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I am reading a paper, and the authors write the following inequality for two functions $v$, $z$, both $\mathbb{R}\to\mathbb{R}$: $$\int_{0}^{t} \left(|v(u)|^2\left(\int_0^u|v(r)|^2dr\right)\left(\int_0^u|z(r)|^2dr\right)\right)du \le \left(\int_0^t|v(u)|^2du\right)^2\left(\int_0^t|z(u)|^2du\right).$$ And I do not see why. I have tried with the Cauchy–Schwarz inequality but failed.

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  • $\begingroup$ Could you please share the link to the paper? $\endgroup$ Commented Apr 17, 2020 at 17:26
  • $\begingroup$ Sorry, not at the moment. $\endgroup$
    – Arastas
    Commented Apr 17, 2020 at 17:34

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All of the functions are non-negative, so by replacing the inner upper limits to $t$ instead of $u$ you are increasing the integrals' value. Then, since the limits don't depend on $u$, use fubini: $$\int\limits_0^t|v(u)|^2\left(\int\limits_0^u|v(r)|^2{\rm d}r\right)\left(\int\limits_0^u|z(\tilde{r})|^2{\rm d}\tilde{r}\right){\rm d}u$$ $$\leq\int\limits_0^t\int\limits_0^t\int\limits_0^t\left(|v(u)|^2|v(r)|^2|z(\tilde{r})|^2\right){\rm d}\tilde{r}{\rm d}r{\rm d}t$$ $$=\int\limits_0^t|v(u)^2|{\rm d}u\cdot\int\limits_0^t|v(r)|^2{\rm d}r\cdot\int\limits_0^t|z(\tilde{r})|^2{\rm d}\tilde{r}$$ Thus getting the required inequality.

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  • $\begingroup$ (+1) All too easy. ;-) $\endgroup$
    – Mark Viola
    Commented Apr 17, 2020 at 18:55
  • $\begingroup$ Thanks! You do not even need the Fubini, the key is the replacing $u$ by $t$. $\endgroup$
    – Arastas
    Commented Apr 17, 2020 at 21:35

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