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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 Schwartz 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 $R(C_1,C_2,\varepsilon)$?

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    $\begingroup$ This just follows from straightforward book-keeping: if $\|\chi_{|x|>R}g\|^2\ge \epsilon^2$ (writing $g=\widehat{f}$) and you need to keep $\|x^k g\|$ below a certain bound, then the best you can do is concentrate the whole mass near $x=R$, so you'd get $C_2\ge \epsilon^2 R^{2k}$, and your inequality follows if $R$ is so large that this fails. $\endgroup$ – Christian Remling Jul 9 '18 at 18:49
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    $\begingroup$ To be more precise, the sharp function $R(C_2, C_1, \epsilon)$ can be found by Christian Remling's argument to be $$ R = \sqrt{\left[1 + \frac{C_2 - C_1}{\epsilon}\right]^{1/k} - 1} $$ $\endgroup$ – Willie Wong Jul 10 '18 at 3:54
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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.

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  • $\begingroup$ The OP also assumes that $\int x^2 |u|^2 \le C$, though, and this does give a bound (see my comment above for this). $\endgroup$ – Christian Remling Jul 9 '18 at 19:41

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