1
$\begingroup$

According to the theory of screw functions and screw lines by John Von Neumann and Issai Schoenberg (see here), any function $F:\mathbb{R} \rightarrow \mathbb{R}$ such that $F(|x_i - x_j|) = \|f(x_i)-f(x_j)\|_2^2$ for some function $f:\mathbb{R} \rightarrow H$ (for some Hilbert space $H$) can be written as:

$F(t) = \int_0^\infty \frac{\sin^2(tx)}{x^2} d\gamma (x)$

where $\int_1^\infty x^{-2} \gamma'(x) dx$ is finite.

Experimentally, I've determined that the function $F(t) = 1-e^{-|t|}$ has the property that $F(|x_i-x_j|)$ can be written as $\|f(x_i)-f(x_j)\|_2^2$. How would I go about finding $\gamma$ for this function (or any such function $F$?).

$\endgroup$
5
  • 5
    $\begingroup$ $e^{-|t|} = C\int_0^\infty \frac{\cos(2tx)}{4x^2+1}dx$ $\endgroup$
    – reuns
    Dec 12, 2019 at 15:25
  • $\begingroup$ How did you come up with this? Apologies if this is all common konwledge. (Also, how would you come up with such a formula given some arbitrary function F?) $\endgroup$ Dec 12, 2019 at 17:21
  • 1
    $\begingroup$ @TimothyChu, you want #207 from this table. $\endgroup$ Dec 12, 2019 at 19:09
  • 1
    $\begingroup$ By (inverse) Fourier transform? $\endgroup$
    – lcv
    Dec 12, 2019 at 19:09
  • $\begingroup$ Ah, great. Thanks! I feel silly for asking such a basic question :). $\endgroup$ Dec 14, 2019 at 3:13

0

Your Answer

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