*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?



Proving that a specific kernel is positive definite
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 $\sum\_i c\_i = 0$.



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> **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.
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*Proof.* Lemma 2.1 in [Harmonic analysis on Semigroups](http://rads.stackoverflow.com/amzn/click/0387909257), by van den Berg, Christensen, Ressel (BCR)



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


