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I am solving some problems in a text, I come across this question. I thought I will not take much of my time on it, but that is not the case.

Question:

Prove that if $\sum \limits_{k=1}^{\infty}\alpha_{k}\phi_{k}$ is convergent whenever $\lim_{k\to \infty}\phi_{k}=0$, then $\sum \limits_{k=1}^{\infty}|\alpha_{k}|\lt \infty$

I am sorry for this question, it seems too simple but I do not know how to tackle it. Please I need just a hint for this.

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    What are $\alpha_k$ and $\phi_k$? Constants? Functionals? What norm are you using?2011-11-05
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    @ZhenLin: I think $\alpha_{k}'s$ are constants, $\phi_{k}'s$ are functionals, but the question is silent about norm.2011-11-05
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    No, I think all of them are real numbers, and convergence is in the usual real-number sense. More sophisticated language: This is a step in the proof that the dual of $c_0$ is $l^1$.2011-11-05
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    Since the question was initially tagged "functional analysis, I have a solution with the principle of uniform boundedness. Define $T_n\colon c_0\to \mathbb R $ as $T_n(\varphi)=\sum_{k=1}^n\alpha_k\phi_k$. Since for any fixed $\varphi\in c_0$ we have that the sequence $\{\sum_{k=1}^n\alpha_k\varphi_k\}_n$ is bounded (as a convergent sequence), and $c_0$ with the uniform norm is complete, we have that $\sup_{n\in\mathbb N}\lVert T_n\rVert<\infty$. But taking for a fixed $n$ $\varphi(k)=\operatorname{sgn}\alpha_k$ for $k\leq n$ and $0$ otherwise, we get $\lVert T_n\rVert=\sum_{k=1}^n|\alpha_k|$.2011-11-05
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    It is probably easiest to attack this in contraposed form: If $\sum |a_k|=\infty$, then construct a sequence $\phi_k$ such that $\sum |a_k|\phi_k =\infty$, yet $\phi_k\to 0$.2011-11-05

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