Reading the questions in this forum, I was interested by the Classical Hardy's inequality:
$$\int_0^{\infty}\left(\frac{1}{x}\int_0^xf(s)ds\right)^p dx\leq \left(\frac{p}{p-1}\right)^p\int_0^{\infty}(f(x))^pdx,$$ where $f\geq0$ and $f\in L^p(0,\infty)$, $p>1$.
I tried to prove that the equality holds if and only if $f=0$ a.e.. The trivial part is that $f=0$ a.e. implies the equality.
I would like to know how to prove the converse.