# How to prove the min function is a kernel [closed]

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

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

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

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