# Why does this integral inequality holds?

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.

• Could you please share the link to the paper? Commented Apr 17, 2020 at 17:26
• Sorry, not at the moment. Commented Apr 17, 2020 at 17:34

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.
• Thanks! You do not even need the Fubini, the key is the replacing $u$ by $t$. Commented Apr 17, 2020 at 21:35