TITLE: Inequality associated with Fourier transform
QUESTION [5 upvotes]: Suppose $$\int_{I_1} x^2|f(x)|^2dx\ge\frac12\int_\Bbb Rx^2|f(x)|^2dx$$ and $$\int_{I_2} x^2|\hat f(x)|^2dx\ge\frac12\int_\Bbb Rx^2|\hat f(x)|^2dx$$
for interval $I_1, I_2$ centered at origin and $f\in\mathcal S(\Bbb R)$.
How to prove that $|I_1||I_2|\ge\frac1{2\pi}$?
I use Cauchy-Schwarz inequality and Plancherel formula but it doesn't work...
Please give some hint.

REPLY [2 votes]: In this answer, it is shown that in $\mathbb{R}^n$ and using the same Fourier Transform that saz mentions,
$$
\|\xi\hat{f}\|_2\|xf\|_2\ge\frac{n}{4\pi}\|\hat{f}\|_2\|f\|_2\tag{1}
$$
Therefore,
$$
\begin{align}
\frac12\|xf(x)\|_2\|\xi\hat{f}(\xi)\|_2
&\le\|xf(x)\|_{L^2(I_1)}\|\xi\hat{f}(\xi)\|_{L^2(I_2)}\tag{2}\\
&\le\frac{|I_1||I_2|}{4}\|f(x)\|_{L^2(I_1)}\|\hat{f}(\xi)\|_{L^2(I_2)}\tag{3}\\
&\le\pi|I_1||I_2|\|xf(x)\|_2\|\xi\hat{f}(\xi)\|_2\tag{4}
\end{align}
$$
$(2)$: Given
$(3)$: $|x|\le\frac{|I_1|}2$ and $|\xi|\le\frac{|I_2|}2$
$(4)$: $(1)$ in $\mathbb{R}^1$
Thus, we get that $|I_1||I_2|\ge\frac1{2\pi}$