Mass-one Ernst decay and uniform Harrison denominator bounds

The metric matrix tends to the Minkowski matrix, with determinant tending to $-1$, while its spatial block tends to the identity. Apply [F4] also to the identity matrix-valued map on the open set of invertible matrices: inversion is continuous at both limiting matrices, so the inverses are bounded on a sufficiently large end. Their entries as functions of the end coordinates are $C^6$ by [F4]. Differentiating $g^{-1}g=I$ repeatedly, with the finite product rule, gives $g^{ij}={\delta }^{ij}+O_6(r^{-1})$, $g^{00}=-1+O_6(r^{-1})$, and $g^{0i}=O_6(r^{-2})$. For the last bound use the block identity $g^{0i}=-g^{00}{g}_{0j}(g_{\mathrm{s}\mathrm{p}}^{-1}{)}^{ji}$; its derivatives have the same order. The volume coefficient is bounded with its corresponding differentiated bounds. Stationarity and [F2] give $F_{ij}=\frac{1}{2}({\mathrm{\partial }}_i{g}_{0j}-{\mathrm{\partial }}_j{g}_{0i})=O_5(r^{-3})$ and $F_{0i}=-\frac{1}{2}{\mathrm{\partial }}_i{g}_{00}=O_5(r^{-2})$. In ${ϵ}_{0i\mu \nu }{F}^{\mu \nu }$ the indices $\mu ,\nu$ are spatial; every $F_{0a}$ contribution to the raised spatial components contains a mixed inverse coefficient $g^{j0}=O_6(r^{-2})$. The remaining terms are $O_5(r^{-3})$. Thus ${\mathrm{\partial }}_i\psi =O_5(r^{-3})$. For a fixed unit vector $\omega$, the smooth real function $h(\rho )=\psi (\rho \omega )$ on $(R_0,\mathrm{\infty })$ has derivative $h^{\mathrm{\prime }}(\rho )={\omega }^i{\mathrm{\partial }}_i\psi (\rho \omega )$ by [F5]. For finite $R>r$, [F6] gives $\psi (R\omega )-\psi (r\omega )={\int }_r^R{\omega }^i{\mathrm{\partial }}_i\psi (\rho \omega )d\rho$. The integrand is bounded in modulus by $C{\rho }^{-3}$ uniformly in $\omega$, and its tail integral from $R$ to any $R^{\mathrm{\prime }}>R$ is at most $C\mathrm{/}(2R^2)$. Hence this integral converges as $R\to \mathrm{\infty }$; using $\psi (R\omega )\to 0$ gives $\psi (r\omega )=-{\int }_r^{\mathrm{\infty }}{\omega }^i{\mathrm{\partial }}_i\psi (\rho \omega )d\rho =O(r^{-2})$ uniformly in direction. For derivatives of order one through six use the already obtained five derivative bounds on the gradient. This proves the two potential estimates.

Spatial derivatives of $r^{-1}$ of order $k$ are bounded by $C_k{r}^{-1-k}$. Since $s_q\ge r^{-1}$, they are bounded by $C_k{s}_q{r}^{-k}$. Induction using derivatives of $s_q{s}_q^{-1}=1$ yields the stated inverse bound; the same reasoning gives $\mathrm{\mid }{\mathrm{\partial }}^{\beta }{s}_q^{-2}\mathrm{\mid }\le C_{\beta }{s}_q^{-2}{r}^{-\mathrm{\mid }\beta \mathrm{\mid }}$. All constants are independent of $q$. In particular $q\mathrm{/}{s}_q$ and $r^{-1}\mathrm{/}{s}_q$, together with derivatives of order $k\le 5$, are $O(r^{-k})$ uniformly. Write $\sigma =1-2\mathrm{/}r+e$ with $e=O_6(r^{-2})$. Then $\frac{1+q-(1-q)\sigma }{2s_q}=1-\frac{q}{s_{q}r}-\frac{(1-q)e}{2s_q}=1+a_q,a_q=O_5(r^{-1})\mathrm{.}$ This follows by the preceding inverse bounds and $s_q^{-1}\le r$, with finite product differentiation. Hence $D_q\mathrm{/}(4s_q^2)=\mathrm{\mid }1+a_q{\mathrm{\mid }}^2=1+O_5(r^{-1})$. Enlarge the common radius so this last error has absolute value below one half; its reciprocal also has uniform differentiated bounds by repeated differentiation of its product with its inverse. This proves both inequalities.

The supplied pole lies in $U$, so the same smooth potential can be evaluated there. Since $T(p)=0$, $f(p)=-g(T(p),T(p))=0$, so $\sigma (p)=i\psi (p)$. Taking the squared modulus gives the asserted exact expression for $d_q$. For $0\le q\le 1$, its first term is between one and four and its second term is between zero and $\psi (p{)}^2$. These are finite positive bounds independent of $q$.