Positive Metric Entropy for the Standard Map at Large Parameters

Proof. Fix k\ge k_0 and let H_k be the invariant Oseledets set where \lambda_+>0. It has positive area by Theorem 1.1, while \det Df_k=1 makes the other exponent -\lambda_+<0 there. The normalized restriction \nu_k=m(H_k)^{-1}m|_{H_k} is an absolutely continuous invariant probability on the compact torus, with no zero Lyapunov exponents. Pesin’s ergodic and Bernoulli decomposition theorems, as stated in [11, Theorems 3.3–3.4], apply to \nu_k and give a positive-\nu_k ergodic component E and its finite cyclic partition. Since \nu_k(E)=m(E)/m(H_k)>0, the component has positive area; normalizing \nu_k on E or on a partition piece agrees with normalizing m. Area preservation gives m(E_j)=m(E)/N.□

Proof. For each bad index choose one witnessing interval. Its length is a positive integer, and the sum of the nonnegative step shortfalls over it exceeds (1-c) times its length. Select disjoint intervals greedily in decreasing order of length, discarding each interval that intersects one already selected. Regard intervals as half-open when deciding disjointness, so that the sums of shortfalls over selected intervals have disjoint sets of steps.

Every discarded interval meets a selected interval of at least its length and is contained in the concentric threefold enlargement of that selected interval. The bad endpoints are therefore covered by these enlargements. An interval of integer length L\ge1 has at most 3L+1 integer points in its threefold enlargement. If the selected lengths are L_1,\ldots,L_s, the number of bad indices is at most 4\sum_r L_r. Disjointness and nonnegativity give (1-c)\sum_rL_r\le\mathcal D. This proves the first statement, and the same selection applied to the forward witnessing intervals proves the second.□

Proof. At index N, Equation (3.5) and the endpoint bound imply

and hence

Since \tau_1=0, \upsilon_1\ge0, and N-1\le n, the total shortfalls of the paths \tau and -\upsilon satisfy

Every summand is nonnegative by Equation (3.7). Apply Lemma 2.1 to \tau to count indices where the first line of Equation (3.8) fails. An individual increment smaller than c costs at least 1-c from the relevant shortfall sum. This counts failures of the second line; applied to -\upsilon, it counts failures of either of the two inequalities in the last line, with a factor of at most two because a single increment can be tested at two neighboring indices. Summing these bounds proves Equation (3.9).□

Proof. Choose a small fixed \beta>0, for example \beta=10^{-3}. By Equation (3.9), choosing \delta_0 sufficiently small and then n_* sufficiently large makes the bad-index count on each path at most \beta n. The total-shortfall estimate also gives, at every index,

Shrink \delta_0 once more and increase n_* so that, for n/3\le j\le n/2, the corresponding \tau_j lies in [n/4,3n/4], while on the minus path

There are at least n/6-2-\beta n good plus indices in this range.

For each such index put a=\tau_j^+. Starting from \tau_1^-=0, let k be the first minus index with \tau_k^-\ge a. We have \tau_{k-1}^-<a\le\tau_k^- and \tau_k^--\tau_{k-1}^-\le1. Thus at least one of these two values is within 1/2 of a. The chosen index is in \{2,\ldots,N-2\}: the large lower bound on a excludes the first two indices, and the displayed lower bound at N-2 provides the upper bound. If the chosen minus index is bad, discard this plus index.

Good plus indices j<j' have \tau_{j'}^+-\tau_j^+\ge c(j'-j). At most two such logarithms can belong to an interval of length one, since 2c>1. A fixed bad minus index can therefore be responsible for at most two discarded plus indices. At least n/6-2-3\beta n candidates remain.

Thin these candidates in increasing plus order by retaining the first one, then retaining the next whose plus logarithm is at least two larger than the last retained logarithm, and continuing. Since successive candidate logarithms are separated by at least c, each retention removes at most three candidates. A fixed positive fraction of n remains; for example, decreasing the fraction if necessary, one may fix c_1=10^{-2}. The matched minus logarithms increase strictly, because each is within 1/2 of the corresponding plus logarithm and the retained plus logarithms have gaps at least two. For two good minus indices, increasing index implies strictly increasing logarithm by Equation (3.8). Strict increase of their logarithms therefore forces strict increase of their indices as well. The matching error is at most 1/2, which implies Equation (3.13).

Each path contains at most n+1 available indices. An increasing paired pattern is determined by a subset of the indices on each path, with the two subsets of equal cardinality paired in order. There are at most 2^{2n+2} such choices. Enlarging an absolute C_1 gives the stated exponential bound.□

Proof of Theorem 3.1. Fix once and for all a sufficiently small \delta>0 allowed by Lemma 3.4. Its constants c_1,C_1,n_* are now fixed. Choose \eta>0 so small that

Obtain \eta' and M_\eta from Lemma 3.5. Increase a fixed lower bound M_0 so that all estimates above hold, that M_0\ge M_\eta, and that

These choices are independent of n, of the center, and of M\ge M_0.

Fix e\in\{0,-1\} and q_e. It is enough to consider patterns containing exactly r=\lceil c_1n\rceil pairs, since a longer pattern can be truncated. For a fixed such increasing pattern, let \mathcal F be the set of \theta admitting one common normalized solution with all of the following properties: the original central pair has norm at most \sqrt2; both endpoint pairs satisfy Equation (3.3); and every pair of selected indices is good and satisfies Equation (3.13). We emphasize that one solution must satisfy all conditions of a pattern at once.

The set \mathcal F is measurable, in fact compact. To see this without selecting solutions, parameterize the original central pair by its entry other than the fixed entry u_e=1. The norm bound puts that parameter in [-1,1]. All later solution entries and all recurrence coefficients depend continuously on this parameter and on the circle coordinate \theta. Each pair entering a logarithm is nonzero, by invertibility of the recurrence. The endpoint, goodness, and matching conditions are closed inequalities on the compact product of the circle with [-1,1]. Their projection is therefore compact.

For \theta\in\mathcal F and 1\le m\le r, define

and let

be an open subset of the circle. Goodness at the later of two selected indices gives, separately on both paths,

Since both index sequences increase strictly,

For fixed M,n, these positive radii also have a positive lower bound: the one-step norm bound gives \abs{t_j}\le M^{j-1}\le M^n at all indices in question, so s_m\ge M^{-n}.

We claim

Select disjoint balls from the family defining V_m as follows. At each stage select a remaining ball whose radius is at least half the supremum of the remaining radii, and remove all remaining balls meeting it. Assign each removed ball to this selected ball. All selected balls are disjoint and have a fixed positive lower radius, so only finitely many can be selected on the circle; the procedure must exhaust the family. Every assigned ball meets its selected ball and has radius at most twice that of the selected ball.

Consider one selected center \theta_* and write s=s_m(\theta_*). The centers of its assigned balls are at circle distance at most 3s/10 from \theta_*. The window of radius s is an injective lift interval by Equations (3.16) and (3.17). All assigned centers belong to \mathcal F, so they are successful for the same selected pair of indices in the local test based at \theta_*. A next-level ball centered at an assigned \theta has radius at most

After rescaling the lifted coordinate by 1/s, the union of all these next-level balls is contained in the \eta'-neighborhood of the successful set in Lemma 3.5, by Equation (3.16). It therefore has unscaled length at most \eta s. Summing over selected balls proves

as claimed. The ball selected at a center is included in its own assigned group, so no next-level balls have been omitted.

Since \mathcal F\subset V_m for every m and \abs{V_1}\le1, iteration gives

There are at most \exp(C_1n) patterns. By Equation (3.15), their union has circle measure at most C\exp(-\gamma n) for a fixed \gamma>0, uniformly in the fixed coordinate q_e. The coordinate pair (q_e,q_{e-1}) has normalized product Lebesgue law. Integrating the conditional bound over q_e and summing the two choices of e therefore gives the same form of bound for the original event, for every n\ge n_*. Every point of that event is covered, by Equation (3.3) and Lemma 3.4. Finally enlarge C to cover the finitely many positive integers n<n_*, whose event probabilities are at most one. Stationarity restores every center i.□

Proof. Let s_l be a most contracted unit input direction for T_{0,l}. Since a single inverse step has norm at most M,

The projection onto the expanded input direction for T_{0,l} therefore gives

The absolute sine between two unoriented lines satisfies the triangle inequality. Summing Equation (5.6), and using Equation (5.3), yields

Resolve s_{B_+} into the singular input directions for T_{0,l}. Using D_l\leq M^l as well as its lower bound gives

The last inequality follows from 1-c+(1-2c)l=0.02-0.96l\leq-0.94l for l\geq1.

The first component of an input vector becomes the second component after one step. Thus Equation (5.7) at l=1 shows that s_{B_+} is nearly vertical. Let t solve the variational recurrence with t_{-1}=0, t_0=1. Horizontal input has a projection bounded away from zero onto the expanded input direction for T_{0,B_+}. For large M, at least one of the two coordinates t_{B_+} and t_{B_+-1} consequently has magnitude at least D_{B_+}/3. Choose the corresponding j. This proves that the two closed choices in Equation (5.5) cover the states under consideration.

The row of T_{0,B_+} selecting q_j has norm at least \abs{t_j}\geq D_{B_+}/3, and its value on s_{B_+} has magnitude at most D_{B_+}^{-1}. Its kernel direction therefore differs from s_{B_+} by O(D_{B_+}^{-2}). Normalize that direction so that its second component is 1, and let U denote its variational solution. Then U_{-1}=1, U_j=0, and, for 0\leq p<j,

Indeed, apply Equation (5.7) at l=p+1 to its main component. The error component is bounded by

because p+1\leq B_+. Normalization changes these bounds only by an absolute factor.

We now solve the nonlinear boundary problem. Let \xi_p be the change of the lifted position q_p, with boundary data \xi_{-1}=r and \xi_j=0. The tridiagonal matrix on 0,\ldots,j-1 with diagonal v(q_p) and adjacent entries -1 has determinant t_j\ne0. Its inverse is

For completeness, the proposed column solves the homogeneous recurrence except at its source. At that source the output is 1, because the Wronskian t_{p+1}U_p-t_pU_{p+1} equals its boundary value 1. Both boundary values of the proposed column are zero. This verifies Equation (5.9) directly.

Subtract the original orbit recurrence from the new one, and move the nonlinear Taylor remainder to the right side. With its sign absorbed into F_p, the boundary problem becomes

Use the finite-dimensional weighted norm

On a ball of fixed radius R_0, the Lipschitz bound for the nonlinear part of Equation (5.10), before weighting its output at index l, is bounded by a constant depending on R_0 times

Here \abs{t_p}\leq M^p and Equation (5.8) were used. After multiplying by M^{0.8(l+1)}, the exponents are, respectively,

Their sums tend uniformly to zero as M tends to infinity, independently of j. The linear term Ur has bounded weighted norm for \abs r\leq1/4. Choose R_0 larger than twice this bound and then choose M large. The right side of Equation (5.10) maps this ball to itself and is a contraction there. Its unique fixed point gives the desired graph on the whole parameter interval.

The same estimates show that I-G\diag(F'_p(\xi_p)) has a uniformly bounded inverse in this weighted norm. Multiplication by the original Dirichlet matrix identifies it with the varied linearized Dirichlet problem. Consequently that problem is invertible, and the finite-dimensional implicit function theorem gives smooth dependence on r. Differentiating Equation (5.10) solves with right side U, which has bounded weighted norm. This proves Equation (5.4). The contraction and its fixed point also depend continuously on the eligible initial data: all the coefficients are continuous there, and the contraction constant and ball are uniform. This proves the asserted continuous dependence.

If j=B_+, the endpoint pair estimate follows directly from the fixed boundary value and Equation (5.4). If j=B_+-1, the recurrence and the fixed value of q_{B_+-1} give \xi_{B_+}=-\xi_{B_+-2}, proving the same estimate without an extra factor of M.

It remains to control transverse distortion. At the varied orbit let t_p(r) again have initial values t_{-1}(r)=0, t_0(r)=1. Invertibility of the varied Dirichlet problem gives t_j(r)\ne0. The graph derivative U_p(r)=\dd q_p(r)/\dd r has boundary values 1 and 0 at indices -1,j and solves that varied problem. Its Green diagonal is therefore t_p(r)U_p(r). Differentiating the determinant of the varied Dirichlet matrix gives the exact identity

The derivative of a diagonal entry is \phi''(q_p(r))U_p(r), explaining the second factor of U_p. Since \abs{t_p(r)}\leq M^p for every varied orbit, Equation (5.4) bounds the absolute value of Equation (5.11) by

Integration over \abs r\leq1/4 proves the bounded-ratio claim.

For backward evolution set \widetilde q_p=q_{-1-p}. The recurrence has the same form in these variables. At step l its ordered pair is the swapped original pair at time -l, and the initial pair is (q_{-1},q_0). Swapping coordinates is an isometry, so the backward growth hypotheses and all norm estimates are identical. The resulting graphs vary q_0 and have small slope in the q_{-1} coordinate.□

Proof of Theorem 5.1. We first find many usable middle times. On the bridge event choose a unit vector realizing \norm{T_{a,b}} and follow its norms along the bridge. The increments of their logarithms to base M are at most 1, and their total shortfall from 1 is at most (1-c_b)n. By Lemma 2.1, if c_b is sufficiently close to 1, at least \eta n indices i in the middle third, for an absolute \eta>0, satisfy

In detail, remove indices admitting a forward interval starting there or a forward interval ending there whose average log increment is less than c. The counting bound removes at most C(1-c_b)n/(1-c) indices. At a retained index, the chosen vector bounds the forward operator norm on either kind of interval from below. On an interval ending at i, use also equality of the forward and inverse operator norms to obtain the second inequality in Equation (5.12). For large n, the integer rounding of the middle third does not affect a fixed positive proportion.

It suffices to establish Equation (5.2), with a uniform constant, when the bridge indicator is replaced by the indicator of Equation (5.12) for one fixed middle index i. Indeed, the bridge indicator is at most (\eta n)^{-1} times the sum of these middle-index indicators; multiplying by the nonnegative weight and summing then costs only an absolute constant.

Shift i to time 0. Put B_+=b-i and B_-=i-a; both are at least n/3-O(1). Apply Lemma 5.2 forward and backward. There are two possible endpoint-coordinate choices in each direction. We work with one fixed pair of choices and eventually sum over the four possibilities. Write Q(x,y) for the lifted forward coordinate held fixed, and F(x,y) for the lifted backward coordinate held fixed. On forward graphs Q_x\ne0; on backward graphs F_y\ne0. For large M, all forward slopes \abs{\dd x/\dd y} and backward slopes \abs{\dd y/\dd x} are less than a common absolute number a_0<1/2. The relevant transverse partial derivatives have uniformly bounded absolute ratios along each graph.

Transport of endpoint observations. Along a forward graph the endpoint pair at time b changes by at most CM^{-0.8(B_+-1)}. A single forward or inverse map is M-Lipschitz on the lifted coordinates. Thus pair differences throughout the window of radius \kappa n about b are at most

The one-step matrix difference is at most CM times the position difference. Since forward and inverse product norms agree, the endpoints of each observation interval may first be put in increasing order. For a product of length at most 2\kappa n, the telescoping product identity therefore bounds its matrix difference by a length factor times

Every product norm compared here is at least 1. The elementary inequalities \abs{\norm A-\norm B}\leq\norm{A-B} and \abs{\log u-\log v}\leq\abs{u-v} for u,v\geq1 consequently give the same exponential control for the normalized logarithms in Equation (5.1). The positive integer interval lengths in their denominators cause no additional loss.

Choose fixed \kappa,\sigma>0 with

The strict margin absorbs the length factor and the bounded powers of M when n is large. After increasing the size requirements to account for the observation bounds, each observation at the forward end changes by at most

along a forward graph. The same statement holds for an observation at the backward end along a backward graph. All these estimates are uniform in the chosen middle index, endpoint-coordinate choices, and observation indices. Increase M,n further so that \delta_n<h/2.

Figure 2 shows the two transverse coordinate families used next. It depicts the change of variables, not the motion of an orbit.

Forward labels and their area. Assume first that H is at the forward end and W at the backward end. Cover the middle-time torus by the finitely many closed squares of a grid of side s_0=1/32 in a fundamental domain. Overlaps on boundaries have area zero and cause no problem for an upper bound. Fix one square I_x\times I_y and choose y_c\in I_y. All position functions in this square and its small neighborhood are computed from the same real lifted recurrence starting at (x,y).

Every forward graph through an eligible point in the square extends over all of I_y, with a margin, since its parameter range has length 1/2 and s_0<1/4. Label it by its intercept s at y=y_c and write it as (X(s,y),y). Two such graphs which meet have the same value of Q and coincide locally because Q_x\ne0. They then coincide on the whole common interval: the coincidence set is both open, by implicit uniqueness, and closed, by continuity. In particular distinct labels have disjoint images at every fixed y.

Let E be the labels of forward graphs through forward-eligible points of this square for which H\geq h. We use closed eligibility conditions, including Equation (5.5). They and H\geq h define a compact subset of the closed square. Continuous dependence of the graph in Lemma 5.2 makes its intercept a continuous function of the initial point. Thus E is compact. The graph is uniquely determined by its intercept, so these constructions define X unambiguously on E\times I_y.

For these labels, every point of the graph over I_y has H>h/2 by Equation (5.14). The identity

and implicit differentiation give

The derivative in Equation (5.15) is interpreted through a local smooth extension when E has empty interior; the existence of that extension is justified below. The distortion assertion of Lemma 5.2 bounds its absolute value above and below by positive constants.

The map (s,y)\mapsto(X(s,y),y) on E\times I_y is injective and has Jacobian comparable to 1. Its image stays in a fixed bounded lift: the small slope bounds the displacement from the square by an absolute constant. Projection of this bounded lift to the torus has bounded multiplicity. Change of variables therefore yields

The factor \abs{I_y}=s_0 has been absorbed into the constant.

A transverse sweep for the weight. Fix s\in E. Let D_s\subseteq I_y consist of the parameters y for which z(y)=(X(s,y),y) lies in the square and is backward-eligible in the fixed backward endpoint choice. This is a compact set. Carry each z(y) along its backward graph to first coordinate u\in I_x, and denote the resulting second coordinate by \Theta_s(u,y). The carry is defined throughout I_x, with a margin.

For each fixed u, the map y\mapsto\Theta_s(u,y) is injective on D_s. If two carried points agreed, their backward graphs would coincide throughout I_x by regular-level uniqueness. A single such graph cannot meet the chosen forward graph in two distinct points: two intersections would satisfy both \abs{\Delta x}\leq a_0\abs{\Delta y} and \abs{\Delta y}\leq a_0\abs{\Delta x}, forcing equality of the points. This proves the asserted injectivity.

The equation defining the carry is

Differentiation with respect to the starting parameter gives

The ratio of the two F_y terms is uniformly bounded above and below along the backward graph. Moreover \abs{F_x/F_y}\leq a_0 at its starting point, since -F_x/F_y is its slope, and \abs{X_y}\leq a_0. Hence the absolute value of Equation (5.17) is uniformly bounded above and below by positive constants.

The endpoint observation estimate gives, for all u\in I_x and y\in D_s,

Apply change of variables to the injective map (u,y)\mapsto(u,\Theta_s(u,y)) on I_x\times D_s. Its image lies in a bounded lift. Since W\geq0, its integral over that image is at most a fixed multiple of \E W. Using Equation (5.17) and then dividing by \abs{I_x} gives

Changes of variables on the eligible subsets. Here are details ensuring that the preceding changes of variables do not require the eligible sets to have interior. Consider first one forward graph with label s. It is a compact regular level arc over I_y and extends with a margin beyond this interval. The partial derivative Q_x is nonzero on the arc. The implicit function theorem in finitely many overlapping charts along the arc therefore continues it smoothly for intercepts in an open neighborhood of s. The continued level is prescribed by Q(s',y_c) for each nearby intercept s'. Uniqueness in the overlapping charts makes these local continuations consistent. Every nearby eligible label follows the same continuation, because its regular graph passes through that intercept. Thus X on E\times I_y is locally the restriction of a smooth function of (s,y), and Equation (5.15) holds at all required points.

For the transverse sweep, fix s and a point of D_s. Its backward graph is likewise a compact regular arc, with F_y\ne0, extending with margin over I_x. The same continuation argument applies to the smoothly varying starting points (X(s,y),y) on the fixed forward graph. It gives a local smooth extension of \Theta_s(u,y) in both variables, agreeing with all nearby eligible carries by uniqueness. The maps used in both area comparisons are consequently restrictions of local smooth diffeomorphisms, with the displayed nonvanishing Jacobians.

A countable cover by such charts can be split into disjoint Borel pieces. On each piece ordinary change of variables applies to its measurable subset; the images of the pieces are disjoint because the full restricted maps are injective. Summing proves the formulas on the compact eligible domains. This also establishes the measurability of all integrands and swept regions used above. No extension of eligibility to a neighborhood is assumed.

Completion of the area comparison. Every point of the fixed square satisfying both eligibility conditions and H>h belongs to a forward graph with label in E. Change variables from x to this label at each y, using Equation (5.15). Drop conditions other than square membership and backward eligibility, which is allowed because the weight is nonnegative. The desired weighted integral over this square and this pair of choices is bounded by

Combining Equations (5.18) and (5.16) bounds this by the right side of Equation (5.2).

If H is at the backward end, begin instead with the labels of the almost horizontal backward graphs and sweep them in the x direction to bound their label measure by \PP(H>h/2). For each such graph, transport its eligible points along the almost vertical forward graphs to estimate W. The proof just given applies with x,y and Q,F interchanged: the graph extents, distortion bounds, small-slope injectivity argument, and local chart construction are the same. This proves the estimate for the other assignment of the two observations as well.

Finally sum over the finitely many squares and the at most four endpoint-coordinate choices. The constants remain uniform. Average over the usable middle indices as explained after Equation (5.12). This establishes Equation (5.2) and completes the proof.□

Proof. Choose a most expanded unit input direction a_j for T_{i,j}; if D_j=1, any singular direction may be used. Set

We claim

To obtain the sine lower bound, use a unit input at the base perpendicular to a_p. Its norm at p is D_p^{-1}, whereas its norm at j is at least D_j|a_j\wedge a_p|. The transfer from p to j therefore has norm at least D_jD_p|a_j\wedge a_p|. The inequalities

give the other two lower bounds; the second uses equality of the norms of a determinant-one matrix and its inverse.

For the upper bound, transport a unit vector from p back to the base. Its components along a_p and a_p^\perp have absolute values at most D_p^{-1} and D_p, respectively. Also

It follows that

proving Equation (6.3). By Equation (2.8), Q_{jp}\le2\max\{Q_{jl},Q_{lp}\}, so C_0=6 is permissible.□

Proof. The triangle inequality and Equation (2.7) give

and similarly on the right. In particular,

Apply Equation (6.2) twice along the chain L',L,R,R' to obtain Equation (6.5). Taking negative base-M logarithms gives Equation (6.6), with any fixed upper bound for 2\log_M C_0 as A. Such a bound is uniform after imposing, for example, M\ge2. Finally, each cut wing has length at most s, which yields Equation (6.7).□

Proof. Use the notation of Lemma 6.2. Both D and \ell are nonnegative, and \ell\le2n. Let \delta>0 be the fixed constant in Theorem 3.1. Choose a fixed H sufficiently large that

This is possible since s\ge\ell/2-1. Also s\le n and 2s\le\ell. Thus Equations (6.4) and (6.7) imply

This is the event bounded by Theorem 3.1, with spacing s. On the part where this integer s is selected, \ell<2s+2. Writing the constants of that theorem as C_{\rm can},\gamma>0, we obtain

The complementary contribution is at most H(1+\E D), hence

Stationarity gives

These identities and Equation (6.9) prove the result. In particular the lower inequality uses only \ell\ge0.□

Proof. First, for every fixed integer a\ge1, we have e_a\to0 as K\to\infty. To see this, fix 0<\zeta<1. The uniform marginal law of each q_i implies

Indeed v(x)=2+K\cos(2\pi x), and the length of the set on which |\cos(2\pi x)|\le(M^{1-\zeta}+2)/K tends to zero. A union bound therefore shows that, with probability tending to one, all a coefficients in the product satisfy |v(q_i)|\ge M^{1-\zeta}. Start the linear recurrence with (u_0,u_{-1})=(1,0). On this event, for M large, induction gives

Consequently g(0,a)/a\ge\log_M(M^{1-\zeta}-1) there. The deficit always lies in [0,1], so \limsup_{K\to\infty}e_a\le\zeta. Letting \zeta\downarrow0 proves the claim.

Choose powers of two n_0\to\infty sufficiently slowly that e_{n_0}\to0, by diagonal selection over the fixed lengths 1,2,4,\ldots. Set

For all sufficiently large K, e_{n_0}<\theta_K<1. By Equation (6.1), there is a first power of two N\ge n_0 for which e_N>\theta_K. It satisfies N>n_0 and e_{N/2}\le\theta_K. Applying Lemma 6.3 once gives

Also \epsilon n_0\ge\theta_K n_0\ge\sqrt{n_0}\to\infty. For fixed m\ge0, iteration of Lemma 6.3 yields

and

For a fixed negative m, the lengths 2^m n_0 and 2^mN are eventually positive integers. Dyadic monotonicity bounds their deficits by e_{n_0}=o(\epsilon) and e_N=\epsilon, respectively. This proves every assertion.□

Proof. By stationarity,

Summation over consecutive dyadic scales gives the exact identity

By Lemma 6.4 and dyadic monotonicity, its nonnegative right side is O_I(\epsilon).

Put a=h(I_l), b=h(I_r), and \bar h=(a+b)/2. At the aligned sampled arrays, 0\le\bar h\le h(I)\le1, and therefore

Write b_q=\E\bigl[(1-g(0,q)/q)^2\bigr] for integer q\ge1. By stationarity and another dyadic telescoping,

Here and in this display only, the unqualified sum is over the scales in Equation (6.19). Since 0\le b_q\le e_q, the last expression is O_I(\epsilon). This proves Equation (6.16).

At any single aligned scale, J(I)\le h(I) and V(I)=(a-b)^2\le a+b. Hence the expectation in Equation (6.17) is at most 2e_{pn}+2e_{pn/2}. Dyadic monotonicity bounds these terms by the corresponding terminal-scale deficits, which are O_I(\epsilon) by Lemma 6.4. At n=n_0 they are o(\epsilon), again by that lemma, including its consequence for negative shifts.□

Proof. Every affine array annihilates these tests. Conversely, suppose they all vanish. For one of the stated intervals I, the equalities J(I)=V(I)=0 say that its two halves have the same speed as I and that their distances add to the full distance. Repeating this argument shows that every edge of every dyadic subdivision of I has distance w_I times its length, with w_I=d(s,t)/(t-s). The triangle inequality gives the same upper bound between any two subdivision nodes. Applying the triangle inequality across the full interval, using those nodes as intermediate points, gives the reverse bound, and therefore equality. Every dyadic point of I is a subdivision node at some level. Apply this argument to [-2^a,2^a] for arbitrarily large integers a. The common pair 0,1 fixes one speed on all these intervals, so the array is globally affine. The ambient bound forces that speed to lie in [0,1].

For the last assertion, each point of F has a positive test. The corresponding open sets cover F, and a finite subcover suffices. Their sum has a positive minimum on the compact set F.□

Proof. For each compact F\subset\mathcal U, combine Equation (6.20) with Equations (6.16) and (6.17) to obtain uniform bounds on \mu_K(F) and \nu_K(F). Using Equation (6.18) instead gives \xi_K(F)=o(1). The latter proves Equation (6.21), since a continuous compactly supported test is bounded and has compact support.

Here is an explicit way to extract the local limits. Fix a compatible metric on X and let a(d)=\dist(d,\mathcal A). Choose a continuous nondecreasing function \chi:[0,\infty)\to[0,1] equal to zero on [0,1/2] and to one on [1,\infty). Set \chi_j(d)=\chi(2^j a(d)). These functions increase to one on \mathcal U, have compact support there, and satisfy \chi_{j+1}=1 on \supp\chi_j. The finite measures \chi_j\mu_K and \chi_j\nu_K have bounded total masses for each j. Weak compactness for finite measures on compact metric spaces, followed by a diagonal extraction, gives weak limits for all of them. Their restrictions agree on the open regions where the corresponding cutoffs are one: this follows by testing against continuous functions supported in such a region, for which the prelimit integrals agree. These compatible restrictions define locally finite measures on \mathcal U. Every compactly supported continuous function is supported where some \chi_j is one, so these are the asserted vague limits.

For completeness, the support restrictions survive despite the possibly unbounded total masses. The arrays g_n satisfy g_n(s,t)\le|s-t|+1/n. Taking logarithms in Equation (6.2), then dividing by n, shows that their version of Equation (6.22) has error at most (\log_M C_0)/n. These errors are bounded uniformly by C/n_0 at every sampled scale. Any fixed strict violation, by a fixed positive margin, is thus absent from every sampled array once K is large. Testing on compact subsets of its open violation set shows that it has zero limiting measure. There are countably many dyadic arguments and positive rational margins, which proves the stated support assertion. For an array with d(s,t)\le|s-t|, the semimetric inequality gives

It therefore has a unique jointly continuous extension to real arguments; all its inequalities extend by continuity.

For fixed r\in\mathcal D, each prelimit law is invariant under \tau_r once n_0r is an integer. Indeed \lfloor n(s+r)\rfloor=\lfloor ns\rfloor+nr, and stationarity applies simultaneously to the array. The map \tau_r is a homeomorphism of X preserving \mathcal A, so it preserves compact support in \mathcal U. Invariance therefore passes to the vague limits.

Finally, multiply any nonnegative test J(I)+V(I) by \chi_j, pass to the vague limit, and use Equation (6.16). Monotone convergence as j\to\infty gives the first bound in Equation (6.23). For \nu use Equation (6.17) in the same way. Although h(I) is not nonnegative everywhere on X, it is nonnegative on the aligned samples and on the supported limiting class X_0; these facts justify the cutoff bounds and monotone convergence for that term as well. In the case I=[0,1] the uncut terminal integral is exactly e_N/\epsilon=1, giving Equation (6.24).□

Proof. The floor convention in Equation (6.13) gives the exact identity Sg_n=g_{2n}. Hence, as finite measures on X,

To pass to the local limits, observe that S^{-1}(\mathcal A) =\mathcal A: multiplication by two permutes the dyadic rationals, so Sd=d_w implies d=d_w. If F\subset\mathcal U is compact, its preimage under S is closed in compact X and disjoint from \mathcal A, hence compact in \mathcal U. Thus S is proper on \mathcal U, and composition with S preserves continuous compactly supported tests there. Vague convergence and Equation (6.21) now give Equation (6.26).

We may henceforth work on X_0\cap\mathcal U, where S is invertible. Applying its inverse to Equation (6.26) and iterating gives, for every b\ge1,

The increasing positive partial sums are dominated by \mu. Their limit \lambda=\sum_{j\ge1}(S^{-j})_*\nu is therefore a locally finite measure. On relatively compact Borel sets define \mu_\infty=\mu-\lambda. The definitions agree on overlaps and extend by exhaustion to a locally finite nonnegative measure; this construction avoids subtracting infinite total masses. Monotone convergence for the positive series yields

Subtract this identity from Equation (6.26) on relatively compact sets to obtain S_*\mu_\infty=\mu_\infty.

For r\in\mathcal D and j\ge1, the identity

and translation invariance of \nu show that every summand of \lambda is invariant under \tau_r. The same holds for their sum and, by local subtraction from the translation-invariant measure \mu, for \mu_\infty. This proves all assertions.□

Proof. First fix r,t\in\mathcal D, t>0. For all sufficiently large K, nr and nt are integers at every scale n\ge n_0 in our sums. The event Equation (7.1) for g_n is contained in the cancellation event of Theorem 3.1, at center nr and spacing nt. Consequently its \mu_K-mass is at most

The last conclusion follows from \epsilon n_0\to\infty in Lemma 6.4. The event is open in X. To pass to a locally finite limit, test its indicator from below by nonnegative continuous functions with compact support in \mathcal U. This shows that its \mu-mass is zero. Taking the countable union over r,t\in\mathcal D and then using continuity of d excludes every real triple: strict inequalities at a real center and spacing persist at nearby dyadic ones.

We record explicitly the weighted consequence of the bridge estimate needed for the second exclusion. Fix a bridge [u,v] with dyadic endpoints and length r=v-u>0. Fix a power-of-two interval I in the allowed window about one endpoint and another such interval I' in the allowed window about the other endpoint. The two endpoints can be interchanged. Put

and fix a threshold \eta>0. These are fixed bounded Lipschitz functions of finitely many shortfalls, and W\ge0. All their endpoints align with the integer times at every scale in the sum once K is large. By Equation (6.17) and Markov’s inequality,

Applying Theorem 5.1 at the integer bridge length rn, summing, and using Lemma 6.5, we obtain

Indeed the first bracketed term is bounded, and the second tends to zero: the exponential sum is at most 2M^{-\sigma r n_0} for all sufficiently large K, whereas 1/\epsilon=o(n_0). All constants and thresholds in this application are fixed before K tends to infinity. The lower bounds on M and on bridge length in Theorem 5.1 therefore hold eventually; no uniformity over varying test intervals is being assumed.

Suppose now that Equation (7.2) fails for some s\in P_d, t\in Q_d. Necessarily s\ne t. Choose nearby distinct dyadic bridge endpoints so that their distance still exceeds c_b times their temporal separation. The endpoints can be chosen close enough to s,t that sufficiently small neighborhoods of these two times fit in the respective bridge windows. Some test J(I)+V(I) with power-of-two length and dyadic endpoints contained in the first neighborhood is positive. To see this, choose such an interval containing s in its interior. If the tests for it and all its dyadic subdivisions vanished, successive subdivision, equal half speeds, and equality in the triangle inequality would make d affine on that interval, contrary to s\in P_d. In the second neighborhood there is similarly a power-of-two interval I' with h(I')>0. In fact exact unit distance between the endpoints of any interval forces exact unit distance on every subinterval, by the triangle inequality and d(x,y)\le|x-y|; hence an interval containing t in its interior cannot have zero shortfall.

Choose a positive rational \eta<h(I'). The resulting strict bridge and shortfall conditions define an open set. By Equation (7.3), its W\mu-mass is zero, using compactly supported continuous tests in \mathcal U as above. Its part with W>0 therefore has zero \mu-mass. There are only countably many choices of the bridge, the two tests, their orientation, and the rational threshold. Every violation is covered by one of these choices, proving Equation (7.2) simultaneously for all real s,t.

Both properties are invariant under every translation and positive dilation of time, including the corresponding division of distances by the dilation factor. The measure inequality (S^{-1})_*\nu\le\mu from Proposition 6.8 consequently transfers them to \nu. All the other component measures are dominated by \mu, so they inherit them as well. This also provides a measurable full-measure set: the exceptional sets were covered by countably many measurable strict tests before extension to real arguments.□

Proof. We first show that a connected component U of \R\setminus P_d is affine with a single speed throughout. Its local speeds agree on overlapping neighborhoods, and hence are a common w\in[0,1] by connectedness. A compact subinterval of U has a sufficiently fine partition x_0<\cdots<x_m such that each consecutive triple lies in a local affine neighborhood. When w=0, the triangle inequality immediately gives d(x_0,x_m)=0. When w>0, propagate additivity inductively. Suppose it holds from x_0 through x_j, and use the four-point inequality of Equation (6.22) at base v=x_j in its limiting form

The inductive additivity gives B_v(x_0,x_{j-1})=2d(x_{j-1},x_j)>0, whereas local additivity gives B_v(x_{j-1},x_{j+1})=0. Apply Equation (7.4) to the latter pair with intermediate point x_0. Nonnegativity of B_v forces B_v(x_0,x_{j+1})=0, exactly the next additivity statement. The first two steps are already locally affine. Thus d(x_0,x_m)=w(x_m-x_0), and varying the compact subinterval proves the claim. In particular P_d\ne\varnothing for a nonaffine array.

Suppose a component U has speed w>c_b. It has at least one finite endpoint s\in P_d, since U=\R would imply affinity. If w<1, every interior point t\in U belongs to Q_d. Continuity at the endpoint gives d(s,t)=w|t-s|, contradicting Equation (7.2). Therefore w=1. Such a component cannot have two finite endpoints: they would both belong to P_d\subset Q_d, and their unit mutual speed would again contradict Equation (7.2). Hence every component of speed greater than c_b is an exterior unit ray; there are at most two.

Let I be the closed complement of these rays. It is the whole line, a closed half-line, or a closed bounded interval. For x<y in I, if [x,y]\cap P_d is empty, the two points lie in a component of speed at most c_b. Otherwise let p and q be the first and last points of [x,y]\cap P_d. The portions from x to p and from q to y have speed at most c_b, by the component claim and continuity. The middle portion satisfies the same bound by Equation (7.2), because p,q\in P_d\subset Q_d. The triangle inequality gives d(x,y)\le c_b(y-x). The only case not yet settled is a bounded slow interval of length zero; Lemma 7.3 below excludes it.□

Proof. For the moment permit L=0. Write x_u=a-u, y_v=b+v, where u,v\ge0, and put D=d(a,b)\le L. Set

If s\le0, the absolute shortfall of [x_u,y_v] is at most L+D\le2L. Suppose s>0. The reverse triangle inequalities imply d(x_u,y_v)\ge u-D-v and d(x_u,y_v)\ge v-D-u, so that s\le\min\{u,v\}. At base a the unit rays and triangle inequalities give

Twice applying Equation (7.4) yields B_a(x_s,y_s)\ge2s-2D. As d(a,x_s)=s and d(a,y_s)\le D+s, it follows that

Use the temporal midpoint m=(a+b)/2 and spacing t_*=s+L/2. Its two wing distances obey

If L>0 and s>3L/\delta, these bounds are strictly greater than (1-\delta)t_* and Equation (7.7) is strictly less than \delta t_*. This is the forbidden triple of Equation (7.1). If L=0, every s>0 produces that same strict contradiction. Thus always s\le3L/\delta. A full crossing interval has absolute shortfall

An interval meeting only part of the slow interval has absolute shortfall at most twice the length of that part. For example, if x<a\le y\le b, then d(x,y)\ge(a-x)-d(a,y)\ge(a-x)-(y-a), giving loss at most 2(y-a). The other orientation is identical. An interval inside [a,b] has loss at most its length, hence at most L, and intervals in the unit rays have zero loss. This proves Equation (7.6) in every case. If L=0, these conclusions would give exact unit distance between every pair of times, an affine array. Nonaffinity therefore forces L>0.□

Proof. Stationarity makes the two half-interval terms in Equation (8.3) have equal expectation. Since h_2(0;g_n)=h_1(0;g_{2n}), summing over the dyadic scales gives exactly

The final term tends to zero by Equations (6.18) and (8.2). Thus these integrals converge to H_\Phi. We must show that this mass does not escape to \mathcal A under vague convergence.

The measures J\mu_K and V\mu_K have uniformly bounded total mass on the compact space X, and the signed measures G\mu_K have uniformly bounded total variation by Equation (8.4). Take simultaneous weak limits

On \mathcal U they agree with J\mu, V\mu, and G\mu respectively: multiply any compactly supported continuous test there by the indicated continuous weight. We show that \eta|_{\mathcal A}=0 by treating all three ranges of affine speeds. Write d_w(s,t)=w|s-t|.

Speeds 0\le w\le c_b. Both \Phi' and \Phi'' vanish at every shortfall 1-w in this compact speed range, including its boundary w=c_b. For each \rho>0 there is an open neighborhood U_\rho of \{d_w:0\le w\le c_b\} on which the Taylor argument for Equation (8.4) gives

Indeed all arguments between the three relevant shortfalls are then uniformly close to [1-c_b,1], where both derivatives vanish. For a continuous test supported in U_\rho, pass this domination to the weak limits. It implies |\eta||_{U_\rho}\le\rho(\lambda_J+\lambda_V)|_{U_\rho}. The latter positive measure is finite on X. Letting \rho decrease to zero shows that \eta has zero restriction to this compact set of affine arrays.

Speeds c_b<w<1. Choose a fixed positive dyadic R so large that the endpoint windows of the bridge [0,R] contain [0,2] and [R,R+1] respectively; for example require \kappa R>2. Put W=J+V and H=h([R,R+1]). For each positive rational \eta_0, the open set

has W\mu_K-mass tending to zero by Equation (7.3). The open-set inequality for weak convergence of positive measures gives (\lambda_J+\lambda_V)(U_{\eta_0})=0. Each d_w in the current speed range belongs to such a set by taking 0<\eta_0<1-w. This countable cover and Equation (8.4) show that \eta vanishes on all intermediate affine speeds. Notice that the threshold is fixed separately for each test; no threshold uniform as w\uparrow1 is needed.

The speed w=1. Near d_1, all three arguments in Equation (8.3) are in the linear part of \Phi, so G=\alpha J. It remains to prove \lambda_J(\{d_1\})=0. Fix an inverse power of two q\in(0,1). For the product over [0,2n], with center n, put

The quantities D/n and \ell/n are continuous tests of g_n, both zero at d_1. Thus the condition 2q-2D/n>\ell/n defines an open neighborhood U_q of d_1. Eventually s=qn is an integer at every scale in the sum. On U_q, the truncation estimate of Lemma 6.2 gives, with its fixed additive constant A,

The shorter-interval test telescopes exactly, by stationarity:

where dyadic monotonicity gives e_{qN}\le e_N=\epsilon. Also \epsilon^{-1}\sum_n n^{-1}=O((\epsilon n_0)^{-1})=o(1). Since J\ge0,

The open-set inequality now gives \lambda_J(U_q)\le q, so \lambda_J(\{d_1\})\le q. First take the fixed weak limit and then let q decrease through inverse powers of two to obtain zero. The local identity G=\alpha J transfers this conclusion to \eta.

These cases cover \mathcal A. Weak convergence on compact X preserves the total mass of the signed measures, so \eta(X)=H_\Phi. As \eta|_{\mathcal A}=0 and \eta|_{\mathcal U}=G\mu, this is Equation (8.6).□

Proof. Nonnegative compact cutoffs exhausting \mathcal U, the terminal prelimit identities, and monotone convergence imply

For each integer m\ge0, subdivision and the triangle inequality give

Dyadic translation invariance of \nu therefore implies \int h_{2^{-m}}(0)\,\dd\nu\le1. The same shortfalls inserted into \Phi have finite integrals, bounded by C_\Phi.

Set A_m=\int\Phi(h_{2^{-m}}(0))\,\dd\nu. Translation invariance at the shift 2^{-j} gives

Absolute convergence of the series implies convergence of A_m. Fatou’s lemma yields

If \nu=0, the series is zero and Equation (8.5) proves the strict conclusion. Otherwise every shape in its full-measure support has a slow interval with nonempty interior by Proposition 7.2 and Lemma 7.3. There is a dyadic z in that interior. For all sufficiently large m, the interval [z,z+2^{-m}] is slow, so \Phi(h_{2^{-m}}(z))=1 by Equation (8.1).

The measurable sets

cover this full-measure support. At least one has positive \nu-measure, and translation invariance gives \nu(E_0)=\nu(E_z)>0. Consequently the subtracted integral in Equation (8.10) is positive. It is finite by Fatou’s lemma and the bound A_m\le C_\Phi. This proves Equation (8.8) in all cases.□

Proof. On an entirely slow shape, each shortfall in Equation (8.3) is at least 1-c_b, so all three capped values equal one and G=0.

Consider next shapes with exactly one unit ray. First take the orientation with a left unit ray and finite interface a. For a bounded interval [A,B] of possible interface locations, choose one fixed dyadic secant strictly to the left of A and a second strictly to the right of B. Their speed ratios on every such shape are respectively 1 and at most c_b. The arrays with these two prescribed closed conditions form a compact subset of X_0 disjoint from \mathcal A: an affine array would have the same speed on both secants. Local finiteness of \mu_\infty on \mathcal U therefore implies finite mass for interfaces in [A,B].

Push the restriction of \mu_\infty to this shape class forward by a, obtaining a measure \lambda on \R finite on bounded intervals. Its invariances, from Equation (8.13), give

The first equality uses S-invariance; the last uses translation by one. Finiteness forces \lambda([0,1))=0, and integer translates covering \R show that this entire shape class has zero mass. The same argument with left and right reversed treats the other one-ray class. Half-open intervals make the argument valid without any advance assumption about endpoint atoms.

It remains to consider the two-ray class. Restrict to the length band

This measure is invariant under dyadic translations. Its left endpoint has finite intensity on bounded intervals, as follows explicitly. Partition any bounded range of possible a into finitely many bins [k/4,(k+1)/4]. In each bin the fixed secant [(k+1)/4,(k+2)/4] lies in [a,b], since L\ge1, and hence has ratio at most c_b. A fixed dyadic secant to the left of the whole bounded range has ratio one. The two conditions again put each bin’s arrays inside a compact subset of \mathcal U. A finite union gives the claimed finite intensity. The right endpoint also has finite intensity on bounded intervals because 1\le b-a<2 bounds a whenever it bounds b.

Neither endpoint intensity has an atom. If the mass at some point were positive, dyadic translation invariance would give the same positive mass to infinitely many distinct points in a bounded interval, contradicting local finiteness. Define the finite number

Every integer interval of length one has this same endpoint mass.

For j\in\Z, let B_j be the two-ray band 2^j\le L<2^{j+1}. Since S divides lengths by two and preserves \mu_\infty,

The bands are disjoint and cover the two-ray class. Integrability of |G| permits reindexing j to -j, yielding

Put F_j=\int\Phi(h_{2^j}(0))\,\dd\mu_*. These quantities are finite. Indeed Lemma 7.3 makes the integrand zero unless [0,2^j] meets [a,b], which requires -2\le a\le2^j; the endpoint intensity on this bounded interval is finite. Translation invariance at the dyadic shift 2^j gives

The absolute convergence in Equation (8.15) implies that F_j has finite limits both as j\to-\infty and as j\to+\infty.

For the fine-scale limit, use the dominating integrable function \one_{\{-2\le a\le1\}} for j\le0. Off the null sets \{a=0\} and \{b=0\}, if a<0<b, then [0,2^j] is contained in the slow interval for all sufficiently negative j, and the capped shortfall equals one. If zero lies outside the slow interval, the small interval is eventually in a unit ray and its shortfall is zero. Dominated convergence therefore gives

The inequality uses L\ge1; possible equality endpoints have zero mass.

For the coarse-scale limit, Equation (7.6) gives 0\le h_t(0)\le2C_*/t throughout this band. Once t is sufficiently large, all these shortfalls lie in the linear region of Equation (8.1), uniformly in the shape. Hence

For the large dyadic integers t=2^j, atomlessness and integer translation invariance make the last mass (t+2)I_0 (the weaker bound (t+3)I_0 would also suffice). Thus

Summing Equation (8.16) and applying Equations (8.17) and (8.18) now proves

This argument also covers I_0=0; in that case integer translates of \{0\le a<1\} show that \mu_*=0. Together with the entirely slow and one-ray cases, this establishes Equation (8.14).□

Proof. Under the supposed zero-exponent sequence, all the preceding measures and identities were constructed. Combining Equations (8.7), (8.8) and (8.14) gives

contradicting the exact balance Equation (8.6).

For every parameter, the top Lyapunov exponent exists almost everywhere and is nonnegative, since the derivative cocycle has determinant one. If positivity on a set of positive area failed at arbitrarily large positive parameters, the exponent would therefore be zero almost everywhere at each member of a sequence tending to +\infty, which has just been excluded. There is consequently a finite k_0 such that, for every k>k_0, the top exponent is positive on a set of positive normalized area. Increasing the threshold gives the same statement for every k\ge k_0. The entropy formula Equation (1.2) gives h_m(f_k)>0 for every such k. In particular the measurable parameter interval [k_0+1,k_0+2] has positive one-dimensional Lebesgue measure and satisfies the required assertion.□