showing a sequence is in $l^1$ [duplicate]

Question

Let $(a_n)_n$ be a sequence in $\mathbb C$ such that for any sequence $(x_n)_n \in c_0$ the sum $\sum a_nx_n$ is convergent. Show that $(a_n)_n$ is in $l^1$

I have shown $(a_n)_n$ is in $l^p$ for all $p>1$. I was trying to figure out what goes wrong for the sequence $(1/n)$ in which case we get a concrete $x_n=1/(log{ }n)$. So how do I do this general case ?

EDIT: Since there were some comments regarding duplicates of some post let me post an approach I was trying We know $c_0^*=l^1$ so if $(a_n)_n$ is not in $l^1$ the functional from $c_0 \rightarrow \mathbb K$ given by $(x_n)_n \mapsto \sum a_nx_n $ is unbounded and hence we will get a sequence of sequences say $(x_{n,k})_n$ such that $ |x_{n,k}|<1/k$ $ \forall n$ but $\sum a_{n,k}x_n > \epsilon \forall k $ (after possibly multiplying by $e^{i\theta_k}$ ) Consider the sequence $(z_n)_n$ given by $z_n=\sum_{k} \frac {x_{n,k}}{k}$. Using DCT it is easy to see that $lim_{n\to \infty} z_n =lim_{n\to \infty} \sum_{k}\frac{x_{n,k}}{k}=0 $ and hence $(z_n)_n\in c_0$ but $\sum a_nz_n> \epsilon \sum 1/k$ which goes to infinity. But the catch is in the last part. Any insight will be helpful.

Answer 1

If you are willing to look at a different approach you can do the following: define $T_N:c_0 \to \mathbb R$ by $T_N(x_n)=\sum\limits_{k=1}^{N}a_kx_k$. Then $\|T_n\| \leq \sum\limits_{k=1}^{N}|a_k|$. By taking $x_k=\frac {|a_k|} {a_k}$ if $a_k \neq 0$ and $0$ if $a_k=0$ we see that $\|T_n\| = \sum\limits_{k=1}^{N}|a_k|$. Since $T_N$ converges at every point of $c_0$ Uniform Boundedness Principle tells you that $sup_n \|T_n\| <\infty$ which proves that $(a_n) \in \ell^{1}$.

Answer 2

If you know that the dual space of $c_0$ is $\ell^1$ then I believe you may be complicating things. By the proof in the linked post, we see that a sequence $(a_n)$ is an element of $\ell^1$ iff $(a_n)$ is an element of the dual of $c_0$ by identifying it with the mapping $c_0\ni(x_n)\mapsto\sum_na_nx_n$. Your assumption on $(a_n)$ is precisely that it is in the dual. If you need to prove it directly, this proof is basically that given in the second part of the question in the linked post.