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Let $A : \ell^1 \to \ell^1$ be an operator where for any $x \in \ell^1$, $Ax = \left(\frac{x_n}{n}\right)_{n \in \mathbb{N}}$.

Question. I'm trying to show that $\text{range}(A)$ is dense subspace of $\ell^1$.

I know that essentially we want to show that the closure of the range is equivalent to $\ell^1$, i.e.

$$\overline{\text{range}}(A) = \{x \in \ell_1 : \exists x_n \in \text{range}(A)~\text{such that}~x_n \to x\} = \ell^1.$$

But I don't really have an idea of how to formally go about showing the above using this definition?

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2 Answers 2

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You show it by demonstrating there is $x_n \in \ell^1$ (and more precisely $Ax_n$ is Cauchy), such that $Ax_n\rightarrow y,$ for any $y\in \ell^1$.

So let $y\in \ell^1$, then

$$\sum |y_i|<\infty$$

We can now have a sequence $x_n:=y_i i1(i<n)\in\ell^1$. Now $Ax_n$ converges to $y$, and the statement is proved.

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  • $\begingroup$ $1(i<n)$ is the characteristic function of $\{i \in \mathbb N: i<n\}$? $\endgroup$
    – azif00
    Mar 25, 2021 at 0:26
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    $\begingroup$ @azif00 yes.... $\endgroup$
    – Dole
    Mar 25, 2021 at 0:27
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Hint: consider sequences with only finitely many nonzero terms.

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