which means if the conditions $S(r^*=+\infty)\le 0$ and $S(r^*=-\infty)\ge 0$ is true for $S$ defined in (\[eq:S-def\]). For GR, it is easy to calculate that $$\begin{aligned}
S=\frac{A(r)}{r}\,,\quad\text{where}  ~~A=1-\frac{2M}{r}\,.\end{aligned}$$ We consider the Schwarzschild solution $$\begin{aligned}
A=B=1-\frac{2M}{r}\,,~~C=r^2\,,\end{aligned}$$ and find that $S(r^*=+\infty)=S(r^*=-\infty)=0$ where $r^*$ is the tortoise coordinate, which allows us to conclude that Schwarzschild spacetime is stable under odd perturbations. For completeness, we derive the potential $V(r)$ from eq. (\[waveeq\]) $$\begin{aligned}
V(r)=\Bigl(1-\frac{2M}{r}\Bigr)\Bigl[\frac{l(l+1)}{r^2}-\frac{6M}{r^3}\Bigr]\,,\end{aligned}$$ which is the Regge-Wheeler potential [@Regge:1957td].

We conclude that the Schwarzschild spacetime does not exhibit either ghost or gradient instability, it is stable and the propagation speeds along both radial and angular directions are equal to the speed of light, $c_r^2=c_\theta^2=1$.

General Relativity conformally coupled to a scalar filed
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Considering a scalar field with non-negative potential in asymptotically flat spacetimes, it is well known that static black holes do not have scalar hair [@Bekenstein:1971hc] (see [@Herdeiro:2015waa] for a recent review). In fact, if the scalar field is not zero, then the equations show that it will diverge at the horizon, implying a divergence of the curvature too. Therefore to obtain a regular black hole solution, one has to consider the trivial solution where the scalar field vanishes and hence 