Suppose $f\in L^2$ and define $g\in L^2$ by $g(x) = \lvert f(x)\rvert$. Based on numerical experiments I believe that
$$ \Big\lVert\hat f\Big\rVert_p\leq\Big\lVert\hat g\Big\rVert_p$$
whenever $p\geq 4$. How can I prove this?
I do have a proof in the case when $p=2n$ for $n\in\mathbb Z$. Then we can write:
$$\left\lVert \hat f\right\rVert_p = \left\lVert \hat f\right\rVert_{2n} = \left\lVert {\hat f}^n\right\rVert_{2}^{1/n}=\left\lVert f^{*n}\right\rVert_{2}^{1/n}$$
(using the convolution power) and likewise $\left\lVert \hat g\right\rVert_p =\left\lVert g^{*n}\right\rVert_{2}^{1/n}$. Then since we have
$$\lvert f^{*n}(x)\rvert=\left\lvert\int_{\sum x_i=x}f(x_0)\dots f(x_{n-1})\right\rvert\leq\int_{\sum x_i=x}\lvert f(x_0)\rvert\dots \lvert f(x_{n-1})\rvert=\lvert g^{*n}(x)\rvert$$
for all $x$, we are done.
So I'm interested in the more general case where $p$ is an arbitrary real number (greater than or equal to $4$).
My numerical experiments were based on the discrete fourier transform. The requirement that $p\geq 4$ comes from the example $f = (1,-1,0)$. Then
$$\left\lVert\hat f\right\rVert_p=\left(\frac{\sqrt{3}^p+\sqrt{3}^p+0^p}3\right)^{1/p}$$
and $$\left\lVert\hat g\right\rVert_p=\left(\frac{2^p+1^p+1^p}3\right)^{1/p}.$$
Thus for this $f$ and $g$, the inequality $ \Big\lVert\hat f\Big\rVert_p\leq\Big\lVert\hat g\Big\rVert_p$ holds only when $p\leq 2$ or $p\geq 4$.
The motivation for this question is to try to tighten the entropic uncertainty principle in quantum mechanics. The theorem would say that for a given position distribution and $n\geq 2$, the Rényi entropy $H_n$ of the momentum distribution is minimized by the wavefunction with constant phase. (I do know how to prove that this wavefunction minimizes the variance of the momentum distribution, so it already gives a tightening of Heisenberg's uncertainty principle.)