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How can I prove that the constant in classical Hardy's inequality is optimal?

$\int_0^{\infty}\left(\frac{1}{x}\int_0^xf(s)ds\right)^p dx\leq \left(\frac{p}{p-1}\right)^p\int_0^{\infty}(f(x))^pdx,$ where $f\geq0$ and $f\in L^p(0,\infty)$.

This inequality fails for $p=1$ and $p=\infty$ ?

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    You may want to look at this, especially Davide's answer: http://math.stackexchange.com/questions/83946/hardys-inequality-for-integrals/95399#953992012-01-15

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In this form you need $p > 1.$ For the constant, take small $\epsilon > 0$ and define $ f(x) = x^{(-1/p) - \epsilon}, \; \; x \geq 1 $ but $ f(x) = 0, \; 0 \leq x < 1. $

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    :-) ${}{}{}{}{}$2012-01-15