A local Penrose inequality for conformal perturbations of Schwarzschild–anti-de Sitter data

Proof. The radial and tangential sectional curvatures of \dd s^2+r^2\sigma are -r''/r and (1-(r')^2)/r^2. Differentiating (2.1) 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 (2.13) 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 (2.14). 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 (2.15) by direct differentiation.□

Proof. Let h=e^{-\beta s}. Since N/r\geq0,

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 (3.2). 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

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

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

Variation of constants for the roots 1,-3 eliminates the growing homogeneous term and gives

For example the particular solution is a linear combination of e^s\int_s^\infty e^{-t}G(t)\,\dd t and e^{-3s}\int_s^\infty e^{3t}G(t)\,\dd t; both integrals converge with the claimed bounds. Since r=C e^s(1+O(e^{-2s})), this is (3.3). 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 3.1 constructs u_1 with homogeneous interior equation. Its derivatives have the asserted asymptotics by (3.4). 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 (3.9), 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 \Hess v-Bv has O(r^{-3}) components with all derivatives. Inductively, every interior source in (3.7) 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}\dd\eta^2+O(r^{-5}) in scaled components, with all derivatives. In each positive degree the only r^{-3} term in (1+\sum\eps^ju_j)^4-1 is 4r^{-3}\zeta_j; products of two solution coefficients are O(r^{-6}). This proves (3.10). 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 \eps; 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

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

For the tensor 4\psi b in dimension three,

and the remaining two terms of the prescribed mass integrand add 8\psi\,\dd V_\mu. Thus the flux difference is 1/(2\pi) times the flux of \psi\grad_bV_\mu-V_\mu\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\grad W_\mu-W_\mu\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 (3.13). Finally L\psi=O(r^{-2\tau}) by (3.6); hence the bulk integrals converge, and T\to\infty proves the identity.

For the asserted boundedness, W_\mu=O(e^s) and \dd V_{g_0}=O(e^{2s})\,\dd s\,\dd\omega. The W_\mu F term is dominated by a constant times

which is integrable by (3.1). 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 3.2 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 (3.14).□

Proof. Let U(R)q denote the pullback by a rotation R\in SO(3), and let \dd R be normalized invariant volume. Define

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 (3.8) gives, for finite type seeds q,\hat q,

Here the elementary product estimate is

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 (3.8) 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 (3.6) leaves an interior source bounded by C\eps^2e^{-2\tau s} and boundary derivative bounded by C\eps. Since \psi=o(e^{-\beta s}), Lemma 3.1 gives

Finite angular type seeds. Suppose first that the needed coefficients have been constructed by (3.7). The first bound (3.20) starts an induction. If

then \psi and P_{j-1} are both O(\eps e^{-\beta s}). For such arguments, the mean value theorem applied to (3.5) gives

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,

and the boundary Taylor error has the same order. Thus \psi-P_j has interior source O(\eps^{j+1}e^{-2\beta s}) and boundary derivative O(\eps^{j+1}). It is o(e^{-\beta s}) at infinity, so comparison improves its weighted norm to O(\eps^{j+1}). This proves all three assertions in (3.16) inductively through the required degree. In particular, it proves the fourth-order assertions in the special case (3.9).

Arbitrary seeds through degree two. Let q_\nu be the approximants of Lemma 3.4, 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 (3.20), the equation for \psi-\eps u_{1,\nu} has source O(\eps^2e^{-2\beta s}) and boundary derivative O(\eps d_\nu+\eps^2). Hence

Expanding the nonlinearities once more, formula (3.8) and (3.21) imply

For clarity, the only quadratic interior discrepancies are \psi^2-\eps^2u_{1,\nu}^2 and \eps^2(Q_q-Q_{q_\nu}); their weighted norms are bounded by the first right-hand side. At the boundary the linear discrepancy is \eps(h_q-h_{q_\nu})/4, and the next one contains \eps h_q\psi-\eps^2h_{q_\nu}u_{1,\nu}, producing exactly the second bound. All remaining terms are cubic with the stated weights. Comparison applied to \psi-\eps u_{1,\nu}-\eps^2u_{2,\nu} now gives

The constants are independent of \nu. Letting \nu\to\infty and using coefficient continuity proves the full three estimates (3.16) for p=2 and the original arbitrary seed.

Mass, area, and the deficit. Apply the bounded functional in Lemma 3.3 to the three remainders. This proves (3.17) directly. The area is the compact-boundary integral

so its expansion follows from the first remainder alone. If A_\eps=A_0+\eps A_1+\eps^2 A_2+O(\eps^3), then

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

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

Lemma 3.4 and the bounded Green functional show that this is a continuous quadratic form on the entire formal seed space \mathscr T_{\tau,\alpha}. Formula (3.23) 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 (3.19).□

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.

The graph equation on finite radial intervals. For a normal variation with speed w, the mean-curvature variation is

Indeed, the induced metric varies by 2w\II, the variation of the unit normal is -\grad_\Sigma w, and differentiation of the second form gives the Hessian term, the ambient curvature term, and the quadratic second-form term. Tracing, including the variation of the inverse induced metric, gives (4.4). A tangential velocity adds the directional derivative of H.

For a background coordinate sphere, H_0=2N/r and

Thus its Jacobi operator is

On degree-\ell spherical harmonics its eigenvalue is

It is strictly positive on every nonconstant mode. In particular, the degree-one eigenvalue is 6m_0/r^3>0. At s=0, using 2m_0=a(1+a^2), the constant eigenvalue is

and hence J_0^0=D, positive on all modes.

Write a graph as s+\rho(s,x;\eps), with \rho_0=0. At degree j, its mean-curvature coefficient is

where \mathcal R_j is known from the metric coefficients and the lower graph coefficients. To make the first sphere minimal, solve this equation at s=0 with right-hand side zero, using D^{-1}. At general s, prescribe any smooth round mean for \rho_j that agrees with this value at zero, and solve the projection of the equation onto the nonconstant modes. The operator (4.5) preserves the splitting into constants and nonconstants, so the latter equation has a unique solution. At each order it is a finite-dimensional linear equation with a smooth invertible coefficient matrix at every finite radius. This produces smooth coefficients on each finite interval. Induction also shows that the resulting graph at zero is exactly the minimal graph jet already constructed there. The constant eigenvalue need not be inverted away from zero.

The flux aspect. For an angular scalar F, write \langle F\rangle_\sigma=(4\pi)^{-1}\int_{S^2}F\,\dd A_\sigma. The asymptotic form (4.1) gives

Here \mu denotes only a flux aspect. To verify the formula directly, let e=G-b. Its leading trace is (C+2E)r^{-3}, and its radial divergence-minus-trace derivative is

For V_0=\sqrt{1+r^2} or V_i=rx_i, let v=1 or x_i, respectively. Both V and \partial_\eta V have leading term rv. The terms (\tr_b e)\dd V-e(\grad_bV,\cdot) add 2Ev r^{-2}+o(r^{-2}) to the normal flux integrand. The total is therefore (2C+6E)v r^{-2}+o(r^{-2}). Multiplying by r^2\dd A_\sigma/(16\pi) proves (4.7). The first differentiated remainder assumed in (4.1) makes every omitted flux integral tend to zero.

Boosted coordinates. Realize hyperbolic space as

Let \mathcal B(d) be the exponential of the Lorentz Lie-algebra matrix with time-space entries d\in\mathbb R^3 and zero spatial rotation part. Thus \mathcal B(0) is the identity. Write its action on future null vectors as

For small d, k>0. In pullback coordinates the old radius is kr+O(r^{-1}), the old radial distance is \eta+\log k+O(r^{-2}), and the old radial unit covector agrees with the new one up to O(r^{-1}) in hyperbolic norm. These statements, including all their needed derivatives, follow by applying the matrix \mathcal B(d) to the displayed hyperboloid coordinates. Since \mathcal B(d) preserves b, the leading coefficients of its pullback of G are

For example, the angular part of \dd\log k has hyperbolic norm O(r^{-1}); it therefore changes neither leading scaled diagonal coefficient in (4.1).

Differentiating (4.8) tangentially and taking Minkowski inner products gives y^*\sigma=k^{-2}\sigma. Consequently \dd A_\sigma(y)=k^{-2}\dd A_\sigma(x), and (1,x)=k\mathcal B(d)^{-1}(1,y). Changing angular variables proves the exact leading-flux transformation

This establishes the needed covariance directly for the specified flux.

The asymptotic CMC equation. For large s, put R=r(s) and \eta_R=\operatorname{arsinh}R. In the coordinates obtained from \mathcal B(d), seek a graph

where d and h are jets with zero base values, and every coefficient of h has zero constant and degree-one components. The parameters d,h are held fixed when computing an individual leaf. We claim, coefficientwise for bounded input jets in fixed finite angular spaces, that

Here are the estimates underlying this claim. In b, the unit normal to (4.11) is

Its divergence gives

Indeed the slope has hyperbolic norm O(R^{-2}), its normalization differs from one by O(R^{-4}), and the second Taylor term in 2\coth(\eta_R+h/R) is O(R^{-4}). These estimates hold for every coefficient of the finite Taylor expansion when the input coefficients are bounded.

On an unshifted sphere the perturbation in (4.1), after boosting, changes the radial lapse by C_dR^{-3}/2+o(R^{-3}). This contributes -C_dR^{-3}+o(R^{-3}) to H. The relative area density changes by E_dR^{-3}+o(R^{-3}); its radial logarithmic derivative contributes -3E_dR^{-3}+o(R^{-3}). The mixed components in the remainder contribute o(R^{-3}), including their tangential-divergence term. For completeness, on a fixed surface the comparison between two metrics b and \widehat g is

It follows by splitting \nabla_X^bY into its tangential and normal parts and using normal orthogonality. After tracing, the linear metric correction depends smoothly on the tangent plane, the background shape operator, and the scaled components of \widehat g-b,\nabla^b(\widehat g-b). The slope of (4.11) tends to zero, and its background shape operator differs by o(1) from that of a coordinate sphere. Its displacement is O(R^{-1}) in radial distance. The leading metric correction on that graph is consequently still -(C_d+3E_d)R^{-3}. Quadratic metric corrections are O(R^{-6}). This proves (4.12).

The estimate just proved also justifies its use at every Taylor order through p. A coefficient of G of degree j can undergo at most p-j spatial differentiations from Taylor composition with the graph and the boost. Mean curvature requires one additional derivative of the metric. Thus the mixed derivative allowance p-j+1 in (4.1) controls the coefficient of every remainder after multiplication by R^3. The base g_0 has the required bounds to all orders. Angular differential operators cause no further restriction within a fixed finite-dimensional angular space. More explicitly, denote the degree-j residual in (4.1) by \mathcal E_j, and set

Then \omega_j(R)\to0. For every n\le p, bounded coefficients of d,h in fixed finite angular spaces give the estimate

Here \mathcal V denotes the finitely many angular spaces and M a bound for the input coefficients; the fixed metric coefficients may also enter the constant. The explicit base residual is O(r^{-5}) with all derivatives and is included in the R^{-1} bound. The preceding graph estimates, the product rule, and the derivative count prove this estimate term by term. It is an estimate for Taylor coefficients, exactly as needed for the recursion.

Solving the rescaled equation. Project R^3(H-2\coth\eta_R) onto the nonconstant modes and set the projection equal to zero, through order p. At the background, the finite-radius linearization preserves the splitting between the h-modes, of degrees at least two, and the three boost modes. Set

A graph variation h/R in \eta has physical normal displacement \alpha_Rh/R. Equations (4.5)–(4.6) give its exact degree-\ell multiplier in the rescaled equation:

An infinitesimal boost has \eta-displacement d\cdot x. Its physical normal displacement is \alpha_Rd\cdot x, so its multiplier is 6m_0\alpha_R. These blocks converge respectively to \ell(\ell+1)-2 and 6m_0, both invertible. Equivalently, the latter sign and coefficient follow by differentiating 2m_0k^{-3} in (4.12), since k=1+d\cdot x+O(|d|^2).

At each Taylor order only fixed finite angular spaces are involved. In particular, the Taylor coefficients at d=0 of the boost map and its compositions are finite sums of coordinate functions and angular derivatives of the given profiles. This assertion concerns the finite Taylor expansion: it does not require a finite harmonic cutoff to be preserved by a finite nonzero boost. One can see the closure directly from the infinitesimal boost field

It follows by differentiating the hyperboloid coordinates. Its Lie derivative, repeated finitely many times, uses only radial derivatives, angular derivatives, and multiplication by degree-one profiles, also for tensor fields. It follows that the inverses of the finite-dimensional blocks above are bounded for all sufficiently large R and converge as R\to\infty.

Now solve successively for the coefficients of h,d. At each order their linear operator is the displayed background linearization. The remaining source depends only on lower solved coefficients and the prescribed metric coefficients. Equation (4.12), with its uniform coefficientwise remainder estimate, shows inductively that this source converges once the lower coefficients do. The converging inverse then gives bounded coefficients of h,d with limits. This starts at degree one and proves the assertion through order p. Smooth dependence on s follows at every finite radius from the smooth finite-dimensional equations.

In original coordinates these spheres are graph jets over the background sphere: the angular map has identity as base and can be inverted formally. Their round mean graph heights are smooth scalar functions of s. Choose the free means in the finite-radius recursion to agree with these functions for all sufficiently large s, and to agree with the minimal graph’s means at zero, using a smooth interpolation. Inductive uniqueness of the nonconstant graph coefficients makes the resulting recursion coincide with the far construction. This gives the family on the whole half-line. Its base is the coordinate-sphere family, so its lapse has base one.

The Hawking mass limit. Let d_\infty be the coefficientwise limiting boost. The degree-one projection of (4.12) gives

Writing

the round average of that same equation gives

The area of (4.11) satisfies

coefficientwise. To see the relevant cancellation, the relative hyperbolic area density is

and the metric correction is O(R^{-3}). The sole possible relative term of order R^{-1} integrates to zero because every coefficient of h has zero round mean.

Since H_b(R)^2-4=4/R^2, direct substitution into (4.2) gives

Therefore

By (4.10), this last scalar is the time component of \mathcal B(d_\infty)^{-1}p(G), whose spatial components vanish. Lorentz invariance and its positive base value m_0 identify it with the formal square root in (4.3).□

Proof. The area and mean-curvature variation formulas give

The latter is a constant on each leaf; any tangential transport term vanishes since H is constant there. For brevity put

Differentiation of (4.2) gives

The Gauss equation and |\II|^2=H^2/2+|\II^\circ|^2 imply

where K_\Sigma is intrinsic Gauss curvature. The sphere jets have \int K_\Sigma\,\dd 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 \Ric-\frac12\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

The operator J is self-adjoint on the closed leaf. Since \langle\delta f\rangle=0 and Jf=H' is constant,

The scalar jet \bar f is invertible because its base value is one. Eliminating the common term gives

Substitution into (4.16) 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\,\dd A_G=0, using the vanishing of all lower coefficients of \delta f. Every term in the bracket in (4.15) starts at degree 2l. At that degree, all exterior factors can therefore be evaluated at the background:

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

and all lower coefficient derivatives vanish.

The integrand in (4.19) is nonnegative after sphere integration. Indeed, on the mean-zero field F,

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 (4.3). For degree 2l, the right-hand side of the integrated (4.19) is an increasing function of T and its limit equals the finite limiting endpoint coefficient. This proves both its convergence and (4.18). 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. The operator D=-\Delta_S+2\kappa/a is strictly positive on all spherical harmonics. Lemma 3.1, applied to the finitely many modes of the boundary datum, gives the unique solution of (5.1); it has v=O(r^{-3}) with the differentiated asymptotics needed below.

The background identities \tr B=3, \Div B=0, and B=\Ric_{g_0}+3g_0 imply, by commuting the derivatives of a function,

Thus Q(v) is TT. Since S is totally geodesic, the tangential trace of \Hess v on S is \Delta_Sv. Moreover B_t(0)=\kappa/a. Taking the tangential trace of Q(v) and using its vanishing full trace therefore yields

Set \omega=N\,\dd v-v\,\dd N. With the averaged convention for the symmetric gradient, \operatorname{sym}\nabla\omega =(\nabla_i\omega_j+\nabla_j\omega_i)/2, the mixed products cancel and the static identity \Hess N=NB gives

Symmetry and vanishing divergence of k consequently give, on M_R=\{0\leq s\leq s(R)\},

On S, N=0 and \dd N=\kappa\,\dd s, so that \omega^\sharp=-\kappa vn. Since \nu=-n there, the inner contribution is +\kappa\int_Svk_{ss}\,\dd A_0. At infinity, |\omega|=O(r^{-2}), and hence the outer contribution is O(r^{-\gamma}) and tends to zero. We obtain

The same calculation with k replaced by Q(v) gives \int_MN|Q(v)|^2\,\dd V_0=\kappa\int_SvDv\,\dd A_0. Expanding the square proves (5.2). Finally, N\dd V_0 has radial size O(r^3)\,\dd s\,\dd A_\sigma, so the integral of N|k|^2 converges because 2\gamma>3. The stronger decay of Q(v) gives all remaining convergence assertions.□

Proof. For a finite-angular-type seed the scalar constraint gives \scal_g+6=\eps^2|q|^2+O(\eps^3). Proposition 4.4, 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

The areas in (5.5) 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\,\dd A_0. Lemma 5.1 now proves (5.6) with precisely the displayed factor and sign.

On a spherical harmonic with -\Delta_\sigma eigenvalue \lambda\geq2, the eigenvalue of J_s^0 is

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 3.5 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 3.4. The formula proved above gives c_2[q_\nu]\geq0, and continuity in \|\cdot\|_\beta gives c_2[q]\geq0. Proposition 3.5 also supplies (5.7) for every prescribed actual branch.□

Proof. First suppose q has finite angular type and c_2[q]=0. Equation (5.6) and the strict positivity just proved imply

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 -(\Hess_Sh_1)^{\circ}: the radial curvature term is pure trace. Since h_1=-v|_S, the vanishing of U|_S implies (\Hess_\sigma(v|_S))^{\circ}=0. For completeness, the divergence of this equation on the unit sphere gives \dd(\Delta_\sigma v+2v)=0. Subtracting the mean reduces it to \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

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

Here the second identity follows by taking the normal component of the velocity of the graph s+\eps h(s,x) in g_0+4\eps u g_0. On the degree-one part, the CMC equation therefore fixes

All formulas in the next calculation are componentwise on that part. The background equations and (5.10) give

Consequently,

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 -(\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,

We already have q=Q(v), and hence (5.6) 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 (5.3) 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 5.2, \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 3.4. 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

and determine a full formal time series recursively from

The coefficients take values in the ring of smooth spatial jets modulo \eps^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(\eps) 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=\tr_\gamma\mathcal K. The time-index Christoffel symbols are

and direct contraction of the curvature gives

Consequently E:=\overline\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 \scal_\gamma+H^2-|\mathcal K|_\gamma^2=-6; tracing (6.6) at t=0 therefore gives \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

Its vanishing divergence, evaluated with the displayed Christoffel symbols, is

This homogeneous system determines each next time coefficient of (e,j) from the preceding ones. All coefficients vanish by induction. Thus

as a formal time series modulo \eps^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 (6.8) 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 \eps. 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

Equivalently one can form Gaussian coordinates about that graph by the Taylor recursion for its normal geodesics and use (6.7). 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

Using -\overline g(n_T,\overline\nabla_{X_i}X_j) gives

The Hessian connection in this formula is taken at fixed time before evaluation at t=T. The induced metric is \widetilde g=\gamma(T)-\dd T\otimes\dd 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 \Hess b-Bb=Q(b). Equation (6.1) gives

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

At the background \partial_t^2\mathcal K(0)=0: differentiating (6.6) there uses \partial_t\gamma(0)=0 and \mathcal K(0)=0. Expanding (6.11) to degree two thus gives

In particular,

Choose the unique decaying solution of

Its right side is O(r^{-6}), with differentiated bounds, so Lemma 3.1 applies and gives w=O(r^{-3}) with the corresponding derivatives. Formula (6.12) gives the asserted decay of k. The degree-two momentum equation in (6.10), together with \tr_{g_0}k=0, gives \Div_{g_0}k=0.

The full transformed trace is O(\eps^3). Its square consequently begins in degree six, whereas

This proves (6.3); maximality at higher orders on the changed slice is not needed.

Finally,

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 \dd s,r\,\dd 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 \dd T\otimes\dd T, proving (6.4). 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-\dd v\otimes\dd v. Substituting q=\Hess v-Bv and using \Hess(v^2)=2v\Hess v+2\dd v\otimes\dd v proves (6.5).□

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=\tr\chi for its expansion. The initial sphere is umbilic in \phi_\eps^4g_0. Moreover the coefficient of \eps in the tracefree tangential part of K_\eps equals the tracefree tangential part of Q(v). Since S is totally geodesic in g_0, that part is (\Hess_{a^2\sigma}(v|_S))^{\tf}. It vanishes for the constant and degree-one modes of v|_S. The vanishing initial expansion and this shear calculation together give

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

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,\eps)=(0,0), spherical symmetry makes \mathcal R_\ell a scalar multiple of the screen identity. Its trace is \overline\Ric(\ell,\ell)=0 by (6.9) and nullness, so this curvature endomorphism is zero. Thus \partial_\lambda\chi(0,0)=0.

It follows from (6.18) that every retained monomial of \chi has total degree at least two in (\lambda,\eps). Raychaudhuri and the identically zero initial expansion then imply

Here O_{\rm tot}(j) means that each monomial \lambda^a\eps^b has a+b\geq j. The order statements are made in the formal series modulo \eps^5. Equivalently one can choose any sufficiently high finite Taylor lift: an Einstein residual of order \eps^5 integrated along a segment \lambda=O(\eps) 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(\eps). 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 (6.14).

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

because \tr_{g_0}k=0. Its zero null expansion gives (6.15).

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 (6.16). The first variation of area at the minimal Taylor leaf vanishes to the retained order. A height difference beginning with \eps^2z consequently has, at degree four, only the base second-variation contribution \frac12\int_S zDz\,\dd A_{g_0}. This is (6.17).

All of these assertions are identities in the joint spatial, \lambda, and \eps Taylor jets at the original boundary. The boundary-jet justification in Lemma 6.2 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

For finite-angular-type P, the choices may retain that property.

Let P^{\tf}=P-\frac12(\tr_\sigma P)\sigma, and identify vector fields and one-forms using \sigma. Define

On the curvature-one sphere, commuting covariant derivatives gives

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=\Div P^{\tf}. To decompose it, solve the scalar equations

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-\dd\varphi-*\dd\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

The one-forms \dd x_i and *\dd x_i correspond to conformal Killing vector fields. Integration by parts against the tracefree tensor P^{\tf} 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

on their angular modes, and set Y^\flat=\dd A+*\dd C. Equation (6.25) gives \Div\mathcal C(Y)=\Div P^{\tf}.

The tracefree tensor P_*=P^{\tf}-\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

Integration yields -\int|\nabla P_*|^2=2\int|P_*|^2, hence P_*=0. Taking h=\frac12\tr_\sigma P-\Div Y proves (6.24). 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

The derivative in a Lie direction is zero because the background scalar curvature is constant. If the scalar-curvature variation of P is constant, (6.24) therefore implies that h consists only of a constant and degree-one modes. A degree-one function h_1 satisfies \Hess_\sigma h_1=-h_1\sigma, so

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+\eps^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 (6.31) 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)\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 6.4 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

shows all terms of the left side in these modes. With B=B_s\dd s^2+B_t r^2\sigma, equality of the radial, mixed, Hessian, and tangential scalar coefficients respectively gives

Terms involving v_0 have only degrees zero or one and do not enter these equations. Solving the last three equations yields

Substitute these expressions in the first equation and use

After cancellation one obtains

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, (6.33) reads

The coefficient on the right is integrable toward infinity, and z\to0. The integral differential inequality, applied backwards between s and S>s, gives

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 6.1. The coefficient vanishings and nonnegativity were proved in (6.20)–(6.22). 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 (6.22) give (6.31) on each interior radial interval. Lemma 6.5 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 3.5. Lemma 3.3 converts the weighted function and equation remainders to the mass remainder. Consequently

in the case under consideration. Its positive coefficient gives strict positivity for all sufficiently small positive \eps, 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.□

Proof. We first prove radiality of the actual solution, rather than merely of its Taylor coefficients. Fix a rotation \mathcal R, and set \widehat\phi=\phi_\eps\circ\mathcal R and w=\widehat\phi-\phi_\eps. Because q is radial, \widehat\phi satisfies the same equation and boundary condition as \phi_\eps. Their difference satisfies

where divided differences of the nonlinearities give, uniformly on M and uniformly in \mathcal R,

Indeed, the derivatives of the interior and boundary nonlinearities with respect to their positive scalar argument z are respectively

The asserted bounds follow from boundedness of q and the given uniform convergence \phi_\eps\to1.

Choose 3/2<\beta<\min(\tau,\sqrt3). Then w=o(e^{-\beta s}), and (7.1) has the form of Lemma 3.1 with V=c_\eps-3, Robin coefficient a_*=c_\eps^*, and zero interior and boundary data. For sufficiently small \eps, uniformly in the rotation, \|V\|_\infty\leq c_\beta and \|a_*\|_\infty\leq\beta/2. The comparison estimate gives w=0, so the actual \phi_\eps is radial.

Write the invariant seed as q=A_q(s)\,\dd s^2+B_q(s)r^2\sigma. Its trace and divergence equations give

Consequently, for a constant C_q\in\mathbb R,

Fix a sufficiently small parameter. Define proper radial distance \ell and the actual area radius R by

Then the actual metric and second-form data are

A dot denotes an \ell-derivative. The momentum and scalar constraints are therefore

The scalar-curvature formula here follows from the sectional curvatures -\ddot R/R and (1-\dot R^2)/R^2 of this warped product.

Define

Direct differentiation, using the momentum equation, gives

by the scalar equation in (7.3). No division by \dot R has been used, so conservation is valid at turning points of R as well. At the boundary the MOTS equation reads

Substitution in (7.4) yields

We identify the same constant with the specified mass flux at infinity. Put \psi=\phi_\eps-1. Its actual radial equation gives

where (7.2) was used in the source estimate. The assumed weighted decay gives \psi,\psi'=O(e^{-\tau s}), and N/r=1+O(e^{-2s}). It follows that

Variation of constants for the roots 1,-3 excludes the growing solution by the prescribed decay and yields, as in Lemma 3.1, a constant c=c(\eps) with

In particular,

Using N^2=1+r^2-2m_0/r, we obtain

Also R^3k^2=O(R^{-3}), so

For a direct flux normalization check, rotational invariance implies p_i=0 for i=1,2,3. Equation (7.6) implies

in scaled components, with its first differentiated asymptotic. For a perturbation fb in dimension three, the mass-flux covector integrand with test function V simplifies to 2(f\,\dd V-V\,\dd f). With f=4cr^{-3}, V=V_0=\sqrt{1+r^2}, and \nu_b=\sqrt{1+r^2}\,\partial_r, its normal component is

The error terms give zero limiting flux. Integrating over a coordinate sphere of area 4\pi r^2 and dividing by 16\pi gives p_0-m_0=8c. Since p_0\to m_0>0 as \eps\to0,

for sufficiently small \eps, by (7.5)–(7.7). This argument includes C_q=0. In that case one can also apply the comparison estimate of Lemma 3.1 to \phi_\eps-1 and conclude \phi_\eps\equiv1 directly.□

Completion of the proof of Theorem 2.1. Fix the seed and the given solution branch. If q is radial, Proposition 7.1 gives exact equality on a sufficiently small parameter interval. Suppose henceforth that q is not radial. Proposition 5.2 gives the actual expansion

If c_2[q]>0, this proves strict positivity of the deficit for every sufficiently small positive \eps.

If c_2[q]=0, Corollary 5.3 gives q=Q(v) with Lv=0 and with only constant and linear angular modes. Its linear part is nonzero, for otherwise q would be radial. These homogeneous modes have the stronger differentiated decay needed for the actual fourth-order expansion in Proposition 3.5. Proposition 6.1 then gives

with all coefficients below order four equal to zero. Again the actual deficit is strictly positive for every sufficiently small positive \eps.

At \eps=0 the background identity m_0=b_*(4\pi a^2) gives equality. The preceding alternatives exhaust all fixed seeds. Choose \eps_0>0 small enough for the applicable alternative and the given branch, with \eps_0\leq\eps_*. This proves the exact inequality on 0\leq\eps<\eps_0, equality for radial seeds, and strict inequality for nonradial seeds at positive parameters in that interval. The size of the interval is allowed to depend on the fixed seed and branch; no uniform positive lower bound for c_2 or c_4 is asserted or required.□