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 $\mathbb Z$: $$||f||= \sum_{n\in \mathbb Z} |\hat{f}(n)| < \infty; \ \text {for} \ f\in A(\mathbb T). $$ 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 for this $f$; I want estimate for norm of $e^{if}$ like this, $||e^{if}||\leq e^{r}$.
Here is my attempt: Note that, $e^{if}= \sum_{n=0}^{\infty}\frac{(if)^{n}}{n !} = \lim_{N\to \infty} \sum _{n=0}^{N} \frac{(if)^{n}}{n!} $; so
$$||e^{if}||= ||\lim_{N\to \infty} \sum _{n=0}^{N} \frac{(if)^{n}}{n!}|| \leq \lim_{N\to \infty} ||\sum _{n=0}^{N}\frac{(if)^{n}}{n!}||$$
My question is: How do I justify the above inequality: the inter change of norm and limit ?
(Once I do that, then I think, $\lim_{N\to \infty} ||\sum _{n=0}^{N} \frac{(if)^{n}}{n!}||\leq \lim {N\to \infty} \frac{||(if)^{n}||}{n !}$ (by triangle inequality of norm)) and then by algebra property ($||fg||\leq ||f||\cdot || g ||$) and norm propery ($||\alpha f||= |\alpha| ||f||$); we get, $\lim_{N\to \infty} ||\sum _{n=0}^{N} \frac{(if)^{n}}{n!}|| \leq \lim_{N\to \infty} \frac{||f||^{n}}{n!}$ and since $||f||\leq r$, we will get, $\lim_{N\to \infty} ||\sum _{n=0}^{N} \frac{(if)^{n}}{n!}||\leq e^{r}$ )
Or, am I missing something completely and I need to change my approach ? Bit vague question: Can we expect to do such things in any Banach algebra or there is something special about $A(\mathbb T)$ ?
In every Banach space absolutely convergent series converge and you have $$\|\sum_{n=0}^\infty x_n\|\le \sum_{n=0}^\infty \|x_n\|.$$ The proof is exactly as in the case of real or complex numbers (the partial sums form a Cauchy sequence and the bound for the norms of the partial sums carries over to the limit). Therefore, your attempt is perfectly okay.