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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.

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    $\begingroup$ See that $0\le 1-L<1$ since $0<L<1$ $\endgroup$
    – Guy Fsone
    Commented Oct 17, 2017 at 12:52
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    $\begingroup$ $\|f(x)-x_0\|=\|f(x)-f(x_0)+f(x_0)-x_0\|\leq \|f(x)-f(x_0)\|+\|f(x_0)-x_0\|\leq Lr+(1-L)r = r$ $\endgroup$
    – Surb
    Commented Oct 17, 2017 at 12:56
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    $\begingroup$ This is the statement of the Banach fixed-point theorem (BFT) for normed spaces. You are looking for a proof of the BFT, not just to use it. As this is a standard topic, many proofs can be found on wikipedia, wikibooks and online scripts. Select one and come back with more specific questions. $\endgroup$ Commented Oct 17, 2017 at 12:56
  • $\begingroup$ Thank you guys for the help $\endgroup$
    – user439387
    Commented Oct 17, 2017 at 12:57

3 Answers 3

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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 $r\in (0,1)$.

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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.

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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 $$x_f = \lim_{k\to\infty}x_k$$

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