Is it true that for $f\in\mathscr{S}(\mathbb{R}^n)$, we have $\widehat{\lvert\nabla f\rvert}=\left\lvert\widehat{\nabla f}\right\rvert$?

Question

Let $f\in\mathscr{S}(\mathbb{R}^n)$ be Schwartz class, then I wish to prove that $$\widehat{\lvert\nabla f\rvert}=\left\lvert\widehat{\nabla f}\right\rvert$$ where $\widehat{\nabla f}$ is defined term-wise. Now, it is not so hard to calculate that $$\left\lvert\widehat{\nabla f}\right\rvert=\lvert\xi\rvert\lvert\widehat{f}(\xi)\rvert$$ but I don't see any way to simplify $$\widehat{\lvert\nabla f\rvert}(\xi) = \int_{\mathbb{R}^n}\sqrt{\lvert\partial_1 f(x)\rvert^2+\cdots+\lvert\partial_n f(x)\rvert^2}e^{-2\pi i x\cdot\xi}\,\mathrm{d}x$$

Answer 1

Say $n=1$ and $g=\nabla f \in \mathscr{S}$. If $g$ crosses zero, say, with $g'\neq 0$ there, then $|g|$ isn't smooth, so its Fourier transform doesn't decay fast at infinity, but $|\hat{g}|$ decays fast at infinity. So they can't be equal.