If $f\in L^p(R)$ with $1\leq p\leq 2$, then Hausdorff-Young inequality implies that the Fourier transform $\widehat{f}\in L^{p'}$, $p'$ is the dual exponent of $p$, and $$ \|\widehat{f}\|_{L^{p'}}\lesssim \|f\|_{L^p}. $$ We note that $f\in L^p$ is equivalent to $|f|\in L^p$, so we find $\widehat{|f|}\in L^{p'}$ and $$ \|\widehat{|f|}\|_{L^{p'}}\lesssim \|f\|_{L^p}. $$ So my question arises, does there exist connection between $\|\widehat{f}\|_{L^{p'}}$ and $\|\widehat{|f|}\|_{L^{p'}}$ ? Precisely, does the following $$ \|\widehat{f}\|_{L^{p'}}\lesssim \|\widehat{|f|}\|_{L^{p'}} $$ holds? Note that $p=2$ it is a trivial by Plancherel theorem.
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$\begingroup$ Have you tried looking in Katznelson's book, in the remarks and examples surrounding his discussion of the Hausdorff-Young inequality? $\endgroup$– Yemon ChoiJan 2, 2015 at 1:19
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$\begingroup$ @YemonChoi Thanks for your comment. Unfortunately, I don't find the answer there. $\endgroup$– Wang MingJan 2, 2015 at 2:08
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$\begingroup$ Have you tried writing $|f(x)|=s(x)f(x)$ (where $s:\mathbb R\to\{-1,1\}$ is the sign of $f$) and studying the convolution that arises from the Fourier transform of $|f|$? I'm not sure if it takes you anywhere, but it might be worth a try. $\endgroup$– Joonas IlmavirtaJan 2, 2015 at 9:33
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No. The inequality $||\hat{f}||_{L^p} \lesssim || \hat{|f|} ||_{L^{p}}$ does not hold for $p \neq 2$. This is, perhaps, easier to see in the case of Fourier series on the circle. A sketch of a contstruction is given in my answer to another mathoverflow quesion here. This example can be modified to work in Euclidean space as well.
answered Jan 2, 2015 at 17:42