# Fixed Point of a affine non-expansive Operator

For a complete normed Space $V$ let $T:V\rightarrow V$ a linear continous map and $x_0 \in V$. Given $\lVert T^4 \lVert <1$, i.e $\lVert T^4x\lVert<\lVert x\lVert \ \ \forall x\in V$, show that the equation $Tx-x=x_0$ has a unique solution.

It's quite easy to show that the solutions is unique: Let $x,y$ be solution to the equation. Hence $T(x-y)=x-y$ and $T^4(x-y)=x-y$, which implies $\lVert T^4\lVert \geq 1$.

But how to show the existence? The Problem is equivalent to the fixed point problem $Ux=Tx-x_0=x$ of the affine Operator $U$. But I don't know if $U$ is a contraction.

$x=(T-I)^{-1}x_0$.
You just need to know that $1$ is not in the spectrum of $T$. The spectral radius of $T$ is $\lim\limits_{n\to\infty}\|T^n\|^{1/n}$, and
$$\lim\limits_{n\to\infty}\|T^n\|^{1/n}=\lim_{n\to\infty}\|T^{4n}\|^{1/(4n)}\leq \lim_{n\to\infty}(\|T^4\|^n)^{1/(4n)}=\|T^4\|^{1/4}<1.$$