Is there a feature mapping for this kernel $k(x,y) = (\frac{\min(x,y)}{\max(x,y)})^2$?

Question

In the following paper: https://perso.crans.org/besson/publis/mva-2016/MVA_2015-16__Kernel_Methods__Homework__Besson_Clement_Zerbib.en.pdf

problem 2, Kernel 9. it is shown that $K(x,y) = \frac{\min(x,y)}{\max(x,y)}$ is a positive definite kernel.

I am asking myself, if there is a feature mapping $\phi$ in a Hilbert space such that

$$\langle \phi(x) , \phi(y) \rangle = K(x,y)^2 = k(x,y)$$

"Motivation": The Lorentz factor in special relativity has the form:

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

If we let $k(v_1,v_2) = (\frac{\min(v_1,v_2)}{\max(v_1,v_2)})^2$ then it is a product of two positive kernels, hence a p.d.k and the Euclidean distance between the two points in Hilbert space is:

$$d(v_1,v_2) := \sqrt{k(v_1,v_1)+k(v_2,v_2) - 2k(v_1,v_2)} = \sqrt{1+1-2\frac{v_1^2}{v_2^2}} = \sqrt{2}\sqrt{1-\frac{v_1^2}{v_2^2}}$$

so the Lorentz factor might be written as:

$$\gamma = \frac{\sqrt{2}}{d(v,c)}$$

where it would be interesting to interpret this further as the reciprocal of the distance between two points in some Hilbert space.

Also note the interesting fact, that:

$$k(v,v)= 1$$

which means that each $v$ lies in unit sphere of some Hilbert space.

(In QM those vectors are associated with pure states.)

(Also asked at MSE in case it is not research related: https://math.stackexchange.com/questions/4181618/is-there-a-feature-mapping-for-this-kernel-kx-y-frac-minx-y-maxx-y )

Answer 1

The native Hilbert-space of $K^2$ is well known.

I assume the domain of $K$ is $\mathbb{R}^{>0}\times\mathbb{R}^{>0}.$ Note that: $$ K(x,y)=\begin{cases} \frac{x}{y} \text{ for } x\leq y\\ \frac{y}{x} \text{ for } x\geq y.\\ \end{cases}$$ Set $x = \exp u$ and $y = \exp v$ to obtain $$ K(x,y)^2=K(\exp 2u, \exp 2v)=\exp\big(-2\lvert u-v\rvert\big)=G_2(u,v).$$

This means $K^2$ is the pullback kernel of $G_2$ by the pullback map $\log.$ For this pullback it is true (see for example V. Paulsen, M. Raghupathi Theorem 5.7) that

• $f\in\mathcal{H}(K^2)$ if and only if $f=g\circ\log$ for $g\in\mathcal{H}(G_2)$
• and $\lVert f\rVert_{K^2}=\lVert g\rVert_{G_2}.$

This specifies $\mathcal{H}(K^2)$ completely in terms of $\mathcal{H}(G_2).$ The space $\mathcal{H}(G_2)$ is equal as a set of functions to the Sobolev Space $W^{1,2}(\mathbb{R})$. It's norm is $$ \lVert g\rVert^2_{G_2}=4 \left[\int_\mathbb{R}g^2(u)\,du+4\int_\mathbb{R}(g'(u))^2\,du\right].$$ (see S. Saitoh, Y.Sawano Theorem 1.7)