TITLE: An inequality by Hardy
QUESTION [7 upvotes]: Young's inequality for convolutions states that if $1 \leq p, q, r \leq \infty$ satisfy
$$\frac{1}{q} + 1 = \frac{1}{p} + \frac{1}{r}$$
for all $f \in L^p(G)$ and all $g \in L^r(G)$ where $g$ and $g'$ have the same $L^r$-norm  and $g'(x) = g(x^{-1})$ ($G$ is a topological group) we have that:
$$\|f * g\|_q \leq \|g\|_r \|f\|_p.$$
Now Grafakos claims we can use this to prove the following inequality due to Hardy:
$$\left ( \int_0^\infty \left ( \frac{1}{x} \int_0^x |f(t)| \, dt \right )^p \, dx \right  )^{1/p} \leq \frac{p}{p - 1} \|f\|_{L^p(0, \infty)}$$
The hint is to consider on the multiplicative group $(\mathbb{R}^+, \frac{dt}{t})$ the convolution of $|f(x)| x^{1/p}$ and ${x^{-1/p'}} 1_{[1, \infty)}$. So if we use this, the RHS is no problem, it is just a direct computation (I can add it if someone wants it for future reference). However, if I compute the convolution I get:
$$\int_0^{x - 1} |f(t)| (y(t - y))^{1/p'} \, dt$$ but I don't see how this is larger (or equal) to the inner integral on the LHS of the inequality. Any suggestions?
Edit: As Willie Wong points out below, the convolution is wrong. It is an multiplicative group, not an additive one.

REPLY [9 votes]: You are doing the convolution wrong. On the multiplicative group $(\mathbb{R}_+, dt/t)$, the convolution is
$$f * g(x) = \int_0^\infty f(y) g(x/y) dy/y$$
(for harmonic analysis on an abelian group, you need to re-interpret the $+$ and $-$ signs in formulae to be the group binary operator). If you plug in, as $g$, the weight function Grafakos suggested, you should get exactly the LHS. 
Just to be more general: let $(G,\mu)$ be an Abelian group with an invariant measure $\mu$, where $\cdot$ denotes the group binary operator, then the convolution of two functions $f,g: G\to \mathbb{R}$ is defined as the function $G\to \mathbb{R}$
$$ f*g(y) = \int_G f(x) g(y\cdot x^{-1})\mu(dx) $$