Proof. Varying the initial velocity in the geodesic equation gives (3). Since \(\mathcal R\) is symmetric, the fundamental matrix \[\Phi(T)= \begin{pmatrix}C(T)&J(T)\\ C'(T)&J'(T)\end{pmatrix}\] is symplectic. In particular, \[ J^*J'=J'^*J, \qquad C^*J'-C'^*J=\mathrm{Id}, \qquad CJ^*=JC^*. \tag{7}\] These identities can also be obtained by differentiating their left sides and using the Jacobi equation; their values at zero give the displayed constants.

For a variation of the initial point by \(a\) with terminal variation zero, the variational field is \(C(s)a+J(s)b\), where \(b=-J(T)^{-1}C(T)a\). The initial gradient of the divided action is the negative initial velocity. Differentiating it in a normal frame at \(x\) therefore gives \(S(T)=J(T)^{-1}C(T)\). The last identity in (7) shows that this matrix is symmetric. Moreover, the first two identities give \[J^*\bigl(C'-J'J^{-1}C\bigr) =J^*C'-J'^*C=-\mathrm{Id}.\] Consequently \[\frac{d}{dT}(J^{-1}C) =J^{-1}\bigl(C'-J'J^{-1}C\bigr) =-J^{-1}J^{-*},\] including the order of the two inverse factors in (4).

For completeness, symplecticity also gives \[\Phi(T)^{-1}= \begin{pmatrix}J'(T)^*&-J(T)^*\\-C'(T)^*&C(T)^*\end{pmatrix}.\] A Jacobi field for the reversed geodesic with initial position zero and initial derivative \(a\) has, in the forward parametrization, terminal data \((0,-a)\). Multiplication by this inverse matrix shows that its value at the other endpoint is \(J(T)^*a\). The reversed Jacobi matrix at time \(T\) is thus \(J(T)^*\). Dividing both matrices by \(T\) as in (3) proves (5); this argument does not require \(J(T)\) to be invertible.

In the radial direction the Jacobi equation has the explicit solutions \(J(s)v=sv\) and \(C(s)v=v\). Thus \(S(T)v=v/T\). For \(v\in I(x)\), taking \(T=1\) gives (6); the case \(v=0\) is immediate. ◻

Proof. The bundle \[\mathcal G=\{(x,v):x\in M,\ v\in G_x\}\] is compact: its defining distance equality is closed, and \(|v|\) is bounded by \(\operatorname{diam}(M)\). For each such vector consider \(\gamma(s)=\exp_x(sv)\) on \([0,1]\). Fix the splitting time \(h=1/4\) and allow the final time \(T\) to vary in \([1/2,1]\). The prefix \(hv\) belongs to \(I(x)\), since it extends by the factor \(1/h=4\) to the original minimizing ray. For the suffix, reverse the ray at \(y_T=\gamma(T)\) and put \[w_T=-(T-h)\dot\gamma(T),\qquad a_T=\frac{T}{T-h}\ge\frac43.\] Then \(\exp_{y_T}w_T=\gamma(h)\) and \(d(y_T,\exp_{y_T}(a_Tw_T))=a_T|w_T|\), so \(w_T\in I(y_T)\). The reversed suffix is therefore before cut; reversal preserves uniqueness and, by Lemma 11, nonconjugacy. The parameter family is compact and its image lies in the open injectivity domain. Thus both piece differentials and their inverses have uniform bounds. Their time lengths are bounded below by \(1/4\).

Let \(C,J\) be the matrices from Lemma 11. Starting at time \(h\), let \(C_h(s),J_h(s)\) have the same initial data and solve the Jacobi equation along \(\gamma(h+s)\). Smooth dependence on \((x,v,T)\) and the preceding nonconjugacy show that \[ J(h),\ J(h)^{-1},\ J_h(T-h),\ J_h(T-h)^{-1},\ C(h),\ J'(h) \tag{12}\] all have uniformly bounded norms. This compactness assertion also includes \(v=0\), for which \(J(s)=s\mathrm{Id}\).

Write \(z=\gamma(h)\) and \(y_T=\gamma(T)\). The Hessian in the joining point of the broken action \[z'\longmapsto \frac{c(x,z')}{h} +\frac{c(z',y_T)}{T-h}\] at \(z\) is \[ D(T)=J'(h)J(h)^{-1} +J_h(T-h)^{-1}C_h(T-h). \tag{13}\] The first term is the terminal Hessian of the first piece; the second is the source Hessian of the second piece. At \(T=1\), the broken action has a minimum at \(z\). To see this directly, the triangle inequality and Cauchy–Schwarz give \[\frac{d(x,z')^2}{2h}+\frac{d(z',y_1)^2}{2(1-h)} \ge\frac{(d(x,z')+d(z',y_1))^2}{2} \ge\frac{d(x,y_1)^2}{2},\] with equality at \(z\). Hence \(D(1)\succeq0\).

Only the second term in (13) depends on \(T\). Applying (4) with initial point \(z\) yields \[ \begin{split} D(1-\delta)&=D(1)+\delta G_\delta,\\ G_\delta&=\frac1\delta\int_{1-\delta}^{1} J_h(s-h)^{-1}J_h(s-h)^{-*}\,ds, \qquad c\mathrm{Id}\preceq G_\delta\preceq C\mathrm{Id}. \end{split} \tag{14}\] The bounds on \(G_\delta\) are precisely the bounds in (12): the smallest eigenvalue of the integrand is \(\|J_h(s-h)\|^{-2}\), and its largest eigenvalue is \(\|J_h(s-h)^{-1}\|^2\). Thus the crossing is positive on the whole space, including any null space of \(D(1)\).

Jacobi propagation through the joining point gives the exact factorization \[ \begin{split} J(T) &=C_h(T-h)J(h)+J_h(T-h)J'(h)\\ &=J_h(T-h)D(T)J(h). \end{split} \tag{15}\] This formula is also valid at \(T=1\) when either of the two full matrices is singular. The splitting point is kept fixed while \(T\) varies; differentiating a moving midpoint is unnecessary. Taking \(h=1/2\) gives the usual midpoint factorization on \(T\in[3/4,1]\). The choice \(h=1/4\) covers all the parameters in the statement with one set of bounds.

Write \(\lambda_1(D)\le\cdots\le\lambda_n(D)\) for the increasing eigenvalues of a positive semidefinite matrix \(D\). The min–max characterization applied to (14) gives \[\lambda_j(D(1))+c\delta \le\lambda_j(D(1-\delta)) \le\lambda_j(D(1))+C\delta.\] Multiplication of a matrix on either side by a uniformly bounded invertible matrix changes each ordered singular value by at most a uniform multiplicative factor. Apply this fact to (15), use \((d\exp_x)_{Tv}=J(T)/T\), and note \(1/2\le T\le1\). At \(T=1\), the singular values of \(J(1)\) are comparable to the eigenvalues of \(D(1)\); at \(T=1-\delta\), they are comparable to the eigenvalues of \(D(1-\delta)\). These observations prove (8) with uniform constants.

We give the Hessian calculation as well, because it identifies the bounded term that must not be discarded at a radial eigenvector. Set \[P=J(h)^{-1}C(h),\qquad E=J(h)^{-*}.\] The source–join mixed Hessian of the first divided action is \(-J(h)^{-1}=-E^*\). For \(T<1\), eliminating the joining-point variation in the Hessian of the broken action gives \[ \frac{A_x(Tv)}{T}=P-E^*D(T)^{-1}E. \tag{16}\] To verify the elimination, write the quadratic form for source variation \(a\) and joining variation \(b\) as \(a^*Pa-2a^*E^*b+b^*D(T)b\). Its unique stationary value in \(b\) is at \(b=D(T)^{-1}Ea\) and has value \(a^*(P-E^*D(T)^{-1}E)a\). This is the Hessian of the stationary composed action. Positivity of \(D(T)\) follows from (14); the composed action is the smooth cost branch of the shortened minimizing ray.

For \(T=1-\delta\), formula (16) gives \[ \begin{split} N_{x,\delta}(v)&=Q_{x,v,\delta} +\delta(\mathrm{Id}-TP),\\ Q_{x,v,\delta}&=T\delta E^*D(T)^{-1}E. \end{split} \tag{17}\] The second form is positive definite. The bounds on \(P,E,E^{-1}\), the min–max inequalities just proved, and inversion of \(D(T)\) give (9) and (10). They imply the norm and lower bounds in (11). The remaining identity follows from (6). In particular, for \(v\ne0\) the radial eigenvalue of \(N_{x,\delta}(v)\) is exactly zero, while the radial singular value of \((d\exp_x)_v\) is one. These two facts are consistent because the comparison with \(Q_{x,v,\delta}\) includes the bounded background \(\delta(\mathrm{Id}-TP)\). ◻

Proof. For \(\theta<1\), multiply the lower bounds in (8) with \(\delta=1-\theta\) and discard the nonnegative terms \(\delta\). The case \(\theta=1\) is immediate. For (19), apply (18) to the longer vector \(\bar p/l\) with shortening factor \(l\), and then use (5). For (20), apply it to \(\bar p\) with shortening factor \(1/l_+\) and use the same reversal identity. Thus the direction of the two inequalities reflects, respectively, a longer and a shorter true reversed ray. ◻

Proof. Every time less than \(T_*\) lies strictly before the cut time of this ray, so \(r\) and \(\tau r\) are in \(I(y)\), and the reverse vector \(p\) is in \(I(x)\). Let \(J(s)\) be the Jacobi matrix based at \(y\) with unit initial derivative. By reversal, \[ L=J(1)^*. \tag{23}\] Apply Lemma 12 to the minimizing vector \(T_*r\), with shortening factor \(1/T_*\). Since \(\delta=(T_*-1)/T_*\ge(T_*-1)/2\), its lower bound gives \[ s_1(J(1))=s_1(y,r)\ge c(T_*-1), \qquad \|J(1)^{-1}\|\le\frac{C}{T_*-1}. \tag{24}\] There is also a uniform bound on \(\|J'(s)\|\) for \(0\le s\le T_*\). Indeed, \(|T_*r|\le\operatorname{diam}(M)\), \(T_*\le2\), and smooth Jacobi evolution on this compact set of initial data bounds its fundamental matrices. Therefore, for \(1\le s\le\tau\), \[ \begin{split} \|J(s)J(1)^{-1}\| &\le 1+\|J(s)-J(1)\|\,\|J(1)^{-1}\|\\ &\le 1+C\frac{s-1}{T_*-1}\le C. \end{split} \tag{25}\]

Write \(S(s)=A_y(sr)/s\). Lemma 11 and (23) give \[ L^*[-S'(s)]L =J(1)J(s)^{-1}J(s)^{-*}J(1)^*. \tag{26}\] The matrix on the right is \(B_sB_s^*\) with \(B_s=J(1)J(s)^{-1}\). Its smallest eigenvalue equals \(\|B_s^{-1}\|^{-2}=\|J(s)J(1)^{-1}\|^{-2}\) and is thus bounded below by a uniform positive constant by (25). This uses the displayed order of the factors and does not commute any Jacobi matrices. Integration over \([1,\tau]\), whose length is \((T_*-1)/2\), proves (21).

If the terminal branch is nonconjugate, \(S\) is defined through time \(T_*\) and \(-S'(s)\) is positive definite on the remaining interval. Adding its integral over \([\tau,T_*]\) preserves the lower bound. Since \(T_*-1=(1-l)/l\ge1-l\), this gives (22). For arbitrary two nonconjugate times on a minimizing ray, integration of \(-S'(s)=J(s)^{-1}J(s)^{-*}\succ0\) proves strict decrease. When the true terminal point is conjugate, only the half-extension formula is used; no undefined terminal Hessian has been differentiated. ◻

Proof. If there is a configuration in the statement then \(B>0\). Set \[a=\min\{a_0,(2B)^{-1}\}.\] We use only the segment \([-a,1]\). First suppose that all its points lie in \(I(x)\). Write \[p(t)=p+tq,\qquad \beta=\frac{|q|^2}{p\cdot q},\qquad L(t)=\mathrm{Id}+t\frac{qq^*}{p\cdot q}.\] The eigenvalues of \(L(t)\) are \(1\) on \(q^\perp\) and \(1+t\beta\) on \(\mathbb Rq\). Hence \[ \|L(t)\|\le1+B,\qquad \|L(t)^{-1}\|\le2 \qquad(-a\le t\le1). \tag{30}\] For \(0<\delta\le1/2\), put \[ N_\delta(t)=-\delta\bigl(A_x((1-\delta)p(t))-\mathrm{Id}\bigr), \qquad F_\delta(t)=L(t)^*N_\delta(t)L(t). \tag{31}\] These matrices are uniformly bounded by Lemma 12 and (30), and \(F_\delta(t)\succeq-C\delta\mathrm{Id}\).

We next prove full matrix convexity of \(F_\delta\). For any fixed \(z\in T_xM\), the vector \[\xi=z-\frac{z\cdot q}{p\cdot q}p\] is perpendicular to \(q\). Moreover, \[\xi-L(t)z=-\frac{z\cdot q}{p\cdot q}p(t).\] The last vector belongs to the kernel of \(N_\delta(t)\) by the radial identity. Symmetry consequently gives the exact equality \[ z^*F_\delta(t)z=\xi^*N_\delta(t)\xi. \tag{32}\] Weak MTW means that \(t\mapsto A_x((1-\delta)p(t))(\xi,\xi)\) is concave whenever \(\xi\perp q\). The negative sign in the definition of \(N_\delta\) therefore makes the right side of (32) convex. Since this holds for every \(z\), \(F_\delta\) is convex in the Loewner order.

This argument remains uniform when \(p\cdot q\) is small. The fixed test vector \(\xi\) itself might be large, but its value modulo the radial kernel is \(L(t)z\). Equations (30) and (32) bound the scalar convex function by \(C|z|^2\), independently of the size of \(\xi\). For any compact interval \(K\subset(-a,1)\), a real convex function bounded in absolute value by \(C|z|^2\) has slopes on \(K\) bounded by \(2C|z|^2/\operatorname{dist}(K,\{-a,1\})\): sandwich each secant slope between secants to the two endpoints. Applying this estimate to basis vectors and their sums and differences shows by polarization that every matrix entry of \(F_\delta\) has a uniform Lipschitz bound on \(K\).

To turn this bounded matrix convexity into a spectral comparison, we argue by contradiction. The spectral profile \(\delta/(s_i+\delta)\) from (10) detects singular values below the radial scale: it stays positive when \(s_i\ll\delta\) and tends to zero when \(s_i\gg\delta\). At a scale separated from the singular values, the rank of a limiting matrix therefore counts those below the scale. A failing ordered singular-value ratio will provide a radial scale between the center and endpoint values. We choose it to separate asymptotically from every singular value at each fixed point of a countable dense set of segment parameters. The resulting positive semidefinite matrix limit has a uniform gap between zero and its positive eigenvalues, forcing constant interior rank. Convexity then prevents rank loss at the endpoint, whereas the failing ratio will force precisely such a loss.

Suppose that no uniform constant in (28) exists. There are configurations \((x_k,p_k,q_k)\) satisfying the same \(a_0,B\), an index \(j\) fixed after passing to a subsequence, and \[ \frac{s_j(x_k,p_k)}{s_j(x_k,p_k+q_k)}\longrightarrow0. \tag{33}\] For now all the segments are in their open injectivity domains, so the singular values are positive. Choose an orthonormal identification of each \(T_{x_k}M\) with \(\mathbb R^n\). No smooth choice in \(k\) is needed. Abbreviate \[s_{i,k}(t)=s_i(x_k,p_k+tq_k),\qquad R_k=\log\frac{s_{j,k}(1)}{s_{j,k}(0)}\longrightarrow\infty.\] Let \(\{t_1,t_2,\ldots\}\) be a fixed countable dense subset of \((-a,1)\), enlarged to include \(0\) and \(1\). We now choose one radial scale that separates every fixed member of this set. Take \(\log\delta_k\) in the middle half of the interval \[[\log s_{j,k}(0),\log s_{j,k}(1)].\] Exclude the intervals of radius \(R_k^{1/3}\) centered at \[\log s_{i,k}(t_m),\qquad 1\le i\le n,\quad 1\le m\le\lfloor R_k^{1/3}\rfloor.\] Their total length is at most \(2nR_k^{2/3}\), while the available middle interval has length \(R_k/2\). For all sufficiently large \(k\) there is a remaining point, which we choose as \(\log\delta_k\). This construction ensures \[\begin{align*} &s_{j,k}(0)/\delta_k\longrightarrow0, \qquad s_{j,k}(1)/\delta_k\longrightarrow\infty, \tag{34}\\ &\left|\log\frac{s_{i,k}(t_m)}{\delta_k}\right| \longrightarrow\infty \quad\text{for every fixed }i,m. \tag{35}\end{align*}\] All singular values on \(\mathcal G\) are uniformly bounded above. Since \(\delta_k\le s_{j,k}(1)e^{-R_k/4}\), we also have \(\delta_k\to0\). In particular these scales are eventually admissible in Lemma 12.

Form the matrices \(F_k(t)\) in (31) using the \(k\)th segment and scale \(\delta_k\). Their uniform bounds and interior Lipschitz bounds give, by successive extraction on compact subintervals, a locally uniform limit \(F(t)\) on \((-a,1)\). Extract also a limit \(E=\lim_kF_k(1)\). Both \(F(t)\) and \(E\) are positive semidefinite, because their approximants are bounded below by \(-C\delta_k\mathrm{Id}\). The limit \(F\) is matrix convex and bounded.

To identify its possible ranks, use (9) to write \[F_k(t)=L_k(t)^*Q_{x_k,p_k+tq_k,\delta_k}L_k(t) +\mathcal E_k(t), \qquad \|\mathcal E_k(t)\|\le C\delta_k.\] The positive form in this expression has its \(i\)th decreasing eigenvalue between uniform multiples of \[ \frac{\delta_k}{s_{i,k}(t)+\delta_k}, \tag{36}\] by (10) and (30). At a fixed dense point \(t_m\), put \(\eta_k=\exp(-R_k^{1/3})\). For all sufficiently large \(k\), depending on \(m\), (35) says that \(s_{i,k}(t_m)/\delta_k\) is at most \(\eta_k\) or at least \(\eta_k^{-1}\). Weyl’s eigenvalue inequality therefore gives one of the two alternatives \[\lambda_i^\downarrow(F_k(t_m)) \ge \frac{c_0}{1+\eta_k}-C\delta_k, \qquad\text{or}\qquad -C\delta_k\le\lambda_i^\downarrow(F_k(t_m)) \le C_0\eta_k+C\delta_k,\] with \(c_0,C_0>0\) uniform in \(i,m,k\). Since \(F_k(t_m)\) already converges, its ordered eigenvalues converge. Their limits are therefore either zero or at least \(c_0\), even if the alternatives initially alternate. No simultaneous rate at all dense points is required. Set \(c_*=c_0/2\).

The ordered eigenvalues of a symmetric matrix are continuous in its entries. Density of the set \(\{t_m\}\) and continuity of \(F(t)\) therefore give \[ \operatorname{spec}F(t)\subset\{0\}\cup[c_*,C_*] \qquad(-a<t<1) \tag{37}\] after decreasing \(c_*\) once if necessary. Indeed, an eigenvalue strictly between zero and \(c_*\) at an interior point would remain in that interval in a neighborhood and contradict the dense-set conclusion. The rank is now locally constant, because a continuous eigenvalue cannot pass from zero to a positive value without entering that gap. It is consequently constant on the connected interval \((-a,1)\).

The endpoint also has to be treated; constancy of the interior rank alone would not suffice. For every fixed vector \(z\), the bounded scalar convex function \(z^*F(t)z\) has a finite limit as \(t\uparrow1\). Polarization yields a matrix limit \(F(1-)\). The gap (37) persists in this limit, so its rank equals the interior rank. For fixed \(s<t<1\), matrix convexity of \(F_k\) says \[F_k(t)\preceq\frac{1-t}{1-s}F_k(s) +\frac{t-s}{1-s}F_k(1).\] First let \(k\to\infty\), then let \(t\uparrow1\); this gives \[ F(1-)\preceq E. \tag{38}\] Positive semidefinite order implies \(\operatorname{rank}E\ge\operatorname{rank}F(1-)\), since the kernel of \(E\) is contained in the kernel of \(F(1-)\).

On the other hand, the first inequality in (34) and the ordering of singular values make the first \(j\) eigenvalues in (36) at \(t=0\) bounded away from zero. Thus \(\operatorname{rank}F(0)\ge j\). The second inequality makes all eigenvalues with decreasing index \(i\ge j\) tend to zero at \(t=1\), so \(\operatorname{rank}E\le j-1\). This contradicts the interior rank constancy and (38). The uniform ordered comparison is proved for open segments.

Finally, replace a closed segment in (27) by \((1-\varepsilon)(p+tq)\), with \(0<\varepsilon<1\). It lies in \(I(x)\) by radial shortening. Its extension interval and the ratio \(|q|^2/(p\cdot q)\) are unchanged, so the same constant applies. Letting \(\varepsilon\downarrow0\) and using continuity of the singular values proves (28) on \(G_x\). Multiplication over \(j\) proves (29). Every bound in the contradiction argument depended only on \((M,g),a_0,B\), which establishes the stated uniformity. ◻

Proof. Set \(q=t_ie\). Expansion of the squared norms in (39) gives \[2ht_i+t_i^2\le L_0D, \qquad \frac{|q|^2}{p\cdot q}=\frac{t_i}{h} \le\frac{L_0D}{2h^2}\le\frac{L_0}{2\kappa}.\] Because \(q\ne0\), \(m_i<1\). The remaining barycenter belongs to \(G_x\) by its convexity and is \[\frac1{1-m_i}\sum_{j\ne i}m_jp_j =p-\frac{m_i}{1-m_i}q.\] The coefficient \(m_i/(1-m_i)\) is at least \(m_i\ge\mu\). Convexity therefore puts \(p+s q\) in \(G_x\) for every \(s\in[-\mu,1]\). Lemma 15 applies with \(a_0=\mu\) and \(B=L_0/(2\kappa)\) and proves the result. No upper bound on \(m_i/(1-m_i)\) is needed: a shorter common opposite extension is enough. ◻

Proof. Write \[h_x(p)=u(x)+\frac12|p|^2+v(\exp_xp),\qquad p\in G_x.\] By Corollary 8, the lifted section \(\{p\in G_x:h_x(p)\leq r\}\) is convex. If \(p,q\) belong to it and \(m=(p+q)/2\), the squared-norm identity gives \[\begin{align*} \frac18|p-q|^2 &=v(\exp_xm)-\frac{v(\exp_xp)+v(\exp_xq)}2 +\frac{h_x(p)+h_x(q)}2-h_x(m) \\ &\leq F(r)+r. \tag{41}\end{align*}\] The exponential map has a common Lipschitz bound on the compact bundle of minimizing velocities. Hence the target diameter of a section is at most \(C\sqrt{F(r)+r}\).

Let \((x,y)\) and \((x',y')\) be contacts of the same dual pair. The common Lipschitz bounds for \(u\) and \(c\) imply \[0\leq u(x)+c(x,y')+v(y')\leq C d(x,x').\] Thus \(y,y'\) belong to a section at \(x\) of that height. For sufficiently small \(d(x,x')\), (41) gives \[d(y,y')\leq C d(x,x')^{\frac12\min\{\beta,1\}}.\] In particular, letting the section height tend to zero shows that every pole has exactly one contact endpoint. The bound proves that this endpoint depends continuously on the pole. The class also contains the reversed pair \((v,u)\) after renormalization, which does not affect its section oscillations. The same reasoning therefore gives a unique continuous contact in the reverse direction. The dual identities give a contact above every point of either copy of \(M\), and the two resulting maps are inverse. They agree almost everywhere with the optimal maps by optimality and uniqueness. Finally, compactness absorbs the remaining, larger distances into the same uniform constant. ◻

Proof. If \(F\) vanished at a positive scale, monotonicity would already give a power bound below that scale. Thus \(F(r)>0\). Fix \(C_F\geq\max\{1,\sup_r F(r)\}\) and set \[f(s)=\log\frac{C_F}{F(e^{-s})}.\] This function is nonnegative and nondecreasing. Failure of every power bound implies \(\liminf_{s\to\infty}f(s)/s=0\); otherwise \(f(s)\geq\beta s\) eventually for some \(\beta>0\). Choose \(N_j\to\infty\) with \(f(N_j)=o(N_j)\), and divide \([N_j/2,N_j]\) into \[m_j=\left\lceil\sqrt{N_j\max\{1,f(N_j)\}}\right\rceil\] equal intervals. Their length \(N_j/(2m_j)\) tends to infinity. Since their total \(f\)-increment is at most \(f(N_j)\), some interval \([A_j,B_j]\) has increment at most \(f(N_j)/m_j=o(1)\). Put \(R_j=e^{-A_j}\) and \(r_j=e^{-B_j}\). Then \[\log(R_j/r_j)=B_j-A_j\longrightarrow\infty,\qquad \frac{F(R_j)}{F(r_j)}=e^{f(B_j)-f(A_j)}\longrightarrow1.\] Choose an actual section with oscillation \(D_j\geq(1-1/j)F(r_j)\), discarding the first few indices. Its extrema are attained, which gives (42). Moreover \(A_j\geq N_j/2\) and \(f(B_j)\leq f(N_j)=o(N_j)\), so \(|\log D_j|=o(A_j)\). This proves (43). Finally \[\log(R_j/D_j^k)=-A_j-k\log D_j=-A_j+o(A_j)\longrightarrow-\infty.\] ◻

Proof. Choose \(\gamma\) so large that \(2/(1+e^{\gamma/8})\leq\varepsilon_0\), and set \[B_c(s)=1-\frac{2}{1+e^{-\gamma(s-5/8)}}.\] This has all the asserted properties. In particular, \(|B_c'|\) has a positive minimum on every fixed compact interval.

For the outer function let \[\beta(z)=\begin{cases}e^{-1/z},&z>0,\\0,&z\leq0,\end{cases} \qquad \chi(z)=\frac{\beta(z)}{\beta(z)+\beta(1-z)}.\] Thus \(\chi\) is a smooth step, equal to zero on \((-\infty,0]\) and to one on \([1,\infty)\). Put \[\kappa=K+1,\qquad A=M_0+3,\qquad L=128A\kappa,\qquad h_0=\frac1{128\kappa},\qquad 0<\eta\leq\min\left\{\frac1{128},\frac{\varepsilon_0}{24L}\right\}.\] Define \[\begin{align*} w(s)&=(L-\kappa s) \chi\!\left(\frac{s+3\eta}{\eta}\right) \left[1-\chi\!\left(\frac{s-h_0}{h_0}\right)\right],\\ J(s)&=1+\frac{\varepsilon_0}2+\int_0^s w(q)\,dq,\\ B_o(s)&=J(s) \left[1-\chi\!\left(\frac{s-(1-\eta)}{\eta}\right)\right]. \end{align*}\] The function \(w\) is nonnegative and supported in \([-3\eta,2h_0]\). Also \[\int_{-3\eta}^0w\leq3\eta(L+3\kappa\eta) \leq6\eta L\leq\varepsilon_0/4,\qquad \int_0^{2h_0}w\leq2Lh_0=2A.\] It follows that \[1+\varepsilon_0/4\leq J(s)\leq1+\varepsilon_0/2+2A.\] On \([-2\eta,h_0]\), one has \(J'=L-\kappa s>0\) and \(J''=-\kappa\). Moreover \[J(h_0)-J(0)=A-\frac1{32768\kappa}>M_0+1.\] Since \(J\) is nondecreasing, the condition \(B_o(s)\leq M_0\) on \([-2\eta,1-\eta]\) forces \(s<h_0\); here \(B_o=J\), proving (49). The function \(J\) is constant beyond \(2h_0\leq1/64\), so (48) also follows. The final factor is identically one up to \(1-\eta\), and decreases to zero only in \((1-\eta,1)\). This proves the remaining assertions, including smoothness at all joins and the uniform height bound. ◻

Proof. Set \(x_\tau=\exp_{x_0}(\tau p_0)\) for \(0\leq\tau\leq T:=8b/D\), and let \(\Pi_\tau\) be parallel transport along this geodesic. For small scales \(T<1/2\). The left cost is exact: \[f_y(\tau):=-c(x_\tau,y)-v(y),\qquad f_0(\tau)=-\frac12(1-\tau)^2P-a,\qquad q_0(\tau)=(1-\tau)\Pi_\tau p_0.\] The initial gaps at \(y_0,y_1\) lie in \([0,r]\), whence \[ P-|p_1|^2=2D+O(r),\qquad |p_0-p_1|^2\leq8(D+r),\qquad P-\langle p_0,p_1\rangle\leq6D \tag{56}\] for small scales; the second inequality is (41) applied using the actual oscillation \(D\) of this section. In particular \(P\geq D\).

First exclude the middle levels. If \(1/16\leq s_y\leq1-\eta\), put \[h_y(\tau)=f_y(\tau)-bB_o(s_y) -f_0(\tau)+bB_o(0).\] Initially \(h_y(0)\leq r-b\). Wherever \(h_y\geq-b/2\), the identity relating values and squared distances gives, for every minimizing log \(q_y\) at the current pole, \[|q_y|^2-|q_0|^2 =-2Ds_y-2b(B_o(s_y)-B_o(0))-2h_y \leq-D/8-b.\] Equations (53) and \(\dot x_\tau=\Pi_\tau p_0\) imply \(h_y'\leq-D/16\) at almost every such time. The positive part of \(h_y+b/2\) is therefore nonincreasing and starts at zero. Thus \(h_y\leq-b/2\) throughout \([0,T]\). No middle-level mountain can be active.

Let \(G_L\) and \(G_R\) be the maxima of \(f_y-bB_o(s_y)\) over the compact sets \(s_y\leq1/16\) and \(s_y\geq1-\eta\), respectively. Both are continuous, and the preceding paragraph shows \(g_b(x_\tau)=\max\{G_L(\tau),G_R(\tau)\}\). Initially \(G_L(0)\leq u(x_0)-b\) and \(G_R(0)\geq f_1(0)\geq u(x_0)-r\). While \(G_R>G_L\), every active endpoint is right and comparison with the left modified mountain gives \[|q_y|^2\leq|q_0|^2-2D(1-\eta)+2bB_o(0) \leq|q_0|^2-D.\] By rationalizing the difference of the square roots in (53), \[\frac d{d\tau}\{g_b(x_\tau)-f_0(\tau)\}\leq-D/2 \quad\text{almost everywhere while }G_R>G_L.\] This difference starts at most \(r\) and is always at least \(-bB_o(0)\geq-2b\). It cannot obey the displayed inequality throughout \([0,T]\). There is therefore a first equality time \(\tau_b\leq T\). At that time both maxima are attained, and the middle-level exclusion gives \(s_{y_-}<1/16\). Throughout the preceding interval, including its endpoint, \[ -bB_o(0)\leq g_b(x_\tau)-f_0(\tau)\leq r. \tag{57}\]

We next anchor the absolute levels. By (51) and (57), \(u(x_\tau)-f_0(\tau)\leq bH_o+r\). The initial-cost upper support and (56) give \[ f_1(\tau)\geq f_0(0)-r+(P-6D)\tau-C_g\tau^2=:\mathcal L(\tau). \tag{58}\] Since \(f_0(\tau)=f_0(0)+P\tau-P\tau^2/2\), it follows that \(f_0-f_1\leq r+6D\tau+C\tau^2=O(b)\) on this interval. Thus both \(y_0,y_1\) lie in an original section of height \(Cb\). For any active endpoint \(y\), \[u(x_\tau)+c(x_\tau,y)+v(y) =u(x_\tau)-g_b(x_\tau)-bB_o(s_y)\leq bH_o.\] All these endpoints belong to the same section of height \(Cb<R\). Its oscillation is at most \((1+\delta_j)D\) and it contains levels \(a,a+D\). Hence every active level lies in \([-\delta_j,1+\delta_j]\), proving in particular the required lower bound for \(s_{y_-}\).

To prove the lower time estimate, keep the winning left endpoint \(y_-\) fixed, and put \[k(\tau)=f_{y_-}(\tau)-bB_o(s_{y_-})-\mathcal L(\tau).\] One has \(k(0)\leq2r-b<-b/2\), while \(k(\tau_b)\geq0\), since \(g_b(x_{\tau_b})\geq f_1(\tau_b)\). Let \(\theta\) be the last time before \(\tau_b\) at which \(k=-b/2\). Then \(k\geq-b/2\) on \([\theta,\tau_b]\). On this interval, using \(s_{y_-}\geq-2\eta\) and \(B_o\geq0\), the inequality defining this last strip gives \[\begin{align*} |q_{y_-}|^2-|q_0|^2 &\leq4\eta D+2r+12D\tau+2C_g\tau^2+b\\ &\leq D \end{align*}\] for all sufficiently small scales. This uses only (45); in particular \(\tau/D\to0\). Rationalizing once more gives \(f_{y_-}'\leq(1-\tau)P+2D\) almost everywhere in the strip. Since \(\mathcal L'=P-6D-2C_g\tau\), we have \(k'\leq10D\) there for small scales. Therefore \[b/2\leq k(\tau_b)-k(\theta)\leq10D(\tau_b-\theta),\] which implies the stated weaker lower bound \(\tau_b\geq b/(32D)\). No upper bound for the template or its derivatives occurs in this last estimate.

Finally, if \(y_i\) is active at the switch, the exact equality of its modified value with \(g_b\) gives \[|q_i|^2=(1-\tau_b)^2P-2Ds_{y_i} -2bB_o(s_{y_i})-2\{g_b(x_b)-f_0(\tau_b)\}.\] Equation (57) and (45) prove (55). ◻

Proof. Use the two active logs at the switch supplied by Lemma 20. Equation (55) and their level bounds show that their squared norms lie on opposite sides of \(P-D/2\). By continuity along their segment, choose \(p\in\partial g_b(x_b)\) with \[|p|^2=P-D/2.\] Theorem 6 says that \(p\) is minimizing and that \(y=\exp_{x_b}p\) is a contact endpoint for \(g_b\). Thus \[E_b(y)=v(y)+bB_c(s_y)+c(x_b,y)+g_b(x_b).\] Minorizing \(g_b\) by the left modified mountain gives the direct bound \[ E_b(y)\geq D(s_y-1/4)+P\tau_b-P\tau_b^2/2 +b\{B_c(s_y)-B_o(0)\}. \tag{61}\] If \(s_y>1/2\), the right side is at least \(D/4-3b\), which is larger than \(c_0b\) for small scales.

It remains to consider \(s_y\leq1/2\). The original support inequality at \(x_0\) yields \[v(y)-a\geq c(x_0,y_0)-c(x_0,y)-r.\] Apply (52) at \(x_b\) with backward displacement \(-\tau_b\Pi_{\tau_b}p_0\) to get \[c(x_b,y)-c(x_0,y) \geq-\tau_b\langle p,\Pi_{\tau_b}p_0\rangle-C\tau_b^2.\] The corresponding cost difference for \(y_0\) is exactly \(-P\tau_b+P\tau_b^2/2\). Therefore \[ \frac{E_b(y)}b\geq B_c(s_y)-B_o(0) +\bigl(P-\langle p,\Pi_{\tau_b}p_0\rangle\bigr)\frac{\tau_b}b -\frac rb-C\frac{\tau_b^2}b. \tag{62}\] The norm choice gives \[P-\langle p,\Pi_{\tau_b}p_0\rangle =\frac{P-|p|^2+|p-\Pi_{\tau_b}p_0|^2}{2}\geq D/4.\] The middle term in (62) is consequently at least \(1/128\). The template difference is at least \(-2\varepsilon_0\), and the last two terms tend to zero uniformly by (45). The numerical choices (46) give the asserted lower bound. The upper bound is (59). ◻

Proof. At \(b=b_+\), (60) makes \(E(b)-\pi(b)\) negative. At \(b=b_-\), it is at most \(b_-\). At \(b_*=c_0b_+/8\), which belongs to the interval for small scales, \[E(b_*)-\pi(b_*)\geq c_0^2b_+/16.\] Since \(b_-/b_+\to0\), these inequalities give a strictly positive interior maximum with a positive separation from both boundary values.

For a fixed scale and \(b\) in its compact interval, \(Q_tg_b\to Q_1g_b=-\widehat w_b\) uniformly as \(t\uparrow1\). This follows directly from the uniform bound for \(|c/t-c|\). The functions \(w_{c,b}\) have a common Lipschitz constant \(L_*\) on that interval, and \[w_{c,b}-\tfrac12L_*^2(1-t)\leq Q_{1-t}w_{c,b}\leq w_{c,b}.\] Thus \(\Phi_t\to E_b-\pi\) uniformly in \((b,z)\), preserving the positive interior maximum when \(t\) is sufficiently close to one.

Proposition 9 gives the asserted regularity and unique pole for \(Q_tg_b\). To check the other term, the scalar function \(q\mapsto q+bB_c((q-a)/D)\) has positive derivative, and its first two derivatives are bounded at the fixed scales. Composing a Lipschitz semiconvex function with this function preserves semiconvexity: the lower Taylor support for \(v\), its Lipschitz bound, and the ordinary second-order Taylor inequality for the scalar function give a common quadratic lower support for \(w_{c,b}\). For completeness, a minimizer in \(Q_\epsilon w_{c,b}(z)\) has \(d(z,y)\leq2L_*\epsilon\), by comparison with \(y=z\). For small \(\epsilon\), all minimizers therefore lie in one convex normal ball, where the Hessian of \(c(z,\cdot)/\epsilon\) dominates the negative semiconvexity bound. Strict convexity gives uniqueness. In one fixed coordinate chart, let \(K_*\) be the semiconvexity constant and let \(\mu/\epsilon\) be a lower Hessian bound for the smooth cost term. The subgradient inequalities at the minimizers \(y,y'\) for nearby \(z,z'\) and the mixed cost derivative bound give \[(\mu/\epsilon-K_*)|y-y'|^2 \leq (C/\epsilon)|z-z'|\,|y-y'|.\] For small \(\epsilon\) this proves local Lipschitz dependence on \(z\). The envelope gradient formula then gives \(C^{1,1}\) regularity. This is uniform in \(b\) on the fixed interval.

At a maximum, the two spatial gradients sum to zero. Their envelope formulas show that the geodesic from \(x\) to \(z\), continued for the remaining time \(1-t\) with the same velocity, reaches \(y\). The pole minimum gives \(p\in\partial g_b(x)\). By Theorem 6, this full vector is minimizing and its endpoint is a support for \(g_b\). Splitting its energy at time \(t\) proves (64) exactly.

Finally put \(l=1+bB_c'(s_y)/D\in(0,1)\). The short final minimization gives its backward velocity in \(\partial w_{c,b}(y)=l\partial v(y)\); the equality follows from the increasing scalar chain rule, which follows in both directions by Taylor expansion and inversion of that scalar function. Dividing by \(l\) gives a minimizing velocity by Theorem 6. The backward velocity to \(x\) is a strict radial shortening of this vector, and is therefore before cut. ◻

Proof. Fix the prefix time and the spatial maximum point \(z\), and abbreviate \(A(b)=Q_tg_b(z)\) and \(C(b)=Q_{1-t}w_{c,b}(z)\). Decrease \(b\) by \(h>0\), and let \(x_h\) be the new pole for \(A(b-h)\). Because \(0\leq g_{b-h}-g_b\leq hH_o\), compactness and uniqueness of the old pole imply \(x_h\to x\). The strict local quadratic growth in Proposition 9 gives, in normal coordinates \(\Delta_h=\exp_x^{-1}x_h\), \[ |\Delta_h|=O(\sqrt h),\qquad 0\leq A(b-h)-A(b)\leq hH_o. \tag{66}\]

Choose one duration \(q\in(t,1)\) and split each active endpoint cost at \(z_i=\exp_x(qp_i)\). The energy inequality \[c(x',y_i)\leq c_q(x',z_i)+c_{1-q}(z_i,y_i)\] is an equality at \(x'=x\). Its first term is smooth near \(x\), because \(qp_i\) is before cut. Its gradient is \(-p_i\) and its Hessian is \(A_x(qp_i)/q\). Combining this upper support with the smooth prefix cost gives \[\begin{align*} A(b-h)&\geq A(b)+hB_o(s_i) +\langle p_i-p,\Delta_h\rangle +\frac12 Q_i(\Delta_h,\Delta_h)+o(|\Delta_h|^2), \tag{67}\\ Q_i&=\frac1t A_x(tp)-\frac1q A_x(qp_i). \end{align*}\] All expansions here concern fixed scales, a fixed time, and a fixed finite representation. Equations (66)–(67) bound each \(\langle p_i-p,\Delta_h\rangle\) above by \(O(h)\). Their positive weighted sum is zero, so each is also bounded below by \(-O(h)\). Hence, for \(E=\mathop{\mathrm{span}}\{p_i-p\}\), \[ |\operatorname{proj}_E\Delta_h|=O(h). \tag{68}\]

On \(E^\perp\), MTW concavity on the simplex of the \(qp_i\) gives \[A_x(qp)\geq\sum_i m_iA_x(qp_i).\] Radial monotonicity of divided action, equivalently Lemma 4, then gives \[ \sum_i m_iQ_i =\frac1t A_x(tp)-\sum_i\frac{m_i}qA_x(qp_i)\geq0 \quad\text{on }E^\perp. \tag{69}\] Average (67). The linear terms cancel. By (66), (68), and (69), its quadratic term is bounded below by \(-o(h)\): the transverse part is nonnegative, mixed terms are \(O(h^{3/2})\), and the remaining part is \(O(h^2)\). The Taylor remainders are \(o(h)\) as well. Thus \[A(b-h)-A(b)\geq h\sum_i m_iB_o(s_i)+o(h).\] The unique center minimizer and the elementary envelope argument (compare the old and new minimizers and use their convergence) give \[C(b-h)-C(b)=-hB_c(s)+o(h).\] Since \(b\) is an interior maximum of \(A+C-\pi\), division of the resulting one-sided comparison by \(h\) and passage to the limit give (65). The argument applies to the arbitrary representation fixed at the start, proving the asserted universal quantifier. ◻

Proof. An active endpoint has original gap \(u-g_b-bB_o(s_i)\in[0,bH_o]\). By (64), the center gap is \(u-g_b+E_b(y)-bB_c(s)\), which is at most \((H_o+2)b\) by (59). This proves the common section assertion and its oscillation bound. Subtracting the active contact equality from (64) gives (70). Since \(E_b(y)>0\) and \(E_b(y)\leq bB_c(s)\), one has \(B_c(s)>0\), and therefore \(s<3/4\).

Let \(m_R\) be the total right weight. Nonnegativity of \(B_o\) and its lower bound by one outside the right region give \[1-m_R\leq\sum_i m_iB_o(s_i)\leq B_c(s)-c_1\leq1-c_1.\] Thus \(m_R\geq c_1\). In particular a right endpoint is present. The common range bound then puts every level above \((1-\eta)-(1+\delta_j)=-\eta-\delta_j\geq-2\eta\). ◻

Proof. Write \(p_i=p+d_i e\). The level inequality gives total right mass at least \(c_1\), as in Lemma 24. A right point has \(s_i-s>1/4-\eta\), hence \[|p_i|^2-|p|^2=2qd_i+d_i^2\leq-D/4.\] It follows that \(d_i<0\) and \(|d_i|\geq D/(8q)\). Every outward point satisfies, by the common level range, \[0<2qd_i+d_i^2\leq2D(1+\delta_j)+\zeta D\leq3D,\] and so \(d_i\leq3D/(2q)\). The barycenter identity \(\sum_i m_id_i=0\) therefore implies \[\sum_{d_i>0}m_id_i\geq\frac{c_1D}{8q}.\] Choose an outward index contributing at least \(1/N\) of this moment. Its upper displacement bound gives \(m_i\geq c_1/(12N)\geq m_*\), proving (72).

The parameter inequality and nonnegativity of the outer template give \(m_iB_o(s_i)\leq1\), proving its cap bound. The lower level bound is already known. The positive norm increment implies \(s_i\leq s+\zeta/2<3/4+1/32<1-\eta\). Thus the selected endpoint lies in the region where (49) applies when \(M_0=2/m_*\). ◻

Proof. The contact relation is closed. Its projections of Borel sets are analytic, hence measurable for the completed measures. The optimal coupling has full mass on \(\mathcal C\). At every differentiability point of either potential its contact endpoint is unique: all minimizing first-variation logs must equal its gradient, and their exponential endpoints therefore coincide. Since both marginals are absolutely continuous, almost every contact pair lies in the sets where both potentials are differentiable. The two contact maps on these sets are inverse to each other.

Write \(\mu=\rho_0\mathop{\mathrm{vol}}\) and \(\nu=\rho_1\mathop{\mathrm{vol}}\). Consequently \(\nu(\mathcal C(A))=\mu(A)\) for sets on the first side, and \(\mu(\mathcal C(A))=\nu(A)\) for sets on the second side: any additional contact image outside the almost-everywhere map image consists, up to a null set, of points where the opposite potential is not differentiable. The density bounds give (75). ◻

Proof. It suffices first to prove the result after decreasing the Hessian by a small positive multiple of the metric. Thus we may arrange strict touching, with \(v-\phi\ge\varepsilon d(\cdot,y)^2\) in a small coordinate ball. The map \[S(z)=\exp_z\mathop{\mathrm{grad}}\phi(z)\] is a local diffeomorphism near \(y\). Indeed its defining equation is \(D_zc(z,S(z))+\mathop{\mathrm{grad}}\phi(z)=0\), and differentiation gives \[ |\det DS(z)| =\sigma_z(\mathop{\mathrm{grad}}\phi(z)) \det\{\mathop{\mathrm{Hess}}\phi(z)+A_z(\mathop{\mathrm{grad}}\phi(z))\}. \tag{77}\] The relative Hessian is positive throughout a sufficiently small ball, and the logs remain before cut.

For \(h>0\), let \(\Omega_h\) be the connected component containing \(y\) of \(\{\phi+h>v\}\) inside that ball. It is relatively compact and shrinks to \(y\). If \(z\in\Omega_h\) and \(x=S(z)\), the cost mountain \[m_x(w)=\phi(z)+c(z,x)-c(w,x)\] is tangent to \(\phi\) at \(z\). On a fixed smaller coordinate ball, \(\phi-m_x\) has positive Hessian, uniformly for \(x\) close to \(S(y)\); thus \(\phi\ge m_x\) there. On \(\partial\Omega_h\) we have \(v-m_x\ge h\), whereas at \(z\) we have \(v(z)-m_x(z)<h\). The minimum of \(v-m_x\) on \(\overline{\Omega_h}\) is therefore attained at an interior point \(z'\). Thus \(\mathop{\mathrm{grad}}m_x(z')\in\partial v(z')\). By Theorem 6, the corresponding local contact is a global cost contact. Its endpoint is \(x\). This proves \[S(\Omega_h)\subset\mathcal C(\Omega_h).\] Use the area formula for the diffeomorphism \(S\) (Simon 2018, chap. 2, Theorem 3.3), followed by Lemma 26, to obtain \[\inf_{\Omega_h}|\det DS|\,\mathop{\mathrm{vol}}(\Omega_h) \le \mathop{\mathrm{vol}}(S(\Omega_h)) \le(\Lambda/\lambda)\mathop{\mathrm{vol}}(\Omega_h).\] Let \(h\downarrow0\) and then restore the original Hessian. ◻

Proof. Use normal coordinates \(x'=\exp_xa\) and set \[B(a,q)=c_t(\exp_xa,z(q)),\quad f(q)=F(z(q)),\quad a(q)=\log_xx(z(q)).\] At the reference point \(a(p)=0\). First variation gives the exact identities \[B(0,q)=\tfrac t2|q|^2,\qquad B_a(0,q)=-q.\] Thus \(B_{qq}=t\mathrm{Id}\), \(B_{qa}=-\mathrm{Id}\) at \((0,p)\). The envelope identity \(Df(q)=B_q(a(q),q)\) and differentiability of the pole imply \[ D^2f=t\mathrm{Id}-R,\qquad R=D_qa(p)=L. \tag{81}\] In particular \(R\) is symmetric. Since \(f(q)\le g(x)+B(0,q)\) with equality at \(p\), we have \(L\ge0\).

Choose a common \(s\in(t,1)\). For each \(j\) define \[y_j(d)=\exp_x(p_j+d),\qquad z_j(d)=\exp_x(s(p_j+d)),\] and the split upper cost \[ C_j(a,d)= \frac{c(\exp_xa,z_j(d))}{s} +\frac{c(z_j(d),y_j(d))}{1-s}. \tag{82}\] Both cost pieces are smooth near \((0,0)\) because they are proper subsegments of a minimizing ray. Energy splitting gives \(c(\exp_xa,y_j(d))\le C_j(a,d)\). At \(a=0\), the chosen geodesic remains minimizing on both pieces for small \(d\), and consequently \[ C_j(0,d)=\tfrac12|p_j+d|^2,\quad (C_j)_a(0,d)=-(p_j+d),\quad (C_j)_{ad}=-\mathrm{Id},\quad (C_j)_{aa}=\frac1sA_x(sp_j). \tag{83}\] The use of the fixed reference pole in \(z_j(d)\) is what makes the mixed derivative exactly \(-\mathrm{Id}\).

The nonnegative function \[g(x(z(q)))+C_j(a(q),0)+w_o(y_j)\] vanishes at \(p\). It is differentiable there, since \(g(x(z(q)))=f(q)-B(a(q),q)\), and its derivative equals \((p-p_j)^*R\). This derivative is zero. Symmetry in (81) proves \(LE=0\).

On \(E^\perp\), MTW concavity on the convex hull of the \(sp_j\), followed by divided-action monotonicity, gives \[Q_s\ge \frac1tA_x(tp)-\frac1s A_x(sp)>0.\] All points used in this inequality are before cut by Theorem 6; equivalently one may invoke Lemma 4. Notice that unsplit endpoints may still be conjugate.

Fix \(i\) and write \(m=m_i\). Given \(d\), choose any linear map \(U:T_xM\to E\), and set \(k=m(d+Ud)\) and \(a=a(p+k)\). In the lower bound defining \(g(\exp_xa)\), move endpoint \(i\) to \(y_i(d)\), keep all other endpoints fixed, replace the costs by (82), and average with the old weights. Write \(\widehat\Psi_i(d)=\Psi_i(d)-\Psi_i(0)\). The resulting inequality is \[f(p+k)\ge B(a,p+k)-\sum_jm_j\{C_j(a,d_j)+w_o(y_j(d_j))\},\] where \(d_i=d\) and \(d_j=0\) for \(j\ne i\). Subtract the equality at \((a,k,d)=(0,0,0)\) and use (81)–(83). The linear terms cancel: the pole terms cancel because \(\sum m_jp_j=p\), and the prefix terms cancel because \(Df(p)=tp\). The remaining inequality is \[ m\widehat\Psi_i(d)\ge \tfrac12L(k,k)+\tfrac12Q_s(a,a) +a\cdot(md-k)+o(|d|^2). \tag{84}\] Here \(a=Lk+o(|d|)\). Since \(md-k=-mUd\) and \(\operatorname{range}L\subset E^\perp\), the mixed term is \(o(|d|^2)\). Hence the lower Hessian is \[m(\mathrm{Id}+U)^*(L+LQ_sL)(\mathrm{Id}+U).\] In fact this equals \(m(L+LQ_sL)\) because \(LU=0\). The latter dominates \(mL\), proving the asserted jet. The linear term is exactly zero independently of any second function or maximum condition.

The map \(d\mapsto y_i(d)\) need not be invertible for the preceding pullback calculation. When \(p_i\) is nonconjugate it is a local diffeomorphism, and \(\tfrac12|p_i+d|^2\) is the smooth chosen upper branch of \(c(x,y_i(d))\). This converts the pullback jet into an endpoint jet. ◻

Proof. Write \(\tau_j=1-t_j\), and let \(y_j\) be the unique final minimizer. Set \[r_j=\tau_j^{-1}\log_{y_j}z_j,\qquad l_j=\varphi_c'(v(y_j))<1.\] The short minimum gives a smooth local lower test for \(w_c\) with gradient \(r_j\). Inverting \(\varphi_c\) gives a local lower test for \(v\) with gradient \(r_j/l_j\). Theorem 6 therefore implies that \[\zeta_j=\exp_{y_j}(r_j/l_j)\] is a true contact endpoint of \(v\), joined by the indicated minimizing ray. In particular \(\widehat x_j=\exp_{y_j}r_j\) is strictly before cut along that ray.

For the moment omit subscripts. Put \(\theta=(1+l^{-1})/2\) and \(\xi=\exp_y(\theta r)\). Split the true support at \(\xi\), using the fraction \(l\theta\) of its unit duration. The resulting upper cost support is \[C(h)=\frac{c(h,\xi)}{l\theta} +\frac{c(\xi,\zeta)}{1-l\theta}.\] It agrees with \(c(h,\zeta)\) at \(h=y\), and \(v(h)\ge v(y)+c(y,\zeta)-C(h)\). Thus \[\psi(h)=\varphi_c\{v(y)+c(y,\zeta)-C(h)\}\] is a smooth local lower support for \(w_c\), with \[ \mathop{\mathrm{grad}}\psi(y)=r,\qquad \mathop{\mathrm{Hess}}\psi(y) =-\theta^{-1}A_y(\theta r) +\frac{\varphi_c''(v(y))}{l^2}rr^*. \tag{89}\] The original true endpoint \(\zeta\) may be conjugate. The point \(\xi\) is halfway into the available extension and is before cut, which is all that this construction uses.

At the fixed scales, \(1-l\) has a positive lower bound and the parameters \((y,r,l,v(y))\) range over a compact set. The first cost piece in \(C\) is uniformly smooth on a fixed neighborhood of \(y\); the second piece is constant in \(h\). Consequently the supports \(\psi\) have uniform smooth bounds and a uniform supporting radius. For sufficiently small \(\tau\), the local infimum \(Q_\tau\psi\) is smooth near \(z\), has minimizer \(y\) at \(z\), and supports \(Q_\tau w_c\) from below there. To justify the last assertion, all minimizers of the latter lie within \(O(\tau)\) of \(z\), by the Lipschitz bound on \(w_c\), and so lie in the supporting neighborhood. For \(\psi\) the local Hessian of the minimizing expression is \(\tau^{-1}\mathrm{Id}+O(1)>0\). Equality and stationary first variation hold at \(y\). The implicit-function theorem and the Schur formula for a minimum give, after parallel identification, \[ \mathop{\mathrm{Hess}}Q_\tau\psi(z)=\mathop{\mathrm{Hess}}\psi(y)+O(\tau). \tag{90}\] All constants in this statement are uniform for the fixed-scale family just described.

Let \(\widehat p=\log_{\widehat x}y\) and \(\widehat L=(d\exp_{\widehat x})_{\widehat p}\). Lemma 14, applied to the minimizing ray through duration \(1/l\), gives \[\widehat L^* \{A_y(r)-\theta^{-1}A_y(\theta r)\}\widehat L \ge c(1-l)\mathrm{Id}.\] Gauss’s Lemma gives \(\widehat L^*r=-\widehat p\). Combining this with (89) yields \[ \widehat L^*\{\mathop{\mathrm{Hess}}\psi(y)+A_y(r)\}\widehat L \ge c(1-l)\mathrm{Id} +\frac{\varphi_c''(v(y))}{l^2} \widehat p\widehat p^*. \tag{91}\] Here \(1-l\ge c b/D\) and \(|\varphi_c''(v(y))|/l^2\le Cb/D^2\).

We finally compare the reference pole with \(\widehat x_j\). The envelope identities give \[x_j=\exp_{z_j}(-t_j\mathop{\mathrm{grad}}F_j(z_j)),\qquad \widehat x_j=\exp_{z_j}(t_j\mathop{\mathrm{grad}}G_j(z_j)).\] They are at distance tending to zero. The pairs \((\widehat x_j,y_j)\) have a uniform fixed-scale extension, so their cost charts and inverse exponential charts are uniformly smooth. Omit subscripts again and put \(U=Q_\tau\psi\), \(Z(q)=\exp_x(tq)\), \(M=D_qZ(p)\), and \(R=\nabla_q^2Z(p)\). Since \(G-U\) has a local minimum at \(z\), its first derivative vanishes and its second-order form is positive semidefinite. Consequently \[H\ge H_\psi :=t\mathrm{Id}+M^*\mathop{\mathrm{Hess}}U(z)M+\langle\mathop{\mathrm{grad}}U(z),R\rangle.\] Only the smooth support \(U\) occurs on the right. Its gradient and Hessian are uniformly bounded at the fixed scales, and (90) identifies its Hessian with that of \(\psi\), up to \(O(\tau)\) after parallel transport.

Write \(\delta=|\mathop{\mathrm{grad}}(F+G)(z)|\) and identify nearby pole tangent spaces by parallel transport. The explicit pole formulas give \(d(x,\widehat x)=O(\delta)\). Moreover \(p=t^{-1}\log_xz\) and \(\widehat p=t^{-1}\log_{\widehat x}z\) differ by \(O_{\mathrm{fixed}}(\delta)\), since the latter ray has a uniform fixed-scale extension. Thus replacing \(x,p\) by \(\widehat x,\widehat p\) in \(M,R\) changes their entries by \(O_{\mathrm{fixed}}(\delta)\). Changing \(t\) to \(1\) then changes them by \(O(\tau)\). Every resulting error in \(H_\psi\) multiplies a bounded derivative of \(U\).

For the full endpoint map \(Y(q)=\exp_{\widehat x}q\), let \(B=DY(\widehat p)\). The identity \(c(\widehat x,Y(q))=|q|^2/2\) on the chosen branch gives \[\mathrm{Id}=B^*A_y(r)B-\langle r,\nabla_q^2Y(\widehat p)\rangle.\] Since \(\mathop{\mathrm{grad}}\psi(y)=r\), the preceding expansions yield \[H\ge H_\psi =B^*\{\mathop{\mathrm{Hess}}\psi(y)+A_y(r)\}B +O_{\mathrm{fixed}}(\tau+\delta).\] Combining this with (91) proves (88). No coordinate error multiplies the unknown Hessian of \(G\). ◻

Proof. The active inequality reads \(\varphi_o(v(h))\ge-c(x,h)-g_b(x)\). For \(1/l<s<1\), put \(\xi_s=\exp_y(sr)\) and form the smooth upper support \[C_s(h)=c(h,\xi_s)/s+c(\xi_s,x)/(1-s).\] It has value \(c(y,x)\) and gradient \(-r\) at \(y\). Since \(\varphi_o\) is increasing, \(\varphi_o^{-1}(-C_s-g_b(x))\) is a smooth lower test for \(v\) with gradient \(r/l\). That log is strictly before cut because it shortens a minimizing ray. It also gives a global contact by Theorem 6, but its before-cut property does not depend on a density equation.

The scaled true relative Hessian of this test is \[ W_s=lA_y(r/l)-s^{-1}A_y(sr) -\frac{\varphi_o''(v(y))}{l^2}rr^*. \tag{92}\] Fix \(s_0\in(1/l,1)\). For \(s\ge s_0\), divided-action monotonicity gives a positive definite lower bound for the difference of the first two terms, independent of \(s\). The last term is positive semidefinite. If the endpoint at duration one were conjugate, the index-form characterization of conjugacy, equivalently Lemma 12, would force an eigenvalue of \(s^{-1}A_y(sr)\) to tend to \(-\infty\) as \(s\uparrow1\). Thus \(\det W_s\to\infty\), while \(W_s\) retains a positive definite baseline. Lemma 27, at the fixed true log \(r/l\), instead gives \[\det W_s\le l^n(\Lambda/\lambda)/\sigma_y(r/l)<\infty.\] This is a contradiction. ◻

Proof. Lemma 30 makes the endpoint coordinate map locally invertible. Its smooth cost branch has endpoint Hessian \(A_y(r)\), and the zero linear coefficient fixes the gradient of the induced lower test of \(w_o\) to be \(r\). Inverting \(\varphi_o\) and using the chain rule shows that (93) is \(l\) times the true relative lower-test Hessian, pulled back by \(B\). Gauss’s Lemma gives the rank-one term displayed there.

The divided-action difference between durations \(1/l\) and \(1\) is strictly positive definite, so \(V_o>0\). Decrease the lower Taylor Hessian by \(\varepsilon\mathrm{Id}\) to obtain a genuine lower test, apply Lemma 27, and let \(\varepsilon\downarrow0\). Taking the determinant of the scaling and pullback gives (94).

For the uniformity assertion, include the minimizing log among the parameters. The family is compact and active contact is a closed condition. Lemma 30 excludes a conjugate limit anywhere in that family; radial shortening excludes true cut limits. The smooth branch matrices and the positive divided-action difference therefore have the claimed uniform bounds by compactness. Multiple ordinary branches cause no difficulty: each specified nonconjugate minimizing branch is a smooth upper branch and has been included among the parameters. ◻

Proof. Adding a fixed quadratic reduces the assertion to a convex function, and subtracting it later restores the displayed formula. In the convex case write the remainder as \(R(h)\), where \(|R(h)|\le\omega(|h|)|h|^2\) for a nondecreasing modulus \(\omega(r)\to0\). For \(r=|h|>0\), a unit vector \(e\), and \(0<\alpha\le1\), the subgradient inequalities at \(h\pm\alpha r e\) give \[|(\zeta-p-Ah)\cdot e| \le \tfrac12\|A\|\alpha r +5\omega(2r)r/\alpha.\] For small \(r\), choose \(\alpha=\sqrt{\max\{\omega(2r),r^2\}}\) and take the supremum over unit \(e\). This proves the uniform assertion. ◻

Proof. Step 1: fixed scales and exact maxima. Throughout the proof the original potential \(v\), the scales, and both templates are fixed. Constants used only to pass limits may depend on all of them. Choose \(t_k\uparrow1\) and interior maxima \((z_k^0,b_k)\) supplied by Lemma 22, for \[F_k(z)=Q_{t_k}g_{b_k}(z),\qquad G_k(z)=Q_{1-t_k}w_{c,b_k}(z).\] The functions are \(C^{1,1}\) in space: Proposition 9 applies to the first, and Lemma 5 to the second. At an exact maximum the velocities align. If \(x_k^0\) and \(y_k^0\) are its two minimizers and \(p_k^0=t_k^{-1}\log_{x_k^0}z_k^0\), then \(y_k^0=\exp_{x_k^0}p_k^0\) along the concatenated minimizing ray. Every finite positive active representation of \(p_k^0\) satisfies the parameter inequality of Lemma 23. The level relations and common section bounds of Lemma 24 hold for these representations.

Step 2: null sets are removed before Jensen sampling. Let \(N\) be the null set for the original potential in (78). For this fixed \(k\), the final minimizer map \(Y_k(z)\) is locally Lipschitz and hence differentiable almost everywhere (Simon 2018, chap. 2, Theorem 1.4). In each coordinate chart the locally Lipschitz area formula (Simon 2018, chap. 2, Theorem 3.3) implies that \[ Z_k=\{z:DY_k(z)\text{ exists},\ \det DY_k(z)\ne0,\ Y_k(z)\in N\} \tag{100}\] is a null set. Indeed the integral of \(|\det DY_k|\) on \(Y_k^{-1}(N)\) is zero. This argument does not require us to know in advance where \(DY_k\) is nonsingular.

Apply the semiconvex maximum Lemma (Crandall et al. 1992, Appendix, Lemmas A.3–A.4) to \(F_k+G_k\) at \(z_k^0\), using a small fourth-order strictness penalty if needed. Its positive-measure conclusion allows us to choose a sequence \(z_{k,\nu}\to z_k^0\) off \(Z_k\) and off all the null sets where \(F_k,G_k\), the prefix pole map, or \(Y_k\) lack the derivatives used below. We have \[ \mathop{\mathrm{grad}}(F_k+G_k)(z_{k,\nu})\longrightarrow0,\qquad D^2(F_k+G_k)(z_{k,\nu})\le \eta_{k,\nu}\mathrm{Id},\quad \eta_{k,\nu}\longrightarrow0 \tag{101}\] in the fixed chart. At every such point choose a representation with at most \(n+1\) active logs, padding by zero weights only for compactness. Lemma 28 applies at each point. Its matrices \(L_{k,\nu}\) are positive semidefinite and annihilate the active differences, exactly; its endpoint lower jets are \(m_{i,k,\nu} L_{k,\nu}\). The matrices \(H_{k,\nu}\) satisfy (87).

Step 3: take the first, fixed-prefix limits. For each fixed \(k\), the local \(C^{1,1}\) bounds are finite. After taking a subsequence in \(\nu\), all configuration parameters, weights, and the two matrices \(H_{k,\nu},L_{k,\nu}\) converge. Denote the limits by \(p_k,p_{i,k},m_{i,k},H_k,L_k\) and the corresponding base points. Discard zero limiting weights. The limiting endpoints are active at the exact maximum, and their weighted sum is \(p_k\). In particular the universal parameter inequality applies to this limiting representation; we have not asserted forward persistence of any prescribed active vertex. We retain \[ L_k\ge0,\qquad L_kE_k=0,\qquad H_k\le L_k, \quad E_k=\mathop{\mathrm{span}}\{p_{i,k}-p_k:m_{i,k}>0\}. \tag{102}\] The quantitative proof of Lemma 29 has an error bounded by a fixed-scale constant times \(1-t_k+|\mathop{\mathrm{grad}}(F_k+G_k)(z_{k,\nu})|\). Passing \(\nu\to\infty\) therefore gives \[H_k\ge c\frac{b_k}{D}\mathrm{Id} -C\frac{b_k}{D^2}p_kp_k^* -C_{\mathrm{fixed}}(1-t_k)\mathrm{Id}.\] Since \(b_k\ge b_->0\), for sufficiently large \(k\) this implies, after reducing \(c\), \[ H_k\ge c\frac{b_k}{D}\mathrm{Id} -C\frac{b_k}{D^2}p_kp_k^*. \tag{103}\]

If \(\dim E_k\ge2\), a nonzero vector in \(E_k\cap p_k^\perp\) contradicts (102) and (103). If \(E_k=0\), all active logs in the representation equal \(p_k\) and all endpoints equal the center. But \(B_c\le B_o\) at every level, contradicting the strict parameter inequality. Thus \(E_k=\mathbb Re_k\). Orient \(e_k\) so \(q_k=p_k\cdot e_k>0\); the same two inequalities give \(q_k^2\ge cD\). Lemma 25 now supplies an index with \(m_{i,k}\ge m_*\), a strict outward displacement, and a bounded template value. We choose an index carrying a fixed share of the outward moment, as permitted by that Lemma, so its displacement has a positive lower bound at the fixed scales. Relabel it as index \(1\).

Step 4: fix the outer branch before bounding any center Hessian. Pass to a subsequence in \(k\) so that the selected configurations and their weights converge. Their centers and selected endpoints remain in the compact level ranges of Lemma 25. The selected limiting weight is at least \(m_*\), its displacement is strictly outward, and its outer template value is below the prescribed cap. By imposing the template conditions under a slightly larger fixed cap, the limiting slope and curvature conditions have strict margins: \(B_o'>0\) and \(B_o''<-K\). The limiting selected pair is active. Lemma 30 excludes conjugacy of its modified branch, using only this activity and the original contact-volume bound. Its shortened true log is before cut. Lemma 31 consequently supplies a neighborhood of this configuration with constants \(\delta>0\) and \(M<\infty\) such that its divided-action gain is at least \(\delta\mathrm{Id}\) and its determinant upper bound is at most \(M\). These constants may depend on the fixed scales and templates.

For every remaining \(k\), return to its already chosen Jensen subsequence and choose \(\nu=\nu(k)\) sufficiently large. We require the configuration and matrices to be within \(1/k\) of their fixed-prefix limits, the selected weight to be at least \(m_*/2\), and the selected endpoint to lie in that neighborhood. We also require both errors in (101), including their transformed versions in (87), to be at most \((1-t_k)^3/k\). All these requirements are compatible because the limit in \(\nu\) is taken with \(k\) fixed. For the rest of the proof, a subscript \(k\) denotes this diagonal sequence of actual Jensen points.

Step 5: the outer test bounds the exact pole matrix. At each diagonal point, Lemma 28 gives the actual lower Taylor Hessian \(H_{o,k}=m_{1,k}L_k\ge0\) at the selected endpoint. Use Lemma 31 to undo the modification and call the resulting form \(V_{1,k}\). The positive outer curvature term and the uniform divided-action baseline give \[ V_{1,k}\ge m_{1,k}L_k+\delta\mathrm{Id},\qquad \det V_{1,k}\le M. \tag{104}\] Every eigenvalue of \(V_{1,k}\) is at least \(\delta\), so its largest eigenvalue is at most \(M\delta^{1-n}\). Since \(m_{1,k}\ge m_*/2\), this bounds \(L_k\) from above. The comparison \(H_k\le L_k+o(1)\mathrm{Id}\) and Lemma 29 now bound \(H_k\) from both sides. There has still been no use of a center density equation.

The center extension gives uniformly nonsingular fixed-scale prefix-log coordinate maps, as in the proof of Lemma 29. Their derivatives and second derivatives are bounded; the gradients of \(G_k\) are bounded by the Lipschitz bound of the datum. Thus the coordinate change in (85) converts the bound on \(H_k\) into \[ \|\mathop{\mathrm{Hess}}G_k(z_k)\|\le C_{\mathrm{fixed}}. \tag{105}\]

Step 6: the chosen centers are good original endpoints. Write \(\tau_k=1-t_k\), \(y_k=Y_k(z_k)\), and \(K_k(z,y)=c(z,y)/\tau_k\). The envelope identity \(\mathop{\mathrm{grad}}G_k=(K_k)_z(z,Y_k(z))\) and differentiability of \(Y_k\) give \[\mathop{\mathrm{Hess}}G_k=(K_k)_{zz}+(K_k)_{zy}DY_k.\] In parallel orthonormal frames, the distance between \(z_k\) and \(y_k\) is \(O(\tau_k)\), and the smooth short-cost expansions are \[(K_k)_{zz}=\tau_k^{-1}\mathrm{Id}+O(\tau_k),\quad (K_k)_{yy}=\tau_k^{-1}\mathrm{Id}+O(\tau_k),\quad (K_k)_{zy}=-\tau_k^{-1}\Pi+O(\tau_k),\] where \(\Pi\) is the appropriate parallel isometry. Solving the preceding identity, using (105), gives \[ DY_k=\Pi^{-1}\{\mathrm{Id}-\tau_k\mathop{\mathrm{Hess}}G_k\}+O(\tau_k^2). \tag{106}\] It is nonsingular for large \(k\). The Jensen points were chosen off the set (100); hence \(y_k\notin N\). This is the promised justification for using the original Alexandrov and Jacobian statements at the centers.

In the following formulas the scalar modifications are evaluated with \(b=b_k\). At these points \(w_c=\varphi_c\circ v\) has an Alexandrov Hessian. The minimizing property supplies a proximal subgradient \[\xi(z)=-(K_k)_y(z,Y_k(z))\in\partial w_c(Y_k(z))\] at every nearby point, even if \(w_c\) is not differentiable there. Apply Lemma 32 at the Alexandrov point \(y_k=Y_k(z_k)\). Since \(Y_k\) is Lipschitz and differentiable at \(z_k\), it gives, in normal coordinates at \(z_k\) and \(y_k\), \[\xi(z_k+h)=\mathop{\mathrm{grad}}w_c(y_k) +\mathop{\mathrm{Hess}}w_c(y_k)DY_k(z_k)h+o(|h|).\] Expanding the smooth function \((K_k)_y\) in its two variables therefore gives \[\{\mathop{\mathrm{Hess}}w_c(y_k)+(K_k)_{yy}\}DY_k+(K_k)_{yz}=0.\] Thus stationary differentiation uses the subgradient expansion even when neighboring endpoints are nonsmooth. Solving it with (106) gives, in parallel frames, \[ \mathop{\mathrm{Hess}}w_c(y_k)=\mathop{\mathrm{Hess}}G_k(z_k)+O(\tau_k). \tag{107}\] In particular these endpoint Hessians are uniformly bounded. The error statement follows directly from the displayed short-cost expansions and the bounded inverse in (106); it does not assume a bound for \(\mathop{\mathrm{Hess}}w_c\) in advance.

Step 7: insert the true center form and take matrix limits. Set \[r_k=\tau_k^{-1}\log_{y_k}z_k,\quad l_k=\varphi_c'(v(y_k)),\quad \widehat x_k=\exp_{y_k}r_k,\quad \widehat p_k=\log_{\widehat x_k}y_k.\] The true center endpoint is \(\zeta_k=\exp_{y_k}(r_k/l_k)\), and it is before cut because \(y_k\notin N\). Identify \(T_{x_k}M\) with \(T_{\widehat x_k}M\) by parallel transport over their short joining segment, and let \(B_k:T_{x_k}M\to T_{y_k}M\) be the exponential differential at \(\widehat p_k\) after that identification. Define \[ V_k=l_k B_k^* \{\mathop{\mathrm{Hess}}v(y_k)+A_{y_k}(r_k/l_k)\}B_k>0. \tag{108}\] The exact chain rule for the scalar modification gives \[\begin{align*} B_k^*\{\mathop{\mathrm{Hess}}w_c(y_k)+A_{y_k}(r_k)\}B_k ={}& V_k +B_k^*\{A_{y_k}(r_k)-l_kA_{y_k}(r_k/l_k)\}B_k \\ &+\frac{\varphi_c''(v(y_k))}{l_k^2} \widehat p_k\widehat p_k^* . \tag{109}\end{align*}\] All vectors on the last line are understood after the same isometric identification. Lemma 14 bounds the middle term below by \(c(1-l_k)\mathrm{Id}\). Equation (107) and the vanishing pole mismatch show that the left side differs from \(H_k\) by \(o(1)\mathrm{Id}\). The verification is the same smooth coordinate comparison as in Lemma 29, now using the bounded unregularized Hessians in place of the smooth support Hessians. This proves \[H_k\ge V_k+c\frac{b_k}{D}\mathrm{Id} -C\frac{b_k}{D^2}p_kp_k^*-o(1)\mathrm{Id}.\] It also bounds \(V_k\) from above, because \(V_k>0\) and \(H_k\) is bounded.

Taking the determinant in (108) and using (78) gives the exact factorization \[ \det V_k =l_k^n\sigma_{\widehat x_k}(\widehat p_k)^2 \frac{|\det D(\exp_{\cdot}\mathop{\mathrm{grad}}v)(y_k)|} {\sigma_{y_k}(r_k/l_k)} \ge c\,\frac{\sigma_{\widehat x_k}(\widehat p_k)^2} {\sigma_{y_k}(r_k/l_k)}. \tag{110}\] The squared factor is the determinant of the Hessian pullback. All scalar derivatives lie between \(1/2\) and \(2\), so the constants in this density inequality are independent of \(K\).

There is no assertion that the limiting endpoint is an Alexandrov point of \(v\). We take limits of matrices at the good endpoints just selected. The center modified rays have a uniform extension at the fixed scales: \(1-l_k\ge c b_-/D>0\), and the ray through \(\widehat x_k\) continues to the true endpoint through duration \(1/l_k\). Compactness of this family of proper minimizing subsegments therefore gives \[\sigma_{\widehat x_k}(\widehat p_k)\ge c_{\mathrm{fixed}}>0.\] Since \(0<V_k\le C_{\mathrm{fixed}}\mathrm{Id}\), equation (110) now implies \[\sigma_{y_k}(r_k/l_k) \ge \frac{c\,\sigma_{\widehat x_k}(\widehat p_k)^2} {\det V_k} \ge c'_{\mathrm{fixed}}>0.\] Continuity of the exponential differential excludes a conjugate limit of the true center rays. An ordinary cut limit may remain; the specified nonsingular branch suffices.

Finally take a subsequence so that \(b_k,H_k,L_k,V_k,V_{1,k}\) and all configuration parameters converge, in a smooth orthonormal trivialization. Discard any additional zero limiting weights. The selected weight remains at least \(m_*\), and its outward displacement retains the positive fixed-scale lower bound from Step 3. The barycenter identity then gives another surviving vertex, so the surviving difference span is still the same nonzero line. The selected modified and true outer branches were already uniformly nonsingular in Step 4; the center modified branch has its fixed-scale extension. Equation (110) gives a positive lower determinant in the limit, so \(V>0\). Equations (80), (87), and (109) give (97).

At each selected outer endpoint in the diagonal sequence, formula (93), with \(H_{o,k}=m_{1,k}L_k\), has a positive divided-action term. Its scalar curvature coefficient satisfies \[-\frac{\varphi_o''(v(y_{1,k}))}{l_{1,k}^2} =-\frac{b_kB_o''((v(y_{1,k})-a)/D)}{D^2l_{1,k}^2} \ge cK b_k/D^2.\] Passing this inequality and (94) along \(V_{1,k}\to V_i\) gives (98) and the other inequality in (99). No lower Taylor jet is asserted at the limiting endpoint. The weights, directions, and section-level estimates pass through the same chosen subsequence and give (95)–(96). Their constants and the cap were fixed before \(K\). Only \(\delta,M\), the rate of the prefix limit, and the Jensen radii depended on the fixed outer template. ◻

Proof. Put \(\beta=Cb/D^2\), increasing \(C\) if necessary, and \(\gamma=cKb/D^2\). Define \[P=\mathrm{Id}-e p^*/q.\] This is the projection onto \(p^\perp\) along \(e\). Since \(LE=0\), \(P^*LP=L\). On \(p^\perp\) the center inequality gives \(H\ge V\), and \(L\ge H\). Thus \[ V_i\ge m_iP^*VP+\gamma p_ip_i^*. \tag{112}\] Also \(H(e,e)\le L(e,e)=0\) implies \(V(e,e)\le\beta q^2\).

Let \(Q\) have orthonormal columns spanning \(p^\perp\), and use the nonsingular basis matrix \([Q,e]\). In this basis write \[[Q,e]^*V[Q,e] =\begin{pmatrix} A&z\\z^*&h\end{pmatrix},\qquad S=h-z^*A^{-1}z>0.\] Then \(S\le h=V(e,e)\le\beta q^2\). The determinant of the right side of (112), in the same basis, is \(m_i^{n-1}\det A\,\gamma(p_i\cdot e)^2\). Taking the ratio cancels the determinant of the nonorthogonal change of basis and gives \[\frac{\det V_i}{\det V} \ge m_i^{n-1}\frac{\gamma(p_i\cdot e)^2}{S} \ge m_i^{n-1}\frac{\gamma}{\beta} \frac{(p_i\cdot e)^2}{q^2}.\] Since \(p_i=p+d_ie\) with \(d_i>0\) and \(q>0\), we have \(p_i\cdot e\ge q\). Now use \(m_i\ge m_*\) and \(\gamma/\beta\ge cK\). ◻

Proof. Suppose otherwise, and take the sequence of scales supplied by that Lemma. Fix all center and weight constants, then choose \(K\) and its outer template, and then take sufficiently small scales to apply Proposition 33. Lemma 34 and (99) imply \[cK\,\frac{\sigma_x(p)^2}{\sigma_y(r/l)} \le \det V_i \le C\,\frac{\sigma_x(p_i)^2}{\sigma_{y_i}(r_i/l_i)}.\] The radial comparison of Corollary 13 gives \[\sigma_y(r/l)\le C\sigma_x(p),\qquad \sigma_{y_i}(r_i/l_i)\ge c\sigma_x(p_i).\] For the selected outward endpoint, (95)–(96) give \(m_i\ge m_*\) and \(q^2\ge cD\). Its norm increment satisfies \(|p_i|^2-|p|^2\le3D\) by Lemma 25. Corollary 16 therefore applies with \(\mu=m_*\), \(\kappa=c\), and \(L_0=3\), and gives \(\sigma_x(p)\ge c'\sigma_x(p_i)\). These parameters were fixed before \(K\). Hence \[cK\sigma_x(p)\le C\sigma_x(p_i)\le C'\sigma_x(p).\] The center modified branch is nonconjugate, so \(\sigma_x(p)>0\). This bounds \(K\) by a constant fixed before it was chosen. Choosing \(K\) larger gives a contradiction. ◻