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.
I'd be greatfull for advice