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Young's inequality for convolution of functions states that for $f\in L^p(\mathbb{R}^d)$ and $g\in L^q(\mathbb{R}^d)$ we have

$\|f\star g\|_r\le\|f\|_p\|g\|_q$

for $p$, $q$, $r$ satisfying

$\frac{1}{p}+\frac{1}{q}=\frac{1}{r}+1.$

Does this inequality hold for sequences? That is, can we replace $L^n(\mathbb{R}^d)$ with $\ell_n$, $n=p,q$ respectively, where convolution of sequences is the discrete convolution?

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    There is Minkowski's inequality: $\|a\ast b\|_{\ell^q}\le \|a\|_{\ell^1}\|b\|_{\ell^q},\ 1\le q\le \infty.$2013-04-26

1 Answers 1

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Yes, Young's inequality can be shown to hold for arbitrary locally compact groups — under suitable integrability assumptions on $f$ and $g$, see Hewitt–Ross, Abstract Harmonic Analysis, I, Theorem (20.18) on page 296 for the precise statement.

If $G$ happens to be abelian, compact, discrete (or, more generally, unimodular) then these assumptions translate to: If $f \in L^{p}$, $g \in L^q$ and $\frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r}$ for $1 \leq p, q, r \leq \infty$ then $f \ast g \in L^r$, and

$\|f \ast g\|_r \leq \|f\|_p\,\|g\|_q.$

Replacing integrals by sums robjohn's argument here carries over painlessly to $\mathbb{Z}$ or $\mathbb{Z}^d$.

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    I see it sorry, q<1 is the problem.2016-10-31