Reverse Hausdorff Young for nonnegative functions The classical Hausdorff-Young inequality states that
$$
\Vert \widehat{f} \Vert_{p'} \leq \Vert f \Vert_p \text{ for } 1 \leq p \leq 2.
$$
For $p=2$, we even have equality due to Plancherel.
If we additionally assume that $f \geq 0$, we also get
$$
\Vert \widehat{f} \Vert_\infty = \widehat{f}(0) = \int f(x) \, dx = \Vert f \Vert_1,
$$
i.e. we get equality in the Hausdorff-Young inequality for $p=1$ also.
My question is, wether this generalizes to $1 \leq p \leq 2$ (at least asymptotically), i.e. do we have
$$
\Vert \widehat{f} \Vert_{p'} \asymp \Vert f \Vert_p \text{ for } 1 \leq p \leq 2 \text{ and } f \geq 0 \text{?}
$$
We can not use interpolation here (at least I do not see it), because the estimate on the "boundary" points (at least at $p=1$) is only valid for $f \geq 0$ (and the whole inequality can also only be valid for those $f$), so that the usual "splitting" (for real interpolation) can not be applied. Similarly, complex interpolation does not seem to work.
But also the classical method for constructing a counterexample does not work, i.e. one can not take something like
$$
f = \sum_{m=1}^n M_{\xi_m} g,
$$
where $M_\xi g (y) = e^{2\pi i \xi y} g(y)$ denotes modulation, because this will violate the non-negativity.
Taking
$$
f = \sum_{m=1}^n T_{x_n} g
$$
does not violate this assumption and I can asymptotically calculate $\Vert f \Vert_p$ in this case (for $\min_{n \neq m} |x_n  - x_m| \to \infty$), but I am unable to calculate the integral
$$
\Vert \widehat{f} \Vert_{p'} = \bigg( \int \big| \sum_{m=1}^n e^{\pm 2\pi i x_n \xi }\big|^{p'} \cdot |\widehat{g}(\xi)|^{p'} \, d\xi \bigg)^{1/p'}.
$$
precisely enough.
Any ideas would be appreciated.
 A: It is an elementary exercise in Banach space theory to show that there is NO operator from $L_p$ into $L_{q}$ that satisfies the inequality when $1\le p \not= 2$ with $p<q<\infty$. First, for such $p$ and $q$ every operator from $\ell_p$ into $L_q$ is strictly singular (because $\ell_p$ is not isomorphic to a subspace of $L_q$). Take any (bounded, linear) $T:\ell_p \to L_q$.  If $\|Te_n\|_q $ does not converge to zero, assume by passing to a subsequence and making a small perturbation that $Te_n$ is a block basis of the Haar basis and hence is unconditional.  By the strict singularity of $T$ there is a normalized block basis $y_k = \sum_{j=n_k}^{n_{k+1}-1} a_j e_j$ s.t. 
$\|Ty_k\|_q\to 0$.  Since $Te_n$ is unconditional, setting $z_k =  \sum_{j=n_k}^{n_{k+1}-1} |a_j| e_j$, we also have $\|Tz_k\|_q \to 0$, and of course $\|z_k\|_p = \|y_k\|_p =1$.  This proves the result, because 
$e_n$$ \mapsto \mu(A_n)^{-1/p} 1_{A_n}$ extends to an isometry from $\ell_p$ into $L_p$ if the $A_n$ are disjoint sets of positive measure.
So your question has nothing to do with the Fourier transform. 
A: If $1<p<2$, then it is not possible to have the inequality
$$
\|f\|_p \lesssim \|\widehat{f}\|_{p'} \quad\quad\quad\quad\quad (1)
$$
for all $f\ge 0$. This follows from the existence of (positive) purely singular measures $\mu$ with $\widehat{\mu}\in L^{p'}$ (in fact, $\widehat{\mu}$ can have power decay). (I used this fact also in my answer to this question; here, however, decay of $\widehat{\mu}$ in an average sense is enough, which is much easier to obtain.)
I think it is intuitively clear that such a $\mu$ refutes (1), but to elaborate some more on this, consider $f_t=\varphi_t*\mu$, with $\varphi_t(x)=(1/t)\varphi(x/t)$ and $\varphi\ge 0$, $\varphi\in C_0^{\infty}$. Then, if we had (1), it would follow that $f_t$ is a Cauchy sequence in $L^p$; however, this sequence converges to $\mu$ in the sense of distributions, and as $\mu$ is singular, clearly $\mu\notin L^p$.
