From a Schwarzschild spatial isometry to the normalized spacetime exterior

Pull back the radial function $U=(1-\mu \mathrm{/}(2s))\mathrm{/}(1+\mu \mathrm{/}(2s))$ through the spatial isometry, and write $a=\mu \mathrm{/}2$ and $\varphi (s)=1+a\mathrm{/}s$. By the coordinate divergence formula in [F7], its radial Laplacian is $({\varphi }^6{s}^2{)}^{-1}{\mathrm{\partial }}_s(s^2{\varphi }^2{\mathrm{\partial }}_{s}U)=0$, since $s^2{\varphi }^2{\mathrm{\partial }}_{s}U=\mu$. Thus $U$ is harmonic, zero at $B$, and tends to one. The difference $w=V-U$ is harmonic and zero on $B$. Outside compact cores both asymptotically Euclidean metrics are uniformly comparable to their Euclidean metrics, so distance from a fixed compact set is bounded above and below by positive multiples of the corresponding radius. The isometry preserves those distances, hence the original end radius and $s$ are comparable. The given end estimates and preservation of gradient norms consequently give $w=O(s^{-1})$ and $\mathrm{\mid }Dw{\mathrm{\mid }}_h=O(s^{-2})$.

Use the spatial isometry to identify $X$ with $(a,\mathrm{\infty })\times S^2$. For integers $j\ge 1$, put ${\epsilon }_j=1\mathrm{/}(j+1)$, $R_j=a+j+1$ and $D_j=\{a+{\epsilon }_j\le s\le R_j\}$. These are compact oriented smooth Riemannian domains entirely in $X$, with two disjoint smooth boundary spheres; they increase and their union is $X$. On a sphere $s=b$, the unit normals are $\pm \varphi (b{)}^{-2}{\mathrm{\partial }}_s$ and the area density is $\varphi (b{)}^4{b}^{2}d\omega$. In a fixed compact strip adjacent to $B$, the $C^2$ regular data give a uniform bound for $\mathrm{\mid }DV{\mathrm{\mid }}_h$, and the explicit $U$ has bounded gradient, hence $\mathrm{\mid }Dw{\mathrm{\mid }}_h\le C$. The radial segments from $s=a$ to $a+\epsilon$ have length ${\int }_a^{a+\epsilon }\varphi (s{)}^{2}ds=O(\epsilon )$. The zero boundary value of $w$ then gives $w=O(\epsilon )$ on the inner face, whose area is uniformly bounded by the displayed formula. Thus the absolute inner flux of $wDw$ is $O({\epsilon }_j)$. The outer area is $O(R_j^2)$, and the bounds in step 1.1 make its absolute outer flux $O(R_j^{-1})$. This constructs the exhaustion and both estimates directly for $w$.

By [F7], ${\mathrm{div}}_h(wDw)=\mathrm{\mid }Dw{\mathrm{\mid }}_h^2$. Apply the divergence theorem [F3] on each $D_j$, using countable choice [F2] and the integration convention [F5]. Step 2.1 bounds the nonnegative energy by ${\int }_{D_j}\mathrm{\mid }Dw{\mathrm{\mid }}_h^{2}d{\mu }_h\le C({\epsilon }_j+R_j^{-1})\to 0$. The compact domains are Borel and ${\mathbf{1}}_{D_j}\mathrm{\mid }Dw{\mathrm{\mid }}_h^2$ are nonnegative Borel functions increasing pointwise to $\mathrm{\mid }Dw{\mathrm{\mid }}_h^2$. Monotone convergence [F6] therefore gives ${\int }_X\mathrm{\mid }Dw{\mathrm{\mid }}_h^{2}d{\mu }_h=0$. A nonzero value of $\mathrm{\mid }Dw{\mathrm{\mid }}_h$ would, by continuity, give an interior coordinate ball with the integrand bounded below by a positive constant, and the positive continuous coordinate density in [F5] gives that ball positive measure. Hence $Dw=0$ everywhere. Connectedness makes $w$ constant, and its zero limit at infinity makes it zero.

Set $\mathcal{R}=s(1+\mu \mathrm{/}(2s){)}^2$. This is strictly increasing from $2\mu$ to infinity on $s>\mu \mathrm{/}2$, and direct differentiation gives $h_{\mathrm{S}}=(1-2\mu \mathrm{/}\mathcal{R}{)}^{-1}d{\mathcal{R}}^2+{\mathcal{R}}^{2}d{\omega }^2$ and $U^2=1-2\mu \mathrm{/}\mathcal{R}$. Taking the product of the spatial isometry with the identity on all $t\in \mathbb{R}$ therefore gives a global isometry of the full static exterior to $-(1-2\mu \mathrm{/}\mathcal{R})d{t}^2+(1-2\mu \mathrm{/}\mathcal{R}{)}^{-1}d{\mathcal{R}}^2+{\mathcal{R}}^{2}d{\omega }^2$. With the parameter and unit-round-metric identifications in [F4], this is precisely the Schwarzschild exterior model, and the map sends ${\mathrm{\partial }}_t$ to ${\mathrm{\partial }}_{x^0}$, preserving the time orientation.

On the same $D_j$, the product identity [F7] and the harmonicity of $U$ give ${\mathrm{div}}_h(UDU)=\mathrm{\mid }DU{\mathrm{\mid }}_h^2$. Apply [F3] with [F2] and [F5]. The sphere normal and area formula of step 2.1 give the flux of $UDU$ through $s=b$, oriented toward increasing $s$, as $4\pi U(b)b^2\varphi (b{)}^2{\mathrm{\partial }}_{s}U(b)=4\pi \mu U(b)$. The inner outward normal has the opposite sign, so ${\int }_{D_j}\mathrm{\mid }DU{\mathrm{\mid }}_h^{2}d{\mu }_h=4\pi \mu (U(R_j)-U(a+{\epsilon }_j))\to 4\pi \mu$. Apply [F6] to ${\mathbf{1}}_{D_j}\mathrm{\mid }DU{\mathrm{\mid }}_h^2$ as in step 3.1 to obtain ${\int }_X\mathrm{\mid }DU{\mathrm{\mid }}_h^{2}d{\mu }_h=4\pi \mu$. These computations on the model are also computations on $X$: an isometry preserves gradient norms, and its metric pullback identity gives $\sqrt{\det (F^{\ast }{h}_{\mathrm{S}})}=\mathrm{\mid }\det dF\mathrm{\mid }\sqrt{\det h_{\mathrm{S}}}\circ F$ in coordinates, exactly the transformation law of the intrinsic density in [F5]. Since $V=U$ on $X$ by step 3.1, comparison with the independently supplied identity ${\int }_X\mathrm{\mid }DV{\mathrm{\mid }}_h^{2}d{\mu }_h=4\pi m$ in [F3] gives $\mu =m$, without presuming invariance of an end-coordinate coefficient.