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 $\mathbb{R}^{2n}$?



Positive Definiteness of a certain function
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 $\sum\_{i,j}a\_{ij}\bar c\_ic\_j=\left\|\sum\_ic\_iU\_{a\_i}g\right\|^2\geqslant 0$.


