In fourier transformation theory, we have the Heisenberg uncertainty principle, i.e. Suppose $\phi$ is a function in [Schwarz space][1] which satisfies the normalizing condition $\int_{R}|\phi(x)|^2dx=1$, then $$\left(\int_{-\infty}^{\infty}x^2 |\phi(x)|^2 dx \right)\left(\int_{-\infty}^{\infty}\xi^2 |\hat{\psi}(x)|^2 dx \right)\geq \frac{1}{16\pi ^2}.$$ Where $\hat f(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi i x \xi}dx.$
Via holder inequality && integration by parts, we can get this conclusion.
Now, we have a familiar question, but seems more interesting. Let the following inequality hold for some intevral $I_1$ and $I_2 \in R$, since $f$ in Schwart Space, such interval must exist. $$\int_{I_{1}}x^2|f(x)|^2dx\geq \frac{1}{2}\int_{-\infty}^{\infty}x^2|f(x)|^2dx$$ $$\int_{I_{2}}\xi^2|\hat f(\xi)|^2d\xi\geq \frac{1}{2}\int_{-\infty}^{\infty}\xi^2|\hat f(\xi)|^2d\xi$$ Denote $L_j$ as the length of $I_j$, $j=1,2$. We want to prove $$L_i L_j\geq \frac{1}{2\pi}$$ It would be more difficult, because we need to find the relationship of the domain, not the integral result. Could you give some hints?
Thanks to everyone!