Proposition 17.10.248AB36536E36D

Proof. After subtracting suitable constants from $\varphi _0$ and $\varphi _1$ and adding back the corresponding affine function of $t$, we may assume that $\varphi _0,\varphi _1\leq \phi $. It follows from proposition 17.10.7 that $\varphi _ t\leq \phi $ for all $t\in (0,1)$. In fact, we have the stronger estimate

We first observe that when $\varphi _0,\varphi _1\in \mathcal{E}(X,\theta ;\phi )$, so is $\varphi _0\land \varphi _1$, see proposition 16.5.49. In particular, the constant subgeodesic $t\mapsto \varphi _0\land \varphi _1$ is a candidate in the Perron envelope (17.10.21). So

By proposition 17.10.13, $(\varphi _ t)_{t\in (0,1)}$ is a subgeodesic.² It follows from proposition 16.5.52 that $\varphi _ t\in \mathcal{E}(X,\theta ;\phi )$ for all $t\in (0,1)$.

Next, we show that as $t\to 0+$, we have $\varphi _ t\xrightarrow {L^1}\varphi _0$. The corresponding result at $t=1$ is similar.

We first argue the special case where $\varphi _0\preceq \varphi _1$. Take a constant $C>0$ such that

Then $(\varphi _0-Ct)_{t\in (0,1)}$ is clearly a candidate in (17.10.21), see example 17.10.6. Therefore, for all $t\in (0,1)$,

It follows that $\varphi _ t\xrightarrow {L^1}\varphi _0$ as $t\to 0+$.

Let us come back to the general case. By corollary 17.10.11, $\{ \varphi _ t:t\in (0,1)\} $ is a relatively compact subset of $\mathrm{PSH}(X,\theta )$ with respect to the $L^1$-topology.

Let $\psi $ be an $L^1$-cluster point of $\varphi _ t$ as $t\searrow 0$, and it suffices to show that $\psi =\varphi _0$.

For each $M\in \mathbb {N}$, we write

Observe that $\varphi _0^ M\in \mathcal{E}(X,\theta ;\phi )$ by proposition 16.5.49. Let $(\varphi _ t^ M)_{t\in (0,1)}$ be the geodesic from $\varphi _0^ M$ to $\varphi _1$. Since $\varphi _0^ M\preceq \varphi _1$, the special case proved above applies to the geodesic $(\varphi _ t^ M)_ t$, and

Moreover, every candidate for the geodesic from $\varphi _0^ M$ to $\varphi _1$ is also a candidate for the geodesic from $\varphi _0$ to $\varphi _1$, so

Letting $t\to 0+$ along a sequence for which $\varphi _ t\to \psi $ in $L^1$, and using corollary 5.14.8, we obtain

almost everywhere, hence everywhere by proposition 5.14.6.

On the other hand, by (17.10.25), $\psi \leq \varphi _0$. So it suffices to show that

as $M\to \infty $, which is shown in proposition 16.5.54.

Now we have shown that $(\varphi _ t)_{t\in [0,1]}$ is a subgeodesic.

Next, take $0\leq a\leq b\leq 1$. We want to show that the restriction $(\varphi _ t)_{t\in [a,b]}$ is the geodesic from $\varphi _ a$ to $\varphi _ b$. We may assume that $a<b$. The argument is the standard balayage argument.

Let $(\psi _ t)_{t\in (a,b)}$ be the geodesic from $\varphi _ a$ to $\varphi _ b$. Since $(\varphi _ t)_{t\in [a,b]}$ is a subgeodesic by proposition 17.10.15, we have $\psi _ t\geq \varphi _ t$ for all $t\in (a,b)$.

We define

Extending it by setting $\psi _ a=\varphi _ a$ and $\psi _ b=\varphi _ b$, the first part of the proof applied to the endpoints $\varphi _ a,\varphi _ b$ shows that $(\psi _ t)_{t\in [a,b]}$ is a subgeodesic from $\varphi _ a$ to $\varphi _ b$. Since $\psi _ t\geq \varphi _ t$ for $t\in (a,b)$, the gluing lemma 17.10.16 shows that $(\eta _ t)_{t\in [0,1]}$ is a subgeodesic from $\varphi _0$ to $\varphi _1$.

Therefore, for all $t\in (0,1)$, we have

In particular, for $t\in (a,b)$, we have

In other words, $(\varphi _ t)_{t\in (a,b)}=(\psi _ t)_{t\in (a,b)}$ is the geodesic from $\varphi _ a$ to $\varphi _ b$.

Finally, assume furthermore that $\varphi _0,\varphi _1\in \mathcal{E}^1(X,\theta ;\phi )$ (resp. $\mathcal{E}^{\infty }(X,\theta ;\phi )$). Thanks to (17.10.26), proposition 16.5.49 and proposition 16.5.52, we find $\varphi _ t\in \mathcal{E}^1(X,\theta ;\phi )$ (resp. $\mathcal{E}^{\infty }(X,\theta ;\phi )$) for all $t\in (0,1)$. $\square$