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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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