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Let $f\in L^1(\mathbb R^n)$. Define operator $T_f(g)=|f|\ast g$ for functions $g$ on $\mathbb R^n$. The set of measurable functions $f$ on $\mathbb R^n$, such that $T_f$ is bounded from $L^p(\mathbb R^n)$ to $L^p(\mathbb R^n)$ is $L^1(\mathbb R^n)$, which is a Banach algebra.

There are locally compact topological (unimodular) groups where the set of such functions can be bigger than $L^1$. (That is a stronger Young's inequality is true there.) The group $G = \mathrm {SL}(2, \mathbb R)$ is an example.

Is this true that for $G = \mathrm{SL}(2, \mathbb {R})$, the set of measurable functions $f$ such that $T_f$ is bounded from $L^p(G)$ to $L^p(G)$ is a Banach algebra?

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You say a stronger Young's inequality is true: what is this stronger inequality, exactly? I thought Young's inequality was the quantitative statement that the operator norm of $T_f$ is bounded by the $L^1$ norm of $f$, not just the statement that $f \in L^1$ implies $T_f$ is bounded. –  Zen Harper Sep 15 '11 at 9:19
[deleted some comments which were based on a misreading] –  Yemon Choi Sep 15 '11 at 9:56
Zen: I think the OP might be referring to the Kunze-Stein theorem, at least when $p=2$. –  Yemon Choi Sep 15 '11 at 9:56
Yes Kunze-Stein is the modified Young's inequality for $SL(2, R)$ (or for any noncompact semisimple groups). For these groups: If $1<p\le 2$ then for any $1\le q<p \le 2$, $\|f\ast g\|_p\le \|f\|_q\|g\|_p$. This shows there are lot of functions other than $L^1$-functions which takes $L^p$ to $L^p$. –  spr Sep 16 '11 at 5:25
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