Inequality of continuous functions with integral . need proof and intuition.

Question

Let $f:R\to R$ be a differentiable function such that $f(0)=0$ and $f'$ continuous on $R$. prove:

$$\int\limits_{0}^1|f(x)f'(x)|dx \le \frac{1}{2} \int\limits_{0}^1(f'(x))^2dx$$

Answer 1

Define

$$F(x) = \int_0^x \lvert f'(t)\rvert\,dt.$$

Since $f'$ is continuous, $F$ is a continuously differentiable function, with $F'(t) = \lvert f'(t)\rvert$ for all $t$. Also, we have $F(x) \geqslant \lvert f(x)\rvert$ for all $x$. Hence

\begin{align} \int_0^1 \lvert f(x)f'(x)\rvert\,dx &\leqslant \int_0^1 F(x)\lvert f'(x)\rvert\,dx \\ &= \int_0^1 F(x)F'(x)\,dx \\ &= \frac{1}{2}\bigl(F(1)^2 - F(0)^2\bigr) \\ &= \frac{1}{2} F(1)^2 \\ &= \frac{1}{2} \biggl(\int_0^1 \lvert f'(x)\rvert\,dx\biggr)^2 \\ &\leqslant \frac{1}{2} \int_0^1 \lvert f'(x)\rvert^2\,dx \tag{CauchySchwarz} \\ &= \frac{1}{2} \int_0^1 f'(x)^2\,dx, \end{align}

the last equality holding because $f'$ is real-valued.