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.

  • $\begingroup$ Possibly something here is related to what you want? $\endgroup$ – 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.

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There are various mathematical formulations of this phenomenon and some of them are quite quantitative.

First, there is the classical Heisenberg uncertainty principle which roughly states that the product of the variances of a functions and its Fourier transform is bounded from below, independently of the function.

Then there is the uncertainty principle of Donoho and Stark. Its discrete form says that for every $x\in\mathbb{C}^N$ 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 $$|T||W|\geq (1- \epsilon_T-\epsilon_W)^2.$$

There is much more to say, e.g. there is Lieb's uncertainty principle for the short-time-Fourier transform, there are results of "time-and-band-limiting operators" by Landau/Pollack and Slepian/Landau. I can recommend the books "Foundations of Time-Frequency Analysis" by Karlheiz Gröchenig and "Ten Lectures on Wavelets" by Ingrid Daubechies for further reading.

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