\subsection{Inverse branches and phase comparison}
\label{sec:comparison}

The statistical argument uses two descriptions of an era sequence.
Its digits count the actual cores. Its phase is measured at the actual
entrance. The digits determine an exact continued fraction, while the
phase satisfies a perturbed inverse recurrence. The following estimates
compare these objects with constants independent of the digits.

Write $\mathcal D=\N_{\geq1}^{\N_0}$ for the space of positive digit
sequences, with its product Borel structure, and let $\theta$ be the
left shift. For a finite word $\mathbf b=(b_0,\ldots,b_{r-1})$ put
$G_{\mathbf b}=g_{b_0}\circ\cdots\circ g_{b_{r-1}}$, with
$G_\varnothing$ the identity.

\begin{lemma}[Contraction of inverse branches]
\label{lem:contraction}
For $a,b\geq1$ and $x,y\geq0$,
\begin{align}
 |g_a(x)-g_a(y)|&\leq|x-y|,\notag\\
 |g_a(g_b(x))-g_a(g_b(y))|&\leq\tfrac14|x-y|.
 \label{eq:pair-contraction}
\end{align}
Consequently, for a positive word of length $r$,
\begin{equation}
 |G_{\mathbf b}(x)-G_{\mathbf b}(y)|
   \leq4^{-\lfloor r/2\rfloor}|x-y|.
 \label{eq:word-contraction}
\end{equation}
\end{lemma}

\begin{proof}
The difference for a single branch is
\[
 g_a(x)-g_a(y)=\frac{y-x}{(a+x)(a+y)}.
\]
For two branches,
\[
 g_a(g_b(x))=\frac{b+x}{a(b+x)+1},
\]
and their difference at $x,y$ is
\[
 \frac{x-y}{(a(b+x)+1)(a(b+y)+1)}.
\]
The first denominator is at least $1$, and each factor in the last
denominator is at least $2$. This proves~\eqref{eq:pair-contraction}.
Pairing successive branches proves~\eqref{eq:word-contraction}, with
one unpaired nonexpanding branch when $r$ is odd.
\end{proof}

\begin{proposition}[The code of a positive digit sequence]
\label{prop:code}
For $\mathbf a=(a_0,a_1,\ldots)\in\mathcal D$ the limit
\begin{equation}
 c(\mathbf a)=\lim_{r\to\infty}G_{(a_0,\ldots,a_{r-1})}(0)
 \label{eq:code}
\end{equation}
exists, lies in $(0,1)\setminus\mathbb Q$, and is Borel as a function
of $\mathbf a$. It satisfies
\begin{equation}
 c(\theta^m\mathbf a)=g_{a_m}(c(\theta^{m+1}\mathbf a)),\qquad
 T^m c(\mathbf a)=c(\theta^m\mathbf a),
 \label{eq:code-recurrence}
\end{equation}
where $T(x)=\{1/x\}$. If two sequences agree in their first $r$ digits,
their codes differ by at most $4^{-\lfloor r/2\rfloor}$.
\end{proposition}

\begin{proof}
Each branch maps $[0,1]$ into $[0,1]$. If $s\geq r$, the approximant
of length $s$ is obtained by applying $G_{(a_0,\ldots,a_{r-1})}$ to a
point of $[0,1]$. Its distance from the approximant of length $r$ is
at most $4^{-\lfloor r/2\rfloor}$ by Lemma~\ref{lem:contraction}.
The approximants are therefore Cauchy and their limit belongs to $[0,1]$.
Passing to the limit through the first branch gives
$c(\mathbf a)=g_{a_0}(c(\theta\mathbf a))$.
This first shows that every code is positive. Applying this positivity
to the shifted sequence shows that every code is strictly smaller than
one. Hence
\[
 1/c(\mathbf a)=a_0+c(\theta\mathbf a),
\]
so its integer part is $a_0$ and its fractional part is
$c(\theta\mathbf a)$. Iteration gives~\eqref{eq:code-recurrence}.

If $c(\mathbf a)$ were rational, each of its iterates under $T$ would
be rational and strictly between zero and one. For a reduced rational
$p/q\in(0,1)$, the reduced denominator of $T(p/q)=\{q/p\}$ is at most
$p<q$. This would give an infinite strictly decreasing sequence of
positive denominators, a contradiction.

Every finite approximant in~\eqref{eq:code} depends on finitely many
discrete coordinates and is Borel. Its pointwise limit is Borel. Finally,
each code with prefix $\mathbf b$ is $G_{\mathbf b}$ of its tail code;
\eqref{eq:word-contraction} gives the claimed bound.
\end{proof}

Fix $\mathbf a\in\mathcal D$ and a phase sequence $y_m\in[0,1]$.
Define its inverse recurrence error by
\begin{equation}
 e_m=|y_m-g_{a_m}(y_{m+1})|.
 \label{eq:inverse-error}
\end{equation}
The trajectory fixes the digit sequence $\mathbf a$ used in this comparison.

\begin{proposition}[Comparison with the prescribed code]
\label{prop:phase-comparison}
Let $z_m=c(\theta^m\mathbf a)$. For every $m\geq0$,
\begin{equation}
 |y_m-z_m|\leq
 \sum_{j=0}^\infty4^{-j}(e_{m+2j}+e_{m+2j+1}).
 \label{eq:phase-comparison}
\end{equation}
If $e_m\to0$, then $y_m-T^m c(\mathbf a)\to0$. If instead the
stronger condition $\sum_m e_m<\infty$ holds, then
\begin{equation}
 \sum_{m=0}^\infty|y_m-T^m c(\mathbf a)|
 \leq\frac83\sum_{m=0}^\infty e_m.
 \label{eq:summable-comparison}
\end{equation}
\end{proposition}

\begin{proof}
Use~\eqref{eq:code-recurrence}, insert the two intermediate inverse
branches, and apply Lemma~\ref{lem:contraction}. The triangle inequality
gives
\[
 |y_m-z_m|\leq e_m+e_{m+1}+\tfrac14|y_{m+2}-z_{m+2}|.
\]
Iterating this inequality $l$ times yields
\[
 |y_m-z_m|\leq
 \sum_{j=0}^{l-1}4^{-j}(e_{m+2j}+e_{m+2j+1})
 +4^{-l}|y_{m+2l}-z_{m+2l}|.
\]
Both phase sequences belong to $[0,1]$, so the last absolute value is
at most one. Also $e_m\leq1$, and the series is dominated by
$2\sum_j4^{-j}$. Letting $l\to\infty$ proves
\eqref{eq:phase-comparison}.

If $e_m\to0$, the right side is bounded by
$(8/3)\sup_{k\geq m}e_k$, which tends to zero.
For summable errors, nonnegativity permits interchange of the two sums.
For each $j$, both shifted sums of errors are at most $\sum_m e_m$.
Thus
\[
 \sum_m\sum_j4^{-j}(e_{m+2j}+e_{m+2j+1})
 \leq2\left(\sum_j4^{-j}\right)\sum_m e_m
 =\frac83\sum_m e_m.
\]
This proves~\eqref{eq:summable-comparison}.
\end{proof}

\begin{corollary}
\label{cor:continuous-transfer}
Under the hypotheses of Proposition~\ref{prop:phase-comparison}, suppose
$e_m\to0$. Then for every $f\in C([0,1])$,
\begin{equation}
 \frac1n\sum_{m=0}^{n-1}
 \bigl(f(y_m)-f(T^m c(\mathbf a))\bigr)\longrightarrow0.
 \label{eq:continuous-transfer}
\end{equation}
\end{corollary}

\begin{proof}
The function $f$ is uniformly continuous on $[0,1]$.
Proposition~\ref{prop:phase-comparison} therefore implies
$f(y_m)-f(T^m c(\mathbf a))\to0$. The Ces\`aro averages of this
sequence tend to zero.
\end{proof}

The role of the two comparisons is now explicit. Phase estimates transfer
continuous observations to the exact code. Frequencies of digit blocks
are observations of the code itself, since its digits are the prescribed
$a_m$. To deduce almost everywhere statements for physical initial data,
the next issue is the measure carried by this coding. In particular,
the Birkhoff typical set must have full measure for the codes produced
by the actual return branches.