The integrability of $\widehat{e^{-|x|^a}}$, $a>0$

Question

Let $f_a(x)=e^{-|x|^a}$, $x\in \mathbb{R}^n$. Then $f \in L^p(\mathbb{R}^n)$ for every $1\leq p\leq \infty$. It is also smooth away from the origin and decays faster than any polynomial as $|x|\rightarrow \infty$.

By the Hausdorff-Young inequality $\widehat{f_a}\in L^p(\mathbb{R}^n)$ for every $2\leq p\leq \infty$.

What about $1\leq p <2$ ?

Let $\phi$ be a bump function supported in the unit ball $\{|x|<1\}$ that is 1 on the ball $\{|x|<1/2\}$. Then $(1-\phi)e^{-|x|^a}$ is a Schwartz function and so is its Fourier transform.

The question then is: For what $p=p(a)\in [1,2[$ do we have
$\widehat{\phi e^{-|x|^a}} \in L^p(\mathbb{R}^n)$ ? This depends the smoothness of $x\mapsto e^{-|x|^a}$. Notice that $\widehat{\phi e^{-|x|^a}}$ is smooth.

If $a\geq d+1$ then the $\widehat{\phi e^{-|x|^a}}$ decays as fast as $|x|^{-(d+1)}$ which makes it an $L^1(\mathbb{R}^n)$ function, and consequently in every $L^p(\mathbb{R}^n)$ by interpolation. What happens for the smaller and more interesting values of $a$ ?

Answer 1

The Fourier transform of $|x|^b$, $b\notin\mathbb Z$, is the function $c|\xi|^{-1-b}$ away from $\xi =0$. See entry 313 of the table here and the discussion in the last column.

Moreover, the large $\xi$ asymptotics won't change if instead we consider a smoothly cut off version $\phi(x)|x|^b$ since we are just taking an extra convolution with the Schwartz function $\widehat{\phi}$.

Now a Taylor expansion $$ \phi e^{-|x|^a} = \phi (1-|x|^a + \ldots +O(|x|^{Na})) $$ shows that $$ ( \phi e^{-|x|^a})\widehat{\:\:\:}( \xi) \simeq |\xi|^{-1-a} \quad\quad |\xi|\to\infty \quad\quad (a\not= 2,3,\ldots). $$ Note here that $|x|^{Na}$ for sufficiently large $N$ will have a number of derivatives at zero, so its Fourier transform will have faster power decay.

It follows that $( \phi e^{-|x|^a})\widehat{\:\:\:} \in L^p$ precisely when $p>1/(1+a)$, and, in particular, the Fourier transform is in $L^p$ for all $p\ge 1$, for any $a>0$.

Answer 2

Heuristics

Since $e^{-|x|^a}$ behaves like $1-|x|^a$ plus better behaved terms close to zero, and since formally the Fourier trasform of $|x|^a$ is proportional to $|\xi|^{-a-n}$, I would expect that the decay at infinity of your object is something like $|\xi|^{-a-n}$, which by the way gives the correct prediction for $a=1$ (in that case, the Fourier transform is explicit and should be $(1+|x|^2)^{-(n+1)/2}$, up to rescaling and multiplicative constants).

So, the answer should be that it lies in $L^1(\mathbb R^n)$ for all $a>0$ (and actually in any $L^p$ with $n/(n+a)<p\leq\infty$). You can most likely turn it into a proof, but I did not work out all the details.

---

An idea for a different proof of $L^p$ estimates.

In a paper by Weinstein et al. it is proved the estimate for the fractional chain rule:

$$ \|D^s F\circ u\|_{L^r}\lesssim \|G\circ u\|_{L^p}\|D^s u\|_{L^q} $$ for $0<s<1$, $1<p,q,r<\infty$, $1/p+1/q=1/r$, $F$ a continuous (maybe locally Lipschitz) function s.t. $F(0)=0$, and $G$ is a non-negative function such that $$ |F(x)-F(y)|\lesssim (G(x)+G(y))|x-y| $$ (often, you can take $G=|F’|$).

You can also find the statement on some lecture notes of Harmonic Analysis by Monica Visan that are available online (chapter 22).

Then, you can write $\phi (x)e^{-|x|^a}$ as a composite function $F\circ u$, where $$ F(t)=\exp(-1/t)\eta(t), $$ $$ u(x)=|x|^{-a}\chi(x), $$

where $\eta$ is a smooth cutoff function that equals $0$ for $t<2\epsilon$ and equals $1$ for $t>4\epsilon$, and $\chi$ is a cutoff function on $\mathbb R^n$ like $\phi$, that equals $1$ for $|x|<(\epsilon)^{-1/a}$.

Now you can try to use the above estimate with some $r\leq 2$ (the larger the better) and prove that $\|D^s\phi e^{-|x|^a}\|_{L^r}$ is finite. By Young inequality, this is implies a weighted estimate for the Fourier transform of your function:

$$ \||\xi|^{s}\widehat{\phi e^{-|x|^a}}\|_{L^{r’}}<\infty. $$

Then you can turn that weighted estimate into an estimate in $L^q$ with $q<r’$ by Hölder’s inequality :)

It could work, since computing fractional derivatives of $|x|^a$ is probably not so hard. I did not try to apply this argument because the calculations are a little messy, and I don’t know if you can go all the way to $L^1$, but if you have enough patience you could try to apply this method to a given example with fixed $n$ and $a$. There are surely better proofs for this simple case, but this method works for much more general compositions of functions, without symmetry or smoothness.

Answer 3

I would like to thank @Lorenzo Pompili and @Christian Remling for their insightful answers.

I found a straight-forward proof in Lemma 2.1 in "C. Miao, B. Yuan, B. Zhang, Well-posedness of the Cauchy problem for the fractional power dissipative equations, Nonlinear Anal. 68 (2008) 461–484."