integrable itself).

(2.14): $1- r^{-1} = 2(p^{-1} - \alpha m^{-1})$

To simplify the statement of the theorem we use the notation $D^\alpha_pX$ in the following sense: for $p\alpha > m$ we take $X = Q^m$ [the unit cube] and $D^\alpha_pX= W^{\alpha,p} Q^m$ ; for $p\alpha < m$ we take $X = \mathbb R^m$ and $D^p_\alpha X =W_{hom}^{\alpha,p}(X)$. . The notation $D^{\alpha_1}_{p_1}(X) D^{\alpha_2}_{p_2}(Y) $ is to be interpreted similarly (four cases are possible here).

THEOREM 2.5 Let $X = Q^{m_1}$ or $X = \mathbb R^{m_1}$, $Y = Q^{m_2}$ or $Y = \mathbb R^{m_2}$ and $T \in D^{\alpha_1}_{p_1}(X) D^{\alpha_2}_{p_2}(Y)$ for $2 \le p_1$ and $p_2 < \infty$. Let $\rho \in M_{r_1}(X)$ and $\tau \in M_{r_2}(Y)$ with $r_i= 1$ for $p_i\alpha_i > m_i$ and $r_i > 1$ for $ρ_i\alpha_i = m_i$ and suppose that (2.14) is satisfied when $ρ_i \alpha_ i < m_i$($i=1, 2$). Then $$s_n(T_{\rho \tau}) \le C n^{-\gamma} N(T \mid D^{\alpha_1}_{p_1}(X)D^{\alpha_2}_{p_2}(Y) ) N^{\frac 1 2} ( \rho \mid M_{r_1}(X) ) N^{\frac 1 2} ( \tau \mid M_{r_2}(Y) ) , $$ $$ \gamma = \frac 1 2+ \frac {\alpha_1} {m_1} + \frac {\alpha_2} {m_2}. $$

Here $M_r(X)$ is the space of all measures, such that the density with respect to the Lesbegue measure is in $L^r$. The operator $T_{\rho \tau}$ is the integral operator with kernel $T$ from $L^2(Y,\tau)$ to $L^2(X,\rho)$ and the norm in the composition of spaces $H( X) H'( Y)$ is the space of all kernels, such that the map from $X$ to the Banach space $H'(Y)$ is in the space $H(X)$.

I am not sure how understandable that is. As I said, the paper is not easy to understand and even the statement is quite hard to tell.

Different idea:

Another possible approach would be to consider the Laplace operator on $S^d$ and to show that it satisfies a Weyl law. Then you could use that the operator $(- \Delta)^{-\frac s 2} $ has decaying eigenvalues and is in some $p$ Schatten class (meaning p-summability of the eigenvalues). Than you can conclude that if $(-\Delta)^{\frac s 2 }K$ is in the Hilbert Schmidt class (square integrable), the operator $T$ itself can be written as a comoposition of a $p$-Schatten class operator and a HS operator. Hence, it is in the $Q$-Schatten class for $\frac 1 q= \frac 1 2 + \frac 1 p$ by Hölder.

  • $\begingroup$ Thanks (upvoted). Indeed, the paper is not accessible, even in the literal sense... $\endgroup$
    – dohmatob
    Apr 26, 2021 at 11:36
  • $\begingroup$ It used to be, when I posted the answer. An easier version of the statement can be found (without proof) as Proposition 11 in link.springer.com/article/10.1007/s00220-020-03907-w $\endgroup$ Apr 27, 2021 at 12:08
  • $\begingroup$ Thanks. It turns out the version of the result stated in the linked paper has been taylored for the perculiar needs of the authors. Since the original paper is not accessible, would you mind kindly restating the original results (Propositions 2.1-2.3, or Theorem 1 of Birman-Solomyak, which ever is sufficient to solve the problem for the cube / sphere) as part of your proposed answer ? $\endgroup$
    – dohmatob
    Apr 27, 2021 at 12:35
  • $\begingroup$ It took me about an hour to write above answer. I hope it helps, but it is really hard to read and not so easy to convey. Especially, if you want all the technical details. $\endgroup$ Apr 27, 2021 at 14:41
  • $\begingroup$ Thanks for the additional input. Unfortunately, this involves too much operator theory than I currently understand, and I have trouble getting concrete eigenvalue estimates for $T_K$ (the object I care about) using any of this. So, for concreteness, when all the dust has settled, does any of these general considerations give eigenvalue estimates for $T_K$ when $K(x,x') := \exp(-\|x-x'\|^\gamma/\sigma)$ (the Laplace kernel on the unit-sphere), say ? Thanks in advance, and please don't feel oblaged in any way to give more input on this problem (though that would be very much helpful). Thanks x2! $\endgroup$
    – dohmatob
    May 8, 2021 at 19:52

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