Proposition 17.16.3.15E98759958F A subgeodesic is continuous in $L^1(X)$ on the interior of its parameter interval. If the interval is $(a,b)$ with $a<b$ finite, and the subgeodesic runs from $u_ a$ to $u_ b$, where $u_ a,u_ b\in \mathrm{PSH}(X,\theta )$, then
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17.16.4
\begin{equation} u_ t\leq \frac{b-t}{b-a}u_ a+\frac{t-a}{b-a}u_ b, \qquad a<t<b. \end{equation}
Moreover, the following upper limits, taken jointly in space and time, hold at the endpoints:
\[ \limsup _{\substack {y\to x\\ t\downarrow a}}u_ t(y)\leq u_ a(x), \qquad \limsup _{\substack {y\to x\\ t\uparrow b}}u_ t(y)\leq u_ b(x). \]
Proof.
For a compact interval $J\Subset I$, choose $a'<\inf J\leq \sup J<b'$ in $I$. By lemma 5.41.1, the path lies below the chord through $u_{a'}$ and $u_{b'}$, hence below a constant on $J$, and by proposition 5.41.3 the function $\inf _{a'<t<b'}u_ t$ is an integrable lower bound. For almost every $x$, the function $t\mapsto u_ t(x)$ is finite and convex, hence continuous, and dominated convergence proves continuity in $L^1$ on $J$.
For the chord bound, choose sequences of times tending to $a$ and $b$ along which the assumed $L^1$ convergence holds almost everywhere. Applying convexity between these times and passing to the limit proves (17.16.4) almost everywhere in $x$, for each $t$, and hence everywhere by lemma 5.24.3.
For the joint upper limit at $a$, note that in (17.16.4) the coefficient of $u_ a$ tends to one, while the term involving $u_ b$ has a vanishing coefficient and a uniform upper bound. Upper semicontinuity of $u_ a$ gives the assertion when $u_ a(x)$ is finite. At a pole of $u_ a$, bound $u_ a$ near $x$ above by an arbitrary real number, take the joint upper limit, and let that number tend to $-\infty $. The argument at $b$ is the same.
$\square$
Pome
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