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.