Does Wightman's unitary $U(\Lambda)$ really exist for Lorentz boost? This question is related to another question here. But I am asking a more fundamental question about the existence of Wightman's unitary $U(\Lambda)$ for Lorentz transformation.
Let $\psi^\alpha$ be a quantum spinor field transforming in the $(\frac{1}{2},0)$ irrep of $SL(2,\mathbb{C})$. That is, under a boost $\Lambda$, with unitary representative $U(\Lambda)$, the field transforms as:
\begin{equation} U(\Lambda)^\dagger \psi^\alpha(x) U(\Lambda) = S(\Lambda)^\alpha_{\space\space\beta} \psi^\beta(\Lambda^{-1}x) \tag{1}\end{equation}
with $S(\Lambda)\in SL(2,\mathbb{C})$.
And the unitary $U(\Lambda)$ also satisfies
$$
U(\Lambda)|\Omega \rangle = |\Omega \rangle \tag{2}
$$
for all Lorentz boost $\Lambda$,  where $|\Omega \rangle $ is the vacuum.
My question is: given that $S(\Lambda)$ for Lorentz boost is NOT unitary, is there really a unitary $U(\Lambda)$ that satisfies both eq. (1) and eq. (2) under Lorentz boost?
The existence of unitary $U(\Lambda)$ is crucial for Wightman axioms. So please DON'T just quote the the Wightman axioms as your answer. And for that matter, please DON'T just quote relativity or Lorentz invariance as your answer. Lorentz invariance implies the existence of $S(\Lambda)$ (known as the non-unitary $K(\Lambda)$ for Lorentz boost), but not necessarily $U(\Lambda)$. Instead, please give a proof of the existence of unitary $U(\Lambda)$.
The proof does not have to be super rigorous, one concrete example for spinor field  under Lorentz boost would suffice. And yes, it has to be specifically for spinor field, with details of $U(\Lambda)$ worked out, rather than a general procedural guidance of how to get $U(\Lambda)$. Note that I am NOT asking for a proof for complex/real scalar field.

Added note:
The references given by various users are pertaining to the scalar field. The whole situation made me doubtful of the proof of unitarity for the spinor field. That is why I am asking the question.
 A: A QFT is said to be relativistic invariant if it realizes a unitary representation of the Lorentz group. Not all QFTs are relativistic, so there is no general proof. Instead, this is a definition: if you have a unitary representation, then your theory is relativistic.
So, how do we construct relativistic theories? The so-called canonical approach is arguably the most useful tool in doing this. The claim is that the QFT is relativistic if you begin with a hermitian and Lorentz invariant Lagrangian. What follows is a sketch of this claim, although the "full proof" requires a whole textbook (cf.ref1).
In the canonical approach to QFT the symmetries under Lorentz transformations give rise to the Noether current (cf. ref1 §7.4)
$$
M_{\mu\nu\rho}\sim\frac{\partial\mathcal L}{\partial \phi_{,\mu}}\delta_{\nu\rho}\phi
$$
where $\phi$ denotes the fields of your theory and $\delta_{\nu\rho}$ denotes a Lorentz variation in the $\nu\rho$ direction.
The generator of Lorentz transformations is, then
$$
J_{\mu\nu}=\int M_{\mu\nu0}\mathrm d\boldsymbol x
$$
which is conserved thanks Noether's theorem.
Finally, the operator that implements Lorentz transformations is, by definition,
$$
U(\Lambda)=\exp\big(\frac i2J_{\mu\nu}\omega^{\mu\nu}\big)
$$
where $\Lambda=e^{\frac i2 \mathcal J_{\mu\nu}\omega^{\mu\nu}}$, with $(\mathcal J_{\mu\nu})_{\rho\sigma}\sim\eta_{\mu[\rho}\eta_{\sigma]\nu}$ the generators of the Lorentz group in the fundamental representation.
Unitarity of $U$ follows from hermiticity of $J$, which in turns follows from hermiticity of $\mathcal L$.
Whether $U(\Lambda)$ leaves $\Omega$ invariant or not is a question of whether the Lorentz symmetry is spontaneously broken or not. As a matter of principle this can happen, so again, there is no proof. We are again in a situation where things become definitions: if $\Omega$ is invariant, then we say that Lorentz is unbroken.
Usually, when constructing $M_{\mu\nu\rho}$ there are ordering ambiguities which can be exploited in order to make $\Omega$ invariant. For example, the Hamiltonian is defined up to a constant, and you can fix it by declaring that $\Omega$ has zero energy. If you can do this for all Lorentz generators then the symmetry is unbroken. If you cannot, then it is broken.
References.

*

*Weinberg S. - Quantum theory of fields, Vol.1. Foundations.

