Question

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`?

Answer 1

`(f(x)-6x+2)^2= f(x)^2+36x^2+4-12xf(x)+4f(x)-24x`
so
`int_0^1f(x)^2dx=int_0^1(f(x)-6x+2)^2dx-36int_0^1x^2dx-4int_0^1dx`
`+12int_0^1xf(x)dx-4int_0^1f(x)dx+24int_0^1xdx>=`
`>=0-36/3-4+12-4+24/2=4`

Explanation:

Just think that `f^2=(f-ax-b)^2+F_(a,b)(x)` where
`F_(a,b)(x)=2axf(x)+2bf(x)-a^2x^2-b^2-2abx`
Then find a and b that maximize `int_0^1F_(a,b)(x)dx`
`int_0^1F_(a,b)(x)dx=2a+2b-a^2/3-b^2-ab`

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