pointwise limit of uniformly bounded sequence in $A(\mathbb T)$ is again in $A(\mathbb T)$?

Question

Let $\mathbb T$ be a circle group, and $\hat{f}(n)= \frac{1}{2\pi}\int_{0}^{2\pi} f(t) e^{-int} dt;$ $(n\in \mathbb Z, f\in L^{1} (\mathbb T)).$
Put $A(\mathbb T)= \{f\in C(\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 $$\|f\|=\|f\|_{A(\mathbb T)}= \sum_{n\in \mathbb Z} |\hat{f}(n)|; (f\in A(\mathbb T)).$$ Suppose $f_{n} \in A(\mathbb T), \ \|f_{n}\|\leq M,$ (Fix, $M>0$) $$f(y)= \lim_{n\to \infty} f_{n} (y), (y\in \mathbb T)$$ and $f$ is continuous.

My Question is: Can we expect, $f\in A(\mathbb T)$ and $\|f\|\leq M$ ?

(If answer is positive, and proof is long, proper reference also will be o.k for me;)

Thanks,

Answer 1

Using $n$ for indices of two kinds is evil. I rename $f_n$ as $f_k$.

Since $\sup_{\mathbb T} |f| \le \|f\|_{A(\mathbb T)}$, the sequence $(f_k)$ is uniformly bounded. The dominated convergence theorem implies $f_k\to f$ in $L^1$. Therefore, for every $n$ we have $\hat {f_k}(n)\to \hat f(n)$ as $k\to\infty$. By Fatou's lemma, $$\sum_{n\in \mathbb Z} |\hat f(n)| \le \liminf_{k\to\infty} \sum_{n\in \mathbb Z} |\hat {f_k}(n)| \le M$$