at every order in $\lambda$ the sum of diagrams (G1) is finite, and also the sum of diagrams (G2) is finite. Thus, the beta function must vanish at large $N$.

Spontaneous Breakdown of Conformal Symmetry {#sponbreak}
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In order to verify that our theories are conformally invariant, we should also make sure that they do not spontaneously break conformal invariance, by a vacuum expectation value for $\phi^2$. For the theory with $\lambda=0$ and $\lambda_6 \neq 0$, this was analyzed in detail in [@Bardeen:1983rv], and it was found that for $\lambda_6 < (4\pi)^2$ such a breaking does not arise. In fact, the effective potential for $\sigma = \phi^2/N$ can be computed exactly for infinite $N$, and it takes the form [@Amit:1984ri] $$V(\sigma) = \frac{N}{6} \left[ (4\pi)^2 - \lambda_6 \right] |\sigma|^3.$$ Thus, for small $\lambda_6$ the only minimum of the effective potential is at the conformal point $\phi^2=0$. We expect that turning on a small coupling $\lambda$, as we analyzed above, will lead to small changes in the coefficient of $|\sigma|^3$ in this effective potential (which can be computed in perturbation theory in $\lambda$), but at least for small $\lambda$ and small $\lambda_6$ it seems clear that there will still be a 