Consider the following function in $\mathbb{R}^3$: $$ f_t(x)=(1+|x|^2)^{-\alpha}e^{-g(x)t},\,\,\,\,\, \text{where}\,\, g(x)=\frac{x^2_1\cdot x^2_2}{1+|x|^2}, $$ where $\frac{1}{2}<\alpha<1$, and the parameter $t$ satisfies $0<t<1$.

I'm interested in the behavior near the origin of the Fourier transform of $f_t(x)$ (in the distributional sense). To be more precisely, let $g_t(\xi)=\mathcal{F}{f_t(x)}$ denote the Fourier transform of $f_t(x)$. Since in the limiting case $t=0$, the Fourier transform can be explicitly computed and we have (using properties of Bessel functions) $$ |g_0(\xi)|\leq C |\xi|^{-3+2\alpha}, \,\,\,\, |\xi|\leq 1. $$

My question is that does this property keep valid uniformly for $0<t<1$? i.e., do we have a constant which doesn't depend on t such that $$ |g_t(\xi)|\leq C |\xi|^{-3+2\alpha}, \,\,\,\, |\xi|\leq 1,\,\,\,0<t<1 ? $$

Edit It seems that one difficulty is that the Fourier transform of $e^{-g(x)t}$ is not quite easy to handle (if $g(x)=|x|^2$, then the case is quite simple). I tried to take the Taylor expansion of this, but it seems to get more troubles. Another attempt is to write $f_t(x)$ in to the following integral form: $$ f_t(x)=\frac{1}{\Gamma(\alpha)}\int_0^{\infty}e^{-(|x|^2+1)s-g(x)t}s^{\alpha}\frac{ds}{s} $$ Then in this case, we have to consider the behavior of the Fourier transform of the "Gaussian like" function $e^{-|x|^2s-g(x)t}$, and show that it's a small "perturbation" of the Fourier transform of the Gaussian $e^{-|x|^2}$. I don't know if there is a good way to study the Fourier transform of this type of functions, which is not radial and not regular enough. Comments and references are welcome.



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