I was given a question in an exam, and it made me realize i don't understand Mercer's condition quite well. I'd be happy for some insight about why my intuition is not right :)
for the question we need to decide if the claim is right or wrong. given the RBF kernel as defined here:
$K(\mathbf{x}, \mathbf{x'}) = \exp(\beta\|\mathbf{x} - \mathbf{x'}\|^2) \>\: \beta >0$
Is there a space F and ψ : X → F for which we can use k above to compute the terms $G_{i j} =\langle ψ(xi),ψ(xj)\rangle$
As for the formal answer they rely on the closure property of PSD Kernels, and show that :
$K(x,x')=exp(\beta||x||^2)\cdot exp(-\beta x^Tx')\cdot exp(\beta||x'||^2)$
and because $-\beta x^Tx'$ is not a PSD kernel so is K.
now there are two main misunderstandings i have. the first is that while
$K(x,x')\space is\space a \space PSD-kernel \longrightarrow f(x)K(x,x')f(y)\space is\space a \space PSD-kernel$
why does it imply that
$f(x)K(x,x')f(y)\space is\space not \space a \space PSD-kernel \longrightarrow K(x,x')\space is\space not \space a \space PSD-kernel$
the second misunderstanding is that looking for the definition of RBF as written in Wikipedia here.
https://en.wikipedia.org/wiki/Radial_basis_function_kernel
if we use the complex function $i\phi(x)$ where's $i^2=-1$ can it be considered a "valid" decomposition to an inner product of k? while it is still not computable because of the infinite dimension I'm not sure if introducing complex numbers is "legal" here and how does it relate to mercer's condition validity or invalidity.
Answer for the first part: since f(x) is invertible in this case by looking at $f^-1(x)K(x,y)f^-1(y)$ if we assume this is a kernel function so is $f(x)f^-1(x)K(x,y)f^-1(y)f(y) = K(x,y)$ which is a contradiction