Mean curvature at identical spacelike graph jets has positive Hessian symbol

Construct the needed finite-regularity extension directly in an ambient smooth coordinate chart. For a boundary parametrization $q(z,r)$ on a small half-box, keep $\tilde{q}=q$ for $r\ge 0$ and, for sufficiently small $r<0$, set $\tilde{q}(z,r)=6q(z,-r)-8q(z,-2r)+3q(z,-3r)$, componentwise. The sums of the coefficients multiplied by $(-j{)}^k$, for $j=1,2,3$, are $1$ for $k=0,1,2$. Consequently every tangential/normal partial derivative of total order at most two has the same limiting value from both sides of $r=0$. Difference quotients across the edge first give the matching first derivatives and then the matching second derivatives; hence $\tilde{q}$ is $C^2$. After shrinking, its values stay in the ambient chart and its derivative retains rank three near the contact point. This gives the required $C^2$ extension.

Choose an orthonormal basis $e_1,e_2,e_3$ of $P$ by finite Gram–Schmidt and put $e_0=n$. Use the map $y\mapsto {\exp }_p(y^a{e}_a)$ as a normal chart: [F2] gives its smoothness and identity differential at zero and [F3] gives a smooth local inverse. The coordinate vectors at zero are therefore $e_a$, so $g_{ab}(0)=\mathrm{diag}(-1,1,1,1{)}_{ab}$. By exponential scaling, each radial coordinate curve $s\mapsto sv$ is a geodesic. At zero [F6] gives ${\mathrm{\Gamma }}^a{}_{bc}(0)v^b{v}^c=0$ for every $v$. Torsion freedom makes the lower indices symmetric, and polarization gives ${\mathrm{\Gamma }}^a{}_{bc}(0)=0$. Metric compatibility then gives ${\mathrm{\partial }}_c{g}_{ab}(0)={\mathrm{\Gamma }}^d{}_{ca}(0)g_{db}(0)+{\mathrm{\Gamma }}^d{}_{cb}(0)g_{ad}(0)=0$. These Lorentzian center identities have been proved here, rather than imported from a Riemannian normal-coordinate theorem.

Choose a smooth ambient chart whose time vector at $p$ is $n$ and whose spatial coordinate vectors span $P$; by continuity the time vector remains future timelike nearby. The projection $\pi (t,x)=x$ restricts to an isomorphism on $P$, because its kernel is the timelike time line and $P$ is spacelike. For either interior parametrization, or the extension from step 1.1, the map $\pi \circ \tilde{q}$ therefore has invertible derivative. Apply [F4] to obtain its $C^2$ local inverse; composing the time component of $\tilde{q}$ with that inverse gives the $C^2$ graph function. Equality of the tangent hyperplanes gives $Df(p)=Dg(p)$; both graphs pass through $p$, so their values agree as well. In a general chart with $\alpha >0$, the time vector is again transverse to $P$, and the same projection argument applies.

In the boundary case the graph projection is a $C^2$ diffeomorphism on each extended parametrization. Its restriction to the rank-two boundary parametrization is thus a rank-two embedded $C^2$ surface in ${\mathbb{R}}^3$. These two projected boundaries agree, since the original boundary images agree. After a spatial linear change choose two projected components with invertible derivative on a boundary parametrization and use [F4] in dimension two to express the boundary as $x^3=\psi (x^1,x^2)$ with $\psi$ of class $C^2$. Each projected interior lies on one side: in the projected parametrization the differential of $x^3-\psi (x^1,x^2)$ vanishes in the two boundary directions and is nonzero in the inward direction, so its sign is fixed for small positive $r$. The common-side hypothesis selects the same sign for both. Reverse the transverse spatial coordinate if necessary, and set $x=\chi (y)=(y^1,y^2,y^3+\psi (y^1,y^2))$. Its explicit inverse subtracts $\psi$ in the third coordinate; it maps a sufficiently small common half-neighborhood $y^3\ge 0$ onto the selected side. The extended graph functions from step 2.1 restrict to the original graphs there.

The normal and mean curvature require only the stated $C^2$ regularity. For an extended graph $f$, put $E_i={\mathrm{\partial }}_i+f_i(x){\mathrm{\partial }}_t$ throughout a small ambient neighborhood. Their Gram matrix $h$ remains positive definite near the graph. With $T={\mathrm{\partial }}_t$ and $h^{-1}=(h^{ij})$, define $u=T-h^{ij}g(T,E_i)E_j$. Then $u$ is orthogonal to the three $E_i$, is nonzero by transversality, and is timelike: the orthogonal complement of a positive three-plane in signature $(-,+,+,+)$ is a negative line. The field $u$ is $C^1$, so normalizing it and choosing its future sign gives a unit $C^1$ ambient extension $\tilde{n}$ of the future graph normal. Metric compatibility gives $g({\mathrm{\nabla }}_{\tilde{n}}\tilde{n},\tilde{n})=\frac{1}{2}\tilde{n}g(\tilde{n},\tilde{n})=0$. Thus the ambient divergence trace on the graph equals $h^{ij}g({\mathrm{\nabla }}_{E_i}n,E_j)$, independently of unit extension. Differentiating $g(n,E_j)=0$ tangentially yields [step 2.1] $H(f)=-h^{ij}g(n,{\mathrm{\nabla }}_{E_i}{E}_j)\mathrm{.}$ All differentiations are valid for $C^1$ fields, including the one-sided limits at a boundary.

Write $E_i^a$ for the components of the graph tangent vectors, with $a=0,1,2,3$. The coordinate connection formula gives ${\mathrm{\nabla }}_{E_i}{E}_j=f_{ij}{\mathrm{\partial }}_t+{\mathrm{\Gamma }}^a{}_{bc}{E}_i^b{E}_j^c{\mathrm{\partial }}_a$. At the shared point and first jet, the two graphs have identical $E_i$, $h^{ij}$ and future normal $n$, and the Christoffel terms are evaluated at the same ambient point. Subtracting the two formulas from step 3.2 therefore cancels every Christoffel term and gives [step 3.2] $H(f)-H(g)=-g(n,{\mathrm{\partial }}_t)h^{ij}(f-g{)}_{ij}\mathrm{.}$ Since $h^{-1}$ is positive definite and $\alpha =-g(n,{\mathrm{\partial }}_t)>0$, this is the asserted positive-definite symbol. For a future timelike ${\mathrm{\partial }}_t$, write ${\mathrm{\partial }}_t=an+V$ with $V\in P$; the future cone in the orthonormal frame with time vector $n$ has $a>0$, and $\alpha =a$. Thus that sufficient orientation condition has precisely the required sign.

Apply the graph construction of step 2.1 in this normal chart. Both graph values at the center and both first derivatives vanish. There ${\mathrm{\partial }}_t=n$, $h_{ij}={\delta }_{ij}$ and $\alpha =1$, so step 4.1 gives $A^{ij}={\delta }^{ij}$. In addition the vanishing Christoffel symbols from step 1.2 and step 3.2 give $H(f)={\sum }_i{f}_{ii}$ and $H(g)={\sum }_i{g}_{ii}$ individually at the center.

For the common spatial flattening $x=\chi (y)$ from step 3.1, let $w=f-g$ and $B=D\chi$ at contact. The first jet of $w$ vanishes, so applying [F9] twice gives $D_y^2(w\circ \chi )=B^T(D_x^{2}w)B$: the terms involving $Dw$ and second derivatives of $\chi$ vanish. Hence the contracted coefficient in the flattened variables is $\tilde{A}=B^{-1}A{B}^{-T}$, which is positive definite because $B$ is invertible. Time is unchanged in this spatial change. The $C^2$ extensions and their restrictions have the same derivatives through order two on the boundary, so the formulas calculate the original boundary mean curvatures and do not depend on the chosen extensions. This proves the common half-neighborhood assertion as well as the interior and normal-coordinate assertions.