How might a Mathematician reformulate this Conjecture?

Question

I have a question pertained to the below conjecture here. However, from the comments, it is clear that the question could be improved using math jargons and symbols for clarity and conciseness. How might a mathematician reformulate this conjecture?

• Given an iterative piecewise linear function, $f$, with domain $\mathbb{Z}$ and range $\mathbb{Z}$ :
• under iteration by $f$, if the trajectories/iterates of the first integers in all the subdomains converge to a cycle;
• then, the trajectories of all integers in the domain of $f$ will converge to a cycle, that is, no trajectory diverges.

Special Conditions: (1) Zero can be included or excluded from $f$. (2) It is assumed that the subdomains and subranges are in arithmetic progression containing non-negative or non-positive integers but not a mix. (3) It is also assumed that if $f$ has only two subfunctions then the cycle must be of length 1 and, of course, $f(x)\neq x$ unless $x$ is the element to which all the trajectories of the first elements converge. (4) Importantly, if one element in a subdomain is greater than its corresponding element in the subrange, then all elements in that subdomain must be greater than their corresponding elements in the subrange. The same is true if one element in a subdomain is less than its corresponding element in the subrange.

$\color{black}{\text{Example 1}}$

$$f(x) = \begin{cases} x+1, & \text{subdoman: 1, 3, 5, 7,...Subrange: 2, 4, 6, 8,...} \\ -x+2, & \text{subdoman: 2, 4, 6, 8,...Subrange: 0,-2,-4,-6...}\\ -x, & \text{subdoman: -1, -3, -5,...Subrange: 1, 3, 5, ...}\\ x+1, & \text{subdoman: -2, -4, -6,...Subrange: -1, -3, -5,...}\\ 0, & \text{subdoman: 0 $\hspace{1.6cm}$ Subrange: 0} \end{cases}$$ Note that the trajectories of the first elements $(1, 2, -1$ and $-2)$ of the subdomains all converge to $0$ implying that no trajectory will diverge. For example, the trajectory of $6$ under $f$ is: $6\to-4 \to-3\to3\to4\to-2\to-1\to1\to2\to0.$

$\color{black}{\text{Example 2}}$

Here is a second example in which the first elements of the subdomains converge to the cycle $-1, 1$ and $0$ is excluded: $$g(x) = \begin{cases} \frac{3}{2}x, & \text{subdoman:2, 4, 6, 8,...Subrange: 3, 6, 9, 12...} \\ \frac{-3}{2}x+\frac{1}{2}, & \text{subdoman:1, 3, 5, 7,...Subrange:-1,-4,-7, -10...}\\ \frac{-1}{2}x+\frac{1}{2}, & \text{subdoman:-1, -3, -5,...Subrange:1, 2, 3, ...}\\ \frac{1}{2}x, & \text{subdoman:-2, -4, -6,...Subrange:-1, -2, -3,...} \end{cases}$$ In this example, the trajectory of 6 is $6\to9\to-13\to7\to-10\to-5\to3\to-4\to-2\to-1\to1.$

Answer 1

If I understand you correctly, you make the following

Conjecture: Let $f\colon \Bbb Z\to\Bbb Z$ be a function. Assume that for some index set $I$, there are $a_i,b_i\in\Bbb Z$, $c_i,d_i\in\Bbb Q$ such that

• $\Bbb Z=\bigcup_{i\in I}(a_i+b_i\Bbb N_0)$
• For all $i\in I$, $n\in\Bbb N_0$, we have $f(a_i+b_in)=c_i+d_in$
• For all $i\in I$, we have $a_ib_i\ge 0$ and $c_id_i\ge 0$
• For all $i\in I$, if $b_i\ne d_i$, then $\frac{a_i-c_i}{b_i-d_i}>0$
• For all $i\in I$, the set $\{\,f^{\circ k}(a_i)\mid k\in\Bbb N_0\,\}$ is finite

Then for all $x\in \Bbb Z$, the set $\{\,f^{\circ k}(x)\mid k\in\Bbb N_0\,\}$ is finite.

The first bullet point describes that we partition $\Bbb Z$ into arithmetic progressions (with singletons allowed); I do note specify that the parts be disjoint for I am sure that the conjecture with and without this condition are equivalent.

The second bullet point expresses the linearity on each of the arithmetic progressions. Note that $b_i=0$ leads to a singleton subdomain (for which necessarily $d_i=0$)

The third bullet point codifies that the subdomains shall be non-negative or non-positive, but not mixed; same for the subranges

The fourth bullet point encodes your requirement (4) (do you see why?)

In the last bullet point speaks about the trajectories of the first elements of subdomains. Note that I use finiteness instead of boundedness or eventual periodicity; all these are equivalent.

In the claim, I also use finiteness instead of boundedness or eventual periodicity.