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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.

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