1
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.