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.



Reverse Hausdorff Young for nonnegative functions
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](https://mathoverflow.net/questions/180424/is-a-polynomial-decay-sufficient-for-a-smooth-function-to-be-in-mathcalfl1/180685#180685); 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 $\mu\notin L^p$.


