An automorphism in a Banach space

1. Dec 22, 2015

Calabi

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.

2. Dec 22, 2015

Samy_A

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?

3. Dec 22, 2015

Calabi

Oh no. Change homomorphisme in automorphism.
And I never say that f is linear.

4. Dec 22, 2015

Samy_A

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
5. Dec 22, 2015

WWGD

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.

6. Dec 22, 2015

Krylov

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.

7. Dec 22, 2015

Samy_A

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
8. Dec 23, 2015

Calabi

Hello guys and girls and sorry for the time of answer.

9. Dec 23, 2015

Calabi

Is it always true please?

10. Dec 23, 2015

Samy_A

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
11. Dec 24, 2015

Calabi

Which property do you refere please?

12. Dec 24, 2015

Calabi

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.

13. Dec 24, 2015

Calabi

But $$k < \frac{1}{||A||^{-1}}$$. We could have $$k \ge 1$$.

14. Dec 24, 2015

Calabi

It's not taht esay.

15. Dec 24, 2015

Samy_A

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.

,

16. Dec 24, 2015

Calabi

I think we want to proove A + f is an automorphisme.

17. Dec 24, 2015

Calabi

As it at been said we have to proove f linear and bijectif.

18. Dec 24, 2015

Samy_A

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.

19. Dec 24, 2015

Krylov

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.

20. Dec 25, 2015

Staff: Mentor

Thread closed for moderation.

21. Dec 27, 2015

Staff: Mentor

@Calabi after discussion with the mentors we recommend that you open up a new thread in the advanced homework section. Please use the template, that will greatly benefit the discussion and add some structure and clarity.

22. Dec 30, 2015

Calabi

Hello sorry I'm late, I accept and i'm sorry for 2 reason : first I was not very active on the conversation.
Then this exercicse is extract from a paper and the teacher made many mistake in the paper.