product of $L^2$ function and its $L^2$ derivative

Question

Let $f$ be absolutely continuous on $[0,x]$ for all $x >0$. $f(0) = 0$ and $f,f' \in L^2([0,\infty))$. Prove that $\int\limits_0^x |ff'| dm\leq \frac{1}{2}(\int\limits_0^x |f'| dm)^2$. The problem is simple if we remove the absolute value, since $(f^2)' = 2ff'$, but because of the absolute value, I am not sure how to proceed.

Answer 1

From absolute continuity, $f(t)=\int_0^tf'$, therefore $$\int_0^x|ff'|=\int_0^x\left|\int_0^tf'(s)f'(t)\,ds\right|\,dt\leq\int_0^x\int_0^t|f'(s)f'(t)|\,ds\,dt.$$ Changing the order of integration, we obtain that $$\int_0^x|ff'|\leq\int_0^x\int_0^t|f'(s)f'(t)|\,ds\,dt=\int_0^x\int_s^x|f'(s)f'(t)|\,dt\,ds,$$ and renaming the integration parameters in the last integral, we have $$\int_0^x|ff'|\leq\int_0^x\int_0^t|f'(s)f'(t)|\,ds\,dt=\int_0^x\int_t^x|f'(t)f'(s)|\,ds\,dt.$$ Hence, $$\begin{align*}2\int_0^x|ff'|&\leq\int_0^x\int_0^t|f'(s)f'(t)|\,ds\,dt+\int_0^x\int_t^x|f'(t)f'(s)|\,ds\,dt\\ & =\int_0^x\int_0^x|f'(t)f'(s)|\,ds\,dt=\left(\int_0^x|f'|\right)^2.\end{align*}$$