Vtyjkyuk

Consider the Lebesgue measure on Borel sets of $(0,infty)$. Prove that for every $fin L^2(0,infty)$,

$|int_0^x f(s) ds|^2leq 2sqrtxint_0^x sqrts |f(s)|^2 ds$.

My trial:

Naturally, we can think of Holder's inequality

$|int_0^xf(s)ds|^2leq left(int_0^infty|chi_[0,infty)||f(s)|dsright)^2 leq |chi_[0,x) |_2^2 |f|^2_2=xint_0^infty|f(s)|^2 ds$

And it is totally different from the desired result. I have no idea how to put $|chi_[0,x)|$ into the integral of $f$ anyway.

What about the Cauchy-Schwarz Inequality? Note that
$$left|int_0^x,f(s),textdsright|^2leqleft|int_0^x,frac1sqrt[4]s,Big(sqrt[4]s,big|f(s)big|Big),textdsright|^2leq left(int_0^x,frac1sqrts,textdsright) ,left(int_0^x,sqrts,big|f(s)big|^2,textdsright),.$$
The equality holds iff there exists a constant $cinmathbbC$ such that $f(s)=dfraccsqrts$ for almost every $sin[0,x]$.

Write $f(s) = frac1sqrt[4]ssqrt[4]sf(s)$ and use Holder:
$$left|int_0^xf(s)dsright|^2 leq
int_0^xs^-1/2dsint_0^1s^1/2f(s)ds =2sqrtxint_0^1sqrtsf(s)ds$$