Proof. The radial and tangential sectional curvatures of \(\mathrm{d}s^2+r^2\sigma\) are \(-r''/r\) and \((1-(r')^2)/r^2\). Differentiating (2) gives \(N'=r+m_0/r^2\) and \(N''=N(1-2m_0/r^3)\). The Ricci eigenvalues are therefore \(-2-2m_0/r^3\) and \(-2+m_0/r^3\), giving (14) and \(R_{g_0}=-6\). The radial and tangential Hessian eigenvalues of \(N\) are \(N''\) and \(N'N/r\), respectively, proving the first identity in (15). Its trace gives \(LN=0\). The divergence identity follows from the contracted Bianchi identity and the constant scalar curvature. At \(S\), \(B_t=\kappa/a\) and \(2\kappa/a>0\), so \(D>0\). Finally \(2m_0=a(1+a^2)\) gives (16) by direct differentiation. ◻

Proof. Let \(h=e^{-\beta s}\). Since \(N/r\geq0\), \[Lh=(\beta^2-2\beta N/r-3)h \leq-(3-\beta^2)h.\] Choose \(c_\beta<(3-\beta^2)/2\). A sufficiently large multiple \(Ah\) satisfies \((L-V)(Ah)\leq-|F|\). For Dirichlet data, enlarge \(A\) to dominate \(|b|\). For Neumann data, enlarge it so that \(-\beta A-b<0\) and \(-\beta A+b<0\) on \(S\). The functions \(Ah-z\) and \(Ah+z\) are positive sufficiently far out. A negative interior minimum contradicts their differential inequalities, because the zeroth-order coefficient of \(L-V\) is negative. A negative minimum on \(S\) is impossible in the Neumann case: the inward derivative at such a minimum is nonnegative, whereas the chosen derivative is negative. This proves (18). For the Robin condition, apply the maximum argument to \(z/h\). At a positive boundary maximum its inward derivative is nonpositive, whereas \(\partial_s(z/h)=(\beta+a_*)(z/h)+b\) on \(S\). Since \(\beta+a_*\geq\beta/2\), a sufficiently large positive maximum is impossible; apply the same argument to \(-z/h\).

For the existence assertion, work in each angular eigenspace. The radial equation is \[ y''+2\frac Nr y'-\left(3+\frac\lambda{r^2}\right)y=F_\lambda. \tag{20}\] First impose the required condition at \(0\) and \(y(T)=0\) on a finite interval. The homogeneous problem has trivial kernel: multiplication by \(r^2y\) and integration gives \[0=-\int_0^T r^2(y')^2\,\mathrm{d}s -\int_0^T(3r^2+\lambda)y^2\,\mathrm{d}s,\] with zero boundary terms. Thus the finite-interval linear problem is solvable. The barrier estimate is independent of \(T\). On each compact interval, the equation bounds \(y'\) and then \(y''\): the mean value theorem supplies a point with bounded \(y'\), and the first-order equation for \(y'\) propagates that bound. A subsequence therefore converges on compact intervals to a solution with \(y=O(e^{-\beta s})\). The same local argument on unit intervals gives \(y'=O(e^{-\beta s})\).

Now \(N/r=1+O(e^{-2s})\) and \(r^{-2}=O(e^{-2s})\). Hence \[y''+2y'-3y=G,\qquad G=O(e^{-\gamma' s}),\qquad \gamma'=\min\{\gamma,\beta+2\}>3.\] Variation of constants for the roots \(1,-3\) eliminates the growing homogeneous term and gives \[y=c e^{-3s}+O(e^{-\gamma' s}),\qquad y'=-3c e^{-3s}+O(e^{-\gamma' s}).\] For example the particular solution is a linear combination of \(e^s\int_s^\infty e^{-t}G(t)\,\mathrm{d}t\) and \(e^{-3s}\int_s^\infty e^{3t}G(t)\,\mathrm{d}t\); both integrals converge with the claimed bounds. Since \(r=C e^s(1+O(e^{-2s}))\), this is (19). The equation gives the second differentiated remainder, and differentiated equations give the rest. Angular differentiations are bounded operations on the fixed finite spaces. Any other solution tending uniformly to zero also satisfies the initial weighted bound: compare on sufficiently long intervals with \(Ah+\delta\), where \(\delta>0\) dominates the far boundary value, and then let \(\delta\downarrow0\). The same ODE improvement applies. The improved decay permits comparison for a difference of two solutions, proving uniqueness. ◻

Proof. Lemma 4 constructs \(u_1\) with homogeneous interior equation. Its derivatives have the asserted asymptotics by (20). In degree two the interior source is a sum of \(u_1^2=O(r^{-6})\) and \(|q|^2=O(r^{-2\tau})\), with \(\min\{6,2\tau\}>3\). Its first radial derivative has the same bound because \(q\in C^{1,\alpha}_\tau\). Finite angular type supplies any fixed number of angular derivatives. The inverse lemma therefore gives the claimed second coefficient and, in particular, its first differentiated asymptotic. No bound on higher radial derivatives of a general seed is asserted here.

For (25), solve the homogeneous Dirichlet problem separately in its four boundary modes. The homogeneous ODE gives \(v=O(r^{-3})\) with all differentiated versions. The scaled components of the connection, \(B\), and all their derivatives are bounded at infinity. It follows that \(\mathop{\mathrm{Hess}}v-Bv\) has \(O(r^{-3})\) components with all derivatives. Inductively, every interior source in (23) is a sum of products containing at least two factors among lower solution coefficients and \(q\). It therefore has \(O(r^{-6})\) decay with all derivatives. Another application of the inverse lemma completes the induction through degree four.

Finally \(g_0-b=2m_0r^{-3}\mathrm{d}\eta^2+O(r^{-5})\) in scaled components, with all derivatives. In each positive degree the only \(r^{-3}\) term in \((1+\sum\epsilon^ju_j)^4-1\) is \(4r^{-3}\zeta_j\); products of two solution coefficients are \(O(r^{-6})\). This proves (26). At \(p=2\), the derivative count is two for coefficient one and one for coefficient two, both already established. At \(p=4\) all the needed derivatives are available from the stronger case. ◻

Proof. First fix \(\epsilon\); no parameter limit is taken in this argument. The assumed weighted regularity gives \(\psi,\nabla^b\psi=O(r^{-\tau})\). The difference between the actual metric perturbation \(\phi^4g_0-g_0\) and \(4\psi b\) consists of terms with scaled components and first derivatives \[O(r^{-2\tau})+O(r^{-\tau-3}).\] Indeed \(\phi^4-1-4\psi=O(\psi^2)\) and \(g_0-b=O_1(r^{-3})\). Since a mass test function and its derivative have size \(O(r)\) and the sphere area has size \(O(r^2)\), these terms give flux errors \[ O(r^{3-2\tau})+O(r^{-\tau})\longrightarrow0. \tag{31}\] For the tensor \(4\psi b\) in dimension three, \[\mathop{\mathrm{div}}_b(4\psi b)-\mathrm{d}\mathop{\mathrm{tr}}_b(4\psi b)=-8\mathrm{d}\psi,\] and the remaining two terms of the prescribed mass integrand add \(8\psi\,\mathrm{d}V_\mu\). Thus the flux difference is \(1/(2\pi)\) times the flux of \(\psi\mathop{\mathrm{grad}}_bV_\mu-V_\mu\mathop{\mathrm{grad}}_b\psi\). Replacing the hyperbolic normal by \(\partial_s\) produces a vanishing error since the relative normal change is \(O(r^{-3})\). For the time test, replacing \(\sqrt{1+r^2}\) by \(N\) also produces a vanishing error, because their difference and its first radial derivative are \(O(r^{-2})\). The sphere area forms are the same.

The \(g_0\) divergence of \(\psi\mathop{\mathrm{grad}}W_\mu-W_\mu\mathop{\mathrm{grad}}\psi\) is \(\psi LW_\mu-W_\mu L\psi\). Apply the divergence theorem to \([0,T]\times S^2\). The outward normal of this integration domain at its inner boundary is \(-\partial_s\), yielding exactly the sign of the boundary term in (29). Finally \(L\psi=O(r^{-2\tau})\) by (22); hence the bulk integrals converge, and \(T\to\infty\) proves the identity.

For the asserted boundedness, \(W_\mu=O(e^s)\) and \(\mathrm{d}V_{g_0}=O(e^{2s})\,\mathrm{d}s\,\mathrm{d}\omega\). The \(W_\mu F\) term is dominated by a constant times \[\|F\|_{2\beta}e^{(3-2\beta)s},\] which is integrable by (17). The \(uLW_i\) term is dominated by \(C\|u\|_\beta e^{-\beta s}\), and \(LW_0=0\). The boundary terms are bounded by the stated norms. For finite coefficient jets, apply the same finite-radius identity coefficientwise. Lemma 5 supplies their differentiated asymptotics, and all nonlinear coefficient products discarded from the flux are \(O(r^{-6})\) with first derivatives. Thus their coefficient flux is also (30). ◻

Proof. Let \(U(R)q\) denote the pullback by a rotation \(R\in SO(3)\), and let \(\mathrm{d}R\) be normalized invariant volume. Define \[k_\nu(R)=c_\nu\left(\frac{1+\mathop{\mathrm{tr}}R}{4}\right)^\nu, \qquad \int_{SO(3)}k_\nu\,\mathrm{d}R=1, \qquad q_\nu=\int_{SO(3)}k_\nu(R)U(R)q\,\mathrm{d}R.\] The expression raised to the power \(\nu\) lies in \([0,1]\) and attains one only at the identity. The normalized densities therefore concentrate there. To verify this directly, outside any fixed neighborhood their base is at most \(1-\delta\), whereas on a smaller neighborhood of positive volume it is at least \(1-\delta/2\); the ratio of the corresponding integrals tends to zero.

The map \(R\mapsto U(R)q\) is continuous in \(\|\cdot\|_\beta\). On a compact part of \(M\) this follows from smoothness. On the tail, the bound \(Ce^{-(\tau-\beta)s}\) is uniform in \(R\), so the strict weight margin controls it. Concentration gives \(\|q_\nu-q\|_\beta\to0\). Rotations preserve \(g_0\), the end background, and \(s\). They commute with trace and divergence and preserve the weighted regularity. Consequently each average is smooth and TT, with the asserted uniform bounds. They also preserve the radial normal, so positivity of \(q_{ss}|_S\) is retained by the nonnegative average.

For fixed \(\nu\), the density is a polynomial in matrix entries. After a change of integration variable, the rotation orbit of \(q_\nu\) depends on translates of this polynomial. These translates lie in a finite-dimensional polynomial space; their coefficients multiply finitely many fixed averaged tensors. Thus \(q_\nu\) has finite angular type.

It remains to check that the coefficient limits are independent of the approximation. On a bounded set in \(\|q\|_\beta\), comparison applied to (24) gives, for finite type seeds \(q,\hat q\), \[\begin{align*} \|u_1[q]-u_1[\hat q]\|_\beta &\leq C\|q-\hat q\|_\beta,\\ \|F_2[q]-F_2[\hat q]\|_{2\beta} +\|B_2[q]-B_2[\hat q]\|_{C^0(S)} &\leq C\|q-\hat q\|_\beta,\\ \|u_2[q]-u_2[\hat q]\|_\beta &\leq C\|q-\hat q\|_\beta. \end{align*}\] Here the elementary product estimate is \[\bigl\||q|^2-|\hat q|^2\bigr\|_{2\beta} \leq(\|q\|_\beta+\|\hat q\|_\beta)\|q-\hat q\|_\beta.\] The comparison constant has no angular-cutoff dependence. These estimates prove existence, uniqueness, and continuity of the limits in the complete weighted uniform space. Formula (24) preserves their linear and quadratic dependence. No derivative convergence of the limiting coefficients is needed in this construction or in the mass functional. ◻

Proof. Initial estimate. The given convergence of \(\psi\) in \(C^{2,\alpha}_\tau\) implies that \(\|\psi\|_{C^0}\) is small. Write \(\mathcal A(\psi)=V_\psi\psi\), extending the quotient by zero at \(\psi=0\). Then \(\|V_\psi\|_{C^0}=o(1)\). Moving this term to the potential in (22) leaves an interior source bounded by \(C\epsilon^2e^{-2\tau s}\) and boundary derivative bounded by \(C\epsilon\). Since \(\psi=o(e^{-\beta s})\), Lemma 4 gives \[ \|\psi\|_\beta=O(\epsilon). \tag{36}\]

Finite angular type seeds. Suppose first that the needed coefficients have been constructed by (23). The first bound (36) starts an induction. If \[\|\psi-P_{j-1}\|_\beta=O(\epsilon^j),\] then \(\psi\) and \(P_{j-1}\) are both \(O(\epsilon e^{-\beta s})\). For such arguments, the mean value theorem applied to (21) gives \[|\mathcal N_{\epsilon,q}(\psi) -\mathcal N_{\epsilon,q}(P_{j-1})| \leq C\epsilon e^{-\beta s}|\psi-P_{j-1}| =O(\epsilon^{j+1}e^{-2\beta s}).\] Taylor substitution into the finite polynomial \(P_{j-1}\) leaves an error of this same order after its coefficients through degree \(j\) are extracted. Every interior term contains at least two decaying factors, counting the factor \(Q_q\) as two. On \(S\), \[|\mathcal B_{\epsilon,q}(\psi) -\mathcal B_{\epsilon,q}(P_{j-1})| \leq C\epsilon|\psi-P_{j-1}|=O(\epsilon^{j+1}),\] and the boundary Taylor error has the same order. Thus \(\psi-P_j\) has interior source \(O(\epsilon^{j+1}e^{-2\beta s})\) and boundary derivative \(O(\epsilon^{j+1})\). It is \(o(e^{-\beta s})\) at infinity, so comparison improves its weighted norm to \(O(\epsilon^{j+1})\). This proves all three assertions in (32) inductively through the required degree. In particular, it proves the fourth-order assertions in the special case (25).

Arbitrary seeds through degree two. Let \(q_\nu\) be the approximants of Lemma 7, and put \(d_\nu=\|q-q_\nu\|_\beta\). Write \(u_{j,\nu},F_{j,\nu},B_{j,\nu}\) for their coefficients. These have uniform bounds in the norms of that lemma. We compare them directly with the given branch for \(q\); no branch for \(q_\nu\) is introduced. Using (36), the equation for \(\psi-\epsilon u_{1,\nu}\) has source \(O(\epsilon^2e^{-2\beta s})\) and boundary derivative \(O(\epsilon d_\nu+\epsilon^2)\). Hence \[ \|\psi-\epsilon u_{1,\nu}\|_\beta \leq C(\epsilon d_\nu+\epsilon^2). \tag{37}\] Expanding the nonlinearities once more, formula (24) and (37) imply \[\begin{align*} \|L\psi-\epsilon^2F_{2,\nu}\|_{2\beta} &\leq C(\epsilon^2d_\nu+\epsilon^3),\\ \|\partial_s\psi|_S-\epsilon B_{1,\nu}-\epsilon^2B_{2,\nu}\|_{C^0(S)} &\leq C\bigl((\epsilon+\epsilon^2)d_\nu+\epsilon^3\bigr). \tag{38}\end{align*}\] For clarity, the only quadratic interior discrepancies are \(\psi^2-\epsilon^2u_{1,\nu}^2\) and \(\epsilon^2(Q_q-Q_{q_\nu})\); their weighted norms are bounded by the first right-hand side. At the boundary the linear discrepancy is \(\epsilon(h_q-h_{q_\nu})/4\), and the next one contains \(\epsilon h_q\psi-\epsilon^2h_{q_\nu}u_{1,\nu}\), producing exactly the second bound. All remaining terms are cubic with the stated weights. Comparison applied to \(\psi-\epsilon u_{1,\nu}-\epsilon^2u_{2,\nu}\) now gives \[\|\psi-\epsilon u_{1,\nu}-\epsilon^2u_{2,\nu}\|_\beta \leq C\bigl((\epsilon+\epsilon^2)d_\nu+\epsilon^3\bigr).\] The constants are independent of \(\nu\). Letting \(\nu\to\infty\) and using coefficient continuity proves the full three estimates (32) for \(p=2\) and the original arbitrary seed.

Mass, area, and the deficit. Apply the bounded functional in Lemma 6 to the three remainders. This proves (33) directly. The area is the compact-boundary integral \[A_\epsilon=\int_S(1+\psi)^4\,\mathrm{d}A_{g_0},\] so its expansion follows from the first remainder alone. If \(A_\epsilon=A_0+\epsilon A_1+\epsilon^2 A_2+O(\epsilon^3)\), then \[A_1=4\int_Su_1\,\mathrm{d}A_{g_0},\qquad A_2=\int_S(4u_2+6u_1^2)\,\mathrm{d}A_{g_0}.\] The mass norm is smooth near \((m_0,0,0,0)\) and \(b_*\) is smooth near \(A_0=4\pi a^2\). Their expansions therefore have the same controlled remainders. The constant deficit is zero. Also \(F_1=0\), \(N|_S=0\), \(\partial_sN|_S=\kappa\), and \(LN=0\), so \[a_{01}=\frac\kappa{2\pi}\int_Su_1\,\mathrm{d}A_{g_0} =b_*'(A_0)A_1.\] The spatial mass components have zero base value and first enter the mass norm quadratically. This proves that the linear deficit coefficient vanishes.

In degree two the scalar expression is explicitly \[ c_2[q]=a_{02} -\frac1{2m_0}\sum_{i=1}^3a_{i1}^2 -b_*'(A_0)A_2-\frac12b_*''(A_0)A_1^2. \tag{39}\] Lemma 7 and the bounded Green functional show that this is a continuous quadratic form on the entire formal seed space \(\mathscr T_{\tau,\alpha}\). Formula (39) uses only the coefficient functions and absolutely convergent Green integrals; it does not assume a geometric integral formula for arbitrary seeds. Rotations leave the area invariant, fix the time mass component, and rotate the three spatial components. Thus \(c_2\) is rotation invariant. The identical finite coefficient argument through degree four proves (35). ◻

Proof. We first solve the Jacobi equations on finite radial intervals. At infinity, hyperbolic boosts account for the order-one degree-one displacement. The asymptotic CMC equation selects a frame in which the spatial mass components vanish. We then choose the free graph means to join those spheres to the minimal first leaf.

Proof. The area and mean-curvature variation formulas give \[A'=AH\bar f,\qquad H'=Jf.\] The latter is a constant on each leaf; any tangential transport term vanishes since \(H\) is constant there. For brevity put \[c=\frac{4\pi}{A}+3-\frac{3H^2}{4}, \qquad Q_H=\frac{\mathop{\mathrm{Scal}}_G+6+|\mathrm{II}^\circ|^2}{2}.\] Differentiation of (41) gives \[ m_H'=\sqrt{\frac A{16\pi}}\frac{HA}{8\pi} (c\bar f-H'). \tag{55}\] The Gauss equation and \(|\mathrm{II}|^2=H^2/2+|\mathrm{II}^\circ|^2\) imply \[J1=K_\Sigma+3-\frac{3H^2}{4}-Q_H,\] where \(K_\Sigma\) is intrinsic Gauss curvature. The sphere jets have \(\int K_\Sigma\,\mathrm{d}A_G=4\pi\) coefficientwise. One can verify this without a limiting argument: the variation of integrated scalar curvature on a closed two-dimensional metric is the integral of a divergence minus its variation tensor paired with \(\mathop{\mathrm{Ric}}-\frac12\mathop{\mathrm{Scal}}\,g\). The latter tensor is zero in dimension two, so that integral is constant under every variation and equals its round value. Applying the same identity to the Taylor coefficients proves the assertion for jets. Consequently \[\langle J1\rangle=c-\langle Q_H\rangle.\]

The operator \(J\) is self-adjoint on the closed leaf. Since \(\langle\delta f\rangle=0\) and \(Jf=H'\) is constant, \[\begin{align*} H'&=\bar f\bigl(c-\langle Q_H\rangle\bigr) +\langle\delta f\,J1\rangle,\\ 0&=\langle\delta f\,Jf\rangle =\bar f\langle\delta f\,J1\rangle +\langle\delta f\,J\delta f\rangle. \end{align*}\] The scalar jet \(\bar f\) is invertible because its base value is one. Eliminating the common term gives \[c\bar f-H' =\bar f\langle Q_H\rangle +\bar f^{-1}\langle\delta f\,J\delta f\rangle.\] Substitution into (55) proves the identity. All calculations may first be made on a compact radial interval. Smooth representatives of the metric and graph jets realize the ordinary geometric variation identities there, and the imposed CMC identities hold to the required Taylor order. ◻

Proof. The first nonzero coefficient \(F\) has zero round mean on each background sphere: extract degree \(l\) from \(\int_{\Sigma_s}\delta f\,\mathrm{d}A_G=0\), using the vanishing of all lower coefficients of \(\delta f\). Every term in the bracket in (54) starts at degree \(2l\). At that degree, all exterior factors can therefore be evaluated at the background: \[A_0=4\pi r^2,\quad H_0=2N/r,\quad \bar f_0=1, \quad\sqrt{\frac{A_0}{16\pi}}H_0=N.\] The same reasoning evaluates the norms, the induced measure, and the Jacobi operator inside the leading terms at the background. Composing the ambient scalar coefficient with a graph displacement changes only higher degrees. It follows that \[ [m_H']_{2l} =\frac{N(s)}{16\pi}\int_{\{s\}\times S^2} \bigl(|k|^2+|U|^2+2FJ_s^0F\bigr)\mathrm{d}A_{g_0}, \tag{58}\] and all lower coefficient derivatives vanish.

The integrand in (58) is nonnegative after sphere integration. Indeed, on the mean-zero field \(F\), \[\int_{\{s\}\times S^2}FJ_s^0F\,\mathrm{d}A_{g_0} =\sum_{\ell\ge1} \left(\frac{\ell(\ell+1)-2}{r^2}+\frac{6m_0}{r^3}\right) \|F_\ell\|_{L^2(r^2\sigma)}^2.\] Only finitely many modes occur, and every displayed eigenvalue is strictly positive at a finite radius. Also \(N(s)>0\) for \(s>0\).

For any finite \(T\), integrate the coefficient identities on \([0,T]\). The first leaf is minimal to the required order, so \(m_H(\Sigma_0)=b_*(A_{\mathrm{leaf}})\) through that order. The coefficientwise endpoint limits exist by (42). For degree \(2l\), the right-hand side of the integrated (58) is an increasing function of \(T\) and its limit equals the finite limiting endpoint coefficient. This proves both its convergence and (57). For lower degrees the integrated derivative is zero, proving their claimed vanishing.

This order of operations uses only smooth jets on finite radial intervals and coefficientwise endpoint limits. In particular, convergence of the derivatives of the asymptotic graph coefficients or of their lapse is not needed to obtain the integral’s convergence. If the integral is zero, each of its continuous nonnegative sphere-integrated summands is zero for every \(s>0\). The positivity just established forces \(k=U=F=0\) there. Smoothness gives their boundary values. ◻

Proof. For a finite-angular-type seed the scalar constraint gives \(\mathop{\mathrm{Scal}}_g+6=\epsilon^2|q|^2+O(\epsilon^3)\). Proposition 13, with \(l=1\), shows that the constant and first coefficients of the mass norm minus \(b_*(A_{\mathrm{leaf}})\) vanish, and that its second coefficient is \[\frac1{16\pi}\int_M N\bigl(|q|^2+|U|^2+2FJ_s^0F\bigr)\,\mathrm{d}V_0.\] The areas in (63) agree through order one, so the first coefficient of the original deficit is also zero. Since \(b_*'(4\pi a^2)=\kappa/(8\pi)\), the contribution to its second coefficient from that area difference is \(-\kappa(16\pi)^{-1}\int_SvDv\,\mathrm{d}A_0\). Lemma 14 now proves (64) with precisely the displayed factor and sign.

On a spherical harmonic with \(-\Delta_\sigma\) eigenvalue \(\lambda\geq2\), the eigenvalue of \(J_s^0\) is \[\frac{\lambda-2}{r^2}+\frac{6m_0}{r^3}>0.\] Thus its quadratic form is strictly positive on mean-zero functions at every finite \(s\), and \(N>0\) for \(s>0\). This proves nonnegativity for finite-angular-type seeds without any boundary sign restriction.

Proposition 8 defines \(c_2\) on the full real space \(\mathscr T_{\tau,\alpha}\) of smooth TT seeds as a continuous, rotation-invariant quadratic form, independently of branch existence and boundary sign, and proves \(c_1=0\). For an arbitrary seed \(q\), take the finite-angular-type TT approximants \(q_\nu\) from Lemma 7. The formula proved above gives \(c_2[q_\nu]\geq0\), and continuity in \(\|\cdot\|_\beta\) gives \(c_2[q]\geq0\). Proposition 8 also supplies (65) for every prescribed actual branch. ◻

Proof. First suppose \(q\) has finite angular type and \(c_2[q]=0\). Equation (64) and the strict positivity just proved imply \[ q=Q(v),\qquad U=0,\qquad F=0 \quad\text{on }s>0. \tag{67}\] Smoothness extends these equalities to the boundary. Coordinate spheres are umbilic under conformal changes of \(g_0\). At the base minimal sphere, the tracefree second-form variation due to a normal height \(h_1\) is \(-(\mathop{\mathrm{Hess}}_Sh_1)^{\circ}\): the radial curvature term is pure trace. Since \(h_1=-v|_S\), the vanishing of \(U|_S\) implies \((\mathop{\mathrm{Hess}}_\sigma(v|_S))^{\circ}=0\). For completeness, the divergence of this equation on the unit sphere gives \(\mathrm{d}(\Delta_\sigma v+2v)=0\). Subtracting the mean reduces it to \(\mathop{\mathrm{Hess}}_\sigma(v-\bar v)=-(v-\bar v)\sigma\). This equation determines its solution along every geodesic from its value and differential at one point, and its solutions are precisely the linear coordinate functions. Thus \(v|_S\) is a constant plus a linear harmonic.

Uniqueness of the decaying Dirichlet problem for \(L\) then excludes every other angular mode throughout \(M\). Each degree-one radial coefficient is a multiple of the same decaying solution of \[ V''+2\frac NrV'-\left(3+\frac2{r^2}\right)V=0. \tag{68}\] Hence \(q\in\mathcal K\). Conversely, every element of \(\mathcal K\) is null, as we now check; the conditions \(U=F=0\) do not impose a further restriction on its degree-one part.

Take \(q=Q(v)\in\mathcal K\). The first conformal coefficient \(u\) also has only constant and linear modes, by its Neumann equation and decaying uniqueness. Let \(h(s,x)\) be the first height coefficient of the CMC spheres, and put \(p=N/r\). Conformal variation of mean curvature and first variation of the lapse give \[ [H]_1=J_s^0h+4(u_s-pu), \qquad [f]_1=h_s+2u. \tag{69}\] Here the second identity follows by taking the normal component of the velocity of the graph \(s+\epsilon h(s,x)\) in \(g_0+4\epsilon u g_0\). On the degree-one part, the CMC equation therefore fixes \[ h=-\frac{2r^3}{3m_0}(u_s-pu). \tag{70}\] All formulas in the next calculation are componentwise on that part. The background equations and (68) give \[p'=\frac{3m_0}{r^3}-\frac1{r^2},\qquad p^2=1+\frac1{r^2}-\frac{2m_0}{r^3},\qquad u''=-2pu'+\left(3+\frac2{r^2}\right)u.\] Consequently, \[\begin{align*} \frac1{r^3}\partial_s\bigl[r^3(u'-pu)\bigr] &=3p(u'-pu)+u''-p'u-pu'\\ &=\left(3+\frac2{r^2}-p'-3p^2\right)u =\frac{3m_0}{r^3}u. \tag{71}\end{align*}\] It follows that \(h_s=-2u\) on the degree-one part, so its lapse fluctuation vanishes. The freely chosen radial mean of \(h\) changes \([f]_1\) only by a radial function, which disappears upon subtracting the mean. Thus \(F=0\) independently of the radial labeling.

At every background sphere, the tracefree second-form variation of these graphs is \(-(\mathop{\mathrm{Hess}}_{r^2\sigma}h)^{\circ}\); the conformal variation and radial curvature terms are pure trace. The constant and linear modes of \(h\) have vanishing tracefree spherical Hessian, so \(U=0\) as well. At \(S\) there is also exact agreement with the prescribed first minimal height: on degree one, \[D_1=\frac{6m_0}{a^3},\qquad h(0)=-\frac{4u_s(0)}{D_1}=-v(0).\] We already have \(q=Q(v)\), and hence (64) proves \(c_2[q]=0\). Existence and uniqueness for each of the four boundary modes show that \(\mathcal K\) has dimension at most four; it has dimension exactly four because (61) and invertibility of \(D\) make the map from these boundary data to \(Q(v)\) injective.

It remains to rule out additional null directions of arbitrary angular type. Let \(\mathcal Z=\{q:c_2[q]=0\}\) in the full smooth TT space. A nonnegative quadratic form has a linear kernel: if \(c_2[q]=0\), nonnegativity of \(c_2[q+th]\) for all real \(t\) forces its polarized bilinear form to vanish on \((q,h)\) for every \(h\). By Proposition 15, \(\mathcal Z\) is also closed in \(\|\cdot\|_\beta\) and invariant under rotations. For \(q\in\mathcal Z\), let \(q_j\) be its polynomial rotational averages from Lemma 7. Each such average is a norm limit of finite linear combinations of rotations of \(q\); therefore \(q_j\in\mathcal Z\). Each \(q_j\) has finite angular type, so the preceding argument puts it in \(\mathcal K\). Since \(q_j\to q\) in norm and the fixed finite-dimensional space \(\mathcal K\) is closed in that norm, \(q\in\mathcal K\). This proves the full assertion without applying a pointwise geometric limit to the CMC fields \(U\) and \(F\). ◻

Proof. We give the formal evolution and its constraint justification before performing the graph calculation. Write \[\overline g=-\mathrm{d}t^2+\gamma(t),\qquad \gamma(0)=g_\epsilon,\qquad \mathcal K(0)=K_\epsilon,\] and determine a full formal time series recursively from \[\begin{align*} \partial_t\gamma&=2\mathcal K,\\ \partial_t\mathcal K &=-\mathop{\mathrm{Ric}}_\gamma-3\gamma -(\mathop{\mathrm{tr}}_\gamma\mathcal K)\mathcal K +2\mathcal K\circ_\gamma\mathcal K. \tag{77}\end{align*}\] The coefficients take values in the ring of smooth spatial jets modulo \(\epsilon^5\). At each time order the right side is a known spatial differential expression in previously determined coefficients, so the recursion is well defined. A substitution \(t=T=O(\epsilon)\) uses only finitely many of these coefficients. Keeping a full time series here ensures that taking time derivatives never treats an omitted time coefficient as zero.

Let \(H=\mathop{\mathrm{tr}}_\gamma\mathcal K\). The time-index Christoffel symbols are \[\overline\Gamma^t_{ij}=\mathcal K_{ij},\qquad \overline\Gamma^i_{tj}=\mathcal K^i{}_j,\] and direct contraction of the curvature gives \[\begin{align*} \overline\mathop{\mathrm{Ric}}_{ij} &=\mathop{\mathrm{Ric}}_{\gamma,ij}+\partial_t\mathcal K_{ij} +H\mathcal K_{ij} -2(\mathcal K\circ_\gamma\mathcal K)_{ij},\\ \overline\mathop{\mathrm{Ric}}_{tt} &=-\mathop{\mathrm{tr}}_\gamma(\partial_t\mathcal K)+|\mathcal K|_\gamma^2, &\overline\mathop{\mathrm{Ric}}_{ti} &=(\mathop{\mathrm{div}}_\gamma\mathcal K)_i-\partial_iH. \tag{78}\end{align*}\] Consequently \(E:=\overline\mathop{\mathrm{Ric}}+3\overline g\) has \(E_{ij}=0\). The initial Hamiltonian and momentum constraints give \(E_{tt}=E_{ti}=0\) at \(t=0\). For completeness, the Hamiltonian constraint is \(\mathop{\mathrm{Scal}}_\gamma+H^2-|\mathcal K|_\gamma^2=-6\); tracing (77) at \(t=0\) therefore gives \(\mathop{\mathrm{tr}}_\gamma\partial_t\mathcal K=-3+|\mathcal K|_\gamma^2\), which proves the asserted \(tt\) equation with the required sign.

The contracted Bianchi identity propagates these remaining equations formally. Indeed, if \(e=E_{tt}\) and \(j_i=E_{ti}\), the trace reversal of \(E\) has components \[\widehat E_{tt}=e/2,\qquad \widehat E_{ti}=j_i,\qquad \widehat E_{ij}=e\gamma_{ij}/2.\] Its vanishing divergence, evaluated with the displayed Christoffel symbols, is \[ \partial_t e=2\mathop{\mathrm{div}}_\gamma j-2He, \qquad \partial_tj_i=\tfrac12\partial_i e-Hj_i. \tag{79}\] This homogeneous system determines each next time coefficient of \((e,j)\) from the preceding ones. All coefficients vanish by induction. Thus \[ \overline\mathop{\mathrm{Ric}}=-3\overline g \tag{80}\] as a formal time series modulo \(\epsilon^5\).

There is no extension hypothesis hidden in this construction. The original coefficient fields are smooth up to \(S\), and their constraint residuals and all one-sided spatial derivatives vanish there. The recursion and (79) therefore also hold in their spatial Taylor jets at \(S\). Each finite calculation below involves only finitely many such derivatives. They may, if desired, be realized in a collar across \(S\) by their Taylor polynomials in \(s\), to an order larger than the differential order of that calculation. Curvature and its residual are formed before substituting a displacement of order \(\epsilon\). Their retained boundary jets vanish, so the substitution is valid even when the leading displacement is negative. This neither constructs nor uses an actual Einstein extension on the other side of \(S\).

Gauss–Codazzi applied as a differential identity to any spacelike graph now gives \[ \mathop{\mathrm{Scal}}_{\rm slice}+6 =|K_{\rm slice}|^2-(\mathop{\mathrm{tr}}K_{\rm slice})^2, \qquad \mathop{\mathrm{div}}\bigl(K_{\rm slice}-(\mathop{\mathrm{tr}}K_{\rm slice})g_{\rm slice}\bigr)=0. \tag{81}\] Equivalently one can form Gaussian coordinates about that graph by the Taylor recursion for its normal geodesics and use (78). Either interpretation needs only differential identities of finite jets.

We next compute the graph second form explicitly. With \(T_i=\partial_iT\), its tangent fields are \(X_i=\partial_i+T_i\partial_t\), and its future unit normal is \[n_T=\frac{\partial_t+\mathop{\mathrm{grad}}_{\gamma(t)}T} {\sqrt{1-|\mathrm{d}T|_{\gamma(t)}^2}}\bigg|_{t=T}.\] Using \(-\overline g(n_T,\overline\nabla_{X_i}X_j)\) gives \[ \widetilde K_{ij} =\left. \frac{\mathcal K_{ij}(t)+(\mathop{\mathrm{Hess}}_{\gamma(t)}T)_{ij} -T_k\bigl(T_i\mathcal K^k{}_j(t) +T_j\mathcal K^k{}_i(t)\bigr)} {\sqrt{1-|\mathrm{d}T|_{\gamma(t)}^2}} \right|_{t=T}. \tag{82}\] The Hessian connection in this formula is taken at fixed time before evaluation at \(t=T\). The induced metric is \(\widetilde g=\gamma(T)-\mathrm{d}T\otimes\mathrm{d}T\).

At the background initial slice, \(\mathcal K=0\) and \(\partial_t\mathcal K=-B\). A first-order time displacement \(b\) therefore changes the second form by \(\mathop{\mathrm{Hess}}b-Bb=Q(b)\). Equation (72) gives \[[\widetilde K]_1=q-Q(v)=0.\] We choose \(w\) to remove the trace at the next degree. This can be done with an explicit scalar source. Since \([\gamma(0)]_1=4ug_0\), the conformal Ricci variation and \(Lu=0\) give \[[\partial_t\mathcal K(0)]_1 =2\mathop{\mathrm{Hess}}u-6ug_0, \qquad [\mathcal K(0)]_2=-2uq.\] At the background \(\partial_t^2\mathcal K(0)=0\): differentiating (77) there uses \(\partial_t\gamma(0)=0\) and \(\mathcal K(0)=0\). Expanding (82) to degree two thus gives \[\begin{align*} k={}&Q(w)-2uq-2v\mathop{\mathrm{Hess}}u+6uvg_0\\ &\quad+2\mathrm{d}u\otimes\mathrm{d}v+2\mathrm{d}v\otimes\mathrm{d}u -2\langle\mathrm{d}u,\mathrm{d}v\rangle_{g_0}g_0. \tag{83}\end{align*}\] In particular, \[\mathop{\mathrm{tr}}_{g_0}k=Lw+12uv-2\langle\mathrm{d}u,\mathrm{d}v\rangle_{g_0}.\] Choose the unique decaying solution of \[ Lw=2\langle\mathrm{d}u,\mathrm{d}v\rangle_{g_0}-12uv, \qquad w|_S=0. \tag{84}\] Its right side is \(O(r^{-6})\), with differentiated bounds, so Lemma 4 applies and gives \(w=O(r^{-3})\) with the corresponding derivatives. Formula (83) gives the asserted decay of \(k\). The degree-two momentum equation in (81), together with \(\mathop{\mathrm{tr}}_{g_0}k=0\), gives \(\mathop{\mathrm{div}}_{g_0}k=0\).

The full transformed trace is \(O(\epsilon^3)\). Its square consequently begins in degree six, whereas \[|\widetilde K|_{\widetilde g}^2 =\epsilon^4|k|_{g_0}^2+O(\epsilon^5).\] This proves (74); maximality at higher orders on the changed slice is not needed.

Finally, \[\gamma(T)-\gamma(0) =2\mathcal K(0)T+\tfrac12\partial_t^2\gamma(0)T^2+\cdots.\] In every retained coefficient the first term is \(O(r^{-6})\), and every subsequent term has at least two decaying time-displacement factors multiplying bounded background time coefficients. The recursion preserves these bounded differentiated estimates: in the scaled coframe \(\mathrm{d}s,r\,\mathrm{d}x\), the background connection and curvature, with their required derivatives, are bounded; each coefficient is obtained by finitely many covariant differentiations, contractions, and inverse-metric Taylor operations from the smooth initial coefficients. The same bounds hold for \(\mathrm{d}T\otimes\mathrm{d}T\), proving (75). An \(O(r^{-6})\) metric coefficient with its first covariant derivative contributes \(O(r^{-3})\) to the integrated mass flux, since the test potentials and their gradients are \(O(r)\) and the sphere area is \(O(r^2)\). Thus the leading asymptotic data, every mass component, and the mass norm are unchanged through degree four.

Since \(\partial_t^2\gamma(0)=-2B\) at the background, the degree-two difference is \(-2vq-Bv^2-\mathrm{d}v\otimes\mathrm{d}v\). Substituting \(q=\mathop{\mathrm{Hess}}v-Bv\) and using \(\mathop{\mathrm{Hess}}(v^2)=2v\mathop{\mathrm{Hess}}v+2\mathrm{d}v\otimes\mathrm{d}v\) proves (76). ◻

Proof. On the original initial slice take the future outward null normal \(\ell=n_{\rm future}+\nu\) to \(S\). Its affinely parametrized null geodesics determine a null hypersurface jet, with parameter \(\lambda\) zero at \(S\). These are Taylor solutions of the geodesic and Jacobi equations for the formal metric \(\overline g\) constructed above. At the initial sphere the expansion is identically zero by the original boundary equation.

Write \(\chi\) for the null second form and \(\theta=\mathop{\mathrm{tr}}\chi\) for its expansion. The initial sphere is umbilic in \(\phi_\epsilon^4g_0\). Moreover the coefficient of \(\epsilon\) in the tracefree tangential part of \(K_\epsilon\) equals the tracefree tangential part of \(Q(v)\). Since \(S\) is totally geodesic in \(g_0\), that part is \((\mathop{\mathrm{Hess}}_{a^2\sigma}(v|_S))^{\circ}\). It vanishes for the constant and degree-one modes of \(v|_S\). The vanishing initial expansion and this shear calculation together give \[ \chi(0,\epsilon)=O(\epsilon^2). \tag{89}\]

The screen space is the positive-definite quotient \(\ell^\perp/\operatorname{span}\{\ell\}\). In a parallel orthonormal frame of this space the null shape operator obeys the optical Riccati equation \[\partial_\lambda\chi=-\chi^2-\mathcal R_\ell, \qquad \partial_\lambda\theta =-|\chi|^2-\overline\mathop{\mathrm{Ric}}(\ell,\ell).\] These equations follow directly from the Jacobi equation: angular Jacobi fields remain orthogonal to \(\ell\), and their derivative matrix times their inverse is the shape operator. At \((\lambda,\epsilon)=(0,0)\), spherical symmetry makes \(\mathcal R_\ell\) a scalar multiple of the screen identity. Its trace is \(\overline\mathop{\mathrm{Ric}}(\ell,\ell)=0\) by (80) and nullness, so this curvature endomorphism is zero. Thus \(\partial_\lambda\chi(0,0)=0\).

It follows from (89) that every retained monomial of \(\chi\) has total degree at least two in \((\lambda,\epsilon)\). Raychaudhuri and the identically zero initial expansion then imply \[ \chi=O_{\rm tot}(2),\qquad \theta=O_{\rm tot}(5). \tag{90}\] Here \(O_{\rm tot}(j)\) means that each monomial \(\lambda^a\epsilon^b\) has \(a+b\geq j\). The order statements are made in the formal series modulo \(\epsilon^5\). Equivalently one can choose any sufficiently high finite Taylor lift: an Einstein residual of order \(\epsilon^5\) integrated along a segment \(\lambda=O(\epsilon)\) contributes only beyond every degree used here. Since the logarithmic screen area density has derivative \(\theta\), its change along such a segment vanishes through degree four.

Intersect this null hypersurface jet with \(t=T\). At the background initial point, the derivative of \(t-T\) along its generator is one. Successive Taylor substitution therefore solves for the intersection parameter, which is \(O(\epsilon)\). The angular projection has identity background differential and likewise has a formal inverse. The resulting cut \(S^\sharp\) is a graph over \(S\) in the changed slice.

The area conclusion also holds for this variable-parameter cut. If \(J_A\) are the transported angular Jacobi fields, its tangent fields are \(J_A+(\partial_A\lambda)\ell\). Nullness and \(\overline g(J_A,\ell)=0\) show that the additional terms do not change the induced metric. Its area density is just the transported screen density evaluated at the variable parameter. This proves (85).

The outward future null normal obtained from the new slice differs from \(\ell\) by a scalar with positive, nonzero background value. Expansion is multiplied by that scalar, so it still vanishes through the needed orders. At leading order \[\mathop{\mathrm{tr}}_{S^\sharp,\widetilde g}\widetilde K =-\epsilon^2k_{ss}|_S+O(\epsilon^3),\] because \(\mathop{\mathrm{tr}}_{g_0}k=0\). Its zero null expansion gives (86).

Both \(S^\sharp\) and the minimal Taylor leaf have zero mean-curvature coefficients through degree one. The difference of their first height coefficients is therefore annihilated by the background operator \(D\), which is invertible. At degree two, their already equal first coefficients make all common nonlinear terms cancel, leaving \(D(h_2^\sharp-h_2^{\min})=k_{ss}|_S\). This proves (87). The first variation of area at the minimal Taylor leaf vanishes to the retained order. A height difference beginning with \(\epsilon^2z\) consequently has, at degree four, only the base second-variation contribution \(\frac12\int_S zDz\,\mathrm{d}A_{g_0}\). This is (88).

All of these assertions are identities in the joint spatial, \(\lambda\), and \(\epsilon\) Taylor jets at the original boundary. The boundary-jet justification in Lemma 19 permits their evaluation at negative leading displacements. Neither the sign of an affine segment nor that of a height coefficient changes the order calculation or the area Hessian. No area inequality for an actually extended manifold is being applied. ◻

Proof. We first prove the infinitesimal decomposition used twice below. For a symmetric tensor \(P\) on the round sphere, there are a vector field \(Y\) and a function \(h\) such that \[ P=\mathcal L_Y\sigma+h\sigma. \tag{95}\] For finite-angular-type \(P\), the choices may retain that property.

Let \(P^{\circ}=P-\frac12(\mathop{\mathrm{tr}}_\sigma P)\sigma\), and identify vector fields and one-forms using \(\sigma\). Define \[\mathcal C(Y)_{ij} =\nabla_iY_j+\nabla_jY_i-(\mathop{\mathrm{div}}Y)\sigma_{ij}.\] On the curvature-one sphere, commuting covariant derivatives gives \[ \mathop{\mathrm{div}}\mathcal C(Y)=(\Delta_\sigma+1)Y, \qquad (\Delta_\sigma+1)\mathrm{d}f=\mathrm{d}(\Delta_\sigma+2)f. \tag{96}\] Here the Laplacian on one-forms is the rough Laplacian \(\nabla^i\nabla_i\). The second identity also holds after applying the parallel Hodge star \(*\) to one-forms.

Put \(\alpha=\mathop{\mathrm{div}}P^{\circ}\). To decompose it, solve the scalar equations \[\Delta_\sigma\varphi=\mathop{\mathrm{div}}\alpha,\qquad \Delta_\sigma\psi=-\mathop{\mathrm{div}}(*\alpha)\] with zero means. Both right sides have zero integral. In finite angular spaces these equations are solved by inversion on the nonconstant spherical modes. The residual \(\omega=\alpha-\mathrm{d}\varphi-*\mathrm{d}\psi\) is closed and co-closed. Commuting derivatives of a closed one-form of zero divergence gives \(\nabla^i\nabla_i\omega=\omega\); integration against \(\omega\) forces \(\omega=0\). We have therefore obtained \[\alpha=\mathrm{d}\varphi+*\mathrm{d}\psi.\]

The one-forms \(\mathrm{d}x_i\) and \(*\mathrm{d}x_i\) correspond to conformal Killing vector fields. Integration by parts against the tracefree tensor \(P^{\circ}\) consequently shows that \(\alpha\) is orthogonal to all of them. It follows that both \(\varphi\) and \(\psi\) have zero degree-one components. Thus we may solve \[(\Delta_\sigma+2)A=\varphi,\qquad (\Delta_\sigma+2)C=\psi\] on their angular modes, and set \(Y^\flat=\mathrm{d}A+*\mathrm{d}C\). Equation (96) gives \(\mathop{\mathrm{div}}\mathcal C(Y)=\mathop{\mathrm{div}}P^{\circ}\).

The tracefree tensor \(P_*=P^{\circ}-\mathcal C(Y)\) is consequently divergence free. Such a tensor vanishes on the round sphere; here is a direct verification. In an orthonormal frame write its components as \(\left(\begin{smallmatrix}a&b\\b&-a\end{smallmatrix}\right)\). Its two divergence equations are precisely the two independent Codazzi equations \(\nabla_i(P_*)_{jk}=\nabla_j(P_*)_{ik}\). Contracting and commuting these equations on the curvature-one sphere gives \[\nabla^i\nabla_i(P_*)_{jk} =2(P_*)_{jk}-\sigma_{jk}\mathop{\mathrm{tr}}_\sigma P_* =2(P_*)_{jk}.\] Integration yields \(-\int|\nabla P_*|^2=2\int|P_*|^2\), hence \(P_*=0\). Taking \(h=\frac12\mathop{\mathrm{tr}}_\sigma P-\mathop{\mathrm{div}}Y\) proves (95). Every operation used above commutes with rotations or inverts a scalar operator within finitely many spherical eigenspaces, so it preserves finite angular type.

The scalar-curvature derivative in the direction \(h\sigma\) is \[(D\mathop{\mathrm{Scal}})_\sigma(h\sigma)=-\Delta_\sigma h-2h.\] The derivative in a Lie direction is zero because the background scalar curvature is constant. If the scalar-curvature variation of \(P\) is constant, (95) therefore implies that \(h\) consists only of a constant and degree-one modes. A degree-one function \(h_1\) satisfies \(\mathop{\mathrm{Hess}}_\sigma h_1=-h_1\sigma\), so \[h_1\sigma =\mathcal L_{-\frac12\mathop{\mathrm{grad}}_\sigma h_1}\sigma.\] Thus \(P\) is a Lie derivative plus a constant multiple of \(\sigma\).

Apply this conclusion to \(P_1\). A formal inverse flow and a scalar scaling remove its first coefficient. Both operations preserve the property that curvature is angularly constant. The resulting metric two-jet has the form \(\sigma+\epsilon^2\widehat P_2\), so its degree-two curvature equation is the same linearized equation, with no quadratic contribution from a first coefficient. The argument just given removes \(\widehat P_2\) by a degree-two flow and scaling. Undoing the two operations proves the stated rounding. This proves only an identity of two-jets, which is exactly what is required. ◻

Proof. Set \(Y_2=x^2-1/3\), a degree-two spherical eigenfunction. Project (102) onto its scalar, gradient, and tangential Hessian modes. Write the contributing radial vector-field coefficient as \(p(s)Y_2\partial_s\), its tangential coefficient as \(d(s)\mathop{\mathrm{grad}}_\sigma Y_2\), and the scalar-multiple coefficient as \(c(s)Y_2\). These are the only generator components contributing to the tested modes. This follows either from the one-form decomposition in Lemma 21 or by integration by parts, using \(\Delta_\sigma Y_2=-6Y_2\). In particular, the coexact component is orthogonal to both the gradient test and the electric Hessian test, and the other spherical eigenspaces are orthogonal as well.

For clarity, the angular identity \[\mathrm{d}x\otimes\mathrm{d}x =\tfrac12\mathop{\mathrm{Hess}}_\sigma Y_2+x^2\sigma\] shows all terms of the left side in these modes. With \(B=B_s\mathrm{d}s^2+B_t r^2\sigma\), equality of the radial, mixed, Hessian, and tangential scalar coefficients respectively gives \[\begin{align*} V'^2+B_sV^2&=2p'+c, &\frac{VV'}2&=p+r^2d',\\ \frac{V^2}2&=2r^2d, &V^2+r^2B_tV^2&=2rNp+r^2c. \tag{103}\end{align*}\] Terms involving \(v_0\) have only degrees zero or one and do not enter these equations. Solving the last three equations yields \[d=\frac{V^2}{4r^2},\qquad p=\frac{NV^2}{2r},\qquad c=\frac{3m_0V^2}{r^3}.\] Substitute these expressions in the first equation and use \[B_s=1-\frac{2m_0}{r^3},\qquad B_t=1+\frac{m_0}{r^3},\qquad N'=r+\frac{m_0}{r^2},\qquad N^2=1+r^2-\frac{2m_0}{r}.\] After cancellation one obtains \[ \left(V'-\frac NrV\right)^2 =\frac{6m_0}{r^3}V^2. \tag{104}\] The local generators have disappeared from this relation. Because the radial interval was arbitrary, it holds throughout the end even though their choices need not agree on overlapping intervals.

For \(z=V/r\), (104) reads \[|z'|=\sqrt{6m_0}\,r^{-3/2}|z|.\] The coefficient on the right is integrable toward infinity, and \(z\to0\). The integral differential inequality, applied backwards between \(s\) and \(S>s\), gives \[|z(s)|\leq |z(S)| \exp\left(\int_s^S\sqrt{6m_0}\,r(t)^{-3/2}\,\mathrm{d}t\right).\] Letting \(S\to\infty\) proves \(z=0\) on the end. The homogeneous degree-one ODE obtained from \(Lv=0\) then gives \(V\equiv0\) by uniqueness. ◻

Completion of the proof of Proposition 18. The coefficient vanishings and nonnegativity were proved in (91)–(93). Suppose that the degree-one part of \(v\) is nonzero and that \(c_4[q]=0\). All three degree-one radial coefficients solve the same decaying homogeneous ODE. Decaying uniqueness makes them proportional, so after a fixed rotation their sum is \(V(s)x\) for one axis and a nonzero radial function \(V\). The vanishing consequences of (93) give (102) on each interior radial interval. Lemma 22 forces \(V=0\), a contradiction. Thus \(c_4[q]>0\).

Finally this is a statement about the actual deficit, not only a formal inequality. The kernel representation gives precisely the finite angular type and differentiated decay needed for the fourth-order expansion in Proposition 8. Lemma 6 converts the weighted function and equation remainders to the mass remainder. Consequently \[\mathfrak D_q(\epsilon)=c_4[q]\epsilon^4+O(\epsilon^5)\] in the case under consideration. Its positive coefficient gives strict positivity for all sufficiently small positive \(\epsilon\), with the interval allowed to depend on the fixed data. The radial kernel, including the zero seed, is handled by the exact argument in the next section. ◻