I am reading J. Glimm, A. Jaffe,"Quantum Physics, A functional integral point of view", Springer, (1987). In chapter 9, Section 4: It is written that
"The renormalizable models are also characterized by the fact that the Sobolev inequality which dominates the interaction part of the Euclidean action by the kinetic (free) part of the action becomes borderline."
I would appreciate if someone could explain this to me using a toy model like $\phi^4$ in dimension 4.
I am not aware of a proof of renormalizability using Sobolev inequalities, nor of the latter using the renormalization group. So I don't think one phenomenon is the cause of the other (renormalizability versus the validity of a Sobolev embedding). Perhaps the numerical coincidence is because of a common cause related to exponents under rescaling.
Consider the Sobolev Inequality controlling the $L^q$ norm in terms of the $W^{k,p}$ norm on a bounded domain (suitable for a UV problem so there is no infrared to worry about). In dimension $n>2$, one needs $\frac{1}{q}\ge \frac{1}{p}-\frac{k}{n}$. The kinetic (plus mass) part of a traditional local scalar bosonic QFT corresponds to $k=1$, $p=2$. So the condition is $$ q\le \frac{2n}{n-2} $$ or rather $$ n-q[\phi]\ge 0 $$ where $[\phi]:=\frac{n-2}{2}$ is the so-called scaling dimension of the field.
For the $\phi^q$ model in $n$ dimensions, the coupling, after a Wilsonian RG step given by dilation by a factor of $L>1$, gets multiplied by $$ L^{n-q[\phi]} $$ at linear order. When $n-q[\phi]>0$, the coupling grows, i.e., is relevant and the QFT is super-renormalizable. When $n-q[\phi]=0$, the coupling is marginal and the QFT is just renormalizable which is the borderline case. When $n-q[\phi]<0$, the coupling decays, i.e., is irrelevant and the QFT is nonrenormalizable.
