# The set of $\ell^2$ sequences dominated by a sequence is equisummable

This is my homework question and I am not sure if my solution is correct.

First, definition of equisummable:

Let $S \subset \ell^2$. The subset $S$ is called equisummable if for any $\epsilon>0$ there exists $N \in \mathbb{N}$ such that $\forall x \in S$ we have $\sum_{i=N}^\infty \, |x_i| ^{2}<\epsilon$.

And the question is

Let $a\in \ell^2$, $S= \{ x \in \ell^2 \,:\, |x_n|\leq |a_n| \text{ for all } n \in \mathbb{N} \}$. Is $S$ equisummable or not?

My answer is yes, it is equisummable.

Otherwise, $\exists \epsilon>0$ such that $\forall N\in \mathbb{N}$ there exists $x \in S$ (depends on $N$) such that $\sum \limits_{i=N}^\infty \, |x_i|^2 \gt \epsilon$. Since $a_n$ is equal or greater than $x_n$ for all $x \in S$, $\forall N\in \mathbb{N}$ we have $\sum_{i=N}^\infty \, |a_i|^2 \gt \epsilon$. This contradicts with the fact that $\sum \limits_{i=1}^\infty \, |a_i|^2 \lt \infty$ (since $a\in \ell^2$).

Is my answer correct? And do I need to explain the last part a little bit more (the reason they contradict)? It seems obvious to me but I am not sure.

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Your proof looks correct to me, but more complicated than it needs to be. In particular, you can completely avoid contradiction in the proof: Since $a \in \ell^2$, given $\varepsilon > 0$, there exists $N$ such that $\sum_{n \geqslant N} |a_n|^2 \lt \varepsilon$. Now show that the same $N$ works for all $x \in S$. Of course, this is the same proof written in a different way, but I hope this is cleaner. –  Srivatsan Dec 26 '11 at 21:26
If you follow Srivatsan’s suggestion, you could write up the new proof as an answer to your own question (though you may have to wait a few hours before you can do that). –  Brian M. Scott Dec 26 '11 at 21:32
It would be good practice to work out exactly why the convergence of $\sum_{i=1}^\infty|a_i|^2$ implies that for each $\epsilon>0$ there is some $N$ such that $\sum_{n\ge N}|a_n|^2<\epsilon$; you’ll need to use the convergence of the sequence of partial sums. –  Brian M. Scott Dec 26 '11 at 21:35

Okay I followed Srivatsan's suggestion and my new answer is as follows:

$a\in \ell^2$, then $\sum \limits_{i=1}^\infty \, |a_i|^2 = L \lt \infty$. Follows, limit $S_k$ as $k \to \infty$ is $L$ where $S_k = \sum \limits_{i=1}^k \, |a_i|^2$.

This implies $\forall \epsilon>0$ $\exists N \in \mathbb{N}$ st $\forall n \geq N$ we have $|S_n-L|\lt \epsilon$. Then,

$$|L-S_n|=\left|\sum \limits_{i=1}^\infty \, |a_i|^2-\sum \limits_{i=1}^n \, |a_i|^2\right| = \left|\sum \limits_{i=n+1}^\infty \, |a_i|^2\right| \lt \epsilon.$$

Since $|a_n| \geq |x_n|$ for all $n$, $\forall n \geq N$ $$\sum \limits_{i=n}^\infty \ |x_i|^2 \lt \sum \limits_{i=n}^\infty \ |a_i|^2 \lt \epsilon.$$

Hence, $S$ is equisummable.

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