Let $G$ be a locally compact group, let $A(G)$ be the Fourier algebra of $G$. We think of $A(G)$ as a subalgebra of $C\_0(G)$.



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> Question 1: Let $f\in A(G)$ be a function that is pointwise positive. Does the function $\sqrt{f}$ belong to $A(G)$?
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The motivation for this Question is the following:



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> Question 2: Given $f\in A(G)$, does the function of absolute values, $|f|$, belong to $A(G)$?
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Since $A(G)$ is closed under passing to complex conjugation, a positive answer to Question 1 would imply a positve answer to Question 2.


Additionally, if $f,|f|\in A(G)$, is there a relation between the norms of $f$ and $|f|$ in $A(G)$?



Root of positive function in Fourier algebra
This won't work. I want to show that we can't take square roots in $A(\mathbb R)$. My function will be of the type
$$
f(x) = \sum h\_n \varphi\left( \frac{x-a\_n}{L\_n}\right) ,\quad\quad\quad\quad (1)
$$
and here the individual summands will have disjoint supports. I will take $h\_n\in\ell^2$, $h\_n\notin\ell^1$. Since the $L^1$ norm of the Fourier transform of $\psi((x-a)/L)$ is independent of $a,L$, the first property will make sure that $\widehat{f^2}\in L^1$, that is $f^2\in A(\mathbb R)$.


So it is now enough to find $\varphi\ge 0$ and $h\_n,a\_n,L\_n$ such that $f\notin A(\mathbb R)$. Fix a $\varphi\ge 0$ that is supported by $[0,1]$, with $\varphi'(0)>0$ and $\varphi$ is smooth otherwise. Then (after normalizing suitably) we will have that $|\widehat{\varphi}(t)|=1/t^2+O(t^{-3})$, and of course $\widehat{\varphi}$ is bounded. Thus the Fourier transform of the $n$th summand of (1) is of the order
$$
\frac{h\_n}{L\_n x^2} + O(h\_nL\_n^{-2}x^{-3}) ,
$$
and it is bounded by $Ch\_nL\_n$. Thus for sufficiently large $B$, this summand on its own on the interval $B/L\_n\le x\le 2B/L\_n$ would make a contribution $\gtrsim h\_n/B$ to the $L^1$ norm of $\widehat{f}$.


We will be done if we can make sure that the other summands cannot completely cancel out this contribution. The summands with $k>n$ cannot contribute more than $CBh\_{n+1}L\_{n+1}/L\_n$, and the ones with $k<n$ can be treated as above, with $h\_n/L\_n$ replaced by $h\_k/L\_k$. So all competing contributions are much smaller if we take $L\_n$'s that converge very rapidly to zero.


