Define the kernel functions for $a\ge 1$,

$$ G_a(t,x) := \frac{C_a t}{t^{1+1/a}+|x|^{1+a}}, \qquad \forall t>0,\: x\in \mathbb{R}\;, $$

where the constant $C_a$ is some normalization constant such that $\int_R G_a(t,x) d x =1$. Clearly, when $a=1$, this kernel function $G_1(t,x)$ is nothing but the Poisson kernel function (see e.g., Yoshida's Functional Analysis, p. 268) with $C_1=1/\pi$, which satisfies the semi-group property:

$$ \int_R G_1(t-s,x-y) G_1(s,y) d y = G_1(t,x) \;. $$

The problem is that whether this is true for general $a>1$? If it is not true in general, is there any possibility to handle the integral like the above convolution? More precisely, are there some ways to calculate or bound from below the following integral:

$$ \int_R G_a(t-s,x-y) G_a(s,y) d y \ge \;? \qquad \forall t>0, x\in R, $$

with $a>1$.

By the way, this kernel function has a nice scaling property:

$$ G_a(t,x) = \frac{1}{t^{1/a}} G_a\left(1,\frac{x}{t^{1/a}}\right)\;. $$

Edit: According to Prof. Hans Engler's comments, here is another related question: what is the inverse Fourier transform of the function

$$ \exp(-|\xi|^a),\qquad a>1. $$

I checked the book "Tables of Integral Transform V.I" (Erdelyi). There are only cases that $a=1$ and $a=2$. This is related to the symmetric stable law.

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    $\begingroup$ Your kernel resembles the function appearing as an upper estimate for the fundamental solution of the Cauchy problem for the space-fractional heat equation (see the book by Eidelman, Ivasyshen and myself, Analytic Methods in the Theory of Differential and Pseudo-differential Equations of Parabolic Type, Birkhauser, 2004). I do not know the origins of your questions, but maybe you should better deal with the space-fractional heat equation where much more is known? $\endgroup$ – Anatoly Kochubei Mar 2 '12 at 19:08
  • $\begingroup$ Thanks Professor Anatoly Kochubei, the problem is indeed from the space-fractional heat equation. I need to estimate the convolutions of these upper bounds. I will have a look of your book. Thanks a lot! :-) $\endgroup$ – Anand Mar 2 '12 at 21:06
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    $\begingroup$ You might also look at recent work by Cabr\'e and Roquejoffre, e.g. arxiv.org/abs/1202.6072 . $\endgroup$ – Hans Engler Mar 3 '12 at 19:48
  • $\begingroup$ Thanks Professor Hans Engler for this reference. It looks very interesting. :-) $\endgroup$ – Anand Mar 3 '12 at 21:04

Let $H_a(t,\xi) = \hat G_a(t,\xi)$ be the spatial Fourier transform of the kernel under study and let $h_a(\xi) = H_a(1,\xi)$. Then $H_a(t,\xi) = H_a(1,t^{1/a}\xi) = h_a(t^{1/a}\xi)$, from the scaling property stated in your question. The postulated semigroup property now implies that for all $\xi$ and all $t, s > 0$ $$h_a(t^{1/a}\xi)h_a(s^{1/a}\xi) = h_a((t+s)^{1/a}\xi) $$ and therefore $$ h_a(t^{1/a}\xi) = e^{A(\xi)t} $$ for some function $A(\xi)$ that may still depend on $a$. Since your kernel is assumed to be isotropic, $A$ can depend only on $|\xi|$. Another scaling argument then shows that $A(\xi) \propto |\xi|^a$ with a negative proportionality constant. Now if $a = 1$, then $A(\xi) \propto -|\xi|$, leading to the Poisson semigroup, which indeed has a kernel of this form. If $a = 2$, then you obtain the heat semigroup, with a very different kernel.

Your question boils down to asking whether for $a \ne 1$, the inverse Fourier transform of $e^{-|\xi|^a}$ is of the form $(1+|x|^{1+a})^{-1}$, up to scaling. This is not true for $a = 2$ and it cannot be true for positive even values of $a$, since in these cases $h_a(\xi)$ is analytic at $\xi = 0$ and hence its inverse Fourier transform must have moments of infinite order. I very much doubt that it is true for other $a \ne 1$.

Consult the book by Zolotarev on stable distributions for more detail. As Dr. Kochubei already stated, the book and papers by Eidelman are also helpful. Finally, there is a website on fractional differentiation and applications at http://www.fracalmo.org/, maintained by F. Mainardi, that may be helpful.

  • $\begingroup$ Thanks Professor Hans Engler. I am also doubtful now about my statement. It might be true only for $a=1$. By the way, do you know what is the Fourier inverse transform of functions $\exp(-|\xi|^a)$ with $a>1$. Thank you very much! :-) $\endgroup$ – Anand Mar 3 '12 at 15:11

In fact, when $0<\alpha<1$, the following estimate has been shown: $$G_{\alpha}(t,x)\leq \frac{t}{{t^{1/\alpha}+x^2}^{n/2+\alpha}}.$$ When $\alpha>1$, a similar result holds by a similar proof. See Miyajima: "Gaussian estimates of order $\alpha$ and $L^p$ spectral independence of generators of $C_0$ Semigroups", Positivity 11 (2007), 15-39.

  • $\begingroup$ Thanks Professor Huang, I will have a look of your reference. $\endgroup$ – Anand Apr 21 '12 at 12:03

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