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.

  • $\begingroup$ 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

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.

  • 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

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 $n=3$:

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


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