Theorem 17.17.5548B64628B04

Proof. By theorem 5.41.9, $U_ t$ is a subgeodesic. The inequalities

give $\sup _ XU_ t\geq t\tau $ for each $\tau <c$, since $\sup _ X\Gamma _\tau =0$, and $\sup _ XU_ t\leq ct$. Letting $\tau \uparrow c$ proves $\sup _ XU_ t=ct$. The formula

shows that $U_ t-ct$ is decreasing in $t$.

Since $\Gamma _\tau \preceq U_ t\preceq \varphi $, monotonicity of the singularity envelope gives

Taking the regularized supremum in $\tau $ proves $P_\theta [U_ t]=\varphi $.

The local envelope theorem in section 5.40 gives $\sup _{\tau <c}\Gamma _\tau =\varphi $ almost everywhere, and at those points the bounds $\Gamma _\tau +t\tau \leq U_ t\leq \varphi +ct$ imply $U_ t\to \varphi $ as $t\downarrow 0$. For any fixed $\tau _0<c$ and $0<t\leq 1$, they also give

Both bounds are integrable, so dominated convergence gives $U_ t\to \varphi $ in $L^1$ as $t\downarrow 0$.

It remains to prove Perron maximality. Fix $0\leq a<b$ and replace $U_ t$ on $(a,b)$ by the Perron segment from $U_ a$ to $U_ b$, with $U_0=\varphi $ when $a=0$; it exists by proposition 17.16.6, because the restriction of $U_ t$ itself runs between these endpoints. The joint upper limits of the Perron segment at $a$ and $b$ allow gluing by lemma 5.24.2, proposition 17.16.3. The resulting subgeodesic $v_ t$ dominates $U_ t$ and equals it outside $(a,b)$, where therefore $v_ t\leq ct$. On $(a,b)$, the chord bound (17.16.4) together with $U_ a\leq ca$ and $U_ b\leq cb$ gives $v_ t\leq ct$ as well.

For $\tau <c$, put $\chi _\tau =\inf _{t>0}(v_ t-t\tau )$. Since $v_ t-t\tau \geq U_ t-t\tau \geq \Gamma _\tau $, proposition 5.41.3 gives $\chi _\tau \in \mathrm{PSH}(X,\theta )$. We claim that

The first inequality was just noted. For the second, let $s>0$. If $s\in [a,b]$, then $v_ b=U_ b$, and since $U_ t-ct$ is decreasing,

If $s\notin [a,b]$, then $\chi _\tau \leq v_ s-s\tau =U_ s-s\tau $. In both cases $\chi _\tau \leq U_ s-s\tau +(b-a)(c-\tau )$. Taking the infimum over $s>0$ and using lemma 5.41.8 proves (17.17.6).

Moreover, $v_ t\leq ct$ gives $\chi _\tau \leq t(c-\tau )$ for every $t>0$, hence $\chi _\tau \leq 0$. By (17.17.6), $\chi _\tau $ has the ordinary type of $\Gamma _\tau $, so it is a competitor for $P_\theta [\Gamma _\tau ]=\Gamma _\tau $. Thus $\chi _\tau =\Gamma _\tau $. Recovering $v_ t$ from its inverse transforms by lemma 5.41.5 gives $v_ t=U_ t$, which proves Perron maximality on every finite interval. $\square$