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Apr 15, 2014 at 19:19 comment added Christian Remling If $X_p\subset B$, then, in particular, any $f$ with $f,\widehat{f}\in L^p\cap L^{\infty}$ would have to satisfy $f\in L^1$ also. I don't see why this would be true (start with a smooth $f\in L^p\cap L^{\infty}$ that is not in $L^1$).
Apr 15, 2014 at 18:01 comment added Nate Eldredge So to rephrase, is every $f \in L^p \cap L^\infty$ the Fourier transform of some complex Borel measure?
Apr 15, 2014 at 13:41 comment added Yemon Choi My first instinct is to try and disprove this indirectly by using a Closed Graph Theorem / Open Mapping Theorem argument, similar to the known standard proof that the Fourier transform is not a surjection from $L^1({\bf R})$ onto $C_0({\bf R})$. In other words, rather than look for an explicit counter-example, try to prove that the norms don't match
Apr 15, 2014 at 11:48 history asked Inquisitive CC BY-SA 3.0