PomeProposition 17.16.6.5D2A9C4885B8 If an admissible subgeodesic exists, the Perron envelope is the largest admissible subgeodesic. If some admissible subgeodesic runs from $u_0$ to $u_1$, so does the envelope, which is then the Perron segment. In that case, for every $[a,b]\subset [0,1]$ with $a<b$, its restriction to $[a,b]$ is the Perron segment between $u_ a^{\mathrm P}$ and $u_ b^{\mathrm P}$.
17.16.5 Perron segments872589A6FBF3
The chord bound survives taking a regularized supremum, and a single competitor attaining the endpoints bounds that supremum from below and forces the same endpoint limits. Fix $u_0,u_1\in \mathrm{PSH}(X,\theta )$ and call a subgeodesic on $(0,1)$ admissible if its joint upper limits at $t=0$ and $t=1$, in the sense of proposition 17.16.3, are at most $u_0$ and $u_1$, respectively. Define the Perron envelope
\[ U=\left(\sup \{ V:V\text{ is an admissible complexification}\} \right)^*, \]
where the upper semicontinuous regularization is taken on $X\times S_{(0,1)}$, and denote the slices of $U$ by $u_ t^{\mathrm P}$. When these slices form a subgeodesic running from $u_0$ to $u_1$, they define the Perron segment from $u_0$ to $u_1$, and we set $u_0^{\mathrm P}=u_0$ and $u_1^{\mathrm P}=u_1$. An affine change of parameter gives the same construction on any finite interval.
Proof. Convexity in $t$ and the upper limits at the endpoints bound every competitor by $(1-t)u_0+t u_1$. This bound is jointly upper semicontinuous on the interior and locally bounded above, so it persists after regularization. The local envelope theorem of section 5.40 shows that $U$ is quasi-plurisubharmonic with $\pi ^*\theta +\mathrm{dd}^{\mathrm{c}}U\geq 0$, and $U$ is still independent of $\Im z$. Since it dominates an admissible competitor, each slice belongs to $\mathrm{PSH}(X,\theta )$. The chord bound gives the joint upper limits at the endpoints, by the argument at the end of the proof of proposition 17.16.3. The envelope therefore belongs to its defining family and is its largest member.
If a competitor $w_ t$ runs from $u_0$ to $u_1$, then
\[ w_ t\leq u_ t^{\mathrm P}\leq (1-t)u_0+t u_1. \]
In particular, almost everywhere,
\[ |u_ t^{\mathrm P}-u_0|\leq |w_ t-u_0|+t|u_1-u_0|, \]
which proves $u_ t^{\mathrm P}\to u_0$ in $L^1$ as $t\downarrow 0$. The argument at $t=1$ is the same.
For $0\leq a<b\leq 1$, let $v_ t$ be the Perron envelope on $(a,b)$ between $u_ a^{\mathrm P}$ and $u_ b^{\mathrm P}$. The original path restricted to $(a,b)$ runs between these endpoints and is admissible by proposition 17.16.3. Thus $v_ t$ also runs between them and dominates $u_ t^{\mathrm P}$. The joint upper limits of $v_ t$ at $a$ and $b$ allow us to glue its complexification to $U$ outside the smaller strip, by lemma 5.24.2; if $a=0$ or $b=1$, the same upper limits show that the glued family is still admissible for $u_0,u_1$. The resulting competitor cannot exceed $U$, which proves $v_ t=u_ t^{\mathrm P}$ on $(a,b)$. $\square$
When the endpoints differ by a bounded function, an explicit lower barrier is available.
Theorem 17.16.7.66E38862017F Suppose that $u_0\sim u_1$, and choose $C\geq 0$ with $u_0-C\leq u_1\leq u_0+C$. Then the Perron segment from $u_0$ to $u_1$ exists, every slice has the ordinary singularity type of $u_0$ and $u_1$, and
\[ u_0-Ct\leq u_ t^{\mathrm P}\leq u_0+Ct, \qquad u_1-C(1-t)\leq u_ t^{\mathrm P}\leq u_1+C(1-t). \]
Moreover, away from the common pole set,
\[ |u_ t^{\mathrm P}-u_ s^{\mathrm P}|\leq C|t-s|, \qquad 0\leq s,t\leq 1. \]
Proof. Consider the barrier
\[ w_ t=\max \{ u_0-Ct,\ u_1-C(1-t)\} . \]
It is a subgeodesic by proposition 17.16.2, and the bounded difference gives $w_0=u_0$ and $w_1=u_1$. Upper semicontinuity gives the joint upper limits at the endpoints. Off the common poles, $|w_ t-u_0|\leq Ct$ and $|w_ t-u_1|\leq C(1-t)$, so $w_ t$ runs from $u_0$ to $u_1$, and proposition 17.16.6 applies.
The lower bounds follow from $w_ t\leq u_ t^{\mathrm P}$, and the upper bounds from the chord estimate; together they give the asserted singularity type. Off the common poles, put $f(t)=u_ t^{\mathrm P}(x)$. For $0<s<t<1$, monotonicity of the slopes of secants of the convex function $f$ gives
\[ -C\leq \frac{f(s)-f(0)}s \leq \frac{f(t)-f(s)}{t-s} \leq \frac{f(1)-f(t)}{1-t}\leq C. \]
The two-sided estimates at the endpoints cover the cases $s=0$ and $t=1$, which proves the Lipschitz bound on $[0,1]$. $\square$
For endpoints of different types, a common lower potential still gives an admissible constant path, so the envelope exists. Attaining the endpoints requires a barrier approaching the prescribed values; proposition 17.16.6 separates this requirement from the existence of the envelope.
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.