TITLE: Hardy's Inequality for Integrals
QUESTION [25 upvotes]: I am trying to prove Hardy's Inequality for integrals:

Let $f:(0,\infty) \rightarrow \mathbb R$ be in $L^p$. Let $F(x) = \frac{1}{x} \int_0 ^x f(t)dt$. Then $F \in L^p$ and $\|F\|_p \leq \frac{p}{p-1} \|f\|_p$. 

I have seen proofs of this which use Haar measure and Fourier analysis, eg. here An inequality by Hardy. However, I am wondering if there is a more elementary proof. 
I have tried to deduce this using Jensen's inequality and Fubini's Theorem as follows:
Jensen's inequality implies that $\left|\frac{1}{x} \int_0 ^x f(t)dt)\right|^p \leq \frac{1}{x} \int_0 ^x |f(t)|^p dt$. Therefore 
$$\int_0 ^\infty |F(x)|^p dx \leq \int_0 ^\infty \frac{1}{x} \int_0 ^x |f(t)|^p dtdx = \int_0 ^\infty \int_t ^\infty \frac{1}{x} |f(t)|^p dx dt.$$
But, this last integral is infinite, and I'm not sure how to get a sharper bound on $|F(x)|^p$. Any suggestions?

REPLY [12 votes]: Dualize - it becomes
$$ \left(\int_0^\infty |F(x)|^p \, dx\right)^{1/p} \le p \left(\int_0^\infty |f(x)|^p \, dx \right)^{1/p} $$
where $$ F(x) = \int_x^\infty \frac{f(t)}t \, dt $$
W.l.o.g. $f(x) \ge 0$, and w.l.o.g. $F(x) \to 0$ as $x \to \infty$.
$$ |F(x)|^p = - p \int_x^\infty F'(t) |F(t)|^{p-1} \, dt $$
Substitute this into the LHS of the first inequality, and reverse the order of integration
$$ - p \int_0^\infty F'(t) |F(t)|^{p-1} \int_0^t \, dx \, dt = p \int_0^\infty f(t) |F(t)|^{p-1} \, dt \le p \|f\|_p \|F\|_p^{p-1} $$
where the last inequality is Hölder.  Now assume $\|F\|_p$ is finite, and divide both sides by $\|F\|_p^{p-1} $.