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It is known that $(x,y)\in \mathbb{R}^2 \mapsto \min(x,y)$ is a positive definite kernel. Can we generalize this result in the following way : Let $X$ be any set and $f,g:X\longrightarrow \mathbb{R}^{+}$ be non-negative functions. Is $$k:(x,y)\in X^2 \mapsto \min(f(x)g(y),f(y)g(x))$$ a positive definite kernel ?

I tried to write $k$ as an integral over indicator functions in order to prove that for all vector $a$ $\sum_{i,j} a_i a_j k(x_i,x_j) \ge 0$, but I could not write this sum as a squared term because $x$ and $y$ appear in both arguments of "min"

Sycorax
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dada
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1 Answers1

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Yes, it is. The proof is as follows,

$$K(x, y) = \min(f(x)g(y), f(y)g(x))=\min(\frac{f(x)}{g(x)}, \frac{f(y)}{g(y)})g(x)g(y)$$ $$\min(\frac{f(x)}{g(x)}, \frac{f(y)}{g(y)})=\int\mathbb{1}_{[0, \frac{f(x)}{g(x)}]} \mathbb{1}_{[0, \frac{f(y)}{g(y)}]} = \langle \mathbb{1}_{[0, \frac{f(x)}{g(x)}]}, \mathbb{1}_{[0, \frac{f(y)}{g(y)}]} \rangle = \langle \phi(x), \phi(y) \rangle$$

By Aronszajn theorem using $\phi(x) = \mathbb{1}_{[0, \frac{f(x)}{g(x)}]}$ , $\min(\frac{f(x)}{g(x)}, \frac{f(y)}{g(y)})$ is positive definite kernel.

$g(x)g(y)$ is trivially a positive kernel as it is a product.

$K(x, y)$ is thus the product of two positive definite kernels, so it is a positive definite kernel.

Sun Haozhe
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