17.17 Test curves and Legendre dualityD5DBDCE92D6A

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. $\square$

Proof. By lemma 5.41.1, for each $x$ the function $t\mapsto u_ t(x)-ct$ is either finite and convex or identically $-\infty $. It is bounded above by zero, so it is decreasing. Since $\sup _ X(u_ t-ct)=0$, theorem 5.25.2, corollary 5.25.3 give

For $\tau <c$, we have $\widehat u_ c\leq \widehat u_\tau \leq u_1-\tau $, so proposition 5.41.3 gives $\widehat u_\tau \in \mathrm{PSH}(X,\theta )$. Also

while the bound $\widehat u_\tau \leq t(c-\tau )$ with $t\to \infty $ shows $\widehat u_\tau \equiv -\infty $ for $\tau >c$. Concavity and monotonicity in $\tau $ hold because $\widehat u_\tau $ is an infimum of affine functions of $\tau $ with negative slopes $-t$.

Fix $\tau <c$ and write $g=\widehat u_\tau $. The inequality $g\leq u_ M-M\tau $, valid for every $M>0$, gives

For $C>0$ put $q_ C=P_\theta (g+C,0)$. Since $g\leq 0$, the function $g$ is a competitor, so $q_ C\not\equiv -\infty $, and

The affine path $q_ C-Ct/M$ is admissible for the endpoints $\varphi $ and $u_ M-M\tau $ on $[0,M]$. Adding $-t\tau $ gives a bijection between the competitor families for the endpoints $\varphi ,u_ M$ and $\varphi ,u_ M-M\tau $, by proposition 17.16.2, so the Perron segment for the latter endpoints is $u_ t-t\tau $. For $0<s\leq 1$ this yields

Taking the infimum over $M>0$ gives $q_ C-Cs\leq g$, and letting $s\downarrow 0$ gives $q_ C\leq g$. Thus $q_ C=g$ for every $C>0$, and letting $C\to \infty $ in the definition of $P_\theta [g]$ proves $P_\theta [g]=g$.

Let $\psi =(\sup _{\tau <c}\widehat u_\tau )^*$; since $\widehat u_\tau =P_\theta [\widehat u_\tau ]\leq \varphi $, we have $\psi \leq \varphi $. Because $u_ t-ct$ is decreasing in $t$, the limit of $u_ t(x)$ as $t\downarrow 0$ exists at every $x$, and an almost everywhere convergent sequence given by the $L^1$ convergence $u_ t\to \varphi $ identifies it with $\varphi $ almost everywhere. At these points, the endpoint identity in lemma 5.41.5 gives

Therefore $\psi =\varphi $ by lemma 5.24.3.

As $\tau \uparrow c$, the transforms decrease to $\widehat u_ c$, since the infima over $t>0$ and over $\tau <c$ commute; this is the convention for the value at the cutoff. Finally, lemma 5.41.5 recovers $u_ t$ from its inverse transforms, also at points where $t\mapsto u_ t(x)$ is identically $-\infty $, and lemma 5.41.8 recovers each test curve from its forward transform. Hence the correspondence is a bijection. $\square$

Proof. Fix $0<\lambda <1$. For $s=\log |z|^2$, let $q_\lambda (s)=\Gamma _\lambda (z)+\log (1+\mathrm{e}^ s)$. In the first region of (17.17.10) this is the affine function $\lambda s+H(\lambda )$, which is tangent to $\log (1+\mathrm{e}^ s)$ at $s_\lambda =\log (\lambda /(1-\lambda ))$, and in the second region it is $\log (1+\mathrm{e}^ s)$. Its derivative is continuous at the junction and nondecreasing, so $q_\lambda $ is convex. By corollary 5.41.2, $\Gamma _\lambda $ is $\theta $-plurisubharmonic away from zero, and it extends across its logarithmic pole by lemma 5.24.7. It vanishes near infinity and has the ordinary type of $\lambda g$.

To prove $\Gamma _\lambda =P_\theta [\lambda g]$, let $v\leq 0$ be a competitor with $v\preceq \lambda g$. On the punctured disk $0<|z|^2<\lambda /(1-\lambda )$, the function

is subharmonic, since $v+\log (1+|z|^2)$ is subharmonic and $\log |z|^2$ is harmonic there. Since $v\preceq \lambda g$, this difference is bounded above near zero, so it extends subharmonically across zero by lemma 5.24.7. Its upper limits at the boundary circle are at most zero, since $v\leq 0$ and $\Gamma _\lambda =0$ there. The maximum principle gives $v\leq \Gamma _\lambda $ on the disk (at zero both have the value $-\infty $), and outside the disk the same inequality follows from $\Gamma _\lambda =0$. Thus every competitor lies below $\Gamma _\lambda $, which is itself a competitor, so $\Gamma _\lambda =P_\theta [\lambda g]$. Since the envelope depends only on the ordinary type, also $P_\theta [\Gamma _\lambda ]=\Gamma _\lambda $. The zero potential is an envelope fixed point as well.

For a fixed $0<|z|<\infty $, put $p=|z|^2/(1+|z|^2)$. The function $\lambda \mapsto \Gamma _\lambda (z)$ is zero for $\lambda \leq p$. For $p<\lambda <1$ its derivative is

It vanishes at $\lambda =p$ and decreases in $\lambda $, which proves that $\lambda \mapsto \Gamma _\lambda (z)$ is concave and decreasing. At $z=0$, the function is zero for $\lambda \leq 0$ and $-\infty $ for $\lambda >0$; at $z=\infty $, it is zero for $\lambda \leq 1$. These functions are concave and decreasing as well. Finally, $\sup _{\lambda <1}\Gamma _\lambda =0$, and the limit as $\lambda \uparrow 1$ is $g$, so the family is an envelope test curve with cutoff one.

For finite nonzero $z$ and $t>0$, the supremum defining $U_ t(z)$ is attained at a parameter $\lambda >p$, namely at

and substituting this value into the first line of (17.17.10) gives (17.17.12). At $z=0$, only the parameters $\lambda \leq 0$ give finite values, and the supremum is zero; at $z=\infty $, the supremum of $t\lambda $ over $\lambda <1$ is $t$. In the coordinate $w=1/z$, the formula reads $\log ((|w|^2+\mathrm e^ t)/(|w|^2+1))$, so the ray is smooth in both charts. $\square$