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Let $f: \mathbb R\to \mathbb R$ be expanding. Then $f$ exhibits sensitive dependence on initial data.

My attempt: It suffices to show that $f$ has a positive Lyapunov exponent. $$\lambda(x_0)=\lim_{n \to \infty}\frac{1}{n}\sum_{i=0}^{n-1}\log|f'(x_i)|$$ Since, $f$ is expanding, $|f'(x_i)|>1.$ Hence, $\log|f'(x_i)|>0$ for all $i=0,1,\cdots,n-1.$

How to proceed?

Did
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Mark
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  • It is false in general that $|f'(x_i)|>1$. But really what you need is a uniform bound, isn't it? – John B Dec 14 '17 at 21:38
  • @JohnB Uniform bound on? – Mark Dec 14 '17 at 21:39
  • First obtain the best lower bound you can using the definition of expanding. You should be able to take it from there. – John B Dec 14 '17 at 21:41
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    @JohnB The definition we were given for expanding map is: A continuous and differentiable function $f: \mathbb R \to \mathbb R$ is said to be expanding if $|f'(x)|>1$ for all $x \in \mathbb R.$ You can check out the definition 2.1.8 here. – Mark Dec 14 '17 at 21:47
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    That's not the usual definition. But in that case the statement that you want to prove is false in general. Are you sure that you have no further conditions? – John B Dec 14 '17 at 21:52
  • @JohnB What is the usual definition? – Mark Dec 14 '17 at 21:54
  • A differentialbe map $f$ is said to be expanding if there exist constants $c>0$ and $\lambda>1$ such that $|(f^n)'(x)|\ge c\lambda^n$ for all $n\ge1$. – John B Dec 14 '17 at 21:55
  • @JohnB No. In fact, we're following this and I noticed that proof of Proposition 2.3.10 is incorrect but from what you say, the statement is also incorrect? – Mark Dec 14 '17 at 22:05
  • Sure, it is easy to give counterexamples. You need something more. But to use a master thesis is peculiar... to say the least. – John B Dec 14 '17 at 22:13
  • Note that your approach is also doomed as an expanding system may not have a well-defined Lyapunov exponent to begin with. For example consider something like $f(x)=\exp(x^2)$. — Also, to provide a counterexample as alluded to by @John B: $f(x)=x-\tfrac{1}{1}$. – Wrzlprmft Dec 15 '17 at 10:52
  • @Mark: Dangit. This is obviously a typo, though admittedly it’s not obvious what the typo is. It should have been $f(x)=x-\frac{1}{x}$. – Wrzlprmft Dec 16 '17 at 08:11

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