TITLE: Proving that a specific kernel is positive definite
QUESTION [6 upvotes]: 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?

REPLY [6 votes]: 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.