Given $\int_0^1 x f(x)dx=0$, show that $\int_0^1|1-f(x)|dx>1/2$

Question

I have seen this statement before, and I would like to use it in a proof I am working on. I do not quite remember the condition on $f$--whether it is just integrable or continuous.

Can someone point out how to prove it?

Answer 1

$\int_0^1 |x(1-f(x))|dx\ge\int_0^1 x(1-f(x))dx=1/2$ so $\int_0^1 |(1-f(x))|dx=\int_0^1 (1/x)|x(1-f(x))|dx> \int_0^{1/2} |x(1-f(x))|dx$. You can figure out why the last strict inequality holds (hint: for some $\alpha<1$, $\int_0^{\alpha} |x(1-f(x))|dx>0$ then scaling the integrand by $1/x$ at least on this interval $[1,\alpha]$ strictly increases the integral).

Answer 2

Hint:

$$ \int_0^1x\big(1-f(x)\big)\,dx=\int_0^1x\,dx=\frac{1}{2}. $$