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How to prove that min(x,y) is a kernel. I got some hints online

$$ min(x,y) = \int_0^\infty 1 [s \leq min(x,y)]ds = \int_0^\infty 1 [s \leq x] 1 [s \leq y]ds $$

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  • $\begingroup$ That looks like a solution to me. What else do you need to show? $\endgroup$ Apr 14 at 1:40
  • $\begingroup$ why is that a kernel? the last integral does not look like a kernel to me $\endgroup$
    – Tim Ge
    Apr 14 at 18:17

1 Answer 1

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I understand you want to show that $K(x,y)=\min(x,y)$ is a positive definite kernel.

Let $c_i,\,c_2,\,\ldots c_n\in\mathbb{C}$ and $x_i,\,x_2,\,\ldots x_n\in\mathbb{R}$. Them $$\sum_{i,j=1}^nc_i\overline{c}_jK(x_i,x_j)=\sum_{i,j=1}^nc_i\overline{c}_j \int_0^\infty 1 [s \leq x_i] 1 [s \leq x_j]ds=\int_0^\infty \left|\sum_{i=1}^nc_i(1 [s \leq x_i])\right|^2ds\geq 0.$$

This means that $K(x,y)$ is indeed positive definite.

You can find some more elaborated results in Elaborating Mercer's theorem (RKHS) on Cameron-Martin space k(x,y)=min(x,y) and Positive definite kernels involving the min function.

You can find more results searching for "\(K(x,y)=\min(x,y)\) reproducing kernel" on SearchOnMath.

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