Invertibility of any bounded operator $R$ st $\|I-R\|<1$

Question

Let $X$ be a real Banach space and $\, R:X \rightarrow X$ a linear bounded operator such that
$$ \Vert I-R \Vert_{\mathcal{B}(X)} < 1 $$ Prove that $R$ is invertible.

I cannot prove surjectivity. Any help? :)

For injectivity:
$\quad$ From $\Vert I-R \Vert_{\mathcal{B}(X)} < 1 $, we get $\quad \forall x \ne 0 \quad \Vert x-Rx \Vert < \Vert x \Vert $.
$\quad$ Suppose now $Rx=0$ and, by contradiction, $x \ne 0$. Then we have: $\Vert x-Rx \Vert = \Vert x \Vert < \Vert x \Vert $. Absurd.
$\quad$ Therefore $\operatorname{Ker} R ={0}$.

Answer 1

If $\|u\|<1, I-u$ is invertible, it inverse is $\sum_{n\geq 0} u^n$, take $u=I-R$, $I-(I-R)=R$ is invertible.

Answer 2

If $S$ is a linear operator such that $\Vert S\Vert_{\mathcal{B}(X)} < 1$ then $I-S$ is invertible with $ U=\sum_{k=0}^\infty S^k$ for inverse.

In your case $I-R$ is having a norm strictly less than one and therefore $R = I -(I-R)$ is invertible. This also relates to the question you posted an hour ago.

From there you can prove that $R$ is surjective...

Answer 3

You can find an inverse map directly. Let $T=I-R$. For every $n\in\mathbb{N}$ we have:

$\sum_{k=0}^n ||T^k||\leq \sum_{k=0}^n ||T||^k\leq\sum_{k=0}^\infty ||T||^k$.

The geometric series on the right hand side converges because $||T||<1$ by assumption. Hence the series $\sum_{k=0}^\infty T^k$ converges absolutely, and since $X$ is a Banach space (which implies the space of bounded linear operators $\mathcal{L}(X)$ is also Banach) the series itself converges to some element $S\in \mathcal{L}(X)$. We claim that $S$ is exactly the inverse of $R$. For every $n\in\mathbb{N}$:

$R(I+T+...+T^n)=(I-T)(I+T+...+T^n)=I-T^{n+1}$

By taking $n\to\infty$ on both sides (again, we use the inequality $||T||<1$) we get $RS=I$. Similarly show that $SR=I$.