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Devaney defines that

a funtion $f:J \to J$ has sensitive dependence on initial conditions if there exists $\delta > 0$ such that, for every $x \in J$ and any neighborhood $N$ of $x$, there exists $y \in N$ and $n \geq 0$ such that $|f^n(x)-f^n(y)| > \delta$.

Here I already see a misspell... I would rather say "exist $y \in N$ and $n \geq 0$", without the "s". But probably here I am wrong.

I was told that this definition is equivalent to

there exists $\delta > 0$ such that for every open $U \subset J$ there exist two points $x, y\in U$ and a number $n \in \mathbb{Z}^+$ such that $|f^n(x)- f^n(y)| \geq \delta$

But I cannot see how. Any hint on a formal proof of the equivalence?

ThePunisher
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1 Answers1

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It is clear that the first definition implies the second because there is the freedom to choose $x$ while in the second definition we have only the existence of $x$.

To prove that the second definition suppose we have a neighborhood $N$ of $x$. There exists $x'$ and $y'$ in $N$ and $n\in \mathbb{Z}^+$ such that $|f^n(x')-f^n(y')|>\delta '$ therefore we have that $|f^n(x')-f^n(x)|+|f^n(x)-f^n(y')| \geq|f^n(x')-f^n(y')| \geq \delta$ by the triangle inequality and from this we have that either $|f^n(x')-f^n(x)| \geq \frac{\delta'}{2}$ or $|f^n(x)-f^n(y')|\geq \frac{\delta'}{2}$ therefore we can set $\delta = \frac{\delta'}{2}$

ThePunisher
  • 1,076