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How can I prove that the following triangular kernel function defined in $[0, 1] \subset R^1$

$k(x, x') = (1 - 2|x-x'|)$

is a positive semidefinite function?

It turns out to be psd function when using a numerical simulation tool (Matlab) by checking psd of a kernel matrix..

I found out that $k(x, x') = (1 - |x-x'|)^+$ is a psd function using Bochner's theorem. Also the lower bound of isotropic kernel $k$ to be psd in $R^1$ is -1..

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    $\begingroup$ Can't you just compute the Fourier transform of $1-2|t|$ (on $[-1,1]$) and check if it's non-negative. $\endgroup$
    – Christian Remling
    Commented Aug 25, 2014 at 14:39
  • $\begingroup$ I checked the Fourier transform and the result wasn't non-negative.. $\endgroup$
    – Sungjoon Choi Samuel
    Commented Aug 25, 2014 at 14:46
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    $\begingroup$ So doesn't that prove that $k$ is not positive definite (since Bochner's theorem gives a characterization)? $\endgroup$
    – Christian Remling
    Commented Aug 25, 2014 at 15:10

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Actually the Fourier coefficients of $1 - 2 |t|$ on $[-1,1]$ are $$ \int_{-1}^1 (1 - 2 |t|) \exp(-\pi i t n)\; dt = \cases{ 0 & for even $n$\cr 8/(n^2 \pi^2) & for odd $n$\cr}$$ so as a kernel on $[0,1]$ this is positive semidefinite, i.e. the operator on $L^2[0,1]$ given by $$ Tf(x) = \int_0^1 k(x,y) f(y)\; dy$$ is positive semidefinite. However, as a kernel on $\mathbb R$ (extended to be $0$ outside $[0,1]\times [0,1]$) it is not positive semidefinite.

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  • $\begingroup$ I have two more questions. Is there any reason why you use the Fourier series instead of the Fourier transform? The second is that why is the (extended) kernel function not psd? Thank you in advance! $\endgroup$
    – Sungjoon Choi Samuel
    Commented Aug 25, 2014 at 18:34

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