Let $f\in L^{1}(\mathbb T)$ and define the Fourier coefficient of $f$ : $\hat{f}(n)=\frac{1}{2\pi} \int _{-\pi}^{\pi} f(t) e^{-int} dt; (n\in \mathbb Z)$.Consider the space, $$A(\mathbb T):= \{f\in L^{1}(\mathbb T): \hat{f}\in \ell^{1}(\mathbb Z), \ \text {that is,} \ \sum_{n\in \mathbb Z} |\hat{f}(n)| < \infty \}.$$ $A(\mathbb T)$ is normed by the $L^{1}-$ norm on $\mathbb Z$: $$||f||_{A(\mathbb T)}= \sum_{n\in \mathbb Z} |\hat{f}(n)| < \infty; \ \text {for} \ f\in A(\mathbb T). $$ We also note that $A(\mathbb T)$ is a Banach algebra under pointwise addition and multiplication.
My Question: Can we expect $\|fg\|_{A (\mathbb T)} \leq C \|f\|_{L^{2}} \|g\|_{A (\mathbb T)},$ ( where $C$ is some constant) for $f,g \in A(\mathbb T)$? If yes, how to prove it?
[We recall, $\|f\|^{2}_{L^{2}}= \int_{0}^{2\pi}|f(t)|^{2} dt= \sum_{n\in \mathbb Z}|\hat{f}(n)|^{2}$]
First, note that $$ \left\|\,f\vphantom{\hat{f}}\,\right\|_{A(\mathbb{T})}=\left\|\,\hat{f}\,\right\|_{\ell^1} $$ Suppose $g(x)=1$. Then $\hat{g}(n)=[n=0]$, and therefore, $\|g\|_{A(\mathbb{T})}=1$. Then, your question asks if $$ \left\|\,f\,\right\|_{A(\mathbb{T})} \le C\left\|\,f\,\right\|_{L^2} $$ which is impossible. For counterexample, we can look at $$ \widehat{f_n}(k)=\left\{\begin{array}{cl} \frac1{|k|}&\text{if }1\le|k|\le n\\ 0&\text{otherwise} \end{array}\right. $$ where $\|f_n\|_{A(\mathbb{T})}=2H_n$, yet $\|f_n\|_{L^2}\le\frac{\pi^2}3$, where $H_n$ is a Harmonic number.