- Suppose $(y_n)_n$ is a sequence in $\mathbb{C}$ with the following property: for each sequence $(x_n)_n$ in $\mathbb{C}$ for which the series $\sum_n x_n$ converges absolutely, also the series $\sum_n \left(x_ny_n\right)$ converges absolutely. Can you then conclude that $(y_n)_n$ is a bounded sequence?
We are told that the series $\sum_n x_n$ converges absolutely, which means that:
$$\forall \varepsilon> 0, \exists N\in\mathbb{C} :\sum_{n=m}^{t} |x_n|<\varepsilon, \forall t>m> N$$
We also know this property of the absolute values:
$$\left|\sum x_n y_n\right|\le\sum|x_n y_n|$$
Now, what I think is that we have to assume that $(y_n)_n$ is bounded so as to assert that the series $\sum_n \left(x_ny_n\right)$ converges absolutely. Otherwise, the latest would not be true.
Assuming that $(y_n)_n$ is bounded means $|y_n|\le B$; more explicitely:
$$\left|\sum\limits_{n}x_n y_n\right|\leq \sum\limits_{n}|x_n y_n| = \sum\limits_n |x_n||y_n|\leq \sup\limits_{k}|y_k|\sum\limits_{n}|x_n|< +\infty.$$
What leads to prove that the series $\sum_n \left(x_ny_n\right)$ converges absolutely:
$$\left|\sum x_n y_n\right|\le\sum|x_n y_n|\le B\sum|x_n| < \epsilon$$
For more details on this proof see Masacroso's question: Show that if $(\sum x_n)$ converges absolutely and $(y_n)$ is bounded then $(\sum x_n y_n)$ converges
My reasoning is highly based on Masacroso's question which works with real numbers, while I do it with complex numbers. I have not read in my book that the property of the absolute values I used does not apply to complex numbers, so I assumed I could do it.
To sum up:
Can you then conclude that $(y_n)_n$ is a bounded sequence?
Yes because if it were not bounded, $\sum_n \left(x_ny_n\right)$ would not converge absolutely.
Actually I do not know if it would not converge neither absolutely nor conditionally, so if someone wanted to shed some light on this as well it would be nice.
Best Answer
You can prove this result by using the Banach-Steinhaus theorem.
More precisely, let $A_n\colon \ell^1\to\mathbb{C}$ be the functional defined by $$ A_n x := \sum_{j=1}^n x_j y_j. $$ As is customary, $\ell^1$ denotes the set of complex sequences $x = (x_1, x_2, \ldots)$ such that $\|x\|_1 := \sum_{j=1}^\infty |x_j| < +\infty$.
Clearly $|A_n x| \leq C_n \|x\|_1$, where $C_n := \max\{|y_1|, \ldots, |y_n|\}$. Hence, $A_n \in (\ell^1)^* = \ell^\infty$ and it is not difficult to check that $\|A_n\|_* = C_n$.
By assumption, for every $x\in\ell^1$ there exists the limit $$ Ax := \lim_n A_n x = \sum_{j=1}^\infty x_j y_j. $$ Then, by the Banach-Steinhaus theorem, $A\in (\ell^1)^*$ and $$ \|A\|_* \leq \liminf_n \|A_n\|_{*} = \sup_{j\in\mathbb{N}} |y_j| < \infty, $$ so that $(y_j)$ is bounded.