Inequality using integrals

Question

Using that

$ u^{2}(x) - u^{2}(0) = \int_{0}^{x}(u^{2}(x))' dx$

and the Cauchy-schwarz inequality, show that if $u(0) = 0$, then exist a constant c such that

$\int_{0}^{b}u^{2}(x)dx < c\int_{0}^{b}(u'(x))^{2}dx $

Someone can give me a hint?

Answer 1

Here's a solution. Don't read past the first 2 lines if you only want a hint.

We have $$\begin{split} \int_0^b u^2(x)dx &=\int_0^b \left(\int_0^x u^\prime(t) dt\right)^2dx \\ &= \int_0^b \left(\int_0^b 1_{[0, x]}(t)u^\prime(t) dt\right)^2dx\\ &\leq \int_0^b \left(\sqrt {\int_0^b (1_{[0, x]}(t))^2dt}\sqrt {\int_0^b (u^\prime(t))^2dt} \right)^2dx \mbox{ by CS}\\ &\leq \int_0^bb\int_0^b (u^\prime(t))^2dt dx\\ &\leq b^2 \int_0^b (u^\prime(t))^2dt \end{split}$$