$\phi$ a pseudoscalar field $\Rightarrow$ $\phi(x^0, \vec 0) = 0$? On page 9 of Relativistic Quantum Mechanics and Introduction to Quantum Field Theory, A. Z. Capri defines scalar and pseudoscalar fields as follows. Under a parity transformation,
$$
{x^0}' = x^0, \quad \vec x' = - \vec x,
$$
$\phi$ is a scalar field if
$$
\phi'(x^0, \vec x) = \phi(x^0, -\vec x) = \phi(x^0, \vec x),
$$
and $\phi$ is a pseudoscalar field if
$$
\phi'(x^0, \vec x) = \phi(x^0, -\vec x) = -\phi(x^0, \vec x).
$$
Okay, so let $\phi_{ps}$ be a pseudoscalar field. Then,
$$
\phi_{ps}(x^0, -\vec 0) = -\phi_{ps}(x^0, \vec 0),
$$
so that $\phi_{ps}$ must be zero at the 3-space origin. This seems like a very strong constraint on $\phi_{ps}$. Moreover, there should be nothing special about our chosen origin, so could we not translate to any other 3-space point and do a parity transformation there as well, hence forcing $\phi_{ps}$ to be zero everywere? By a similar argument, a scalar field $\phi_s$ must be constant across space (but not across time). This all seems way off, so I must be missing something, but what?
 A: There is a distinction between the dynamics of a system being invariant under a symmetry, and a given solution of the dynamical equations being invariant under that symmetry. For instance, Maxwell's equations are Lorentz invariant, but it's not a problem that the electric field around a point charge has a preferred frame where it is static. Or to give another example, a massless scalar field $\mathcal{L}=-\frac{1}{2}(\partial_\mu \phi)^2$ has a shift symmetry $\phi\rightarrow \phi+c$, but of course for completely trivial reasons it is not possible to find a solution to the theory that is invariant under adding a constant.
If a pseudo-scalar field appears in the action, it means that if $\phi(x,t)$ is a solution of the equations of motion, then so is $-\phi(-x,t)$ (provided you consistently apply the parity transformation to all other fields and sources). But it does not mean that any given solution satisfies $\phi(x,t)=-\phi(-x,t)$, which as you point out would be a very strong condition.
At the quantum level, a symmetry implies the existence of relationships among correlation functions.
