Let $f\in L^{1}(\mathbb T)$ and define the Fourier coefficient of $f$ : $\hat{f}(n)=\frac{1}{2\pi} \int _{-\pi}^{\pi} f(t)\, \mathrm{e}^{-int} dt; (n\in \mathbb Z)$.Consider the space, $$A(\mathbb T):= \{f:\mathbb T \to \mathbb R : \hat{f}\in \ell^{1}(\mathbb Z), \ \text {that is,} \ \sum_{n\in \mathbb Z} |\hat{f}(n)| < \infty \}.$$ $A(\mathbb T)$ is normed by the $L^{1}-$ norm on $\mathbb Z$: $$||f||= \sum_{n\in \mathbb Z} |\hat{f}(n)| < \infty; \ \text {for} \ f\in A(\mathbb T). $$ We also note that $A(\mathbb T)$ is a Banach algebra under pointwise addition and multiplication.
Fix $r\in (0, \infty)$, and let $f\in A(\mathbb T)$ such that $||f|| \leq r$; and so, $\mathrm{e}^{if}\in A(\mathbb T)$ and $||\mathrm{e}^{if}||\leq \mathrm{e}^{||f||}\leq \mathrm{e}^{r}$.
My question are: (1) Does there exists $f\in A(\mathbb T)$ such that $||f||=r$ and $||\mathrm{e}^{if}||= \mathrm{e}^{r}$ ? (2) Otherwise, How to we get a sequence $(f_{n})\subset A(\mathbb T)$ with $||f_{n}||\leq r$ such that $\lim_{n\to \infty} ||\mathrm{e}^{i f_{n}}||= \mathrm{e}^{r}$ ?
Edit:(Note that if we allowed complex valued functions in $A(\mathbb T)$; then the answer is given below by prof. Y S)
Thanks,
Question $(1)$. Try $f(t)=-i$ (a constant function).