# Support of functions in Fourier domain

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

• 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. Commented Jul 9, 2018 at 18:49
• To be more precise, the sharp function $R(C_2, C_1, \epsilon)$ can be found by Christian Remling's argument to be $$R = \sqrt{\left[1 + \frac{C_2 - C_1}{\epsilon}\right]^{1/k} - 1}$$ Commented Jul 10, 2018 at 3:54

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.
• 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). Commented Jul 9, 2018 at 19:41