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in closed form. We prove below that in the case that the black hole mass exceeds the critical mass M$_{0}$ the roots of the equation g$_{00}$ ( r ) = 0 are simple. This result will play an important role in the following considerations. First we need to clarify the nature of the critical mass M$_{0}$ , which turns out to be the minimal mass to have horizons. Therefore, if we employ the horizon equation to define a function M = M ( r$_{H}$ ), where r$_{H}$ is a solution of the equation g$_{00}$ ( r ) = 0, we obtain 

<img src='content_image/370992.jpg'>

Thus, having considered the derivative dM ( r$_{H}$ ) /dr$_{H}$ , we can look for r$_{0}$ such that the latter vanishes, i.e. dM/dr$_{H}$ | r 0 = 0. We do this in order to define M$_{0}$ ≡ M ( r$_{0}$ ). One can easily show that the above derivative vanishes if and only if 

From the properties of the gamma function one can easily derive the following result 

and as a consequence 

where r$_{m}$ ∈ [ 0 , r$_{H}$ ]. Eq. (5) can now be written as 

<img src='content_image/370990.jpg'>

<img src='content_image/370989.jpg'>

<img src='content_image/370988.jpg'>

<img src='content_image/370991.jpg'>

which admits a unique solution if and only if r$_{H}$ = r$_{m}$ = r$_{0}$ . We can conclude that there exists a unique horizon radius r$_{0}$ , corresponding to the critical mass M$_{0}$ . With this in mind we can establish the following lemma 

Lemma 1 Let M > M$_{0}$ and 0 < r − < r$_{+}$ such that g$_{00}$ ( r ± ) = 0 . Then, r − and r$_{+}$ are simple zeroes of (3). 

Proof. We give the proof for a generic solution r$_{H}$ of the horizon equation g$_{00}$ ( r$_{H}$ ) = 0. Therefore r$_{H}$ corresponds either to r$_{+}$ or r − . Notice that if the limit 

<img src='content_image/370987.jpg'>

