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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.

  • 0
    possible duplicate of [Hardy's inequality](http://math.stackexchange.com/questions/99671/hardys-inequality)2012-01-17
  • 1
    Instead of re-posting your question, you should go back and edit your old question (there is a little link at the bottom of the question that says "edit").2012-01-17
  • 1
    I am afraid this is spam. The user after having "read" through the forum failing to notice the extensive editing that takes place in the site is hard to believe.2012-01-17
  • 0
    okay, I'm new here... :(2012-01-17

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