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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?

Rudy the Reindeer
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mpiktas
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    Yes, Young's inequality is true for convolution on locally compact groups (not necessarily abelian), in particular $\mathbb{Z}$. See e.g. Theorem 20.18 on page 296 of Hewitt-Ross, Abstract Harmonic Analysis, I but that's serious overkill. The slick argument given by robjohn here should carry over without any pain. – t.b. Oct 05 '11 at 11:54
  • Thanks! I wonder can I cite this theorem in the article, or should I prove this result myself with the hints given. After I posted the question I found the result proven in Bogachev's book, his argument is similar to robjohn. – mpiktas Oct 05 '11 at 12:23
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    I would say that Young's inequality is standard and straightforward enough and thus very likely to fall victim to a referee in the publication process anyway, so I wouldn't waste time to write up the proof in an article I'm writing for publication unless I was really unable to find it in the desired form in the literature. You seem to have two references already, so the statement with a reference should be amply sufficient. – t.b. Oct 05 '11 at 12:26
  • There is Minkowski's inequality: $$|a\ast b|{\ell^q}\le |a|{\ell^1}|b|_{\ell^q},\ 1\le q\le \infty.$$ –  Apr 26 '13 at 01:13

1 Answers1

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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 $\mathbb{Z}^d$.

Community
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t.b.
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  • Over a year late, but I just came across this post. Summation is just integration over a discrete set :-) – robjohn Oct 31 '12 at 16:58
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    Sorry for fishing an old thread. But doesnt that mean, that $l^P=l^1$ (settheoretic) whenever $p\in[1,2]$? Since $l^1\subset l^p$ we would get by convolution with $e^0$, where $e^0_z=1$ for $z=0$ and $e^0_z=0$ for $z\neq 0$ $$e^0\ast f(z)=f(z)~\text{for all}~z$$ and so $$|f|_1=|e^0\ast f|_1\leq|e^0|_q|f|_p=|f|_p.$$ Thanks for reply. –  Oct 31 '12 at 10:23
  • @Burn: Are you talking about $\ell^1$ and $\ell^p$? In that case, $\ell^1\subset\ell^p$ for $p\ge1$, but the reverse containment is not true. – robjohn Oct 31 '12 at 17:02
  • @robjohn: yeah, right :) Thanks, hi and bye, Theo. – t.b. Nov 29 '12 at 14:27
  • @robjohn in that case what is wrong in the argument of @user47666? – BelowZero Oct 31 '16 at 13:26
  • I see it sorry, $q<1$ is the problem. – BelowZero Oct 31 '16 at 13:33