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.. 



How can I prove that the negative biased triangular kernel is positive semidefinite
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. 


