# On a weaker condition of summability for Fourier series

The Wiener algebra $W:=W(\mathbb{T}^n)$ on the torus is defined as the algebra of all continuous fonctions $f$ on $\mathbb{T}^n$ such that $(\widehat f(k))_{k\in \mathbb{Z}^n} \in \ell^1(\mathbb{Z}^n)$. This is equivalent to say that the family $(\widehat f(k) e_k)_{k\in \mathbb{Z}^n}$ (where $(e_k)$ is the Fourier basis) is absolutely summable in the Banach space $C(\mathbb{T}^n)$.

Define $W'$ the set (containing $W$) of all integrable functions $f$ on the torus such that $(\widehat f(k) e_k)_{k\in \mathbb{Z}^n}$ is summable (but not necessarily absolutely) in $C(\mathbb{T}^n)$.

Question : Do we have $W=W'$ ? If not, has this space $W'$ been explored?

• @Fedor : summability in a normed space makes sense for any family indexed by any set I. The sum of a summable family $(x_i)_{i\in I}$ (for $I$ strictly countable) can be obtained as the sum of any series $\sum x_{\sigma(n)}$ where $\sigma : \mathbb{N}\to I$ is a bijection. But, if you prefer, you can consider already the question for $n=1$. – Phil-W Feb 27 '17 at 9:24
• yes, the family $(x_i)$ is summable if and only if all those series (depending on $\sigma$) converge, and in that case, they all converge to the same sum. – Jon-S Feb 27 '17 at 11:14
• @Fedor : sorry I meant : no. The family does not have to be absolutely summable for that, summability itself is enough for independence on $\sigma$. In the case of a finite dimensional normed space, summability and absolute summability coincide, but it's not the case in an infinite dimensional Banach space (where we only have one implication : absolute summability implies summability). – Jon-S Feb 27 '17 at 11:28
• ah, I see, this is a series in a Banach space, not of real numbers! You are correct, of course. – Fedor Petrov Feb 27 '17 at 11:30

## 1 Answer

If a series $\sum c(k) e_k$ is summable in $C({\mathbb T}^n)$, in particular, the series $\sum c(k) e_k(0)=\sum c(k)$ is summable (in $\mathbb{C}$), thus $\sum |c(k)|<\infty$.