## Proof of Young’s convolutions inequality for a general measure on $\mathbb R^d$

Is Young's inequality true for an arbitrary measure on $\mathbb R^d$? If so, where can I find a proof of it? In particular, where can I find the proof of the discrete version (i.e the version for $\ell^p$ spaces) of this inequality?

Here is the statement of the inequality (from Wikipedia):

Suppose $f$ is in $L^p(\mathbb R^d)$ and $g$ is in $L^q(\mathbb R^d)$ and $$\frac{1}{p} + \frac{1}{q} = \frac{1}{r} + 1$$

with $1\leq p, q, r\leq \infty$

then $$|| f * g || _r \leq ||f||_p ||g||_q.$$

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It strikes me as unlikely that there is a generalization to arbitrary measures. Lebesgue measure is translation invariant and the definition of convolution implicitly uses translation. – Robin Chapman May 30 2010 at 6:28
The analogous result holds for locally compact groups with unimodular Haar measure. See for example section 20 in chapter 5 of Hewitt and Ross: books.google.com/… In particular this gives a version for the sequence spaces indexed by Z. – Jonas Meyer May 30 2010 at 7:08

Note that if your measure is not translation-invariant, the convolution product is not commutative.

This allows for a simple counterexample with $r=q=\infty$, $p=1$, $d\mu(x)={\bf 1}_{[0,1]}(x) dx$.

Define $f*g(x)= \int f(x-s)g(s)d\mu(s) \ (\neq g*f(x))$.

Take $g\equiv 1, \ f={\bf 1}_{[1,2]}$. This gives $||f||_1=0$, but the spreading in the convolution makes the left term non zero.

$\max_x \ \ \int_0^1 {\bf 1}_{[1,2]}(x-s)ds =1$.

Just add $\varepsilon e^{-|x|}dx$ to $d\mu$ to get a counterexample with a measure of full support. I think that the reasonable setting for a Young inequality is the case of a translation invariant measure on a locally compact group. Jonas Mayer gave a reference (Hewitt and Ross) for that case in the comments.

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It shouldn't be hard to find a counterexample for any measure that is not translation invariant – Deane Yang May 30 2010 at 15:21
@Yang. The Young inequality holds for the Dirac measure. It is just the not very interesting bound $|f(0)g(0)|\leq |f(0)||g(0)|$. So maybe there are counterexamples for any non-atomic measure of full support that is not translation invariant. We can also ask when it holds if we allow a multiplicative constant in the inequality. – coudy May 30 2010 at 18:34

In SHARPNESS IN YOUNG'S INEQUALITY FOR CONVOLUTION by JOHN J. F. FOURNIER the author proves that, under certain conditions on the underlying locally compact unimodular group, the Young inequality can be improved with a constant $C<1$ in front of the RHS of the standard inequality