# Sharkovsky's Theorem and Triangular Functions

I'm trying to prove that Sharkovsky's Theorem

Let $$\vartriangleleft$$ denote the Sharkovsky ordering given (informally) by $$\underbrace{1\vartriangleleft 2 \vartriangleleft 4\vartriangleleft 8\vartriangleleft ...}_{\text{Powers of 2}} \vartriangleleft...\vartriangleleft\underbrace{...\vartriangleleft28\vartriangleleft20\vartriangleleft 12}_{\text{4x Odd numbers}} \vartriangleleft \underbrace{ ...\vartriangleleft14\vartriangleleft10\vartriangleleft 6}_{\text{2x Odd numbers}}\vartriangleleft\underbrace{ ...\vartriangleleft7\vartriangleleft5\vartriangleleft 3}_{\text{Odd numbers}},$$

and let $$I$$ be a compact non-degenerate interval with $$f:I\to I$$ a continuous function on $$I$$. Suppose $$m\vartriangleleft n.$$ Then if $$x$$ is a $$f$$-periodic point with primitive period $$n$$ (denoted $$p_f(x)=n$$), then there exists $$y\in I$$ such that $$p_f(y)=m$$.

also holds for triangular functions $$f:I^2\to I^2$$, functions $$f$$ such that the first coordinate is dependent only on the first argument, i.e. there exists continuous $$g$$ such that $$\pi_1(f(x,y))=g(x), \forall x\in I$$, for canonical projection $$\pi_1$$.

For fixed $$x\in I, k\in$$ N, I define $$F_{x,k}(y) = \pi_2(f^k(y)), \forall y\in I$$.

My first step is to show that, given $$x\in I$$ such that $$p_g(x)=k$$, there exists $$y\in I$$ such that $$p_f(x,y)=k$$. To do this I use the intermediate value theorem on $$h(y):= F_{x,k}(y)-y$$, as this will find a fixed point for $$F_{x,k}$$. Clearly we have that if (for $$I=[a,b]$$) either $$h(a)=0$$ or $$h(b)=0$$ we are done.

My First Problem: Clearly I need to show that either $$h(a)>0$$ and $$h(b)<0$$ or vice versa. I proceed by contradiction: suppose that $$h(y)$$ is non-zero for all $$y\in I$$. Then I need to show that if $$h(a),h(b)>0$$, we have a contradiction. I am unsure how to proceed.

My Second Problem: Given the first claim, and having shown that $$F^l_{x,k}(y)=F_{x,lk}(y)$$ and that necessarily $$p_f (x,y)=p_g(x)p_{F_{x,k}}(y)$$, it remains to conclude that Sharkovsky's theorem holds for such triangular $$f:I^2\to I^2$$. To do this I first suppose that $$m\vartriangleleft p_g(x)=k.$$ Then we have that there is $$\hat x\in I$$ such that $$p_g(\hat x)=m$$ and so the first claim finds us the point. (Also if $$m= p_g(x)$$ the result follows again by Claim 1 trivially)

The second case is where $$k=p_g(x)\vartriangleleft m$$. My suspicion is that I then need to consider $$k$$ in the form $$k=2^\alpha p$$ for odd $$p$$ and do some case analysis on $$p$$ and $$\alpha$$, likely using the fact that $$k$$ divides $$p_f(x,y)$$ to simplify the cases somewhat. However I have little doubt there will be a need to use Sharkovsky's theorem on some function $$I\to I$$, but I see not how to use either $$g$$ or $$F_{x,k}$$ to get the result from here.

Any help with either of these two arguments would be greatly appreciated.