# Fixed point theorem

1. Aug 10, 2016

### Calabi

1. The problem statement, all variables and given/known data
Let be $f \in C^{1}(\mathbb{R}^{n}, \mathbb{R}^{n})$ and $a \in \mathbb{R}^{n}$ with $f(a) = a$. I'm looking for a suffisent and necessar condition on f that for all $(x_{n})$ define by $f(x_{n}) = x_{n+1}$, then $(x_{n})$ converge.

2. Relevant equations
$f(a) = a$

3. The attempt at a solution

If all sequences define like this converge it's necessarly on a.
We've got the following result : if A is a complex matrix, then the seuqences define by $X_{n+1} = AX_{n}$ converge if and only if the spectral radius $\rho{A} < 1$.

So I think we should care about the spectral radius of the jacobian of f.

By whrogting $f(a + x_{n}) = a + df(a)x_{n} + ||x_{n}||\epsilon(x_{n})$ but I get nothing.

Have you got an idea please?

Thank you in advance and have a nice afternoon.

2. Aug 10, 2016

### Calabi

It is necessar beacause if all sequences define by f converge it's in $a$ then We can choose $(x_{n})$ no stationnar if $df(a)$ is invertible : by the local reverse theorem we can find a neighbourhood $V$ of a with $f_{|V} : V \rightarrow V$ is injective then it is enough to choose $x_{0} \in V - \{a\}$(then if for a certain n $x_{n} = a$ we easily show that necessarly $x_{0} = a$.). so $x_{n + 1} = f(a + x_{n} - a) = a + df(a).(x_{n} - a) + ||x_{n} - a||\epsilon(x_{n} - a)$(whith $\epsilon(x) \rightarrow_{x \rightarrow 0} 0$.). $\forall \epsilon > 0$ we can choose n enough high as $||x_{n} - a|| < \epsilon$ and $||x_{n+1} - a|| \leq ||x_{n} - a||$(indeed, by convergence hypothesis, $\exists N \in \mathbb{N} | n > N \Rightarrow ||x_{n} - a|| < \epsilon$, then if forall n integer > N we have $||x_{n+1} - a|| > ||x_{n} - a||$ then it doesn't converge in 0 which is absurd.). so $||\rho(df(a))|| ||x_{n} - a|| \leq |||df(a)||| ||x_{n} - a|| \leq ||x_{n + 1} - a|| + ||x_{n} - a|| ||\epsilon(x_{n} - a)|| \leq ||x_{n} - a|| + ||x_{n} - a|| ||\epsilon(x_{n} - a)||$ so as $(x_{n})$ is not stationnar $||\rho(df(a))|| \leq 1 + \epsilon$ so $||\rho(df(a))|| \leq 1$.
If the jacobian is non invertible I don't know.

Last edited: Aug 10, 2016
3. Aug 10, 2016

### Calabi

4. Aug 10, 2016

### Calabi

We can choose $(x_{n})$ no stationnar if $df(a)$ is invertible : by the local reverse theorem we can find a neighbourhood $V$ of a with $f_{|V} : V \rightarrow V$ is injective then it is enough to choose $x_{0} \in V - \{a\}$(then if $x_{n} = a$ we easily show that necessarly $x_{0} = a$.).

5. Aug 11, 2016

### Calabi

Could you help me?