Proposition 17.7.461A0C5DD7757

Proof. By lemma 17.7.3, for every $x\in X$ the function $t\mapsto u_ t(x)$ is convex on $(0,1)$, and either identically $-\infty $ or finite and continuous; it is finite for almost every $x$; and every slice $u_ t$ is $\theta $-psh. Since a sequence converging in $L^1(X)$ has a subsequence converging almost everywhere, we can choose sequences $t_ j\downarrow 0$ and $s_ j\uparrow 1$ with $u_{t_ j}\to u_0$ and $u_{s_ j}\to u_1$ almost everywhere. For fixed $t$ and large $j$, convexity on $[t_ j,s_ j]$ gives

and letting $j\to \infty $ gives $u_ t\leq (1-t)u_0+tu_1$ almost everywhere. Both sides are $\theta $-psh, so the inequality holds everywhere.

Let $D$ be a connected coordinate chart of $X$ on which $\theta =\mathrm{dd}^{\mathrm{c}}h$ with $h$ smooth. The function $U+h\circ p$ is plurisubharmonic on $D\times S=D\times ((0,1)+\mathrm i\mathbb R)$ and invariant under imaginary translations, so corollary 5.12.2, with $V=(0,1)$, shows that $w+h=\inf _{0<t<1}(u_ t+h)$ is plurisubharmonic or identically $-\infty $ on $D$. For almost every $x$, the function $t\mapsto u_ t(x)$ is finite and convex, hence bounded below on $(0,1)$ by lemma 17.7.2(2). Thus $w$ is finite almost everywhere, the alternative $-\infty $ is excluded on every chart, and $w$ is $\theta $-psh. Since $w\leq u_ t$ for all $t$ and $u_ t\to u_0$, $u_ t\to u_1$ in $L^1(X)$, we get $w\leq u_0$ and $w\leq u_1$ almost everywhere, hence everywhere. Thus $w$ is a competitor for the obstacle $\min \{ u_0,u_1\} $, and $w\leq P_\theta (u_0,u_1)$.

The bounds $w\leq u_ t\leq \max \{ u_0,u_1\} $ hold for all $t$, with integrable outer functions. For almost every $x$ the function $t\mapsto u_ t(x)$ is finite and continuous on $(0,1)$, so dominated convergence gives continuity of $t\mapsto u_ t$ from $(0,1)$ to $L^1(X)$; continuity at the endpoints is part of the definition. The image of $[0,1]$ under a continuous map is compact. Now fix $x\in X$. If $t\mapsto u_ t(x)$ is identically $-\infty $, its limits at $0$ and $1$ are $-\infty $. Otherwise it is finite and convex and bounded above by $\max \{ u_0(x),u_1(x)\} <\infty $, so by lemma 17.7.2(3), applied to $t\mapsto u_ t(x)$ and to $t\mapsto u_{1-t}(x)$, it has finite limits at $0$ and $1$. By the displayed bound these limits are at most $u_0(x)$ and $u_1(x)$. Since the pointwise limits exist everywhere and $u_{t_ j}\to u_0$, $u_{s_ j}\to u_1$ almost everywhere, they equal $u_0$ and $u_1$ almost everywhere.

It remains to prove equality quasi-everywhere; we treat the endpoint $0$, and the endpoint $1$ follows by reversing the parameter. Fix $s\in (0,1)$ and a sequence $t_ j\downarrow 0$ with $t_ j<s/2$. Consider the value at $t=0$ of the chord of $t\mapsto u_ t$ through the parameters $t_ j$ and $s$:

Where $u_ s$ is finite, the function $t\mapsto u_ t(x)$ is finite and convex, so by lemma 17.7.2(3) $a_ j$ increases with $j$ and tends to $\lim _{t\to 0}u_ t$; this limit equals $u_0$ almost everywhere and is at most $u_0$ everywhere. The functions

are combinations of $u_{t_ j}$ and $u_ s$ with positive coefficients of sum two, hence $2\theta $-psh, since $2\theta +\mathrm{dd}^{\mathrm{c}}(\alpha u_{t_ j}+\beta u_ s)=\alpha \theta _{u_{t_ j}}+\beta \theta _{u_ s}$ for $\alpha ,\beta \geq 0$ with $\alpha +\beta =2$. They increase with $j$ and are bounded above by $u_0+u_ s$. Their pointwise limit is $\lim _{t\to 0}u_ t+u_ s$, which equals $u_0+u_ s$ almost everywhere; hence their regularized upper envelope, a $2\theta $-psh function by theorem 5.10.2 which agrees almost everywhere with the pointwise limit, is $u_0+u_ s$ itself. By corollary 5.47.11(2), applied in charts after adding smooth local potentials of $2\theta $, the pointwise limit $\lim _{t\to 0}u_ t+u_ s$ equals $u_0+u_ s$ quasi-everywhere. Removing the pole set of $u_ s$ proves the assertion at $0$. $\square$