Let $\mathcal F$ be the Fourier transform. I would like to understand whether being in a Sobolev space implies that the Fourier transform of a function is necessarily supported on a compact ball up to some controlled error:


The Sobolev space $W^{k,2}(\mathbb R)$ with $k>0$ is defined by all the $L^2$ functions $f$ such that $ \int\_{\mathbb R} (1+\left\lvert x \right\rvert^2)^{k} \left\lvert \mathcal F(f)(x) \right\rvert^2 \ dx< \infty.$


Let $C\_1\le C\_2$ be given. Consider all functions with $\left\lVert f \right\rVert\_{L^2}=C\_1$ and $ \int\_{\mathbb R} (1+\left\lvert x \right\rvert^2)^{k} \left\lvert \mathcal F(f)(x) \right\rvert^2 \ dx\le C\_2.$


Let $\varepsilon>0. $For which $k$ does there exist an $R(C\_2,C\_1,\varepsilon)>0$ such that


$\left\lVert \mathcal F(f) 1\_{B(0,R)^C} \right\rVert< \varepsilon?$


Ideas:


If $k>0.5$ then one can use the Cauchy-Schwarz inequality to construct such an $R$.
If $C\_2=C\_1$ then $\left\lvert \mathcal F(f)(x) \right\rvert=0$ almost surely and the statement is true for all $k>0.$


So the only question is: Does there exist for $k \in (0,0.5]$ and $C\_2>C\_1$ such a radius $R(C\_1,C\_2,\varepsilon)$?



Support of functions in Fourier domain
No. You may ask the same equivalent question for a function in $L^2$. A function $u$ belongs to $L^2$ means measurability and
$$
\int \vert u(x) \vert^2 dx<+\infty,
$$
which implies
$
\lim\_{R\rightarrow +\infty}\int\_{\vert x\vert\ge R} \vert u(x) \vert^2 dx=0,
$
but the rate of convergence can be arbitrarily slow.


