Let $G$ be a locally compact but not compact topological group. I'm looking for an example to show that for an arbitrary $p>1$, there exist some $f,g\in L^{p}(G)$ such that $f*g\notin L^{p}(G)$.
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$\begingroup$ Is $G=\mathbb{R}$ sufficient for you, or you want an example in any non-C, LCTG $G$ ? $\endgroup$– Pietro MajerAug 7, 2013 at 13:50
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$\begingroup$ Yes,$\mathbb R$ is good. $\endgroup$– hassan zakiAug 7, 2013 at 14:01
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2$\begingroup$ Then any bounded $f=g$, decaying at infinity like $|x|^{-a}$ will do it, for a convenient choice of $0 < a < 1$ depending on $p$. $\endgroup$– Pietro MajerAug 7, 2013 at 17:41
1 Answer
In the case of $G=\mathbb{R}$, define for instance, for all $\alpha >0$ $$f_\alpha(x):=\frac{x^{\alpha-1}}{\Gamma(\alpha)}\chi_{[1,+\infty)}(x).$$ So $f_\alpha\in L^p(\mathbb{R})$ if and only if $\alpha < 1- 1/p$. Also, for $\alpha >0$ and $\beta >0$ we have $$f_\alpha\ast f_\beta(x)=\frac{x^{\alpha+\beta-1}}{\Gamma(\alpha)\Gamma(\beta)}\int_{1/x}^{1-1/x}t^{\alpha-1}(1-t)^{\beta-1}dt= \big(1+o(1)\big) f_{\alpha+\beta}(x), $$ as $x\to+\infty$. In particular, for any $p>0$, any $\alpha$ between $\frac{1}{2}\big(1- \frac{1}{p}\big)$ and $1- \frac{1}{p}$ satisfies $f_\alpha\in L^p(\mathbb{R})$, $f_\alpha\ast f_\alpha\notin L^p(\mathbb{R})$.