Showing that $a_n \in \ell^{\infty}$

Question

Let $a_n \subset \mathbb C$ is a sequence

Show that if $\sum_{n=1}^{\infty} a_nx_n \lt \infty$ $\forall x_n \in \ell^1$ then $a_n \in \ell^{\infty}$

What did I do :

$x_n \in \ell^1 \Rightarrow \sum_{n=1}^{\infty} |x_n| \lt \infty $

Since $\mathbb C$ is a Banach space $\sum_{n=1}^{\infty} x_n \lt \infty $ and from Leibnitz rule $x_n \to 0$

$\sum_{n=1}^{\infty} a_nx_n \lt \infty \Rightarrow$ $a_nx_n \to 0$ i.e. $\forall \varepsilon \gt 0$$ |a_nx_n| \lt \varepsilon $

$|a_n||x_n|=|a_nx_n| \lt \varepsilon \Rightarrow$ $|a_n| \lt \varepsilon /|x_n| $ and $|x_n|$ is a finite number thus terms of $a_n$ are bounded. So $a_n \in \ell^{\infty}$

I’m not sure about my writtens. It seems wrong to me. Could someone fix me please? Thanks a lot

Answer 1

Hint: try contraposition. Suppose $(a_n)_n\notin \ell^\infty$, and show that there is a sequence $n_k$ such that $\lvert a_{n_k}\rvert\geq k^2$. Using that, find a sequece $(b_n)_n\in \ell^1$ such that $\sum a_nb_n=\infty$.