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;-) 

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, $f=g\ast h \in L^{2}\ast L^{2}(\mathbb T)$. 
