 Solution

Take $M>0$, set $b=Q^2/M$, and denote the squared areal radius by $C(r)=r(r-b)$. The positive-area exterior has $r>b$. The stationary norm is $-f(r)$, where $f(r)=1-2M/r$, so its candidate Killing horizon is at $r=2M$. There it has areaA regular horizon must have $b<2M$. The curvature singularity then lies strictly inside the horizon, and ingoing coordinates extend smoothly across it because $C(2M)>0$. In the future interior $f<0$, the gradient of $r$ is timelike and future causal curves move to decreasing $r$, rather than back to infinity. Thus the regular black hole condition isThe equality case has zero horizon area and a curvature-singular null limiting surface. It is often included in the family as a singular extremal solution, but it is not a regular black hole. If that limiting nomenclature is used, the family condition is $Q^2\le 2M^2$ with this essential qualification. This geometry has the same spherical form as the magnetically charged dilaton black hole; no unprinted field equation is needed for the horizon comparison.