Why is QFT not even unitary prior to renormalisation? The Hamiltonian is Hermitian. That should've been enough to make it unitary. But infinite amplitudes mean it's not even unitary. One could say that this is because we're dealing with a crazy Hilbert space having a continuum of variables. But the theory isn't unitary even after discretization! The amplitudes after regularization/discretization become finite but are still larger than 1. So again, why isn't it unitary?
 A: The core problem that arises in QFT is that, if you want to be rigorous, expressions like $\phi(x)^4$ do not actually exist when the $\phi(x)$ is a quantum field, i.e. an operator-valued distribution, because there is no general unique theory of multiplying distributions. The UV divergences that we usually have to renormalize away can in some formulations (causal perturbation theory, also sometimes called Epstein-Glaser renormalization) be directly related to a choice of how to define the point-wise product of such distributions.
So an interacting Hamiltonian of e.g. $\phi^4$-theory isn't "Hermitian", it's nonsense, $+\lambda\phi^4$ isn't a description for a self-adjoint operator, it's a chiffre for $\phi^4$-theory and the usual ways (including renormalization!) to extract QFT predictions from such a formal Hamiltonian that isn't actually an operator on anything in a mathematically rigorous sense.
The way physicists often work with QFT is contradictory - on the one hand we pretend this is mostly like quantum mechanics, and on the other hand we must avoid fallacious statements like "the Hamiltonian is self-adjoint so time evolution cannot possibly diverge" - and that is probably a large part of why it is so hard to put QFT on mathematically rigorous footing.
