Inequality associated with Fourier transform 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.
 A: First of all, note that the claim does not hold for $f=0$. In the sequel, we assume that $f$ is not identical zero.
Set $I_1=[-a,a]$, $I_2=[-b,b]$ for $a,b>0$. Since $f$ is a Schwartz function, hence in particular bounded, we can assume $$\int f^2(x) \, dx = 1.$$
By the uncertainty principle and the assumption on the integrals,
$$\begin{align*} \frac{1}{16 \pi^2} &\leq \left( \int_{\mathbb{R}} x^2 \cdot f(x)^2 \, dx \right) \cdot \left( \int_{\mathbb{R}} x^2 \cdot \hat{f}(x)^2 \, dx \right) \\ &\leq 4 \left( \int_{-a}^a x^2 \cdot f(x)^2 \, dx \right) \cdot \left( \int_{-b}^b x^2 \cdot \hat{f}(x)^2 \, dx \right) \end{align*}$$
Now find a upper bound (depending on $a$, $b$) for the remaining integrals using that $\|f\|_{L^2}=1$ and Plancherel's identity.
Remark Note that there are several definitions for the Fourier transform. The constant $\frac{1}{16\pi^2}$ refers to the Fourier transform
$$\hat{f}(\xi) = \int f(x) \cdot e^{-2\pi \imath \, x \cdot \xi} \, dx$$
Depending on the definition in your book, you might have to modify this constant.
A: 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 $|I_1||I_2|\ge\frac1{2\pi}$
