Renormalization in strongly coupled theories without Lagrangian description In a weakly coupled theories, it is quite transparent how the flow works since we have a path integral and then we integrate out fields to see the behaviour at different length scales. But if I don't have a Lagrangian or the theory is strongly coupled, this procedure won't make a lot of sense. In that case how do we see the flow. Please correct me if anything that I have stated is wrong. Apologies in advance. Thanks !
 A: RG flows which come from perturbing Lagrangians often lead to strong coupling when we approach the IR. But it is indeed interesting to consider the cases when we already do not have a Lagrangian for the starting UV theory. Sometimes that's because we know literally nothing about the theory, in which case making sense of renormalization is the least of our troubles. So I will focus on the case where the UV fixed point is well understood but nevertheless non-Lagrangian.
Unperturbed correlation functions $\left < \mathcal{O}_1 \dots \mathcal{O}_n \right >_0$ therefore have an abstract definition and we can consider what happens to them after adding $\exp \left [ g_0 \int d^d x \, \mathcal{O}(x) \right ]$ to the would-be "path integral". An especially useful observable to consider is the two-point function of $\mathcal{O}$ itself. At leading order,
\begin{align}
\left < \mathcal{O}(x_1) \mathcal{O}(x_2) \right > &= \sum_{n = 0}^\infty \frac{g_0^n}{n!} \int dy_1 \dots \int dy_n \left < \mathcal{O}(x_1) \mathcal{O}(x_2) \mathcal{O}(y_1) \dots \mathcal{O}(y_n) \right >_0 \\
&\approx \left < \mathcal{O}(x_1) \mathcal{O}(x_2) \right >_0 + g_0 \int dy \, \left < \mathcal{O}(x_1) \mathcal{O}(x_2) \mathcal{O}(y) \right >_0.
\end{align}
We can now plug in the known form of a conformal three-point function and apply a point splitting regulator to the integral to get
\begin{equation}
g_0 \int_{|y - x_i| > \Lambda^{-1}} dy \, \frac{\lambda_{\mathcal{O}\mathcal{O}\mathcal{O}}}{(|x_{12}| |y - x_1| |y - x_2|)^\Delta}.
\end{equation}
This shows that if $\Delta > d$ (meaning $\mathcal{O}$ is irrelevant), we can take the short distance cutoff to zero and still get a convergent integral. In more Wilsonian language, these operators will contribute negative powers of $\Lambda_1$ and $\Lambda_2$ when we evolve an effective field theory between two scales with $\Lambda_2 \ll \Lambda_1$. Their effects at $\Lambda_1$ will therefore be unimportant in determining the long wavelength physics.
For $\Delta < d$, the situation is reversed and the integral will diverge as we take $\Lambda \to \infty$. It therefore needs to be renormalized by writing $g_0 = Z_g g$ and letting $g$ run according to some scheme. This is not necessarily easy, because of how complicated position space correlators can be at high orders, but https://arxiv.org/abs/1603.04444 and https://arxiv.org/abs/1703.05325 are good reviews. These deal with perturbations to the Ising model which is strongly coupled. In 2d it is exactly solved while in 3d it is known with enough numerical accuracy for the above approach to work.
