$T:X\rightarrow X$ bounded linear operator ($X$ Banach space), $||T||\leq 1$ and $T(x_0)=x_0$, then $x_0\notin \{x-Tx \mid x\in X \}$

Question

Assume $T:X\rightarrow X$ is a bounded linear operator from a Banach space $X$ to itself, with $||T||\leq 1$. Assume there exists some $x_0\in X (x_0\neq 0)$ such that $T(x_0)=x_0$.

I need to show that $x_0\notin \{x-Tx \mid x\in X \}$, but I don't know how. I have tried to derive a contradiction from $$||x_0-(x-Tx)||=0$$ by constantly substituting $x_0=T(x_0)$ etc. but without succes.

Any hints would be helpfull.

Answer 1

If $x_0=x-Tx$ then $Tx_0=x_0$ implies $x_0=Tx-T^2x$, and hence $x_0=T^2x-T^3x$, etc. Adding the first $n$ oof these equations gives $$x_0=\frac{x-T^nx}{n}.$$Now use the fact that $||T||\le 1$...

Answer 2

If $x_0=x-Tx$ for some $x\in X$, then $Tx-T^2x=x-Tx$. This simplifies to $0=(T-2I)^2x$, so either $x=0$ (but it can't be, because $x_0\neq0$) or $2\in\sigma(T)$. But since $\|T\|\leq1$, any point in $\sigma(T)$ has magnitude at most $1$. Contradiction.

Answer 3

(not enough reputation for a comment)

I might be wrong but I believe your assertion is incorrect.

Since X is a Banach space and $\|T\|<1$, $Id_X-T$ is invertible (cf. Neumann series), thus surjective. It follows $X=\mathrm{Im}(Id_X-T)=\{x-Tx~|~x\in X\}$.

EDIT: Seems like I made a mistake, can someone please point me to it?