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There is a conclusion: For any $x\in \mathbb R^\mathbb N$, we denote by $A_x$ the set $$A_x= \{a\in \mathbb R^\mathbb N:\sum_n x(n)\alpha(n)~\text{converges}\},$$ then for $y,x_1,x_2,\dots,x_k \in \mathbb R^\mathbb N$ we have: $$ \bigcap_{n=1}^k A_{x_n} \subset A_y~ \Leftrightarrow ~ y(n)=\sum_{m=1}^k x_m(n)L_m(n) \quad(n\geq N_0) $$ where $L_m \in \mathbb R^\mathbb N $ is a sequence of bounded variation ($\sum_n |L_m(n+1)-L_m(n)| < \infty$) for any $m=1,2,\dots,k$ and $N_0$ is a large enough constant.

Actually, I think this conclusion has already been proved by someone; who has seen anything related to this conclusion?? Thanks.

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  • $\begingroup$ Do you have a proof for this result? $\endgroup$ Commented Aug 4, 2021 at 15:14
  • $\begingroup$ Yes, I have a proof of my own. However, I think this result has been finished by someone earlier. $\endgroup$
    – Bo Peng
    Commented Aug 5, 2021 at 10:05

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