I am not interested in Fourier analysis, but I would like to know whether the following claim is true or not for practical purposes:
Let $f$ be the Fourier transform of $g$. Then $|f|^2$ is the Fourier transform of $|g|^2$.
As I've said, I am not interested in rigorous details, so you can assume functions $f$ and $g$ are perfectly well-behaved. I would really appreciate it if you shared with me any theorem or counterexample related to the above statement. Thanks in advance.
On
So, I believe I have a valid counter-example, so I hope you can check this quickly. I picked the "simplest" functions whose FT I know, $f(x) = e^{-\alpha x^2} \mapsto g(\mu) = \dfrac{1}{\sqrt{2 \alpha}}e^{-\mu^2/(4\alpha)}$. We have that $|f|^2 = e^{-2\alpha x^2} \mapsto F(\mu) = \dfrac{1}{\sqrt{4 \alpha}}e^{-\mu^2/(8\alpha)}$ which is not equal to $|g|^2 = \dfrac{1}{2 \alpha}e^{-\mu^2/(2\alpha)}$. Pick $\alpha = 1$ for example and you have a clear disagreement. Considering this counter-example is so simple, I would be very curious if someone could help classify functions which satisfy this property, or something of that nature.
Let me denote $\widehat{f} = \mathcal{F}(f) = \int_{\Bbb R^d} e^{-2i\pi \,x\cdot y} \,f(y)\,\mathrm dy$ the Fourier transform. Then you are looking to solutions of the equation $$\tag{1}\label{1} |\mathcal{F}(f)|^2 = \mathcal{F}(|f|^2) $$
As the example of mildboson shows, this property is of course false in general. It is of course true if $f=0$.
One can tell however more than that. Let me assume that $f\in L^2$ is a solution of the above Equation \eqref{1}. Then by Plancherel theorem $$\tag{2}\label{2} \int_{\Bbb R^d} |\mathcal{F}(|f|^2)| = \int_{\Bbb R^d} |\mathcal{F}(f)|^2 = \|f\|_{L^2}^2 < \infty $$ so $|f|^2$ has an integrable Fourier transform, which implies that $|f|^2 \in C^0_0$ (the set of continuous functions that vanish at infinity) and so $|f|\in C^0_0$. In particular $|f|(0)$ is well defined. Moreover, up to replacing $f$ by $C\,f$ for some constant $C$, one can assume without loss of generality that $f(0)=1$. By the Fourier inversion theorem and Equation \eqref{1}, one get even more that $$ 1 = |f|^2(0) = \int_{\Bbb R^d} \mathcal{F}(|f|^2) = \int_{\Bbb R^d} |\mathcal{F}(f)|^2 = \|f\|_{L^2}^2. $$ By Equation \eqref{1} and the properties of the Fourier transform, we also see that $$ \|\widehat{f}\|_{L^\infty}^2 = \| |\mathcal{F}(f)|^2 \|_{L^\infty} = \| \mathcal{F}(|f|^2)\|_{L^\infty} \\ \leq \int_{\Bbb R^d} |f|^2 = \mathcal{F}(|f|^2)(0) \leq \|\widehat{f}\|_{L^\infty}^2. $$ Hence, we proved that $\|\widehat{f}\|_{L^\infty} = \|f\|_{L^2} = f(0) = 1$. Moreover, since $|f|^2\in L^1$, it follows that $|\mathcal{F}(f)|^2 = \mathcal{F}(|f|^2)\in C^0_0$ and so $|\widehat{f}|\in C^0_0$, in particular $|\widehat{f}|(0)$ is well defined and $$ |\widehat{f}|^2(0) = \mathcal{F}(|f|^2)(0) = \int_{\Bbb R^d} |f|^2 = 1. $$ so $|\widehat{f}|(0) = 1$. Notice also that $1=f(0)\leq \|f\|_{L^\infty}$ and by Equation \eqref{2}, $\|f\|_{L^\infty}^2 = \||f|^2\|_{L^\infty} \leq \|\mathcal{F}(|f|^2)\|_{L^1} = 1$.
To summarize, any function $f\in L^2$ verifying Equation \eqref{1} should verify $$ \boxed{\|\widehat{f}\|_{L^\infty} = |\widehat{f}|(0) = \|f\|_{L^2} = \|f\|_{L^\infty} = |f(0)|}. $$
In the particular case when $f\geq 0$, still assuming without loss of generality that $f(0)=1$, then $$ \int_{\Bbb R^d} |f| = \int_{\Bbb R^d} f = \widehat{f}(0) \leq |\widehat{f}|(0) $$ so $f\in L^1$ and $0\leq \int f = \widehat{f}(0) = |\widehat{f}|(0) = 1$, which implies that $f\,(1-f)$ is a nonnegative function such that $$ \int_{\Bbb R^d} f\,(1-f) = \int_{\Bbb R^d} f - \int_{\Bbb R^d} |f|^2 = 0 $$ and so $f\,(1-f) = 0$, that is $f^2 = f$, so $f$ is the characteristic function of some set (i.e. for almost every $x\in\Bbb R^d$, $f(x) \in \{0,1\}$). But we already proved that $f\in C^0_0$, so $f$ is the constant function $0$ or the constant function $1$. Since $f\in L^2$, we deduce that $f = 0$. So there is no nontrivial nonnegative function verifying your equation!
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