ScalingStacks

Proof. The idea is to regard $\bar{\phi}_{m,w}+\langle m,x\rangle$ as approximately defining a locally convex function on $\partial\Delta_{\lambda}^{\vee}$ in the sense of Def. 3.22, and then the problem is essentially to prove an effective version of the extension property (cf. Prop. 3.27). We will outline the main modifications.

We will produce $u$ by mimicking the Legendre duality construction in Prop. 3.19. For $p\in\Delta$, define

$$u^{*}(p)=\sup_{x\in\partial\Delta_{\lambda}^{\vee}}\{\langle x,p\rangle-(\bar{\phi}_{m,w}+\langle m,x\rangle)\},$$

where it is tacitly understood that $\bar{\phi}_{m,w}+\langle m,x\rangle$ is defined only over $\partial\Delta_{\lambda}^{\vee}\cap U^{\infty}_{w}$, and the sup is taken over all choices of $m,w$ whenever $\bar{\phi}_{m,w}$ is defined. Since $\bar{\phi}_{m,w}$ are uniformly bounded on $\partial\Delta_{\lambda}^{\vee}$, we see $\left\lVert u^{*}\right\rVert_{C^{0}(\Delta)}\leq C$. We then define a convex function $u$ on $N_{\mathbb{R}}$ by another Legendre transform

$$u(x)=\sup_{p\in\Delta}\{\langle p,x\rangle-u^{*}(p)\},$$

which is admissible because $u^{*}$ is bounded. By the same reasoning in Prop. 3.19, on $\partial\Delta_{\lambda}^{\vee}\cap U^{\infty}_{w}$,

$$u(x)\leq\bar{\phi}_{m,w}+\langle m,x\rangle+Cs^{-1/2}.$$

We are only left to show

$$u(x)\geq\bar{\phi}_{m,w}+\langle m,x\rangle-Cs^{-1/2},$$

which amounts to showing that there exists $p\in\Delta$, such that for any $y\in\partial\Delta_{\lambda}^{\vee}$,

$$\bar{\phi}_{m^{\prime},w^{\prime}}(y)+\langle m^{\prime},y\rangle\geq\bar{\phi}_{m,w}(x)+\langle m,x\rangle+\langle p,y-x\rangle-Cs^{-1/2}.$$

Notice our setting enjoys the discrete symmetry. This last step is the effective version of Prop. 3.27, and the proof is basically the same. ∎