If $f\in L^p(R)$ with $1\leq p\leq 2$, then HausdorffYoung 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.

$\begingroup$ Have you tried looking in Katznelson's book, in the remarks and examples surrounding his discussion of the HausdorffYoung inequality? $\endgroup$ – Yemon Choi Jan 2 '15 at 1:19

$\begingroup$ @YemonChoi Thanks for your comment. Unfortunately, I don't find the answer there. $\endgroup$ – Wang Ming Jan 2 '15 at 2:08

$\begingroup$ 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 Ilmavirta Jan 2 '15 at 9:33
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.