A charged RN background does not amplify a neutral test scalar

Write $x=r_{\ast }$ and use the choices in [A1]--[A2]. By [F3], $x$ ranges over all of $\mathbb{R}$ and $dr\mathrm{/}dx=f$. The charged radial equation [F1] becomes $u^{\mathrm{\prime }\mathrm{\prime }}(x)+({\omega }^2-V(x))u(x)=0,V(x):=\frac{f(r(x))f^{\mathrm{\prime }}(r(x))}{r(x)}\mathrm{.}$ The definition gives $dr\mathrm{/}dx=f$ and $f(r)=(r-r_{+})(r-r_{-})\mathrm{/}{r}^2$. Since $r_{+}>M>r_{-}$ and $r_{+}{r}_{-}=Q^2$, we have $f(r)>0$ and $Mr>M{r}_{+}>Q^2$ for $r>r_{+}$, so $f^{\mathrm{\prime }}(r)=2(Mr-Q^2)\mathrm{/}{r}^3>0$ and $V>0$. Thus $V$ is smooth and real. Moreover, ${\int }_{\mathbb{R}}\mathrm{\mid }V(x)\mathrm{\mid }dx={\int }_{r_{+}}^{\mathrm{\infty }}\frac{f^{\mathrm{\prime }}(r)}{r}dr={\int }_{r_{+}}^{\mathrm{\infty }}\frac{2(Mr-Q^2)}{r^4}dr<\mathrm{\infty }\mathrm{.}$

Define $u_{-}$ by the Volterra equation $u_{-}(x)=e^{-i\omega x}+\frac{1}{\omega }{\int }_{-\mathrm{\infty }}^x\sin \text{\hspace{0.17em}⁣}(\omega (x-y))V(y)u_{-}(y)dy\mathrm{.}$ Iteration gives a uniformly convergent Neumann series: the absolute value of its $n$th iterated term is at most $\mathrm{\parallel }V{\mathrm{\parallel }}_{L^1(\mathbb{R})}^n\mathrm{/}({\omega }^{n}n!)$, using $\mathrm{\mid }\sin z\mathrm{\mid }\le 1$ for real $z$ and the ordered integration region. The sum is bounded on $\mathbb{R}$; differentiating the integral equation shows it is a smooth solution of the equation in step 1.1. The tail integrals tend to zero as $x\to -\mathrm{\infty }$, and differentiation under the integral gives $u_{-}(x)=e^{-i\omega x}+o(1),u_{-}^{\mathrm{\prime }}(x)=-i\omega e^{-i\omega x}+o(1)(x\to -\mathrm{\infty })\mathrm{.}$

Since $V{u}_{-}\in L^1(\mathbb{R})$, expanding the sine in step 2.1 into exponentials and replacing the upper integration limit by $+\mathrm{\infty }$ gives $u_{-}(x)=a{e}^{-i\omega x}+b{e}^{i\omega x}+o(1),u_{-}^{\mathrm{\prime }}(x)=-i\omega a{e}^{-i\omega x}+i\omega b{e}^{i\omega x}+o(1)$ as $x\to +\mathrm{\infty }$, where $a=1-\frac{1}{2i\omega }{\int }_{\mathbb{R}}{e}^{i\omega y}V(y)u_{-}(y)dy,b=\frac{1}{2i\omega }{\int }_{\mathbb{R}}{e}^{-i\omega y}V(y)u_{-}(y)dy\mathrm{.}$ The same tail estimate after differentiating the integral equation proves the displayed derivative asymptotic.

For $j=\mathrm{Im}(\overline{u_{-}}{u}_{-}^{\mathrm{\prime }})$, the real-coefficient equation in step 1.1 gives $j^{\mathrm{\prime }}=0$. The horizon asymptotics in step 2.1 yield $j=-\omega$, while step 3.1 yields $j=\omega (\mathrm{\mid }b{\mathrm{\mid }}^2-\mathrm{\mid }a{\mathrm{\mid }}^2)$ at infinity; the two oscillatory cross terms have real sum and contribute no imaginary part. Hence $\mathrm{\mid }a{\mathrm{\mid }}^2=\mathrm{\mid }b{\mathrm{\mid }}^2+1$, so $a\mathrm{\ne }0$. Put $u=u_{-}\mathrm{/}a$, $R=b\mathrm{/}a$, and $T=1\mathrm{/}a$. Then $u$ has the asymptotics in [F3], with unit incoming amplitude and only the ingoing branch at the horizon. The separated-wave substitution in the proof of [F1] is reversible here: $Y$ is a nonzero constant, $A_{\mathrm{i}\mathrm{n}}$ and $fr$ are nonzero, and each displayed step is an equality. Therefore $\psi =A_{\mathrm{i}\mathrm{n}}{e}^{-i\omega t}Yu\mathrm{/}r$ solves ${\mathrm{\square }}_{g,A}\psi =0$ and is an admitted scattering solution for any fixed nonzero allowed $A_{\mathrm{i}\mathrm{n}}$.

Apply [F2] to this constructed solution. Since $e=0$ and $T=1\mathrm{/}a\mathrm{\ne }0$, $\mathrm{\mid }R{\mathrm{\mid }}^2=1-\mathrm{\mid }T{\mathrm{\mid }}^2=1-\mathrm{\mid }a{\mathrm{\mid }}^{-2}<1.$ Thus this nonzero-charge background admits a neutral mode whose reflected radial flux is less than the incident flux, contradicting the universal amplification statement.