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I have the following question:

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

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

where the last inequality follows from the Cauchy-Schwarz inequality applied to the sequences $\lvert a_{j}\rvert^{2}$ and $\lvert b_{j}-b_{j}^{n}\rvert^{2}$ and using the fact that $\ell^{2} \subseteq \ell^{4}$.

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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