A source point is sensitive to initial conditions?

Question

Is it true that if a point is a source for a one-dimensional (smooth) map $f(x)$ (i.e.: $x$ satisfies $|f'(x)| > 1$), then this point is sensitive to initial conditions (i.e.: for every $\epsilon > 0$, there is an $\bar{x}$ and a $d>0$ such that $|f^k(\bar{x}) - f^k(x)|\geq d$ for some integer $k$)?

Intuitively, it seems that yes, because if you "deviate" a bit for every such point, you can be dragged too far by further iterations.

Answer 1

Is it true that if a point is a source for a one-dimensional (smooth) map $f(x)$ (i.e.: $x$ satisfies $|f'(x)| > 1$), then this point is sensitive to initial conditions (i.e.: for every $\epsilon > 0$, there is an $\bar{x}$ and a $d>0$ such that $|f^k(\bar{x}) - f^k(x)|\geq d$ for some integer $k$)?

I presume that wanted to say something along the lines of:

For every $ε>0$ and $d>0$*, there are $\bar{x}$ and $k$ such that $\left | \bar{x}-x \right | < ε$ and $|f^k(\bar{x}) - f^k(x)|\geq d$.

^{* possibly restricting $d$ to the size of your dynamics}

Then $|f'(x)| > 1$ does not suffice because you also have to look at the derivative at other points. For a counterexample, consider the map:

$$ f(x) = \begin{cases} 0 & \text{if } x≤0\\ x^2 & \text{if } 0≤x≤1\\ 0 & \text{if } 1≤x\\ \end{cases}$$

For $\tfrac{1}{2}<x<1$, you have $|f'(x)| > 1$, but $f^k(x)$ quickly converges to $0$.

However, if the geometric average of $|f'(x)|$ is larger than one on your trajectory, you do indeed have sensitivity to initial conditions. This is typically expressed in terms of the maximum Lyapunov exponent, which for a map is:

$$ λ = \lim_{n→∞} \frac{1}{n} \sum_{k=0}^n \ln \left( f'\left( f^k(x) \right) \right)$$