Heisenberg inequality for the Fourier transform

Remarks

For a Schwartz function $f$ and $a = b = 0$, integrate by parts: $\int \abs{f}^2\, dx = -\int x\, \frac{d}{dx}\abs{f(x)}^2\, dx = -2 \operatorname{Re} \int x f(x) \overline{f'(x)}\, dx$. The Cauchy–Schwarz inequality bounds this by $2 \norm{x f}_2 \norm{f'}_2$. By the differentiation rule and the Plancherel theorem, $\norm{f'}_2 = 2\pi \norm{\xi \hat f}_2$. Squaring gives the inequality; translation in $x$ and modulation by $e^{2\pi i b x}$ give general $a$ and $b$, and a density argument gives all $f$ in $L^2(\R)$. Equality in Cauchy–Schwarz forces $f' = -2c x f$, whose solutions are Gaussians.

If one of the two integrals is infinite and $f \ne 0$, then the other is positive, so the left side is $+\infty$ and the inequality holds.

Example: for $f(x) = e^{-\pi x^2}$, one has $\hat f = f$, $\norm{f}_2^2 = 2^{-1/2}$ and $\int x^2 e^{-2\pi x^2}\, dx = \frac{1}{4\pi} 2^{-1/2}$, so the left side equals $\frac{1}{16\pi^2} \cdot \frac12 = \frac{\norm{f}_2^4}{16\pi^2}$.

Going deeper

For a student

The proof is a commutator argument: the operators of multiplication by $x$ and of differentiation do not commute, and Cauchy–Schwarz turns this into a lower bound. The same scheme gives the Robertson uncertainty relation for any two self-adjoint operators. The inequality also follows from the entropic uncertainty principle, since among densities with a given variance the Gaussian has the largest entropy. In $\R^n$ the analogous bound is $\norm{\abs{x} f}_2 \norm{\abs{\xi} \hat f}_2 \ge \frac{n}{4\pi} \norm{f}_2^2$.