By Schwartz-inequality and Riesz–Fischer theorem, one can deduced that, 
 $$L^{2}(\mathbb T) \ast L^{2}(\mathbb T) = A(\mathbb T)(:= \{f\in L^{1}(\mathbb T): \sum\_{n\in \mathbb Z} |\hat{f}(n)| < \infty \}).$$


My question is:


Let $f \in L^{2}(\mathbb T) \ast L^{2}(\mathbb T)$ that is, $f=g \ast h$ for some $g, h \in L^{2}(\mathbb T)$.


Can we expect $|f| \in L^{2}(\mathbb T)\ast L^{2} (\mathbb T)$, that is, $|f|$ is also can be factorize as a convolution of two $L^{2}(\mathbb T)$ functions; in other words, $|f|$ also has absolutely convergent Fourier series; 


Or, we can produce counter example ?


(Note that, there exist $f \in L^{1}(\mathbb T)$ such that, $|f|\in A(\mathbb T)$ but $f\not\in A(\mathbb T);$ for instance, take $f(x)= x, \ \text {for} \ x\in [-\pi, \pi]$; $\hat{f}(n)= \frac {(-1)^{n+1}}{in}$ , for $n\neq 0$ and
$\hat{|f|}(n) = \frac {2} {\pi} \frac {(-1)^{n} - 1}{n^{2}}$, for $n\neq 0$.)




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Thanks to math fraternity;-) 



If $f$ is non-prime, can we say $|f|$ is also a non-prime; in convolution algebra?
The answer is no. Let $A$ be the space of real values functions with absolutely convergent
Fourier series, and $F$ is a real function on some interval $I$.


Theorem (Katznelson). If $F\circ f\in A$ for all $f\in A$ with range in $I$, then $F$
is analytic on $I$.


Taking $F(x)=|x|$ we obtain that there is $f\in A$ such that $|f|$ is not in $A$.


References:


J-P Kahane, Series de Fourier absolument convergentes, Springer NY&, 1970, Ch. VI, sections 5-6.


Y Katznelson, Sur les fonctions operant sur l'algebre des series de Fourier
absolument convergentes, C. R., 247 (1958) 404-406.


