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All of the sequences, mentioned in this post, are real sequences. $l_1$ is the set of real sequences such that the sum of absolute value of the terms converges and $l_2$ is the set of real sequences such that sum of squares of the terms converges.

My question is: for which fixed $(\alpha_n)_{n=1}^{\infty}$, the mapping $(x_n)_{n = 1}^{\infty} \in l_2 \mapsto (\alpha_nx_n)_{n=1}^{\infty}\in l_1$ is well-defined. My guess is that the mapping is well-defined if and only if $(\alpha_n)_{n=1}^{\infty} \in l_2$. I prove one direction of the proposition as follows:

Assume $(\alpha_n)_{n=1}^{\infty} \in l_2$. By Hölder's Inequality, we have $\sum_{n= 1}^{\infty}|x_n|| \alpha_n| \le (\sum_{n= 1}^{\infty}x_n^2)^{\frac{1}{2}}(\sum_{n= 1}^{\infty}\alpha_n^2)^{\frac{1}{2}}< \infty$. I need help with the other direction of the proposition.

Klomanad
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    @milore I think I'm wrong. What I said has not disproved the converse. – fwd Jan 07 '23 at 12:36
  • @fwd. In order to show that the converse is false we need to find $(\alpha_n) \not \in l_2$ such that for all $ x_n \in l_2$, $\sum|x_n||\alpha_n|$ converges. According to the way you write, I take $(\alpha_n)_n = (1)_n$ and $(x_n)_n = (\frac{1}{n})_n$ and these two fails to disprove the converse. Can you be more specific. – Klomanad Jan 07 '23 at 12:55
  • I am very confident that the criteria for $(\alpha_n)$ is correct. – Klomanad Jan 07 '23 at 12:57
  • Ups, that's true! I will stop commenting unless I find some time to think about this... –  Jan 07 '23 at 13:47
  • see https://math.stackexchange.com/questions/37647/if-sum-a-n-b-n-infty-for-all-b-n-in-ell2-then-a-n-in-ell2 – Evangelopoulos Foivos Jan 08 '23 at 19:15

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