A problem on the Wiener Algebra We say that $f\in C(\mathbb{T})$ is in the Wiener Algebra $A(\mathbb{T})$ if the series of Fourier coefficients $$\sum|\widehat{f}(n)|$$ is convergent (absolutely).
It is verified that this sum defines a norm ,which we symbolize $\Vert{f}\Vert_{A(\mathbb{T})}$.
The point is to prove that if $f_n \in A(\mathbb{T})$ with $\Vert{f_n}\Vert_{A(\mathbb{T})} \leq 1$ converges uniformly on $f$ then $f$ is in the Wiener Algebra and moreover $\Vert{f}\Vert_{L^1} \leq 1$.
My ideas:
 Uniform converge justifies that $\int f_n(s) e^{-iks}ds$ goes to $\int f(s)e^{-iks}ds=\widehat{f}(k)$.
Then for a fixed $k$,$r>0$ we can choose a subsequence $f_{n_j}$ with $|\widehat{f}_{n_j}(k)-\widehat{f}(k)|\leq \frac{r}{2^k}$ and use the triangle inequality to prove the first claim.
Now i find it hard to answer the claim about the norm.One thought that comes to mind is to make use of a functional analysis result namely $$\Vert {f}\Vert=sup\{|x^*(f)|,x^* \in B_{C(\mathbb{T})^*}\}$$ which must be the regular Borel measures on the circle .
Does $$m_k(A)=\int_A e^{-iks}dm(s)$$ define such a measure and is it sufficient (because of some density argument ) to prove the claim for the norm??? 
Any thoughts would be great.
 A: Here is a proof using the Banach-Alaoglu theorem for the $A(\mathbb{T})$-membership part, as the norm estimate is already mentioned in the comments:
Since $A(\mathbb{T}) \cong \ell^1(\mathbb{Z}) = c_0(\mathbb{Z})^*$ and $f_n$ is a sequence in the unit ball of $A(\mathbb{T})$, which is weak-$\ast$-sequentially compact (as $c_0(\mathbb{Z}) = \{x \in \mathbb{C}^{\mathbb{Z}} \mid \lim_{n \to \pm\infty} x_n = 0\}$ is separable), there exists a subsequence $f_{n_k}$ and some $g \in A(\mathbb{T})$ with $||g||_{A(\mathbb{T})} \leq 1$ such that $ \lim_{k\to \infty} \langle h, f_{n_k} \rangle = \langle h, g \rangle$ for any $h \in c_0(\mathbb{Z})$.
Picking $h$ as unit basis vectors, we see that the Fourier coefficients of $f_{n_k}$ converge pointwise to the Fourier coefficients of $g$.
But since $f_{n_k}$ converges uniformly to $f \in C(\mathbb{T})$, the Fourier coefficients of $f_{n_k}$ also converge pointwise to the Fourier coefficients of $f$.
Hence $\hat{g}(n) = \hat{f}(n)$ for all $n \in \mathbb{Z}$ and $f \in A(\mathbb{T})$.
