17.13 Ross–Witt Nyström correspondenceC9D15D724069

Proof. Fix $x\in X$. We want to establish (17.13.6) at $x$. We distinguish two cases. First suppose that $\Gamma _{\tau }(x)=-\infty $ for all $\tau <\Gamma _{\max }$ and hence all $\tau \in \mathbb {R}$. In this case, we have $\Gamma _ t^*(x)=-\infty $ for all $t>0$. Therefore, (17.13.6) follows trivially.

Otherwise, by 17.13.4, we know that $\Gamma ^*_ t(x)=\infty $ for all $t<0$. The relative interior of the domain of $t\mapsto \Gamma ^*_ t(x)$ is contained in $(0,\infty )$. Therefore, (17.13.6) follows from theorem 21.11.24, proposition 17.12.16. $\square$

Proof. We first observe that for each $x\in X$, we have

Let $R=\{ a+\mathrm{i}b\in \mathbb {C}:a>0,b\in \mathbb {R}\} $. We consider

Let $\pi \colon X\times R\rightarrow X$ be the natural projection. Observe that the upper semicontinuous regularization $G$ of $F$ is $\pi ^*\theta $-psh by proposition 5.14.2. It suffices to show that $F=G$. We let

We want to argue that $E=\varnothing $. Since $F$ is invariant under translations in the imaginary direction of $R$, the same is true for its upper semicontinuous regularization $G$. So $E$ can be written as $B\times \mathrm{i}\mathbb {R}$ for some set $B\subseteq X\times (0,\infty )$. Since $E$ is a pluripolar set by proposition 5.14.5, it has zero Lebesgue measure. Hence, $B$ has zero Lebesgue measure. For each $x\in X$, write

By Fubini's theorem, $B_ x$ has vanishing $1$-dimensional Lebesgue measure for all $x\in X\setminus Z$, where $Z\subseteq X$ is a subset of measure $0$. We may assume that $Z\supseteq \{ \Gamma _{\Gamma _{\max }}=-\infty \} $ so that for $x\in X\setminus Z$, $\Gamma ^*_ t(x)\neq -\infty $ for all $t>0$.

For any $x\in X\setminus Z$, both $t\mapsto F(x,t)$ and $G(x,t)$ are convex functions with values in $\mathbb {R}$ on $(0,\infty )$. They agree almost everywhere, hence everywhere by their continuity. It follows that for $x\in X\setminus Z$, we have $B_ x=\varnothing $.

By proposition 17.13.5, for any $x\in X$, we have

On the other hand, let

By Kiselman's principle proposition 5.14.15, $\chi _{\tau }\in \mathrm{PSH}(X,\theta )$. But on $X\setminus Z$, we already know that $\Gamma _{\tau }=\chi _{\tau }$ for all $\tau <\Gamma _{\max }$. By proposition 5.14.6,

Now we conclude that $F(x,t)=G(x,t)$ by corollary 21.11.26. $\square$

Proof. Fix $t>0$. We already know that $\Gamma ^*_ t\in \mathrm{PSH}(X,\theta )$ by 17.13.7. It suffices to show that

From (17.13.2) and proposition 16.8.28, we know that

This completes the proof. $\square$

Proof. Since $\Gamma _{\tau }\leq 0$ for all $\tau <\Gamma _{\max }$, we have

for all $t>0$. Conversely, for each $\tau <\Gamma _{\max }$,

Letting $\tau \to \Gamma _{\max }$ gives the reverse inequality. The final assertion follows directly from the formula

This completes the proof. $\square$

Proof. It follows from 17.13.7, (17.13.2) and proposition 5.14.2 that the restriction of $\Gamma ^*$ to each compact subinterval of $(0,\infty )$ is a subgeodesic. By corollary 17.13.9, for any $t>0$, $\Gamma ^*_ t\in \mathcal{E}(X,\theta ;\phi )$.

First observe that as $t\to 0+$, we have

By 17.13.10 and proposition 5.17.3, it suffices to show each $L^1$-cluster point $\psi \in \mathrm{PSH}(X,\theta )$ of $\Gamma _ t^*$ as $t\to 0$ is equal to $\phi $.

To see this, first observe that by (17.13.2), for any fixed $t>0$,

Therefore, $\psi \leq \phi $. On the other hand, for any fixed $\tau <\Gamma _{\max }$, by (17.13.2), we have

for any $t>0$. So $\psi \geq \Gamma _{\tau }$ almost everywhere and hence everywhere by proposition 5.14.6. It follows that $\psi \geq \phi $. Therefore, $\psi =\phi $.

Assume that $\Gamma ^*$ is not a geodesic ray. Then we can find $0\leq a<b$ such that $(\Gamma ^*_ t)_{t\in (a,b)}$ differs from the geodesic $(\eta _ t)_{t\in (a,b)}$ from $\Gamma ^*_ a$ to $\Gamma ^*_ b$. The existence of $(\eta _ t)_ t$ is guaranteed by proposition 17.10.24. We consider the subgeodesic $(\ell _ t)_{t>0}$ given by $\ell _ t=\eta _ t$ for $t\in (a,b)$ and $\ell _ t=\Gamma ^*_ t$ otherwise. Note that $\ell $ is a subgeodesic due to the gluing lemma 17.10.16.

Consider the Legendre transform

Then $\Gamma '_{\tau }\geq \Gamma _{\tau }$ and $\Gamma '_{\tau }\in \mathrm{PSH}(X,\theta )\cup \{ -\infty \} $ by proposition 5.14.15 for all $\tau \in \mathbb {R}$.

We claim that

Observe that $\Gamma '_{\tau }\equiv -\infty $ when $\tau >\Gamma _{\max }$ by 17.13.10. Thus it suffices to consider $\tau \leq \Gamma _{\max }$. Put

For any $s\in [a,b]$, the monotonicity of $t\mapsto \Gamma _ t^*-t\Gamma _{\max }$ from 17.13.10 gives

Since $\ell _ b=\Gamma _ b^*$, we get

For $s\notin [a,b]$, we have $\ell _ s=\Gamma _ s^*$, hence

Taking the infimum over all $s>0$ gives

Therefore, (17.13.13) follows. In particular, for any $\tau <\Gamma _{\max }$, we have $\Gamma '_{\tau }\sim \Gamma _{\tau }$. Moreover, $\sup _ X\ell _ t\leq t\Gamma _{\max }$ for all $t>0$: this is clear outside $(a,b)$ from 17.13.10, and on $(a,b)$ it follows from the convexity of subgeodesics applied to the geodesic $(\eta _ t)_{t\in (a,b)}$. Hence, for $\tau <\Gamma _{\max }$,

Letting $t\to 0+$ gives $\Gamma '_{\tau }\leq 0$. It follows from the fact that $\Gamma _{\tau }$ is a model potential that $\Gamma _{\tau }=\Gamma '_{\tau }$ for all $\tau <\Gamma _{\max }$. Therefore, by theorem 21.11.24, we have $\Gamma ^*_ t=\ell _ t$ for all $t>0$, which is a contradiction. $\square$

Proof. Note that it follows from proposition 5.14.15 that $\ell ^*_{\tau }\in \mathrm{PSH}(X,\theta )\cup \{ -\infty \} $ for all $\tau \in \mathbb {R}$. It is clear that $\mathbb {R}\ni \tau \mapsto \ell ^*_{\tau }$ is a decreasing and concave function.

By proposition 17.10.39,

Set $C=\sup _ X\ell _1$. By proposition 17.10.39, $\sup _ X(\ell _ t-Ct)=0$ for all $t>0$. For each $x$ outside a pluripolar set, the function $t\mapsto \ell _ t(x)-Ct$ is convex on $(0,\infty )$ and bounded above by $0$, hence it is decreasing. By proposition 5.14.6, the same monotonicity holds as an inequality of $\theta $-psh functions. Thus $(\ell _ t-Ct)_{t>0}$ is a decreasing net in $\mathrm{PSH}(X,\theta )$.

Its infimum is not identically $-\infty $, since the family is normalized by $\sup _ X(\ell _ t-Ct)=0$ and is compact in $L^1$ by proposition 5.17.3. Therefore

by proposition 5.14.2.

On the other hand, for $\tau >\sup _ X \ell _1$, the same argument shows that

Therefore, $\ell ^*_{\tau }\in \mathrm{PSH}(X,\theta )$ if and only if $\tau \leq \ell ^*_{\max }\coloneqq \sup _ X \ell _1$.

We claim that $(\ell ^*_{\tau })_{\tau <\ell ^*_{\max }}$ is a test curve. We first observe that for $\tau <\ell ^*_{\max }$, we have

Therefore,

Also observe that for any $\tau \leq \ell ^*_{\max }$ and any $t>0$, we have

Letting $t\to 0+$, we find that for any $\tau \leq \ell ^*_{\max }$, we have

Fix $\tau <\ell ^*_{\max }$. We want to argue that

First we claim that for any $C>0$, we have

The $\leq $ direction is trivial. We argue the reverse inequality, which reduces to

Since $\phi $ is model and $(\ell ^*_{\tau }+C)\land V_{\theta }\leq 0$, it suffices to show that

which follows from (17.13.16). Therefore, (17.13.19) is established. Thanks to (17.13.17), we have the obvious inequality

for any $C>0$. Therefore, in order to prove (17.13.18), it remains to argue that for any $C>0$,

For this purpose, let us consider the following geodesics: For any $M>0$ and $t\in [0,1]$, let

It is clear that at $t=0,1$, we have $\ell ^{2,M}_ t\leq \ell ^{1,M}_ t$. Hence, the same holds for all $t\in [0,1]$. In particular, for any fixed $s\in (0,1]$, we have

for all $M>0$. Taking infimum with respect to $M>0$, we find

Since $s\in (0,1]$ is arbitrary, we conclude (17.13.20).

Finally, we check the condition at $-\infty $. Set

The family $(\ell ^*_{\tau })_{\tau <\ell ^*_{\max }}$ is locally uniformly bounded from above, so $\psi \in \mathrm{PSH}(X,\theta )$. Thanks to corollary 17.10.9, outside a pluripolar set, the function $t\mapsto \ell _ t(x)$ is a finite closed convex function on $[0,\infty )$ with value $\phi (x)$ at $t=0$. Applying the one-variable Legendre duality gives

Hence $\psi =\phi $ by proposition 5.14.6. In particular, $\ell ^*_{-\infty }=\phi $ has positive mass. $\square$

Proof. Step 1. We first establish (17.13.22).

It follows from 17.13.11 that the forward map is well-defined. The inverse map is given by (17.13.14). We show that the inverse map is also well-defined. Given $\ell \in \mathcal{R}(X,\theta ;\phi )$, we know from 17.13.15 that $\ell ^*\in \mathrm{TC}(X,\theta ;\phi )$. We need to show that $\ell ^*\in \mathrm{TC}(X,\theta ;\phi )$.

The equality $\ell ^*_{-\infty }=\phi $ was proved in 17.13.15. Therefore, $\ell ^*\in \mathrm{TC}(X,\theta ;\phi )$ as expected.

The two operations are inverse to each other thanks to corollary 21.11.26. Hence, (17.13.22) is established.

Step 2. Next we consider the bounded situation. Namely, we want to establish the second half of (17.13.23).

Suppose that $\Gamma \in \mathrm{TC}^{\infty }(X,\theta ;\phi )$. Take $\tau _0\in \mathbb {R}$ so that $\Gamma _{\tau }=\phi $ for all $\tau \leq \tau _0$. It follows from (17.13.2) that

for all $t>0$. Therefore, $\Gamma ^*_ t\sim \phi $ for all $t>0$ and hence $\Gamma ^*\in \mathcal{R}^{\infty }(X,\theta ;\phi )$.

Conversely, suppose that $\ell \in \mathcal{R}^{\infty }(X,\theta ;\phi )$. Thanks to proposition 17.10.37, there is a constant $C>0$ such that

Therefore, according to (17.13.14), we have

if $\tau \leq -C$. Therefore, $\ell ^*_{\tau }=\phi $ for all $\tau \leq -C$.

Step 3. We establish (17.13.24) and the first half of (17.13.23).

Step 3.1. We reduce to the case where $\Gamma _{\max }=0$.

Suppose that we define

Then $\Gamma '\in \mathrm{TC}(X,\theta ;\phi )$ as well and for all $t>0$,

Therefore,

by (16.5.13). Using (17.12.12), we also have

Therefore, it suffices to establish (17.13.24) for $\Gamma '$ in place of $\Gamma $.

Step 3.2. We assume that $\Gamma _{\max }=0$ and $\Gamma \in \mathrm{TC}^{\infty }(X,\theta ;\phi )$. We prove (17.13.24).

For $N\in \mathbb {Z}_{>0}$, $M \in \mathbb {Z}$, we introduce the following:

We first claim that for all $t>0$, $N\in \mathbb {Z}_{>0}$ and $M\in \mathbb {Z}_{<0}$,

Assuming this, let us prove (17.13.24).

Fixing $N$, let $M=\lfloor 2^{N}\Gamma _{\min }\rfloor $. Recall that $\Gamma _{\min }$ is defined in (17.12.11). Then repeated applications of (17.13.25) yield

Since $M \leq 2^{N}\Gamma _{\min }$, we have

Using (16.5.13), we can continue to write

We now notice that we have Riemann sums on both the left and right of the above inequality. Moreover, as $N\to \infty $, the potentials $\Gamma _ t^{*,N,0}$ increase to $\Gamma _ t^*$, hence their energies converge to $E^\phi _\theta (\Gamma _ t^*)$ by proposition 16.5.37. Using proposition 17.12.9, it is possible to let $N \to \infty $ and obtain

So (17.13.24) follows as desired.

It remains to argue (17.13.25). Fix $t>0$, $N\in \mathbb {Z}_{>0}$ and $M\in \mathbb {Z}$. By proposition 16.5.36,

Clearly $\Gamma _ t^{*,N,M+1} \geq \Gamma _ t^{*,N,M}$. Moreover, since $\mathbb {R}\ni \tau \mapsto \Gamma _{\tau }+t\tau $ is concave, we notice that

and on $U_ t$ we have

We also note that $U_ t$ is $\mathcal{F}$-open by corollary 5.15.19. So from the lower bound in (17.13.27), we have

where on the second line, we applied (17.13.28) and proposition 13.11.4, on the third line, we applied the fact that $\theta _{\Gamma _{(M+1)/2^ N}}^ n$ is supported on the set (cf. theorem 16.3.12)

When $(M+1)/2^ N=0$, we use proposition 16.4.3 to see that $P_\theta [\Gamma _0]=\Gamma _0$, since $\Gamma _0$ is the decreasing limit of the model potentials $\Gamma _\tau $ as $\tau \nearrow 0$. Thus theorem 16.3.12 applies to this endpoint as well.

We have deduced the first inequality in (17.13.25). Next, we apply the upper bound part in (17.13.27) and compute similarly

We conclude the latter half of (17.13.25).

Step 3.3. We assume that $\Gamma _{\max }=0$. Now $\Gamma \in \mathrm{TC}(X,\theta ;\phi )$ only.

For each $\epsilon >0$, we introduce $\Gamma ^{\epsilon }\in \mathrm{TC}^{\infty }(X,\theta ;\phi )$ as follows:

1. 1E08B0569766

   Let $\Gamma ^{\epsilon }_{\max }=0$, and

2. 70863808F7A7

   we set

   \[ \Gamma ^{\epsilon }_{\tau }= \begin{cases} \phi ,& \textup{if }\tau \leq -\epsilon ^{-1};\\ P_{\theta }\left[ (1+\epsilon \tau )\Gamma _{\tau }-\epsilon \tau \phi \right],& \textup{if }\tau \in \left(-\epsilon ^{-1},0\right). \end{cases} \]

The concavity and monotonicity in $\tau $ follow from proposition 16.4.2. The value at $-\infty $ is $\phi $. Moreover, $\Gamma ^\epsilon _\tau =\phi $ for $\tau \leq -\epsilon ^{-1}$, so $\Gamma ^\epsilon \in \mathrm{TC}^\infty (X,\theta ;\phi )$.

It follows from corollary 16.9.38 and corollary 16.9.14 that for each $\tau <0$, the potentials $\Gamma ^{\epsilon }_{\tau }$ decrease to $\Gamma _{\tau }$ as $\epsilon \searrow 0$. Applying proposition 16.4.4, we have

for all $\tau <0$. Hence, by the monotone convergence theorem and Step 3.2, we find

where the last equality follows from (17.13.26). Furthermore, according to proposition 21.11.36, we have

for all $t>0$. Note that we do not have to take the closure of the right-hand side since it is automatically upper semicontinuous in $t$.

Now suppose that $\Gamma \in \mathrm{TC}^1(X,\theta ;\phi )$. Then by (17.13.29), as $\epsilon \to 0+$, $(\Gamma _ t^{\epsilon *})_{\epsilon }$ is a decreasing Cauchy net in $\mathcal{E}^1(X,\theta ;\phi )$ and hence by theorem 17.10.63 for each $t>0$,

where we have applied (17.13.26) and (17.13.29). Hence, $\Gamma ^*\in \mathcal{R}^1(X,\theta ;\phi )$. Moreover, (17.13.24) follows.

Conversely, suppose that $\Gamma ^*\in \mathcal{R}^1(X,\theta ;\phi )$. Then (17.13.29) implies that

Hence, $\Gamma \in \mathrm{TC}^1(X,\theta ;\phi )$. $\square$

Proof. This follows from 17.13.10 and theorem 17.13.21. $\square$

Proof. Fix $x\in X$, $\tau <\ell ^*_{\max }$ and $f\in \mathcal{O}_{X,x}$. Fix a Kähler form $\omega $ on $X$.

Step 1. We first assume that $f$ lies in the right-hand side of (17.13.36).

Given any $y\in X$, it follows from (17.13.14) that there is $t_0>0$ with

Observe that $t\mapsto \ell _ t-t\ell ^*_{\max }$ is decreasing in $t$ by corollary 17.13.31, it follows that for $t\in [t_0,t_0+1]$, we have

Since $\tau < \ell ^*_{\max }$ and $t_0\geq t-1$, we deduce that

Take a sufficiently small open neighborhood $U$ of $x$ such that

Applying (17.13.37), we deduce that

Therefore, $f\in \mathcal{I}(\ell ^*_{\tau })_ x$.

Step 2. Assume that $f\in \mathcal{I}(\ell ^*_{\tau })_ x$.

It follows from the strong openness theorem theorem 5.16.21 (cf. (17.12.5)) that $f\in \mathcal{I}(\ell ^*_{\tau +\epsilon })_ x$ for some small enough $\epsilon >0$ with $\tau +\epsilon <\ell ^*_{\max }$. Take a sufficiently small open neighborhood $U$ of $x$ such that

We compute

Therefore, $f$ lies in the right-hand side of (17.13.36). $\square$

Proof. Fix $\tau <\ell ^*_{\max }$. We first observe that it suffices to show that

Assuming (17.13.40) for the moment. By theorem 17.13.21, $\ell ^*\in \mathrm{TC}(X,\theta ;\phi )$. In particular, $\ell ^*_\tau $ is a model potential and

Since $\phi $ is model as well, proposition 16.3.41 gives

Since $\ell ^*_{\tau }\leq \phi \leq 0$, we have

Therefore, (17.13.39) follows from (17.13.40).

In order to prove (17.13.40), it suffices to show that

Take $x\in X$ so that $(\ell _ t(x))_{t\geq 0}$ is finite and right-differentiable at $t=0$. Note that this condition holds quasi-everywhere. Suppose that $\ell ^*_{\tau }(x)=\phi (x)$. Then

Therefore, the $\supseteq $ direction in (17.13.41) follows. Conversely, suppose that $\dot{\ell }_0(x)\geq \tau $, then

Together with $\ell ^*_\tau \leq \phi $, this gives $\ell ^*_\tau (x)=\phi (x)$. Therefore, the $\subseteq $ direction in (17.13.41) follows as well. $\square$

Proof. Set $\Gamma =\ell ^*$. By theorem 17.13.21, we have $\Gamma \in \mathrm{TC}^1(X,\theta ;\phi )$ and

By proposition 17.13.38, for every $\tau <\ell ^*_{\max }$,

Hence (17.12.12) gives

For $\tau >\ell ^*_{\max }$, we have $\ell ^*_{\tau }\equiv -\infty $ by 17.13.15. Then (17.13.40) shows that

The layer-cake formula therefore gives (17.13.43). The final assertion follows from proposition 17.10.65, since $E_{\theta }^{\phi }(\ell _0)=E_{\theta }^{\phi }(\phi )=0$. $\square$