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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.

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  • $\begingroup$ The second question is answered "no": $T$ could be a non-Fredholm isometry, for example. $\endgroup$ Mar 29, 2012 at 19:04

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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 $X$.

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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.

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  • $\begingroup$ I would have entered this as a comment if I could have, as I am not sure enough about it to make it an answer, but I am unable to comment. Oh well, good luck with this question! $\endgroup$ Mar 29, 2012 at 20:17

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