Hardy's inequality again
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 $p=\infty$ ?
Solution 1:
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 $$ f(x) = 0, \; 0 \leq x < 1. $$