Linear operator image is not closed

Question

Studying functional analysis, I have to prove that the image of the operator

$S:\ell^1(\mathbb{N})\rightarrow \ell^1(\mathbb{N})$ given by $(S\xi)_n=\xi_n/n$ is not closed, even though $S$ is limited.

Well, I thinked about the sequence

$(\xi_{nj})_j\subset\ell^1(\mathbb{N})$ where

$(\xi_{nj})_n$ has the first $n$ entries equals to $1$, the image of this tend to the sequence $(1,1/2,\cdots, 1/n,\cdots)$ and this is not in $\ell^1(\mathbb{N})$.

But this is wrong because I has prooved, in fact, that $\ell^1(\mathbb{N})$ is not complete. To prove the request I need find a sequence that converges to an element of $\ell^1(\mathbb{N})$ and is not at the image.

At my mind, however, $S$ is surjective, I cannot understand.

Many thanks.

Answer 1

Note that

• $S(1,0,0,0,\ldots)=(1,0,0,0,\ldots)$;
• $S\left(1,\frac12,0,0,\ldots\right)=\left(1,\frac1{2^2},0,0,\ldots\right)$;
• $S\left(1,\frac12,\frac13,0,\ldots\right)=\left(1,\frac1{2^2},\frac1{3^2},0,\ldots\right)$

and so on. So, $S\bigl(\ell^1(\mathbb N)\bigr)$ contains the sequence$$\left(\left(1,\frac1{2^2},\frac1{3^2},\ldots,\frac1{n^2},0,0,\ldots\right)\right)_{n\in\mathbb N},$$but not its limit, which is $\left(\frac1{n^2}\right)_{n\in\mathbb N}$. So, $S\bigl(\ell^1(\mathbb N)\bigr)$ is not a closed subset of $\ell^1(\mathbb N)$.