Understanding why ##(y_n)_n## is a bounded sequence

In summary: You can also use ##k^4##, ##k^5##, etc. The point is to pick a sequence that grows faster than the subsequence of ##y_n##. This guarantees that the product will also grow and therefore the series will not converge absolutely.
  • #1
JD_PM
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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?

Well, I am trying to understand two of the answers I got on Math Stack Exchange for this (see for more details: https://math.stackexchange.com/questions/3103987/concluding-whether-y-n-n-is-a-bounded-sequence ).

- In one of the answers, David C. Ullrich states:

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##).

Here I do not see why ##y_n## being unbounded implies that there is a subsequence ##y_{n_k}## with ##|y_{n_k}|>k^3##

- In another answer, Rigel states:

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.

I have been reading about Banach–Steinhaus theorem (https://en.wikipedia.org/wiki/Uniform_boundedness_principle) but still do not see how the theorem is used to see if ##(y_j)## is bounded. Please either explain the general idea or recommend a book where I could read about it. I am currently using Rudin's and Abbott's (note I am a beginner).

Thanks.
 
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  • #2
The sequence can't be bound by 1, there has to be an element such that ##|y_k|>1##. Take the first one as first element of your subsequence. The sequence can't be bound by 23, there has to be an element such that ##|y_l|>2^3##, even if you look at l>k only . Take that as second element of your subsequence. The sequence can't be bound by 33, there has to be an element such that ##|y_m|>3^3##, even if you look at m>l only . Take that as third element of your subsequence. And so on.
 
  • #3
mfb said:
The sequence can't be bound by 1, there has to be an element such that ##|y_k|>1##. Take the first one as first element of your subsequence. The sequence can't be bound by 23, there has to be an element such that ##|y_l|>2^3##, even if you look at l>k only . Take that as second element of your subsequence. The sequence can't be bound by 33, there has to be an element such that ##|y_m|>3^3##, even if you look at m>l only . Take that as third element of your subsequence. And so on.

Sorry but I still do not understand the whole process. I get that if we suppose ##y_n## is unbounded then there has to be an element such that ##|y_k|>1##. But once you start with 'The sequence can't be bound by 23' I get lost. I mean, why the 23 factor?
 
  • #4
It is an arbitrary choice. I picked it because the answer you quoted uses k3 as comparison: 13, 23, 33. There are many more sequences that work.
 
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Related to Understanding why ##(y_n)_n## is a bounded sequence

1. What does it mean for a sequence to be bounded?

A bounded sequence is one in which all the terms of the sequence fall within a certain range or bound. This means that there is some number, called a bound, that is greater than or equal to all the terms in the sequence. Alternatively, there may be a bound that is less than or equal to all the terms in the sequence.

2. How can I tell if a sequence is bounded?

To determine if a sequence is bounded, you can graph the terms of the sequence or look for a pattern in the sequence. If the terms of the sequence appear to approach a certain value or remain within a certain range, then the sequence is likely bounded. Additionally, you can use mathematical techniques such as the limit comparison test or the ratio test to determine if a sequence is bounded.

3. Why is it important to understand why a sequence is bounded?

Understanding why a sequence is bounded can provide valuable insights into the behavior of the sequence. Bounded sequences have certain properties that can be used to analyze and manipulate them in various mathematical and scientific contexts. Additionally, understanding why a sequence is bounded can help in determining the convergence or divergence of the sequence.

4. What are some examples of bounded sequences?

Some examples of bounded sequences include the sequence of even numbers, the sequence of decimal values of a repeating decimal, and the sequence of values of a convergent geometric series. These sequences all have a finite or infinite bound that all the terms of the sequence fall within.

5. Can a sequence be bounded and unbounded at the same time?

No, a sequence cannot be both bounded and unbounded at the same time. A sequence is either bounded or unbounded, there is no in-between. If a sequence has a bound, then it is bounded. If a sequence does not have a bound, then it is unbounded.

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