I know from this question that this is true for Hilbert spaces. But is the result true for Banach spaces as well?
So suppose $x_{n}$ is a bounded sequence and $(I-T)(x_{n})$ converges. Then upto a subsequence, we have that $Tx_{n}$ converges, and hence $x_{n}$ converges to some $x$. Thus, we have that $(I-T)(x_{n})\to (I-T)(x)$. This would be enough to show that the range is closed but for the fact that we assumed $x_{n}$ to be bounded in the first place.
So what if $(T-I)(x_{n})$ converges and $x_{n}$ is unbounded.
Is the result true? And if it is, then how do I prove it?
Just realized this:
We can follow the steps of the linked question.
Firstly, by the Fredholm theory, we have that $\dim(\ker(T-I))<\infty$ and a finite dimensional subspace is always complemented in a Banach Space (Actually true in any normed linear space).
Thus we have that $X=\ker(T-I)\oplus Y$ for some closed subspace $Y$.
we can now follow the exact steps in the linked question and conclude that if $||(T-I)(u_{n})||\to 0$ for some sequence of unit vectors in $Y$, then $u_{n}\to 0$ which contradicts the unit norm of $u_{n}$.
Thus $T-I$ as an operator from the Banach space $Y$ will have a lower bound and hence will be closed.