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i have the following question:

let $(a_{j})$ be a sequence in $l^{2}$ and $(b_{j}^{n})$ be a sequence of sequences in $l^{2}$ such that $(a_{j}b_{j}^{n})$ is an $l^{2}$ sequence for each $n$. Suppose further that $(b_{j}^{n})$ converges to a sequence $(b_{j})$ in $l^{2}$ and also that $(a_{j}b_{j}^{n})$ converges in $l^{2}$. I would like to prove that $(a_{j}b_{j}^{n})$ actually converges to $(a_{j}b_{j})$. This is what I have done:

$$ \sum\limits_{j=1}^{\infty}| a_{j}b_{j} - a_{j}b_{j}^{n} |^{2} = \sum|a_{j}|^{2}|b_{j}-b_{j}^{n}|^{2} \le \sqrt{\sum |a_{j}|^{4}}\sqrt{\sum_{j=1}|b_{j}-b_{j}^{n}|^{4}}$$

where the last inequality follows by Cauchy-Schwartz inequality applied to the sequences $|a_{j}|^{2}$ and $|b_{j}-b_{j}^{n}|^{2}$ and using the fact that $l^{2} \supseteq l^{4}$.

Now as $(b_{j}^{n})$ converges to $(b_{j})$ we can make this term as small as we like.

However, I am not sure whether I am correct in all this analysis. Please point out the mistakes if any in the above.

Thanks for all the help.

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1 Answer 1

up vote 3 down vote accepted

Your proof is correct, but the problem is much simpler and does not require so many assumptions. Consider that

  • Convergence in $\ell^2$ norm implies coordinate-wise convergence
  • Coordinate-wise limit is unique, if it exists

So, if you have $x^n\to x$ coordinate-wise and $x^n\to y$ in $l^2$ norm, it follows that $x=y$, and therefore $x^n\to x$ in $l^2$ norm.

In your case, $a_jb_j^n \to a_j b_j $ coordinate-wise. You also know that $a_jb_j^n$ converges to something in $l^2$. The conclusion is that something is $(a_jb_j)$.

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Thanks! Sorry I cannot upvote. –  Maverick Jun 30 '13 at 6:48
@Maverick But you can (if you wish) mark the answer as accepted by clicking the checkmark to the left of the answer. See Why should we accept answers? –  ˈjuː.zɚ79365 Jun 30 '13 at 7:05
Now I can vote up and did so. And also accepted. Is there any other form of etiquette that I am expected to follow? –  Maverick Jul 2 '13 at 5:27
@Maverick Hm, maybe using proper capitalization in your questions. :) Try it next time. –  ˈjuː.zɚ79365 Jul 2 '13 at 5:29

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