Is the Schwarzschild black hole unphysical? To obtain the Schwarzschild metric from Einstein equations of general relativity, we suppose that the energy density is a distribution :
$$ \rho (\vec{r}) = M \delta(\vec{r})$$
The Schwarzschild radius, corresponding to the horizon, is $R_s = 2 M$, in units $G=c=1$. 
A physical interpretation of this, is that you cannot put more energy than $M$ in a sphere of radius $R_s$. 
The way I understand this, is because the total energy of the black hole, sum of its mass energy (positive) and its auto-gravitational energy (negative), should remain non-negative.
However, for the Schwarzschild black hole, for each radius $r$ between $0$ and $R_s$, the total mass in the sphere of radius r, is :
$$ m(r) = \int_0^r  \rho(\vec{u}) d^3u =\int_0^r  \delta(\vec{u}) d^3u = M$$
So it seems to violate the principles cited above, and so the Schwarzschild black hole should be unphysical.
If it makes sense, a physical black hole (static, with spherical symmetry)  should have a  mass/energy density $\rho (r)$, such that, whatever the value of $r$ is, we have the inequality (in units $G=c=1$): 
$$m(r) = \int_0^r \rho (u) 4\pi u^2du \le \frac{r}{2}$$
 A: I can't improve on Luboš' comment, but I would add that a Schwarzschild black hole is indeed unphysical because it is time independant. A Schwarzschild black hole has existed for an infinite time and this is obviously unphysical. However we expect the Schwarzschild metric to be an excellent approximation to a real black hole.
A: The horizon is not "progressively created" in the collapsing spherical shell (Oppenheimer-Snyder model for black hole formation), it just appears to an exterior observer once the shell falls through. The source of the Schwarzschild metric is not the singularity, but can be thought of as distributed on the horizon.
The unphysical thing about Schwarzschild is that it is unstable to perturbations. if you set it rotating, or give it a bit of charge, the interior changes completely, opening a second Cauchy horizon, and a bridge to another exterior. This property is the main issue with Schwarzschild. It has too much symmetry, so it's singularity is spacelike, it is an instant of time where all observers end.
The timelike singularity in a rotating/charged black hole repels massive matter, and only serves as an end/begin point for null geodesics, the paths of light rays. There is no way to get rid of the null-singularity by the singularity theorem--- the null rays must change from focusing to defocusing at some point, and this must be at a singular place.
The question of whether the singularity inside a generic black hole is spacelike or timelike is open, with Penrose saying "spacelike" and everyone else aping him, and me saying "timelike" and I'm pretty confident. The idea was that the Cauchy horizon would turn into a singularity under a perturbation, but it doesn't in any clear way in simulations, and the AdS/CFT understanding gives more insight into the expected behavior of charged black holes, and suggest they should emit cold matter that falls into them.
A: My take is that the Schwarzschild black hole is indeed physical in the following sense. The S-metric has a timelike Killing vector, which indicates that the solution is symmetric with respect to time, hence, the "static" solution. 
In GR, I generally interpret physically real objects to be scalars, that is, quantities that cannot be transformed away by any coordinate transformations, and indeed the $r=0$ singularity is a feature of all S-metrics. 
The generic feature to determine if the black hole is physically plausible or not is the calculation of the Kretschmann scalar, $K = R^{abcd} R_{abcd} = constant/r^6$, and because this is an invariant quantity, one must conclude that indeed the r=0 singularity is physically significant. 
