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

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;)

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By dominated convergence, $\hat f_n$ converges pointwise to $\hat f$. The claim then follows from Fatou's lemma. – Terry Tao Apr 30 '14 at 16:38
@prof.TT; thanks a lot; but would you please tell me; (If I understood correctly, for applying dominated convergence; put, $g_{n}(t)=f_{n}(t)e^{-ikt}, t\in \mathbb T$ fix $k\in \mathbb Z$, and we need to find, $h\in L^{1} (\mathbb T)$ such that $|g_{n}(t)| \leq h(t)$; my question is, how should I take $h$ ?; or am I missing something completely ?; ) Thanks a lot; – Inquisitive May 1 '14 at 4:37
@DivyangBhimani $|g_n(t)| \leq M$ for all $n$ and all $t$. – Yemon Choi May 1 '14 at 12:16
@YC; thanks; would you please tell me, how it follows ? – Inquisitive May 1 '14 at 12:20
In fact, I've just noticed that this question was cross-posted on MSE math.stackexchange.com/questions/775600 In future, please don't post simultaneously on both sites, or at least add a link on each version to the other version. – Yemon Choi May 5 '14 at 0:40