We first prove

Recall that we have normalized $\varphi$ so that $\max_{i\in I}\varphi(e_{i})=0$. We may therefore assume that $\mathcal{X}_{0}$ has at least two components. Note that $\pi_{*}G=\sum_{j}b_{j}\varphi(e_{j})E_{j}$ by definition. For each component $E_{i}$ the projection formula shows that

which is non-negative since $\pi^{*}\theta_{\mathcal{X}}+G$ and $\pi^{*}\mathcal{A}$ are nef and $\pi^{*}E_{i}\in\Div_{0}(\mathcal{Y})$ is effective. It follows that

for all $i$. Note that $E_{i}\cdot E_{j}\cdot\mathcal{A}^{n-1}\geq 0$ for all $i\neq j$, and

for all $i$, since $\mathcal{X}_{0}$ has connected support and contains at least two components.

Now pick $i_{0},\dots,i_{M}$ such that $\varphi(e_{i_{0}})=0$, $\varphi(e_{i_{M}})=\min_{i\in I}\varphi(e_{i})$, and $e_{i_{m}}$ and $e_{i_{m+1}}$ are connected by a $1$-dimensional face, so that $E_{i_{m}}\cdot E_{i_{m+1}}\cdot\mathcal{A}^{n-1}\geq 1$. Writing

and applying (6.2) to $i_{0},i_{1},\dots,i_{M-1}$, we get by induction

which proves (6.1).

We shall need the following construction, see Figure 1. Let $\sigma=\sigma_{J}$ be a face of $\Delta$ and $L\subset I$ the set of vertices of $\Delta$ contained in $\sta_{\Delta}(\sigma)$. Consider a rational point $v$ in the relative interior of $\sigma$. Given $0<\varepsilon<1$ rational and $j\in L$ set $e_{j}^{\varepsilon}:=\varepsilon e_{j}+(1-\varepsilon)v$. We shall define a projective simplicial subdivision $\Delta^{\prime}=\Delta^{\prime}(\varepsilon,v)$ of $\Delta$.

To define $\Delta^{\prime}$, first consider a polyhedral subdivision $\Delta^{\varepsilon}=\Delta^{\varepsilon}(v)$ of $\Delta$ leaving the complement of $\sta_{\Delta}(\sigma)$ unchanged. The set of vertices of $\Delta^{\varepsilon}$ is precisely $(e_{i})_{i\in I}\cup(e_{j}^{\varepsilon})_{j\in L}$ and the faces of $\Delta^{\varepsilon}$ contained in $\sta(\sigma)$ are of one of the following types:

• •

  if the convex hull $\Conv(e_{j_{1}},\dots,e_{j_{m}})$ is a face of $\Delta$ containing $\sigma$, then $\Conv(e_{j_{1}}^{\varepsilon},\dots,e_{j_{m}}^{\varepsilon})$ is a face of $\Delta^{\varepsilon}$;

• •

  if $\Conv(e_{j_{1}},\dots,e_{j_{m}})$ is a face of $\Delta$ contained in $\sta(\sigma)$ but not containing $\sigma$, then both $\Conv(e_{j_{1}},\dots,e_{j_{m}})$ and $\Conv(e_{j_{1}},\dots,e_{j_{m}},e_{j_{1}}^{\varepsilon},\dots,e_{j_{m}}^{\varepsilon})$ are faces of $\Delta^{\varepsilon}$.

In a neighborhood of $v$, note that the subdivision $\Delta^{\varepsilon}$ is obtained by scaling $\Delta$ by a factor $\varepsilon$. More precisely, consider the affine map $\psi^{\varepsilon}:\sta(\sigma)\to\sta(\sigma)$ defined by $\psi^{\varepsilon}(w)=\varepsilon w+(1-\varepsilon)v$. Then $\sigma^{\varepsilon}:=\psi^{\varepsilon}(\sigma)$ is the face of $\Delta^{\varepsilon}$ containing $v$ in its relative interior, and $\psi^{\varepsilon}(\sta_{\Delta}(\sigma))=\sta_{\Delta^{\varepsilon}}(\sigma^{\varepsilon})$. In particular, even though $\Delta^{\varepsilon}$ is not simplicial in general, all polytopes of $\Delta^{\varepsilon}$ containing $\sigma^{\varepsilon}$ are simplicial.

We claim that $\Delta^{\varepsilon}$ is projective. To see this, write $v=\sum_{j\in J}s_{j}e_{j}$, with $s_{j}>0$ rational and $\sum s_{j}=1$. For $j\in J$, define a linear function $\ell_{j}$ on $\sum_{i\in I}\mathbf{R}_{+}e_{i}\supset\Delta$ by $\ell_{j}(\sum t_{i}e_{i})=-t_{j}/s_{j}$ and set $h=\max\{\max_{j\in J}\ell_{j},-(1-\varepsilon)\}$. A suitable integer multiple of $h$ is then a strictly convex support function for $\Delta^{\varepsilon}$ in the sense of §3.5.

Now define $\Delta^{\prime}=\Delta^{\prime}(\varepsilon)$ as a simplicial subdivision of $\Delta^{\varepsilon}$ obtained using repeated barycentric subdivision in a way that leaves $\sta_{\Delta^{\varepsilon}}(\sigma^{\varepsilon})$ unchanged. By [KKMS, pp.115–117], $\Delta^{\prime}$ is still projective.

Note that $\sigma^{\prime}:=\sigma^{\varepsilon}$ is the face of $\Delta^{\prime}$ containing $v$ in its relative interior, For $j\in L$ set $e^{\prime}_{j}=e^{\varepsilon}_{j}$. These are the vertices of $\Delta^{\prime}$ contained in $\sta_{\Delta^{\prime}}(\sigma^{\prime})$.

Let $\tau$ be a face of $\Delta$. Our aim is to prove by induction on $\dim\tau$ that the $C^{0,1}$-norm of $\varphi$ on $\tau$ is bounded by $C$. Recall that the $C^{0,1}$-norm is defined as the sum of the sup-norm and the Lipschitz norm; see Appendix A.

The case $\dim\tau=0$ is settled by (6.1), so let us assume that $\dim\tau>0$. By Proposition 5.9 the restriction of $\varphi$ to $\tau$ is piecewise affine and convex. It therefore admits directional derivatives, and we set as in Appendix A

for $v,w\in\tau$.

Let us say that a codimension $1$ face of $\tau$ is opposite to a vertex when it is the convex hull of the remaining vertices of $\tau$. This notion is well-defined since $\tau$ is a simplex.

Granting this result, let us explain how to conclude the proof. By induction we have $\sup_{\partial\tau}|\varphi|\leq C$. The convexity of $\varphi$ and the inductive assumption imply $D_{v}\varphi(e)\leq\varphi(e)-\varphi(v)\leq C$ for any $e,v\in\partial\tau$. By Proposition 6.3 this gives $|D_{v}\varphi(e)|\leq C$ for any vertex $e$ of $\tau$ and any rational point $v$ in the relative interior of the face $\sigma$ opposite to $e$ such that $\varphi|_{\sigma}$ is affine near $v$. By density, and since $\varphi$ is piecewise affine, the same bound holds for any $v$ in the relative interior of $\sigma$. We conclude by Proposition A.1 that the $C^{0,1}$-norm of $\varphi|_{\tau}$ is bounded by $C$, completing the proof of Theorem 6.1.