In this article, the authors study the time behaviour of the velocity-velocity correlation function of a particle in a gas. If the gas is immersed in $d$ spatial dimensions, they find that
$$ C(t)=\frac{\langle v(0)_i v(t)_i \rangle}{\langle v(0)_i^2 \rangle} \sim t^{-d/2} $$
where the average refers to an equilibrium ensemble. In two dimensions ($d=2$), the integral over a long time of $1/t$ doesn't exist. On this basis, they conclude that the self-diffusion coefficient $D$ does not exist in two dimensions, so that "conventional hydrodynamics does not exist in two dimensions". This is because thanks to the Green-Kubo formula
$$ D \propto \int_0^{\infty} C(t) \, dt $$
the diffusion coefficient $D$ is logarithmically divergent for $d=2$.
Is this a well-known result? Does it imply that we can not apply Navier stokes for a gas in two dimensions? (i.e. interacting particles constrained on a surface can not be described in the long-wavelength limit by Navier-Stokes hydrodynamics).
As the authors of this other paper comment ($d=2$): Physically the long-time tails $~1/t$ of the correlation functions are caused by the slowly decaying hydrodynamic modes. Kinetically this is due to the possibility of recollisions, i. e. , collisions between two particles that have collided before. They lead to a much slower decay of the initial state of a particle than if they are excluded since they can still "remind" the particle of its initial state after many collisions have taken place.
Those two papers are from the '70s, which is the situation today? Can we really apply Navier Stokes (or the diffusion equation) when $d=2$ and $d=1$?
Note: clearly we can assume that Navier Stokes works in $d=2$, and we can numerically solve it (see this answer), my question is if it can be justified from a more fundamental kinetic approach: for sure in $d>2$ we can derive it from kinetic theory, but how about $d<3$?
Note: interestingly, not even the Boltzmann equation seems to be good for some 2D fluids (and Navier stokes may be obtained from Boltzmann), see this answer.