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An automorphism in a Banach space

  1. Dec 22, 2015 #1
    Hello I've got a problem : let be a normed vectorial space E, N and A an continue automorphism.
    I suppose E is complete. So by the banach theorem
    $$A^{−1}$$
    is continue.
    So now let be f a k lipshitz application with
    $$k<\frac{1}{||A^{−1}||}$$.
    .
    I'd like to show that f + A is an homomorphism.
    I don't even know how to start.

    Have you got any idea please?

    Thank you in advance and have a nice afternoon:oldbiggrin:.
     
  2. jcsd
  3. Dec 22, 2015 #2

    Samy_A

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    I don't understand.
    If f is a Lipschitz function (from E to E), f + A will be continuous, and f + A will also be linear if f is linear.

    What am I missing?
     
  4. Dec 22, 2015 #3
    Oh no. Change homomorphisme in automorphism.
    And I never say that f is linear.
     
  5. Dec 22, 2015 #4

    Samy_A

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    Ok, you changed the question.

    Anyway, an automorphism on the Banach space E would be a bounded linear invertible operator. A necessary condition for f + A to be an automorphism is f being linear.

    EDIT: if A is an automorphism of a Banach space, and B is a linear operator on that space satisfying ##\|A-B\|<\frac{1}{\|A^{−1}\|}##, then B is invertible too.
    That would imply that in your case f + A will be an automorphism if f is linear.
     
    Last edited: Dec 22, 2015
  6. Dec 22, 2015 #5

    WWGD

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    Like Samy_A said, linearity is necessary, unless you consider some structure other than the vector space/linear structure on your space, which does not seem to be what you are doing. EDIT basically, automorphisms of a space are maps (from the space to itself here) that respect/preserve the structure of the space. A Banach space is by definition a linear/vector space, so automorphisms are maps that must preserve this structure.
     
  7. Dec 22, 2015 #6

    Krylov

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    I think it may be that the OP is trying to merely prove that ##A + f## is invertible, not that it is an automorphism. For this, it is indeed not needed that ##f## is linear.
     
  8. Dec 22, 2015 #7

    Samy_A

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    I tried to mimic the proof for the basic linear case: if ##T## is a linear operator satisfying ##\|T\|<1##, then ##I-T## is invertible.
    So the hypothesis is: If ##f## is a Lipschitz function satisfying ##\forall x,y: \ \|f(x)-f(y)\|<=k\|x-y\|##, with ##k<1##, then ##I-f## is invertible.
    In the linear case one constructs the inverse with the series ##\sum_{n=0}^\infty T^n(x)##, that obviously converges for linear T.
    Does the similar series for the Lipschitz function, ##\sum_{n=0}^\infty f^n(x)##, also converge? It does if ##f(0)=0##, but not in general.
     
    Last edited: Dec 22, 2015
  9. Dec 23, 2015 #8
    Hello guys and girls and sorry for the time of answer.
     
  10. Dec 23, 2015 #9
    Is it always true please?
     
  11. Dec 23, 2015 #10

    Samy_A

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    What is true is that if ##T## is a linear operator on a Banach space satisfying ##\|T\|<1##, then ##I-T## is invertible.
    (A specific case of a more general property of unital Banach algebras.)
     
    Last edited: Dec 23, 2015
  12. Dec 24, 2015 #11
    Which property do you refere please?
     
  13. Dec 24, 2015 #12
    Forget what I said : in a Banch space the absolute convegrence make the convergence, so let be a Banach algebra : let be x with $$||x||< 1$$, as $$\Sigma ||x||^{n}$$ converge the $$\sigma x^{n}$$ converge,
    and $$(x - Id)$$ is invertible his inverse is $$\Sigma x^{n}$$.

    Since E is a banach space so all all the continious endomorphisme are a Banach algebra.
     
  14. Dec 24, 2015 #13
    But $$k < \frac{1}{||A||^{-1}}$$. We could have $$k \ge 1$$.
     
  15. Dec 24, 2015 #14
    It's not taht esay.
     
  16. Dec 24, 2015 #15

    Samy_A

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    If ##T## is a linear operator on a Banach space satisfying ##\|T\|<1##, then ##I-T## is invertible.
    You can deduce from this (see it as an exercise):
    if A is an automorphism of a Banach space, and B is a linear operator on that space satisfying ##\|A-B\|<\frac{1}{\|A^{−1}\|}##, then B is invertible too.
    It would be easier if you stated clearly what you want to prove. For now we are just trying to guess what you really mean. Strictly speaking your question as stated has been answered.

    ,
     
  17. Dec 24, 2015 #16
    I think we want to proove A + f is an automorphisme.
     
  18. Dec 24, 2015 #17
    As it at been said we have to proove f linear and bijectif.
     
  19. Dec 24, 2015 #18

    Samy_A

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    Two different answers to the question "what do you want to prove".
    Then we are back to where we were two days ago. A + f will be an automorphism if and only if f is linear (given the properties of A and f in the OP).

    No idea where this comes from. There is no way one can prove f is linear and/or bijective using only the properties of f given in the OP.
     
  20. Dec 24, 2015 #19

    Krylov

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    Could a moderator please clean this thread up? It has become quite unreadable, notwithstanding the amount of patience that Samy_A has displayed. We require a clear problem statement from the OP, written in a single post and in clear English. Only then can others help.
     
  21. Dec 25, 2015 #20

    Dale

    Staff: Mentor

    Thread closed for moderation.
     
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