ScalingStacks

Proof. We need to prove the characterisation in Prop. 3.19. Without loss of generality $L_{\lambda}(x)$ is achieved by $\langle m^{0},x\rangle+\lambda(m^{0})$. We need to find $p\in\Delta$, such that $u(y)-u(x)\geq\langle p,y-x\rangle.$ For this we study the gradient of the function $u_{m^{0}}$ on the various $w$-charts.

First, notice for $x^{\prime},y^{\prime}$ on the face $\{L_{\lambda}=\langle m^{0},\rangle+\lambda(m^{0})\}$, namely the convex hull of $w_{1},\ldots w_{n+1}$, the vector $y^{\prime}-x^{\prime}$ is parallel to the face, and by convexity of $u_{m^{0}}$ the directional derivative $\nabla u_{m^{0}}\cdot(y^{\prime}-x^{\prime})$ is monotone along the path from $x^{\prime}$ to $y^{\prime}$, so must be maximized at $y^{\prime}$. In particular we consider such line segments on the face parallel to $w_{i}-w_{j}$ for $i,j\geq 1$. By the discrete symmetry, $\nabla u_{m^{0}}\cdot(w_{j}-w_{i})$ must be zero on the plane of reflection bisecting the face. Thus for $i,j\geq 1$, $i\neq j$, the subset of the face

$$\{\nabla u_{m^{0}}\cdot w_{i}\geq\nabla u_{m^{0}}\cdot w_{j}\}\cap\{L_{\lambda}=\langle m^{0},\rangle+\lambda(m^{0})\}$$

agrees exactly with the half of the face containing $w_{i}$. Therefore the subset of face

$$\{\nabla u_{m^{0}}\cdot w_{i}\geq\nabla u_{m^{0}}\cdot w_{j},\forall j\geq 1\}$$

is exactly the intersection of $\text{Star}(w_{i})$ with the face. Without loss of generality $x$ lies in $\text{Star}(w_{1})$.

We follow the notation in the proof of Prop. 3.26. In the $w_{1}$-chart, denote the gradient of $u_{m^{0}}$ as $\vec{p}$, so that for $y$ in the $w_{1}$-chart,

$$u_{m^{0}}(y)-u_{m^{0}}(x)\geq\vec{p}\cdot(y-x)_{w_{1}}.$$

A priori $\vec{p}$ lives in $M_{\mathbb{R}}/\mathbb{R}m^{0}$. We lift $\vec{p}$ to $M_{\mathbb{R}}$ by demanding $\langle\vec{p},w_{1}\rangle=0$, so by the above discussion $\langle\vec{p},w_{i}\rangle\leq 0$ for $i\geq 1.$ Define $p=\vec{p}+m_{0}$, then $\langle p,w_{i}\rangle\leq 1$ for all $i\geq 1$. We regard $p\in M_{\mathbb{R}}$ as the gradient of $u$ at $x$, and write $p=\nabla u$ as a function of $x$. This construction can be made on other faces as well, and on the intersection of two faces the definitions are compatible.

We claim $p\in\Delta$: it suffices to show $\langle p,w_{0}\rangle\leq 1$. Notice $w_{0}=-\sum_{1}^{n+1}w_{i}=\sum_{i=2}^{n+1}(w_{1}-w_{i})-(n+1)w_{1}$. Consider the line segment in the face joining $x$ to the boundary of the face in the direction $\sum_{i=2}^{n+1}(w_{1}-w_{i})$, which stays inside $\text{Star}(w_{1})$, and along which $\nabla u\cdot\sum_{i=2}^{n+1}(w_{1}-w_{i})$ increases, or equivalently $\langle\nabla u,w_{0}\rangle$ increases. But the boundary of the face $\{L_{\lambda}=\langle m^{0},\rangle+\lambda(m^{0})\}$ lies also on a different face, and we can use the information from this new face to deduce $\langle\nabla u,w_{0}\rangle\leq 1$ there.

By construction for $y$ in the $w_{1}$-chart,

$$u(y)-u(x)\geq\langle\vec{p}(x),(y-x)_{w_{1}}\rangle+\langle m_{0},y-x\rangle=\langle\nabla u(x),y-x\rangle.$$

We claim that in fact $u(y)-u(x)\geq\langle\nabla u(x),y-x\rangle$ holds for all $y\in\partial\Delta_{\lambda}^{\vee}$. We are left to check for $y$ on the face $\{L_{\lambda}=\langle m^{1},\rangle+\lambda(m^{1})\}$, namely the complement of the $w_{1}$-chart. Consider the $w_{i}$-chart for $i>1$. We can write according to the decomposition $N_{\mathbb{R}}=(m^{1})^{\perp}\oplus\mathbb{R}w_{i}$, that

$$y-x=(y-x)_{w_{i},m^{1}}+\langle m^{1},y-x\rangle w_{i}.$$

By local convexity, in the $w_{i}$-chart $u_{m^{1}}$ is convex, so there is some $\vec{p}^{\prime}$, such that for any $y^{\prime}$ in the $w_{i}$-chart

$$u_{m^{1}}(y^{\prime})-u_{m^{1}}(x)\geq\vec{p}^{\prime}\cdot(y^{\prime}-x)_{w_{i},m^{1}}.$$

But a gradient vector of $u_{m^{1}}$ at $x$ is $\nabla u(x)-m^{1}$, so we may take $\vec{p}^{\prime}=\nabla u(x)-m^{1}$. Thus

$$u(y)-u(x)\geq\vec{p}^{\prime}\cdot(y-x)_{w_{i},m^{1}}+\langle m^{1},y-x\rangle=\langle\nabla u(x),y-x\rangle-\langle\vec{p}^{\prime},w_{i}\rangle\langle m^{1},y-x\rangle.$$

Now $\langle m^{1},y-x\rangle\geq 0$ as in the proof of Prop. 3.26, and $\langle\vec{p}^{\prime},w_{i}\rangle\leq 0$ by $\nabla u\in\Delta$. This implies $u(y)-u(x)\geq\langle\nabla u(x),y-x\rangle$ as required.

We have verified the characterisation in Prop. 3.19, hence the extension property. ∎