Let $G_\beta$, $0 < \beta < n$, denote the Bessel kernel whose Fourier transform is defined by the formula $$ \widehat{G}_\beta(\xi) = (1+|\xi|^2)^{-\beta/2} =: \langle \xi\rangle^{-\beta}. $$ Let $\phi$ be a smooth compactly supported function such that $\phi(x) = 1$ for $|x|\le 1/4$, $\phi \ge 0$ everywhere, and $\operatorname{supp}\phi\subset B(0,1/2) = \{x\in\mathbb R^n:|x|\le 1/2\}$ (we may assume $\phi$ is radial for convenience if necessary). I would like to show that the tempered distribution $$ u := (1-\phi)G_\beta $$ is actually a Schwartz function, or even simply that it is a function with rapidly decaying Fourier transform. This is a claim in a paper that is made in passing without proof, and I have tried to prove it myself for a while now without much luck.
I think it is a cute result if it is true because I would not have expected even $G_\beta$ to be rapidly decaying, rather I thought it would behave like $|x|^{\beta-n}$ for large values of $|x|$. I think it is possible to show that, assuming $u$ is a function, that $u$ is rapidly decaying by considering that $\Delta^{N}\langle\xi\rangle^{-\beta}$ is integrable if $N\ge 10n$, say, ($\Delta$ is the Laplacian) and then "integrating by parts" $$ u(x) = c_N|x|^{-2N}(1-\phi(x))\int \Delta^N\langle \xi\rangle^{-\beta} e^{2\pi ix\cdot\xi}\,d\xi. $$ But this is somewhat sketchy, and I think that as far as proving that $u$ is a function with rapidly decaying Fourier transform, the smoothness of $u$ is more important. Any help is appreciated!
Intuition.
In dimension $1$, this is just the fact that the Fourier transform of $(1+|x|^2)^{-1}$ is $\pi\,e^{-2\pi|x|}$
This one is important in physics, called the Yukawa potential. Again, this is because it is the Fourier transform of $(1+|x|^2)^{-1}$, but this time in dimension $3$.
General bounds
You have a nice representation formula for Bessel potentials that you can find in Stein's book Singular Integrals and Differentiability Properties of Functions, in the form $$ G_\beta(x) = \frac{\omega_\beta}{2} \int_0^\infty t^{(\beta-n)/2-1}\,e^{-\pi t}\,e^{-\pi|x|^2/t}\,\mathrm d t $$ where $\omega_\beta = \frac{2\,\pi^{\beta/2}}{\Gamma(\beta/2)}$ (that is $\omega_\beta = |\Bbb S^{\beta-1}|$ if $\beta\in\Bbb N$). This is easy to prove by using the following formula $$ \frac{2}{\omega_\beta\,r^{\beta/2}} = \int_0^\infty t^{\frac{\beta}{2}-1} \,e^{-\pi r t}\, \mathrm d t $$ which follows by the simple change of variable $s = \pi\,r\,t$ in the left-hand side integral and the definition of the Gamma function. Then the formula for $G_\beta$ follows by taking $r = 1+|x|^2$ and the formula of the Fourier transform of a Gaussian.
With this formula, upper bounds on $G_\beta$ are easier to get. For example, if you want the large $x$ behavior, you can do the change of variable $t\to t\,|x|$ to get $$ G_\beta(x) = \frac{\omega_\beta}{2\,|x|^{(n-\beta)/2}} \int_0^\infty t^{(\beta-n)/2-1}\,e^{-\pi|x|(t+1/t)}\,\mathrm d t \\ = \frac{\omega_\beta\,e^{-2\pi|x|}}{2\,|x|^{(n-\beta)/2}} \int_0^\infty t^{(\beta-n)/2-1}\,e^{-\pi|x|(\sqrt{t}-1/\sqrt{t})^2}\,\mathrm d t $$ and so if $|x|\geq 1$ $$ G_\beta(x) \leq \frac{\omega_\beta\,e^{-2\pi|x|}}{2\,|x|^{(n-\beta)/2}} \int_0^\infty t^{(\beta-n)/2-1}\,e^{-\pi(\sqrt{t}-1/\sqrt{t})^2}\,\mathrm d t. $$ Since the integral on the right-hand-side of the above formula is converging and does not depend on $x$, you can write this as $$\boxed{ G_\beta(x) \leq C_{n,\beta}\,\frac{e^{-2\pi|x|}}{|x|^{(n-\beta)/2}}} $$ when $|x|\geq 1$ (or $|x|\geq 1/4$ as well ...).
Getting higher derivatives is not more difficult. Indeed, using the first formula for $G_\beta$ and differentiating $k$ times gives a formula of the form $$ \nabla^k G_\beta(x) = \frac{\omega_\beta}{2} P_k(x) \int_0^\infty t^{(\beta-n)/2-1}\,Q_k(1/t)\,e^{-\pi t}\,e^{-\pi|x|^2/t}\,\mathrm d t $$ for some polynomials $P_k$ and $Q_k$ of degree $k$. The same manipulations as above will then give you when $|x|\geq 1$ a bound of the form $$\boxed{ |\nabla^k G_\beta(x)| \leq C_{n,\beta,k}\,\frac{e^{-2\pi|x|}}{|x|^{(n-\beta-2k)/2}}} $$
The above bound on $G_\beta$ is not optimal, working a bit more you can normally get a bound of the form $$ G_\beta(x) \leq C_{n,\beta}\,\frac{e^{-2\pi|x|}}{|x|^{(n-\beta+1)/2}} $$ when $|x|\geq 1$. In particular, this agrees with the particular cases $G_{n+1}$ and $G_{n-1}$ given above. These functions are also related to Kummer's hypergeometric functions for which there are known asymptotic developments, that you can use if you want an even more precise asymptotic behavior.