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.