Proposition 17.7.9DD04D189A4D3

Proof. Assertion (1) holds because enlarging the endpoints enlarges the family of competitors, and (2) because adding the affine function $(1-t)c_0+tc_1$, respectively reversing the parameter, is a bijection between the families of competitors.

For (3), the geodesics $G^ k=G(u_0^ k,u_1^ k)$ decrease with $k$ by (1) and dominate $G=G(u_0,u_1)$. Their complexifications decrease to a function which dominates the complexification of $G$ and so is not identically $-\infty $ on any chart; by theorem 5.10.1, applied in charts after adding smooth local potentials of $p^*\theta $, it is $p^*\theta $-psh. It is invariant under imaginary translations, and its slices $\widetilde G_ t$ are the pointwise limits of $G^ k_ t$. For every $k$ we have $G_ t\leq \widetilde G_ t\leq G^ k_ t\leq (1-t)u_0^ k+tu_1^ k$, so $G_ t\leq \widetilde G_ t\leq (1-t)u_0+tu_1$. As $t\to 0$, both outer functions converge to $u_0$ in $L^1(X)$, hence so does $\widetilde G_ t$; similarly at $t=1$. Thus $(\widetilde G_ t)$ is a competitor for the Perron segment between $u_0$ and $u_1$, so $\widetilde G\leq G$, and therefore $\widetilde G=G$.

For (4), let $u_1\leq u_0$ and $0<r<s<1$, and put $\delta =(s-r)/(1-r)$. The reparametrized curve $t\mapsto G_{\delta +(1-\delta )t}$ is a subgeodesic by lemma 17.7.5. Its value at $t=0$ is $G_\delta \leq (1-\delta )u_0+\delta u_1\leq u_0$, and its value at $t=1$ is $u_1$. It is therefore a competitor for the segment between $u_0$ and $u_1$, and at $t=r$, where $\delta +(1-\delta )r=s$, it gives $G_ s\leq G_ r$. For the endpoints, $G_ t\leq (1-t)u_0+tu_1\leq u_0$, and the constant curve $u_1$ is a competitor, so $G_ t\geq u_1$.

For (5), note that $u_1\leq u_0+C$ and $u_0\leq u_1+C$ everywhere, because inequalities between quasi-psh functions that hold quasi-everywhere hold everywhere (section 16.11). Hence the affine curves $u_0-Ct$ and $u_1-C(1-t)$ are competitors, so that $G_ t\geq \max \{ u_0-Ct,u_1-C(1-t)\} $. On the other hand $G_ t-u_0\leq t(u_1-u_0)\leq Ct$ and $G_ t-u_1\leq (1-t)(u_0-u_1)\leq C(1-t)$ by the upper bound of theorem 17.7.8. For (6), the upper bound $u_0+tc$ is itself an affine competitor. $\square$