We consider the convolution product:
\begin{align*} \varphi(\xi)=(1+|\xi|^2)^{2/3}*\frac{e^{-|\xi|^2}}{\pi^{d/2}} \end{align*}
for all $\xi\in\mathbb{R}^d$ such that $|\xi|\ge1$.
I want to prove that there existe a constant $c>0$ such that
$| \varphi(\xi)|\le (1+|\xi|^2)^{2/3}$
for all $\xi\in\mathbb{R}^d$ such that $|\xi|\ge1$.
Thanks
In explicit terms, $$ \frac{\varphi(\xi)}{(1+|\xi|^2)^{2/3}} = \frac{1}{\pi^{d/2}}\int_{\mathbb{R}^d}\left(\frac{1+|\xi-z|^2}{1+|\xi|^2}\right)^{2/3} e^{-|z|^2}\,d\mu $$ and in order to prove that the RHS is bounded by an absolute constant is it enough to split $\mathbb{R}^d$ as $\{|z|\leq |\xi|\}\cup\{|z|>|\xi|\}$. Since $|\xi|\geq 1$, $$ \int_{|z|\leq |\xi|}\left(\frac{1+|\xi-z|^2}{1+|\xi|^2}\right)^{2/3} e^{-|z|^2}\,d\mu\leq 4^{2/3}\int_{\mathbb{R}^d}e^{-|z|^2}\,dz =C_0$$ and similarly $$ \int_{|z|> |\xi|}\left(\frac{1+|\xi-z|^2}{1+|\xi|^2}\right)^{2/3} e^{-|z|^2}\,d\mu\leq \int_{\mathbb{R}^d}(3|z|^2)^{2/3}e^{-|z|^2}\,dz =C_1$$ so $$ \frac{\left|\varphi(\xi)\right|}{(1+|\xi|^2)^{2/3}}\leq C_0+C_1.$$ Sharper bounds can be derived from Young's convolution inequality.