<s>[INST] $\phi^4$-theory: interpreting the RG flow I'm referring to these lectures on the Renormalization Group and more precisely the figure of the RG flow for the $\phi^4$-theory on page 18.
The Lagrangian in the notation used in the text is 
$$\mathcal L = -\frac{1}{2}(\partial_\mu \phi)^2 - V(\phi)\quad \text{with } \quad V(\phi)=\sum_n \mu^{d-n(d-2)}\frac{g_{2n}}{(2n)!}\phi^{2n}\,, $$
where $\mu$ is the hard cut off.
For convenience here is the pertinent figure for the RG flow for dimension $d=4-\epsilon$ with $\epsilon>0$

As explained in the text: 

"F describes a massless interacting
  theory that interpolates between a free theory in the UV and the Ising Model in the IR"

My question is very simple: 
How the line F can represent massless theories as the coupling constant $g_2=m$ along that line is non-zero....?
 [/INST] I have done more direct approach to this task in first order in $\epsilon$. Let's start with direct calculation of propagator:
$$
 \langle \phi(p) \phi(-p) \rangle = \frac{1}{p^2+m^2} - \frac{4 \cdot 3 \cdot g}{(p^2 + m^2)^2} \int^\Lambda \frac{d^{4-\epsilon}k}{(2\pi)^{4-\epsilon}} \frac{1}{k^2+m^2} + \dots
 $$
$$
 \int^\Lambda \frac{d^{d}k}{(2\pi)^d} \frac{1}{k^2+m^2} = \frac{Vol(S^{d-1})}{(2\pi)^d} \int_0^\Lambda \frac{k^{d-1} dk}{k^2 + m^2}
 =
 \frac{ \pi^{d/2}}{(2\pi)^d \Gamma(d/2)}
 \int_0^\Lambda \frac{(k^2)^{d/2-1} d(k^2)}{k^2 + m^2}
 $$
Using $g\sim \epsilon$, we fix $d=4$:
$$
 \int^\Lambda \frac{d^{d}k}{(2\pi)^d} \frac{1}{k^2+m^2} =  \frac{1}{(4\pi)^2} \left(\Lambda^2 - m^2\ln \frac{\Lambda^2 + m^2}{m^2}\right)
 $$
Expanding obtained correlation function:
$$
 \langle \phi(p) \phi(-p) \rangle \approx \frac{1}{p^2+m^2}\left[1 - \frac{4\cdot 3 \cdot g\Lambda^2}{p^2 + m^2} \frac{1}{(4\pi)^2} \left(1 - \frac{m^2}{\Lambda^2}\ln \frac{\Lambda^2}{m^2}\right) \right] 
 \approx \frac{1}{p^2} \left[1 - \frac{4\cdot 3 \cdot g\Lambda^2/ (4\pi)^2 + m^2}{p^2} \right]
 $$
So in critical WF point $m^2_{\star}  \approx  -\frac{1}{6} \Lambda^2 \epsilon$ ,
$\tilde{g}_{\star} = \frac{2\pi^2}{9}\epsilon$ we have:
$$
 \langle \phi(p) \phi(-p) \rangle_{\star} = \frac{1}{p^2} + O(\epsilon^2)
 $$
We have massless particles on curve (in first order in $\epsilon$):
$$ m^2 = -  3  g\Lambda^2/ 4\pi^2$$
This line obviously connects Gaussian fixed point ($m^2 = g =0$) and Wilson-Fisher fixed point.
Also
$$
\frac{d}{ds}\left(4\cdot 3 \cdot g\Lambda^2/ (4\pi)^2 + m^2\right) =0 
$$
on RG equations in first order in $\epsilon$.
$$
 \begin{cases}
  \frac{dm^2}{ds} = 2\mu^2 + \frac{3}{2\pi^2}  g \Lambda^2\\
  \frac{dg}{ds} = \epsilon g - \frac{9}{2\pi^2} g^2
 \end{cases}
 $$
I will appreciate a lot if somebody explains to me the relation of such calculation to
Abdelmalek Abdesselam's answer.
</s>