Making the Fourier transform quantitative

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.

• Possibly something here is related to what you want? – Nate Eldredge Oct 15 '18 at 4:07

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.
Then there is the uncertainty principle of Donoho and Stark. Its discrete form says that for some $$x\in\mathbb{C}^N$$ it holds that product of the number of non-zero entries in $$x$$ and the number of non-zero entries in its Fourier transform $$\hat x$$ is greater than $$N$$. The continuous version is a bit more complicated: Call some function $$f$$ $$\epsilon$$-concentrated on a set $$T$$ if there is some $$g$$ with support in $$T$$ such that $$\|f-g\|_2\leq \epsilon$$ (i.e. the $$L^2$$-norm of $$f$$ outside of $$T$$ is smaller than $$\epsilon$$. The uncertainty principle says that for $$f$$ and $$\hat f$$ with unit norm such that $$f$$ is $$\epsilon_T$$ concentrated on some set $$T$$ and $$\hat f$$ is $$\epsilon_W$$ concentrated on some set $$W$$ it holds that $$|T||W|\geq (1- \epsilon_T-\epsilon_W)^2.$$