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Mar 13, 2013 at 14:51 vote accept Joël
Mar 13, 2013 at 14:51 vote accept Joël
Mar 13, 2013 at 14:51
Mar 12, 2013 at 13:19 comment added Joël Thanks a lot, Yemon. I need to read more about this notion of Fourier Algebra but that seems to be the right concept for answering my question.
Mar 11, 2013 at 21:56 answer added Marc Palm timeline score: 5
Mar 11, 2013 at 21:19 comment added Yemon Choi Correction to my first comment: the space of such $f$ would contain the Fourier algebra but wouldn't in general be equal to it. However, for the OP's purposes, that still would imply density in $L^1(G)$.
Mar 11, 2013 at 21:18 comment added Yemon Choi Also, IIRC, if one is looking for possible counterexample then one candidate might be the ax+b group (over the reals). The paper I have in mind is by Zep/Diep ams.org/mathscinet-getitem?mr=364539 although I think Gelfand+Neumark also did the relevant calculation.
Mar 11, 2013 at 21:12 comment added Yemon Choi This is merely a comment for now as my brain is too flu-addled to check details properly, but I think that for separable Type I , the space of $f$ such that $\pi(f)$ is trace class for Plancherel-a.e. $\pi$ should be isomorphic, via the operator-valued Fourier transform, to the Fourier algebra of $G$. This is dense in $C_0(G)$ for the sup norm and hence dense in $L^1(G)$.
Mar 11, 2013 at 19:45 history asked Joël CC BY-SA 3.0