1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ This holds as seen by substitution $\endgroup$ Commented Jul 9, 2014 at 18:43

1 Answer 1

3
$\begingroup$

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 $\sum_{i,j}a_{ij}\bar c_ic_j=\left\|\sum_ic_iU_{a_i}g\right\|^2\geqslant 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.