I was reading a Navier-stokes paper by Kato, and he affirm that if $k\geq0$ than $$\mathcal{F}^{-1}(| \xi |^ke^{-|\xi|^2})\in L^1(\mathbb{R}^2)\cap L^\infty(\mathbb{R}^2)$$
(here $\mathcal{F}^{-1}$ denotes the inverse Fourier Transform).
Does anyone knows how to do it or where I can find something to help me with this?
On
If what @amsmath said on the comments is true,i.e. $|\xi|^ke^{|\xi|^2}\in \mathcal{S}$ (here $\mathcal{S}$ denotes the Schwarz space). Then:
Note that $\mathcal{F}$ is an isomorphism of $\mathcal{S}$ into itself (you can see this result in corollary 8.28 from Folland's Real Analysis). Hence $\mathcal{F}^{-1}(|\xi|^ke^{-|\xi|^2})\in\mathcal{S}$.
Using the fact that $\mathcal{S}\subset L^1\cap L^\infty$, you conclude the result.
Index.
0: Introduction.
Denoting $G(x):=e^{-|x|^2}$ and $f:=|\nabla|^k G$, the non-trivial part is to show that $f\in L^1(\mathbb R^d)$. Here $$ |\nabla|^k h(x):= \mathcal F^{-1} |\xi|^k \hat{h}(\xi).$$
1: Integrating by parts. Here we assume $d=1$. Since $f\in L^\infty$, if we can show that $f(x)=O(|x|^{-2})$ as $|x|\to \infty$ then we can conclude that $f\in L^1$. Now integrating by parts we have $$\begin{split} f(x)&=\int_{-\infty}^\infty |\xi|^ke^{-|\xi|^2} e^{ix\xi}\, d\xi \\ & = \frac{-1}{ix}\left[k\int_{-\infty}^\infty |\xi|^{k-2}\xi e^{-|\xi|^2}e^{i x \xi}\, d \xi - 2 \int_{-\infty}^\infty |\xi|^k\xi e^{-|\xi|^2} e^{i x \xi}\, d\xi\right].\end{split}$$ We claim that both summands in the square bracket are $O(|x|^{-1})$. If $k>1$, this is immediately proven by redoing the integration by parts as before. However, if $0<k< 1$, the first summand poses a problem, because $$\tag{1}\int_{-\infty}^\infty |\xi|^{k-2}\xi e^{-|\xi|^2}e^{i x \xi}\, d \xi = \frac{-(k-1)}{ix}\text{pv}\int_{-\infty}^\infty |\xi|^{k-2} e^{-|\xi|^2}e^{i x \xi}\, d\xi + O(|x|^{-1}), $$ where we have a nonintegrable singularity in the integrand and that's why the integral has to be taken in principal value sense.
At first sight, it is unclear whether the integral is convergent. However we know that the Fourier transform of $|\xi|^{k-2}$ is, up to multiplicative constants, $|x|^{1-k}$ (see here). On the other hand, the Fourier transform of $e^{-|\xi|^2}$ is $e^{-|x|^2}$ up to irrelevant constants. So, rewriting the pv-integral in (1) as a convolution, we find (up to multiplicative constants) $$ |x|^{1-k} \ast e^{-|x|^2} =\int_{-\infty}^\infty |x-y|^{1-k}e^{-|y|^2}\, dy, $$ and here it is clear that the integral on the right-hand side is convergent, thanks to the strong decay of $e^{-|y|^2}$. This proves our claim.
The case $k=1$ is still untreated. In this case, integration by parts produces a Dirac delta: $$\int_{-\infty}^\infty |\xi|^{-1}\xi e^{-|\xi|^2}e^{i x \xi}\, d \xi = \frac{-2}{ix}\int_{\infty}^\infty \delta(\xi)e^{-|\xi|^2}e^{i x \xi}\, d\xi + O(|x|^{-1}), $$ and the integral on the right-hand side evaluates to $1$. The proof is complete.
2: Sobolev embedding.
This is a conditional answer, since it relies on the Sobolev embedding (!) (see below), and I am not entirely sure it is correct for a reason that I will explain below.
Now, if the following inequality is true: $$\tag{!} \| f\|_{L^1(\mathbb R^d)}\le C_d \| |\nabla|^{-d/2} f \|_{L^2(\mathbb R^d)}, $$ then we can argue that $$ \|f\|_{L^1(\mathbb R^d)}^2 \le C_d^2\||\nabla|^{k-d/2} G\|_{L^2(\mathbb R^d)}^2 =C_d^2 \int_{\mathbb R^d} |\xi|^{2k-d}e^{-2|\xi|^2}\, d\xi, $$ and the right hand side is a convergent integral for all $k > 0$.
WHY AM I NOT SURE? The Sobolev embedding (!) is never stated with a negative power of $|\nabla|$ in the right-hand side. I should see if the proof extends to the case at hand.
3: Failed attempt. Here follows a naïve attempt of mine, which fails at the label (WRONG). I am not deleting it because I am convinced that we learn from errors at least as much as we learn from correct answers.
In comments, amsmath suggests that this follows from $|\xi|^k e^{-|\xi|^2}$ being a Schwartz function, which is not entirely correct because the function fails to be smooth at $\xi=0$ when $k$ is not an even integer. However, there is a simple workaround. Consider a smooth cutoff function $\zeta$ that equals $1$ in a neighborhood of $0$ and is compactly supported. Write $$ f= \mathcal F^{-1}(\zeta|\cdot|^k e^{-|\cdot|^2}) + \mathcal F^{-1}((1-\zeta)|\cdot|^k e^{-|\cdot|^2})=f_0 + f_1.$$ Now $f_1$ is the inverse Fourier transform of a genuine Schwartz function and so it is itself a Schwartz function. On the other hand, $$ \begin{split} \lVert f_0\rVert_1 &= \int_{\mathbb R^d} \left\lvert \int_{\mathbb R^d} \zeta(\xi)|\xi|^k e^{-|\xi|^2}e^{i x\xi}\, d\xi\right\rvert\,dx \\ &\le R^k \int_{\mathbb R^d} \left\lvert \int_{\mathbb R^d} \zeta(\xi) e^{-|\xi|^2}e^{i x\xi}\, d\xi\right\rvert\,dx \ \text{(WRONG)}\\ &= R^k\lVert \mathcal F^{-1} (\zeta e^{-|\cdot|^2)}\rVert_1 \end{split} $$ where $R$ is a constant that depends only on the support of $\zeta$. Since $\mathcal F^{-1} (\zeta e^{-|\cdot|^2})$ is again a genuine Schwartz function, its $L^1$ norm is finite and the proof is complete.