Question: Let $f,g:[0,1]\to[0,1]$ be two continuous self-maps of the interval and let $f\times g:[0,1]^2\to[0,1]^2,$ $(x,y)\mapsto(f(x),g(y))$ be the product system. Then "Period Three Implies Chaos" applies to $f\times g$, while it's not clear if Sharkovskii's Theorem does ($\dagger$). Hence two natural questions arise:
- For which continuous $f,g$ does Sharkovskii's Theorem apply to $f\times g$?
- For arbitrary continuous $f,g$, how much can one refine "Period Three Implies Chaos" to Sharkovskii's Theorem for $f\times g$?
Here are some related definitions and theorems for context (see e.g. Devaney's A First Course in Chaotic Dynamical Systems, Ch.11 for further details):
Definition: We define the Sharkovskii's Ordering on $\mathbb{Z}_{\geq1}$ to be the total ordering
$$2^0 3\triangleleft 2^0 5\triangleleft\cdots\triangleleft 2^1 3\triangleleft 2^1 5\triangleleft\cdots \triangleleft 2^3 \triangleleft 2^2 \triangleleft 2^1 \triangleleft 2^0.$$
Theorem (Sharkovskii): Let $f:I\to I$ be a continuous self-map of a nonempty connected subset of $\mathbb{R}$. Then if $f$ has a periodic point of least period $k$ (a $k$-periodic point), then $\forall l:k\triangleleft l \implies$ $f$ has a $l$-periodic point.
Corollary ("Period Three Implies Chaos" (P3C)): Let $f:I\to I$ be a continuous self-map of a nonempty connected subset of $\mathbb{R}$. If $f$ has a $3$-periodic point, then it has a periodic point of any least period.
Theorem (Converse of Sharkovskii's Theorem): For any $l\in\mathbb{Z}_{\geq1}$ and any nonempty connected subset $I$ of $\mathbb{R}$, there is a continuous $f:I \to I$ which has an $l$-periodic point but no $k$-periodic points for any $k\triangleleft l$.
($\dagger$) Let us see why P3C applies to anonymous products. Suppose $(a,b)$ is a $3$-periodic point of $f\times g$. Then $f^3(a)=a,g^3(b)=b$. $a$ and $b$ cannot both be fixed points of their respective systems. Let $n\in\mathbb{Z}_{\geq1}$. If both $a$ and $b$ are $3$-periodic, applying P3C to $f$ and $g$, we see that there are $x\in X$ and $y\in Y$ such that $x$ and $y$ are $n$-periodic. Hence $(x,y)$ is an $n$-periodic point of $f\times g$. Else, wlog suppose $b$ is a fixed point of $g$. Applying P3C to $f$, we find some $n$-periodic point $x$. Then $(x,b)$ is an $n$-periodic point of $f\times g$.
Added 01/30/23: In an earlier version of this question I claimed that Sharkovskii does not apply to anonymous products, after Lutz Lehmann pointed out a mistake in the argument I had for it I was not able to come up with a working counterexample.