# Show that Hardy's inequality holds iff $f=0$ alomost everywhere

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.

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possible duplicate of Hardy's inequality – Alex Becker Jan 17 '12 at 0:51
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"). – Alex Becker Jan 17 '12 at 0:52
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. – user21436 Jan 17 '12 at 0:57
okay, I'm new here... :( – user23069 Jan 17 '12 at 0:57

Suppose that $f \geq 0$ on $(0,\infty)$ and $f \in L^p, 1 < p < \infty$ and define $F(x)=\frac{1}{x} \int_0^x f(t)\,dt$. Following a suggestion in Rudin (8.14) we write $F(x)=\frac{1}{x} \int_0^x t^{-\alpha} t^{\alpha}f(t)\,dt$ for $\alpha =\frac{1}{pq}$ and apply Holder's inequality to get an upper bound of $$F(x) \leq \frac{1}{x}\left(\int_0^x t^{-\alpha q}\, dt\right)^{\frac{1}{q}}\left(\int_0^x t^{\alpha p}f^p(t)\,dt\right)^\frac{1}{p}.$$
The only inequality we applied there is Holder's and so equality holds if and only if it holds in Holder's inequality. It is not hard to show that this is the case iff the functions used in Holder's are linearly dependent if both sides of the inequality are not zero. But, if both sides of the inequality are not zero (which is iff $f = 0$ by positivity) then this implies that $\lambda t^{\alpha p}f^p(t)=t^{-\alpha q}$ for some $\lambda$ and a.e. $t$. But no power of t is integrable on $(0,\infty)$.