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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

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

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

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

then we must have

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

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

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

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

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

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

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

completing the square, we obtain

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

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

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

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

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

Using (2) we have

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

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

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

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

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

On the other hand, by differentiating the Fourier analytic identity

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(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.