beta function vanishes to all orders, since there are no contributing diagrams; it is indeed well-known that this coupling is exactly marginal in the large $N$ limit [@Bardeen:1983rv] (see [@Moshe:2003xn] for a review). However, for finite $\lambda$ there do exist divergent planar diagrams. The vanishing of the $\lambda^4$ term in at this order is due (in our gauge choice) to a non-trivial cancelation between the diagrams (A2) and (A3). There is also a large $N$ divergence proportional to $\lambda^2 \lambda_6$ arising from (A1), that exactly cancels in the planar limit with the similar contribution from the anomalous dimension of $\phi^i$. In fact, one can show that, at large $N$, contributions to the beta function can have either zero or one $(\phi^2)^3$ vertices, and that the planar diagrams contributing with a single $(\phi^2)^3$ vertex are the same as the diagrams contributing to the anomalous dimension of $\phi^2$. Thus, the large $N$ beta function takes the form $$\beta_{\lambda_6}(\lambda,\lambda_6) = b\, \gamma_{\phi^2}(\lambda)\lambda_6 + f(\lambda) + O(1/N),$$ where $\gamma_{\phi^2}$ is the anomalous dimension of $\phi^2$ and $b$ is a constant. In the next subsection we argue that both this anomalous dimension and the beta function $\beta_{\lambda_6}$ vanish in the large $N$ limit, so that 