Recall that a function $f: \mathbb{R}^d \longrightarrow \mathbb{C}$ is positive definite, iff for all numbers $N$ and $x_1, \dots, x_N \in \mathbb{R}^d$, the matrix $(a_{ij})$ with entries $$a_{ij} = f(x_j - x_i)$$ is positive semi-definite.
My question is: Let $g \in L^2(\mathbb{R}^n)$. Is the function $$f(z, y) =\int_{\mathbb{R}^n} g(x - y) \overline{g(x-z)} d x$$ positive definite on $\mathbb{R}^{2n}$?
Define a unitary representation $U$ of $\mathbf R^{2n}$ on $L^2(\mathbf R^n)$ by $(U_{z,y}g)(w)=g(w+z-y)$. (Special case for $G=\mathbf R^n$ of the so-called two-sided regular representation of $G\times G$ on $L^2(G)$.) Then we have \begin{align} (g,U_{z,y}g)&=\int_{\mathbf R^n}\overline{g(w)}g(w+z-y)\,dw\\ &=\int_{\mathbf R^n}\overline{g(x-z)}g(x-y)\,dx\\ &=f(z,y). \end{align} Thus $f$ is a diagonal matrix coefficient of a unitary representation, and as such is well known to be positive definite: indeed $a_{ij}= f(a_j-a_i)=(U_{a_i}g,U_{a_j}g)$ gives $\sum_{i,j}a_{ij}\bar c_ic_j=\left\|\sum_ic_iU_{a_i}g\right\|^2\geqslant 0$.
