Banach fixed point question Let $f: \mathbb R^n \rightarrow \mathbb R^n$ with arbitrary norm $\|\cdot\|$. It exists a $x_0 \in \mathbb R^n$ and a number $r \gt 0 $ with 
$(1)$ on $B_r(x_0)=$ {$x\in \mathbb R^n: \|x-x_0\| \leq r$}  $f$ is a contraction with Lipschitz constant L
$(2)$ it applies $\|f(x_0)-x_0\| \le (1-L)r$
The sequence $(x_k)_{k\in\mathbb N}$ is defined by $x_{k+1}=f(x_k).$
How do I show that $x_k \in B_r(x_0) \forall k \in \mathbb N$?
How do I show that $f$ has a unique Fixed point $x_f$ with $x_f = \lim_{k \rightarrow\infty} x_k$?
I know this has something to do with Banach but I am totally clueless on how to prove this. Any help is welcome. Thanks. 
 A: For $x\in B_r(x_0)$, we have
$$\|f(x)-x_0\|=\|f(x)-f(x_0)+f(x_0)-x_0\|\leq \|f(x)-f(x_0)\|+\|f(x_0)-x_0\|\\
\leq L\|x-x_0\|+(1-L)r \leq Lr+(1-L)r=r$$
thus $f(x)\in B_r(x_0)$.
To show that $\lim_{k\to\infty}x_k=x_f$ and $f$ has a unique fixed point in $B_r(x_0)$, you can indeed use the Banach fixed point theorem on $f|_{B_r(x_0)}$ provided that $r\in (0,1)$.
A: As for the unique fixed point, one shows that the iteration sequence is related to a geometric sequence with factor $L$ and uses that to show that it is a Cauchy sequence. Completeness of the space gives existence, contractivity the uniqueness of the fixed point.
A: Hint: Prove by induction 
Assume that $x_k\in B_r(x_0)$ then you have 
$$ \|x_{k+1} -x_0\| =  \|f(x_k) -f(x_{0}) +f(x_{0}) -x_0\| \\\le \|f(x_k) -f(x_{0})\|+\|f(x_{0}) -x_0\| \\ \le L
 \|x_k -x_{0}\|+\|f(x_{0}) -x_0\|    $$
That is 
$$ \|x_{k+1} -x_0\| \le L \|x_k -x_{0}\|+\|f(x_{0}) -x_0\|   .$$
Using the assumption that, 
$$\|f(x_0)-x_0\| \le (1-L)r$$
we get,
$$ \|x_{k+1} -x_0\| \le L\|x_{k} -x_0\|+(1-L)r  \tag{E}$$
Now since $x_0\in B_r(x_0)$ we assume if we Assume that $x_k\in B_r(x_0)$ . 
Then from the estimate above we have 
$$ \|x_{k+1} -x_0\| \le L\|x_{k} -x_0\|+(1-L)r\le (1-L +L)r= r  $$
Hence $$x\in B_r(x_0)~~~\forall~~ k$$

Replacing $x_k$ by $y$ in (E) in  the above prove we see that for all $ y\in B_r(x_0)$ we obtain
  $$ \|f(y) -x_0\| \le L \|y -x_{0}\|+(1-L)r  \le Lr +(1-L)r= r  .$$

Therefore $$f :B_r(x_0) \to B_r(x_0)$$ is a contraction on closed set $B_r(x_0)$(which is therefore complete). From the fix point theorem 
$f$ has a fix point $x_f$ and its satisfies 
$$x_f = \lim_{k\to\infty}x_k$$
