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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.

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It might be something with the underscores; see meta.math.stackexchange.com/questions/107/…. By the way, I can't quite parse your first paragraph grammatically either. For all $f$ and $g$ satisfying this or that, what holds? –  Hans Lundmark Oct 17 '10 at 12:52
    
Oops, I was too busy trying to correct the math that I forgot to add what the conclusion was. I couldn't fix the error so I wrote it in a different way. –  Jonas Teuwen Oct 17 '10 at 13:12
    
OK, that's much better! But I think it would be even easier to read if your first three sentences were merged into one: "...states that if XXX satisfy XXX, then for all XXX, where XXX and XXX, we have that XXX." –  Hans Lundmark Oct 17 '10 at 13:40
    
Okay, done that. –  Jonas Teuwen Oct 17 '10 at 13:52
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Just a comment, if $\mu$ "the" left invariant the Haar measure on $G$ is right invariant too (which is the case e.g. if $G$ is abelian) $G$ is called unimodular and then $g$ and $g'$ have the same norm - a good reference on this is Loomis book "Abstract harmonic analysis". So this is more common than $f*g=g*f$ which happens only when $G$ is abelian (e.g. $G=SL(2,\mathbb{R})$ is unimodular). –  AD. Dec 16 '11 at 14:45

1 Answer 1

up vote 8 down vote accepted

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 $G\to \mathbb{R}$

$$ f*g(y) = \int_G f(x) g(y\cdot x^{-1})\mu(dx) $$

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Err, thanks! I got confused by the "usual" convolution, let me see if I can obtain the result now. –  Jonas Teuwen Oct 17 '10 at 13:51
    
Yes, it works. Thanks a lot. I couldn't spot the error, that's why I posted it here. –  Jonas Teuwen Oct 17 '10 at 14:06
    
No worries; glad to have helped. –  Willie Wong Oct 17 '10 at 14:24
    
@Willie Wong: You mean $f*g(y) =\int_0\infty f(x)g(y/x)dx/x$. –  AD. Oct 18 '10 at 5:36
    
@AD.: Thanks for keeping me honest. :) –  Willie Wong Oct 18 '10 at 11:13

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