# Dual spaces of complex sequences, show the second member is in the dual space

I'm having trouble with some of (ok, most of) the exercises in my 1st-year-master's functional analysis class, so here's one of them, hoping someone can help me out:

If a sequence $(b_n)$ is absolutely convergent, and $\sum_na_nb_n$ is convergent, how can I show that $(a_n)$ is a bounded sequence? The idea is with dual spaces, as the dual of the space of absolutely convergent sequences is the space of bounded sequences, but I don't know how to show this.

In a similar question with $(b_n)$ a sequence converging to $0$, we showed $(a_n)$ was absolutely convergent by showing that $(a_nb_n)$ converged absolutely, then creating a sequence of functions to apply the Banach Steinhaus theorem.

I think I should do something similar here, but I can't show that $(a_nb_n)$ converges absolutely, so the proof I already know doesn't work.

Does anyone have a few hints? I don't think my professor is going to help us with this one, and he's extremely frightening and hard to understand (to add to understanding problems, my classes are in my 2nd language), and I can't seem to find anything about this problem on the internet. (It's always the sequences $a_n$ and $b_n$, then show the series $(a_nb_n)$ converges, which doesn't help here). Thank you to anyone who can help!

• Do you see why the set of bounded sequence's is contained in the dual of $l_1(\mathbb{C})$? – Moss Mar 25 '13 at 21:31
• Yes, I need to show that the sequence (an) is bounded, and thus contained in l-infinity(C), which is, in my understanding, the dual of l1(C). – JKH Mar 25 '13 at 22:13

You wrote:

If a sequence $(b_n)$ is absolutely convergent, and $\sum_na_nb_n$ is convergent, how can I show that $(a_n)$ is a bounded sequence?

You can’t: $\left\langle\frac1{n^2}:n\in\Bbb Z^+\right\rangle$ is absolutely summable, and $$\sum_{n\ge 1}\left(\big((-1)^nn\big)\cdot\frac1{n^2}\right)=\sum_{n\ge 1}\frac{(-1)^n}n=-\ln 2$$ converges, but $\left\langle(-1)^nn:n\in\Bbb Z^+\right\rangle$ is not bounded. (For that matter, if $b_n=0$ for each $n$, the sequence $\langle a_n:n\in\Bbb Z^+\rangle$ could be anything.)

Perhaps you meant to show this:

If $\sum_{n\ge 1}a_nb_n$ converges for each absolutely summable sequence $\langle b_n:n\in\Bbb Z^+\rangle$, then $\langle a_n:n\in\Bbb Z^+\rangle$ is bounded.

If so, assume that $\langle a_n:n\in\Bbb Z^+\rangle$ is unbounded, and construct an absolutely summable sequence $\langle b_n:n\in\Bbb Z^+\rangle$ such that $\sum_{n\ge 1}a_nb_n$ diverges.

• I'm sorry, I missworded the question, the second answer is the one I was looking for. $(b_n)$ is any sequence in l1(C). Thank you, I didn't realize the answer could be so simple, I was too busy trying to wrap my head around our duality theorems and use of the Banach-Steinhaus theorem. – JKH Mar 26 '13 at 8:16
• @JKH: You’re welcome. – Brian M. Scott Mar 26 '13 at 8:20