Is there an example for a function f in L^{1}(\mb{R}) such that Fourier transform of f is also in L^{1}(\mb{R}), but neither f nor its Fourier transform is in the Wiener's space?
Hans G. Feichtinger
I am sure there is, but cannot answer it "offhand", hgfei
It should be equivalent to the non-existence of some inequalities
(you cannot control the Wiener amalgam norm by the sum of the L1 norm
of f plus the L1-norm of f^). hgfei
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