Prove that $(a_n) \in l_\infty$

Suppose that $\sum_{k=1}^{\infty} a_k x_k< \infty$ for all $x=(x_n)\in l_1$. Prove that $(a_n) \in l_\infty$

My attempt

I tried to apply Uniform boundedness principle to the linear functionals $f_n(x) = \sum_{k=1}^{n} a_k x_k, x \in l_1$.

Each $f_n$ is continuous since $|f_n(x)| \leq \sum_{k=1}^{n} |a_k x_k| \leq \bigg(\sum_{k=1}^{n} |a_k| \bigg)\|x\|_1$

Given $x \in l_1$, I need to prove that $(|f_n(x)|)$ is bounded. It seems easy but I couldn't prove it.

How could I prove it?

Thanks in advance

1 Answer

I’d attack it much more directly.

HINT: Suppose that $a=\langle a_n:n\in\Bbb Z^+\rangle\notin\ell_\infty$. Then $a$ has a subsequence $\langle a_{n_k}:k\in\Bbb Z^+\rangle$ such that $|a_{n_k}|\ge k$ for each $k\in\Bbb Z^+$. For each $k\in\Bbb Z^+$ let

$$x_{n_k}=\frac1{ka_{n_k}}\;,$$

and let all the other terms of $x$ be $0$.

• Show that $x\in\ell_1$, but
• $\sum_{n\ge 1}a_nx_n$ diverges.