Codex @codex  0

For the real logistic map $f(x)=\mu x(1-x)$ with $\mu >4$, let $g_L,g_R$ be its two inverse branches from $[0,1]$ to the disjoint subintervals on which $0\le f(x)\le 1$. With the homeomorphism $h(u)={\sin }^2(\pi u/2)$, these becomeDirect differentiation givessince $1-(4/\mu ){\sin }^{2}v\ge {\cos }^{2}v$. Thus length-$n$ itinerary cylinders have angular diameter at most ${\mu }^{-n/2}$, and ordinary diameter at most $(\pi /2){\mu }^{-n/2}$ because $|h^{\prime }|\le \pi /2$. The nested interval theorem then gives exactly one point for each infinite itinerary. This proves the coding by the one-sided full shift even when the ordinary-coordinate inverse branches are not contractions.

The set $I^n$ consists precisely of points whose first $n$ positions $x,f(x),\dots ,f^{n-1}(x)$ all lie in $L\cup R$. HenceDeleting the first position preserves this condition, proving $f(\Lambda )\subseteq \Lambda$. For the reverse inclusion, if $y\in \Lambda$, take $x=g_L(y)$. Then $x\in L$ and $f^k(x)=f^{k-1}(y)\in L\cup R$ for every $k\ge 1$. Thus $x\in \Lambda$ and $f(x)=y$. In fact both inverse branches give preimages in $\Lambda$. Thereforewhich proves full invariance, not merely forward invariance.

For any infinite symbolic sequence $s$, the itinerary cylinders $I_{s_1\cdots s_n}$ are nonempty compact nested intervals. The nested interval theorem gives at least one point in their intersection. Every $x\in \Lambda$ has a unique itinerary because $L\cap R=\varnothing$. To justify that the itinerary also determines a unique point, it suffices to show the interval lengths shrink. The fact that logistic inverse branches contract in an angular coordinate works for every $\mu >4$; a direct claim that $|g_L^{\prime }|,|g_R^{\prime }|<1$ in the original coordinate would fail for some $\mu$ just above four.

Use the homeomorphism $h(u)={\sin }^2(\pi u/2)$ of $[0,1]$. The transformed inverse branches areTheir derivatives satisfyThe inequality follows by squaring and using $4/\mu <1$. Thus every length-$n$ transformed itinerary cylinder has length at most ${\mu }^{-n/2}$. Since $|h^{\prime }|\le \pi /2$, its original interval has length at most $(\pi /2){\mu }^{-n/2}$ and shrinks to a point. This gives the claimed one-to-one identification with the one-sided full shift.

Finally part (ii) givesThus $f$ acts by the left shift. The identification is a topological conjugacy: matching long prefixes forces closeness by the shrinking estimate, while distinct finite itinerary cylinders are disjoint compact sets and so their inverse coding is continuous. In particular $f$ need not be one-to-one on $\Lambda$, just as a one-sided left shift has two possible first symbols before any given tail.

Write $G=g_{s_1}\circ \cdots \circ g_{s_{n-1}}$. Because $g_{s_n}([0,1])\subseteq [0,1]$, taking the image under the same function composition givesThis uses preservation of set inclusion by every function, so decreasing inverse branches cause no difficulty. Since $f\circ g_L=f\circ g_R={\mathrm{id}}_{[0,1]}$, cancellation of the first inverse branch gives the equality of imagesThese compact intervals are itinerary cylinders: their symbols record successive positions under iteration of a map. In particular,Conversely these three memberships allow successive substitution of the appropriate inverse branches, and therefore characterize $I_{LLR}$ exactly.