a range of an operator from $l^p$ to $l^p$ Let T be an operator $T:l^2→l^2$ defined by $T(x_n )=x_n/(n^2+1)$ , then the range of this operator is not closed ..
Iam trying to find a sequence in $l^2$  under $T$ which its limit is not in $l^2$ ..
My attempt is as the following:
Let $(x_n )=(1,1,1,…,1,0,0,0,..)$ then $(x_n)$ in $l^2$ because $∑|x_n |^2=0<+∞$ 
But $T(x_n )=(1,1/5,1/10,…,0,0,0,..)$ in $l^2$ for the same previous reason 
Now Iam not sure about the next steps :
The limit of this sequence is $(1,1/5,1/10,1/17,…)$ which is not in $l^2$  is that right?
 A: The coordinate-wise limit of $T(x_n)$ as $n\to\infty$ is $y\equiv((n^2+1)^{-1})_{n\in\mathbb N}$, which is in $\ell^2$, because $$\sum_{n=1}^{\infty}\left(\frac{1}{n^2+1}\right)^2\leq \sum_{n=1}^{\infty}\frac{1}{n^2+1}\leq\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}<\infty.$$ Now, $y$ is the limit of $T(x_n)$ also in $\ell^2$ as $n\to\infty$ (and this is what really matters), since $$\|y-T(x_n)\|_{\ell^2}^2=\sum_{k=n+1}^{\infty}\left(\frac{1}{k^2+1}\right)^2\to 0\qquad\text{as $n\to\infty$},$$ given that we’re dealing with the tail of a convergent series. But the only possible candidate for a sequence $x$ such that $y=T(x)$ is $x=(1,1,1,\ldots)$, which is not in $\ell^2$.
Conclusion: $y$ is in the closure of $T(\ell^2)$, but not in $T(\ell^2)$, so that the range of $T$ is not closed.
A: Notice that $\operatorname{Im} T$ is dense in $\ell^2$ so it cannot be closed.
Let $e_n$ be the $n$-th canonical vector in $\ell^2$. We have $e_n = T\Big((n^2+1)e_n\Big) \in \operatorname{Im} T$.
Therefore $\operatorname{span}\{e_n\}_{n\in\mathbb{N}} \subseteq \operatorname{Im} T$.
If $\operatorname{Im} T$ were closed, we would have:
$$\ell^2 = \overline{\operatorname{span}\{e_n\}_{n\in\mathbb{N}}} \subseteq \overline{\operatorname{Im} T} = \operatorname{Im} T   \implies \operatorname{Im} T = \ell^2$$
But this is certainly not the case since for example $\left(\frac1n\right)_{n=1}^\infty \in \ell^2 \setminus \operatorname{Im} T$. The only candidate $(x_n)_{n=1}^\infty$ such that $T(x_n)_{n=1}^\infty = \left(\frac1n\right)_{n=1}^\infty$ is the sequence $\left(n + \frac1n\right)_{n=1}^\infty$, which is not in $\ell^2$.
A: The definition operator $T:E\to F$ between normed vector spaces having closed range is that the set $T(E)$ is a closed subspace $F$. It does not mean that there is a sequence $x_n$ in $E$ so that "the limit of $T(x_n)$ does not lie in $F$", that statement is meaningless in this context.
For your counter example the image $T(x_n)$ actually does converge in $\ell^2$. The limit is $(1,1/2,1/5,1/10,...,1/(n^2+1),...)$, which is an element of $\ell^2$ as $\sum_n\frac1{(n^2+1)^2}<\infty$.
In order to verify that an operator between normed vector spaces does not have closed range you need to find a convergent sequence in $T(E)$ so that the limit point is not the image of any point in $E$.
Your example works here. Note that
$$x_n=(1,1/2,..,1/(n^2+1),0...)\in T(\ell^2)$$
as $x_n$ is the image of $(1,...,1,0,...)$ which is in $\ell^2$. This sequence has a limit in $\ell^2$, namely the vector written above. However there is no pre-image under $T$, as such a pre-image would necessarily have "$1$" in all components, but no such vector exists in $\ell^2$.
