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 $\mathbb{R}^d$?
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asked Jun 13, 2017 at 1:05
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1 Answer
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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.
answered Jun 13, 2017 at 11:15
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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 $\Omega$ $\endgroup$ Commented Jun 13, 2017 at 13:40
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1$\begingroup$ isn't $B_1=B_{1/2}-B_{1/2}$? $\endgroup$ Commented Jun 13, 2017 at 17:18
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$\begingroup$ @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? $\endgroup$ Commented Jun 13, 2017 at 21:13