The following problem arose in a question I recently asked : given a (possibly non abelian) compact group $G$ and a neighbourhood $U$ of the identity in $G$, can we always find a function $f : G \mapsto \mathbb{R}$, which vanishes outside $U$, whose Fourier transform is nonnegative, and which satisfies $\hat f(1) \neq 0$ ?
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I didn't think about the non Abelian case, but for a commutative group say like $\mathbb{R}$ you can do the following. Take a function with support in $\frac{1}{2} U$ and symmetric with respect to the origin. Its convolution with itself answers your question.
answered Oct 28, 2012 at 15:31
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$\begingroup$ 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. $\endgroup$ Commented Oct 28, 2012 at 16:04