Let $X$ be a Banach space and let $T$ be a bounded operator acting on $X$. Suppose for each linearly independent unbounded sequence $(x\_n)$ in $E$, the sequence $(Tx\_n)$ is unbounded. Must $T$ be automatically Fredholm? (it has of course finitie-dimensional kernel).


EDIT: Matthew's answer 'no' is sufficient to me.


EDIT2: This question might be deleted.



Unbounded sequences in Banach spaces
We can assume WLOG $X$ is infinite-dimensional. Since $\text{Ker}(T)$ is finite-dimensional, $\text{Ran}(T)$ is infinite-dimensional.
I claim there is $C > 0$ such that for every $x \in X$, $\|Tx\| \ge C \|x\|$.
 If not, there a sequence $x\_n \in X$ with $\|x\_n\| = n$ and $\|Tx\_n\| < 1/n$. 
 If necessary perturbing the $x\_n$ slightly,
we can assume $x\_n$ are linearly independent, and this will contradict your condition.


Now this implies that $T$ is one-to-one, and that $\text{Ran}(T)$ is closed (because if $T x\_n$ converges, $x\_n$ is Cauchy and thus converges, and $\lim Tx\_n = T(\lim x\_n)$). However, $T$ is not necessarily Fredholm: it could be an isomorphism onto a closed proper subspace of $X$.


