Proposition 17.10.71880C1FBD88F

Proof. Let $\Phi $ be the complexification of $(\varphi _ t)_{t\in (0,1)}$.

For each $x\in X$, the map

is either subharmonic or constantly $-\infty $, as follows from definition 17.10.2 1 and proposition 5.13.17. In the latter case, the convexity of $[0,1] \ni t\mapsto \varphi _ t(x)$ is trivial. In the former case, the convexity on the interval $(0,1)$ follows from proposition 5.13.12.

In order to verify the convexity at the boundary, let us fix $s\in (0,1)$. We need to show that

for all $x\in X$. Thanks to proposition 5.14.6, it suffices to prove this for almost all $x$.

Choose sequences $a_ j\searrow 0$ and $b_ j\nearrow 1$ such that $\varphi _{a_ j}\to \varphi _0$ and $\varphi _{b_ j}\to \varphi _1$ almost everywhere. Outside a set $Z\subseteq X$ of zero Lebesgue measure, all these convergences hold and all the values involved are finite. For $x\in X\setminus Z$ and $j$ large enough that $a_ j<s<b_ j$, the convexity on $(0,1)$ gives

Letting $j\to \infty $ gives (17.10.8) almost everywhere.

Let us prove the last assertion. Let

By Kiselman's principle proposition 5.14.15, we know that $\varphi \in \mathrm{PSH}(X,\theta )\cup \{ -\infty \} $. Since the complexification $\Phi $ is not identically $-\infty $, Fubini's theorem shows that for almost every $x\in X$, the function $t\mapsto \varphi _ t(x)$ is finite for almost every $t\in (0,1)$. For such an $x$, the convexity just proved implies that $t\mapsto \varphi _ t(x)$ is a finite convex function on $(0,1)$, hence it is bounded from below on $(0,1)$. Thus $\varphi \not\equiv -\infty $, and so $\varphi \in \mathrm{PSH}(X,\theta )$.

For any $t\in (0,1)$, using the convexity established above, we have

It follows that $\varphi \leq \varphi _0$ and $\varphi \leq \varphi _1$ almost everywhere and hence everywhere by proposition 5.14.6. Our assertion follows. $\square$