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.
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 $X$.
Not sure about this, but I think this has something to do with the compactness of T? In any case, T is not necessarily a Fredholm operator.
