Explicit persistence under the stated forward and conjugated-inverse bounds: coded chaos, hyperbolic fixed points and transverse homoclinic points

Constants: by [F1], $d\le 2\mathrm{/}13$, $2d\le 4\mathrm{/}13$, and $d\mathrm{/}(1-d\theta )\le 1\mathrm{/}6$; also $R\ge 13$, $l\ge 7$, and $R-l=\sqrt{a}+3-(\sqrt{a}-3)=6>0$. Setting $\nu :=2d\le 4\mathrm{/}13<1$.

Coded sequence: for $s\in X_{\pm }$ define $T_s$ on $B_R$ by $(T_{s}q{)}_n:=M_{s_n}(q_{n-1},q_{n+1})$. The values lie in $s_n[l,R]\subseteq [-R,R]$ by [F1], so $T_s$ maps $B_R$ into itself; and $\mathrm{\parallel }{T}_{s}q-T_s{q}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2d\mathrm{\parallel }q-q^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}=\nu \mathrm{\parallel }q-q^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$ by [F1]. By [F4] and $\nu <1$, $T_s$ has exactly one fixed point $q(s)\in B_R$; and $q_n(s)\in s_n[l,R]$, so $\mathrm{sign}{q}_n(s)=s_n$. The margins of [F1] give $R-\mathrm{\mid }{q}_n(s)\mathrm{\mid }\ge \frac{3}{4}$ and $\mathrm{\mid }{q}_n(s)\mathrm{\mid }-l\ge \frac{3}{4}$.

Orbits: with $v_n:=V(q_n(s),q_{n+1}(s))$ of [F2], clause 4 of [F1] states that $P_{u,w}(v)=0$ is equivalent to $B(u,V(u,v))=V(v,w)$. With $(u,v,w)=(q_{n-1}(s),q_n(s),q_{n+1}(s))$, clause 1 gives $P_{u,w}(v)=0$, hence $B(q_{n-1}(s),v_{n-1})=v_n$; and the defining identity of $V$ gives $A(q_{n-1}(s),v_{n-1})=q_n(s)$. Hence $G(q_{n-1}(s),v_{n-1})=(q_n(s),v_n)$, i.e. $G(z_{n-1}(s))=z_n(s)$ for all $n$, and clause 1 is proved.

Locality and continuity: iterating $T_s$ from $0$, one proves by induction on $m$ that $(T_s^{m}0{)}_n$ depends only on the symbols $s_j$ with $\mathrm{\mid }j-n\mathrm{\mid }\le m-1$, exactly as in the unperturbed case, and that $\mathrm{\parallel }{T}_s^{m}0-q(s){\mathrm{\parallel }}_{\mathrm{\infty }}\le {\nu }^m\mathrm{\parallel }{T}_{s}0{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}(1-\nu )\le 2R{\nu }^m$ by [F4] and clause 3 of A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point; hence for $s,t$ agreeing on $\mathrm{\mid }j\mathrm{\mid }\le m$ one has $\mathrm{\mid }{q}_n(s)-q_n(t)\mathrm{\mid }\le 4R{\nu }^m$ for $\mathrm{\mid }n\mathrm{\mid }\le 1$. By [F2], $\mathrm{\mid }{\mathrm{\partial }}_{q}V\mathrm{\mid }\le \epsilon \mathrm{/}(1-\epsilon )\le \frac{1}{5}$ and $\mathrm{\mid }{\mathrm{\partial }}_{q^{\mathrm{\prime }}}V\mathrm{\mid }\le 1+\epsilon \mathrm{/}(1-\epsilon )\le \frac{6}{5}$, so the mean value inequality gives $\mathrm{\mid }V(q_0(s),q_1(s))-V(q_0(t),q_1(t))\mathrm{\mid }\le \frac{1}{5}\cdot 4R{\nu }^m+\frac{6}{5}\cdot 4R{\nu }^m=\frac{7}{5}\cdot 4R{\nu }^m$; with $C_h:=\frac{12}{5}=1+\frac{1}{5}+\frac{6}{5}$ one therefore has $\mathrm{\parallel }{h}_G(s)-h_G(t){\mathrm{\parallel }}_2\le \mathrm{\Delta }{q}_0+\mathrm{\Delta }V\le 4R{C}_h{\nu }^m$ whenever $s,t$ agree on $\mathrm{\mid }j\mathrm{\mid }\le m$. Given ${\epsilon }^{\mathrm{\prime }}>0$, choose $m$ with $4R{C}_h{\nu }^m<{\epsilon }^{\mathrm{\prime }}$ and $\delta :=2^{-m}$; if $D_{\pm }(s,t)<2^{-m}$ then ([F5], cylinder property) $s,t$ agree for $\mathrm{\mid }j\mathrm{\mid }\le m$, so $\mathrm{\parallel }{h}_G(s)-h_G(t){\mathrm{\parallel }}_2<{\epsilon }^{\mathrm{\prime }}$. This proves continuity of $h_G$.

Constant codes: for each sign $u$, the map $\rho \mapsto M_u(\rho ,\rho )$ maps the closed interval $u[l,R]$ into itself by [F1] and is a contraction with constant $2d\le 4\mathrm{/}13<1$; hence it has exactly one fixed point ${\rho }_u\in u[l,R]$, and the constant sequence $q_n:={\rho }_u$ is a fixed point of $T_{\bar{u}}$, so $q(\bar{u})$ is constant and $z_{\bar{u}}=h_G(\bar{u})=({\rho }_u,V({\rho }_u,{\rho }_u))$ is a fixed point of $G$.

Injectivity: if $h_G(s)=h_G(t)$ then $z_0(s)=z_0(t)$; applying the bijection $G$ forward and $G^{-1}$ (which exists because $G$ is a diffeomorphism onto its image, and $z_0(s)$ lies in that image as $G(z_{-1}(s))$) backward gives $z_n(s)=z_n(t)$ for all $n\in \mathbb{Z}$ by induction; hence $q_n(s)=q_n(t)$ for all $n$, and $s_n=\mathrm{sign}{q}_n(s)=\mathrm{sign}{q}_n(t)=t_n$ by [F1], so $s=t$.

Conjugacy: define ${\tilde{q}}_n:=q_{n+1}(s)$. Then $\tilde{q}$ is a fixed point of $T_{S_{\pm }s}$, because $(T_{S_{\pm }s}\tilde{q}{)}_n=M_{s_{n+1}}(q_n(s),q_{n+2}(s))=q_{n+1}(s)={\tilde{q}}_n$ by [F1]; by uniqueness $\tilde{q}=q(S_{\pm }s)$, i.e. $q_n(S_{\pm }s)=q_{n+1}(s)$. Hence $G(h_G(s))=G(q_0(s),v_0(s))=(q_1(s),v_1(s))=(q_0(S_{\pm }s),V(q_0(S_{\pm }s),q_1(S_{\pm }s)))=h_G(S_{\pm }s)$: the conjugacy identity, and $G({\mathrm{\Lambda }}_G)={\mathrm{\Lambda }}_G$ follows as in the unperturbed case because $S_{\pm }$ is bijective.

Hyperbolicity: put $M:=DG(z_{\bar{u}})$, $E:=M-DF(z_{\bar{u}})=\left(\begin{array}{cc}{a}_{11}& a_{12}\\ b_1& b_2\end{array}\right)$ and $v_u:=V({\rho }_u,{\rho }_u)$. At $z_{\bar{u}}$ the six closeness hypotheses give $\mathrm{\mid }{a}_{11}\mathrm{\mid },\mathrm{\mid }{a}_{12}\mathrm{\mid },\mathrm{\mid }{b}_1\mathrm{\mid }\le \epsilon$ and $\mathrm{\mid }{b}_2\mathrm{\mid }\le \epsilon$, while [F2] gives $\mathrm{\mid }{v}_u-{\rho }_u\mathrm{\mid }\le \epsilon$ and $\mathrm{\mid }{v}_u\mathrm{\mid }\le R+1$. Since $DF(x,y)=\left(\begin{array}{cc}0& 1\\ -1& 2y\end{array}\right)$, expanding trace and determinant of $M=DF(z_{\bar{u}})+E$ gives $\mathrm{tr}M=2v_u+a_{11}+b_2,\det M=1+a_{11}(2v_u+b_2)+a_{12}-b_1-a_{12}{b}_1,$ so $\mathrm{\mid }\mathrm{tr}M-2{\rho }_u\mathrm{\mid }\le 4\epsilon$ and $\mathrm{\mid }\det M-1\mathrm{\mid }\le (2R+4)\epsilon +2{\epsilon }^2\le (2l+16)\epsilon +2{\epsilon }^2$. Because $\epsilon \le \frac{1}{6}$ and $l\ge 7$, $\mathrm{\mid }\mathrm{tr}M\mathrm{\mid }\ge 2l-4\epsilon \ge 2l-\frac{2}{3},\mathrm{\mid }\det M\mathrm{\mid }\le 1+\frac{2l+16}{6}+\frac{2}{36}=\frac{l}{3}+\frac{67}{18},$ and the discriminant is bounded below by ${\mathrm{tr}}^{2}M-4\det M\ge (2l-\frac{2}{3}{)}^2-4(1+\frac{2l+16}{6}+\frac{2}{36})=4l^2-4l-\frac{130}{9}>0,$ since $4l^2-4l=168$ at $l=7$ and $l\mapsto 4l^2-4l$ is increasing for $l\ge 1$. Hence the eigenvalues are real. Let ${\lambda }_{+}$ be the one of larger modulus; then $\mathrm{\mid }{\lambda }_{+}\mathrm{\mid }\ge \frac{1}{2}\mathrm{\mid }\mathrm{tr}M\mathrm{\mid }\ge l-\frac{1}{3}>1$, and $\mathrm{\mid }{\lambda }_{-}\mathrm{\mid }=\frac{\mathrm{\mid }\det M\mathrm{\mid }}{\mathrm{\mid }{\lambda }_{+}\mathrm{\mid }}\le \frac{2\mathrm{\mid }\det M\mathrm{\mid }}{\mathrm{\mid }\mathrm{tr}M\mathrm{\mid }}\le \frac{2(\frac{l}{3}+\frac{67}{18})}{2l-\frac{2}{3}}<1,$ the last inequality being equivalent to $\frac{73}{18}<\frac{2l}{3}$, which holds because $l\ge 7$ and $\frac{14}{3}>\frac{73}{18}$. Thus $\mathrm{\mid }{\lambda }_{+}\mathrm{\mid }>1>\mathrm{\mid }{\lambda }_{-}\mathrm{\mid }\ge 0$: $z_{\bar{u}}$ is hyperbolic.

Tail convergence: suppose $s_n=u$ for all $\mathrm{\mid }n\mathrm{\mid }\ge N$, and let $m\ge 0$. For $n\ge N+m+1$ every $j$ with $\mathrm{\mid }j-n\mathrm{\mid }\le m-1$ satisfies $j>N$, so $(T_s^{m}0{)}_n=(T_{\bar{u}}^{m}0{)}_n$; for $n\le -N-(m+1)$ every such $j$ satisfies $j<-N$, so again $(T_s^{m}0{)}_n=(T_{\bar{u}}^{m}0{)}_n$. Hence, by the uniform error estimate of step 3.2 and $q(\bar{u})$ constant by step 3.3, $\mathrm{\mid }{q}_n(s)-{\rho }_u\mathrm{\mid }\le 4R{\nu }^m$ whenever $\mathrm{\mid }n\mathrm{\mid }\ge N+m+1$. Letting $m\to \mathrm{\infty }$ gives $q_n(s)\to {\rho }_u$ as $n\to \pm \mathrm{\infty }$; consequently $z_n(s)\to ({\rho }_u,V({\rho }_u,{\rho }_u))=z_{\bar{u}}$ in both directions, by continuity of $V$ ([F2]).

Backward orbit under $H$: put $w:=\mathrm{\Sigma }(z_{\ast })$ and define $w_n:=\mathrm{\Sigma }(z_{-n}(s))$ for $n\in \mathbb{Z}$. Since $q_j(s)\in [-R,R]$ and $\mathrm{\mid }{v}_j(s)-q_{j+1}(s)\mathrm{\mid }\le \epsilon \le \frac{1}{6}$ by [F1] and [F2], every $z_j(s)$ lies in $\mathcal{Q}$, so every $w_n$ lies in the domain where $H=\mathrm{\Sigma }\circ G^{-1}\circ \mathrm{\Sigma }$ is defined. The orbit identity of step 3.1 gives $G^{-1}(z_{-n}(s))=z_{-n-1}(s)$; hence $H(w_n)=w_{n+1}$, so $(w_n{)}_{n\in \mathbb{Z}}$ is the corresponding full $H$ orbit. Its coordinates are $(w_n{)}_1=v_{-n}(s)$ and $(w_n{)}_2=q_{-n}(s)$, and for every $n$, $\mathrm{\mid }(w_n{)}_1\mathrm{\mid }\in [l+\frac{7}{12},R-\frac{7}{12}],$ because $\mathrm{\mid }{q}_{j+1}(s)\mathrm{\mid }\in [l+\frac{3}{4},R-\frac{3}{4}]$ by [F1] and $\mathrm{\mid }{v}_j(s)-q_{j+1}(s)\mathrm{\mid }\le \frac{1}{6}$ by [F2]; the sign of $(w_n{)}_1$ may depend on $n$. Set $x_n:=(w_n{)}_1$. By definition of the inverse branch $V^H$ of the first component $H_1$ of $H$, the orbit relations $H_1(w_{n-1})=(w_n{)}_1$, $H_2(w_{n-1})=(w_n{)}_2$ say $V^H(x_{n-1},x_n)=(w_{n-1}{)}_2$ and $H_2(x_{n-1},V^H(x_{n-1},x_n))=V^H(x_n,x_{n+1})$; by clause 4 of [F1] applied to $H$, this gives $P_{x_{n-1},x_{n+1}}^H(x_n)=0$. Since $x_n\in \mathrm{sign}(x_n)[l,R]$, clause 1 of [F1] applied to $H$ yields $x_n=M_{\mathrm{sign}{x}_n}^H(x_{n-1},x_{n+1})(n\in \mathbb{Z})\mathrm{.}$

Topology: let $C\subseteq X_{\pm }$ be closed. A sequence in $C$ has a subsequence converging in the sequentially compact space $X_{\pm }$ ([F5] and the isometry $\phi$ of [F6]), and its limit lies in $C$ because closed sets are sequentially closed by the choice-free direction of [F6]; hence $C$ is sequentially compact. The continuous image $h_G(C)$ is sequentially compact: given a sequence $(h_G(x_k))$ with $x_k\in C$, pick a subsequence $x_{k_j}\to x\in C$ and use continuity of $h_G$; hence $h_G(C)$ is sequentially closed (any convergent sequence in it has a subsequence converging to a point of $h_G(C)$, and limits are unique by [F6]), and by the ${\mathrm{A}\mathrm{C}}_{\omega }$ direction of [F6] it is closed in ${\mathbb{R}}^2$. Thus $h_G$ is a closed map, so $h_G^{-1}$ is continuous. Under the Axioms of Countable and Dependent Choice, $X_{\pm }$ is compact by [F5] and $h_G(X_{\pm })={\mathrm{\Lambda }}_G$ is a continuous image of it, hence compact.

Stable curve for $G$: let $s$ be nonconstant with $s_n=u$ for $\mathrm{\mid }n\mathrm{\mid }\ge N$, and set $y_n:=q_{n+1}(s)$ for $n\ge 0$, $b_{\ast }:=q_0(s)$. Then $y_n=M_{s_{n+1}}(y_{n-1},y_{n+1})$ for $n\ge 0$ with $y_{-1}:=b_{\ast }$, and $\mathrm{\mid }{y}_n\mathrm{\mid }\in [l+\frac{3}{4},R-\frac{3}{4}]$ by [F1]. Apply Weighted half-tail contractions for $q_n=M_{s_n}(q_{n-1},q_{n+1})$ give a unique $C^1$ curve with tangent slope bounded by $c\mathrm{/}(1-q)$ with $\theta =1\mathrm{/}2$, $c=d\le 2\mathrm{/}13$, $q=d(\theta +{\theta }^{-1})\le 5\mathrm{/}13<1$, $J=[-R,R]$, $\bar{\rho }=\frac{1}{2}$, ${\delta }_{\ast }=\frac{1}{4}$, and update $M_{s_{n+1}}$. The half-orbit margin follows from the root margins, $[b_{\ast }-{\delta }_{\ast },b_{\ast }+{\delta }_{\ast }]\subseteq J$ follows from $\mathrm{\mid }{q}_0(s)\mathrm{\mid }\le R-\frac{3}{4}$, and ${\delta }_{\ast }<(1-q)\bar{\rho }\mathrm{/}d$ because $d\le 2\mathrm{/}13$. By [F3] there is a $C^1$ family $u(b)$ with $\mathrm{\mid }{d}_0\mathrm{\mid }\le d\mathrm{/}(1-q)\le 1\mathrm{/}4$, and the phase curve $Z(b):=(b,V(b,y_0+u_0(b)))$ through $Z(b_{\ast })=(q_0(s),V(q_0(s),q_1(s)))=z_{\ast }$ has tangent $(1,T)$ with $T={\mathrm{\partial }}_{q}V(b_{\ast },y_0)+{\mathrm{\partial }}_{q^{\mathrm{\prime }}}V(b_{\ast },y_0)d_0,\mathrm{\mid }T\mathrm{\mid }\le \frac{\epsilon }{1-\epsilon }+\frac{1}{1-\epsilon }\cdot \frac{1}{4}\le \frac{1}{5}+\frac{6}{5}\cdot \frac{1}{4}=\frac{1}{2}\le \frac{3}{4}\mathrm{.}$ Thus the tangent lies in $C_{3\mathrm{/}4}^s$. The recurrence coordinates remain in $[-R,R]$ by the weighted-ball construction; their phase points $(x_n,V(x_n,x_{n+1}))$ lie in $\mathcal{Q}\subseteq U$ because $\mathrm{\mid }V(x_n,x_{n+1})\mathrm{\mid }\le R+\epsilon <R+1$. Step 3.1 therefore gives every forward iterate without leaving the domain of $G$. Every point $Z(b)$ has forward orbit tracking that of $z_{\ast }$, which converges to $z_{\bar{u}}$ by step 4.4; hence these points lie in $W_G^s(z_{\bar{u}})$ near $b_{\ast }$.

Sensitivity: let $s\in X_{\pm }$ and let $r>0$. Choose $m\ge 0$ such that $4R{C}_h{\nu }^m<r$, using $0<\nu <1$ from step 1.1, and define $t\in X_{\pm }$ by $t_j=s_j$ for $j\mathrm{\ne }m+1$ and $t_{m+1}=-s_{m+1}$. Since $s,t$ agree on $\mathrm{\mid }j\mathrm{\mid }\le m$, the estimate in step 3.2 gives $\mathrm{\parallel }{h}_G(s)-h_G(t){\mathrm{\parallel }}_2<r$. At coordinate $0$, the codes $S_{\pm }^{m+1}s$ and $S_{\pm }^{m+1}t$ have opposite signs; by [F1] their coded first coordinates have opposite signs and absolute values at least $l+\frac{3}{4}$. Therefore $\parallel h_G(S_{\pm }^{m+1}s)-h_G(S_{\pm }^{m+1}t){\parallel }_2\ge 2(l+\frac{3}{4})>2l\mathrm{.}$ Iterating the conjugacy of step 4.2 identifies these two points with $G^{m+1}(h_G(s))$ and $G^{m+1}(h_G(t))$. Thus for every point of ${\mathrm{\Lambda }}_G$ and every radius $r>0$ there is a point within distance $r$ whose forward iterates are separated by at least $2l$, proving sensitivity with constant $2l$ under the definition in [F6].

Mixing and dense periodic points: $G$ is $C^2$ on an open set containing ${\mathrm{\Lambda }}_G$, so its restriction to ${\mathrm{\Lambda }}_G$ is continuous. Under ${\mathrm{A}\mathrm{C}}_{\omega }$, $h_G$ is a homeomorphism by step 5.1. If $U,V\subseteq {\mathrm{\Lambda }}_G$ are nonempty and relatively open, then $h_G^{-1}(U)$ and $h_G^{-1}(V)$ are nonempty open subsets of $X_{\pm }$. The shift $S_{\pm }$ is topologically mixing by [F5], so there is $N$ such that $S_{\pm }^n(h_G^{-1}(U))\cap h_G^{-1}(V)\mathrm{\ne }\mathrm{\varnothing }$ for all $n\ge N$. By the conjugacy in step 4.2, this gives $G^n(U)\cap V\mathrm{\ne }\mathrm{\varnothing }$ for all $n\ge N$, proving mixing. The same conjugacy and injectivity show that $h_G$ maps periodic shift codes exactly to periodic points of $G{\mathrm{\mid }}_{{\mathrm{\Lambda }}_G}$; since periodic shift codes are dense by [F5] and $h_G$ is a homeomorphism, these periodic points are dense in ${\mathrm{\Lambda }}_G$.

Unstable curve for $G$: shift the $H$-reference to $y_n^{\mathrm{\prime }}:=x_{n+1}$ $(n\ge 0)$ with $y_{-1}^{\mathrm{\prime }}:=x_0=v_0(s)$, and apply Weighted half-tail contractions for $q_n=M_{s_n}(q_{n-1},q_{n+1})$ give a unique $C^1$ curve with tangent slope bounded by $c\mathrm{/}(1-q)$ to $H$ with $\theta =1\mathrm{/}2$, $c=d$, $q\le 5\mathrm{/}13$, $J=[-R,R]$, $\bar{\rho }=\frac{1}{2}$, and ${\delta }_{\ast }=\frac{1}{4}$. The reference recurrence holds by step 4.5. Its margins satisfy $\mathrm{\mid }{y}_n^{\mathrm{\prime }}\mathrm{\mid }\ge l+\frac{3}{4}-\epsilon \ge l+\frac{7}{12}>l+\frac{1}{2}$ and $\mathrm{\mid }{y}_n^{\mathrm{\prime }}\mathrm{\mid }\le R-\frac{3}{4}+\epsilon \le R-\frac{7}{12}<R-\frac{1}{2}$; also $\mathrm{\mid }{y}_{-1}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }{v}_0(s)\mathrm{\mid }\le R-\frac{7}{12}$, so the margin and parameter interval conditions hold. Moreover ${\delta }_{\ast }<(1-q)\bar{\rho }\mathrm{/}d$ because $d\le 2\mathrm{/}13$. By [F3] we obtain a $C^1$ recurrence curve ${\gamma }_H(b)=(b,y_0^{\mathrm{\prime }}+u_0(b))$; composing with the $H$-phase map gives $Z^H(b):=(b,V^H(b,y_0^{\mathrm{\prime }}+u_0(b)))$ through $Z^H(y_{-1}^{\mathrm{\prime }})=(v_0(s),V^H(v_0(s),v_{-1}(s)))=w$, with tangent $(1,T^H)$ and $\mathrm{\mid }{T}^H\mathrm{\mid }\le 1\mathrm{/}5+(6\mathrm{/}5)(1\mathrm{/}4)=1\mathrm{/}2$. The reference orbit $w_n=\mathrm{\Sigma }(z_{-n})$ converges to $\mathrm{\Sigma }(z_{\bar{u}})$ by step 4.4, so the tracked points lie in $W_H^s(\mathrm{\Sigma }(z_{\bar{u}}))$. Thus ${\gamma }_u(t):=\mathrm{\Sigma }(Z^H(v_0(s)+t))$ for sufficiently small $\mathrm{\mid }t\mathrm{\mid }$ is a $C^1$ curve through $z_{\ast }$ in $W_G^u(z_{\bar{u}})$, since $G^{-n}(\mathrm{\Sigma }\xi )=\mathrm{\Sigma }{H}^n(\xi )\to \mathrm{\Sigma }(\mathrm{\Sigma }(z_{\bar{u}}))=z_{\bar{u}}$. Its tangent is $\mathrm{\Sigma }(1,T^H)=(T^H,1)\in C_{3\mathrm{/}4}^u$. Finally $z_{\ast }\mathrm{\ne }{z}_{\bar{u}}$ because $h_G$ is injective by step 4.1 and $s\mathrm{\ne }\bar{u}$.