I am undergraduate Physics student and understand that this is a professional mathematics forum. But due to perhaps broader interest, I hope this question is suitable for this website.


I understand the Fourier transform is a Hilbert space isometry 
$F:L^2(\mathbb R^d) \rightarrow L^2(\mathbb R^d).$


In Quantum Mechanics we are told that intuitively the Fourier transform transforms narrow signals into extended ones, I would like to make this precise. 


The best example which illustrate this, is when one considers the Fourier transform on Schwartz distributions, $F(\delta\_x)(k)=e^{ikx},$ but this is very unquantitative.


A good decay measure in "Fourier space" seem to be Sobolev spaces since 


$$\left\lVert f \right\rVert\_{H^s(\mathbb R^d)} = \left( \int\_{\mathbb R^d} \left\lvert F(f) \right\rvert^2 (1+\vert x \vert^2)^{s} \ dx \right)^{1/2}$$


Let us fix a signal $f \in L^2(\mathbb R^d)$ of unit norm. 


I call the signal $f$ to be $\varepsilon,\delta-$localized if there is a ball $B(x,\delta)$ such that $\left\lVert f1\_{B(x,\delta)} \right\rVert\_{L^2} \ge 1-\varepsilon$.


I would say that the intuitive explanation of the Fourier transform should then imply that if $f$ is in the $H^s$ Sobolev space, with $s>0$ and $\left\lVert f \right\rVert\_{H^s} \le k$ then for any $\delta(k)>0$ there is $\varepsilon'(\delta(k))>0$ such that $f$ cannot be $\varepsilon'(\delta(k)),\delta(k)-$localized. 
Is this true? 
Can one find this $\varepsilon'$ explicitly?


The above question is motivated by the statement that if a function is in $H^s$ with bounded norm, then it decays sufficiently rapidly in Fourier space that it cannot be too localized in real space.



Making the Fourier transform quantitative
Notice that $\|f1\_{B(0,\delta)}\|\_{L^2}\to 0$ as $\delta\to 0$. So the answer to the question is no because it should reasonably include the requirement $\varepsilon'<1$. -- Rapid decrease of the Fourier transform $F(f)$ implies smoothness of $f$ but, without further assumptions, it does not imply localization.


