Regular static boundary geometry and positivity of the end coefficient

The boundary limit in [F1] is $D^{2}V=0$. Because $V{\mathrm{\mid }}_B=0$ and $dV\mathrm{\ne }0$ there, $DV$ is nonzero and normal to $B$. If $n$ is the inward unit normal, positivity of $V$ inside implies $n(V)>0$, so $DV=n(V)n$ on $B$. For tangent fields $A,C$, the Hessian formula gives $D^{2}V(A,C)=n(V)h(D_{A}n,C)$. Metric compatibility gives $h(D_{A}C,n)=-h(C,D_{A}n)$, and hence $\mathrm{I}\mathrm{I}(A,C)=-h(D_{A}n,C)n$. These identities remain valid for the stated boundary regularity: the inverse metric is $C^2$, its Christoffel coefficients are $C^1$, and $n=DV\mathrm{/}\mathrm{\mid }DV\mathrm{\mid }$ on $B$ is $C^1$. In a boundary chart, two extensions of a tangent field differ by coefficients vanishing on $B$; their tangential derivatives also vanish there. The formula $D_{A}W=A(W^i){\mathrm{\partial }}_i+{\mathrm{\Gamma }}^i{}_{jk}{A}^j{W}^k{\mathrm{\partial }}_i$ therefore proves extension independence along $B$. Orthogonal projection onto the line spanned by $n$ gives exactly the normal-valued form in [F3]. Since $n(V)>0$ and $D^{2}V(A,C)=0$, this form vanishes, proving total geodesicity in the pointwise sense of [F3].

Take the collar $c$ from [F4] and fix $0<t_0<1$. The strip $c(B\times [0,t_0])$ is compact. Choose $R_{\ast }$ so large that it and the compact core lie inside the region bounded by the end sphere $S_{R_{\ast }}$. For $R\ge R_{\ast }$ let $K_R$ be the complement in $\bar{X}$ of the end region $r>R$; it is compact by the compact-core hypothesis. Set $R_j=R_{\ast }+j$, ${\epsilon }_j=t_0\mathrm{/}(j+1)$ and $D_j=K_{R_j}\setminus c(B\times [0,{\epsilon }_j))$. These are compact smooth domains entirely in $X$, with disjoint boundary faces $S_{R_j}$ and $B_{{\epsilon }_j}=c(B\times \{{\epsilon }_j\})$. They satisfy $D_j\subset D_{j+1}$ and ${\bigcup }_j{D}_j=X$: the end cutoff eventually includes each point, and the shrinking collar eventually excludes no fixed interior point. The product rule gives ${\mathrm{div}}_h(VDV)=\mathrm{\mid }DV{\mathrm{\mid }}_h^2+V{\mathrm{\Delta }}_{h}V=\mathrm{\mid }DV{\mathrm{\mid }}_h^2$. Applying [F5] on $D_j$ therefore expresses its energy integral as the sum of its outer and inner fluxes.

For tangent $A$, metric compatibility and the Hessian formula give $A(\mathrm{\mid }DV{\mathrm{\mid }}_h^2)=2D^{2}V(A,DV)=0$ on $B$. Thus $\mathrm{\mid }DV{\mathrm{\mid }}_h^2$ is locally constant on $B$, and is constant on each connected component. Its value is positive there because $dV\mathrm{\ne }0$.

The pullback metrics $c_s^{\ast }h$ on the compact $B$, for $0\le s\le t_0$, depend continuously on $(p,s)$ and are positive definite. Relative to $c_0^{\ast }h$ their area ratios are uniformly bounded by compactness. The latter has finite area by a finite boundary atlas and bounded continuous coordinate densities. Hence ${\mathrm{Area}}_h(B_s)\le C$. Also $V(c(p,s))={\int }_0^s{\mathrm{\partial }}_t(V\circ c)(p,t)dt=O(s)$ uniformly, and $\mathrm{\mid }DV{\mathrm{\mid }}_h$ is uniformly bounded on the strip by the $C^2$ data. With either unit normal on $B_s$, the absolute flux of $VDV$ is consequently at most $Cs$, so the inner flux for $D_j$ tends to zero. These estimates use the smooth collar and continuity on its compact strip, not a smooth-metric theorem at $B$.

The differentiated end conditions give ${\mathrm{\partial }}_{r}V=m\mathrm{/}{R}^2+O(R^{-3})$, $DV=O(R^{-2})$ in the end Cartesian frame, the outward unit normal ${\nu }_h={\mathrm{\partial }}_r+O(R^{-1})$ in that frame, and $d{A}_h=(1+O(R^{-1}))R^{2}d\omega$. Since $V=1+O(R^{-1})$, the outer flux is ${\int }_{S_R}Vh(DV,{\nu }_h)d{A}_h=4\pi m+O(R^{-1})$. Combining this with the vanishing inner flux yields ${\lim }_j{\int }_{D_j}\mathrm{\mid }DV{\mathrm{\mid }}_h^{2}d{\mu }_h=4\pi m$.

The sets $D_j$ are Borel, and $f_j={\mathbf{1}}_{D_j}\mathrm{\mid }DV{\mathrm{\mid }}_h^2$ are nonnegative measurable functions on the measure space $(X,\mathcal{B}(X),{\mu }_h)$ of [F6]. They increase pointwise to $\mathrm{\mid }DV{\mathrm{\mid }}_h^2$ by step 1.2. Monotone convergence [F7] and step 3.1 give ${\int }_X\mathrm{\mid }DV{\mathrm{\mid }}_h^{2}d{\mu }_h=4\pi m$, in particular a finite integral. To see strict positivity, fix a boundary point, where $\mathrm{\mid }DV{\mathrm{\mid }}_h>0$. Continuity supplies an interior coordinate ball on which $\mathrm{\mid }DV{\mathrm{\mid }}_h^2$ is bounded below by a positive constant. A smaller ball has positive ${\mu }_h$-measure by the positive continuous coordinate density in [F6]. The energy is therefore strictly positive, and $m>0$ follows.