ScalingStacks

Proof. Let $\langle m,w\rangle=1$, and consider the function $u_{m}$ on the chart $\partial\Delta_{\lambda}^{\vee}\cap U_{w}^{\infty}$. Given $x$ in the chart, we need to find $\vec{p}$ such that

$$u_{m}(y)-u_{m}(x)\geq\vec{p}\cdot(y-x)_{w},$$

where $\vec{p}$ is a covector, and $(y-x)_{w}$ refers to the representation of $y-x$ in the local coordinates $x^{m_{1}},\ldots x^{m_{n}}$; after identifying $x^{m_{1}},\ldots x^{m_{n}}$ as coordinates on the plane $m^{\perp}=\{\langle m,x^{\prime}\rangle=0\}$, we may regard $(y-x)_{w}$ as an element of $m^{\perp}$, and according to the decomposition $N_{\mathbb{R}}=m^{\perp}\oplus\mathbb{R}w$,

$$y-x=(y-x)_{w}+\langle y-x,m\rangle w.$$

Since the convexity of $u_{m}$ and $u_{m^{\prime}}$ are equivalent in the $w$-chart if $\langle m,w\rangle=\langle m,w\rangle=1$, we may assume $L_{\lambda}(x)$ is attained by $\langle m,x\rangle+\lambda(m)$. By the extension property and Prop. 3.19, there is some $p\in\Delta$, such that

$$u(y)-u(x)\geq\langle p,y-x\rangle,$$

hence

$$u_{m}(y)-u_{m}(x)\geq\langle p-m,y-x\rangle=\langle p-m,(y-x)_{w}\rangle+\langle y-x,m\rangle\langle p-m,w\rangle.$$

Since $p\in\Delta$, we have $\langle p,w\rangle\leq 1=\langle m,w\rangle.$ Since $L_{\lambda}(x)$ is attained by $\langle m,x\rangle+\lambda(m)$, and the polytope $\Delta_{\lambda}^{\vee}$ lies in the half space $\{\langle m,\rangle+\lambda(m)\leq 0\}$, we have

$$\langle m,y\rangle+\lambda(m)\leq 0=\langle m,x\rangle+\lambda(m).$$

Combining the above

$$u_{m}(x)-u_{m}(y)\geq\langle p-m,(y-x)_{w}\rangle+\langle y-x,m\rangle\langle p-m,w\rangle\geq\langle p-m,(y-x)_{w}\rangle,$$

so we have produced $\vec{p}$ as required. ∎