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Oct 28, 2012 at 16:05 vote accept user25235
Oct 28, 2012 at 16:05
Oct 28, 2012 at 16:04 comment added user25235 An unexpectedly simple solution ; thanks ! I think this still applies in the noncommutative case (the Haar measure is left and right invariant) : If $VV \subset U$, $g_1$ is any map vanishing outside $V$, with nonzero mean, and $g_2 : x \mapsto g_1^{*}(x^{-1})$, then $f = g_1 \star g_2$ should work.
Oct 28, 2012 at 15:31 history answered Abdelmalek Abdesselam CC BY-SA 3.0