There is a fixed number $\delta_{1}>0$, such that for $s\gg 1$ and any $x\in\mathcal{A}^{s}_{\lambda}$ (or $x\in\mathcal{A}^{\infty}_{\lambda}$), there is a simplex $\sigma$ in the triangulation of $\Delta$, verifying $\langle x,m\rangle+\lambda(m)<L_{\lambda}(x)-\delta_{1}$ for $m\in\Delta_{\mathbb{Z}}\setminus\sigma$.