Considering the function $f:\mathbb{R} \to \mathbb{C}$, with $\left f(x) \right=1$ for all $x\in \mathbb{R}$.
Considering $g:\mathbb{R} \to \mathbb{C}$ with $\int_{\infty}^{\infty}{\leftg(x)\right^2dx}=1$
I am interested on properties of the amplitude of the Fourier Transform of the product of $f$ and $g$:
$A(k)=\leftFT(f(x)g(x))\right$
Is there any constraint on $A$ apart from the fact that A is real positive? Considering a fixed $g$, is it possible to attain any $A$ simply by changing $f$?
Thank you for any help



If $g$ happens to be in $L^1$, then the amplitude of the Fourier transform of $fg$ is bounded by the $L^1$ norm of $g$, for any unimodular $f$. This is the only restriction from above since you can always choose $f$ so that $fg\ge 0$, thus bringing the (essential) supremum of $\widehat{fg}$ up to $\g\_{L^1}$. Another part of the question is how small we can make $A$. I guess "arbitrarily small", but don't have a proof. (Except for special case: if $g$ is in $L^1$, then we can chop it into pieces with disjoint supports and small $L^1$ norm, and then use $f$ to move the Fourier transforms of pieces far from one another.) 


The only thing you can say is $\int_{\infty}^{+\infty}A(k)^2dk=1$, since the Fourier transform is an isometry of $L^2(\mathbb{R})$. Any function $A(k)$ with $\int_{\infty}^{+\infty}A(k)^2dk=1$ is the Fourier transform of a function of the form $f\cdot g$. Let $h$ be the inverse Fourier transform of $A(k)$, so that $A=\hat h$ and $\int_{\infty}^{+\infty}h(x)^2dx=1$. We can write $h=g\cdot f$ with $g=h/f$; then $$ \int_{\infty}^{+\infty}g(x)^2dx=\int_{\infty}^{+\infty}h(x)^2dx=1 $$ and $A=\widehat{f\cdot g}$ 

