# Fourier transform: can we expect $\|\mathcal F( | \mathcal F f |)\|_{L^\infty} \lesssim \|f\|_{L^\infty}$?

If $\mathcal Ff$ denotes the Fourier transform on functions $f:\mathbb R^d\to \mathbb R$, is the following inequality true, for say Schwartz functions $f$?

$$\|\mathcal F( | \mathcal F f |)\|_{L^\infty} \le C \|f\|_{L^\infty}$$

There are functions for which $\mathcal Ff>0$(e.g. a Gaussian) for which this is true with an equality.

I am trying to prove some commutator estimates but I am getting this suboptimal result with $\|\mathcal F( | \mathcal F f |)\|_{L^\infty}$ instead of $\|f\|_{L^\infty}$.

The simple bound $\|\mathcal F g \|_{L^\infty} ≤ \|g\|_{L^1}$ is not enough because after applying it to get rid of the norm, I obviously want to then 'go the other way' to put back $L^\infty$ on the right hand side.

We can't expect this, note that as per $f$ Schwartz we can say $$||\mathcal{F}(|\mathcal{F}(f)|)||_\infty \geq \mathcal{F}(|\mathcal{F}(f)|)(0)=||\mathcal{F}(f)||_1$$
So your desired inequality would imply $||\mathcal{F}(f)||_1 \lesssim ||f||_\infty$.
This is certainly false in general, e.g. the indicator on $[-1,1]$ has fourier transform not in $L^1$, this also implies it is false for Schwartz functions as we can approximate this indicator by smooth compact support functions.