Why is it that the interacting fields cannot be decomposed into Fourier modes like free fields? In quantum field theory, a free scalar field can be decomposed into Fourier modes. But why is that, an interacting field cannot have Fourier decomposition? Can we not decompose any arbitrary reasonably well-behaved function $f(x)$ into Fourier integral? 
Free fields satisfy linear equations while interacting fields obey nonlinear equation of motion. But what says that the solution of a nonlinear equation cannot be broken into a Fourier integral?
 A: You can of course write the 4-dimensional Fourier transform
$$ f(x) = \int \frac{d^4 p}{(2\pi)^4} e^{i p_{\mu} x^{\mu}} \tilde{f}(p). $$
The form of the Lagrangian doesn't affect this expression. However, the interacting action can no longer be written as a Fourier integral.
Take a free theory for example:
$$ S[\phi] = \int d^4 x \left( \frac{1}{2} \partial_{\mu}\phi \partial^{\mu} \phi - \frac{1}{2} m \phi^2 \right) = \int d^4 p \left( - \frac{1}{2} p^2 \tilde{\phi}(p)^2 - \frac{1}{2} m \tilde{\phi}(p)^2 \right). $$
The free action can be written as a momentum integral because of the unitarity of the Fourier transform (Parseval's theorem).
Now consider an interacting action ($\phi^4$ theory for example):
$$ S[\phi] = \int d^4 x \left( \frac{1}{2} \partial_{\mu}\phi \partial^{\mu} \phi - \frac{1}{2} m \phi^2 - \frac{\lambda}{4!} \phi^4 \right) =$$
$$ \int d^4 p \left( - \frac{1}{2} p^2 \tilde{\phi}(p)^2 - \frac{1}{2} m \tilde{\phi}(p)^2 \right) + V_4, $$
where $V_4$ can not be written as the momentum integral. Thus, the local in spacetime interaction is not local in the momentum space.
However, it is not true that it is not used in interacting QFT. The formalism of perturbation theory is based on the momentum-space Feynman rules. Its just that interactions are nonlocal in Fourier space, and can only be accounted by perturbing the free theory.
