[This post was typeset using a LaTeX to WordPress-HTML converter kindly provided to me by Luca Trevisan.]

Many properties of a (sufficiently nice) function \({f: {\mathbb R} \rightarrow {\mathbb C}}\) are reflected in its Fourier transform \({\hat f: {\mathbb R} \rightarrow {\mathbb C}}\), defined by the formula

$$\hat f(\xi) := \int_{-\infty}^\infty f(x) e^{-2\pi i x \xi}\ dx. \ \ \ \ \ (1)$$

For instance, decay properties of \({f}\) are reflected in smoothness properties of \({\hat f}\), as the following table shows:

If \({f}\) is… then \({\hat f}\) is… and this relates to…
Square-integrable square-integrable Plancherel’s theorem
Absolutely integrable continuous Riemann-Lebesgue lemma
Rapidly decreasing smooth theory of Schwartz functions
Exponentially decreasing analytic in a strip
Compactly supported entire and at most exponential growth Paley-Wiener theorem

Another important relationship between a function \({f}\) and its Fourier transform \({\hat f}\) is the uncertainty principle, which roughly asserts that if a function \({f}\) is highly localised in space, then its Fourier transform \({\hat f}\) must be widely dispersed in space, or to put it another way, \({f}\) and \({\hat f}\) cannot both decay too strongly at infinity (except of course in the degenerate case \({f=0}\)). There are many ways to make this intuition precise. One of them is the Heisenberg uncertainty principle, which asserts that if we normalise

$$\int_{{\mathbb R}} |f(x)|^2\ dx = \int_{\mathbb R} |\hat f(\xi)|^2\ d\xi = 1$$

then we must have

$$(\int_{\mathbb R} |x|^2 |f(x)|^2\ dx) \cdot (\int_{\mathbb R} |\xi|^2 |\hat f(\xi)|^2\ d\xi)\geq \frac{1}{(4\pi)^2}$$

thus forcing at least one of \({f}\) or \({\hat f}\) to not be too concentrated near the origin. This principle can be proven (for sufficiently nice \({f}\), initially) by observing the integration by parts identity

$$\langle xf, f' \rangle = \int_{\mathbb R} x f(x) \overline{f'(x)}\ dx = - \frac{1}{2} \int_{\mathbb R} |f(x)|^2\ dx$$

and then using Cauchy-Schwarz and the Plancherel identity.

Another well known manifestation of the uncertainty principle is the fact that it is not possible for \({f}\) and \({\hat f}\) to both be compactly supported (unless of course they vanish entirely). This can be in fact be seen from the above table: if \({f}\) is compactly supported, then \({\hat f}\) is an entire function; but the zeroes of a non-zero entire function are isolated, yielding a contradiction unless \({f}\) vanishes. (Indeed, the table also shows that if one of \({f}\) and \({\hat f}\) is compactly supported, then the other cannot have exponential decay.)

On the other hand, we have the example of the Gaussian functions \({f(x) = e^{-\pi a x^2}}\), \({\hat f(\xi) = \frac{1}{\sqrt{a}} e^{-\pi \xi^2/a }}\), which both decay faster than exponentially. The classical Hardy uncertainty principle asserts, roughly speaking, that this is the fastest that \({f}\) and \({\hat f}\) can simultaneously decay:

Theorem 1 (Hardy uncertainty principle) Suppose that \({f}\) is a (measurable) function such that \({|f(x)| \leq C e^{-\pi a x^2 }}\) and \({|\hat f(\xi)| \leq C' e^{-\pi \xi^2/a}}\) for all \({x, \xi}\) and some \({C, C', a > 0}\). Then \({f(x)}\) is a scalar multiple of the gaussian \({e^{-\pi ax^2}}\).

This theorem is proven by complex-analytic methods, in particular the Phragmén-Lindelöf principle; for sake of completeness we give that proof below. But I was curious to see if there was a real-variable proof of the same theorem, avoiding the use of complex analysis. I was able to find the proof of a slightly weaker theorem:

Theorem 2 (Weak Hardy uncertainty principle) Suppose that \({f}\) is a non-zero (measurable) function such that \({|f(x)| \leq C e^{-\pi a x^2 }}\) and \({|\hat f(\xi)| \leq C' e^{-\pi b \xi^2}}\) for all \({x, \xi}\) and some \({C, C', a, b > 0}\). Then \({ab \leq C_0}\) for some absolute constant \({C_0}\).

Note that the correct value of \({C_0}\) should be \({1}\), as is implied by the true Hardy uncertainty principle. Despite the weaker statement, I thought the proof might still might be of interest as it is a little less “magical” than the complex-variable one, and so I am giving it below.

— 1. The complex-variable proof —

We first give the complex-variable proof. By dilating \({f}\) by \({\sqrt{a}}\) (and contracting \({\hat f}\) by \({1/\sqrt{a}}\)) we may normalise \({a=1}\). By multiplying \({f}\) by a small constant we may also normalise \({C=C'=1}\).

The super-exponential decay of \({f}\) allows us to extend the Fourier transform \({\hat f}\) to the complex plane, thus

$$\hat f(\xi + i \eta) = \int_{\mathbb R} f(x) e^{-2\pi i x \xi} e^{2\pi \eta x}\ dx$$

for all \({\xi, \eta \in {\mathbb R}}\). We may differentiate under the integral sign and verify that \({\hat f}\) is entire. Taking absolute values, we obtain the upper bound

$$|\hat f(\xi + i \eta)| \leq \int_{\mathbb R} e^{-\pi x^2} e^{2\pi \eta x}\ dx;$$

completing the square, we obtain

$$|\hat f(\xi + i \eta)| \leq e^{\pi \eta^2} \ \ \ \ \ (2)$$

for all \({\xi, \eta}\). We conclude that the entire function

$$F(z) := e^{\pi z^2} \hat f(z)$$

is bounded in magnitude by \({1}\) on the imaginary axis; also, by hypothesis on \({\hat f}\), we also know that \({F}\) is bounded in magnitude by \({1}\) on the real axis. Formally applying the Phragmen-Lindelöf principle (or maximum modulus principle), we conclude that \({F}\) is bounded on the entire complex plane, which by Liouville’s theorem implies that \({F}\) is constant, and the claim follows.

Now let’s go back and justify the Phragmén-Lindelöf argument. Strictly speaking, Phragmén-Lindelöf does not apply, since it requires exponential growth on the function \({F}\), whereas we have quadratic-exponential growth here. But we can tweak \({F}\) a bit to solve this problem. Firstly, we pick \({0 < \theta < \pi/2}\) and work on the sector

$$\Gamma_\theta := \{ re^{i\alpha}: r > 0, 0 \leq \alpha \leq \theta \}.$$

Using (2) we have

$$|F(\xi + i\eta)| \leq e^{\pi \xi^2}.$$

Thus, if \({\delta > 0}\), and \({\theta}\) is sufficiently close to \({\pi/2}\) depending on \({\delta}\), the function \({e^{i\delta z^2} F(z)}\) is bounded in magnitude by \({1}\) on the boundary of \({\Gamma_\theta}\). Then, for any sufficiently small \({\epsilon > 0}\), \({e^{i\epsilon e^{i\epsilon} z^{2+\epsilon}} e^{i\delta z^2} F(z)}\) (using the standard branch of \({z^{2+\epsilon}}\) on \({\Gamma_\theta}\)) is also bounded in magnitude by \({1}\) on this boundary, and goes to zero at infinity in the interior of \({\Gamma_\theta}\), so is bounded by \({1}\) in that interior by the maximum modulus principle. Sending \({\epsilon \rightarrow 0}\), and then \({\theta \rightarrow \pi/2}\), and then \({\delta \rightarrow 0}\), we obtain \({F}\) bounded in magnitude by \({1}\) on the upper right quadrant. Similar arguments work for the other quadrants, and the claim follows.

— 2. The real-variable proof —

Now we turn to the real-variable proof of Theorem 2, which is based on the fact that polynomials of controlled degree do not resemble rapidly decreasing functions.

Rather than use complex analyticity \({\hat f}\), we will rely instead on a different relationship between the decay of \({f}\) and the regularity of \({\hat f}\), as follows:

Lemma 3 (Derivative bound) Suppose that \({|f(x)| \leq C e^{-\pi a x^2 }}\) for all \({x \in {\mathbb R}}\), and some \({C, a > 0}\). Then \({\hat f}\) is smooth, and furthermore one has the bound \({|\partial_\xi^k \hat f(\xi)| \leq \frac{C}{\sqrt{a}} \frac{k! \pi^{k/2}}{(k/2)! a^{(k+1)/2}}}\) for all \({\xi \in {\mathbb R}}\) and every even integer \({k}\).

Proof: The smoothness of \({\hat f}\) follows from the rapid decrease of \({f}\). To get the bound, we differentiate under the integral sign (one can easily check that this is justified) to obtain

$$\partial_\xi^k \hat f(\xi) = \int_{\mathbb R} (-2\pi i x)^k f(x) e^{-2\pi i x \xi}\ dx$$

and thus by the triangle inequality for integrals (and the hypothesis that \({k}\) is even)

$$|\partial_\xi^k \hat f(\xi)| \leq C \int_{\mathbb R} e^{-\pi a x^2} (2\pi x)^k\ dx.$$

On the other hand, by differentiating the Fourier analytic identity

$$\frac{1}{\sqrt{a}} e^{-\pi \xi^2/a} = \int_{\mathbb R} e^{-\pi a x^2} e^{-2\pi i x \xi}\ dx$$

\({k}\) times at \({\xi = 0}\), we obtain

$$\frac{d^k}{d\xi^k}(\frac{1}{\sqrt{a}} e^{-\pi \xi^2/a})|_{\xi=0} = \int_{\mathbb R} e^{-\pi a x^2} (2\pi i x)^k\ dx;$$

expanding out \({\frac{1}{\sqrt{a}} e^{-\pi \xi^2/a}}\) using Taylor series we conclude that

$$\frac{k!}{\sqrt{a}} \frac{(-\pi/a)^{k/2}}{(k/2)!} = \int_{\mathbb R} e^{-\pi a x^2} (2\pi i x)^k\ dx$$

◻

Using Stirling’s formula \({k! = k^k (e+o(1))^{-k}}\), we conclude in particular that

$$|\partial_\xi^k \hat f(\xi)| \leq (\frac{2 \pi}{e a}+o(1))^{k/2} k^{k/2} \ \ \ \ \ (3)$$

for all large even integers \({k}\) (where the decay of \({o(1)}\) can depend on \({a, C}\)).

We can combine (3) with Taylor’s theorem with remainder, to conclude that on any interval \({I \subset {\mathbb R}}\), we have an approximation

$$\hat f(\xi) = P_I(\xi) + O( \frac{1}{k!} (\frac{2 \pi}{e a}+o(1))^{k/2} k^{k/2} |I|^k )$$

where \({|I|}\) is the length of \({I}\) and \({P_I}\) is a polynomial of degree less than \({k}\). Using Stirling’s formula again, we obtain

$$\hat f(\xi) = P_I(\xi) + O( (\frac{2 \pi e}{a}+o(1))^{k/2} k^{-k/2} |I|^k ) \ \ \ \ \ (4)$$

Now we apply a useful bound.

Lemma 4 (Doubling bound) Let \({P}\) be a polynomial of degree at most \({k}\) for some \({k \geq 1}\), let \({I = [x_0-r,x_0+r]}\) be an interval, and suppose that \({|P(x)| \leq A}\) for all \({x \in I}\) and some \({A>0}\). Then for any \({N \geq 1}\) we have the bound \({|P(x)| \leq (CN)^k A}\) for all \({x \in NI := [x_0-Nr, x_0+Nr]}\) and for some absolute constant \({C}\).

Proof: By translating we may take \({x_0=0}\); by dilating we may take \({r=1}\). By dividing \({P}\) by \({A}\), we may normalise \({A=1}\). Thus we have \({|P(x)| \leq 1}\) for all \({-1 \leq x \leq 1}\), and the aim is now to show that \({|P(x)| \leq (CN)^k}\) for all \({-N \leq x \leq N}\).

Consider the trigonometric polynomial \({P(\cos \theta)}\). By de Moivre’s formula, this function is a linear combination of \({\cos(j \theta)}\) for \({0 \leq j \leq k}\). By Fourier analysis, we can thus write \({P(\cos \theta) = \sum_{j=0}^k c_j \cos(j \theta)}\), where

$$c_j = \frac{1}{\pi} \int_{-\pi}^\pi P(\cos \theta) \cos(j \theta)\ d\theta.$$

Since \({P(\cos \theta)}\) is bounded in magnitude by \({1}\), we conclude that \({c_j}\) is bounded in magnitude by \({2}\). Next, we use de Moivre’s formula again to expand \({\cos(j \theta)}\) as a linear combination of \({\cos(\theta)}\) and \({\sin^2(\theta)}\), with coefficients of size \({O(1)^k}\); expanding \({\sin^2(\theta)}\) further as \({1 - \cos^2(\theta)}\), we see that \({\cos(j \theta)}\) is a polynomial in \({\cos(\theta)}\) with coefficients \({O(1)^k}\). Putting all this together, we conclude that the coefficients of \({P}\) are all of size \({O(1)^k}\), and the claim follows. ◻

Remark 1 One can get slightly sharper results by using the theory of Chebyshev polynomials. (Is the best bound for \({C}\) known? I do not know the recent literature on this subject. I think though that even the sharpest bound for \({C}\) would not fully recover the sharp Hardy uncertainty principle, at least with the argument given here.)

We return to the proof of Theorem 2. We pick a large integer \({k}\) and a parameter \({r > 0}\) to be chosen later. From (4) we have

$$\hat f(\xi) = P_r(\xi) + O( \frac{r^2}{ak} )^{k/2}$$

for \({\xi \in [-r,2r]}\), and some polynomial \({P_r}\) of degree \({k}\). In particular, we have

$$P_r(\xi) = O( e^{-br^2} ) + O( \frac{r^2}{ak} )^{k/2}$$

for \({\xi \in [r,2r]}\). Applying Lemma 4, we conclude that

$$P_r(\xi) = O( 1 )^k e^{-br^2} + O( \frac{r^2}{ak} )^{k/2}$$

for \({\xi \in [-r,r]}\). Applying (4) again we conclude that

$$\hat f(\xi) = O( 1 )^k e^{-br^2} + O( \frac{r^2}{ak} )^{k/2}$$

for \({\xi \in [-r,r]}\). If we pick \({r := \sqrt{\frac{k}{cb}}}\) for a sufficiently small absolute constant \({c}\), we conclude that

$$|\hat f(\xi)| \leq 2^{-k} + O( \frac{1}{ab} )^{k/2}$$

(say) for \({\xi \in [-r,r]}\). If \({ab \geq C_0}\) for large enough \({C_0}\), the right-hand side goes to zero as \({k \rightarrow \infty}\) (which also implies \({r \rightarrow \infty}\)), and we conclude that \({\hat f}\) (and hence \({f}\)) vanishes identically.