A note on multiplier convergent series

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• 作者：Jan Spěvá ; k ^{jan.spevak@ujep.cz" class="auth_mail" title="E-mail the corresponding author}
• 关键词：primary ; 46A35 ; secondary ; 46A11
• 刊名：Topology and its Applications
• 出版年：2016
• 出版时间：15 September 2016
• 年：2016
• 卷：211
• 期：Complete
• 页码：28-37
• 全文大小：342 K

文摘

Given a topological ring R   and F⊂R^N$\mathcal{F}\subset R^{\mathbb{N}}$ a (formal) series ∑_n∈Nx_n${\sum }_{n\in \mathbb{N}}{x}_n$ in a topological R-module E   is 18e0892aed1a66c606df4d" title="Click to view the MathML source">F$\mathcal{F}$multiplier convergent in E   (respectively 18e0892aed1a66c606df4d" title="Click to view the MathML source">F$\mathcal{F}$multiplier Cauchy in E  ) provided that the sequence e05a5b0236">$\{{\sum }_{i=0}^{n}r(i)x_i:n\in \mathbb{N}\}$ of partial sums converges (respectively, is a Cauchy sequence) for every sequence function r∈F$r\in \mathcal{F}$. In this paper we investigate for which G⊂R^N$\mathcal{G}\subset R^{\mathbb{N}}$ every 18e0892aed1a66c606df4d" title="Click to view the MathML source">F$\mathcal{F}$ multiplier convergent (Cauchy) series is also G$\mathcal{G}$ multiplier convergent (Cauchy). We obtain some general theorems about the Cauchy version of this problem. In particular, we prove that every Z^N${\mathbb{Z}}^{\mathbb{N}}$ multiplier Cauchy series is already R^N${\mathbb{R}}^{\mathbb{N}}$ multiplier Cauchy in every topological vector space. On the other hand, we construct examples that in particular show that a Z^N${\mathbb{Z}}^{\mathbb{N}}$ multiplier convergent series need not to be even Q^N${\mathbb{Q}}^{\mathbb{N}}$ multiplier convergent and that there are topological vector spaces containing non-trivial Q^N${\mathbb{Q}}^{\mathbb{N}}$ multiplier convergent series that do not contain non-trivial R^N${\mathbb{R}}^{\mathbb{N}}$ convergent series. As a consequence of this example, there are topological vector spaces containing the topological group Q^N${\mathbb{Q}}^{\mathbb{N}}$ (and thus Z^N${\mathbb{Z}}^{\mathbb{N}}$ and Z^(N)${\mathbb{Z}}^{(\mathbb{N})}$ as well) that do not contain the topological vector space R^N${\mathbb{R}}^{\mathbb{N}}$. On the contrary, it was proved in [3], that a sequentially complete topological vector space that contains the topological group Z^(N)${\mathbb{Z}}^{(\mathbb{N})}$ must already contain the topological vector space R^N${\mathbb{R}}^{\mathbb{N}}$. Hence our example demonstrates, that in the latter result, the condition of sequential completeness can not be weakened by assuming that the space in question contains the topological group Z^N${\mathbb{Z}}^{\mathbb{N}}$ (which is the sequential completion of Z^(N)${\mathbb{Z}}^{(\mathbb{N})}$).