If for some $n\in\mathbb{N}$, $T^{n+1}\left( X\right) =T^{n}\left( X\right)$ do we have $T^{n}\left(X\right) $ closed? Let $X$ be a Banach space, and let $T$ be a bounded operator on $X$ such
that for some $n\in 
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\mathbb{N}
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$, $T^{n+1}\left( X\right) =T^{n}\left( X\right) $.
Do we have $T^{n}\left( X\right) $ closed ?
 A: This is such a good question. Even though I do not have a complete answer yet, I feel the need to share what I figured out so far.
As I mentioned in the first comment, we can assume $n=1$. I also figured that the contrary to your statement is equivalent to the following.
Equivalent statement: There exists an incomplete normed vector space $X$ and some surjective $T\in B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence.
Proof: Contrary to your statement $\implies$ Equivalent statement: Let us have a Banach space $X$ and some $T\in B(X)$ with $T^2(X)=T(X)$, while $T(X)$ is not closed. Then $T(X)$ is an incomplete normed vector space, and $T|_{T(X)}\in B(T(X))$ is surjective. Furthermore, for every Cauchy sequence $\{x_i\}$ in $T(X)$ there is a limit $x\in X$, so $Tx_i\to Tx\in T(X)$.
Proof: Equivalent statement $\implies$ Contrary to your statement: Let us have an incomplete normed vector space $X$ and some surjective $T\in B(X)$ such that every Cauchy sequence gets mapped to a convergent sequence. Let $\overline{X}$ denote the completion of $X$. Define $S\in B(\overline{X})$ by $S(\{x_i\}):=\lim Tx_i$. Then $S(S(\overline{X}))=S(\overline{X})=X$ and $X$ is not closed in $\overline{X}$.
I am not quite sure whether this equivalence is useful, but I did already find some incomplete normed vector space for which you can prove there is no surjective bounded linear operator that maps every Cauchy sequence to a convergent sequence. Namely, the vector space $(d,\|\cdot\|_p)$ of all finite sequences of real numbers.
Proof: Let $T\in B(d)$ be surjective. Then we have some sequence $\{x_i\}$ such that $Tx_i=e_i$. We have that $$y_i:=\sum_{k=1}^i\frac{x_k}{\|x_k\|k^2}$$ is a Cauchy sequence, and $\{Ty_i\}$ is not convergent.
The reason this proof does not work for general incomplete normed vector spaces is that we can not simply conclude that $\{Ty_i\}$ is not convergent. I am still not sure whether there exists a surjective $T\in B(C[0,1],\|\cdot\|_2)$ that maps Cauchy sequences to convergent sequences, for instance. This is as far as I got so far.
