Characterization of the non-negative definite functions $f(x,y)$  Hello,
The common definition of the non-negative definite functions is as follows:
Definition 1: A continuous complex-valued function $f(x)$ is called non-negative definite, if for any real numbers $x_1,\dots,x_m$ and complex numbers $\xi_1,\dots,\xi_m$, one has 
$$
\sum_{k,j=1}^m f(x_k-x_j) \xi_k \bar{\xi}_j \ge 0\:.
$$
See wikipedia for example. 
For my use, I need the following more general definition:
Definition 2: A continuous complex-valued function $f(x,y)$ is called non-negative definite, if for any real numbers $x_1,\dots,x_m$ and complex numbers $\xi_1,\dots,\xi_m$, one has 
$$
\sum_{k,j=1}^m f(x_k,x_j) \xi_k \bar{\xi}_j \ge 0\:.
$$
For the first definition, there is a Characterization by Bochner's theorem, which is discussed in the post. 
My question is, for the second definition, is there a characterization like Bochner's theorem? A necessary condition is that $f(x,y)$ is Hermitean in the sense that
$$
f(x,y) = \bar{f}(y,x)\:.
$$
Clearly, we need more conditions to guarantee that $f(x,y)$ is non-negative definite. Does any one know a way to determine whether a given function  $f(x,y)$ is non-negative definite or not? 
Thank you very much for any hints! 
Anand

EDIT:
Thanks Mikael de la Salle for the useful link. After reading, I still have some problems.
As one might know, Definition 1 above can be extended to distributions. For example, a distribution $F$ is positive-definite, if for every $\psi\in C_c^\infty(R^d)$, 
$$
\left(F,\psi *\widetilde{\psi}\right)\ge 0,
$$
where $\widetilde{\psi}(x)=\psi(-x)$. So it is natural to think that Definition 2 can also be extended to, say, $C_c^\infty(R^d)\otimes C_c^\infty(R^d)$, or $C_c^\infty(R^{2d})$. The condition in the book Kazhdan's property (T) that the kernel function should be continuous seems too restrictive. Are there some more general statements?

Second EDIT:
Here is a more explicit statement of the problem:
Theorem: Every translation-invariant positive-definite Hermitian bilinear functional $B(\phi,\psi)$ on $C_c^\infty(R^d)$ has the form
$$
B(\phi,\psi)= \int \hat{\phi}(\lambda) \overline{\hat{\psi}(\lambda)} d \mu(\lambda)
$$
where $\mu$ is some positive tempered measure and $\hat{\psi}$, $\hat{\phi}$ are the Fourier transforms, respectively, of $\psi$ and $\phi$.
See Gel'fand Volumn 4 P.169 Theorem 6. . My problem is
   What happens if we remove the "translation--invariant" condition?

Thanks a lot!
Anand
 A: Such a function is called function of positive type. For characterizations, see for example appendix C in the book Kazhdan's property (T) by Bekka, de la Harpe and Valette, available freely on Bachir Bekka's webpage.
A: On a Lie group, every positive-definite distribution is the distributional derivative of some continuous, positive-definite function. See Aarnes, Johan F.
Distributions of positive type and representations of Lie groups. 
Math. Ann. 240 (1979), no. 2, 141–156. 
A: In light of Anand's comments on translation invariant kernels of the form $f(x-y)$, I thought I'd mention the class of completely monotone functions $\left((-1)^k f^{(k)}(x) \ge 0\right)$---used in the famous Bernstein and Widder theorem
This theorem allows us to characterize completely monotone functions of the form $k(x,y) = f(x+y)$, by showing that equivalently they must be Laplace transforms of a positive measure.
A: It seems to me that if you drop the translation invariance condition on $B(\phi,\psi)$, then you simply end up with a symmetric quadratic form $B(\phi)=B(\phi,\phi)$. In that case the main simplification I can think of is the spectral representation:
$$ B(\phi) = \int_{\operatorname{spec}(B)} ~ \lambda ~ \mathrm{d}\mu_\phi(\lambda) , $$
where $\operatorname{spec}(B)$ is the spectrum of $B$ and $\mathrm{d}\mu_\phi$ is the spectral measure associated to $\phi$.
The spectral theory of quadratic forms is discussed in section VIII.6 of Reed & Simon's Methods of Modern Mathematical Physics (vol. 1). The main theorem is (VIII.15): If $B$ is a semibounded quadratic form, then it is the quadratic form $B(\phi)=(\phi,A\phi)$ of a unique self-adjoint operator $A$. The inner product in your case is the obvious one on $L^2(\mathbb{R})$ and the spectral representation of $B$ is obtained directly from the spectral representation of $A$. Also, since your $B$ is positive semidefinite, the spectrum will also be bounded $\operatorname{spec}(B)\ge 0$.
When $B$ is translation invariant, the unitary transformation that takes $A$ to its diagonal form is automatically the Fourier transform and $\mathrm{d}\mu_\phi(\lambda) = |\hat{\phi}(\lambda)|^2\mathrm{d}\lambda$ (well, up to a measurable reparametrization of $\lambda$ really), where $\hat{\phi}(\lambda)$ is the Fourier transform of $\phi(x)$. In the generic case, short of explicitly diagonalizing $B$ or $A$, I don't see a simple way of obtaining $\mathrm{d}\mu_\phi$.
