Everything below is to be viewed in the Euclidean setting with $d$ dimensions and all measures are understood to be Borel measures.
The usual problem of Quantum Field Theory is to make sense of Feynman Path Integrals like
\begin{equation} \int \exp \left[- \frac{m^2}{2} \int \phi^2 + \frac{Z}{2} \int \phi \Delta \phi - \lambda \int \phi^4 \right] \mathrm{d} \phi \end{equation}
where the outer integral is supposed to run over some space of test functions. It is well-known that this does not really work. What one can do, however, is for $\lambda = 0$ to produce a Gaussian probability measure on e.g the space $\mathcal{S}'$ of tempered distributions. More generally this works for any continuous positive definite bilinear form on $\mathcal{S}$. Hence, for such theories we know what we are doing!
With $\lambda \neq 0$ this is no longer the case. In order to ensure the finiteness of expressions of the form $\int \phi^4$ we would like $\phi$ to live in $\mathcal{S}$. To this end, we pick two non-negative mollifiers $\chi, \xi \in \mathcal{D}$ with $\xi \left( 0 \right) = 1$. Furthermore, for any $\epsilon > 0$ and $\Lambda > 0$, define
\begin{aligned} \chi_\epsilon \left( x \right) &= \epsilon^{-d} \chi \left( \frac{x}{\epsilon} \right) \\ \xi_\Lambda \left( x \right) &= \xi \left( \frac{x}{\Lambda} \right) \\ \end{aligned}
and consider the continuous (? - it should certainly be Borel measurable) mapping
\begin{aligned} \mathcal{M} \left( \xi_\Lambda \right) \mathcal{C} \left( \chi_\epsilon \right) : \mathcal{S}' &\to \mathcal{S} \\ T &\mapsto \xi_\Lambda \cdot \left( \chi_\epsilon \ast T \right) \, . \end{aligned}
We may now take a Gaussian measure $\mu$ on $\mathcal{S}'$ corresponding to some bilinear theory and regularize it as $\left[ \mathcal{M} \left( \xi_\Lambda \right) \mathcal{C} \left( \chi_\epsilon \right) \right]_\ast \mu$ to a measure on $\mathcal{S}$. One can then define measures
\begin{equation} \nu_{\epsilon, \Lambda} = \exp \left[ -\lambda \int \phi^4 \right] \cdot \left[ \mathcal{M} \left( \xi_\Lambda \right) \mathcal{C} \left( \chi_\epsilon \right) \right]_\ast \mu \end{equation}
normalize them, push them to $\mathcal{S}'$ again and study their behaviour as $ \left( \epsilon, \Lambda \right) \to \left( 0, \infty \right)$. But this also does not work as it just produces the expected divergences coming from attempting to multiply distributions. Instead, one has to give up on keeping the model parameters $m, Z, \lambda$ constant and let them depend on $\epsilon$ and $\Lambda$ (i.e on cutoffs) instead. One might then hope that for some nice $\left( \epsilon, \Lambda \right)$-dependence a limit (or at least a cluster point) $\nu_{0, \infty}$ indeed exists.
From a Wilsonian point of view, we would however like to take $\nu_{0, \infty}$ and integrate out degrees of freedom that are cut off by $\epsilon > 0$ and $\Lambda > 0$. We may do this by fixing a family $\left( \alpha_{\epsilon, \Lambda} \right)_{\epsilon, \Lambda > 0}$ of Gaussian probability measures on $\mathcal{S}'$ satisfying the following properties for all $\epsilon, \delta, \Lambda, \kappa > 0$:
- $\alpha_{\epsilon, \Lambda} \gg \nu_{0, \infty}$
- $\alpha_{\epsilon, \Lambda} \ast \alpha_{\delta, \kappa} = \alpha_{\epsilon+\delta, \frac{\lambda \kappa}{\Lambda + \kappa}}$ (or someway similar)
Then we obtain a renormalization group equation
\begin{equation} \frac{\mathrm{d} \nu_{0, \infty}}{\mathrm{d} \alpha_{\epsilon, \Lambda}} \left( T \right) = \int_{\mathcal{S}'} \frac{\mathrm{d} \nu_{0, \infty}}{\mathrm{d} \alpha_{\epsilon + \delta, \frac{\Lambda \kappa}{\Lambda + \kappa}}} \left( T + R \right) \mathrm{d} \alpha_{\delta, \kappa} \left( R \right) \end{equation}
for all $\epsilon, \delta, \Lambda, \kappa > 0$.
Are these lines of thought correct? And presuming the existence of some nice QFT corresponding to $\nu_{0, \infty}$, can we always find such a family $\alpha_{\epsilon, \Lambda}$?
EDIT: I removed the other questions as they turned out not to make too much sense.
PS: The above is pretty much as far as my level of mathematics goes. Please try to not go too much further.
