Fixed point of a map iteration sequence Let $(X, \lVert \cdot \lVert)$ be a Banach space and $M \subset X$ an nonempty, closed subset. 
For positive numbers $a_n$ the sum $\sum_{n_0}^{\infty}a_n$ converges.
For the map $A: M \rightarrow M$ it holds $\forall x,y\in M$, $n \in \mathbb{N}$ that 
$$\lVert A^nx-A^ny \lVert \leq a_n \lVert x-y \lVert$$
Show that


*

*A has exactly one fixed point $x^* \in M$

*$x^*$ is the limit of the series $x_n=A^nx_0$ for any $x_0 \in M$.

*The error estimate holds true: $$\lVert x^*-x_n\lVert\leq(\sum_{k\geq n}a_k)\lVert x_1-x_0\lVert$$

 A: $1.$
Since the sum converges, $a_n\to 0$ for $n \to \infty$.
So $\exists N \in \mathbb{N}:0<a_N\leq \gamma <1 $
Which would yield $\lVert A^Nx-A^Ny \lVert \leq \gamma \lVert x-y \lVert$, which would make $A^N$ strongly contractive and thus would guarantee an unique fixed point $x^* \in M$ for $A^N$.
Then we have $0\leq d(Ax^*,x^*)=d(A A^Nx^*, A^Nx^*)=d(A^NAx^*,A^Nx^*)\leq \gamma d(Ax^*, x^*)$ 
and since $\gamma \in (0,1)$ it follows that $d(Ax^*, x^*)=0$, which proves the that $x^*$ is a fixed point of $A$.
Assume there exists a different fixed point of $A$ called $y^*$. 
Then $Ay^*=y^* \implies A^my^*=y^*$ for all natural $m$ including $N$, which would contradict the contractivity of $A^N$, because it guarantees an unique fixed point of $A^N$.
$2.$ 
Let $m=rN+i$ for natural $r,i$ with $0\leq i <N$ and $N$ from $1.$
For any $x_0 \in M$ we have 
$$d(A^mx_0, x^*)=d(A^{rN+i}x_0, A^Nx^*)\leq \gamma d(A^{(r-1)N+i}x_0, x^*) \leq \gamma^r d(A^ix_0, x^*) $$
and since $d(A^ix_0, x^*)$ is bounded (fixed), for $m \to \infty$ we get $r \to \infty$ and thus $\gamma^r \to 0$ since $\gamma \in (0,1)$ proving that $lim_{n \to \infty} A^nx_0=x^*$ for any $x_0 \in M$.
$3.$ 
$$\lVert x^*-x_n\lVert\leq(\sum_{k\geq n}\lVert x_k-x_{k+1}\lVert)$$ by triangle inequality and since $$\lVert x_k-x_{k+1}\lVert=\lVert A^kx_0-A^{k+1}x_0\lVert=\lVert A^kx_0-A^{k}Ax_0\lVert=\lVert A^kx_0-A^kx_1\lVert \leq a_k \lVert x_0-x_1\lVert$$ 
one gets the desired 
$$\lVert x^*-x_n\lVert\leq(\sum_{k\geq n}\lVert x_k-x_{k+1}\lVert)\leq(\sum_{k\geq n}a_k)\lVert x_1-x_0\lVert$$
