ScalingStacks

Proof. Let $x\in\mathcal{A}_{\lambda}^{\infty}$. If $m=0\in\Delta$ achieves the maximum $L_{\lambda}(x)$, then $x\in\partial\Delta_{\lambda}^{\vee}$. If not, then the maximum is achieved by at least two $m\in\partial\Delta$, so $x\in\mathcal{A}_{\lambda,\sigma}^{\infty}$ for some $\sigma\subset\partial\Delta$ with $\dim\sigma\geq 1$. ∎