PomeProposition 17.16.2.1A5A4BBF82E3 Subgeodesics are preserved by adding an affine function of $t$, by a positive affine reparametrization of time, by finite maxima, and by convex combinations. For $h\in C^\infty (X,\mathbb R)$, replacing $\theta $ by $\theta +\mathrm{dd}^{\mathrm{c}}h$ and $u_ t$ by $u_ t-h$ also preserves them.
17.16.1 Subgeodesics and endpoint bounds8B8AAB4B5DD3
As in lemma 5.41.1, for a nonempty open interval $I\subset \mathbb R$ write
\[ S_ I=\{ z\in \mathbb C:\operatorname{Re}z\in I\} ,\qquad \pi :X\times S_ I\longrightarrow X, \]
where $\pi $ is the projection. A family $u_ t\in \mathrm{PSH}(X,\theta )$, $t\in I$, is a subgeodesic if its complexification $U(x,z)=u_{\operatorname{Re}z}(x)$ is quasi-plurisubharmonic and satisfies $\pi ^*\theta +\mathrm{dd}^{\mathrm{c}}U\geq 0$ on $X\times S_ I$. For finite $a<b$ and $u_ a,u_ b\in \mathrm{PSH}(X,\theta )$, a subgeodesic on $(a,b)$ runs from $u_ a$ to $u_ b$ if $u_ t\to u_ a$ in $L^1(X)$ as $t\downarrow a$ and $u_ t\to u_ b$ as $t\uparrow b$.
The elementary operations below produce lower barriers. Convexity in $t$, given by lemma 5.41.1, and the minimum principle of section 5.41 then give the continuity and upper bounds needed for the Perron envelope.
Proof. Affine functions of $\operatorname{Re}z$ have zero complex Hessian, and a positive affine change of time is induced by a holomorphic affine change of the strip coordinate. Convex combinations preserve the inequality between currents, and the change $(\theta ,u_ t)\mapsto (\theta +\mathrm{dd}^{\mathrm{c}}h,u_ t-h)$ preserves the current itself. Maxima are covered by lemma 5.24.1. $\square$
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
3E2013A813CD
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. $\square$
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