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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.

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  • $\begingroup$ What if $f\equiv 1?$ $\endgroup$
    – zhw.
    Nov 12, 2015 at 23:01
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    $\begingroup$ 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. $\endgroup$
    – Simon S
    Nov 12, 2015 at 23:01
  • $\begingroup$ I thought it is implied in the inequality that $f$, $xf(x)$ and $f'$ are all in $L^2$. $\endgroup$
    – Exort
    Nov 12, 2015 at 23:10
  • $\begingroup$ Usually I think we are more explicit. But assuming everything is $L^2$, then my hint gives the answer. $\endgroup$
    – Simon S
    Nov 12, 2015 at 23:11
  • $\begingroup$ Oh I see! Thanks Simon. $\endgroup$
    – Exort
    Nov 12, 2015 at 23:15

1 Answer 1

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For well behaved $f$ (and usually suppressing limits of integration):

$$\int_{-\infty}^\infty xff' \ dx = \left[\frac 12 xf^2 \right]_{-\infty}^\infty - \int \frac 12 f^2 \ dx = - \frac 12 \int f^2 dx$$

Now applying Cauchy-Schwarz,

$$\left( \int |f|^2 \ dx \right)^2 = 4 \left( \int xff' \ dx \right)^2 \leq 4 \left( \int |xf|^2 \ dx \right) \left( \int |f'|^2 \ dx \right)$$

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