# How to show that $I-T$ is surjective, if $\|T\|<1$?

I got stuck on an exercise in page 258, Real analysis(4ed),H.L. Royden et al:

Let $X$ be Banach space and $T \in \mathcal{L}(X,X)$ have $\|T\|<1$. Show that $I-T$ is an isomorphism.

$\mathcal{L}(X,X)$ is the collection of all bounded linear operators from $X$ to $Y$. $I$ is the identity function. I know that since $T$ is contractive, $I-T$ must be one-to-one. $\|(I-T)(u)\|\leq(1+\|T\|)\|u\|$ and $\|(I-T)^{-1}(u)\|\leq\frac{1}{(1-\|T\|)}\|u\|$ together imply that $I-T$ and its inverse are both continuous. But I have no clue how to show that $I-T$ is surjective.

Added: Finally, with copious hints, I got it. Thank you again!

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Well, $(I-T)^{-1}$ exists, so... – Andrés E. Caicedo Jun 25 '13 at 5:30
Try showing the infinite sum $1+T+T^2+T^3+\ldots$ converges. – Jeff Tolliver Jun 25 '13 at 5:35
@AndresCaicedo: Thank you for your hint. Since it and its inverse are continuous, its domain should be clopen and unbounded. But how to proceed? – Metta World Peace Jun 25 '13 at 5:38
@MettaWorldPeace You seem to be missing the forest for the trees. Forget all the other details and think about what $I-T$ being invertible means. – Ragib Zaman Jun 25 '13 at 5:40
Here is the proof for the $(I-T)^{-1}$ exists. – Mhenni Benghorbal Jun 25 '13 at 5:42

Since $\lVert T\rVert \lt 1$, the series $S = \sum_{n=0}^\infty T^n$ converges because $\lVert S\rVert \leq \sum_{n=0}^\infty \lVert T\rVert^n = \frac{1}{1-\lVert T\rVert}$ shows its absolute convergence.
A direct computation (geometric series) shows that $$(1-T)S = (1-T)\sum_{n=0}^\infty T^n = 1 +\sum_{n=1}^\infty T^n -\sum_{n=0}^\infty T^{n+1} = 1$$ and similarly $S(1-T) = 1$. This implies that $(1-T)$ is invertible with inverse $S = \sum_{n=0}^\infty T^n$, and it follows that $(1-T)$ is both injective and surjective.