This is an interesting question, and although I don't know a rigorous answer, we can discuss some typical cases.

Usually, the inverse exists, but the cases where this inverse does not exist are not necessarily pathological (sound models can have the problem that the inverse does not exist).

For standard field theories (say, $\phi^4$, O(N) models, classical spins models, ...), generically the inverse exists, and this can be shown order by order in a loop expansion (I don't know if this has been proven at all order, but in standard textbooks, this is shown to order 1 or 2). However, the inverse will not exist necessarily for all $\phi_{cl}$, especially in broken symmetry phases. Indeed, an ordered phase is characterized by $$\bar\phi_{cl}=\lim_{J\to 0 } \phi_{cl}[J]=\lim_{J\to 0 }W'[J]\neq 0 ,$$
where $\bar \phi_{cl}$ is the equilibrium value of the order parameter. Therefore, you cannot inverse the relationship $\phi_{cl}[J]$ for $\phi_{cl}\in [0,\bar \phi_{cl}]$ ($\phi_{cl}[J]$ generically increases when $J$ increases).

Furthermore, there are cases where the inverse is simply not defined, because $\phi_{cl}[J]={\rm const}$ for all $J$. This is usually the case when the field has no independent dynamics without a source. For instance, if you take a single quantum spin at zero temperature, the only dynamics is given by external magnetic field (here in the $z$ direction) $$\hat H= -h.\sigma_z.$$
With $h>0$, the ground state is always $|+\rangle$, and the "classical field" $\phi_{cl}(h)=\langle \sigma_z\rangle=1/2$ for all $h$, and the Gibbs free energy (the Legendre transform of the free energy with respect to $h$, which is essentially the effective action) does not exist.

This post imported from StackExchange Physics at 2014-04-13 14:08 (UCT), posted by SE-user Adam