# Given f:[0,1]->RR an integrable function such that int_0^1f(x)dx=int_0^1 xf(x)dx= 1 prove that int_0^1f(x)^2dx ge 4?

Oct 28, 2016

${\left(f \left(x\right) - 6 x + 2\right)}^{2} = f {\left(x\right)}^{2} + 36 {x}^{2} + 4 - 12 x f \left(x\right) + 4 f \left(x\right) - 24 x$
so
${\int}_{0}^{1} f {\left(x\right)}^{2} \mathrm{dx} = {\int}_{0}^{1} {\left(f \left(x\right) - 6 x + 2\right)}^{2} \mathrm{dx} - 36 {\int}_{0}^{1} {x}^{2} \mathrm{dx} - 4 {\int}_{0}^{1} \mathrm{dx}$
$+ 12 {\int}_{0}^{1} x f \left(x\right) \mathrm{dx} - 4 {\int}_{0}^{1} f \left(x\right) \mathrm{dx} + 24 {\int}_{0}^{1} x \mathrm{dx} \ge$
$\ge 0 - \frac{36}{3} - 4 + 12 - 4 + \frac{24}{2} = 4$

#### Explanation:

Just think that ${f}^{2} = {\left(f - a x - b\right)}^{2} + {F}_{a , b} \left(x\right)$ where
${F}_{a , b} \left(x\right) = 2 a x f \left(x\right) + 2 b f \left(x\right) - {a}^{2} {x}^{2} - {b}^{2} - 2 a b x$
Then find a and b that maximize ${\int}_{0}^{1} {F}_{a , b} \left(x\right) \mathrm{dx}$
${\int}_{0}^{1} {F}_{a , b} \left(x\right) \mathrm{dx} = 2 a + 2 b - {a}^{2} / 3 - {b}^{2} - a b$

Calculate gradient and put it = 0. You obtain $a = 6 , b = - 2$