Nonuniqueness with local conservation for the hard-sphere Boltzmann equation

Proof. Put g=p-\xi. Reflection sends g to g-2(g\cdot\omega)\omega. On either hemisphere of the normal sphere, the change from the polar angle of \omega to the reflected direction shows that |g\cdot\omega|\,\dd\omega pushes forward to (|g|/4)\,\dd\sigma. The two hemispheres therefore give (|g|/2)\,\dd\sigma, where \sigma\in S^2 is the outgoing relative direction. Equivalently, for every nonnegative test function \psi, the one-output measure is

Its mass is 2\pi|p-\xi|. The affine map in this integral multiplies surface area by |p-\xi|^2/4, so the density on its image sphere is 2/|p-\xi|.

The same sphere is the zero set of f(v)=(v-p)\cdot(\xi-v). On that sphere, |\nabla_v f(v)|=|p-\xi|. The coarea formula consequently identifies the measure in (3.2) with 2\delta(f(v))\,\dd v. Integration against the two incoming densities and Tonelli’s theorem give (3.1). The usual pre/post-collisional change of variables identifies this incoming-pair expression with the gain in (1.1). Coincident incoming velocities have zero collision rate. The same identities can first be integrated against a velocity measure c(\dd p), proving the asserted extension.□

Proof. Set r=|\bar v|, n=\bar v-\bar p, and e_n=n/|n|. Except on a set of transverse measure zero, n\ne0. For fixed p_3,\xi_3, the delta in the transverse variables is line measure divided by |n|, on a line whose signed distance from zero is

The regular part of its integral over |\xi-v|\leq R is at most 2R/|n|. The singular part is bounded by its integral over the full line, namely

where a_q=\int_{\R}(1+s^2)^{-q/2}\,\dd s<\infty because q>1. Infinite values on a zero-distance slice are harmless in the nonnegative iterated integral.

For every interval J\subset[-R,R] of length at most \ell,

If 2R|B|\leq|A|, the integrand is at most C_q|A|^{1-q}, and the assertion follows from \ell\leq2R. Otherwise, integration about the possible root gives an upper bound C_q|B|^{1-q}\ell^{2-q}; since 1-q<0 and |B|>|A|/(2R), this proves (3.6). When A=0 the displayed upper bound is understood in the extended sense.

We next prove

It suffices to consider a disk of radius d. If d\leq r/2, then |n|\geq r/2 and e_n\cdot e_{\bar v}\geq1/3, which proves the bound. If d>r/2, use polar coordinates for n. Its range lies in |n|<r+d<3d. The factor |n|^{-1} cancels the radial Jacobian, while

because q<2. Division by the disk area gives C_q/d\leq2C_q/r. The point mass at zero satisfies the same estimate directly. Finally, the Gaussian is a probability mixture of these disks: for d>0 its mixture density in the disk radius a is

whose integral is one. This proves the Gaussian assertion as well.

Integrating (3.5) in \xi_3, applying (3.6), and then (3.7), bounds the expression in (3.4) by

Both fractions are bounded for q>1. These bounds are independent of p_3, so integration against the arbitrary probability \rho preserves them. Taking the entire axial range of length 2R also gives (3.3).□

Proof. By positivity it suffices to treat an elementary sheet. Put k=\xi-v and w=p+k. On the collision plane,

The same identity implies \langle v\rangle\leq\langle p\rangle\langle\xi\rangle. Since \langle p\rangle^m\leq C_m e^{|p|^2}, the kernel is bounded by

The tilted cold measure retains a product majorant. For 0<d\leq1/8,

This follows by dividing the two Gaussian densities: their ratio is 4\exp[-(3/(4d^2)-2)|\bar p|^2], and 3/(4d^2)-2\geq46. The assertion at d=0 has the evident interpretation. The axial tilted measure \widetilde\rho(\dd a)=e^{2a^2}\rho(\dd a) has mass at most L, and satisfies

Consequently e^{2|p|^2}c(\dd p) is bounded by 4L times a product of a centered transverse Gaussian and an axial probability. Moreover, its restriction to |p|>u is bounded by CLe^{-cu^2} times a sum of at most two such product probabilities. To see the last assertion, split into |\bar p|>u/\sqrt2 and |p_3|>u/\sqrt2, and apply (3.14) and (3.15), respectively. The axial probabilities obtained by normalization may have large support; Lemma 3.2 permits this.

On the shell in (3.11), either |p|>R/4 or |w|>R/2, because |p|+|k|\leq2|p|+|w|. In the first case the preceding tilted-measure tail contributes CLe^{-cR^2}. In the second case e^{-|w|^2}\leq e^{-R^2/4}. In either case the remaining integral has |k|\leq2R and is bounded by C_qR^2 using Lemma 3.2. This proves (3.11). The base region |p|+|k|\leq2 is bounded by the same geometric lemma and the total tilted mass. Summing the dyadic shells proves (3.10) and also the finiteness of (3.9).□

Proof. The dyadic estimate (3.11) proves (3.16). Write k=\xi-v. We may consequently work on |p|+|k|\leq R and remove that restriction at the end of each argument. On this region the exponential and polynomial factors in (3.12) are bounded by a constant depending on R.

For (3.17), if |\bar v|>2R+1, then |\bar v-\bar p|\geq|\bar v|/2 and |\bar\xi|\geq|\bar v|-R. The transverse line estimate, followed by the axial integration over a length at most 2R, gives C_R/|\bar v|. If |v_3|>2R+1, integrate the delta in \xi_3 instead. Its divisor is |v_3-p_3|\geq|v_3|/2, and

For completeness, if |a|\leq2R the disk is contained in the disk of radius 3R about zero; if |a|>2R, then |\zeta|>|a|/2>R on the disk, giving the same bound. We obtain C_R/|v_3|. At least one of the two outgoing components becomes large with |v|, proving (3.17).□

Proof. The Gaussian majorant is a product measure. Its transverse image is \mathcal G_d, and its axial image is the probability \rho=\Phi_*(\pi^{-1/2}e^{-a^2}\,\dd a). There is no need to differentiate or invert \Phi. From (3.19),

Even under the wider axial probability (2\pi)^{-1/2}e^{-a^2/2}\,\dd a, the integral of the right side is

We may therefore take L_A=2e^{16A^2} for both the original and the wider parameter laws. The original image has the required sheet domination.

On the parameter tail,

After normalization the right side is 2^{3/2}e^{-B^2/2} times a Gaussian probability with parameter widths enlarged by \sqrt2. Its transverse velocity width is at most \sqrt2/16<1/8, and its axial image has the exponential moment just proved. Lemma 3.4 applied to this dominating image gives (3.20).

For a probability mixture, mix the dominating sheets just constructed; this is again a member of \mathcal S_{L_A}. If the mixing measure has mass B_0>0, normalize it by B_0 and retain that factor in the domination amplitude. If B_0=0, the mixed image is zero. The same argument applied to the wider Gaussian majorants of the component tails proves the asserted factor in (3.20). Thus the estimates are uniform over families of mixtures whose masses B_0 have a common finite bound independent of the tail cutoff. Masks taking values in [0,1] only decrease the measures, whether imposed before or after the pushforward.□

Proof. The integrand is symmetric in p,\xi. On the collision plane, |v|^2\leq|p|^2+|\xi|^2, so at least one incoming speed is at least |v|/2. It suffices, at a factor of two, to retain |\xi|\geq|v|/2. Formula (3.12) now gives

because \langle v\rangle^m/\langle\xi\rangle^m\leq2^m.

Again write k=\xi-v and w=p+k. On R<|p|+|k|\leq2R, R\geq1, either |p|>R/4 or |w|>R/2. Consequently

In the second alternative this uses the finite supremum of s^m e^{-s^2/2}; in the first it follows directly from the polynomial factor.

The measure W(p)\ind_{|p|\leq2R}\,\dd p is dominated by

Its transverse factor is a positive mixture of centered disks. Indeed, for D>0,

almost everywhere. Upon normalizing the disks, the mixture has total mass \pi D^2+2\pi D^{2-q}/(2-q). The axial factor has mass 4R. For D=2R the resulting product mixture therefore has total mass at most C_qR^3.

Discard e^{-|w|^2/2} and the incoming-speed restriction in (3.23). Apply Lemma 3.2 to each normalized product in this mixture, with |k|\leq2R. In conjunction with (3.24), this yields the shell estimate

The notation on the left denotes the restricted nonnegative integral from (3.23), including its fixed symmetry factor. The base region is bounded by the same disk-mixture argument with R=1. Summation over dyadic R proves the result for m>5.□

Proof. The case B=0 is immediate. Assume B>0. We first establish

As in the proof of Lemma 3.7, symmetry reduces the integral to |\xi|\geq|v|/2, up to a factor of two. The shell estimate (3.25) then removes |p|+|\xi-v|>R with an error tending to zero uniformly in v as R\to\infty. On the retained region, the right side of (3.23) is bounded by a constant times

If |\bar v|>2R+1, the transverse divisor is at least |\bar v|/2, and W(\xi) is bounded on the retained region. The transverse line integral and the axial interval of length 2R give an upper bound C_R/|\bar v| after integration in p. If |v_3|>2R+1, integrating the delta in \xi_3 instead and using (3.18) gives C_R/|v_3|. Here \int_{|p|\leq R}W(p)\,\dd p<\infty, since q<2. At least one output component tends to infinity with |v|. First letting |v|\to\infty and then R\to\infty proves (3.27).

Choose M so large that BP(H,H)\leq\varepsilon H on |v|>M. It remains to prove the desired estimate for |v|\leq M. On this output ball, the part of the normalized quadratic gain with |p|+|\xi|>L tends uniformly to zero as L\to\infty. Indeed, this deletion is symmetric in the two inputs. After the same reduction to |\xi|\geq|v|/2, it implies

so (3.25) bounds its cost by a summable tail. The condition j\leq BH transfers this estimate to P(H,j).

Next retain both inputs in |p|,|\xi|\leq L. The Gaussian and polynomial factors in the normalized quadratic integrand are then bounded by a constant depending on M,L, leaving

For 0<\delta<1, the restriction |\bar p|<\delta is dominated in its p variable by the product measure

Its transverse factor is a positive mixture of centered disks, by the layer-cake formula used in Lemma 3.7, and its total mass is

After normalization, (3.3), with radius at least L+M, therefore bounds the cost of this restriction by C_{M,L}(\delta^2+\delta^{2-q}). Symmetry gives the same bound for |\bar\xi|<\delta. These estimates are uniform even when the output velocity approaches its axis. Again j\leq BH transfers them to the mixed gain.

We have proved that, for a suitable compact set

the truncations H_K=H\ind_K and j_K=j\ind_K satisfy

uniformly over 0\leq j\leq BH. The function H_K is bounded and supported in the ball of radius L. By symmetry of P, put j_K outside the plane integral in (3.1). The area of a plane section of that ball is at most \pi L^2, so

For 0<\rho<1, the part with |v-p|<\rho is at most C_{B,L,\delta}\rho^2, because j_K is bounded. The complement is at most C_{L,\delta}\rho^{-1}\int j. Finally, H has a positive lower bound on |v|\leq M off its null axis. Choose \rho small enough that the near part is at most \varepsilon H/2 there. Absorbing the other fixed factors into C_{B,\varepsilon} proves (3.26). The cutoff assertion follows from P^b\leq P on nonnegative inputs.□

Proof. An original integer vector (k,m,\ell) has transverse coordinates ((m-\sqrt2 k)/\sqrt3,\ell). If \ell\ne0, its transverse length is at least one. Otherwise (k,m)\ne(0,0), and

The numerator is a nonzero integer. Since rotation preserves length, this proves the claim.□

Proof. Integration over one period of the lattice sum equals integration of each unshifted prototype over \mathbb R^3_x. The transverse area is O(s^2), while

The velocity factor has integral one and support in \{|v|<2\}. Thus every fixed velocity moment of a sign prototype is O(s^2). On its positive set the logarithm of the density splits into

The first term is O(1+|\log s|), and \int S|\log S|<\infty, \int J|\log J|<\infty. Its absolute entropy is therefore O(s^2(1+|\log s|)). The corresponding series over the geometric scales converges. Disjointness proves the asserted integrability and convergence for the cold sum. The background is bounded by \mu, and its velocities satisfy |v|\le R_\mu, giving the datum claims.

For a seed b=h^0_{N,T,\lambda}, scaling x gives

These constants are uniform because \lambda m_T stays in a fixed bounded interval and r|\log r| is bounded on such an interval. The preexisting truncated density f=u^0+\sum_{j\le N}c_i^0 satisfies f\le C s_N^{-31}. For f,b\ge0, integration of 1+\log(f+r) over 0<r<b gives

Indeed, the positive logarithm is bounded by its terminal value, while \int_0^b\log^-(f+r)\dd r\le b(1+|\log b|). The entropy change caused by the seed is consequently O(s_N^4(1+|\log s_N|)). Its reference terms are controlled by its mass and energy. This proves all convergence assertions.

Removing the hole and the velocity tail from \mu costs precisely the removed mass in \mathcal H_\mu. The inserted jets have bounded velocities, so their contribution is bounded by the preceding absolute entropy, mass, and energy sums, which tend to zero with s_0. The same is true uniformly for the seeds. The stated order of choices therefore makes the initial relative entropy arbitrarily small.□

Proof. The gain estimate P(H,H)\le CH and

make the gain and loss locally Lipschitz in the weighted norm. For the colored system use the norm of the pointwise sum of the absolute values of the components, and the sum of absolute differences for two solutions. All quadratic estimates close in this norm. The initially disjoint lattice copies have finite total norm, and periodic covariance is preserved by uniqueness. Picard iteration in free coordinates gives the local solution; the loss integrating factor and positive gain iteration give nonnegativity. Repetition on short subintervals gives continuation and the difference estimate. These arguments use the integral equation along flights and require no norm continuity of spatial translation for measurable data. On an interval with bounded norm, the increments along flights are Lipschitz in that norm. Since free transport is isometric, each component’s norm is continuous. Transfers only regroup finitely many scales and preserve the total density.

The weight H has every polynomial moment and belongs to L^{1+\epsilon}(\mathbb R^3) for some \epsilon>0. Hence the cutoff equation has integrable gain and loss with all polynomial weights on every bounded interval under consideration. Integrating its path formula and using particle interchange and the elastic substitution yields conservation of the collision invariants. The same calculation with a smooth spatial test gives each local balance, including its initial trace.

Here are details of the entropy calculation at a fixed approximation. Put g(v)=e^{-2|v|^2} and F_\epsilon=F+\epsilon g. Differentiate F_\epsilon\log(F_\epsilon/\rho)-F_\epsilon+\rho along flights; its derivative is Q^b(F,F)\log(F_\epsilon/\rho). Replacing Q^b(F,F) by Q^b(F_\epsilon,F_\epsilon) changes its integral by a quantity tending to zero. Indeed, the operator difference is O(\epsilon)H on the fixed interval, while

The right-hand side is integrable against H, and the resulting error is O(\epsilon(1+|\log\epsilon|)). For the shifted density the usual symmetric calculation is now justified by this integrability. The logarithm of either Maxwellian reference is a collision invariant. Symmetrization therefore gives minus the cutoff dissipation. The regularized entropies converge by dominated convergence using the same bounds, and Fatou’s lemma applies to the nonnegative dissipation. This proves (4.10). The integrated path formula includes every time section, so the conservation and entropy statements hold at every time, also when colors are transferred.□

Proof. Within a fixed copy, the radial margins are larger than the corresponding enlargement radii. Adjacent radial cores have gap c_\eta(s_j^2+s_{j+1}^2), whereas the sum of their enlargement radii is one quarter of that gap. Opposite stripe patterns at a single scale have gap s_j^2/4, larger than the sum c_\eta s_j^2/2 of their enlargement radii. This proves the assertions for one copy.

For distinct copies let S_* =\max\{s_i,s_{i'}\}. An intersection would force

By (4.2), the first quantity is at least c/(1+|n-n'|)\ge c' S_*^{1/2}. For sufficiently small s_0 this contradicts 4S_*. The choice is uniform over all pairs of scales and all lattice copies.□

Proof. Put A=(a_s^\sigma)'(\tau)\ge0. Differentiation of (5.2) gives

Thus \D|\xi|^2=-2U_z\xi_3^2. The distance functions are locally Lipschitz along every flight, and

With r=(D_y,D_z), differentiation of the logarithm of the barrier therefore yields

The last two terms are at least -|\D r|^2/(8d_s^2L), by completing a square. Since |\D r|\le|\xi| and 2-t/L\ge1, this proves the derivative estimate. The large axial derivative U_z has a favorable sign.

For a flight starting in the core at x^0, write v=a_s^\sigma(z_c^0)e_3+d_sw, |w|\le1. At x=x^0+tv, the increasing function p\mapsto p-a_s^\sigma(z_c-tp) has derivative at least one, value d_sw_3 at p=v_3, and value zero at p=U. Consequently

The distance exponent in (5.4) is at most t/(8L), and its velocity exponent is at most two. A fixed C_m, depending only on \|J\|_\infty, thus gives f_i\le m_i. Since L_sd_s=d_0s^{19/2}\ll s^2, the transverse displacement is less than the enlargement radius. The axial speed is below two, so the flight also stays inside the axial interval of (5.1).

At fixed (t,x), the majorant is a scalar multiple of a Maxwellian centered at Ue_3. Detailed balance gives P(m_i,m_i)=m_i\nu_{m_i}. Its mass is at most CK/s, and its first moment about Ue_3 is at most CKd_s/s, uniformly for t\le L_s. The inequality |v-p|\le|\xi|+|p-Ue_3| gives (5.6).

Finally, along a flight starting on the closed initial spatial support, both distance functions vanish initially and are locally Lipschitz, so their squared-distance-over-time terms tend to zero. Away from that support the barrier tends to zero. These observations give its path trace and allow the differential inequalities to be integrated from zero. Boundary conventions for the masks change only null sets; no spatial indicator is differentiated.□

Proof. Reduce s_0 so that \epsilon_s\le1 at every scale. Then \mathcal U_i\le2m_i, and Lemma 5.2 gives

Since 1+z^2\ge c(1+z), a fixed large C_* makes \D\mathcal U_i\ge P^b(\mathcal U_i,\mathcal U_i). Dropping the nonnegative loss in the equation for c_i permits positive gain comparison from the initial trace.

For completeness, this comparison is valid in the measurable solution class of Proposition 4.3. At a fixed approximation m_i\le C_sH throughout the interval, and z_i=(c_i-\mathcal U_i)_+ has a pathwise upper integral comparison from zero with source 2P^b(\mathcal U_i,z_i)+P^b(z_i,z_i). The bilinear H-norm estimate and Gronwall’s inequality give \|z_i\|_H=0. One may first integrate from a positive time and then use the trace established in Lemma 5.2.

For the second assertion, e^{-\kappa(t-t_1)}\mathcal U_i is a supersolution for \D+\kappa: its linear derivative has factor e^{-\kappa(t-t_1)}, whereas its quadratic gain has the smaller factor e^{-2\kappa(t-t_1)}. Its value at t_1 dominates c_i(t_1). The same comparison proves (5.8).□

Proof. The transverse Jacobian is d_s^2, while differentiation of the implicit equation for p_3 gives

Multiplying by the density in (4.4) proves (5.9). Its coefficient is at most C/(s+t), since 0\le S\le1 and \alpha\gamma>0. A fixed product Gaussian dominates J. Moreover, the monotonicity argument in Lemma 5.2 gives |p_3-U|\le d_s|w_3|. The Gaussian parameter estimate (Lemma 3.6) therefore supplies the asserted sheet domination, with constants independent of the axial compression.

The remainder is a Gaussian centered at Ue_3, of width comparable to d_s, whose mass is at most C\epsilon_s K/s=C L_sd_s/s^2. Its transverse Gaussian width is within the sheet class and its axial exponential moment is uniformly bounded because |U|\le1+\alpha. Since L_sd_s/s^2=d_0s^{15/2} and t\le s^{-1/2}, this also gives the last inequality of (5.10). Uniform Gaussian moments and tails, together with (5.7), prove (5.11).□

Proof. Free cores vanish off their enlargements. For a nonmain label either D_y\ge(c_\eta/4)s^2, or |z_c|>a+2t+b_{\mathrm{sp}}, in which case D_z\ge b_{\mathrm{sp}} because |U|<2. Completing the velocity square in the remainder gives

Here the Gaussian rate is at least eight once the widths are fixed small, and its centers stay bounded. Half the distance exponential is at most \exp[-c/(s^{16}L_st)]. Since t\le L_s, this is bounded by e^{-c/s^{15}}, which absorbs every polynomial in 1/s. If also t\le1, it absorbs arbitrary powers of t and s.

It remains to sum the other half over lattice copies. Uniformly,

The Gaussian distance scale d_s\sqrt{L_st} is bounded, and the central spatial region has diameter O(1+L_s). The lattice has unit separation, hence at most C(R+1)^3 points in a ball of radius R. Summing in successive shells gives at worst a fixed polynomial loss in 1/s. The extracted exponential absorbs this loss and the factor s^{-31}. Finally sum over the geometric scales. This proves both claims, including their stated uniformity.□

Proof. For (5.13), use f_i\le m_i and the same Gaussian lattice sum as in Lemma 5.5, without extracting a small distance factor. The prefactor and the number of central copies are polynomial in 1/s. Completing the velocity square bounds the remaining velocity factor by Ce^{-4|v|^2}\le CH(v). For (5.14), the transverse area at scale s_j is O(s_j^2), and its axial length while active is O(1+L_{s_j}). Projection onto the torus does not increase volume. Summation gives C\sum_j(s_j^2+s_j^{3/2})\le Cs_0^{3/2}.□

Proof. The distance D_0 is 1-Lipschitz. Fix 4<\beta'<\beta_0 and a positive absolute constant c, and consider

At almost every point of a free path,

The last inequality is completion of the square in D_0/t. At fixed (t,x), B is a Maxwellian in velocity. Energy conservation in a collision therefore gives P(B,B)=B\nu_B, and, as long as \beta_0-C_1t\ge\beta'>4 and e^{C_1t}\le2,

Choose C_1 large enough that the preceding lower bound for \D\log B dominates C(1+|v|), and then choose t_1\le1/32 small enough to maintain both restrictions. These choices use only \beta_0 and the collision normalization.

The trace of B along a path starting outside the hole dominates u^0. Indeed, if D_0(x)=0, then D_0(x+tv)\le t|v|, so the distance factor tends to one. If D_0(x)>0, its trace is zero, as is u^0(x,v). The differential inequality can be integrated from a positive time and then passed to this trace. The positive-gain comparison used for Proposition 5.3, with the nonnegative loss omitted, yields u\le B. This uses only absolute continuity along paths at positive times and boundedness in the H norm; it does not differentiate the initial spatial mask. Since e^{C_1t} is bounded, (6.1) follows.

For t\le1/32, the axial interval defining \mathcal O_i(t) has half-length

Its transverse distance from the corresponding lifted axis is at most s_0. Thus every point of \mathcal O(t) lies within distance s_0 of the segment defining the hole. Since the hole is its open 2\delta neighborhood, the triangle inequality gives D_0\ge2\delta-s_0\ge\delta when s_0\le\delta.□

Proof. At each (t,x) there is at most one main color. Denote it by a, taking a=0 if no main color is present, and put r=c-a. The main-color measure bound in Lemma 5.4 and the sheet gain estimates, including Lemma 3.6, give

By (5.12), for each M>0,

uniformly for sufficiently small s_0. The bath bound gives u\le CH. The quadratic gain bound of Lemma 3.7 proves the first estimate in (6.3), absorbing the bounded terms since t+s_N\le2.

If a\ne0, (6.2) improves the bath estimate to u\le C e^{-c\delta^2/t}H at this position. The ordered cross-color sum is bounded by 2P(a,r)+P(r,r), so positivity yields

The exponential is bounded by any prescribed power of t, with a constant depending on that power and on \delta. Taking M=A proves the second estimate. When a=0, only the last two types of terms remain, and the same conclusion follows.□

Proof. For the dormant solution, the function Y(t)=C_dt^A is a supersolution of (6.4) if

because (t+s_N)^{-1}\le t^{-1}. At each fixed N, comparison begins at the regular initial section t=0, where both functions vanish. Choosing an initial interval on which C_dt^A<1/2 closes the condition Z\le1. The cold and bath bounds are finite in the H norm at each fixed approximation. Thus the continuation criterion of Proposition 4.3 also proves existence on that interval and gives (6.7).

Reduce T_0 until this interval contains [0,T_0] and C_dT_0^{A-p-1}<1. Then, for 0<t\le T\le T_0,

At t=0 the ratio is zero. Hence the dormant solution stays strictly below the cap. At the other endpoint, \lambda=L_0, the ratio at time zero equals L_0\|\chi\|_H>1.

Fix T,N, and the cutoff. Let I be the set of \lambda\in[0,L_0] whose solutions exist on [0,T] and remain strictly below the cap there. Continuous dependence in the local theory and continuation at the terminal section make I relatively open. It contains 0 and does not contain L_0. Let \lambda_* be the right endpoint of the connected component of I containing zero. Along a sequence \lambda_k\uparrow\lambda_* within that component, all hot parts are bounded by the cap. The cold and bath estimates bound the remaining colors uniformly in k in the fixed-approximation norm. The difference estimate in Proposition 4.3 therefore makes these solutions Cauchy, uniformly on [0,T] in free coordinates. Their limit is the solution with parameter \lambda_* and satisfies the non-strict cap inequality.

The norm Z is continuous in time, and the cap denominator is positive, so the displayed ratio has a maximum. If that maximum were less than one, continuous dependence and continuation would place a neighborhood of \lambda_* in I, contrary to its endpoint property. Its maximum is therefore one. Since 0 has a neighborhood in I, while L_0\|\chi\|_H>1 also excludes a neighborhood of L_0, one has 0<\lambda_*<L_0. A final reduction of T_0 makes the cap at most one on [0,T] for every T\le T_0.□

Proof. On [t_b,t_1], the nonmain bound (5.12) gives r\le\varepsilon(s_0)H, with \varepsilon(s_0)\to0. The main-color sheet amplitude is bounded by C/t_b. Consequently all terms of \mathcal S containing a nonmain factor are bounded by C_{t_b}\varepsilon(s_0)H, after enlarging \varepsilon if needed.

For the remaining main–bath term, define a deterministic majorant

Here the sum includes all scales, so it bounds every finite selection; at each position it has at most one nonzero summand. By Proposition 5.3 and (6.1),

The sheet gain estimates give \mathcal S_{\mathrm{mb}}\le C_{t_b}H. The uniform large-output estimate (3.17) gives

with essential suprema in the velocity variable. To handle velocities near the axis, fix q'\in(1,q) and write

Since U_{\mathrm{bath}}\le C_{q'}H_{q'}, the same sheet lemma with exponent q' implies \mathcal S_{\mathrm{mb}}\le C_{t_b,q'}H_{q'}. Hence, for 0<r\le1,

Both estimates include the Gaussian-parameter measures in Lemma 5.4, by Lemma 3.6.

It remains to consider the compact velocity region |v|\le R, |\bar v|\ge r. The spatial support of \mathcal S_{\mathrm{mb}} is contained, for every s\le t_1, in the projection of the fixed union

where B^2_{s_0} is the transverse disk of radius s_0. A lifted free path with |v|\le R and duration at most t_1 can meet only C_R of these cylinders, uniformly in its endpoint: their centers must lie in a fixed bounded enlargement of a segment of length at most Rt_1, and the lattice has a fixed separation. For any one cylinder, its transverse projection is crossed in time at most 2s_0/|\bar v|\le2s_0/r. Thus the normalized integral of \mathcal S_{\mathrm{mb}} along such a path is bounded by C_{t_b,R}s_0/r. Combining the three velocity regions gives

First take R large, then r small, and finally s_0 small. This proves (6.8).

All estimates above concern the common pointwise majorant (6.9), rather than values chosen separately on individual paths. To pass an almost-everywhere inequality to path integrals, use free coordinates (s,x_0,v)\mapsto(s,x_0+sv,v). This map preserves measure, so Fubini’s theorem gives the inequality at almost every time on almost every free path. Integration on those paths then defines representatives at every terminal section. Since free transport preserves \|\cdot\|_H, the resulting bound is precisely the essential norm in (6.8) at each fixed terminal time. The construction applies on any existing solution interval, so this lemma does not presuppose continuation to t_1.□

Proof. We choose an intermediate time t_b\in(0,t_1) and then require T_0<t_b. At the cap horizon, both the selected and dormant solutions satisfy

using only T,s_N\le1; no relation between T and s_N is needed. For T\le t\le t_b, (6.4) and (s+s_N)^{-1}\le s^{-1} give, while Z\le1,

Indeed multiplication of the scalar comparison by t^{-C_2} integrates the source to at most C_{A,\delta}t^{A-C_2}/(A-C_2).

From t_b onward, retain the source inside the path integral rather than using its pointwise bound. The same hot gain estimates and Lemma 6.4 give

Consequently, as long as Z\le1,

where the supremum is restricted to the current interval of existence until continuation has been established.

The order of choices now matters. Since A>C_2, choose t_b so that

Since p>C_2, reduce T_0<t_b, within the range of Proposition 6.3, so that

Finally take \bar s_0 small enough for all preceding cold and bath estimates and for

The last choice is possible by Lemma 6.4. Equation (6.14) is then strictly below \zeta/2. The same first two choices make (6.13) strictly below \zeta/2 on [T,t_b], and the cap gives the same bound on [0,T].

These estimates close the condition Z\le1 with a strict margin. At each fixed approximation the cold sum and bath have bounded H norm through t_1, so the finite-evolution continuation criterion excludes a finite endpoint before t_1. This proves existence and (6.12) uniformly in all the stated indices. The dormant estimate was obtained independently on its initial interval and is preserved under this continuation.□

Proof. Contain V in a ball of radius R. The change of variables z=x-\tau v gives a factor \tau^{-3}. The image ball meets at most C(1+\tau R)^3 translates of a fixed bounded fundamental cell of \Lambda. Periodicity therefore gives

which proves (7.5). In particular, its constant does not grow when \tau tends to infinity.

For L\ge1 let

Then \delta_L\to0, and (7.5) implies, for \tau\ge\rho,

For 0\le\tau<\rho, the same integral is at most A_*M_H. Consequently the expression in (7.6) is at most

and that in (7.7) is at most

Choose first \rho small, then L large, and finally a_* small. Both upper bounds tend to zero in this order. The same proof works when the time integrals are restricted to an interval of existence.□

Proof. Fix two closed velocity balls V_1,V_2 of positive radius with \dist(V_1,V_2)>3, and put V=V_1\cup V_2. All constants until the choice of the entropy tolerance may depend on these fixed balls. Since b\ge1 and \mu is a Maxwellian,

At a fixed (t,x) write d=(\mu-F)_+. Positivity of the gain and F\ge\mu-d imply

By the plane formula (3.1), the Gaussian integral over each plane is bounded, so

For |p-v|<r, use d\le\mu\le1 to bound this integral by Cr^2; for |p-v|\ge r, it is at most Cr^{-1}D_F(t,x). Choose r>0 once so that the first contribution to 2P(\mu,d) is at most m_0/2. With m_*=m_0/2 and a fixed constant C, we obtain

The second inequality follows directly from \min\{b,|v-p|\}\le |v-p|\le C\langle p\rangle on V.

Let \Delta=t_1/2, fix t\ge t_1, and write X_\tau=x-(t-\tau)v. The Duhamel formula for the total density, whose value is continuous through retirements, gives

Using (7.12), e^{-a}\ge1-a for a\ge0, and the upper bound 1 for a survival factor, the integrand is bounded below by

This inequality remains valid when its right side is negative.

Integrate (7.13) over s and over either ball V_k. The positive constant contributes

Choose 0<\rho<\Delta/2. For t-\tau\ge\rho, (7.5) and (7.9) give

The term D_F(s,X_s) always has t-s\ge\Delta/2. Its integrated cost is therefore C_\rho\varepsilon_E. For the double time integral in (7.13), the portion with t-\tau\ge\rho has the same type of bound, whereas the remaining portion costs at most C(B)\rho by (7.10). Thus

After B has been fixed, choose \rho to make the last term at most \min(a_1,a_2)/4, and then choose \varepsilon_E to make the middle term at most the same quantity. It follows that

For any velocity v, one of the balls has distance at least 1 from v: otherwise their mutual distance would be less than 2. On that ball \min\{b,|v-p|\}\ge1. Consequently

Only the entropy tolerance, and not this value of \kappa, depended on B.□

Proof. Choose the entropy tolerance small enough for Lemma 7.2, so that the damping constant \kappa and (7.16)–(7.18) are available. Define

Since w\le F and j\le BH,

Write C_\mu=\|\mu\|_H<\infty and A_B=2C_\mu+B. The inequality w\le\mu+j, bilinearity, and positivity give

The frequency deficit has the bound

Here \mu(v)\langle v\rangle\le CH(v).

Between retirements, subtracting the stationary Maxwellian equation from (7.2) gives

The chain rule for the positive part on almost every free flight therefore gives an upper comparison for j with damping \kappa. Using Lemma 3.8 in (7.22), and then (7.23), (7.16), and (7.17), shows that for each \epsilon>0 its positive source is at most H(v)R(s,x), where

Constants have been enlarged to absorb A_B. At a retirement time, the increase of j is at most the increase of w, because (a+d)_+-a_+\le d for d\ge0. Thus its increments are bounded by the same J_sH as in (7.18).

We also need to estimate the excess already present in the bath at t_1. Put j_u=(u-\mu)_+ and m_{j_u}=\int j_u\dd v for 0\le s\le t_1. The bath starts below \mu, so j_u(0)=0; Lemma 6.1 gives the fixed bound j_u\le B_0H. Since u\le F,

The bath equation is (\D+\nu_F^b)u=P^b(u,u). Subtracting the Maxwellian equation and applying the same two difference estimates as above gives, now discarding the nonnegative damping, a source bounded by H(v)R_u(s,x), where

The cold density appears only through the frequency deficit, which is controlled by D_-. No positive source involving a cold gain occurs in this bath equation.

At t_1, the hot bound gives j(t_1)\le j_u(t_1)+\zeta H. Continue the same backward free flight through t_1, and write X_s=x-(t-s)v. The two comparisons, together with all intervening retirements, imply

We have dropped damping from the initial and retirement contributions; doing so also allows the bath comparison to run along that one flight all the way back to zero.

We claim that the velocity integral of the right side times H can be made smaller than any prescribed number \tau>0, uniformly in (t,x) on the bootstrap interval. The constant terms contribute at most

For the remaining terms, all the spatial functions in question have a fixed pointwise upper bound once B is fixed: m_j\le BM_H, m_{j_u}\le B_0M_H, D_-\le\int\langle v\rangle\mu\dd v, and \ind_{\mathcal O(s)}\le1. Their spatial integrals are respectively bounded by \varepsilon_E, \varepsilon_E, \varepsilon_E, and Cs_0^{3/2}, the last by (5.14). Apply Lemma 7.1 to each term, with its coefficient in (7.24) or (7.26). For clarity, the choices are made in the following order. First choose \epsilon so that its contribution to (7.28) is less than \tau/4; all the coefficients C_{B,\epsilon} and C_{B_0,\epsilon} are then fixed. For each of the finitely many averaged terms, choose the time deletion length \rho and the truncation H_L in the proof of that lemma so that the endpoint and velocity-tail errors, including its coefficient, have total less than \tau/4. Next make \varepsilon_E and s_0 sufficiently small for the remaining averaged errors to total less than \tau/4. Finally take \zeta small and, if necessary, reduce s_0 further to make the other terms in (7.28) less than \tau/4. All these choices are uniform in the endpoint time. Thus

The same calculation on a pre-retirement section includes only the jumps that have already occurred, so it does not require a post-jump bootstrap hypothesis.

It remains to select the particular smallness threshold that controls the gain. Apply Lemma 3.8 once more, now with \epsilon'=1/(2A_B). Write its constant as C_{B,\epsilon'} and set

Choose the parameters in the preceding argument to give (7.29) for this \tau. Then

Equation (7.22) and P^b(\mu,\mu)\le P(\mu,\mu) prove (7.19). The latter has finite H norm because P(\mu,\mu)=\mu\nu_\mu is a Gaussian times a function of at most linear growth.□

Proof. On a bootstrap interval let Z_w(t)=\|w(t)\|_H. The Duhamel formula for (7.2), with all retirement increments included, and (7.11), (7.16), (7.17), and (7.19), give

Take s_0 small enough that \theta_0,J_0\le1 and discard the first exponential in the last integrand. We obtain

The integral version of Gronwall’s inequality therefore gives

The constants defining B_* were fixed before B.

Here is the complete order of parameter choices. Fix the kernel and jet parameters, the narrow width factor, the Maxwellian \mu, and the short time t_1 as in the preceding sections; also fix the early growth exponent p there. The bath estimate fixes B_0. The proof of Lemma 7.2 fixes \kappa from \mu, t_1, and the two chosen balls, without imposing its entropy tolerance yet. The cold estimates then fix C_3,C_4 uniformly for sufficiently small s_0, while C_5=\|P(\mu,\mu)\|_H+1 is already fixed. Choose

For this B, prescribe the entropy tolerance required by Lemma 7.2, and the possibly smaller entropy tolerance, hot tolerance \zeta, and upper bounds on s_0 required by Lemma 7.3. Include \theta_0,J_0\le1 among these requirements.

To realize the entropy tolerance, take the hole radius \delta small and the bath velocity cutoff R_\mu large. Their contributions to the initial relative entropy can then be made arbitrarily small. With these parameters fixed, the jet and seed contributions become arbitrarily small as s_0\to0, as proved in (4.12) and its derivation. To realize the prescribed \zeta, use Proposition 6.5: choose its auxiliary time t_b and horizon upper bound T_0 after \delta,\zeta, and then take s_0 sufficiently small for its source estimate. Decrease s_0 further to meet the entropy, separation, cold-gain, and retirement requirements already listed. Reducing T_0 when required by the early estimates preserves all later estimates. These are fixed construction parameters; none depends on N, on the chosen cap solution, or on its horizon T\in(0,T_0].

For completeness, the strict estimate closes the bootstrap even at retirement times. At t_1 it starts by (7.3). On an interval with no retirement, the characteristic equations make the weighted norm continuous, and (7.32) bounds it by B_*<B-3 as long as it is at most B. Thus it cannot first reach B there. At a retirement time L_s, apply the argument only up to the left-hand section. The cold comparison on that interval and (7.18) give

This starts the bootstrap on the next interval without having assumed its conclusion across the jump. The total density and its entropy bound are unchanged by this relabeling.

Finally, at each fixed approximation only finitely many scales occur. The cold barriers bound the sum of the active colors in H norm on every finite interval, with constants allowed to depend on that approximation. The uniform bound on w therefore bounds the total H norm on every such interval. The cutoff local existence and continuation argument from the construction excludes a finite maximal existence time. There are only finitely many retirements at that approximation, after which F=w. This proves global existence and (7.30), with all the stated auxiliary estimates.□

Proof. Localize to a compact time-space set and a compact velocity neighborhood of \supp\psi, using smooth cutoffs. Denote the localized functions by q_n, extend them by zero, and set r_n=(1+\partial_t+v\cdot\nabla_x)q_n. The equation shows that r_n is uniformly integrable in L^1 and has support in a fixed bounded set of time and velocity, and also of space in the Euclidean case. Spatial cutoff derivatives cause no difficulty because velocity has been localized. Integration from a time preceding the support gives

The damped inverse R is an L^1 contraction, since free transport preserves Lebesgue measure.

Truncate r_n at height L, preserving its sign, to obtain r_n^{(L)}. Uniform integrability implies

The velocity averages of R(r_n-r_n^{(L)}) are therefore uniformly small in L^1. It remains to prove compactness for a bounded L^2 family of right-hand sides.

Take Fourier transform in (t,x), with Fourier series in the torus case. For u_n=Rr_n^{(L)}, Cauchy–Schwarz yields

If \supp\psi\subset\{|v|\le R_0\}, integration in the direction of k gives I_\psi\le C_\psi/|k| for k\ne0. When |\sigma|>2R_0|k|, the denominator instead gives I_\psi\le C_\psi/(1+|\sigma|^2). Thus I_\psi\to0 uniformly as |\sigma|+|k|\to\infty. Plancherel’s theorem shows that the averages have uniformly small high-frequency L^2 tails. Convolution with a fixed smooth approximate identity consequently approximates them uniformly in L^2. On every compact time-space set these convolutions are bounded and equicontinuous, by Cauchy–Schwarz applied to translates of the convolution kernel and its derivatives. They are relatively compact there. This proves local L^2, and hence local L^1, compactness at fixed L; the uniform L^1 error as L\to\infty completes the proof. The argument applies unchanged to the dual lattice of a rotated torus.□

Proof. The supports of the free cores lie in their disjoint spatial enlargements through physical time r_N; no retirement occurs on this interval for large N. The exact Jacobian formula (9.9), or its version before centering a periodic copy, gives the moment estimate because r_N/(r_Nt+s_j)\le1/(t+l_N). The sheet gain estimate gives the second bound.

In the upper cold estimate, the main-color remainder contributes at most C r_Ns_j^{15/2}H to its gain against H. The sum of the nonmain contributions has a vanishing Gaussian majorant after multiplication by r_N. Uniformly over all positions, therefore,

More precisely this inequality follows by applying the positive gain to the nonnegative upper error in C_{r_N}\le A_N+R_N, where P(R_N,H)\le o(1)H; it does not assert a signed total-variation comparison at all scales. On the present time range h_{r_N}\le2H. Drop every nonnegative loss in (9.11), use P^{b_N}\le P, and bound the positive bath, hot, and source terms exactly as in (9.12). This proves (9.16) with an error independent of t,x,v.

All finite approximations obey their equations in free coordinates. Fubini’s theorem gives the comparisons along almost every free path and on full-measure endpoint sets at every specified terminal time. The countably many approximation indices and lattice translations can be treated simultaneously. This is the form of the comparison used in the next section.□

Proof. Use the free-core measure from Lemma 9.2. More explicitly, if the main retained label at (r_nt,r_nx) is i, set

and otherwise set A_n=0. The disjointness in Lemma 5.1 holds on the whole physical interval [0,r_n] for all sufficiently large n, and no color is retired there. Free transport therefore defines A_n on [0,1], regardless of U_n. Equation (9.16) gives (10.2). Its error already includes the bath, the source divided by m_{T_n}=T_n^p, the rescaled hot self-gain, and the difference between actual and free cold cores.

The enlargement of a main transverse annulus remains between its original radial boundaries. Thus, relative to its period point, \varrho<|y|<e^\eta\varrho, where \varrho=s/r_n. The change of velocity variables in (5.9), multiplied by r_n, is exactly (10.3); equivalently, this is the parameter formula in (9.9) before velocity testing. For 0\le S\le1,

Indeed, multiplying by t+\varrho bounds the \varrho-term by K and the t-term by K/(\alpha\gamma). The smallest scale is \ell_n, so the amplitude is at most C/(t+\ell_n).

The implicit axial map depends only on w_3, and its monotonicity gives |p_3|\le1+\alpha+d_s|w_3|. The smooth compactly supported J has a fixed product Gaussian majorant. Lemma 3.6 therefore dominates each core by a sheet, with the preceding amplitude and a fixed axial moment bound. The gain bound in Lemma 3.4 proves (i). The factor 4 in (10.4) is the product of the linear gain coefficient 2 and the coefficient 2 in the collision-plane formula of Lemma 3.1. Lemma 3.5 now gives (iii); Lemma 3.6 also controls Gaussian parameter tails before the compressed axial change of variables. For the exact free cores the parameters already lie in \supp J. All these estimates allow measurable masks bounded by one, which can only decrease the nonnegative integrals.□

Proof. At any one gain, a removed normalized tail has uniformly small mass by Lemma 10.2. Bound the other four gains and the input by (10.6), and sum over the five possible locations of a removed tail. This proves the first assertion. The same argument at the first gain, using its uniform decay for large outgoing velocities, proves (i).

Retain bounded endpoint velocities and bounded increments. Then all velocities are bounded by a fixed V, and every collision position is within distance V of the endpoint, since the total free-flight time is at most one. Suppose

We show that the first kernel becomes uniformly small on these paths. A main core of dilated scale \varrho\ge M has normalized mass at most C/M, by (10.3) and the sheet bound. Here M can be chosen arbitrarily large, independently of n. For a main core with \varrho\le M, let q/r_n, q\in\Lambda, be its period point. Write (y,z) for the first collision position relative to this point. Its transverse coordinate obeys |y|\le e^\eta M, while its distance to this point is at least d_n-V. Consequently |z|\to\infty, uniformly over all encountered cores of these scales. For bounded p_3,t, their parameter amplitudes satisfy

uniformly for 0<\varrho\le M. In fact, when |z-tp_3|\ge Z, the right side is at most 4K\varrho^{-1}e^{-2\gamma Z/\varrho}, whose supremum tends to zero as Z\to\infty. Applying the sheet bound with this vanishing amplitude proves the assertion for \varrho\le M. The remaining gains have bounded mass. First let n\to\infty, then M\to\infty, and finally remove the initial tail error. This proves (ii).

It follows that a sequence with a positive lower bound for its five-fold integral has bounded velocities and bounded distance from r_n^{-1}\Lambda, after subselection. Choose period points q_n/r_n realizing a bounded distance, and translate the endpoints by these periods. The resulting positions have a convergent subsequence. Such translations preserve both the kernels and the periodic hot inputs used below.

We record also the geometry along the translated bounded paths. Every physical main enlargement for times r_nt\le r_n lies in a fixed ball of radius less than 1/2 about its period point, for sufficiently small s_0,r_n. Distinct points of \Lambda are at least distance one apart. A path in a bounded dilated set lies at physical distance O(r_n) from the chosen origin, so for all large n it can encounter only that copy. Its encountered scales satisfy s/r_n\le C, by the transverse annular restriction. Thus the largest physical width along these paths is O(r_n^{10}), which tends to zero.□

Proof. By (3.16), we may work on |p|+|\xi-v|\leq R and remove that restriction at the end of each argument. On this region the exponential and polynomial factors in (3.12) are bounded by a constant depending on R.

For (10.10), on |\bar\xi|\geq\delta the factor W(\xi) is bounded by 1+\delta^{-q}. The regular line estimate and (3.7) therefore give

For (10.11), integrate in \xi_3 first, using |v_3-p_3|^{-1}\leq\delta^{-1} and (3.18). The bound is C_{R,\delta}/W(v)\leq C_{R,\delta}|\bar v|^q. These are uniform in the axial law. For (10.12), discard the bounded exponential and polynomial factors and apply (3.4). This also proves (10.13) after the tail removal.

It remains to prove (10.14). Fix the truncation radius R. The part |\bar p|>r_0/4 is uniformly negligible as the transverse widths tend to zero. Indeed,

and Lemma 3.4 applies to the wider Gaussian for all sufficiently small d. Now retain |\bar p|\leq r_0/4. Write n=\bar v-\bar p, A=(n/|n|)\cdot\bar v, and B=(v_3-p_3)/|n|, as in Lemma 3.2. Then

If |v_3-p_3|\leq r_0^2/(8R), then for |\xi_3-v_3|\leq R,

The transverse collision line stays at distance greater than r_0/3 from zero, and cannot meet |\bar\xi|<\delta when \delta<r_0/3. In the complementary case, |v_3-p_3|^{-1}\leq8R/r_0^2. Integrating first in \xi_3 bounds the small-input contribution by

First let the widths tend to zero, then \delta tend to zero, and finally remove the truncation. This proves the last assertion.□

Proof. The profile satisfies

For fixed t_1,t_2,z_1, the function F(w)=w-u_{\varrho,\sigma}(t_2,z_1-\Delta w) has derivative F'(w)\ge1. Its root is u_{\varrho,\sigma}(t_1,z_1): the defining implicit equation shows that this velocity is unchanged along the straight segment between the two points. Integrating F'\ge1 proves (10.15).

Fix 0<\delta<\alpha. If u_{\varrho,\sigma} is within \delta of an endpoint speed, there is nothing to prove. Otherwise

It follows that

First choose \delta, then \varrho small. This proves the uniform assertion in (ii). Finally, the function

has derivative at least one. Its values at u_{\varrho,\sigma} and at the perturbed p_3 are 0 and d_sw_3. This proves (10.16).□

Proof. Remove normalized parameter and increment tails at an arbitrarily small cost, using Lemma 10.2. On the retained two-gain paths, all velocities and positions lie in fixed compact sets, the times lie in [a,1], and every active scale satisfies 0<\varrho\le M. In particular the largest possible physical width is d_0(r_nM)^{10}\to0.

Write t_1,x_1=(y_1,z_1) for the first collision point, and \xi for its input velocity. The next point has axial coordinate z_2=z_1-(t_1-t_2)\xi_3. All estimates below are uniform when t_1,x_1 vary in their retained compact sets. At the first gain, Lemma 10.4 supplies a common modulus \omega(\rho)\downarrow0 for the mass of any axial interval of length \rho; the interval’s center is arbitrary. It also says that for each fixed b>0, the first-kernel mass on \{|\bar\xi|>b\} tends to zero as \bar v_n\to0.

Here is an order of choices that handles all scales at the second point. Fix an error \epsilon>0, including the fixed factors from the two time integrations and the bound C/a. Choose \delta>0 so that the sum of the first-kernel masses of the \delta-neighborhoods of the five numbers

is at most \epsilon. This is possible uniformly because there are five intervals and their lengths are 2\delta. Lemma 10.6 then gives \varrho_0>0, chosen uniformly on the retained compact sets, so that every profile with \varrho<\varrho_0 is within \delta/4 of its corresponding three collapsed speeds.

The geometric mesh has a uniformly bounded number J_0 of scales in [\varrho_0,M]. Choose b_0>0 so small that the first-kernel mass of the union of the b_0-neighborhoods of

is at most \epsilon. For example it suffices that 2J_0\omega(2b_0)\le\epsilon, after incorporating the fixed amplitude bound into \omega. This list may depend on n; its cardinality bound does not. Its centers may depend on t_1,z_1, which are held fixed when the first kernel is integrated. Thus neither dependence weakens the interval bound.

Let E_n(t_1,z_1) be the union of the neighborhoods of the five collapsed speeds in (10.17) and the neighborhoods of the finite-scale profile speeds just chosen. Outside E_n, every retained cold speed at the second point has an axial mismatch bounded below by a fixed positive number for all large n. For \varrho\ge\varrho_0, this follows from (10.15) and (10.16), with lower bound b_0/2. For a constant collapsed speed it follows immediately, with lower bound \delta/2. For the remaining branch use the exact identity

the factor t_1/t_2\ge1 preserves the lower bound. The errors \delta/4 in small-scale approximation and the vanishing thermal errors can be absorbed. We have therefore obtained, on the complement of E_n,

for every retained second-core parameter.

The second-kernel axial mismatch estimate in Lemma 10.4 now tends uniformly to zero as |\bar\xi|\to0. Choose b>0 so small that this second kernel has mass at most \epsilon whenever |\bar\xi|\le b and the displayed mismatch holds. For this fixed b, the first-kernel mass on |\bar\xi|>b tends to zero. Its mass on E_n is at most 2\epsilon, while the remaining first-kernel mass is bounded. Combining the three regions, then removing the initial tail error, gives a bound tending to zero as \epsilon\downarrow0. All choices preceded the final passage n\to\infty, and were uniform in the two times. This proves the lemma.

In a longer integral, all factors following these two gains are bounded by (10.6); the same proof applies.□

Proof. Write L=L^\perp+\lambda e. The map L^\perp\mapsto n\times L^\perp multiplies two-dimensional area by |n|^2. Integration in \lambda contributes \int\zeta=1, proving (10.18) and the support and density bounds. For successive gains, fix all previous parameters before applying the identity. The current normal is then fixed; Fubini’s theorem for nonnegative functions justifies iteration. No derivatives of the variable normals enter this conditional change of variables.□

Proof. Boundedness and convergence in measure on the finite-volume set B imply \|F_n\|_{L^1(\R^d)}\to0: for every \eta>0, bound the integral by \eta|B|+M|\{|F_n|>\eta\}|.

Fix \epsilon>0. The critical set in K has measure zero, so choose an open neighborhood E of it with |E\cap K|<\epsilon. The compact set K\setminus E consists of regular points. At such a point q_0, choose d input coordinates giving a nonzero minor of D\Phi(q_0). Let \pi be projection onto the complementary k-d coordinates, and define

The matrix A_0 is invertible. Choose a convex ball V\Subset\mathcal O about q_0 so small that, for all sufficiently large n,

Integration on the line segment between q,q'\in V gives

Thus G_n is injective on V, and its Jacobian has a positive lower bound c_V independent of n. The inverse function theorem consequently makes it a diffeomorphism onto its image.

Cover K\setminus E by finitely many such balls V_j. The change-of-variables formula gives

When k=d, the zero-dimensional factor has value one. The integral on E\cap K, including its density, is at most AM\epsilon. Multiplying the regular-chart estimates by A and summing over the finite cover gives

Let \epsilon\downarrow0. This proves (10.19). The argument used the densities only through |a_n|\le A, so arbitrary measurable masks and restrictions may be included in them without any convergence assumption.□

Proof. For t>0, the derivative with respect to u in the defining implicit equation is

The implicit analytic function theorem gives an analytic profile for every z\in\R. Uniqueness, which also follows from this strict monotonicity, identifies the local profiles on overlaps. The recursive formulas (10.20) are consequently analytic. They make sense even without ordering the positive times: a negative formal free-flight duration merely changes the point at which the next analytic profile is evaluated. This supplies the connected open domain stated in the lemma.

We exhibit one rank-seven point. Fix t_5=\tau with 0<\tau<t_0, and set all five collision times equal to \tau. After a rotation about the velocity axis, write v_0=(R,0,z), R>0. All collision positions now coincide at x_0-(t_0-\tau)v_0, independently of the angular parameters. Thus write the five axial velocities simply as p_i=a_i e_3; their values a_i may be arbitrary.

Choose

They satisfy k_1\perp n_1, k_2\perp n_2, and span the transverse plane. Put

and choose a sufficiently small nonzero number \epsilon and

Then k_3\perp n_3, its axial component is nonzero, and \bar v_3\ne0. Hence k_1,k_2,k_3 are linearly independent. Writing \bar v_3=(X,Y), take

The fourth increment is perpendicular to n_4, and

Thus n_4,n_5 are nonparallel. All the normals in this construction are nonzero, and every chosen increment can be represented by L_i=-n_i\times k_i/|n_i|^2. Choose L_5=0.

At the equal-time configuration,

All derivatives of x_5 in angular directions vanish. Keeping t_5 fixed, its first three time columns are

Derivatives of the v_i, including derivatives of the position-dependent p_i, are multiplied by zero free-flight durations in this formula. In the final velocity block,

because L_5=0 and the p_i have zero angular position derivatives at equal times. The sum of the planes n_4^\perp+n_5^\perp is \R^3. Choose three angular columns spanning this sum. Together with the three time columns above, they give an invertible 6\times6 block for (x_5,v_5): its upper-right block is zero and both diagonal blocks are invertible. Appending the t_5 column and target coordinate gives a nonzero 7\times7 minor.

This minor is a nontrivial analytic function on the connected domain. Its zero set has Lebesgue measure zero. For completeness, at any zero of a nontrivial analytic function some derivative of finite order is nonzero; otherwise its Taylor series vanishes on a neighborhood, and analytic continuation through overlapping balls in the connected domain makes the function identically zero. Choosing a derivative of minimal nonzero order puts the point on a regular zero set of a derivative of one lower order. There are countably many such derivatives, and each regular zero set is covered by countably many smooth hypersurface charts. Their union is null. Rank failure is contained in the zero set of our particular minor, so it is null as well. Continuity also shows directly that the witness persists at nearby strictly ordered times. It need not obey any spatial mask.

For the prelimit maps replace p_i by (10.3) and include w_i\in\R^3 among the parameters. On compact sets the scales have positive lower bounds, d_{r_n\varrho_i}\to0, and all implicit derivatives have the same positive lower bound one. Differentiating the implicit equations, or applying their implicit function formulas, gives C^1 convergence in the times, positions, and all bounded L_i,w_i. The limit is \Phi, independent of the fifteen spectator w_i-coordinates. Its rank is still seven almost everywhere on the enlarged parameter domain, by Fubini. Lemma 10.9 now proves the final assertion.□

Proof. Fix an arbitrarily small error tolerance. Remove the tails of Lemma 10.3. All path velocities and positions are now in fixed bounded sets. At every encountered main core, s\le Cr_n, so its transverse width tends uniformly to zero.

Starting from a positive lower bound for the endpoint transverse speed, apply the nonaxial small-input estimate in Lemma 10.4 successively at gains 1,\ldots,5. Choose positive numbers b_1,\ldots,b_5 so that deleting |\bar v_i|<b_i costs less than the prescribed error at step i. At each step the previous lower bound has already been fixed; the later gain masses are bounded by (10.6). Thus this is a finite successive choice, followed by n\to\infty. The sum of the five deleted contributions is arbitrarily small. Write b>0 for a lower bound on all retained transverse speeds, including v_0. For large n, every retained normal v_{i-1}-p_i has length at least b/2.

Next delete collision points with |y_i|<\rho. Conditional on all earlier parameters, y_i is an affine function of t_i with derivative \bar v_{i-1}. Consequently

Integrate the current kernel using its mass bound C/a, and bound the later kernels by (10.6). Summing over the five steps bounds this deletion by C_a\rho/b. Choose \rho>0 after the speed thresholds. The retained scales now belong to a compact subinterval [\varrho_-,\varrho_+]\subset(0,\infty). Mesh convergence gives a fixed finite list of scales on this interval, with positive limits, after an immaterial enlargement of its endpoints. There are consequently only finitely many five-tuples of scales and signs to consider.

The tested time t_5 must remain inside the weak-limit interval. If U>1, it already lies in the compact interval [a,1]\Subset(0,U). If U=1, delete t_5>1-\delta. Ordering forces all five times into (1-\delta,\tau_n), whose length is at most \delta. This costs at most

Choose \delta>0 to make this as small as desired. No convergence at the event time is being asserted.

We also remove a thin neighborhood of the boundary of the ordered time region. Put t_0=\tau_n, and discard paths on which t_5-a<\kappa or t_{i-1}-t_i<\kappa for some 1\le i\le5. Integrating velocities successively with the bound C/a, their total contribution is at most C_a\kappa: each of the six time strips has five-dimensional volume at most \kappa in the unit cube. Choose \kappa>0 after the preceding tolerances. On the retained set all consecutive time gaps are at least \kappa, and the lower time exceeds a+\kappa. Since \tau_n\to1, the retained times lie in a fixed compact subset of 0<t_5<\cdots<t_1<1 for all large n.

Fix one retained scale/sign tuple. Drop all spatial masks in constructing its maps, and retain the masks in its nonnegative integrand. Each plane increment is bounded, all normals are bounded above and below, and each cold parameter is bounded. Apply Lemma 10.8 successively. The integral becomes an integral over five times, fifteen L-coordinates, and fifteen cold parameter coordinates. Its support lies in a fixed compact subset of (0,\infty)^5\times\R^{30}, and its density is uniformly bounded: the core amplitudes are at most C/a, the factors |n_i|\zeta(L_i\cdot n_i/|n_i|) are bounded, and all retained ratios H(v_i)/H(v_{i-1}) are bounded. The masks, time ordering, and cut conditions are arbitrary measurable factors between zero and one in this density.

The target (t_5,x_5,v_5) on the actual support lies in a fixed compact set B inside the open weak-limit region, away from the velocity axis. Set F_n=f_n\ind_B\ind_{\{a\le t\le\tau_n\}} and extend it by zero; it is bounded by two and tends to zero in measure. The maps converge in C^1 to the map in Lemma 10.10. That lemma and Lemma 10.9 show that the integral for this tuple tends to zero. Notice that the entire parameter compact need not map into B; only the support of the measurable density must do so.

There are finitely many tuples, so their sum tends to zero. The total limsup of the original integral is bounded by the arbitrarily small errors from the preceding deletions. Letting those tolerances tend to zero proves the lemma.□

Completion of the proof of Proposition 10.1. Assume l=0 and g=0, as in the reduction above. Choose endpoints (x_n,v_n) at time \tau_n in the common full-measure set fixed above such that f_n(\tau_n,x_n,v_n)\ge1/2. For each fixed a, the residual in (10.8) tends to zero along any subsequence of these endpoints. Indeed, Lemma 10.3 first excludes large velocities and escaping positions. Otherwise choose convergent endpoints after a periodic translation. If the limiting velocity is axial, the first two gains vanish by Lemma 10.7, and the last three have bounded mass by (10.6). If it is nonaxial, use (10.7) in Lemma 10.11. This covers every subsequence; hence the residual tends to zero for the original endpoint sequence as well.

Taking the limsup in (10.8) at these endpoints gives

Finally let a\downarrow0. The right side tends to zero, a contradiction. Throughout this argument U was allowed to be infinite, and the case U=1 was handled by (10.21). This proves the proposition.□

Proof. Set G=2P(\widetilde B,f). In free coordinates f^\sharp(t,x,v)=f(t,x+tv,v), the equation becomes

On bounded tubes its right side is locally integrable, so f^\sharp has absolutely continuous time sections for almost every (x,v) and local L^1 time traces. The rate bounds make \nu^\sharp integrable along every bounded characteristic segment in a compact positive-time tube. The one-dimensional integrating-factor formula therefore consists of the freely transported initial density multiplied by its survival factor, plus the time integral of the nonnegative gain with the same survival factor. Repeated substitution into the gain gives (11.3), by (11.1) and (11.2). Every remainder is nonnegative. Restricting the histories, or the initial density, preserves the lower bound. Tonelli’s theorem justifies the substitutions. On bounded portions with total gain rate at most C, the j-collision term is bounded by C^j(b-a)^j/j! times the initial mass.

These formulas apply to point-velocity partners. Indeed Lemma 3.1 gives a density for the gain against each such partner, with mass (11.2). Thus free flight and each gain preserve absolute continuity when the starting measure has a density. Null spatial sets in representatives of the coefficients do not affect the integrated formula.□

Proof. Set s=|v|^2, z=v_3, and R=\sqrt{(s+1)^2-4z^2}. For either partner sign the squared child speed has the law

This follows by expanding the square in (11.1): a coordinate of a uniform point on S^2 is uniform on [-1,1]. The maximum is at most s+1, so there is no exit for s\le24. For 24<s\le25, the exit probability, if positive, is

The inequality uses R\le s+1 and s-49<0. If R=0, the squared speed is deterministic and at most 25. Mixing the signs preserves the bound.

A Markov history of rate \lambda has the survival factor e^{-\int\lambda}, and one rate times jump-law factor per collision. Multiplication by e^{\int(\lambda-\nu_0)} changes this exactly to (11.3), restricted to E. Both history sums converge absolutely on finite intervals because their rates are bounded. Finally 4\pi\le\nu_0(v)\le24\pi on E, and 2p(v)-1\ge12/13, giving (11.5).□

Proof. Given v\in E, choose a sign with |v-\sigma e_3|\ge1. Use this partner for the first collision, and select an output within a small fixed distance \delta of \sigma e_3. Its probability has a uniform positive lower bound: the incident distance lies in [1,6], and a cap of fixed Euclidean radius \delta at that endpoint has area probability bounded below uniformly on the corresponding diameter spheres. Such first outputs lie strictly in E.

Use partner -\sigma e_3 next. If the first output is exactly \sigma e_3, the second sphere is the unit sphere; restrict its output u near e_1. For the third collision use partner e_3. The last output is

At u=e_1, n_0=-(e_1+e_3)/\sqrt2, this output is zero. The two angular derivatives in n span (e_1+e_3)^\perp. The vector e_3 is tangent to the preceding unit sphere at e_1, and direct differentiation gives

Thus the map of the four angular coordinates in the last two collisions to the last velocity has rank three.

Take the two coordinates of n and the coordinate of u with derivative e_3 as distinguished variables, and hold the fourth coordinate in a short interval. The inverse function theorem, with this fourth coordinate and the first output as parameters, gives neighborhoods on which the distinguished-variable map is one-to-one, its Jacobian and inverse Jacobian are bounded, and its image contains a fixed ball B(0,r_0). Shrink these neighborhoods and \delta so this holds for every first output within \delta of \sigma e_3, for both signs. Sphere area in these coordinates has a positive lower density. Change of variables in the three distinguished coordinates, and integration in the fourth, give a fixed positive multiple of volume on B(0,r_0).

Force the three times into three fixed, disjoint intervals inside (0,1), and prohibit other jumps. The first selected relative speed is at least 1, the second is close to 2, and the third close to \sqrt2. In the reference Markov process the selected transition intensity is

It has positive uniform lower density on the patches just specified. The no-additional-jump factor is at least e^{-C_0}. Integrating the times proves (11.6). Keeping only the first two collisions, selecting a small patch near e_1 in the second, and prohibiting later jumps proves the last assertion.□

Proof. The normalized measure on the right side of (11.6) is a common part of positive mass \epsilon in all rows of P_1. Removing this common part shows that the map \mathfrak m\mapsto\mathfrak m P_1 contracts total variation distance by 1-\epsilon. Probability measures form a complete space in that metric, so iteration yields a unique fixed point \pi and uniform geometric convergence. Since P_t commutes with P_1, \pi P_t is another fixed point of P_1, hence equals \pi. Contractivity of Markov kernels gives the estimate between integer times.

Negation of velocity interchanges the two signs and preserves E, so uniqueness of \pi makes it symmetric and its mean zero. Consequently \sup_{v\in E}|\mathbb E_vV_t|\le Ce^{-ct}. The Markov property and bounded velocities imply

uniformly over starting states. Integrating this estimate on [0,A]^2 bounds the second moment of \int_0^A V_s\dd s by CA.□

Proof. Put A_0=(1-\varkappa)A-1. On the first interval of length A_0, require the reference Markov flight to have integrated velocity of norm at most Ad\varkappa/4. Lemma 11.5 and Chebyshev’s inequality give probability at least 1/2, uniformly in the start when A is sufficiently large. Over the next unit interval require the last velocity to lie in B(e_1,d/4). The conditional probability is at least a fixed \varepsilon>0, by Lemma 11.4. Require no jump during the final interval of length \varkappa A, whose conditional probability is at least e^{-C_0\varkappa A}.

The errors in displacement from \varkappa Ae_1 on the three intervals are at most Ad\varkappa/4, D, and Ad\varkappa/4. For A>4D/(d\varkappa) their sum is less than 3Ad\varkappa/4. The final velocity lies strictly in E. Lemma 11.3 bounds the branching weight below by

Its exponent is positive by (11.8). Choose L large enough that this exceeds 8, and all the preceding lower bounds on A hold throughout I_L.□

Proof. Fix a reference start and duration (v,A)\in E\times I_L. The zero-collision term has weight e^{-\nu_0(v)A}\le1. Discarding it from the family in Lemma 11.6 leaves weight greater than 7. Monotone convergence in the positive history count permits restricting to finitely many counts j\ge1 while retaining weight greater than 7. Every postcollision velocity lies strictly in E almost surely: the intersection of a nondegenerate diameter sphere with \partial E has sphere area zero, and a degenerate collision has zero rate. This includes the final velocity. No inward margin is required for the initial velocity. The displacement inequality is strict. Ordered times may be kept apart and away from the endpoints. For each retained collision count and sign sequence, parameterize the ordered times as fractions of A and partition the angular parameter space into finitely many measurable pieces lying in sphere coordinate charts. Inner regularity of the resulting finite parameter measures allows a further restriction to compact sets with total weight greater than 6. These compact sets have positive distances from the forbidden time coincidences and from every postcollision speed boundary. They also have a positive margin in |Z_A-\varkappa Ae_1|<d\varkappa A, since the reference histories obey the stronger inequality (11.9). We call these finitely many compact parameter sets a test for (v,A).

The reference paths and weights are continuous in these finite parameters and in (v,A). The fixed compact tests therefore retain weight greater than 5 and at least half their original positive margins in a sufficiently small relative neighborhood of that start and duration in E\times I_L. These neighborhoods cover the compact set E\times I_L. A finite subcover gives finitely many tests, a maximum count J_{\mathrm{br}}, and uniform positive margins.

Use the same initial velocity, signs, directions, and fractional times for each perturbed history and its reference history. Their first flights have speed at most D, including boundary starts. The sphere map satisfies

Induction bounds the velocity error after j collisions by j\varepsilon_* and the displacement error by AJ_{\mathrm{br}}\varepsilon_*. This comparison does not compare the medium at two nearby positions: its value at the actual perturbed position is close to the same constant reference partner. A sufficiently small \varepsilon_* preserves all the strict velocity and displacement margins. The actual paths thus have speed at most D, so they stay within the region on which closeness was assumed. Formally this follows by induction up to the first possible failure; the preserved margins exclude such a failure.

The relative-speed errors and full loss-rate errors on these histories are bounded by C_{J_{\mathrm{br}}}\varepsilon_*. The density error gives the same bound for the gain-rate factors. The interval has bounded length, so the error in the integrated loss and its exponential is bounded by C_{J_{\mathrm{br}},L}\varepsilon_*. Finite products of the bounded gain factors then show that (11.3) changes by at most C_{J_{\mathrm{br}},L}\varepsilon_* uniformly on each test. The tests have finite parameter measure. Decreasing \varepsilon_* makes their integrals greater than 3. All coefficient estimates were pointwise; measurability is sufficient.□

Proof. Choose t_{\mathrm{br}} so that (1-\delta_K)(1-d)\varkappa t_{\mathrm{br}}>1. Then every speed-D excursion in (11.13) stays outside the hole for every l\in[0,1] and every t\ge t_{\mathrm{br}}. Put \Delta=a_{\rm br}t. At each starting point of \mathcal B_t, apply Lemma 11.7 using the simultaneous space–time scaling (11.15). Translating back its displacement restriction gives

The starting radius d\varkappa t, followed by the triangle inequality, places the endpoint in \mathcal B_{t+\Delta}. The velocity belongs to E: the selected positive-count histories have a strictly interior final velocity. Their first flight need only have speed at most D, exactly as in the medium comparison. History weights are unchanged by this scaling: each gain rate is divided by \rho_0, each time differential is multiplied by \rho_0, and the integrated loss is invariant. The selected histories therefore have total weight greater than 3 uniformly over starts in \mathcal B_t.

Use the kernel obtained by integrating all histories with 1\le j\le J_{\mathrm{br}} that satisfy the block’s velocity and displacement restrictions. The finite compact tests above certify its weight greater than 3; no choice of a test for each starting point is needed. For each integer m\ge1, integrate this positive kernel against \min\{g(t),m\}\ind_{\mathcal B_t}. This is a bounded, integrable portion of the trace, so Lemma 11.2 applies. Letting m\to\infty and using monotone convergence gives (11.19).□

Proof. At some positive time a_0, a bounded, integrable portion of g(a_0) has positive mass. Such a portion is obtained by restricting to bounded phase-space sets and truncating the density. The velocity axis has Lebesgue measure zero, so a further portion of positive mass satisfies

For a sufficiently long but finite free flight of length s,

on this portion. Its survival is bounded below by a positive constant by (11.4). Lemma 11.2 gives positive mass in a bounded region strictly outside the hole at time a_1=a_0+s. This freely transported portion remains bounded and integrable.

Choose h>0 small so that paths whose initial speed is at most R and whose later speeds are at most 3 remain in a bounded region with |y|>l+1 during [a_1,a_1+h]. Both sign densities have strictly positive lower bounds there. To see this directly, the time, axial coordinate, and transverse radius are bounded, the transverse radius has a positive lower bound, and e^{-\eta}|y|\le\varrho\le|y|. The argument of S is therefore bounded, its numerator is positive, and the denominator in (9.3) is bounded above. The rates on the selected bounded-speed paths also have finite upper bounds.

Confine three collisions to this short interval. Choose a reference sign with |v-\sigma e_3|\ge1 for the first collision. The actual relative speed is at least 1-\alpha\ge3/4. Choose its output in a sufficiently small cap about the actual partner. The cap probability has a positive lower bound for inputs bounded by R. Since the actual partner is within \alpha of \sigma e_3, this first output lies in the reserved first-output neighborhood. Next use the signs -\sigma and +1, and the compact angular tests defining \alpha_{\mathrm{acc}}. Their final velocity lies in (11.12). The second output stays near e_1, so the second and third relative speeds have positive lower bounds; all postcollision speeds are at most 3. Put the three times in disjoint subintervals and prohibit other collisions. Positive lower bounds for the selected gain factors, angular probabilities, time-interval lengths, and survival give strictly positive total weight. The sphere-map comparison used here requires only uniform partner closeness, and therefore remains valid for measurable coefficients.

There is consequently a positive portion at time a_2=a_1+h, with bounded spatial support and |v-\varkappa e_1|<d\varkappa/4. Truncating its density gives a bounded, integrable subportion of positive mass. Let R_2 bound |x-\varkappa a_2e_1| on that portion. After free flight to any sufficiently large finite time t,

Its speed is strictly less than D. For each finite t, survival remains strictly positive by (11.4). The zero-collision term of Lemma 11.2 proves the assertion.□

Proof of Proposition 11.1. Suppose that a nonzero density g solves (9.5) for all positive times and obeys the cap. Lemma 11.9 gives a time t_0\ge t_{\mathrm{br}} with positive mass m_0 in \mathcal B_{t_0}.

Apply Lemma 11.8 successively. At t_n=(1+a_{\rm br})^nt_0, induction gives

The cap and l\le1, on the other hand, give

This contradicts (11.18) as n\to\infty. The positive mass m_0 and starting time may depend on the particular density; their sizes are immaterial to this comparison. The parameters governing the blocks were fixed before the mesh and l were chosen. This rules out every nonzero global density in the proposition. Applying this conclusion to the nonzero limits of Proposition 10.1 gives U<\infty.□

Proof. The early upper bounds for the dormant approximations are independent of the approximation index. Their sum is locally integrable, so testing against nonnegative functions passes these bounds to F^{\mathrm d}. We estimate them on a compact set on which

By (6.7), the hot upper bound is C(Tt)^M H(v), for a fixed M>p+1. Division by T m_T=T^{p+1} makes it tend to zero uniformly relative to H on this compact set.

The bath bound (6.1) is exponentially small here. Indeed Tx lies inside the fixed initial hole with D_0(Tx)\ge\delta for all sufficiently small T. Hence its normalized contribution is bounded by

which tends to zero.

For the cold upper bound, a main spatial enlargement at transverse position Ty, with y bounded, has scale s\le C_R T. For small T, only the copy centered at the origin can be main on this compact set: any other such copy would have a lattice center with transverse distance at most s_0+o(1) from the origin and axial distance at most a+b_{\rm sp}+o(1)<1/2, contradicting the unit minimum length of a nonzero lattice vector. The other copies are covered by the nonmain bound.

The transverse Gaussian in the main-color bound of Proposition 5.3, together with |\bar v|\ge\varepsilon, gives

The supremum of the right side over 0<s\le C_R T tends to zero faster than every power of T. At each point there is at most one main color. Lemma 5.5 bounds the remaining sum by C_M(Tt)^M e^{-4|v|^2} for every fixed M>0. Taking M>p+1 handles the same normalization for this sum. All three components therefore vanish in local L^1, as claimed.□

Proof of Theorem 1.2. Take the datum F_0 defined in (4.5). Lemma 4.2 gives its finite mass, energy, and absolute entropy integral, as well as bounded velocity support. For every 0<T\le T_0, Proposition 6.3 gives cap approximations, and Proposition 8.1 gives a global entropy limit satisfying all the required local conservation laws and having datum F_0. The dormant limit F^{\mathrm d} has these same properties.

Suppose there are horizons T_k\downarrow0 at which a cap limit equals F^{\mathrm d} almost everywhere on (0,T_k). For each fixed T_k, choose approximations along the subsequence converging to that particular common limit. Let K_j be a compact exhaustion of \mathcal V, and choose nonnegative smooth compact functions \zeta_j on \mathcal V with \zeta_j\ge1 on K_j. A diagonal choice of indices N_k can ensure both s_{N_k}/T_k\to0 and

where F_k is the chosen finite solution.

Here is the diagonal choice explicitly. At stage k, require s_{N_k}/T_k<1/k and make the differences between the integrals of the normalized approximation and normalized common limit against \zeta_1,\ldots,\zeta_k smaller than 1/k. Each test, when written in physical coordinates at the fixed horizon T_k, is a fixed smooth test of finite norm; periodizing its spatial factor gives a test on the torus. Thus fixed-horizon weak convergence permits these finitely many requirements. Lemma 12.1 makes the integrals of the common limit tend to zero. Since the approximations are nonnegative and \zeta_j\ge1 on K_j, their integrals over K_j tend to zero. This proves (12.1) without a convergence rate uniform in T_k.

Choose a cap time t_{*,k} of each selected solution and set r_k=t_{*,k}+s_{N_k}. Proposition 9.1 gives, after subselection,

locally in spacetime on 0<t<U. By Proposition 10.1, g\not\equiv0; by Proposition 11.1, U<\infty. Consequently

This positive lower bound permits comparison of the two dilations. Since F_k=c_k+u_k+h_k\ge h_k through its cap horizon,

Fix a compact subset K of (0,U)\times\R^3_x\times\{\bar v\ne0\}. For all sufficiently large k, the images of K under (t,x,v)\mapsto(\rho_kt,\rho_kx,v) lie in one fixed compact subset K' of \mathcal V. The four-dimensional space–time Jacobian is \rho_k^4, so (12.2) yields

Thus g=0 off the velocity axis. The cap 0\le g\le(t+l)H makes g a density, and the axis has Lebesgue measure zero. Hence g=0 almost everywhere, contradicting Proposition 10.1.

It follows that for all sufficiently small horizons no cap limit can equal the fixed dormant limit on the corresponding early interval. Choose one such cap limit. Both it and F^{\mathrm d} are global entropy solutions with the same datum and all local conservation laws, and they differ on a set of positive spacetime–velocity measure. Rotating back from the lattice \Lambda to \mathbb Z^3 preserves these properties and gives the assertion on \mathbb T^3.□