Prove that a bounded operator $A$ on Banach space is invertible (bijective) if $\left\|I - A\right\| < 1 $ holds? I can't seem to make any progress with the following problem:
If $ A $ is a bounded operator on Banach space $ X $ and if $ \lVert I - A \rVert < 1 $, prove that $ A $ is invertible/bijective.
This seems to be a known result, but I wasn't able to found a source.
 A: Hint:
Use the real number analogue to gain some intuition: If $|x|<1$, then
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n.$$

Set $B:=I-A$. Then $A=I-B$, so our intuition says that
$$ ``\ A^{-1} = \frac{1}{I-B} = \sum_{n=0}^\infty B^n.\ "$$
(Quotes in the above expression are there to indicate that it is nonsense. I include the expression anyway because sometimes nonsense can help us prove things.)
Now show that $\sum_{n=0}^\infty \|B^n\|<\infty$. Since the space of bounded operators on a Banach space is complete, this implies that $\sum_{n=0}^\infty B^n$ converges in $X$. Then show that $(I-B)(\sum_{n=0}^\infty B^n)=(\sum_{n=0}^\infty B^n)(I-B)=I$, which will complete the proof that $A$ is invertible and
$$
A^{-1} = \sum_{n=0}^\infty (I-A)^n.
$$
A: For another approach that avoids series expansions, and gives a more general result, note that it suffices to show that $A$ and $A^*$ are bounded below. 
So, suppose that $A$ is $\text{any}$ linear operator, and $B$ is an invertible linear operator such that 
$\left \| A-B \right \|\cdot \left \| B^{-1} \right \|<1.$ Then,
$\left ( \left \| B^{-1} \right \| \right )^{-1}\cdot \left \| x \right \|=\left ( \left \| B^{-1} \right \| \right )^{-1}\cdot \left \|B^{-1}\cdot Bx \right \|\le \left \| Bx \right \|\le\\  \left \| \left ( A-B \right )x \right \|+\left \| Ax \right \| \le \left \| A-B \right \|\cdot \left \| x \right \|+\left \| Ax \right \|\Rightarrow \\ \left \| Ax \right \|\ge \left ( \left ( \left \| B^{-1} \right \| \right )^{-1}-\left \| A-B \right \| \right )\cdot \left \| x \right \|>\left \| x \right \|,\ $ 
from which we conclude that $A$ is bounded below. 
But, $\left \| \left ( A-B \right )^{*}y^{*} \right \|\le \left \| A-B \right \|\cdot \left \| y^{*} \right \|$ and similarly for $(B^{-1})^{*}$ so that 
$\left \| A^*-B^* \right \|\cdot \left \| (B^{-1})^* \right \|<1$ and so the same calculation as above shows that $A^*$ is also bounded below.
