I need some help with this question relating periodic points and dependence on initial conditions:

$\bullet\;$Let I be a real interval, and $g:I\to I$ a derivable function. If $x_0$ is an $m$-periodic point of $g$, such that


then $g$ has not dependence on initial conditions at $x_0$.

Note: We say $g$ has dependence on initial conditions at a point $a$ if there is a constant $\eta>0$ such that $\forall\mu>0$, and $\forall z\in I$ with $|a-z|<\mu$, then $\exists m\in\mathbb{N}$ with $|g^m(a)-g^m(z)|>\eta$.

Thanks a lot for any help.

  • 1
    $\begingroup$ Hint: If the derivative of a function $f$ at $x_0$ has absolute value $<1$, this means that $f$ decreases distances near $x_0$, since if $x,y$ are very close to $x_0$, then $f(x)-f(y)\approx f'(x_0)(x-y)$. Can you make this precise using the mean value theorem? $\endgroup$ – froggie Feb 24 '14 at 20:10
  • $\begingroup$ Using the mean value theorem,I would get $|g^m(x)-g^m(x_0)|=|(g^m)'(\alpha)||x-x_0|$, for some $\alpha$ between $x_0$ and $x$ $\endgroup$ – Mark_Hoffman Feb 24 '14 at 20:43

Let $\eta>0$ be given.

Since $|(g^m)'(x_0)|<1$, we see that $|(g^m)'(x)|<1$ for all $x$ sufficiently close to $x_0$. That is to say, we can find a number $\theta<1$ and a small interval $J = [x_0-\delta, x_0+\delta]$ around $x_0$ such that $|(g^m)'(x)|<\theta$ for all $x\in J$. Of course, I can choose $\delta$ as small as desired, and in particular I can assume $\delta<\eta$.

Claim: First note that $J$ is invariant under $g^m$, i.e., $g^m(J)\subset J$. Indeed, if $x\in J$, then by the mean value theorem $$|g^m(x) - x_0| = |g^m(x) - g^m(x_0)|\leq \theta|x - x_0|\leq \theta\delta<\delta,$$ so $g^m(x)\in J$.

As a consequence of the claim and the chain rule, note that for every $x\in J$, one gets that $|(g^{km})'(x)| \leq \theta^k$ for all $k\geq 1$.

Let $M = \max_{r = 0,\ldots, m-1}\max_{x\in J} |(g^r)'(x)|$. For any $n\geq 1$, we can write $n$ uniquely as $n = km + r$, where $r\in\{0,\ldots, m-1\}$. Thus by the chain rule, for $x\in J$ we get $$|(g^n)'(x)| = |(g^r)'(g^{km}x)||(g^{km})'(x)|\leq M\theta^k.$$ Since $M$ is just a constant and $\theta<1$, for $k$ large enough this quantity is $<1$. Thus we have derived:

Fact: For all $n$ sufficiently large, $|(g^n)'(x)|<1$ on $J$.

In particular, by the mean value theorem, it follows that for $n$ sufficiently large, say $n\geq N$, we have $|g^n(x) - g^n(x_0)| < |x - x_0|<\delta<\eta$ for $x\in J$. Moreover, since $g$ is continuous, up to further shrinking $\delta$, I can also arrange so that $|g^n(x) - g^n(x_0)|<\eta$ for all $n<N$. In this way I have found a neighborhood $J$ of $x_0$ such that $|g^n(x) - g^n(x_0)|<\eta$ for all $x\in J$ and all $n\geq 1$. This is exactly the desired property.

  • $\begingroup$ Very clear and detailed explanation, froggie! Thanks a lot! $\endgroup$ – Mark_Hoffman Feb 24 '14 at 22:26

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