Syracuse Maps as Non-singular Power-Bounded Transformations and Their Inverse Maps

: The Hopf Decomposition provides a partition of $\mathbb{N}$ into sets $C$ and $D$, so that the restriction of $V$ to $C$ is conservative and $C$ is $V$-absorbing. Since $\mathbb{N}$ is countable, $C$ also is, so it can contain at most countable many cycles. This proves (1). Let $D_{1}=\bigcup_{k=1}^{\infty}V^{-k}(C)\backslash C$. Let $D_{2}=\mathbb{N}\backslash[C\cup D_{1}]$. Then, $V^{-1}(C\cup D_{1})=V^{-1}(C)\cup\bigcup_{k=1}^{\infty}V^{-k-1}(C)\backslash C=C\cup D_{1}$. Therefore, $V^{-1}(D_{2})=D_{2}$, and (2) and (3) are proven. Finally, given any $x\in\mathbb{N}$, a bounded trajectory means that there exists $n$ so $V^{n}(x)\in C$, so either $x\in C$ or $x\in D_{1}$. Given an unbounded trajectory of the point $y$, the set $\{y\}$ is wandering, and $V^{m}(y)\notin C$ for any natural number $m$. Thus, $y\in D_{2}$ and (4) is proven. ∎

$\Rightarrow$) Consider the power-bounded system $(\mathbb{N},2^{\mathbb{N}},V,\mu)$. By contradiction, let there exist some $x$ so $V^{k}(x)$ is never in a cycle. Then, for all $k\neq l$, $k,l>0$, $V^{k}(x)\neq V^{l}(x)$. Let $\mu(x)=\delta>0$. Further, since the measure is finite, $\mu(\bigcup_{i=0}^{\infty}V^{i}(x))=\sum_{i=1}^{\infty}\mu(V^{i}(x))=\epsilon>0$, implying that $\mu(V^{k}(x))\rightarrow 0$ as $k\rightarrow\infty$. Given any $M\in\mathbb{R}_{+}$, we may take some large $N$ such that $\mu(V^{N}(x))<\delta/M$. Hence, $V^{-N}(V^{N}(x))>\delta>M\mu(V^{N}(x))$, contradicting that the system is power bounded since $M$ was arbitrary.

$\Leftarrow$) Let this property hold. Since the space is countable, there are at most countable many cycles, and every point is in a pre-image of a cycle. Let the cycles be the sets $C_{1},C_{2},...$. First consider $C_{1}$. The cycle must be of finite length, N, so we construct a measure on $\mathbb{N}$. Define a function $\mu_{1}$, which will have a measure value at each specified point, and set the measure of any $A\in 2^{\mathbb{N}}$ to be the sum of the measures of the points in $A$. Let $\mu_{1}(c)=\frac{1}{2N}$ for any $c\in C_{1}$, and let so that $\mu_{1}(C_{1})=\frac{1}{2}$. Next, there are at most countably many points in $V^{-1}(C_{1})\backslash C_{1}$. The infinite case implies the finite case, so we consider this case. Enumerate these points as $\{c_{1},c_{2},c_{3}...\}$. Set $\mu_{1}(c_{i})=2^{-i-3}$, so that $\mu_{1}(V^{-1}(C_{1})\backslash C_{1})\leq 2^{-2}$. Next, consider $V^{-1}(c_{j})$, which does not contain $c_{j}$ nor any other point with a defined measure since such a case would generate a cycle. It again may be enumerated as $\{c_{j,1},c_{j,2},c_{j,3}...\}$, since the set is at most countably infinite. Set $\mu_{1}(c_{j,i})=2^{(-j-2)+(-i-2)}$ so that $\mu_{i}(V^{-1}(c_{j}))\leq 2^{-j-3}$ and so $\mu_{1}(V^{-1}(V^{-1}(C_{1})\backslash C_{1}))=2^{-3}$. Repeat inductively over the pre-images generated by this set, and set all points not in $\bigcup_{i=0}^{\infty}V^{-i}(C_{1})$ to have measure zero.

The process above gives that $\mu_{1}(\bigcup_{i=0}^{\infty}V^{-i}(C_{1}))=\mu_{1}(C_{1})+\mu_{1}(V^{-1}(C_{1})\backslash C_{1})+\mu_{1}(V^{-1}(V^{-1}(C_{1})\backslash C_{1}))+...\leq\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$. Repeat this inductively again over the $C_{i}$, and we may generate a new measure by the countable sum of measures. Let $\mu=\sum_{i=1}^{\infty}2^{-i-1}\mu_{i}$, so $\mu(\mathbb{N})\leq 1$ and $\mu$ is a finite measure. Further, every point in $\mathbb{N}$ is in one of the $\bigcup_{i=0}^{\infty}V^{-i}(C_{j})$, so every point has nonzero measure under $\mu$. Next, we need to show that the measure allows our point transformation to be power bounded. By construction, $\mu_{i}(V^{-1}(A))\leq\mu_{i}(A)$ for any set $A$ such that $A\cap C_{i}=\emptyset$. For a set $B$ intersecting the cycle $C_{i}$, separating $B=(B\cap C_{i}^{C})\cup(B\cap C_{i})$ gives $\mu_{i}(V^{-n}(B))\leq 2^{-n}\mu_{i}(B\cap C_{i}^{C})+2\mu_{i}(B\cap C_{i})\leq 2\mu_{i}(B)$ for all $n\in\mathbb{N}$. Therefore, $\mu(V^{-n}(A))=\sum_{i=1}^{\infty}2^{-i-1}\mu_{i}(V^{-n}(A))\leq\sum_{i=1}^{\infty}2^{-i}\mu_{i}(A)=2\mu(A)$. The measure $\mu$ is finite and power-bounded in $\mathscr{L}^{1}$, as well as equivalent to the counting measure, as desired.

∎

For arbitrary $n\in{\mathbb{N}}_{2}$, it may be written as $3p+2$ be definition, and thus as $3(p+1)-1$. Using the Fundamental Theorem of Arithmetic, $p+1$ has the desired representation $3^{a-1}2^{b}h$. ∎

This relies on a couple building blocks of concepts that led to the chain structure in the Collatz map. First, the family structure gives a node of the form $p^{\alpha}2^{\beta}k-l$ (where $k$ is neither a multiple of $2$ or $p$) so that $V(p^{\alpha}2^{\beta}k-l)=p^{\alpha+1}2^{\beta-1}k-l$. This requires that $l$ be odd, and so

Hence, $\frac{r-pl}{2}=-l$ and $r=(p-2)l$. Since $l$ is an integer, $\frac{r}{p-2}$ is, and since $|r|<p$, $r=\pm(p-2)$. Note that these calculations also show that elements in the class of $\frac{r}{2-p}\,\,(mod\,\,p)$ have two pre-images. These are, in fact, the only nodes with $2$ pre-images.

This case where $r=\pm(p-2)$ gives a further sub-class of maps with the family structure. Next, we most consider the connection of families to create chains. Using the previous notation, let $m$ be the largest positive integer so $2^{m}$ divides $p^{\alpha+\beta}k+\frac{r}{2-p}$. We would require that

Such a condition would connect two families, since $V^{m+1}(p^{\alpha+\beta}k+\frac{r}{2-p})$ would then have a representation $p^{\alpha_{2}}2^{\beta_{2}}k_{2}-\frac{r}{p-2}$.

The above equation says that

Since $p$ is odd, this occurs exactly when

which reduces to the trivial

We thus have that the families connect for this form of map whenever they occur. ∎