Fourier transform and Heisenberg's inequality. I can show the following lemma.

Lemma. If $f$ is real-valued and continuously differentiable on $\mathbb{R}$, then$$\left(\int |f|^2\,dx\right)^2 \le 4\left(\int |xf(x)|^2\,dx\right)\left(\int |f'|^2\,dx\right).$$By integration by parts,
  $$\int f^2(x)\,dx=xf^2(x)-\int x(f^2(x))'\,dx=xf^2(x)-2\int xf(x) f'(x)\,dx.$$
  If $\lim_{x\to\pm\infty} xf(x)^2 = 0$ then the inequality easily follows by using Cauchy–Schwarz inequality.

Question. How do I see Heinsenberg's inequality is true: there exists $c > 0$ such that if $a$, $b \in \mathbb{R}$ and $f$ is in $L^2$, then$$\left(\int (x - a)^2 |f(x)|^2\,dx\right)\left(\int (u - b)^2 |\widehat{f}(u)|^2\,du\right) \ge c\left(\int |f(x)|^2\,dx\right)^2?$$What's the best constant $c$?
 A: Define the Fourier Transform as
$$
\hat{f}(\xi)=\int_{\mathbb{R}^n}f(x)e^{-2\pi ix\cdot\xi}\,\mathrm{d}x\tag{1}
$$
In this answer, it is shown that in $\mathbb{R}^n$
$$
\|\xi\hat{f}\|_2\|xf\|_2\ge\frac{n}{4\pi}\|\hat{f}\|_2\|f\|_2\tag{2}
$$
If we define $\tau_yf(x)=f(x+y)$, then
$$
\begin{align}
\widehat{\tau_yf}(\xi)
&=\int_{\mathbb{R}^n}f(x+y)e^{-2\pi ix\cdot\xi}\,\mathrm{d}x\\
&=e^{2\pi iy\cdot\xi}\int_{\mathbb{R}^n}f(x)e^{-2\pi ix\cdot\xi}\,\mathrm{d}x\\
&=e^{2\pi iy\cdot\xi}\hat{f}(\xi)\tag{3}
\end{align}
$$
Then The Plancherel Theorem and Fourier Inversion give
$$
\begin{align}
\|(\xi-b)\hat{f}\|_2\|(x-a)f\|_2
&=\|\xi\tau_b\hat{f}\|_2\|x\tau_af\|_2\\
&=\|\xi e^{-2\pi ia\cdot\xi}\tau_b\widehat{\tau_af}\|_2\|x\tau_a f\|_2\\
&=\|\xi\tau_b\widehat{\tau_af}\|_2\|x\widehat{\widehat{\tau_a f}}\|_2\\
&=\|\xi\tau_b\widehat{\tau_af}\|_2\|xe^{-2\pi ib\cdot x}\widehat{\tau_b\widehat{\tau_a f}}\|_2\\
&=\|\xi\tau_b\widehat{\tau_af}\|_2\|x\widehat{\tau_b\widehat{\tau_a f}}\|_2\\
&\ge\frac{n}{4\pi}\|\tau_b\widehat{\tau_af}\|_2\|\widehat{\tau_b\widehat{\tau_a f}}\|_2\\
&=\frac{n}{4\pi}\|\hat{f}\|_2\|f\|_2\tag{4}
\end{align}
$$
