# Show that if operator $T$ is such that $||I-T||<1$ , then $T$ is bijective.

I came across this statement in a proof and I can't figure out why its true, could someone point out why (or give a hint). Thanks.

Suppose that $T:X\to X$ is a bounded linear operator that maps a Banach space $X$ to itself such that

$$||I-T||<1,\quad\quad(*)$$

where $I$ denotes the identity on $X$, and the $||\cdot||$ the induced operator norm. Then $T$ is bijective.

In case anyone comes across this later on here's a follow up on Karene's answer.

It follows from $(*)$ that

$$S_n:=\sum_{i=0}^n(I-T)^i$$

is bounded for all $n$ and that the sequence $(S_n)$ is Cauchy. Since $X$ is complete, the space of linear bounded operators on $X$ is complete as well. Thus $S_n$ tends to some linear bounded operator $S$ as $n$ tends to infinity. Then, it's easy to verify that

$$(I-T)S=S-I\Rightarrow TS=I$$

and similarly to show that $ST=I$.

Show that $I+(I-T)+(I-T)^2+...$ converges and is the inverse of $T$.
The proof is as you would do it for numbers but using $||.||$ instead of the absolute value.