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 $G=c=1$):
$$m(r) = \int_0^r \rho (u) 4\pi u^2du \le \frac{r}{2}$$