Polynomial growth of Fourier transform

Question

I have a function $f$ on $\mathbf{R}$ which is smooth, bounded, and in $L^{1+\epsilon}(\mathbf{R})$ for all $\epsilon>0$. I wish to conclude that the Fourier transform $\widehat f$ has polynomial growth rate, or even that $|f(x)| = O((1+\epsilon)^x)$ for all $\epsilon>0$. Is something like this known? Any references would be appreciated!

Answer 1

The function $$ h(x)=\frac{1}{\sqrt{2\,\pi}}\,\Bigl(\frac{\sin(x/2)}{x/2}\Bigr)^2 $$ is in $L^p$ for all $p\ge1$ and its Fourier transform is the triangle function: $$ \hat h(\xi)=\max(1-|\xi|,0). $$ Let $$ f(x)=\sum_{n=1}^\infty2^n\,2^{-n^2}\,e^{-inx}h(2^{-n^2}x). $$ If $p>1$ then $$ \|f\|_p\le\Bigl(\sum_{n=1}^\infty2^n\,2^{-n^2(1-1/p)}\Bigr)\|h\|_p<\infty. $$ On the other hand $$ \hat f(\xi)=\sum_{n=1}^\infty2^n\,\hat h\bigl(2^{n^2}(\xi-n)\bigr) $$ and $\hat f(n)=2^n$, so that $\hat f$does not have polynomial growth.