\documentclass[11pt]{article} \usepackage[T1]{fontenc} \usepackage{amsmath,amssymb,amsthm} \usepackage[margin=1in]{geometry} \usepackage{microtype} \usepackage{array} \usepackage{needspace} \usepackage[hidelinks]{hyperref} \hypersetup{ pdftitle={Quadratic cylinder energy and integrability of the Syracuse law}, pdfauthor={Lucas Valbuena} } \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \theoremstyle{remark} \newtheorem*{remark}{Remark} \newcommand{\Zthree}{\mathbb Z_3} \newcommand{\E}{\mathbb E} \newcommand{\Prob}{\mathbb P} \newcommand{\one}{\mathbf 1} \newcommand{\norm}[2]{\lVert #1\rVert_{#2}} \DeclareMathOperator{\Law}{Law} \title{Quadratic cylinder energy and integrability\\of the Syracuse law} \author{Lucas Valbuena} \date{8 October 2026} \begin{document} \maketitle \begin{abstract} Let \(\mu\) be the stationary probability measure on \(\Zthree\) for the random affine maps \(x\mapsto(3x+1)/2^a\), where \(\Prob(a=j)=2^{-j}\) for \(j\geq1\). With respect to additive Haar probability \(\lambda\), let \(\rho_k\) be the density obtained by spreading each depth-\(k\) cylinder mass uniformly over that cylinder. We prove \[ \int_{\Zthree}\rho_k^2\,d\lambda \leq24k+8k^2\leq32k^2\qquad(k\geq1). \] The proof combines offset injectivity, an elementary count of distinct integer quotients, and normalized geometric second moments. Together with Tao's fine-scale mixing estimate, the bound gives a single density \(\rho=d\mu/d\lambda\) in every \(L^p(\lambda)\), \(1\leq p<2\). A support-overlap argument applied to the stationary equation shows that \(\rho\notin L^2(\lambda)\). These statements concern Haar integrability of the random affine law. \end{abstract} \section{The stationary law and the results} Write \(\lambda\) for additive Haar probability on \(\Zthree\), so \(\lambda(x+3^k\Zthree)=3^{-k}\). For \(a\geq1\), put \[ \Phi_a(x)=\frac{3x+1}{2^a}. \] If \(a_1,a_2,\ldots\) are independent with \(\Prob(a_i=j)=2^{-j}\), the series \begin{equation}\label{eq:random-series} X=\sum_{j=0}^{\infty}3^j2^{-(a_1+\cdots+a_{j+1})} \end{equation} converges in \(\Zthree\). Its law \(\mu\) is stationary for the maps \(\Phi_a\): separating the first exponent in the series gives \(X\overset{d}{=}\Phi_{a_1}(X')\), where \(X'\) has law \(\mu\) and is independent of \(a_1\). This is the Syracuse law described in \cite[Remark~1.13]{Tao}. Its cylinder probabilities are \[ p_k(v)=\mu(v+3^k\Zthree),\qquad v\in\mathbb Z/3^k\mathbb Z, \] and we set \begin{equation}\label{eq:rho-energy} \rho_k(x)=3^k p_k(x\bmod3^k),\qquad N_k=\int\rho_k^2\,d\lambda =3^k\sum_{v\bmod3^k}p_k(v)^2. \end{equation} In particular, \(\rho_0=1\) and \(N_0=1\). All norms and integrals below use the full additive Haar probability \(\lambda\). \Needspace{6\baselineskip} \begin{theorem}\label{thm:main} The Syracuse law has the following properties. \begin{enumerate} \item For every integer \(k\geq1\), \begin{equation}\label{eq:main-energy} N_k\leq24k+8k^2\leq32k^2. \end{equation} \item There is a nonnegative density \(\rho\), with \(\mu=\rho\lambda\) and \(\int\rho\,d\lambda=1\), such that \[ \rho\in L^p(\lambda),\qquad \sup_{k\geq0}\norm{\rho_k}{p}<\infty \quad\text{for every }1\leq p<2. \] The same density \(\rho\) works for all these exponents. \item This density does not belong to \(L^2(\lambda)\). \end{enumerate} \end{theorem} The energy proof is independent of the mixing estimate. The density and its subcritical integrability use Tao's Proposition~1.14~\cite{Tao}. Offset injectivity, which enters the energy argument, is the content of his Lemma~6.2; Section~6 of that paper also uses separation of offsets in collision estimates. We include the elementary decoding and moment arguments needed here. These estimates are with respect to Haar measure. Embedded ordinary integers form a countable Haar-null set, so the estimates do not supply an integer trace bound or a conclusion about individual Collatz orbits. \section{Words, numerators, and fixed-total fibers} For a word \(w=(a_1,\ldots,a_k)\) of positive integers, use chronological composition: \[ \Phi_w=\Phi_{a_k}\circ\cdots\circ\Phi_{a_1},\qquad A_i=a_1+\cdots+a_i,\quad A_0=0,\quad A=A_k. \] Direct expansion gives \begin{equation}\label{eq:word-formula} \Phi_w(x)=\frac{3^k x+B(w)}{2^A},\qquad B(w)=\sum_{i=0}^{k-1}3^{k-1-i}2^{A_i}. \end{equation} Define the positive real translation quantity and the residue \begin{equation}\label{eq:translation-residue} Y(w)=\frac{B(w)}{3^k},\qquad r_k(w)=2^{-A}B(w)\pmod{3^k}. \end{equation} Here the inverse of \(2^A\) is taken modulo \(3^k\) in the second expression. The image \(\Phi_w(\Zthree)\) is the cylinder \(r_k(w)+3^k\Zthree\). The series~\eqref{eq:random-series}, reduced modulo \(3^k\), is the endpoint of a length-\(k\) composition with the reverse ordering of the exponents. Reversal preserves their product distribution. Consequently, \begin{equation}\label{eq:fair-words} p_k(v)=\sum_{\substack{w\in\mathbb N_{>0}^k\\r_k(w)=v}}2^{-A(w)}. \end{equation} The sum is absolutely convergent since \(\sum_w2^{-A(w)}=(\sum_{a\geq1}2^{-a})^k=1\). \begin{lemma}[Fixed-total decoding]\label{lem:decoding} For fixed \(k\geq1\) and \(A\), the numerator \(B(w)\) determines a word \(w\) with \(A(w)=A\). For a fixed residue \(v\), the map \[ w\longmapsto q(w):=\left\lfloor\frac{B(w)}{3^k}\right\rfloor \] is therefore injective among words with \(A(w)=A\) and \(r_k(w)=v\). Moreover, \(q(w)\) is a nonnegative integer and \(q(w)\leq Y(w)\). \end{lemma} \begin{proof} For \(k=1\), the total fixes the only exponent. For \(k\geq2\), \eqref{eq:word-formula} gives \[ B(w)-3^{k-1}=2^{a_1}B(a_2,\ldots,a_k). \] The numerator on the right after removing \(2^{a_1}\) is odd: its first summand is odd and its remaining summands are even. Thus \(a_1=v_2(B(w)-3^{k-1})\). Division recovers the suffix numerator, and induction with total \(A-a_1\) recovers the suffix. This is the fixed-total numerator form of offset injectivity \cite[Lemma~6.2]{Tao}. On the fiber with total \(A\) and residue \(v\), every numerator has the same remainder modulo \(3^k\), namely \(2^A v\). Equal quotients therefore imply equal numerators and then equal words. The last assertion follows from the definition of the floor. \end{proof} \begin{lemma}[Quotient count]\label{lem:quotient-count} If \(q_1,\ldots,q_m\) are distinct nonnegative integers and \(Y_j\geq q_j\), then \[ m^2\leq2\sum_{j=1}^m(1+Y_j). \] In particular, if \(m_A(v)\) counts any chosen subset of the fixed-\((k,A,v)\) fiber, then \begin{equation}\label{eq:fiber-count} m_A(v)^2\leq 2\sum_{\substack{w\text{ in the chosen fiber}}}(1+Y(w)). \end{equation} \end{lemma} \begin{proof} Order the distinct integers increasingly. The \(j\)-th is at least \(j-1\), so \[ \sum_{j=1}^m(1+Y_j)\geq\sum_{j=1}^m j =\frac{m(m+1)}2\geq\frac{m^2}2. \] Apply Lemma~\ref{lem:decoding} to obtain~\eqref{eq:fiber-count}. \end{proof} \section{The energy estimate} Fix \(k\geq1\). Introduce the probability distribution on words \begin{equation}\label{eq:biased-law} \pi_k(w)=3^k4^{-A(w)} =\prod_{j=1}^k(3\cdot4^{-a_j}). \end{equation} It is normalized because \(\sum_{a\geq1}3\cdot4^{-a}=1\). This distribution will be used only to calculate a moment; the cylinder probabilities in~\eqref{eq:fair-words} retain their fair weights \(2^{-A(w)}\). \begin{lemma}[Energy reduction]\label{lem:energy-reduction} One has \begin{equation}\label{eq:energy-moment} N_k\leq\frac4k\E_{\pi_k}\bigl[A^2(1+Y)\bigr]. \end{equation} \end{lemma} \begin{proof} First restrict to words with \(A\leq M\), where \(M\geq k\). For each residue \(v\), denote the corresponding partial sum in \eqref{eq:fair-words} by \(p_{k,M}(v)\). Every word has \(A\geq k\). The elementary estimate \[ \frac1{A^2}\leq2\left(\frac1A-\frac1{A+1}\right)\qquad(A\geq1) \] implies \(\sum_{A=k}^{M}A^{-2}\leq2/k\). Weighted Cauchy--Schwarz and~\eqref{eq:fiber-count} now give \begin{align*} p_{k,M}(v)^2 &=\left(\sum_{A=k}^{M}2^{-A}m_A(v)\right)^2\\ &\leq\frac2k\sum_{A=k}^{M}A^2 4^{-A}m_A(v)^2\\ &\leq\frac4k\sum_{\substack{w:\,A(w)\leq M\\r_k(w)=v}} A(w)^2 4^{-A(w)}(1+Y(w)). \end{align*} Sum over \(v\) and multiply by \(3^k\). Each word has exactly one residue, and hence \[ 3^k\sum_v p_{k,M}(v)^2 \leq\frac4k\sum_{A(w)\leq M}\pi_k(w)A(w)^2(1+Y(w)). \] Let \(M\to\infty\). The left side tends to \(N_k\), and the nonnegative sums on the right increase to the displayed expectation. The next lemma shows that this expectation is finite. \end{proof} \begin{lemma}[Normalized second moments]\label{lem:moment} For \(k\geq1\), \begin{equation}\label{eq:moment-bound} \E_{\pi_k}\bigl[A^2(1+Y)\bigr]\leq6k^2+2k^3. \end{equation} \end{lemma} \begin{proof} The identity \(\sum_{a\geq1}a^2z^a=z(1+z)/(1-z)^3\), for \(00\) there is a finite constant \(C_A\) such that \begin{equation}\label{eq:mixing} \norm{\rho_n-\rho_m}{1}\leq C_A m^{-A} \qquad(n\geq m\geq1). \end{equation} To check the normalization, consistency of the cylinder masses gives \[ \norm{\rho_n-\rho_m}{1} =\sum_{v\bmod3^n} \left|p_n(v)-3^{m-n}p_m(v\bmod3^m)\right|. \] This is exactly the oscillation in the cited proposition. It follows that \(\rho_k\) is Cauchy in \(L^1(\lambda)\). Let \(\rho\) be a nonnegative representative of its limit. The limit has integral one. For every cylinder \(D\) of depth \(m\), consistency gives \(\int_D\rho_k\,d\lambda=\mu(D)\) when \(k\geq m\). Passing to the \(L^1\) limit proves the same identity for \(\rho\). Cylinders generate the Borel sigma algebra, so the finite measures \(\rho\lambda\) and \(\mu\) are equal. Fix \(1\frac{4p-2}{2-p}. \] Then the exponent in~\eqref{eq:interpolation} is at most \(-2\). With \(D_p=C_A^{(2-p)/p}320^{(p-1)/p}\), finite telescoping and Minkowski give, for every \(n\geq1\), \begin{equation}\label{eq:uniform-lp} \norm{\rho_n}{p} \leq\norm{\rho_1}{p}+\sum_{k=1}^{n-1}\norm{\delta_k}{p} \leq\norm{\rho_1}{p}+2D_p. \end{equation} Here \(\rho_1\) is a bounded cylinder function, and \(\sum_{k\geq1}k^{-2}\leq2\) follows from the same telescoping estimate used in the energy proof. The already established \(L^1\) convergence implies convergence in measure to \(\rho\). The Fatou property for convergence in measure therefore gives \[ \norm{\rho}{p}\leq\liminf_{n\to\infty}\norm{\rho_n}{p} \leq\norm{\rho_1}{p}+2D_p. \] Indeed, one may choose a subsequence realizing the liminf, pass to a further subsequence converging almost everywhere, and apply Fatou's lemma to the \(p\)-th powers. This step uses the same \(L^1\) limit for each \(p\). The case \(p=1\) was established above, and \(\rho_0=1\) supplies the remaining finite level. This proves part~(2) of Theorem~\ref{thm:main}. \section{The stationary equation and the critical exponent} Let \[ J(y)=\frac{3y+1}{2},\qquad D=J(\Zthree)=\{x\in\Zthree:x\equiv2\pmod3\}. \] Multiplication by \(2\) preserves \(\lambda\), while \(J_*\lambda=3\lambda|_D\). For a measurable function \(h\), write its extension along this branch as \[ (Lh)(x)= \begin{cases} h((2x-1)/3),&x\in D,\\ 0,&x\notin D. \end{cases} \] Thus, whenever the integral is defined, \begin{equation}\label{eq:lift-integral} \int Lh\,d\lambda=\frac13\int h\,d\lambda, \end{equation} and, for integrable \(h,g\), \begin{equation}\label{eq:lift-l1} \norm{Lh-Lg}{1}=\frac13\norm{h-g}{1}. \end{equation} The geometric weights yield the exact finite-level recursion \begin{equation}\label{eq:finite-recursion} \rho_{k+1}(x) =\frac12\rho_{k+1}(2x)+\frac32(L\rho_k)(x) \qquad(k\geq0). \end{equation} To see this, separate \(a=1\) from \(a\geq2\) in the outermost exponent of the random series. The first event has probability \(1/2\) and applies \(J\) to the depth-\(k\) tail. Conditional on \(a\geq2\), the exponent \(a-1\) has the original geometric law, and the output is half the corresponding depth-\((k+1)\) variable. The two changes of variables have Haar density factors \(3\) and \(1\), respectively, which gives~\eqref{eq:finite-recursion}. Passing to the \(L^1\) limit in~\eqref{eq:finite-recursion}, using the invariance under doubling and~\eqref{eq:lift-l1}, proves \begin{equation}\label{eq:stationary-density} \rho(x)=\frac12\rho(2x)+\frac32(L\rho)(x) \quad\text{for almost every }x. \end{equation} For an explicit estimate, the \(L^1\) norm of the difference between the two sides is at most \[ \frac32\norm{\rho_{k+1}-\rho}{1} +\frac12\norm{\rho_k-\rho}{1}, \] which tends to zero. We use the following finite-measure observation. \begin{lemma}[Positive support overlap]\label{lem:support} Let \(f,a,b\) be nonnegative measurable functions on a finite measure space \((\Omega,\nu)\). Suppose \(f=a+b\) almost everywhere, \[ \nu\{a>0\}=\nu\{f>0\},\qquad 0<\int b\,d\nu<\infty, \] and \(ab\) is integrable. Then \(\int ab\,d\nu>0\). \end{lemma} \begin{proof} Since \(a\leq f\) almost everywhere, \(\{a>0\}\) is contained in \(\{f>0\}\) up to a null set. Equality of their finite measures makes these sets equal up to a null set. If \(\int ab\,d\nu=0\), nonnegativity and integrability imply \(ab=0\) almost everywhere. But wherever \(b>0\), the equation \(f=a+b\) gives \(f>0\), hence \(a>0\) outside a null set. This forces \(b=0\) almost everywhere, contradicting its positive integral. \end{proof} Suppose now that \(\rho\in L^2(\lambda)\), and put \[ a(x)=\frac12\rho(2x),\qquad b(x)=\frac32(L\rho)(x). \] Both functions belong to \(L^2\), their product is integrable, and~\eqref{eq:stationary-density} gives \(\rho=a+b\) almost everywhere. Doubling preserves Haar measure, so \[ \lambda\{a>0\}=\lambda\{\rho>0\}. \] Equation~\eqref{eq:lift-integral} and \(\int\rho\,d\lambda=1\) give \(\int b\,d\lambda=1/2\). Lemma~\ref{lem:support} therefore implies \(\int ab\,d\lambda>0\). On the other hand, the same changes of variables yield \[ \int a^2\,d\lambda=\frac14\int\rho^2\,d\lambda, \qquad \int b^2\,d\lambda=\frac34\int\rho^2\,d\lambda. \] Squaring \(\rho=a+b\) and integrating would then give \[ \int\rho^2\,d\lambda =\int\rho^2\,d\lambda+2\int ab\,d\lambda, \] a contradiction. This proves part~(3) of Theorem~\ref{thm:main}. Since \(\lambda\) is a probability measure, \(L^p(\lambda)\subseteq L^2(\lambda)\) for every real \(p\geq2\). Thus, among real exponents \(p\geq1\), the density belongs to \(L^p(\lambda)\) exactly when \(p<2\). The support argument uses no ergodicity statement and no claim that \(\rho\) is positive almost everywhere on a prescribed ambient set. \clearpage \section*{Formalization record} The following table records the corresponding Lean endpoints. The energy proof uses the native reverse ordering of the word exponents; an explicit reversal identity connects its numerator to \eqref{eq:word-formula}. Its moment proof uses the same geometric second moments through a direct recursion on native words. \begin{center} \small \begin{tabular}{@{}>{\raggedright\arraybackslash}p{0.23\textwidth} >{\raggedright\arraybackslash}p{0.52\textwidth} >{\raggedright\arraybackslash}p{0.19\textwidth}@{}} \hline Statement & Lean endpoint & Recorded status\\ \hline Native quadratic energy & \nolinkurl{FairEnergy.residue_energy_le_linear_quadratic} & Strict compile and kernel check passed\\[5pt] Haar quadratic energy & \nolinkurl{canonicalPadicRho_sq_integral_le_thirtytwo_quadratic} & Strict compile and kernel check passed\\[5pt] One subcritical density & \nolinkurl{canonicalSyracuseMeasure_has_subcritical_density} & Strict compile and kernel check passed\\[5pt] Uniform finite-level \(L^p\), \(1