Is $f$ is non-prime, Can we say $|f|$ is also non-prime ; in convolution algebra? 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 $n\neq 0$.)

Thanks to math fraternity;-) 
 A: We may phrase the problem as
$$\textit{Is it true that $f\in A(\mathbb{T})$ imply $|f|\in A(\mathbb{T})$?}$$
This was proved to be false by Y. Katznelson (Sur les fonctions operant sur l’algebre des  series de Fourier absolument convergentes, C.R.A.S. Paris, 247 (1958), 404–406). The result is that Levy's theorem cannot be improved - that is analytic functions are the only functions that operate on $A(\mathbb{T})$.
A: Let  $f\in L^{2} \ast L^{2}$. Then $f=g\ast h$ for some $g, h \in L^{2}$; So 
$$\sum_{n\in \mathbb Z} |\hat {f} (n)| = \sum_{n\in \mathbb Z} |\hat {g}(n) \hat {h}(n)|\leq \|\hat {g}\|_{\ell^{2}} \|\hat{h}\|_{\ell^{2}} <  \infty;$$
second inequality by Schwartz-inequality and and third by Riesz–Fischer theorem; hence $f\in A(\mathbb T)$.
On the other hand, let $f\in A(\mathbb T)$; construct $c_{n}= |\hat{f}(n)|^{\frac {1} {2}}$ and $d_{n}= |\hat{f}(n)|^{\frac {1}{2}} e^{i \text{arg} \hat {f} (n)}$; note that, $c_{n}, d_{n} \in \ell^{2}(\mathbb Z)$; so by Reisz-Fisher Theorem, there exists $g, h \in L^{2}(\mathbb T)$ such that $\hat{g}(n)=c_{n}$ and $\hat {h}(n)= d_{n}$; therefore by uniqueness theorem, $f=g\ast h \in L^{2}\ast L^{2}(\mathbb T)$. 
