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Show that if $f$ and $g$ are continuous positive definite functions in $\mathbf{R}^1$, then $f(x)g(y)$ is positive definite on $\mathbf{R}^2$. I just wanted to check if the approach I'm using is correct,

I defined the matrix $A_{j,k}=f(x_j-x_k)$ and $B_{j,k}=g(y_j-y_k)$ are positive definite functions for all $x_1 \cdots x_N$ and $y_1 \cdots y_N$. Would this imply that $A_{j,k} B_{j,k}=f(x_j-x_k)g(y_j-y_k)$ is also positive definite?

Also, how exactly do I use Bochner's Theorem to show the following?

Really appreciate the help, thanks!

Anon
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  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Mar 31 '20 at 01:11
  • Indeed yes. Do you know Bochner's theorem about continuous psd functions? – kimchi lover Mar 31 '20 at 01:16
  • @kimchilover yeah i do know bochner's theorem, but wasn't too sure on how to implement the theorem for this! – Anon Mar 31 '20 at 01:26
  • If you don't want to use Bochner's Theorem, you can also prove that given two pd matrices $A_{i,j}$ and $B_{i,j}$, their entry-wise product $C_{i,j} = A_{i,j}B_{i,j}$ is also p.d. Just decompose $A = S^\ast S$ and and reorder. – Adrián González Pérez Mar 31 '20 at 11:37
  • @AdriánGonzález-Pérez I'm quite curious on how to apply Bochner's theorem for this question. It would be great if someone could help me with that. – Anon Mar 31 '20 at 23:30
  • Sure, I can add a short answer if you will. – Adrián González Pérez Apr 01 '20 at 07:42

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Bochner theorem states that $f$ is positive definite iff $f = \widehat{\mu}$, with $\mu$ a positive Radon measure. If $f_1$, $f_1$ are positive definite then: $$ f_1 \cdot f_2 = \widehat{\mu_1} \cdot \widehat{\mu_2} = \big(\mu_1 \ast \mu_2 \big)^\wedge, $$ and the convolution of two positive measures is positive.

Adrián González Pérez
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