8
$\begingroup$

Let $$\Phi(x) = \int_{\mathbf{R}^n} e^{-|y|^n +i (x,y)} dy.$$ Is $\Phi$ positive everywhere in $\mathbf{R}^n$?

Could someone helps me answer this question or gives a reference for it? Thanks.

$\endgroup$
  • $\begingroup$ For n=2 the answer is yes. $\endgroup$ – user1688 Jan 15 '16 at 8:04
  • $\begingroup$ For n=1 too.... $\endgroup$ – Jean Duchon Jan 15 '16 at 12:33
17
$\begingroup$

Here is a full characterization.

Theorem. The function $e^{-|x|^a}$ is positive definite for $0\le a \le 2$ and is not positive definite for $a>2$. Thus, its Fourier transform is positive and non positive for the said ranges.

Proof. The claim $a=0$ is trivial. Assume $0<a<2$. Then, it can be shown that \begin{equation*} -|x|^a = C_a\int_{-\infty}^\infty \frac{\cos (xt) - 1}{|t|^{a+1}}dt, \end{equation*} where $C_a$ is a constant depending on $a$. Since $\cos(xt)$ is a positive definite function, we see that $-|x|^a$ is conditionally positive definite (because of the $-1$ term). Hence, $\exp(-|x|^a)$ is positive definite, and consequently by Bochner's theorem, it's FT is positive.

For $a>2$, it is easy to construct numerical examples where the associated function is not positive definite, and hence its FT is not positive. Carlo Beenakker's answer gives an example.

To obtain a formal proof of this, here's an outline by contradiction. In particular, suppose that for some $a > 2$, the kernel $e^{-|x-y|^a}$ is positive definite, $|x-y|^a$ is negative definite. Thus, by appealing to Schoenberg's theorem, it must be the case that $d(x,y) := |x-y|^{a/2}$ is a metric on $\mathbb{R}$. Choosing $x,y,z=(0,1,2)$ and comparing $d(x,y)=d(y,z)=1$ but $d(x,z)=2^{a/2}>2$, a contradiction to the triangle inequality.

Reference. Chapter 5, Positive definite matrices, R. Bhatia. Princeton University Press, 2007.

$\endgroup$
  • 1
    $\begingroup$ Not sure if you answered the right question. Note that dimension of the domain is also $n$. $\endgroup$ – Dirk Jan 15 '16 at 14:44
  • $\begingroup$ @Dirk: I answered the question written in the title...I did not look at the main text :-) $\endgroup$ – Suvrit Jan 15 '16 at 14:47
  • $\begingroup$ I guess $|x|$ is the euclidean norm in n dimensions... $\endgroup$ – Dirk Jan 15 '16 at 14:54
  • $\begingroup$ Does not Schoenberg's theorem work in $\mathbb{R}^n$? $\endgroup$ – Fedor Petrov Jan 15 '16 at 14:54
  • $\begingroup$ @FedorPetrov thanks (for your rhetoric question!) all of what I said goes thru trivially to $R^n$...incl. Schoenberg's theorem $\endgroup$ – Suvrit Jan 15 '16 at 15:13
15
$\begingroup$

no, it is not positive everywhere; notice that the integral only depends on $|x|$, so we may orient the vector $x$ along the $x_1$ axis; going to hyperspherical coordinates we have

$$\Phi_n(x)=\frac{2\pi^{(n-1)/2}}{\Gamma[\tfrac{1}{2}(n-1)]}\int_0^\infty dr \int_0^\pi d\phi\; e^{-r^n}\cos\bigl(|x|r\cos\phi\bigr)r^{n-1}(\sin\phi)^{n-2}$$

I checked that it becomes negative for $n=3$:

$$\Phi_3(x)=\frac{4\pi}{|x|}\int_0^\infty e^{-r^3}r\sin(|x|r)\,dr$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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