Oh, right. This is an important point indeed. But, considering that one such decomposition is chosen, the above scenario is accurate, right? My point is that, usually, it is stated that, because of the above, the problem becomes the study of the RG map I defined in my post. But I was confused on how to recover the initial information once this map is studied.

Oh, I think I understand your point now. Because of (\ref{4}), $Z(\varphi) = Z_{n}(\varphi)$ where $Z_{n}(\varphi) = \int d\mu_{C_{n}}(\zeta_{n})e^{-RG^{(n-1)}(\varphi+\zeta_{n})} = (\mu_{C_{n}}*e^{-RG^{(n-1)})}(\varphi) = e^{-RG^{(n)}(\varphi)}$, so for sufficiently large $n$ the fixed point is attained and $Z(\varphi) = e^{-V^{*}(\varphi)}$, right?

Thanks for the answer Carlo! Just a clarification: what is $Z^{*}(\phi)$? You mean $Z(\varphi) = \int d\mu_{C}(\psi)Z^{*}(\psi+\varphi)$?

@IgorKhavkine Maybe I could use the following: if $\mathcal{H} = L^{2}(\mathbb{R}^{3};\mathbb{C}^{2}) \cong L^{2}(\mathbb{R}^{3}\times \{+1,-1\})$ it seems that $\mathcal{H}_{n} \cong L^{2}(\mathbb{R}^{3}\times \{+1,1\})\otimes \cdots \otimes L^{2}(\mathbb{R}^{3}\times \{+1,-1\}) \cong L^{2}(\mathbb{R}^{3n}\times \{+1,-1\}^{n})$? (Don't know for sure if the last isomorphism holds tho).The latter seems more useful to define $\varphi(x,\sigma)$ and $\varphi^{\dagger}(x,\sigma)$ since $x \in \mathbb{R}^{3}$ and $\sigma$ is a spin variable, probably taking values $\{+1,-1\}$.

I'm gonna add more information to the post.

I'm approaching this as a mathematician. But I know it's hard to dodge the physics behind it. But I'm really interested in rigorous approaches to both areas.

In addition, I know people are using $C^{*}$-algebra to study statistical mechanics as well. But I see this as a reflection of the fact that statistical mechanics has some strong connections to QFT. Don't know if this reasoning is accurate, tho.