PomeProposition 17.10.15.E5064A497AB8 Let $(\varphi _ t)_{t\in [0,1]}$ be a subgeodesic in $\mathrm{PSH}(X,\theta )$. Then for any $0\leq a\leq b\leq 1$, the segment $(\varphi _{t})_{t\in [a,b]}$ is a subgeodesic.
Proposition 17.10.15E5064A497AB8
Proof. It suffices to show that
\[ \varphi _{tb+(1-t)a}\xrightarrow {L^1}\varphi _ a,\quad \varphi _{tb+(1-t)a}\xrightarrow {L^1}\varphi _ b \]
as $t\to 0+$ and $t\to 1-$ respectively. In other words, we need to show that for any $c\in (0,1)$, we have
\[ \varphi _ t\xrightarrow {L^1}\varphi _ c \]
as $t\to c$. For this purpose, observe that by proposition 17.10.7,
\[ \adjustlimits \sup _ X \inf _{s\in (0,1)}\varphi _ s\leq \sup _{X}\varphi _ t\leq \left( \sup _ X \varphi _0 \right) \lor \left( \sup _ X\varphi _1 \right) \]
for any $t\in (0,1)$. Therefore, $\{ \varphi _ t\} _{t\in (0,1)}$ is a relatively compact family with respect to the $L^1$-topology on $\mathrm{PSH}(X,\theta )$ by proposition 5.17.3. It suffices to show that any cluster point $\psi $ of $\varphi _ t$ as $t\to c$ is equal to $\varphi _ c$. By corollary 5.14.8 and the convexity proposition 17.10.7, we have $\varphi _ c=\psi $ almost everywhere and hence everywhere by proposition 5.14.6. $\square$
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