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Last edited by hbghlyj 2022-10-21 08:44Cantor sets and bounded orbits in certain “tent” maps常数$A>1$, 函数$$f_{A}(x)=A·\left(1-2\left|x-\frac{1}{2}\right|\right)$$的轨道是$[0,1]$上的Cantor集.
证明:
$f_A$ 的逆映射有两个分支 $f_1$、$f_2$:
$$f_{1}(x)=\frac{x}{2 A}$$
$$f_{2}(x)=-\frac{x}{2 A}+\frac{1}{2 A}$$映射 $f_1$、$f_2$ 是仿射压缩（即导数绝对值小于 1 的仿射映射）。
$f_1$是$[0,1]$到$\left[0, \frac{1}{2 A}\right]$的双射, $f_2$是$[0,1]$到$\left[1-\frac{1}{2 A}, 1\right]$的双射.
$f_1$的像与$f_2$的像的交集为空.
Define $I_{a_0 a_1 \ldots a_k}=f_{a_0 a_1 \ldots a_k}(I)$. It follows easily by induction that if $\mathbf{a}=$ $\left(a_0 a_1 \ldots a_k\right) \neq \mathbf{b}=\left(b_0 b_1 \ldots b_k\right)$, then $I_{a_0 a_1 \ldots a_k} \cap I_{b_0 b_1 \ldots b_k}=\emptyset$. Moreover, the length of $I_{b_0 b_1 \ldots b_k}$ is $\left(\frac{1}{2 A}\right)^{k+1}$.
Let
\[
F_k=\bigcup I_{a_0 \ldots a_k} .
\]
Then,
\[
F=\bigcap_k F_k
\]
is a Cantor set.
Let $B$ denote the set of points $x$ whose forward orbits are bounded. We claim
\[
B=F \text {. }
\]
If $x \in F$, then $f_A^{k+1}(x) \in I$ for each $k \geq 0$, so $F \subset B$.
Now, we prove the converse: $B \subset F$.
Write $f=f_A$.
Note that if $x$ is any point which is not in $I$, then $f^n(x) \rightarrow-\infty$.
So, any $x \in B$ must lie in $I$.
Let
\[
J=I \backslash\left(I_1 \bigcup I_2\right)
\]
Then, if any forward iterate $f^k(x)$ of $x$ gets into $J$, then $f^{k+1}(x)$ is not in $I$, so $f^n(x) \rightarrow \infty$.
Now, let $x$ be a point whose forward orbit is bounded. Then, first $x \in I$.
If $x \notin F$, then it is in some interval in the complement of $F$ in $I$.
Any such interval must have the form $f_{a_0 a_1 \ldots a_k}(J)$ for some finite sequence of $a_i=1$ or 2 .
If follows that $f^{k+1}(x) \in J$. Hence, $f^n(x) \rightarrow-\infty$ as $n \rightarrow \infty$.
This proves that $B \subset F$ as required. QED. |
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