# Subspace in $I-T$ for bounded linear maps

Let $X$ be a Banach space and let $T:X \rightarrow X$ be a bounded linear map.Show that the range of $I - T$ contains the subspace $$Y_T = \{x \in X: \limsup_{n\rightarrow \infty} n^2\|T^nx\| < \infty \}$$

I have really no good idea how to solve this, any hints would be nice!

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Let $x\in Y_T$then we have some $c>0$ and $N\in\mathbb{N}$ such that $n^2\Vert T^n x\Vert<C$ for all $n>N$. This guarantees that the series $$\sum\limits_{n=0}^\infty \Vert T^n x\Vert$$ converges. Since $X$ is Banach this implies that the series $$\sum\limits_{n=0}^\infty T^n x$$ converges. Denote its sum by $y$, then $$Ty= T\left(\sum\limits_{n=0}^\infty T^n x\right)= T\left(\lim\limits_{N\to\infty}\sum\limits_{n=0}^N T^n x\right)= \lim\limits_{N\to\infty}T\left(\sum\limits_{n=0}^N T^n x\right)= \lim\limits_{N\to\infty}\sum\limits_{n=0}^N T^{n+1} x=\\ \lim\limits_{N\to\infty}\sum\limits_{n=1}^{N+1} T^n x= \sum\limits_{n=1}^{\infty} T^n x=\sum\limits_{n=0}^{\infty} T^n x - x=y-x$$ Hence $$x=y-Ty=(I-T)y\in \mathrm{Im}(I-T)$$ The rest is clear.
How do we know that it is well definied for any k? When you check the straightforward part what happens to the $-T^\infty$ part in the telescopic sum? – Johan Dec 26 '12 at 13:02