Proof. We first estimate the input series in the fixed product Riemannian metric, then pass to source coordinates. The chain rule and \(\|DG\|\le B\) give \[ \|D(G^m)\|\le B^m,\qquad \|DT_m\|\le B^m. \tag{11}\] The second bound uses the real projection \(\operatorname{pr}_1\), whose derivative has norm one everywhere. In particular, it requires no choice of output chart for \(G^m\).

For an integer \(k\ge1\), let \[\Phi_k(z)=\chi(z)(z/|z|)^k\quad(z\in\mathbb C\setminus\{0\}), \qquad \Phi_k(0)=\Phi_k(\infty)=0.\] These are smooth complex-valued functions. The support of \(\chi\) lies in a fixed compact annulus, on which the angular map \(z\mapsto z/|z|\) and its first derivative are bounded. The product and chain rules therefore give a constant \(C_\chi\), independent of \(k\), such that \[ \|\Phi_k\|_\infty\le1,\qquad \|D\Phi_k\|_\infty\le C_\chi(1+k). \tag{12}\] The extensions vanish on fixed neighborhoods of both poles.

Write the \(m\)th summand, including its prefactor, as \[F_{\ell,m}(t,z)=\frac{L^{1-m}}{10} \alpha(t)\beta_\ell(T_m(t,z))\Phi_{2^{m+1}}(z).\] Its differential is \[\begin{align*} DF_{\ell,m}=\frac{L^{1-m}}{10}\bigl(& \alpha'(t)\beta_\ell(T_m)\Phi_{2^{m+1}}\,dt\\ &+\alpha(t)\beta_\ell'(T_m)\Phi_{2^{m+1}}\,DT_m\\ &+\alpha(t)\beta_\ell(T_m)D\Phi_{2^{m+1}}\bigr). \end{align*}\] Since there are finitely many \(\beta_\ell\), their first derivatives have a common bound. Equations (11) and (12) yield a constant \(C\), independent of \(\ell\) and \(m\), with \[ \|F_{\ell,m}\|_\infty+\|DF_{\ell,m}\|_\infty \le C L^{1-m}\bigl(1+B^m+2^{m+1}\bigr). \tag{13}\] The prediction and angular derivatives occur in separate product-rule terms; their bounds are added. Each of the three resulting geometric series converges by (6).

For completeness, cover \([0,100]\times\Sigma\) by finitely many relatively compact source coordinate neighborhoods \(V_j\) whose closures lie in larger coordinate charts \(W_j\subset\mathbb R\times\Sigma\). If \(\kappa_j\) is the chart on \(W_j\), the norm of \(D\kappa_j^{-1}\) on \(\kappa_j(\overline{V_j})\) is bounded by a constant \(C_j\) independent of \(m\). Thus the coordinate derivatives of \(F_{\ell,m}\circ\kappa_j^{-1}\) satisfy (13) with the additional fixed factor \(C_j\). The series of functions and of all their first coordinate derivatives consequently converge uniformly on every such neighborhood. The usual differentiation theorem for series now applies: on a smaller coordinate ball, integrate the first derivatives of the partial sums along line segments and pass to the uniform limit. This identifies the derivative of the limit with the continuous limit of the derivatives. Hence each \(U_\ell\) is \(C^1\) on the fundamental strip.

This proof also covers data with arbitrarily late predicted visits. At a datum for which \(\beta_\ell(T_m)=0\) for every \(m\), each summand and its first derivative vanish: a differentiable nonnegative function has derivative zero at any zero. Uniform convergence of the derivative tails ensures continuity even when nearby nonzero summands have indices tending to infinity. No regularity of a selected hitting time is needed.

Because \(\alpha\) vanishes near \(0\) and \(100\), the input and all its first derivatives vanish on neighborhoods of the periodic seams. The periodic extension is therefore \(C^1\). Because \(\chi\) vanishes near zero and infinity, the same holds in fixed neighborhoods of the two poles, so the angular expression introduces no coordinate singularity. We also have the uniform bound \[ |U_\ell(t,z)|\le\frac1{10}\sum_{m=1}^{\infty}L^{1-m} =\frac{L}{10(L-1)}. \tag{14}\]

The clock component of (10) is smooth and well-defined modulo \(100\). Each register component is \(C^1\) and takes its values in the fixed disk of radius \(3L+L/(10(L-1))\) in the affine chart. This includes all source values \(v_\ell=\infty\), since \(S\) is smooth and zero near infinity.

It remains to check the main component at \(z=\infty\). Set \[C_0(t,v_1,\ldots,v_q)=20\sum_{\ell=1}^q\bar\beta_\ell(t)S(v_\ell), \qquad |C_0|\le60q.\] This is a smooth function of all its variables. In the source inverse coordinate \(w=1/z\) and the corresponding inverse coordinate for the output, the main component is \[ w'=\frac{w^2}{1+C_0w^2}. \tag{15}\] For \(|w|<[2(60q+1)]^{-1/2}\) the denominator has modulus greater than \(1/2\), uniformly in \(t\) and the registers. Formula (15) therefore extends smoothly to \(w=0\), with value zero. At all finite \(z\), including \(z=0\), the main component is already smooth in the affine chart. These checks establish the asserted global \(C^1\) regularity of \(f\). ◻

Proof. Monotonicity gives a finite limit \(t_*\), and hence \(h(t_n,z_n)=t_{n+1}-t_n\to0\). Since \(0\le b(t_n)\le h(t_n,z_n)\), continuity and the zero set of \(b\) imply \(t_*=50+100k\). For all sufficiently large \(n\) the clock therefore avoids every readout support, so \(z_{n+1}=z_n^2\). Moreover, \(|z_n|\ge3/4\), including \(z_n=\infty\), would give \(e(z_n)=1\) and \(h(t_n,z_n)=1\). Thus eventually \(z_n\) is finite with \(|z_n|<3/4\). Starting at such a time \(N\) after all readouts have ceased, \[|z_{N+r}|=|z_N|^{2^r}\longrightarrow0.\] The registers may have arbitrary behavior; all their values are included in the definition of \(Z_0\). ◻

Proof. Let \(r\) be the first index in the passage with \(s_r\ge2\). The preceding reading is below \(2\), so \(2\le s_r<3\). At time \(r\) one has \(a(s_r)=0\) and \(\alpha(s_r)=0\), whence \(U_\ell(s_r,z_r)=0\) and \[v_{r+1,\ell}=0\qquad(1\le \ell\le q).\] This conclusion holds even if any earlier register was at infinity. Until preparation starts, the inputs remain zero, and \(S(0)=0\) keeps all registers zero.

On \((8,12)\) the clock speed is exactly one. Its readings in that interval therefore belong to a single translate of the integers, which gives \(|I|\le5\). The first reading at least \(9\) belongs to \([9,10)\) by (17); there \(\alpha=1\). Similarly, the first reading at least \(76\) belongs to \([76,77)\), where the plateau-covering property of the \(\beta_\ell\) supplies a coefficient equal to one. All these crossings occur in the passage because \(s_n\) eventually reaches \(100\).

Suppose now that \(i\in I\) and \(|z_i|\le1/2\). As long as \(s_n\in[8,50]\) and \(|z_n|\le1/2\), there is no readout, \(e(z_n)=0\), and \[z_{n+1}=z_n^2,\qquad s_{n+1}=s_n+b(s_n)\le50.\] The clock cannot decrease, and \(|z_{n+1}|\le1/4\le1/2\). These two conditions thus persist by induction from time \(i\), keeping \(s_n\le50\) forever. This contradicts the assumed complete passage. This also excludes the endpoint \(|z_i|=1/2\) and proves (18). ◻

Proof. Fix \(i\in I\), and let \(r_{70}\) and \(r_{82}\) be the first indices after \(i\) with \(s_{r_{70}}\ge70\) and \(s_{r_{82}}\ge82\), respectively. Before \(r_{70}\) all readout coefficients vanish. The actual pair \((s_n,z_n)\) therefore follows \(G\) exactly from time \(i\) through time \(r_{70}\). At the latter time \(s_{r_{70}}\in[70,71)\). Throughout \([70,82]\) one has \(b=1\), so both the actual and predicted clock increments are one, independently of their respective main coordinates. Starting with the equal clocks at time \(r_{70}\), induction gives \[ T_{n-i}(s_i,z_i)=s_n\qquad(i<n\le r_{82}), \tag{20}\] including after any main-coordinate additions. The predicted clock is nondecreasing at all later times. Once it reaches \(82\), it cannot return to any support of the nonperiodic \(\beta_\ell\), all of which lie in \((74,80)\).

The support of a fixed \(\beta_\ell\) lies in an interval of length less than one. If \(\beta_\ell(s_j)>0\), the next clock reading is \(s_j+1\), strictly beyond that support interval. Monotonicity then rules out any later positive value of this \(\beta_\ell\); the same argument rules out two positive values in either order. Equation (20) identifies every possible predicted hit with an actual hit in this passage, and later predicted hits are impossible. This proves (19). ◻

Proof. Put \(i_0=\min I\) and, for each \(i\in I\), define \[c_i=\frac{\beta_\ell(s_j)}{10}\alpha(s_i)\chi(z_i) \left(\frac{z_i}{|z_i|}\right)^{2^{j-i+1}}.\] The phases are defined by (18). Lemma 6 and (9) give the exact input \[ U_\ell(s_i,z_i)=L^{1+i-j}c_i\qquad(i\in I). \tag{22}\] The input at time \(i\) first enters \(v_{i+1,\ell}\); it receives \(j-i-1\) subsequent factors of \(L\) before \(v_{j,\ell}\) is read. Thus its coefficient would be \(L^{1+i-j}L^{j-i-1}=1\) if the register remained in the region where \(S\) is the identity. We verify that region condition directly.

For every integer \(u\) with \(i_0\le u\le j\), define the candidate value \[w_u=L^{u-j}\sum_{\substack{i\in I\\i<u}}c_i.\] The triangle inequality and \(|I|\le5\) give \[ |w_u|\le\frac{|I|}{10}L^{u-j}\le\tfrac12. \tag{23}\] By Lemma 5, \(v_{i_0,\ell}=0=w_{i_0}\). For \(i_0\le u<j\) the clock lies between the preparation and readout intervals, so \(a(s_u)=L\). The input is zero unless \(u\in I\), and otherwise it is \(L^{1+u-j}c_u\) by (22). Consequently \[w_{u+1}=Lw_u+U_\ell(s_u,z_u).\] If \(v_{u,\ell}=w_u\), the bound (23) implies \(S(v_{u,\ell})=w_u\), so the actual register update gives \(v_{u+1,\ell}=w_{u+1}\). Induction proves \(v_{u,\ell}=w_u\) all the way through \(u=j\). Every preparation index precedes \(j\), hence \(w_j=\sum_{i\in I}c_i\), which proves (21) and the stated bounds.

This induction concerns only the particular register whose readout is at \(j\). Another register can already have been read and can subsequently leave the unit disk. By Lemma 6, its readout coefficient is zero for the rest of this passage. Registers do not enter one another’s updates, so such later nonlinear behavior has no effect on the induction just proved. ◻

Proof. Set \(\theta=z_{i_0}/|z_{i_0}|\). Before the first readout, the main coordinate squares exactly, so it stays nonzero and its phase at time \(n\) is \(\theta^{2^{n-i_0}}\). In particular this holds at every preparation index \(i\in I\), since all preparation precedes all readouts.

Proceed chronologically through the readout indices. Suppose the phase at time \(j\) is \(\theta^{2^{j-i_0}}\) and \(z_j\ne0\). For every \(i\in I\) the phase appearing in (21) is \[\left(\frac{z_i}{|z_i|}\right)^{2^{j-i+1}} =\left(\theta^{2^{i-i_0}}\right)^{2^{j-i+1}} =\theta^{2^{j-i_0+1}},\] which is the phase of \(z_j^2\). Lemma 7 applies separately to each \(\ell\) with \(\beta_\ell(s_j)>0\). The other channels have zero coefficient in the main update. Thus that update is exactly \[ z_{j+1}= \left( |z_j|^2+ 2\sum_{\ell=1}^q\beta_\ell(s_j)^2 \sum_{i\in I}\alpha(s_i)\chi(z_i) \right)\theta^{2^{j-i_0+1}}. \tag{25}\] One factor of \(\beta_\ell(s_j)\) comes from the input and the other from the readout in (10); both are essential to this coefficient. Every added coefficient is nonnegative, and the baseline \(|z_j|^2\) is strictly positive. This proves nonvanishing, phase doubling, and the square lower bound at this readout, even if all its inputs vanish. Between readouts there is exact squaring. Induction therefore proves (24) through the end of the passage, including simultaneous readouts of several channels and successive readouts of different channels.

If some preparation value already has \(|z_i|>1\), take \(N=i\). Otherwise \(1/2<|z_i|\le1\) for every \(i\in I\). Choose \(i_*\) with \(\alpha(s_{i_*})=1\), supplied by Lemma 5. The prescribed cutoff satisfies \(\chi(z_{i_*})=1\), including when \(|z_{i_*}|=1\). The same lemma supplies a readout \(j_*\) and a channel \(\ell_*\) with \(s_{j_*}\in[76,77]\) and \(\beta_{\ell_*}(s_{j_*})=1\). In (25), the single pair \((\ell_*,i_*)\) therefore adds \(2\) to the radius, and all other terms have the same phase. Hence \[|z_{j_*+1}|\ge |z_{j_*}|^2+2>2.\] Taking \(N=j_*+1\) proves the claim; this index is still in the passage because \(s_{j_*+1}\le78<100\). ◻

Proof of Proposition 3. If the lift is bounded, Lemma 4 proves the asserted coordinate limits. In a product metric the distance to \(Z_0\) consequently tends to zero, irrespective of the registers.

If \(z_0=\infty\), it remains at infinity, all inputs vanish, and the clock advances by one at every step. A translate of the integer lattice meets the reset interval \([2,5]\) modulo \(100\). At such a step \(a=0\), so every register becomes zero, regardless of its initial value, and stays zero. The orbit then lies in \(Z_\infty\).

It remains to consider an unbounded lift with finite \(z_0\). Choose any complete passage and an index \(N\) with \(|z_N|>1\) given by Lemma 8. Its square lower bound continues through the rest of that passage. We spell out why it also holds at every later time. After the current readout region \((74,80)\), all readout multipliers vanish until the next period’s region \((174,180)\) in the same translated coordinate. In particular, the main variable squares exactly across the period boundary and throughout the next reset and preparation intervals. An old register, however large, cannot alter \(z\) there. It is erased by the next reset before any new preparation input is entered.

Every later period is a complete passage by the unbounded-lift assumption. Lemma 8 therefore applies afresh from its first preparation index through its end. The intervening indices, before that first preparation, have no readout and give exact squaring. These intervals cover every time after \(N\), and hence \[ |z_{N+r}|\ge |z_N|^{2^r}\qquad(r\ge0). \tag{26}\] This argument does not require \(\chi\) to vanish immediately after the radius exceeds one. New inputs in later periods still have nonnegative weights \(\alpha\chi\), satisfy the register bound, and give the aligned additions in (25). Thus they preserve (26) even while \(\chi\) remains nonzero.

It follows that \(z_n\to\infty\) in \(\Sigma\). Since \(\chi\) vanishes in a neighborhood of infinity, there is a time after which \(\chi(z_n)=0\) at every step. All inputs \(U_\ell(t_n,z_n)\) are then zero. A later reset, which exists because the clock lift is unbounded, makes every register zero; the recurrence and \(S(0)=0\) keep them zero forever. The distance to \(Z_\infty\) therefore tends to zero.

We have proved the assertion in a product metric. All compatible metrics on the compact space \(M\) are uniformly equivalent, so the same convergence in distance holds for any such metric. The conclusion is pointwise in the initial state; no uniform entry time is asserted. ◻