Differentiability problem

**Pinkk** · April 21st 2010, 08:07 PM

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

Okay, so it's pretty clear that I have to use the mean value theorem and so there exists $c\in \mathbb{R}$ such that $\frac{f(b)-f(a)}{b-a}=f'(c)$ for $a,b\in \mathbb{R}$ . Now it seems that I should just replace $a,b$ with $s_{n},s_{n-1}$ and I will arrive at the hint. However, how can I guarantee that for any $n$ , $s_{n}\ne s_{n-1}$ ? 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.

**Drexel28** · April 21st 2010, 08:20 PM

Originally Posted by Pinkk

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

Okay, so it's pretty clear that I have to use the mean value theorem and so there exists $c\in \mathbb{R}$ such that $\frac{f(b)-f(a)}{b-a}=f'(c)$ for $a,b\in \mathbb{R}$ . Now it seems that I should just replace $a,b$ with $s_{n},s_{n-1}$ and I will arrive at the hint. However, how can I guarantee that for any $n$ , $s_{n}\ne s_{n-1}$ ? 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 $n\in\mathbb{N}$ be arbitrary. We may assume WLOG that $s_n< s_{n+1}$ . For, if $s_n=s_{n+1}$ 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 $\xi\in(s_n,s_{n+1})$ we have that $|f(s_{n+1})-f(s_n)|=|s_{n+2}-s_{n+1}|\leqslant |f'(\xi)||s_{n+1}-s_n|\leqslant a|s_{n+1}-s_n|$

**Pinkk** · April 21st 2010, 08:53 PM

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?

**Drexel28** · April 21st 2010, 09:20 PM

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 $m,n\in\mathbb{N},\text{ }m\geqslant n$ .

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

**Pinkk** · April 21st 2010, 09:28 PM

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

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

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