The Majorana condition and C violation Is the Majorana condition
$$ 
\psi = \psi^c = C \overline{\psi}^T,
$$
general?
The point is often made that Majorana particles should be defined by CPT symmetry and not C as generally theories do not have C symmetry. Does this mean that the Majorana condition is wrong and we ought to use
$$
\psi = \psi^{CPT}.
$$
In which case, why does so much literature, for instance neutrino theory papers use the former definition and not the latter?
 A: I think this question mixes up notions involving particles and fields in quantum field theory. 
Recall that at the classical level, charge conjugation may be defined for fields simply by complex conjugation, up to a possible additional linear transformation. If a theory has such a symmetry, it is usually straightforward to produce a quantum operator $\hat{C}$ that acts on particle states, and we define the antiparticle of a one-particle state to be its image under $\hat{C}$.
If the Lagrangian does not have the classical charge conjugation symmetry for the fields, it may be difficult or even impossible to define an appropriate $\hat{C}$ operator at the quantum level. (Sometimes one can define a $\hat{C}$ that isn't conserved, $[\hat{C}, \hat{H}] \neq 0$, but in many cases, such as in the Standard Model, there's just no way to define a $\hat{C}$-like operator.) 
The paper you linked points out that this is problematic if $\hat{C}$ is used to define antiparticles. However, we always have the operator $\widehat{CPT}$, so we may use it to define the word "antiparticle" at the quantum level. This is a completely separate issue from the definition of a Majorana fermion at the level of classical fields. Unlike $\hat{C}$, complex conjugation is always defined, so there's no problem.
For much more about discrete symmetries, see here. For specifics on charge conjugation, see here.
