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I'm studying Lyapunov exponents. Let $f:\mathbb{R}\to\mathbb{R}$ be a smooth map, we define the Lyapunov exponent as $$L(x) := \limsup_{n\to\infty} \frac{1}{n}\sum_{i=0}^{n-1} \log\left|f'(f^{i}(x))\right|. $$

I have a dumb question that I can't answer it:

Question If $L(x)<0$, is it true that there exist $\delta>0$, such that $$|x-y|<\delta \ \Rightarrow\ \lim_{n\to\infty}|f^n(x) - f^n(y)|=0\ \ \ \ \ ? $$

where $f^i (x) = (f\circ \cdots\circ f)(x)$, $i$ times.

Or at least someone can inform me what is the notion of stability generated by the Lyapunov exponent?

Matheus Manzatto
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1 Answers1

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It is not true.

For any $\epsilon\in (0,1]$, construct a smooth function $f_\epsilon: [0,1]\to \mathbb{R}$ such that:

$f_\epsilon(0) = 0$;

$f'_\epsilon(0) = 1/2$;

$f_\epsilon^{(k)}(0) = 0$ for all $k\geq 2$ (I use $f_\epsilon^{(k)}$ to denote the $k$-th derivative);

$x/2\leq f_\epsilon(x)\leq 1$ for all $x\in[0,1]$;

$f_\epsilon(x)=1$ for all $x\in [\epsilon,1]$;

if $\epsilon=1$, then we also need the condition $f_\epsilon^{(k)}(1) = 0$ for all $k\geq 1$.

Now consider $f$ such that for any non-negative integer $k$ and any $x\in [0,1]$, $f(2k\pm x) = 2k+2 \pm f_{4^{-k}}(x)$. (For $x<-1$, let $f(x)$ be $1$, for example.) It is smooth due to the conditions on $f_\epsilon$.

Then $L(0) = -\ln 2$, but for any $y\in (0,1)$, $\lim_{n\to\infty}|f^n(y) - f^n(0)|=1$. It is because $f^n(0) = 2n$, but if $2^{-k}\leq y\leq 1$, then $2k+4^{-k}\leq f^k(y)\leq 2k+1$, then $f^{k+1}(y)=2k+3$, and $f^n(y)=2n+1$ for all $n>k$.

Litho
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