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.

  • $\begingroup$ Right, very helpful. Thanks! $\endgroup$ – Yung-Hsiang Huang Apr 5 '18 at 9:56
  • $\begingroup$ @ Kore-N I think you get $\sqrt {2R}$ as the coefficient in the first term, instead of $R$. Also, I am unable to see how you choose $R$. I minimized the expression over all $R \in (0,\infty)$ and I couldn't get the right inequality. Could you please add more details to teh lst step in your proof? $\endgroup$ – Kavi Rama Murthy Apr 5 '18 at 10:15
  • $\begingroup$ @KaviRamaMurthy Hi, sorry: I forgot the root. But it comes in front of both terms. I simply chose $R = \frac{\| xf\|_{L^2}}{\| f\|_{L^2}}$ $\endgroup$ – Kore-N Apr 5 '18 at 10:33
  • $\begingroup$ @ Kore-N thanks. I did get the value of R after posting above comment. Your proof is really nice. $\endgroup$ – Kavi Rama Murthy Apr 6 '18 at 12:30

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