Homeomorphism in a Banach space

[Calabi]

Hello,

1. Homework Statement
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 $$k < \frac{1}{||A^{-1}||}$$.

Homework Equations

$$k < \frac{1}{||A^{-1}||}$$

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.
Could you help me please?

Thank ou in avnace and have a nice afternoon.

 

[Samy_A]

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

 

[Calabi]

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

 

[Samy_A]

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

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.

 

[Calabi]

OK I'm going to think about it.

 

[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.

 

[Samy_A]

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.

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.

 

[S.G. Janssens]

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

 

[Samy_A]

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

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

 

[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 $$x - f(x) = E$$ which 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.

 

[Calabi]

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

 

[Samy_A]

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.

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).

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 $$x - f(x) = E$$ which proove the surjectivity.

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.

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.

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.

 

[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 $\leq \frac{1}{1-k}$.

 

[fresh_42]

This shows that the inverse of $I−f$ is a Lipschitz function with Lipschitz constant $≤\frac{1}{1−k}$

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?

 

[Samy_A]

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 $q>0$. How does this turn into an upper bound?

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 $\|Tx'-Ty'\| \leq \frac{1}{1-k}\|x'-y'\|$.

 

[fresh_42]

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

 

[Samy_A]

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

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.