I was recently required to put together a detailed proof of Hardy's uncertainty principle in a trial lecture, a part of my PhD defense, and as such scoured the internet for good proofs. As is often the case with classical proofs, time has not been kind to the original arguments and while others have summarized parts of the proof, a detailed description is hard to come by. Before our LLM overlords make this sort of expository work superfluous (you might argue it already is) I thought I would put my version up online for others to use in the (not too distant) future.
At its core, Hardy's uncertainty principle places limits on how aggressively a function and its Fourier transform can decay at infinity and says that with this notion of localization, the Gaussian function is optimal. Later work has extended the theorem to generalizations of the Fourier transform and higher dimensions and used the theorem to place limits on solutions to PDEs such as the Schrödinger equation. For this post, we'll just stick to a straightforward proof of the theorem based on the Phragmén-Lindelöf principle which is similar to the original proof given by Hardy.
Let \( f \in L^2(\mathbb{R}) \) satisfy \( |f(x)| \leq C_1 e^{-\pi a x^2} \) and \( |\hat{f}(\omega)| \leq C_2 e^{-\pi b \omega^2} \). Then
We will start off with the case $ab = 1$ as the \( ab > 1 \) case follows from it. Note first that we can without loss of generality assume that \( C_1 = C_2 = 1 \) by dividing \( f \) by e.g., \( C_1 + C_2 \). Next we show that we can set \( a = 1 \) without loss of generality. Indeed, assume the theorem holds true for \( a = 1 \) and let \( g(x) = f(x/\sqrt{a}) \), then \( \hat{g}(\omega) = \sqrt{a} \hat{f}(\sqrt{a}\omega) \) and the bounds on \( f \) imply that $$ |g(x)| \leq e^{-\pi x^2},\qquad |\hat{g}(\omega)| \leq \sqrt{a}e^{-\pi ab \omega^2}. $$ Applying the theorem to \( g \), we get that \( g(x) = Ce^{-\pi x^2} \) which in turn yields that \( f(x) = Ce^{-\pi a x^2} \) which is what we wished to show. From now on we are free to assume that \( |f(x)| \leq e^{-\pi x^2} \) and \( |\hat{f}(\omega)| \leq e^{-\pi \omega^2} \).
Next up we will extend \( \hat{f} \) to an entire function on the complex plane. Specifically, we define $$ F(z) = \int_{\mathbb{R}} f(u)e^{-2\pi i z u} \,du $$
This standard fact from complex analysis follows from this lemma based on Morera's theorem.
Let \( f : \mathbb{R} \times \mathbb{C} \to \mathbb{C} \) be a measurable function such that
That our integrand is analytic in $z$ is clear. To construct our $f_K$, note that we trivially have the bound $$ |f(u)| |e^{-2\pi i z u}| \leq e^{-\pi u^2} e^{2\pi u \operatorname{Im}(z)} \leq e^{-\pi u^2} e^{2\pi u |z|} $$ Now for $z \in K$ we can bound $|z|$ uniformly by the compactness of $K$ by e.g., $M_K = \sup_{z \in K} |z|$. Integrability follows from the fact that $e^{-\pi u^2}$ dominates the $e^{2\pi u M_K}$ factor.
We now move to bounding \( |F| \). Using \( |f(x)| \leq e^{-\pi x^2} \) and \( |e^z| = e^{\operatorname{Re}(z)} \), we get $$ |F(x+iy)| \leq \int_{\mathbb{R}} |f(u)| |e^{2\pi i z u}|\,du \leq \int_{\mathbb{R}} e^{-\pi u^2 + 2 \pi y u}\,du $$ The quantity in the exponent can be written as $$ -\pi u^2 + 2 \pi y u = -\pi(u^2 - 2y u + y^2) + \pi y^2 = -\pi(u - y)^2 + \pi y^2, $$ and hence, $$ |F(x+iy)| \leq \int_{\mathbb{R}} e^{-\pi u^2 + 2 \pi y u}\,du = e^{\pi y^2} \underbrace{\int_{\mathbb{R}} e^{-\pi(u-y)^2}\,du}_{= 1} = e^{\pi y^2}. $$ Next up we will define the function \( G(z) = e^{\pi z^2} F(z) \). Clearly, \( G \) is entire just like \( F \). Moreover, \( |G| \) is bounded by \( 1 \) along the real and imaginary axes. Indeed, by \( |F(x)| \leq e^{-\pi x^2} \), $$ |G(x)| = |e^{\pi x^2}| |F(x)| \leq 1 $$ and by \( |F(iy)| \leq e^{\pi y^2}, \) $$ |G(iy)| = |e^{\pi (iy)^2}| |F(iy)| \leq 1. $$ Our goal is to show that $G$ is bounded on the entire complex plane so that we may apply Liouville's theorem and conclude that \( G(z) = C \) as rearranging \( G(z) = e^{\pi z^2} F(z) \) would then ultimately allow us to conclude that \( f \) is of the desired Gaussian form.
However, showing boundedness of \( G \) along the axes is not enough to show boundedness on the entire complex plane. Our tool for extending this will be the Phragmén-Lindelöf principle.
Let \(S = \{ z \in \mathbb{C} : \alpha < \arg z < \beta \}\) be a sector of \( \mathbb{C} \) and \( F \) a function which is analytic in \( S \) that is continuous on \( \overline{S} \). If \( |F(z)| \leq M \) on \( \partial S \) and $$ |F(z)| \leq C e^{c |z|^\rho} $$ for some \( c, C > 0 \) and \( \rho < \frac{\pi}{\beta - \alpha} \), then \( |F(z)| \leq M \) on \( \overline{S} \).
Let's start off by verifying the type of growth \( G \) has on \( \mathbb{C} \). Using the \( |F(x+iy)| \leq e^{\pi y^2} \) result and \( |e^z| = e^{\operatorname{Re}(z)} \), we find that $$ |G(z)| = |G(x+iy)| = |e^{\pi (x+iy)^2}| |F(x+iy)|\leq |e^{\pi (x^2 + 2ixy - y^2)}| e^{\pi y^2} = e^{\pi x^2} \leq e^{\pi |z|^2} $$ which in the language of the theorem means that \( \rho = 2 \). This means that we cannot apply the theorem to the first quadrant with \( \alpha = 0 \) and \( \beta = \pi/2 \) because \( \frac{\pi}{\beta - \alpha} = 2 \not > \rho=2 \). To get around this we will need to consider a slightly narrower sector. Specifically, for a parameter \( \delta > 0 \), define $$ S_\theta = \{ z \in S : 0 < \arg z < \theta \} $$ where \( \theta = \theta(\delta) \) is dependent on \( \delta \). Our function \( G \) will not be bounded on \( \partial S_\theta \) so we will define an additional function \( H_\delta \) which is. Specifically, we choose \( H_\delta(z) = e^{i\delta z^2} G(z) \). On the real axis, we trivially have the bound $$ |H_\delta(x)| = |e^{i\delta x^2}| |G(x)| \leq 1. $$ The remainder of the boundary \( \partial S_{\theta(\delta)} \) may be parametrized by the ray \( z = re^{i\theta} \). Along it, using \( |G(x+iy)| \leq e^{\pi x^2} \) and \( |e^z| = e^{\operatorname{Re}(z)} \), we have that $$ \begin{cases} |G(z)| \leq e^{\pi r^2 \cos(\theta)^2},\\ |e^{i\delta z^2}| = e^{-\delta r^2 \sin(2\theta)} \end{cases}\qquad |H_\delta(re^{i\theta})| \leq \exp(r^2 \cos\theta(\pi\cos\theta - 2 \delta \sin\theta)) $$ by collecting factors. Now in order for this quantity to be bounded along the ray, we want the exponent to be negative. Since \( r^2 \cos\theta > 0\) on the entire quadrant, this means that we want for $$ \begin{align} \pi\cos\theta - 2 \delta \sin\theta &< 0\\ \iff \pi\cos\theta &< 2 \delta \sin\theta\\ \iff \frac{\pi}{2\delta} &< \tan\theta. \end{align} $$ To enforce this, we can set $\theta(\delta) = \arctan\frac{\pi}{\delta}$. With this choice of $\theta(\delta)$, $H_\delta$ is bounded along the ray. Before applying Phragmén-Lindelöf, we need to assure ourselves that we still have $\rho = 2$ growth. Indeed, $$ |H_\delta(z)| = |e^{i\delta z^2}| |G(z)| \leq e^{\delta |z|^2} e^{\pi |z|^2} \leq e^{2\pi |z|^2} $$ using the growth bound we had already established for $|G|$. Now since $\rho = 2 < \frac{\pi}{\theta(\delta)-0}$, we are free to apply Phragmén-Lindelöf and conclude that $|H_\delta(z)| \leq 1$ for all $z \in S_{\theta(\delta)}$ for all $\delta > 0$. To lift this to a bound on $G$ on $S_{\pi/2}$, fix a $z_0 \in S_{\pi/2}$ and $\varepsilon > 0$. Since $\theta(\delta) \to \pi/2$ as $\delta \to 0$, there exists a $\delta_0 > 0$ such that $\delta < \delta_0 \implies z_0 \in S_{\theta(\delta)}$. By the continuity of the function $z \mapsto e^{i\delta z^2}$, we can find another $\delta_1$ such that $$ \delta < \delta_1 \implies |e^{-i\delta z_0^2} - 1| < \varepsilon. $$ For $\delta < \delta_0, \delta_1$, we can use the bound $|H_\delta(z_0)| \leq 1$ to conclude that $$ |H_\delta(z_0)| = |e^{i \delta z_0^2}| |G(z_0)| \leq 1 \implies |G(z_0)| \leq |e^{-i \delta z_0^2}| < 1+\varepsilon. $$ Since $\varepsilon$ was arbitrary, we conclude that $|G(z_0)| \leq 1$. The bound further holds on $\overline{S_{\pi/2}}$ since we already established that $|G(iy)| \leq 1$.
At this point, all that remains is to show that $G$ is bounded on the other three quadrants.
Define $H_\delta(z) = e^{-i\delta z^2}$, $\theta(\delta) = \pi - \arctan\frac{\pi}{\delta}$ and $S_{\theta(\delta)} = \{ z \in \mathbb{C} : \theta(\delta) < \arg z \leq \pi \}$. Then clearly $H_\delta$ is still bounded by $1$ along the real axis and its growth is bounded by $e^{2\pi |z|^2}$ for the same reason as for the $H_\delta$ in the first quadrant. As for bounding $H_\delta$ along the ray $re^{i\theta(\delta)}$, we have that $$ \begin{cases} |G(z)| \leq e^{\pi r^2 \cos(\theta)^2},\\ |e^{-i\delta z^2}| = e^{\delta r^2 \sin(2\theta)} \end{cases}\qquad |H_\delta(re^{i\theta})| \leq \exp(r^2 \cos\theta(\pi\cos\theta + 2 \delta \sin\theta)). $$ This time, $\cos\theta < 0$ so a negative exponent is contingent on $$ \begin{align} \pi\cos\theta + 2 \delta \sin\theta &> 0\\ \iff \pi\cos\theta &> -2 \delta \sin\theta\\ \iff \frac{\pi}{2\delta} &< -\tan\theta. \end{align} $$ Our choice of $\theta(\delta) = \pi - \arctan\frac{\pi}{\delta}$ satisfies this inequality since $\tan(\pi-x) = -\tan(x)$. From here the same Phragmén-Lindelöf and $\varepsilon-\delta$ arguments apply to bound $H_\delta$ and in turn $G$ from above by $1$.
Here we again choose $H_\delta(z) = e^{i\delta z^2}$ but $\theta(\delta) = \pi + \arctan\frac{\pi}{\delta}$ and $S_{\theta(\delta)} = \{ z \in \mathbb{C} : \pi < \arg z < \theta(\delta) \}$. Bounding along the real and imaginary axes follows the standard formula. Along the $\theta$ ray we have $$ \begin{cases} |G(z)| \leq e^{\pi r^2 \cos(\theta)^2},\\ |e^{i\delta z^2}| = e^{-\delta r^2 \sin(2\theta)} \end{cases}\qquad |H_\delta(re^{i\theta})| \leq \exp(r^2 \cos\theta(\pi\cos\theta - 2 \delta \sin\theta)) $$ just as in the first quadrant. Since $\cos\theta$ is negative inside the third quadrant, a negative exponent requires $$ \begin{align} \pi\cos\theta - 2 \delta \sin\theta &> 0\\ \iff \pi\cos\theta &> 2 \delta \sin\theta\\ \iff \frac{\pi}{2\delta} &< \tan\theta \end{align} $$ and it can be easily verified that this is satisfied by our choice of $\theta$. The remaining details are identical to those of the first and second quadrants.
For the last quadrant, we choose $H_\delta(z) = e^{-\delta z^2}$, $\theta(\delta) = -\arctan \frac{\pi}{\delta}$ and $S_{\theta(\delta)} = \{ z \in \mathbb{C} : \theta(\delta) < \arg z < 0 \}$. We can proceed by the same arguments to bound $G$ on this quadrant as well.
Having bounded $G$ by $1$ over all of $\mathbb{C}$, Liouville's theorem allows us to conclude that $G$ is the constant function $G(z) = C$. Since $G(x) = e^{\pi x^2} \hat{f}(x) = C$, this means that $$ \hat{f}(x) = Ce^{-\pi x^2} \implies f(x) = Ce^{-\pi x^2} $$ which finishes the proof of the $ab = 1$ case.
Now for the $ab > 1$ case, assume that we have a function $f$ such that $|f(x)| \leq e^{-\pi a x^2}$ and $|\hat{f}(\omega)| \leq e^{-\pi b \omega^2}$. Then setting $a_0 = \frac{1}{b} < \frac{ab}{b} = a$, it holds that $|f(x)| \leq e^{-\pi a_0 x^2}$ and $|\hat{f}(\omega)| \leq e^{-\pi b \omega^2}$. As such, we can apply the $ab=1$ version of Hardy's uncertainty principle since $a_0 b = 1$ and conclude that $$ f(x) = Ce^{-\pi a_0 x^2}. $$ However, combining this fact with that $|f(x)| \leq e^{-\pi a x^2}$, we can conclude that $$ |Ce^{-\pi a_0 x^2}| \leq e^{-\pi a x^2} \implies |C| \leq e^{-\pi (a-a_0) x^2} $$ for all $x \in \mathbb{R}$. Since $a > a_0$, this can only hold true if $C = 0$ meaning that $f(x) = 0$ for all $x$, which is what we wished to show.