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Show that if for a given sequence{$\alpha_i$} the series $$\Sigma_i \alpha_i x_i $$ is converge for all {$x_i$}$\in l^1$, Then we should have {$\alpha_i$}$\in l^\infty$.

we should show that $Sup \ \alpha_i$ is finite, But at first I think it is better to work with contrapositive and I see this post

Is there a constructive proof of this characterization of $l^2$?

I try to construct such sequences for this problem but they don't work! So I think if we use the same idea for this problem we will have some problems. So my problem is how to construct such sequences for contrapositive or if we someone have another ideas.

a.p
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2 Answers2

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If you are not keen on constructive solution there is a standard procedure for many results of this type using UBP (Uniform Boundedness Principle).

Define $T_N:\ell^{1} \to \mathbb R$ by $T_N(x_i)= \sum\limits_{k=1}^{n} \alpha_ix_i$.

Verify that $T_N$ is a bounded linear functional with $\|T_N\|=\max \{|\alpha_i|: 1 \leq i \leq N\}$.

By hypothesis $\{T_Nx: N \geq 1\}$ is bounded for each fixed $x \in\ell^{1}$. By UBP we have $sup_N \|T\|_N <\infty$ which means $(\alpha_n) \in \ell^{\infty}$.

Note: There is a constructive proof of UBP!

Kavi Rama Murthy
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Here is one way to construct a sequence for a contrapositive argument.

Assume that $\{\alpha_i\}$ is not bounded. Then for every $n\in\mathbb{N}$ there exists $i_n$ such that $|\alpha_{i_n}|\geq n^3$, and we can choose the $i_n$'s so that $i_1<i_2<\cdots i_n<\cdots$. Define a sequence $\{x_i\}$ by $\frac{|\alpha_{i_n}|}{\alpha_{i_n}}\frac{1}{n^2}$ if $i=i_n$ for some $n$, and $x_i=0$ otherwise. Then the sequence $\{\alpha_ix_i\}$ is just the sequence $$|\alpha_{i_1}|,\frac{|\alpha_{i_2}|}{2^2},\cdots, \frac{|\alpha_{i_n}|}{n^2},\cdots$$ and by the definition of $i_n$ the general term is larger than $\frac{1}{n}$, so that $\sum_{i=1}^{\infty}\alpha_ix_i$ diverges, but $\sum_{i=1}^{\infty}|x_i|$ converges.

uniquesolution
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