1. ## Re: Contraction mapping theorem

Originally Posted by Ackbeet
Ah. The conditions in your theorems ARE Lipschitz conditions of a particular kind. Look here for a Lipschitz condition definition. You can see that the conditions in your theorems are that the function is Lipschitz with Lipschitz constant in the interval (0,1). If you look at the first property in the Properties section of the wiki, you will see that an everywhere differentiable function is Lipschitz if and only if its derivative is bounded, and it tells you what the Lipschitz constant is in that case: the least upper bound of the magnitude of the derivative! So there's a direct tie-in to the conditions of your theorems. Does that help?

So your logic would work like this:

1. The function is everywhere differentiable.
2. The magnitude of the derivative is bounded by 1 - epsilon everywhere on the interval, for some epsilon greater than zero.
3. Therefore, the function is Lipschitz with a Lipschitz constant in the interval (0,1).
4. Therefore, the conditions of the theorem you've been given apply.
5. Therefore, the conclusion of the theorem holds.

And you're done. The major step here is # 2. What's epsilon?

I got a basic idea on the item but still don't understand how do we know that our derivative is bounded by 1! If I understood well the magnitude of the derivative is bounded by epsilon which is 1 in our case?

Thanks again for all your help and time, you are so helpful. But since I can't use things that I have never been taught about in lectures I think I should use just the counterexample using an approximation. I really appreciate the time you spent for answering my questions!

Btw I gave a proof for the inequality you asked me in the other post, can you see it to tell me if you find it good?

Thank you very much!!!

2. ## Re: Contraction mapping theorem

Originally Posted by Darkprince
I got a basic idea on the item but still don't understand how do we know that our derivative is bounded by 1!
You don't know that ahead of time. That's what you need to prove. Just take the derivative, and find its maximum on the interval using good ol' first-semester calculus techniques.

If I understood well the magnitude of the derivative is bounded by epsilon which is 1 in our case?
I meant that, for some $\displaystyle \epsilon>0,$ the magnitude of the derivative is bounded by $\displaystyle 1-\epsilon.$ That ensures that the Lipschitz constant is not identically 1, which wouldn't work for your theorem.

Thanks again for all your help and time, you are so helpful. But since I can't use things that I have never been taught about in lectures I think I should use just the counterexample using an approximation. I really appreciate the time you spent for answering my questions!

Btw I gave a proof for the inequality you asked me in the other post, can you see it to tell me if you find it good?

Thank you very much!!!

3. ## Re: Contraction mapping theorem

Originally Posted by Ackbeet
You don't know that ahead of time. That's what you need to prove. Just take the derivative, and find its maximum on the interval using good ol' first-semester calculus techniques.

I meant that, for some $\displaystyle \epsilon>0,$ the magnitude of the derivative is bounded by $\displaystyle 1-\epsilon.$ That ensures that the Lipschitz constant is not identically 1, which wouldn't work for your theorem.
Oh so because we find that the derivative is bounded by 1 then in our case epsilon=0 and not >0 so the condition fails and it is not a Lipschitz constant!

4. ## Re: Contraction mapping theorem

Originally Posted by Darkprince
Oh so because we find that the derivative is bounded by 1 then in our case epsilon=0 and not >0 so the condition fails and it is not a Lipschitz constant!
I don't think I'd agree with that. You need to do the work of maximizing the derivative on the interval. What is its maximum value?

5. ## Re: Contraction mapping theorem

Originally Posted by Ackbeet
I don't think I'd agree with that. You need to do the work of maximizing the derivative on the interval. What is its maximum value?
(-e-2)/4? since f'(x)=(-e^(x)-2)/4?

6. ## Re: Contraction mapping theorem

Originally Posted by Darkprince
(-e-2)/4? since f'(x)=(-e^(x)-2)/4?
The maximum of $\displaystyle h(x)=|f'(x)|=\left|\frac{-e^x-2}{4}\right|$ is either of calculus type (that is occurs at a point such that $\displaystyle h'(x)=0$) or is at an end point of the interval.

Well it is not of calculus type (can you prove that?) so we try the end points and find the maximum occurs where $\displaystyle x=1$ and is $\displaystyle (e+2)/4>1$

CB

Page 2 of 2 First 12