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JustWannaKnow
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The ultraviolet limit as a limiting case of the renormalization group flow?

In his paper Constructive Renormalization Theory, V. Rivasseau describes the idea of Wilson's approach of solving path integrals step by step. In section 1.4, page 5, however, there is a statement which I do not follow. He discusses a $\phi^{4}$ theory and decomposes the covariance into a sum: \begin{eqnarray} C_{0}(p) = \int_{1}^{\infty}e^{-\alpha(p^{2}+m^{2})}d\alpha \quad \mbox{and} \quad C^{j}(p) = \int_{M^{-2j}}^{M^{-2(j-1)}}e^{-\alpha(p^{2}+m^{2})}d\alpha \tag{1}\label{1} \end{eqnarray} in such a way that: \begin{eqnarray} C_{\rho}(p) = \sum_{j=0}^{\rho}C^{j}(p) \tag{2}\label{2} \end{eqnarray} He proceeds to explain how the partition function can be obtained by performing interativelly $\rho+1$ (convolution) integral. At some point, he states the following:

We see that constructing the ultraviolet limit is the same as finding a bare action $S_{\rho}(\phi)$ such that the effective action, or renormalized action $S_{0}(\phi)$ converges as $\rho \to \infty$.

I might be missing something, but I don't fully understand this statement. First, it seems that, in the limit $\rho \to \infty$ one recovers the the regularized theory, because (\ref{1}) is a telescoping series with UV cutoff and as $j \to +\infty$ one ends up with regularized covariance \begin{eqnarray} C_{0}(p) = \frac{1}{p^{2}+m^{2}}e^{-(p^{2}+m^{2})}. \tag{3}\label{3} \end{eqnarray} In other words, I don't see how it is possible to take the UV limit using this decomposition into steps as a limiting case. Second, it is a little bit odd for me to say that one finds $S_{\rho}(\phi)$ so that $S_{0}(\rho)$ exists when $\rho \to \infty$. The flow is going backwards, from $\rho$ to zero, how can you then $\rho \to \infty$ afterwards? Is it just because $S_{0}$ should depend on $\rho$ somehow?

JustWannaKnow
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