Renormalization of mass: does it change sign from high temperature to low temperature? Consider a Landau Ginzburg theory for ferromagnets with Hamiltonian
$$H=\int d^{D} x \frac{1}{2}(\nabla\phi(x))^{2} + \frac{1}{2} \mu^{2} \phi^{2}(x) + \frac{\lambda}{4!}\phi^{4}(x)$$
I can compute the 2 point correlation function via Dyson formula and I get
$$G(k)= \dfrac{1}{\mu^{2}+k^2 - \Sigma(k)}$$
at first order in $\lambda$ we have 
$$\Sigma = -\frac{\lambda}{2} \int^{\Lambda} d^{D}q \dfrac{1}{q^{2}+\mu^2}$$ 
with $\Lambda$ my UV cutoff (I do not need $\Lambda$ to go to infinity, but I immagine it is somehow big).
Critical phenomenon occur when $G(k=0)$ diverges and it seems that , since $\Sigma$ is negative, we have it when $\mu^2$ is negative: that seems reasonable, we have the critical temperature of gaussian model when $\mu^2=0$ so it makes sense that critical temperature of a non mean field model is at a lower temperature. 
Still, I wasn't sure of this computation, because we used perturbation theory and Dyson formula assuming the ground state was $\phi(x) \equiv 0$ which is true if $\mu^2 \geq 0$. Instead, if $\mu^2 < 0$ the ground state is $\phi_0 \equiv \sqrt{\frac{-6\mu^2}{\lambda}}$.
Anyway, I thought it would be useful to make the same computation in the low temperature region: this time I write the Hamiltonian as 
$$H=\int d^{D} x H_0(\phi_0) +\frac{1}{2}(\nabla\psi(x))^{2} + \frac{1}{2} \nu^{2} \psi^{2}(x) + \frac{\lambda_3}{3!}\psi^{3}(x)+\frac{\lambda}{4!}\psi^{4}(x)$$
where $\psi(x)= \phi(x) - \phi_0$, $\nu^2=-2\mu^2$ and $\lambda_3= \sqrt{-6 \mu \lambda}$
This time Dyson formula looks
$$G(k)= \dfrac{1}{\nu^{2}+k^2 - \Sigma '(k)}$$
$$\text{and}$$
$$\Sigma '(k=0) = -\frac{\lambda}{2} \int^{\Lambda} d^{D}q \dfrac{1}{q^{2}+\nu^2} + \frac{\lambda}{2} \int^{\Lambda} d^{D}q \dfrac{1}{q^{2}+\nu^2}\dfrac{3\nu^2}{q^{2}+\nu^2}$$ 
with the second term emerging from the new diagrams with two 3-line vertices 
My problem is that, since the second term looks smaller than the first one, the only thing that has changed is that instead of $\mu^2$ we have $\nu^2=-2\mu^2$ and so this time Dyson formula seems to say that criticality occur at $\mu^2 > 0$ which contraddicts the previous computation
Can anyone find where the error(s) is(are) in all of this?
EDIT: you can find these formulas on G. Parisi Statistical Field Theory pages 184-185 
 A: The issue here is that the expectation value of the field, $\phi_0$, also gets a correction at one-loop. You should be careful to include this if you also include the loop corrections you've indicated.
One way to set this up is to write $\phi = \phi_0 + \psi$ with
$$
\phi_0 = \sqrt{\frac{- 6 \mu^2}{\lambda}} \left( 1 + \delta \phi_0 \right),
$$
where $\delta \phi_0$ is a power series in $\lambda$. We now determine by $\delta \phi_0$ demanding that $\langle \psi \rangle = 0$ at every order in perturbation theory, which sets $\phi_0$ to be the actual expectation value of $\phi$. With this setup, to leading order the Hamiltonian density is now (dropping a constant)
$$
\mathcal{H} = \frac{1}{2}(\nabla\psi(x))^{2} + \frac{1}{2} \nu^{2} \psi^{2} + \frac{\lambda_3}{3!}\psi^{3}+\frac{\lambda}{4!}\psi^{4} + v^3 \sqrt{\frac{3}{\lambda}} \delta \phi_0 \, \psi + \frac{3}{2} v^2 \delta \phi_0 \, \psi^2 + \cdots
$$
(I'm dropping a ton of terms which won't contribute at one-loop).
We now calculate:
$$
\langle \psi \rangle = -v^3 \sqrt{\frac{3}{\lambda}}  \delta \phi_0 - \frac{3 v}{2} \sqrt{\frac{\lambda}{3}}\int^{\Lambda} d^{D}q \dfrac{1}{q^{2}+v^2},
$$
which determines $\delta\phi_0$:
$$
\delta\phi_0 = - \frac{\lambda}{2 v^2} \int^{\Lambda} d^{D}q \dfrac{1}{q^{2}+v^2}.
$$
The fact that this is negative works with your intuition about how corrections to the free theory should look - we get more disordered when we include interactions.
We now compute the self-energy for the $\psi$ field, now taking the above results into account. We find
$$
G(k)^{-1} = k^2 + \nu^2 + 3 v^2 \delta\phi_0 + \frac{\lambda}{2} \int^{\Lambda} d^{D}q \dfrac{1}{q^{2}+\nu^2} - \frac{\lambda}{2} \int^{\Lambda} d^{D}q \dfrac{1}{q^{2}+\nu^2}\dfrac{3\nu^2}{q^{2}+\nu^2}.
$$
Then, plugging in the value for $\delta \phi_0$, we finally have
$$
G(k)^{-1} = k^2 + \nu^2 - \lambda \int^{\Lambda} d^{D}q \dfrac{1}{q^{2}+\nu^2} - \frac{\lambda}{2} \int^{\Lambda} d^{D}q \dfrac{1}{q^{2}+\nu^2}\dfrac{3\nu^2}{q^{2}+\nu^2}.
$$
As you can see, the sign of the offending term has been flipped, and everything should work out.
