Subextremal RN scalar mode potential and conditional energy

Put $d:=\sqrt{m^2-Q^2}>0$. By [F6] the specified oriented integral defines $r_{\ast }$ with $d{r}_{\ast }\mathrm{/}dr=f^{-1}$; its integrand is smooth on the exterior, so $r_{\ast }$ is smooth. Then $r_{+}-r_{-}=2d$, $r_{+}{r}_{-}=Q^2$, and $f(r)=\frac{(r-r_{+})(r-r_{-})}{r^2}>0(r>r_{+}),f^{\mathrm{\prime }}(r)=\frac{2(mr-Q^2)}{r^3}\mathrm{.}$ Since $mr>m{r}_{+}=m^2+md>Q^2=m^2-d^2$, we have $f^{\mathrm{\prime }}(r)>0$ throughout the exterior. Also $f^{\mathrm{\prime }}(r_{+})=(r_{+}-r_{-})\mathrm{/}{r}_{+}^2>0$, so near $r_{+}$ the positive function $f(r)$ is bounded above and below by positive multiples of $r-r_{+}$. Therefore the specified integral diverges to $-\mathrm{\infty }$ as $r\downarrow r_{+}$; as $r\to \mathrm{\infty }$, $f(r)\to 1$, so it diverges to $+\mathrm{\infty }$. Thus $r_{\ast }$ is strictly increasing and onto $\mathbb{R}$. Its derivative is nonzero everywhere, so [F7] gives a smooth local inverse at every point; strict increase makes these local inverses agree with the global inverse. It is therefore a smooth coordinate on the whole exterior.

In the polar chart, $\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}=r^2\sin \theta$, $g^{tt}=-f^{-1}$, $g^{rr}=f$, and $g^{AB}=r^{-2}{\gamma }^{AB}$. Applying the divergence formula in [F2] to $g^{ab}{\mathrm{\partial }}_b\psi$ gives ${\mathrm{\square }}_g\psi =-f^{-1}{\mathrm{\partial }}_t^2\psi +\frac{1}{r^2}{\mathrm{\partial }}_r(r^{2}f{\mathrm{\partial }}_r\psi )+\frac{1}{r^2}{\mathrm{\Delta }}_{{\gamma }_{S^2}}\psi \mathrm{.}$ Substitution of $\psi =\varphi Y$ and the specified eigenvalue gives $-f^{-1}{\varphi }_{tt}+\frac{1}{r^2}{\mathrm{\partial }}_r(r^{2}f{\varphi }_r)-\frac{\mathrm{\ell }(\mathrm{\ell }+1)}{r^2}\varphi =0.$ This scalar coefficient vanishes because its product with $Y$ vanishes for all angles and ${\int }_{S^2}{Y}^{2}d{\omega }_{S^2}=1$. The regularity needed for these differentiations is [F8].

With $U_{\mathrm{\ell }}=r\varphi$ and ${\mathrm{\partial }}_{r_{\ast }}=f{\mathrm{\partial }}_r$, direct differentiation gives $r^{2}f{\varphi }_r=r{\mathrm{\partial }}_{r_{\ast }}{U}_{\mathrm{\ell }}-f{U}_{\mathrm{\ell }}$. Multiply the equation in step 1.2 by $fr$, then $\frac{f}{r}{\mathrm{\partial }}_r(r^{2}f{\varphi }_r)=\frac{1}{r}{\mathrm{\partial }}_{r_{\ast }}(r{\mathrm{\partial }}_{r_{\ast }}{U}_{\mathrm{\ell }}-f{U}_{\mathrm{\ell }})={\mathrm{\partial }}_{r_{\ast }}^2{U}_{\mathrm{\ell }}-\frac{f{f}^{\mathrm{\prime }}}{r}{U}_{\mathrm{\ell }}\mathrm{.}$ Together with $-r{\varphi }_{tt}=-{\mathrm{\partial }}_t^2{U}_{\mathrm{\ell }}$, and using ${\mathrm{\partial }}_{r_{\ast }}f=f{f}^{\mathrm{\prime }}$, this yields the stated one-dimensional equation and $V_{\mathrm{\ell }}=f[\mathrm{\ell }(\mathrm{\ell }+1)\mathrm{/}{r}^2+f^{\mathrm{\prime }}\mathrm{/}r]$. The derivatives are justified by [F8].

The factorization in step 1.1 gives $f>0$, and $f^{\mathrm{\prime }}>0$ there, so $V_{\mathrm{\ell }}>0$ for every $\mathrm{\ell }\ge 0$. The formula is smooth for $r>r_{+}$; it tends to $0$ at $r\downarrow r_{+}$ because $f\to 0$ while the bracket stays finite, and it tends to $0$ at infinity because $f\to 1$ and both bracket terms tend to $0$. Thus $V_{\mathrm{\ell }}$ is bounded as a continuous function on the $r_{\ast }$ line. Under the additional decay hypothesis, apply [F3] with $N=1$, $V=V_{\mathrm{\ell }}$, and $S=0$ to obtain $E_{\mathrm{\ell }}^{\mathrm{\prime }}(t)=0$. Finally, $d{r}_{\ast }=dr\mathrm{/}f$ and ${\mathrm{\partial }}_{r_{\ast }}{U}_{\mathrm{\ell }}=f{\mathrm{\partial }}_r{U}_{\mathrm{\ell }}$ give $\int \mathrm{\mid }{\mathrm{\partial }}_{r_{\ast }}{U}_{\mathrm{\ell }}{\mathrm{\mid }}^{2}d{r}_{\ast }=\int f\mathrm{\mid }{\mathrm{\partial }}_r{U}_{\mathrm{\ell }}{\mathrm{\mid }}^{2}dr$, whose coefficient vanishes at the horizon. This proves the stated degeneracy and no stronger horizon estimate.