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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.


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? – Zhen Lin Nov 5 '11 at 14:08
@ZhenLin: I think $\alpha_{k}'s$ are constants, $\phi_{k}'s$ are functionals, but the question is silent about norm. – Hassan Muhammad Nov 5 '11 at 14:17
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 $l^1$. – GEdgar Nov 5 '11 at 14:29
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 $\lVert T_n\rVert=\sum_{k=1}^n|\alpha_k|$. – Davide Giraudo Nov 5 '11 at 14:30
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 $\phi_k\to 0$. – Henning Makholm Nov 5 '11 at 14:31

This is a great question!. Here's a brief sketch.

Assume that $\sum |a_k| = \infty$. Consider $\phi_k$ s.t. if $\sum_{n_k}^{n_{k+1}} |a_j| < k$, then $\phi_k (n) = \frac{1}{k}$ for $n_k < n < n_{k+1}$. Btw, $k \in \mathbb{N}$ throughout.

What happens to $\sum a_j \phi_k$?

I note in addition that the sign of the $\phi_k$ should match their respective $a_j$.

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Good (+1). Perhaps a bit clearer: consider the sequence $\{n_k\in\mathbb{R}\}$ so that $\displaystyle\sum_{n_k\le n<n_{k+1}}|a_n|\ge1$. Let $\phi_n=\frac{1}{k}\operatorname{signum}(a_n)$ for $n_k\le n<n_{k+1}$. – robjohn Nov 5 '11 at 15:27

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