Concluding whether $(y_n)_n$ is a bounded sequence

Question

• 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.

Answer 1

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.

Answer 2

Suppose $y_{(\cdot)}$ is unbounded. Then no matter how much is initially ignored, $$\{y_n:n\ge N\}$$ remains unbounded. Thus you can create subsequences of $y_{(\cdot)}$ that can wildly go to infinity. But you have little control over the angle $\theta_{(\cdot)}$. This wont matter since we are interested in absolute convergence.

Can you think of a way to use your assumption to create a contradiction?

Answer 3

Yes, in fact you only need to assume that $\sum x_ny_n$ converges, perhaps not absolutely. (In fact it's enough to assume just that the sequence $(x_ny_n)$ is bounded for every absolutely convvergent series $\sum x_n$.)

Suppose $y_n$ is unbounded. There is a subsequence $y_{n_k}$ with $|y_{n_k}|>k^3$. Define $x_n$ by saying $$x_{n_k}=1/k^2,$$ $x_n=0$ if $n\ne n_k$. Then $\sum x_n$ converges absolutely, but $\sum y_n x_n$ diverges, since the terms do not even tend to $0$. (Because $|y_{n_k}x_{n_k}|>k$.)