the identity For $f=1-x-\lambda r^2>0$, $P$ vanishes only when $m+1=0$, which is not realised, because $m=k^2-n>0$. Further, if $x=2/(n+1)$ at $r=r_h$ or $r=r_c$, $P'$ does not vanish there. Hence, $f^2/P$ is positive for $r_h<r<r_c$ and vanishes at $r=r_h$ and $r_c$. From this and the regularity condition , we find that the left-hand side of the above integral identity vanishes. Therefore, if $U_X$ is non-negative, we can conclude that there exists no stationary regular scalar perturbation. From , we see that this condition is satisfied if $m\ge n-1$. Although this inequality may not hold for a generic Einstein space $\K^n$ with $K=1$, in the most important case, in which $\K^n=S^n$, we have the necessary inequality, because $m$ takes the discrete values $m=(l-1)(l+n)$ with $l\ge2$: To summarise, we have proved the following proposition.

*For a de Sitter-Schwarzschild black hole, there exists no stationary perturbation that is regular in the region $r_h\le r\le r_c$, except for the exceptional vector perturbations.*

From the above arguments, it follows that near the static and spherically symmetric limit, the rotating version of the de Sitter-Schwarzschild solution with the same number of parameters as the Myers-Perry solution is the only regular stationary vacuum solution for $\Lambda>0$. In four 