1
$\begingroup$

Let $B$ denote the unit ball in $\mathbb{R}^d$, and suppose $f\colon B\rightarrow\mathbb{C}$ has the property that for every $n\geq1$ and $x_1,\ldots,x_n\in\mathbb{R}^d$ with $\|x_i-x_j\|<1$, the $n\times n$ matrix $[f(x_i-x_j)]_{ij}$ is positive semidefinite. Can $f$ be extended to a positive definite function over $\mathbb{R}^d$?

$\endgroup$

1 Answer 1

1
$\begingroup$

This is true for $d=1$ (M. Krein) but not true for $d>1$ (W. Rudin). For a criterion of extension see

O. Jorgensen, R. Niedzialomski, Extension of positive definite functions, J. Math. Anal. Appl. 422 (2015), no. 1, 712–740.

$\endgroup$
3
  • 1
    $\begingroup$ I thought of Rudin's result, but are the hypotheses not slightly different? In the Krein and Rudin results one considers $f$ defined on $\Omega-\Omega$ for some domain $\Omega$ $\endgroup$
    – Yemon Choi
    Jun 13, 2017 at 13:40
  • 1
    $\begingroup$ isn't $B_1=B_{1/2}-B_{1/2}$? $\endgroup$ Jun 13, 2017 at 17:18
  • $\begingroup$ @PietroMajer Good point, I was being slow. Let me be continue being slow: the framework of Krein and Rudin is that $[f(x_i-x_j)]_{ij}$ should be PSD whenever all the $x_i$ come from $\Omega$. This condition is clearly implied by Dustin's condition; is it clear that they are equivalent? $\endgroup$
    – Yemon Choi
    Jun 13, 2017 at 21:13

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.