# Prove that $\int_{0}^{1}|f(x)f'(x)|dx\leq\frac{1}{2}\int_{0}^{1}|f'(x)|^2dx$

Let $f$ be a continuously differentiable function on $[0,1]$ and $f(0)=0$. Prove that $$\int_{0}^{1}|f(x)f'(x)|dx\leq\frac{1}{2}\int_{0}^{1}|f'(x)|^2dx$$

Thank you!

• Do Cauchy-Schwarz on $f(x)=\int_0^xf'$ to get $|f(x)|\leq \sqrt{x}\sqrt{\int_0^1f'^2}$. Plug this into $\int_0^1|ff'|$ and do Cauchy-Schwarz again. – Julien Jul 28 '13 at 21:00

Without loss of generality, we can assume that $f' \geqslant 0$ everywhere (and hence $f \geqslant 0$ too, and $f$ is monotonically non-decreasing) - otherwise consider $F(x) = \int_0^x \lvert f'(x)\rvert\,dx$. Then
$$\int_0^1 f(x)f'(x)\,dx = \int_0^1 \frac12 \frac{d}{dx} f(x)^2\, dx = \frac12 f(1)^2.$$
$$f(1) = \int_0^1 f'(x)\, dx = \int_0^1 1\cdot f'(x)\,dx \leqslant \sqrt{\int_0^1 1^2\,dx}\sqrt{\int_0^1 f'(x)^2\,dx} = \sqrt{\int_0^1 f'(x)^2\,dx}$$
by the Cauchy-Schwarz inequality. Squaring and dividing by $2$ yields the proposition.