Theorem 16.8.321CA975F04423

Proof. 2 $\implies $ 1. This follows from proposition 17.10.24.

1 $\implies $ 2. Let $(\varphi _ t)_{t\in (0,1)}$ be a subgeodesic from $\varphi _0$ to $\varphi _1$.

Step 1. We assume that $\varphi _0\geq \varphi _1$.

We first replace the given subgeodesic by the geodesic from $\varphi _0$ to $\varphi _1$. Note that the geodesic is necessarily a subgeodesic as well by 17.10.22.

Observe that $t\mapsto \varphi _ t$ is decreasing due to 17.10.28.

Let $\varphi _ t=\varphi _1$ for all $t>1$. Then by the gluing lemma 17.10.16, we find that $(\varphi _ t)_{t\geq 0}$ is a subgeodesic ray.

Next, we consider the Legendre transform

It follows from Kiselman's principle proposition 5.14.15 that $\Gamma _{\tau }\in \mathrm{PSH}(X,\theta )\cup \{ -\infty \} $. Note that for $\tau >0$, we clearly have $\Gamma _{\tau }\equiv -\infty $. On the other hand, for $\tau \leq 0$,

By Legendre inversion (cf. corollary 21.11.26), for $t>0$,

Fix a Kähler form $\omega $ on $X$. It follows from proposition 16.8.28 that for each $t>0$,

The right-hand side is independent of $t$. Here by adding $\omega $, we no longer have to worry about the possibility where $\Gamma _{\tau }$ has vanishing mass.

Write

By proposition 16.8.28 again, we find that

as well. So $\varphi _0\sim _ P \varphi _1$.

Step 2. We prove the general case.

Observe that $(\varphi _ t\lor \varphi _1)_{t\in (0,1)}$ is a subgeodesic from $\varphi _0\lor \varphi _1$ to $\varphi _1$. By Step 1,

Hence $\varphi _0\preceq _ P \varphi _1$ by 16.8.13. The converse is proved similarly. Hence 2 follows. $\square$