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Most theoretical papers concerning kernels assume that they are given a positive definite kernel. In this question, we want to show that a specific kernel is positive definite.

We are interested in the following kernel defined for $0\leq x,y \leq 1$: $$K(x,y) = (x+y)^{3/2} - |x-y|^{3/2}.$$ Numerical simulations have shown that this kernel is positive definite, that is, for all non trivial $\phi : (0,1) \mapsto \mathbb{R}$, we have the following inequality: $$ \int_0^1 \int_0^1 K(x,y) \phi(x) \phi(y) > 0.$$ Can this fact be proved? Where should I look for references?

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We show below a slightly more general claim (to simplify notation, I'll write only in terms of matrices). $\newcommand{\reals}{\mathbb{R}}$

Def. We say a kernel $\psi: X \times X \to \reals$ is negative definite (nd) if $\sum_{ij}c_ic_j\psi(x_,x_j) \le 0$ for all $c$ such that $\sum_i c_i = 0$.

Lemma. Let $\psi$ be cpd. Fix some $x_0 \in X$. Then, $$k(x,y) = \tfrac12[\psi(x,x_0)+\psi(y,x_0) - \psi(x,y)-\psi(x_0,x_0)]$$ is positive definite.

Proof. Lemma 2.1 in Harmonic analysis on Semigroups, by van den Berg, Christensen, Ressel (BCR)

Corollary. Let $0 \le \alpha < 2$. Then, $$k(x,y) := (x+y)^\alpha-|x-y|^\alpha,$$ is a positive definite kernel on $[0,1] \times [0,1]$.

Proof. First, observe that $-(x+y)^\alpha$ is nd (Corollary 2.11, BCR). Also, $|x-y|^\alpha$ is nd (Thm. 2.2, BCR). Consider, now $$\psi(x,y) = -(x+y)^\alpha + |x-y|^\alpha.$$

Using the lemma above it follows that $$k(x,y) = \psi(x,0)+\psi(y,0) - \psi(x,y)-\psi(0,0) = -\psi(x,y)$$ is positive definite.

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  • $\begingroup$ Thank you very much for this precise answer and this awesome reference. I will definitely read (BCR) thoroughly. $\endgroup$
    – cs89
    Commented May 14, 2014 at 11:34
  • $\begingroup$ You're welcome! That book is great, you'll enjoy it. $\endgroup$
    – Suvrit
    Commented May 14, 2014 at 15:45
  • $\begingroup$ Sorry to come back to this. I have read (BCR) and I think I understand the proofs. My initial question included a strict inequality. I am having a hard time proving it, because I think one cannot derive the strict positivity of an integral operator from properties of the matrices. Indeed, working on matrices implies that you must take some kind of a limit to recover results on the integral operator. And, of course, this limit will not preserve any kind of strict inequality you may have obtained on the matrices. $\endgroup$
    – cs89
    Commented May 22, 2014 at 12:57
  • $\begingroup$ Indeed, proving strict positive definiteness requires more work (apparently, I only read the first sentence of your original question and the formula for the kernel function but did not notice the strict inequality!) Once I get a chance, I'll try to update with pointers that may help in establishing the strictness. $\endgroup$
    – Suvrit
    Commented May 22, 2014 at 16:56
  • $\begingroup$ $-(x+y)^\alpha$ is nd when $\alpha \in [1,2]$ only, right? $\endgroup$
    – user72012
    Commented Mar 31, 2020 at 21:19

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