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Math Help - Differentiability problem

  1. #1
    Senior Member Pinkk's Avatar
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    Differentiability problem

    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.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    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|
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  3. #3
    Senior Member Pinkk's Avatar
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    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?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    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..
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  5. #5
    Senior Member Pinkk's Avatar
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    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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