# Prove an integral inequality: $\left(\int|f|^2dx\right)^2\le 4\left(\int|xf(x)|^2dx\right)\left(\int|f'|^2dx\right)$

If $f$ is real-valued and continuously differentiable on $\mathbb{R}$, prove that $$\left(\int|f|^2dx\right)^2\le 4\left(\int|xf(x)|^2dx\right)\left(\int|f'|^2dx\right)$$

Attempt: I tried the Cauchy-Schwarz inequality as well as the Plancherel theorem, but none of them seems to work.

• What if $f\equiv 1?$
– zhw.
Nov 12, 2015 at 23:01
• We need some other hypotheses on $f$. For instance, sufficient conditions to ensure $\lim_{x\to\pm\infty} xf(x)^2 = 0$. In which case $$\int_{-\infty}^\infty xff' \ dx = \left[\frac 12 xf^2 \right]_{-\infty}^\infty - \int \frac 12 f^2 \ dx = - \frac 12 \int f^2 dx$$ and we can apply C-S. Nov 12, 2015 at 23:01
• I thought it is implied in the inequality that $f$, $xf(x)$ and $f'$ are all in $L^2$. Nov 12, 2015 at 23:10
• Usually I think we are more explicit. But assuming everything is $L^2$, then my hint gives the answer. Nov 12, 2015 at 23:11
• Oh I see! Thanks Simon. Nov 12, 2015 at 23:15

For well behaved $f$ (and usually suppressing limits of integration):
$$\int_{-\infty}^\infty xff' \ dx = \left[\frac 12 xf^2 \right]_{-\infty}^\infty - \int \frac 12 f^2 \ dx = - \frac 12 \int f^2 dx$$
$$\left( \int |f|^2 \ dx \right)^2 = 4 \left( \int xff' \ dx \right)^2 \leq 4 \left( \int |xf|^2 \ dx \right) \left( \int |f'|^2 \ dx \right)$$