${\left( \int_{0}^{1} f(x)\right)}^2 \le 2\int_{0}^{1} x{(f(x))}^2$

Question

Let f be a non-decreasing, integrable function defined on [0, 1]. Show that $${\left( \int_{0}^{1} f(x)\right)}^2 \le 2\int_{0}^{1} x{(f(x))}^2$$

I tried using by parts , but it is of no help,

${\left( \int_{0}^{1} f(x)\right)}^2 \le\int_{0}^{1} {(f(x))}^2$

Any hints?

Answer 1

Assume first that $f$ is nonnegative.

Let $F(x)=\left(\displaystyle\int_{0}^{x}f(t)dt\right)^{2}$, then \begin{align*} F'(x)=2\int_{0}^{x}f(t)dtf(x)\leq 2x(f(x))^{2}, \end{align*} so \begin{align*} F(1)-F(0)=\int_{0}^{1}F'(x)dx\leq\int_{0}^{1}2x(f(x))^{2}dx. \end{align*}