Characterization of positive-definite kernels via Fourier transforms of measures A function $x\in\mathbb{R}\rightarrow f(x)\in\mathbb{C}$ is said to be positive semidefinite if, for every integer $n$ and any collection $\{x_j\}_{j=1,\dots,n}\subset\mathbb{R}$, the matrix $\left(f(x_j-x_k)\right)_{j,k}$ is positive semidefinite; that is, if the inequality
\begin{equation}
\sum_{j,k=1}^n z_j\overline{z_k}f(x_j-x_k)\geq0
\end{equation}
holds for every $\{x_j\}_{j=1,\dots,n}\subset\mathbb{R}$ and $\{z_j\}_{j=1,\dots,n}\subset\mathbb{C}$. By Bochner's theorem, continuous positive semidefinite functions can be uniquely reconstructed as the Fourier transform of a finite real measure $\mu$:
\begin{equation}
f(x)=\int e^{-isx}\,\mathrm{d}\mu(s).
\end{equation}
Now, more generally consider a two-variable function (kernel) $(x,y)\in\mathbb{R}^2\mapsto F(x,y)\in\mathbb{C}$. Similarly as above, the kernel is said to be positive semidefinite if, for every integer $n$ and any collection $\{x_j\}_{j=1,\dots,n}\subset\mathbb{R}$, the matrix $\left(F(x_j,x_k)\right)_{j,k}$ is positive semidefinite, that is,
\begin{equation}
\sum_{j,k=1}^n z_j\overline{z_k}F(x_j,x_k)\geq0
\end{equation}
for every choice of the parameters.
Clearly, translation-invariant positive semidefinite kernels ($F(x+w,y+w)=F(x,y)$) are (uniquely identifiable with) positive semidefinite functions. In this regard, it would be natural to expect a generalization of Bochner's theorem to positive semidefinite kernels, my guess being the following: there exists a finite real measure $\mu$ on $\mathbb{R}^2$ such that
\begin{equation}
F(x,y)=\iint e^{-i(sx+s'y)}\mathrm{d}\mu(s,s').
\end{equation}
Indeed, it is easy to show that such functions are always positive semidefinite; the case of a translation-invariant kernel would be recovered by assuming $\mu$ to be supported on the line $s'=-s$.
However, I was not able to find any such characterization for positive semidefinite kernels. Is there any counterexample? Or, is this characterization actually true? I am aware of the link between such kernels and reproducing kernel Hilbert spaces, but it is not obvious to me how to reconstruct such an integral expression from them.
 A: A partial answer to my question can be found in this paper by J. Buescu, A.C. Paixao, F. Garcia, and I. Lourtie. They address the particular case of $L^2(\mathbb{R}^2)$ kernels, that is,
\begin{equation}
\iint|F(x,y)|^2\,\mathrm{d}x\,\mathrm{d}y<\infty,
\end{equation}
whence their Fourier transform $\hat{F}(s,t)$ is well-defined and $L^2(\mathbb{R}^2)$ as well. Then it is easy to show that $F(x,y)$ is positive definite if and only if $\hat{F}(s,-t)$ is; that is, $F(x,y)$ is positive definite if and only if it can be written as
\begin{equation}
F(x,y)=\frac{1}{(2\pi)^2}\iint e^{-\mathrm{i}(sx-ty)}\hat{F}(s,t)\,\mathrm{d}s\,\mathrm{d}t.
\end{equation}
Importantly, while being a positive semidefinite kernel, $\hat{F}(s,t)$ is not positive-valued; $F(x,y)$ can indeed be written as the Fourier-Stieltjes transform (up to a sign) of the complex-valued absolutely continuous measure $\hat{F}(s,t)\,\mathrm{d}s\,\mathrm{d}t$, whose density defines a positive semidefinite kernel.
In the general case, relaxing the $L^2(\mathbb{R}^2)$-finiteness assumption, I'm still not aware of analogous results; I would expect something similar with non-absolutely continuous measures, but at the moment that is just a guess.
