Root of positive function in Fourier algebra 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 $C_0(G)$.

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

The motivation for this Question is the following:

Question 2: Given $f\in A(G)$, does the function of absolute values, $|f|$, belong to $A(G)$?

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 $A(G)$?
 A: To complement Christian Remling's nice concrete explanation, let me just add that more is known.
Your 2nd question can be rephrased as asking: does the function $x\mapsto |x|$ ``operate in'' the Fourier algebra? This kind of question used to be of interest to people working on Banach algebras of functions. For the Fourier algebra of a LCA group the answer is: a function which operates in A(G) for G abelian must be real-analytic. This is (apparently) proved in
MR0116185 (22 #6980)
H. Helson, J.-P. Kahane, Y. Katznelson, W. Rudin
The functions which operate on Fourier transforms.
Acta Math. 102 1959 135–157.
In particular, this implies that your Q2, and hence your Q1, have negative answers. Of course for particular $f$ you may be able to do better by the holomorphic functional calculus for Banach algebras.
You can find some generalizations to Fourier algebras of nonabelian groups by looking at the citations as listed in MathSciNet. My impression is that in most cases one just lifts the result from the abelian case by using Herz's restriction theorem.
A: 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.
