Uniform Bi-Holder Transport from Weak MTW

Proof. Varying the initial velocity in the geodesic equation gives (3.1). Since \mathcal R is symmetric, the fundamental matrix

is symplectic. In particular,

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 (3.5) shows that this matrix is symmetric. Moreover, the first two identities give

Consequently

including the order of the two inverse factors in (3.2).

For completeness, symplecticity also gives

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 (3.1) proves (3.3); 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 (3.4); the case v=0 is immediate.□

Proof. The bundle

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

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 3.1, 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 3.1. 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

all have uniformly bounded norms. This compactness assertion also includes v=0, for which J(s)=s\Id.

Write z=\gamma(h) and y_T=\gamma(T). The Hessian in the joining point of the broken action

at z is

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

with equality at z. Hence D(1)\succeq0.

Only the second term in (3.11) depends on T. Applying (3.2) with initial point z yields

The bounds on G_\delta are precisely the bounds in (3.10): 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

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 (3.12) gives

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 (3.13), 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 (3.6) 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

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

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 (3.12); the composed action is the smooth cost branch of the shortened minimizing ray.

For T=1-\delta, formula (3.14) gives

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 (3.7) and (3.8). They imply the norm and lower bounds in (3.9). The remaining identity follows from (3.4). 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(\Id-TP).□

Proof. For \theta<1, multiply the lower bounds in (3.6) with \delta=1-\theta and discard the nonnegative terms \delta. The case \theta=1 is immediate. For (3.17), apply (3.16) to the longer vector \bar p/l with shortening factor l, and then use (3.3). For (3.18), 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,

Apply Lemma 3.2 to the minimizing vector T_*r, with shortening factor 1/T_*. Since \delta=(T_*-1)/T_*\ge(T_*-1)/2, its lower bound gives

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,

Write S(s)=A_y(sr)/s. Lemma 3.1 and (3.21) give

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 (3.23). 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 (3.19).

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 (3.20). 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

We use only the segment [-a,1]. First suppose that all its points lie in I(x). Write

The eigenvalues of L(t) are 1 on q^\perp and 1+t\beta on \R q. Hence

For 0<\delta\le1/2, put

These matrices are uniformly bounded by Lemma 3.2 and (3.28), and F_\delta(t)\succeq-C\delta\Id.

We next prove full matrix convexity of F_\delta. For any fixed z\in T_xM, the vector

is perpendicular to q. Moreover,

The last vector belongs to the kernel of N_\delta(t) by the radial identity. Symmetry consequently gives the exact equality

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 (3.30) 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 (3.28) and (3.30) 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 (3.8) 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 (3.26) 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

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 \R^n. No smooth choice in k is needed. Abbreviate

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

Exclude the intervals of radius R_k^{1/3} centered at

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

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 3.2.

Form the matrices F_k(t) in (3.29) using the kth 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\Id. The limit F is matrix convex and bounded.

To identify its possible ranks, use (3.7) to write

The positive form in this expression has its ith decreasing eigenvalue between uniform multiples of

by (3.8) and (3.28). At a fixed dense point t_m, put \eta_k=\exp(-R_k^{1/3}). For all sufficiently large k, depending on m, (3.33) 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

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

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 (3.35) persists in this limit, so its rank equals the interior rank. For fixed s<t<1, matrix convexity of F_k says

First let k\to\infty, then let t\uparrow1; this gives

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 (3.32) and the ordering of singular values make the first j eigenvalues in (3.34) 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 (3.36). The uniform ordered comparison is proved for open segments.

Finally, replace a closed segment in (3.25) by (1-\eps)(p+tq), with 0<\eps<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 \eps\downarrow0 and using continuity of the singular values proves (3.26) on G_x. Multiplication over j proves (3.27). 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 (3.37) gives

Because q\ne0, m_i<1. The remaining barycenter belongs to G_x by its convexity and is

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 3.5 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

By Corollary 2.7, 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

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

Thus y,y' belong to a section at x of that height. For sufficiently small d(x,x'), (4.1) gives

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

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

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

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 (4.2). 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 (4.3). Finally

□

Proof. Choose \gamma so large that 2/(1+e^{\gamma/8})\leq\eps_0, and set

This has all the asserted properties. In particular, |B_c'| has a positive minimum on every fixed compact interval.

For the outer function let

Thus \chi is a smooth step, equal to zero on (-\infty,0] and to one on [1,\infty). Put

Define

The function w is nonnegative and supported in [-3\eta,2h_0]. Also

It follows that

On [-2\eta,h_0], one has J'=L-\kappa s>0 and J''=-\kappa. Moreover

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 (4.9). The function J is constant beyond 2h_0\leq1/64, so (4.8) 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:

The initial gaps at y_0,y_1 lie in [0,r], whence

for small scales; the second inequality is (4.1) 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

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,

Equations (4.13) 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

By rationalizing the difference of the square roots in (4.13),

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,

We next anchor the absolute levels. By (4.11) and (4.17), u(x_\tau)-f_0(\tau)\leq bH_o+r. The initial-cost upper support and (4.16) give

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,

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

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

for all sufficiently small scales. This uses only (4.5); 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

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

Equation (4.17) and (4.5) prove (4.15).□

Proof. Use the two active logs at the switch supplied by Lemma 4.4. Equation (4.15) 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

Theorem 2.5 says that p is minimizing and that y=\exp_{x_b}p is a contact endpoint for g_b. Thus

Minorizing g_b by the left modified mountain gives the direct bound

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

Apply (4.12) at x_b with backward displacement -\tau_b\Pi_{\tau_b}p_0 to get

The corresponding cost difference for y_0 is exactly -P\tau_b+P\tau_b^2/2. Therefore

The norm choice gives

The middle term in (4.22) is consequently at least 1/128. The template difference is at least -2\eps_0, and the last two terms tend to zero uniformly by (4.5). The numerical choices (4.6) give the asserted lower bound. The upper bound is (4.19).□

Proof. At b=b_+, (4.20) 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,

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

Thus \Phi_t\to E_b-\pi uniformly in (b,z), preserving the positive interior maximum when t is sufficiently close to one.

Proposition 2.8 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

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 2.5, this full vector is minimizing and its endpoint is a support for g_b. Splitting its energy at time t proves (4.24) 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 2.5. 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 2.8 gives, in normal coordinates \Delta_h=\exp_x^{-1}x_h,

Choose one duration q\in(t,1) and split each active endpoint cost at z_i=\exp_x(qp_i). The energy inequality

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

All expansions here concern fixed scales, a fixed time, and a fixed finite representation. Equations (4.26)–(4.27) 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=\Span\{p_i-p\},

On E^\perp, MTW concavity on the simplex of the qp_i gives

Radial monotonicity of divided action, equivalently Lemma 2.3, then gives

Average (4.27). The linear terms cancel. By (4.26), (4.28), and (4.29), 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

The unique center minimizer and the elementary envelope argument (compare the old and new minimizers and use their convergence) give

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 (4.25). 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 (4.24), the center gap is u-g_b+E_b(y)-bB_c(s), which is at most (H_o+2)b by (4.19). This proves the common section assertion and its oscillation bound. Subtracting the active contact equality from (4.24) gives (4.30). 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

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 4.8. A right point has s_i-s>1/4-\eta, hence

It follows that d_i<0 and |d_i|\geq D/(8q). Every outward point satisfies, by the common level range,

and so d_i\leq3D/(2q). The barycenter identity \sum_i m_id_i=0 therefore implies

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 (4.32).

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 (4.9) 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\vol and \nu=\rho_1\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 (5.1).□

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\eps d(\cdot,y)^2 in a small coordinate ball. The map

is a local diffeomorphism near y. Indeed its defining equation is D_zc(z,S(z))+\grad\phi(z)=0, and differentiation gives

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

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 \grad m_x(z')\in\partial v(z'). By Theorem 2.5, the corresponding local contact is a global cost contact. Its endpoint is x. This proves

Use the area formula for the diffeomorphism S [15, Chapter 2, Theorem 3.3], followed by Lemma 5.1, to obtain

Let h\downarrow0 and then restore the original Hessian.□

Proof. Use normal coordinates x'=\exp_xa and set

At the reference point a(p)=0. First variation gives the exact identities

Thus B_{qq}=t\Id, B_{qa}=-\Id at (0,p). The envelope identity Df(q)=B_q(a(q),q) and differentiability of the pole imply

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

and the split upper cost

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

The use of the fixed reference pole in z_j(d) is what makes the mixed derivative exactly -\Id.

The nonnegative function

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 (5.7) proves LE=0.

On E^\perp, MTW concavity on the convex hull of the sp_j, followed by divided-action monotonicity, gives

All points used in this inequality are before cut by Theorem 2.5; equivalently one may invoke Lemma 2.3. 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 (5.8), and average with the old weights. Write \widehat\Psi_i(d)=\Psi_i(d)-\Psi_i(0). The resulting inequality is

where d_i=d and d_j=0 for j\ne i. Subtract the equality at (a,k,d)=(0,0,0) and use (5.7)–(5.9). 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

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

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

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 2.5 therefore implies that

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

It agrees with c(h,\zeta) at h=y, and v(h)\ge v(y)+c(y,\zeta)-C(h). Thus

is a smooth local lower support for w_c, with

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}\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,

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 3.4, applied to the minimizing ray through duration 1/l, gives

Gauss’s Lemma gives \widehat L^*r=-\widehat p. Combining this with (5.15) yields

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

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

Only the smooth support U occurs on the right. Its gradient and Hessian are uniformly bounded at the fixed scales, and (5.16) identifies its Hessian with that of \psi, up to O(\tau) after parallel transport.

Write \delta=|\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

Since \grad\psi(y)=r, the preceding expansions yield

Combining this with (5.17) proves (5.14). 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

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 2.5, but its before-cut property does not depend on a density equation.

The scaled true relative Hessian of this test is

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 3.2, 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 5.2, at the fixed true log r/l, instead gives

This is a contradiction.□

Proof. Lemma 5.5 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 (5.19) 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 \eps\Id to obtain a genuine lower test, apply Lemma 5.2, and let \eps\downarrow0. Taking the determinant of the scaling and pullback gives (5.20).

For the uniformity assertion, include the minimizing log among the parameters. The family is compact and active contact is a closed condition. Lemma 5.5 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

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 4.6, for

The functions are C^{1,1} in space: Proposition 2.8 applies to the first, and Lemma 2.4 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 4.7. The level relations and common section bounds of Lemma 4.8 hold for these representations.

Step 2: null sets are removed before Jensen sampling. Let N be the null set for the original potential in (5.4). For this fixed k, the final minimizer map Y_k(z) is locally Lipschitz and hence differentiable almost everywhere [15, Chapter 2, Theorem 1.4]. In each coordinate chart the locally Lipschitz area formula [15, Chapter 2, Theorem 3.3] implies that

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 [3, 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

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 5.3 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 (5.13).

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

The quantitative proof of Lemma 5.4 has an error bounded by a fixed-scale constant times 1-t_k+|\grad(F_k+G_k)(z_{k,\nu})|. Passing \nu\to\infty therefore gives

Since b_k\ge b_->0, for sufficiently large k this implies, after reducing c,

If \dim E_k\ge2, a nonzero vector in E_k\cap p_k^\perp contradicts (5.28) and (5.29). 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=\R e_k. Orient e_k so q_k=p_k\cdot e_k>0; the same two inequalities give q_k^2\ge cD. Lemma 4.9 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 4.9. 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 5.5 excludes conjugacy of its modified branch, using only this activity and the original contact-volume bound. Its shortened true log is before cut. Lemma 5.6 consequently supplies a neighborhood of this configuration with constants \delta>0 and M<\infty such that its divided-action gain is at least \delta\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 (5.27), including their transformed versions in (5.13), 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 5.3 gives the actual lower Taylor Hessian H_{o,k}=m_{1,k}L_k\ge0 at the selected endpoint. Use Lemma 5.6 to undo the modification and call the resulting form V_{1,k}. The positive outer curvature term and the uniform divided-action baseline give

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)\Id and Lemma 5.4 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 5.4. 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 (5.11) converts the bound on H_k into

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 \grad G_k=(K_k)_z(z,Y_k(z)) and differentiability of Y_k give

In parallel orthonormal frames, the distance between z_k and y_k is O(\tau_k), and the smooth short-cost expansions are

where \Pi is the appropriate parallel isometry. Solving the preceding identity, using (5.31), gives

It is nonsingular for large k. The Jensen points were chosen off the set (5.26); 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

at every nearby point, even if w_c is not differentiable there. Apply Lemma 5.7 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,

Expanding the smooth function (K_k)_y in its two variables therefore gives

Thus stationary differentiation uses the subgradient expansion even when neighboring endpoints are nonsmooth. Solving it with (5.32) gives, in parallel frames,

In particular these endpoint Hessians are uniformly bounded. The error statement follows directly from the displayed short-cost expansions and the bounded inverse in (5.32); it does not assume a bound for \Hess w_c in advance.

Step 7: insert the true center form and take matrix limits. Set

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

The exact chain rule for the scalar modification gives

All vectors on the last line are understood after the same isometric identification. Lemma 3.4 bounds the middle term below by c(1-l_k)\Id. Equation (5.33) and the vanishing pole mismatch show that the left side differs from H_k by o(1)\Id. The verification is the same smooth coordinate comparison as in Lemma 5.4, now using the bounded unregularized Hessians in place of the smooth support Hessians. This proves

It also bounds V_k from above, because V_k>0 and H_k is bounded.

Taking the determinant in (5.34) and using (5.4) gives the exact factorization

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

Since 0<V_k\le C_{\mathrm{fixed}}\Id, equation (5.36) now implies

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 (5.36) gives a positive lower determinant in the limit, so V>0. Equations (5.6), (5.13), and (5.35) give (5.23).

At each selected outer endpoint in the diagonal sequence, formula (5.19), with H_{o,k}=m_{1,k}L_k, has a positive divided-action term. Its scalar curvature coefficient satisfies

Passing this inequality and (5.20) along V_{1,k}\to V_i gives (5.24) and the other inequality in (5.25). 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 (5.21)–(5.22). 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

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

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

Then S\le h=V(e,e)\le\beta q^2. The determinant of the right side of (5.38), 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

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 5.8. Lemma 5.9 and (5.25) imply

The radial comparison of Corollary 3.3 gives

For the selected outward endpoint, (5.21)–(5.22) give m_i\ge m_* and q^2\ge cD. Its norm increment satisfies |p_i|^2-|p|^2\le3D by Lemma 4.9. Corollary 3.6 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

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.□