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Fixed point theorem

  1. Aug 10, 2016 #1
    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:oldbiggrin:.
     
  2. jcsd
  3. Aug 10, 2016 #2
    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
  4. Aug 10, 2016 #3
    But what about the suffisient condition please? Maybe it's wrong.
     
  5. Aug 10, 2016 #4
    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##.).
     
  6. Aug 11, 2016 #5
    Could you help me?
     
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