stellar evolution - death of a red dwarf star / minimum mass needed for a white dwarf?

Stars that have a mass lower than about $0.5 M_{odot}$ will not ignite helium in their cores, in an analogous fashion to the way that stars with $M<8M_{odot}$ have insufficiently massive cores that never reach high enough temperatures to ignite carbon.

The cause in both cases is the onset of electron degeneracy pressure, which is independent of temperature and allows the core to cool at constant pressure and radius. [A normal gas would contract and become hotter as it loses energy!]

The end result for a $0.5M_{odot}$ star will be a helium white dwarf with a mass (depending on uncertain details of the mass-loss process) of around $0.2M_{odot}$. Such things do exist in nature now, but only because they have undergone some kind of mass transfer event in a binary system that has accelerated their evolution. The collapse to a degenerate state would be inevitable even for the lowest mass stars (which would of course then be very low-mass white dwarfs). As an inert core contracts it loses heat and cools - a higher density and lower temperate eventually lead to degenerate conditions that allow the core to cool without losing pressure.

The lowest mass stars ($<0.3 M_{odot}$) do get there via a slightly different route - they are fully convective, so the "core" doesn't exist really, it is always mixed with the envelope. They do not develop into red giants and thus I guess will suffer much less mass loss.

The remnant would be a white dwarf in either case and is fundamentally different from a brown dwarf both in terms of size and structure, because it would be made of helium rather than (mostly) hydrogen. This should have an effect in two ways. For the same mass, the brown dwarf should end up bigger because the number of mass units per electron is smaller (1 vs 2) and also because the effects of a finite temperature are larger in material with fewer mass units per particle - i.e. its outer, non-degenerate layer would be more "puffed up". NB: The brown dwarfs we see today are Jupiter-sized, but are still cooling. They will get a bit smaller and more degenerate.

A simple size calculation could use the approximation of an ideal, cold, degenerate gas. A bit of simple physics using the virial theorem gives you
$$ left(frac{R}{R_{odot}}right) simeq 0.013left(frac{mu_e}{2}right)^{-5/3} left(frac{M}{M_{odot}}right)^{-1/3},$$
where $mu_e$ is the number of atomic mass units per electron.
Putting in appropriate numbers I get $0.32 R_{Jup}$ for a $0.07M_{odot}$ Helium white dwarf versus $1.01 R_{Jup}$ for a $0.07M_{odot}$ completely degenerate Hydrogen brown dwarf (in practice it would be a bit smaller because it isn't all hydrogen).

However, it would be interesting to see some realistic calculations of what happens to a $0.07M_{odot}$ brown dwarf versus a $0.08M_{odot}$ star in a trillion years or so. I will update the answer if I come across such a study.

EDIT: I knew I'd seen something on this. Check out Laughlin et al. (1997), which studies the long-term evolution of very low-mass stars. Low-mass stars do not pass through a red giant phase, remain fully convective and can thus convert almost all their hydrogen into helium over the course of $10^{13}$ years and end up cooling as degenerate He white dwarfs.