PomeTheorem 17.17.5.548B64628B04 The forward transform of an envelope test curve with cutoff $c$ is an envelope ray with the same initial potential and slope $c$. Moreover, $U_ t-ct$ decreases as $t$ increases.
17.17.4 Perron maximalityFA178FC8EDED
The slope and the singularity envelope of the transformed ray follow directly from the normalization of the test curve. To prove Perron maximality, we replace the transform on a finite interval by the Perron segment with the same endpoints. The estimate (17.17.6) keeps each inverse transform in its original ordinary singularity type, where the envelope condition forces equality.
Proof. By theorem 5.41.9, $U_ t$ is a subgeodesic. The inequalities
\[ \Gamma _\tau +t\tau \leq U_ t\leq \varphi +ct \]
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
\[ U_ t-ct=\sup _{\tau <c}\bigl (\Gamma _\tau -t(c-\tau )\bigr ) \]
shows that $U_ t-ct$ is decreasing in $t$.
Since $\Gamma _\tau \preceq U_ t\preceq \varphi $, monotonicity of the singularity envelope gives
\[ \Gamma _\tau =P_\theta [\Gamma _\tau ]\leq P_\theta [U_ t] \leq P_\theta [\varphi ]=\varphi . \]
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
\[ \Gamma _{\tau _0}-|\tau _0|\leq U_ t\leq \varphi +|c|. \]
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
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17.17.6
\begin{equation} \Gamma _\tau \leq \chi _\tau \leq \Gamma _\tau +(b-a)(c-\tau ). \end{equation}
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,
\[ \chi _\tau \leq U_ b-b\tau \leq U_ s-s\tau +(b-s)(c-\tau ). \]
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. $\square$
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