1. ## Differentiability problem

Let $\displaystyle f$ be differentiable on $\displaystyle \mathbb{R}$ with $\displaystyle a=\sup \{|f'(x)|; x\in \mathbb{R}\}< 1$. Select $\displaystyle s_{0}\in \mathbb{R}$ and define $\displaystyle s_{n}=f(s_{n-1})$ for $\displaystyle n\ge 1$. Prove that $\displaystyle (s_{n})$ is a convergent sequence. Hint: To show $\displaystyle (s_{n})$ is Cauchy, first show that $\displaystyle |s_{n+1} - s_{n}| \le a|s_{n} - s_{n-1}|$ for $\displaystyle n\ge 1$.

Okay, so it's pretty clear that I have to use the mean value theorem and so there exists $\displaystyle c\in \mathbb{R}$ such that $\displaystyle \frac{f(b)-f(a)}{b-a}=f'(c)$ for $\displaystyle a,b\in \mathbb{R}$. Now it seems that I should just replace $\displaystyle a,b$ with $\displaystyle s_{n},s_{n-1}$ and I will arrive at the hint. However, how can I guarantee that for any $\displaystyle n$, $\displaystyle s_{n}\ne s_{n-1}$? Is there something I am missing that guarantees this, or is my actual approach incorrect so far? And even if I arrive at the hint, how would I proceed exactly? Thanks.

2. Originally Posted by Pinkk
Let $\displaystyle f$ be differentiable on $\displaystyle \mathbb{R}$ with $\displaystyle a=\sup \{|f'(x)|; x\in \mathbb{R}\}< 1$. Select $\displaystyle s_{0}\in \mathbb{R}$ and define $\displaystyle s_{n}=f(s_{n-1})$ for $\displaystyle n\ge 1$. Prove that $\displaystyle (s_{n})$ is a convergent sequence. Hint: To show $\displaystyle (s_{n})$ is Cauchy, first show that $\displaystyle |s_{n+1} - s_{n}| \le a|s_{n} - s_{n-1}|$ for $\displaystyle n\ge 1$.

Okay, so it's pretty clear that I have to use the mean value theorem and so there exists $\displaystyle c\in \mathbb{R}$ such that $\displaystyle \frac{f(b)-f(a)}{b-a}=f'(c)$ for $\displaystyle a,b\in \mathbb{R}$. Now it seems that I should just replace $\displaystyle a,b$ with $\displaystyle s_{n},s_{n-1}$ and I will arrive at the hint. However, how can I guarantee that for any $\displaystyle n$, $\displaystyle s_{n}\ne s_{n-1}$? Is there something I am missing that guarantees this, or is my actual approach incorrect so far? And even if I arrive at the hint, how would I proceed exactly? Thanks.
Let $\displaystyle n\in\mathbb{N}$ be arbitrary. We may assume WLOG that $\displaystyle s_n< s_{n+1}$. For, if $\displaystyle s_n=s_{n+1}$ we're done and if the opposite inequality is true then in the following method we need only interchange the two values. So, we know that for some $\displaystyle \xi\in(s_n,s_{n+1})$ we have that $\displaystyle |f(s_{n+1})-f(s_n)|=|s_{n+2}-s_{n+1}|\leqslant |f'(\xi)||s_{n+1}-s_n|\leqslant a|s_{n+1}-s_n|$

3. Okay, but I'm afraid of that there exists sequences that repeat indefinitely and therefore do not converge. However after constructing a few examples it seems that if it does repeat then the derivative at at least one point is greater than or equal to one. Is there a way to definitely prove this/is it not really important/does it come up in the proof/whatever? Thanks.

Also, how do I proceed with the hint exactly?

4. Originally Posted by Pinkk
Okay, but I'm afraid of that there exists sequences that repeat indefinitely and therefore do not converge. However after constructing a few examples it seems that if it does repeat then the derivative at at least one point is greater than or equal to one. Is there a way to definitely prove this/is it not really important/does it come up in the proof/whatever? Thanks.

Also, how do I proceed with the hint exactly?
So, you need merely note that if $\displaystyle m,n\in\mathbb{N},\text{ }m\geqslant n$.

Then, $\displaystyle |s_{m}-s_{n}|\leqslant a|s_{m-1}-s_{n-1}|\leqslant\cdots\leqslant a^n|s_{m-n}-s_0|$...so..

5. I'm sorry, I don't see how that follows. $\displaystyle m$ does not necessarily equal $\displaystyle n+1$, so yeah, don't get how you came to that conclusion, and if that's true I don't see how $\displaystyle a^{n}|s_{m-n}-s_{0}|$ would help, sorry.

Because what I get is that $\displaystyle |s_{n+1} - s_{n}| \le a^{n}|s_{1}-s_{0}|$ and so $\displaystyle |s_{m} -s_{n}|\le (m-n)a^{n}|s_{1}-s_{0}|$, and after that I'm stuck.