Our functional analysis instructor assigned us a problem which is as follows:
Let $(a_n)$ be a sequence of real numbers such that for each real sequence $(x_n)\to 0$ the sum $\sum\limits_{n=1}^\infty a_nx_n$ converges.Then show that $\sum\limits_{n=1}^\infty a_n$ is absolutely convergent.
He has also given us some hints.The hint is as follows:
Define, $T_N:c_0\to \mathbb C$ by,
$T_N(x)=\sum\limits_{n=1}^N a_nx_n$ where $x\in c_0$
and apply uniform boundedness principle on the family $\{T_N\}_{N\in \mathbb N}$.
But in order to apply uniform boundedness principle our operators need to be continuous and pointwise.I have shown pointwise boundedness by observing $|T_N(x)|=|\sum\limits_{n=1}^N a_nx_n|\leq B_x$ for all $N$, as the series $\sum\limits_{n=1}^\infty a_nx_n$ converges for each $x=\{x_1,x_2,...\}\in c_0$.But I am not able to show that $T_N$ is continuous for all $N\in \mathbb N$.Can someone give me a clue?
The problem is made up for practicing the uniform boundedness principle (the Banach-Steinhaus theorem). But it can be solved directly.
Assume by contradiction that $\sum |a_n|=\infty. $ Therefore there exists an increasing sequence of indices $\{n_k\}_{k=1}^\infty $ such that $n_1=1$ and $$\sum_{n=n_{k}}^{n_{k+1}-1} |a_n|\ge 1,\quad k=1,2,\ldots $$ Let $a_n=|a_n|e^{i\varphi_n},$ where $\varphi_n$ denotes the argument of the complex number $a_n.$ Consider the sequence $\{x_n\}_{n=1}^\infty$ $$x_n= {1\over k} e^{-i\varphi_n},\quad {\rm for}\ n_k\le n<n_{k+1}$$ The sequence $\{x_n\}_{n=1}^\infty $ tends to $0.$ Moreover $$\displaylines{\sum_{n=1}^\infty a_nx_n=\sum_{k=1}^\infty \sum_{n=n_k}^{n_{k+1}-1} a_nx_n\\ =\sum_{k=1}^\infty \sum_{n=n_k}^{n_{k+1}-1} {1\over k}|a_n|\ge \sum_{k=1}^\infty {1\over k}=\infty}$$ hence the series $\sum a_nx_n$ is divergent.