a function $\tilde F(x)$ that is regular at $x=0$ and $\tilde F(0)=1$.

If we require that $F^t$ be bounded at $r=\infty$, which is weaker than the asymptotic condition , $B$ should vanish. Under this condition, from , we find that $F^t$ diverges at the horizon. Therefore, there exists no regular and bounded stationary vector perturbation other than the exceptional modes.

Scalar perturbations {#scalar-perturbations-1}
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As explained in §\[subsec:BasicEqs:Scalar\], we can assume that $m=k^2-n>0$ for $K=1$. Therefore, we can express $k^2$ in terms of $l>1$ as Then, can be written in terms of $x=(r_0/r)^{n-1}$ as where This equation can be solved explicitly again, and its general solution is given by with where $\nu=l/(n-1)$, and $A$ and $B$ are arbitrary constants. Inserting this solution for $Y$ into , we obtain with From these expressions, we find that the asymptotic condition is satisfied only when $B=0$ if $l>1$. However, when $B=0$, $Y$ diverges at $x=1$, and the regularity condition is not satisfied. Actually, the same conclusion is obtained from the weaker condition that $F$ and $F^a_b$ are bounded at infinity. Hence, there exists no regular and bounded stationary scalar perturbation.

The following proposition summarises the main result obtained in this section.

*For the 