I see the following inequality from a paper
$\| f\|_{L^1(\mathbb{R})}^2 \leq C\| xf\|_{L^2(\mathbb{R})} \| f\|_{L^2(\mathbb{R})} $.
Although the authors say it's elementary. I try some hours and can't prove it.
I try to use Planchel theorem to interpret the $xf(x)$ term, but I don't know how to use the inversion Fourier formula to estimate the $L^1$ norm.
Another easy observation is that we may assume $f$ is radially decreasing by using Steiner symmetrization.
I appreciate any discussion and suggestion.
I think you overthought this problem. You can solve everything with standard inequalities. The idea is that $xf(x)^2$ takes care of the norm on large scales, while $f(x)^2$ takes care of the norm on small scales. You can estimate: $$ \int_{\mathbb{R}} |f(x)| dx = \int_{B(0,R)} |f(x)| dx+ \int_{B(0,R)^c} \frac{1}{|x|}|xf(x)| dx \\ \lesssim \sqrt{R} \|f\|_{L^2} + \frac{1}{\sqrt{R}} \|xf\|_{L^2} \lesssim \sqrt{\| f\|_{L^2} \cdot \| x f\|_{L^2}} $$ where we used C.S. inequality in the first step and chose a suitable $R$ in the second step.