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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.

Aweygan
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Pmorphy
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  • If you are more careful with the first claim, you can show $|f|{A(\mathbb{T})} \le 1$. Then write $|f|{L^1}$ with $f$ expressed in terms of its Fourier series and use the triangle inequality. – Nate Eldredge Aug 02 '17 at 18:53
  • The idea is that $| f|{L^1} = \langle f, \frac{f}{|f|} \rangle = \sum_n \widehat{f}(n) \overline{\widehat{\frac{f}{|f|}}(n)}$. Or even simpler if $|f|{A([0,1])} \le 1$ then $|f(x)| = |\sum_n \widehat{f}(n) e^{-2i \pi nx} | \le |f|{A([0,1])}$ and hence $|f|{L^1([0,1])} \le 1$ – reuns Aug 02 '17 at 18:59

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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 $f \in A(\mathbb{T})$.

Andrei Kh
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