# Homeomorphism in a Banach space

1. Dec 30, 2015

### Calabi

• Member warned about posting with no effort shown
Hello,

1. The problem statement, all variables and given/known data

Let be E a banach space, A a continuous automorphsim(by the banach theorem his invert is continus too.). and f a k lipshitzian fonction with $$k < \frac{1}{||A^{-1}||}$$.

2. Relevant equations
$$k < \frac{1}{||A^{-1}||}$$

3. The attempt at a solution

I have to show that A + f is bijectiv and is invert is continuous too.

I have no real clue for that in the moment.

Thank ou in avnace and have a nice afternoon.

Last edited by a moderator: Jan 5, 2016
2. Dec 30, 2015

### Samy_A

Thanks to @Krylov , here is a tip: Banach fixed-point theorem.

3. Jan 5, 2016

### Calabi

Hello and good year. Could you be more precise please?

4. Jan 5, 2016

### Samy_A

Good year.

This tip is really all you need to solve the problem. You could look up what the Banach fixed-point theorem exactly says, and then see how it applies to this problem.

5. Jan 5, 2016

### Calabi

OK I'm gonna think about it.

6. Jan 8, 2016

### Calabi

Hello and have a nice year : first : $$||A^{-1} o f|| < ||A^{-1}|| ||f|| < kA^{-1} < 1$$ sol let be h in the continious fonction forme E to E with $$h o (Id + A^{-1}f) = (Id + A^{-1}of) o h = Id$$. Since A linear we've got $$(A + f) o h = A \Leftrightarrow (A+f)oh o A^{-1} = Id$$. I don't know if that mean there's an inverse at right.
At least my invert is continous.

7. Jan 8, 2016

### Samy_A

I'm not sure I understand this: for instance; where does this h come from?

Maybe someone else understand what you did and can comment, in that case just disregard what follows.

The way I would approach this (using @Krylov 's tip, but maybe not in the way he meant it):
First consider the simple case where A=I (the identity) and Lip(f)=k<1, and prove that I-f is invertible and the inverse is Lipschitz.

Proving that I-f is injective is easy (and yields the Lipschitz constant for the inverse of I-f).
EDIT: see the following posts.

In order to prove that I-f is surjective, use the Banach fixed-point theorem.
Hint: take y∈E, and consider the mapping Y: E→E defined by Y(x)=y+f(x). Can we apply the Banach fixed-point theorem to Y? And if so, what does it say about the surjectivity of I-f?

Again, maybe your proof is correct, if so, just disregard this.

Last edited: Jan 8, 2016
8. Jan 8, 2016

### Krylov

Samy's suggestion is just fine, but I don't think it's necessary to consider surjectivity and injectivity separately.

9. Jan 8, 2016

### Samy_A

Oh yes, you get injectivity from the uniqueness of the fixed point.

10. Jan 12, 2016

### Calabi

Hello every body the fact is i'm not sure of my first equality. It's not even sur that $$||f||$$ exist(exept if f is nul in 0.).
At least I try something(even if i think it could be true.). anyway.
Let suppose i - f is not injctive and let's consider x in y in E with $$x \neq y and x - f(x) = y - f(x)$$
We get y = x which is absurde. Let's take this by the beginning : we want to show that y as got an antecedant by I - f.
Let's create $$\phi(x) = y + f(x)$$. It's clearly k < 1 lipshitzian. So let's applic the banach theorem, we find x in E with $$x - f(x) = E$$ wich proove the surjectivity.

Since it's proove let's we can wright $$A + f = A o (I - A^{-1}of)$$
and by the hypothesis we've got $$A^{-1} o f$$ $$k||A^{-1}|| < 1$$ lipshitzian.
So it's conclude.

11. Jan 12, 2016

### Calabi

Oh ther's also to proove that the inverse of I - f lipshitzian

12. Jan 12, 2016

### Samy_A

You get $x-f(x)=y-f(y)$, not $x - f(x) = y - f(x)$, so this part of your proof is not correct (or at least incomplete).
You probably meant that the Banach fixed-point theorem gives you a $x\in E$ for which $x-f(x)=y$ (not $=E$ as you wrote). This part is correct.
Yes, this way you get the general case from the particular case where A=I. You have the sign wrong in $A + f = A o (I - A^{-1}of)$, but that is no big deal, as what you proved is equally true for $-f$.

You still have two issues left:
1) Injectivity (though that is almost done with what you have, the hint has already been given previously).
2) Show that the inverse is Lipschitz too.

Last edited: Jan 12, 2016
13. Feb 24, 2016

### Samy_A

As it came up elsewhere, I'd like to add to this thread the proof that the inverse of $I-f$ is a Lipschitz function (under the assumptions of post #7).
Take $x, y \in E$. Then, since $f$ is Lipschitz with Lipschitz constant $k \lt 1$, $\|f(x)-f(y)\|\leq k\|x-y\|$
Using the triangle inequality, one gets:
$\|(I-f)(x)-(I-f)(y)\|=\|x-y-(f(x)-f(y))\| \geq \|x-y\|-\|f(x)-f(y)\| \geq \|x-y\| -k\|x-y\|=(1-k)\|x-y\|$.
This shows that the inverse of $I-f$ is a Lipschitz function with Lipschitz constant $\leq \frac{1}{1-k}$.

Last edited: Feb 25, 2016
14. Feb 24, 2016

### Staff: Mentor

Maybe I'm just too blind to see, but I didn't get it.
We have $||(I-f)(x)-(I-f)(y)|| \geq q ||x-y||$ for a $1>q>0$. How does this turn into an upper bound?

15. Feb 24, 2016

### Samy_A

I should have added the following.

Set $T=(I-f)^{-1}$.
Take $x', y' \in E$. Set $x=Tx', y=Ty'$
As shown above, we have $\|(I-f)(x)-(I-f)(y)\| \geq (1-k)\|x-y\|$.
This give $\|x'-y'\| \geq (1-k)\|Tx'-Ty'\|$, or $\|Tx'-Ty'\| \leq \frac{1}{1-k}\|x'-y'\|$.

Last edited: Feb 25, 2016
16. Feb 24, 2016

### Staff: Mentor

Thanks. I've just forgotten the "inverse of" part before $I-f$.
Edit: you should have written $(I-f)^{-1}$ for dummies like me

17. Feb 25, 2016

### Samy_A

I agree (not about the "dummies" ).
Let's rewrite it in terms of $T=(I-f)^{-1}$. (As it is shorter and clearer.)

For $x \in E$, $(I-f)Tx=x$, so that $Tx=x+f(Tx)$

Take any $x,y \in E$:
$\|Tx-Ty\|=\|x+f(Tx)-y-f(Ty)\|=\|x-y+f(Tx)-f(Ty)\| \leq \|x-y\|+\|f(Tx)-f(Ty)\| \\ \leq \|x-y\| + k\|Tx-Ty\|$
Or $\|Tx-Ty\|-k\|Tx-Ty\| =(1-k) \|Tx-Ty\| \leq \|x-y\|$.
Hence $\|Tx-Ty\| \leq \frac{1}{1-k}\|x-y\|$, proving that $T=(I-f)^{-1}$ is a Lipschitz function.