17.10.40 The metrics on the spaces of potentials and rays770FD61FCF7B

Proof. Step 1. We first assume that $\varphi \geq \psi $. Then

Step 2. We prove the general case.

By Step 1, we have

Adding the two inequalities together, we conclude (17.10.44). $\square$

Proof. There are two facts to prove, as in the two steps below.

Step 1. Let $\varphi ,\psi \in \mathcal{E}^1(X,\theta ;\phi )$. Assume that $d_1(\varphi ,\psi )=0$. Then we will show that $\varphi =\psi $.

We first treat the case $\psi \leq \varphi $. Then $d_1(\varphi ,\psi )=E_\theta ^\phi (\varphi )-E_\theta ^\phi (\psi )=0$, and proposition 16.5.30 gives

We conclude that $\psi =\varphi $ using the domination principle theorem 13.13.43.

In general, put $\gamma =\varphi \land \psi $. Since

both terms on the right vanish. Applying the comparable case to the pairs $(\varphi ,\gamma )$ and $(\psi ,\gamma )$ gives $\varphi =\gamma =\psi $.

Step 2. Let $\varphi ,\psi ,\gamma \in \mathcal{E}^1(X,\theta ;\phi )$. We prove the triangle inequality:

This can be translated to

We just have to compute using proposition 17.10.43:

where the second line follows from 16.5.14. $\square$

Proof. By the plurifine locality of non-pluripolar products proposition 13.11.4 1, we can write

The desired equality follows by comparing these terms. $\square$

Proof. Step 1. We first prove the right-hand part of (17.10.49).

Thanks to proposition 16.5.36 and 16.3.3, we have

Similarly,

Adding these inequalities up, we find

Step 2. We prove the left-hand part of (17.10.49).

We claim that

For this purpose, we compute directly

and (17.10.50) follows.

We now estimate the left-hand side of (17.10.50):

where the second inequality follows from proposition 16.5.36.

Similarly,

where the third inequality follows from the multi-linearity of the non-pluripolar product proposition 13.11.4 6.

Adding these estimates up, we find

Therefore, combined with (17.10.50) we conclude

By symmetry, we also get a similar expression after exchanging $\varphi $ and $\psi $. Adding these inequalities together, we find

The left-hand part of (17.10.49) then follows. $\square$

Proof. By theorem 17.10.48 and 17.10.47, we have

This completes the proof. $\square$

Proof. When $\sup _ X \varphi \leq 0$, the right-hand part of (17.10.54) is trivial. Since $\varphi \leq \phi $, we find

where the last equality follows from proposition 16.3.21.

We can therefore assume that $\sup _ X \varphi > 0$. In this case, the left-hand part of (17.10.54) is trivial. Take a Kähler form $\omega \geq \theta $ on $X$. Then thanks to theorem 16.3.12, we can find a constant $C>0$ so that

Thanks to proposition 5.17.3, there is a constant $C'>0$, independent of the choice of $\varphi $, so that

We estimate

The right-hand part of (17.10.54) then follows from theorem 17.10.48. $\square$

Proof. We compute

This completes the proof. $\square$

Proof. The direct implication is trivial.

Conversely, assume that both conditions are true. Let $(x_ j)_{j>0}$ be a Cauchy sequence in $E$. We want to prove that $(x_ j)_ j$ converges. By passing to a subsequence, we may assume that

For $k,j\geq 1$, let

Then $(y_ j^ k)_ j$ is decreasing, and

So $(y_ j^ k)_ j$ is a decreasing Cauchy sequence. Define

Then

We also need to check that the sequence $(y^ k)_ k$ is increasing. For each $j\geq 1$, we have

Equivalently,

Letting $j\to \infty $ and using the continuity of $\land $ from 17.10.58, we obtain

Thus $y^ k\leq y^{k+1}$. So $(y^ k)_ k$ is an increasing Cauchy sequence. Let

For every $j\geq 0$, we have

with the empty sum understood as $0$. Indeed, this is trivial for $j=0$, and the induction step follows from

and

Letting $j\to \infty $, we get

So $(x_ k)_ k$ converges to $y$. $\square$

Proof. As we have seen in 17.10.45 and proposition 17.10.43, the triple

is a rooftop metric space. Hence thanks to proposition 17.10.59, it remains to show that each increasing or decreasing Cauchy sequence in $\mathcal{E}^1(X,\theta ;\phi )$ converges.

We first consider an increasing sequence $(\varphi _ i)_{i>0}$ in $\mathcal{E}^1(X,\theta ;\phi )$. The Cauchy property simply means that

We claim that

is the $d_1$-limit of the sequence. We first observe that $(\sup _ X \varphi _ i)_{i>0}$ is bounded, as a consequence of 17.10.53. Therefore, proposition 16.5.53 guarantees that $\varphi \in \mathcal{E}^1(X,\theta ;\phi )$, and hence our assertion follows from proposition 16.5.37.

Next we consider a decreasing sequence $(\varphi _ i)_{i>0}$ in $\mathcal{E}^1(X,\theta ;\phi )$. The Cauchy property simply means that

We claim that

is the $d_1$-limit of the sequence. Equivalently, we need to show that $\varphi \in \mathcal{E}^1(X,\theta ;\phi )$ and

We first observe that $(\sup _ X \varphi _ i)_{i>0}$ is bounded, as a consequence of 17.10.53. Therefore, $\varphi \in \mathrm{PSH}(X,\theta ;\phi )$ and $\varphi _ i\xlongrightarrow {L^1} \varphi $, see corollary 5.17.5. We claim that $\varphi \in \mathcal{E}^1(X,\theta ;\phi )$. For this purpose, we may assume that $\varphi _1\leq 0$. Since $(\varphi _ i)_ i$ is decreasing, $\varphi _ i\leq \varphi _1\leq 0$ for all $i>0$. As $\varphi _ i\preceq \phi $ and $\phi $ is model, each $\varphi _ i$ is a candidate for $P_\theta [\phi ]=\phi $, so $\varphi _ i\leq \phi $. Thus, using the energy formula (16.5.11) and the fact that all mixed terms are non-positive, we get for each $i>0$,

We can then apply 16.5.45, with $\gamma =\phi $, to conclude that $\varphi \in \mathcal{E}^1(X,\theta ;\phi )$. Our assertion (17.10.61) then follows from proposition 16.5.20. We conclude the proof. $\square$

Proof. We first note that as a consequence of 17.10.53, $(\sup _ X \varphi _ i)_{i>0}$ is bounded. Let $C_ n$ be the constant in theorem 17.10.48. After replacing $(\varphi _ i)_{i>0}$ by a subsequence, we may assume that

We first construct $(\psi _ i)_{i>0}$. For this purpose, it suffices to define

for each $i>0$. It follows from proposition 16.5.53 that $\psi _ i\in \mathcal{E}^1(X,\theta ;\phi )$ for each $i>0$. For $k\geq 0$, set

Iterating corollary 17.10.51 gives

Since $\psi _ i^ k$ increases almost everywhere to $\psi _ i$, letting $k\to \infty $ and using proposition 16.5.37, we get

Thus $\psi _ i\to \varphi $ in $d_1$. Since $(\psi _ i)_ i$ is decreasing, the monotone case in the proof of theorem 17.10.60 shows that $\psi _ i$ converges almost everywhere to $\varphi $, hence pointwise by proposition 5.14.6.

Next we construct $(\eta _ i)_{i>0}$. For each $i>0$ and $k\geq 0$, we let

Then

Since $\eta _ i^ k\leq \varphi _ i$, the preceding estimate gives

The sequence $(\eta _ i^ k)_{k\geq 0}$ is decreasing, and the monotonicity of $E_{\theta }^{\phi }$ shows that the sequence $\left(E_{\theta }^{\phi }(\eta _ i^ k)\right)_{k\geq 0}$ is decreasing and bounded from below, hence convergent. Thus, for $k\leq l$,

and $(\eta _ i^ k)_{k\geq 0}$ is a decreasing Cauchy sequence. Therefore, the monotone case in the proof of theorem 17.10.60 shows that

is the $d_1$-limit of $(\eta _ i^ k)_{k\geq 0}$ and proposition 16.5.37 shows that

By construction, $\eta _ i\leq \varphi _ i$ and $(\eta _ i)_ i$ is increasing. Therefore, $\varphi $ is the $d_1$-limit of the increasing sequence $(\eta _ i)_{i>0}$. As we have seen in the proof of theorem 17.10.60, this implies that $\varphi $ is also the almost everywhere limit of $(\eta _ i)_{i>0}$. $\square$

Proof. Let $(\varphi _ j)_{j>0}$ be a sequence in $\mathcal{E}^1(X,\theta ;\phi )$ with $d_1$-limit $\varphi \in \mathcal{E}^1(X,\theta ;\phi )$. We wish to show that

It suffices to prove that every subsequence has a further subsequence along which (17.10.64) holds. Thus, after replacing $(\varphi _ j)_{j>0}$ by a subsequence, let $(\eta _ j)_ j$ and $(\psi _ j)_ j$ be the monotone sequences supplied by 17.10.62. From the construction, $\eta _ j\leq \varphi _ j\leq \psi _ j$. By proposition 16.5.37,

The monotonicity of $E_{\theta }^{\phi }$ then squeezes $E_{\theta }^{\phi }(\varphi _ j)$, proving (17.10.64). $\square$

Proof. By an affine change of parameter, we may assume that $[a,b]=[0,1]$.

Step 1. Assume first that $\varphi _0,\varphi _1\in \mathcal{E}^{\infty }(X,\theta ;\phi )$. Let $Y=X\times \mathbb {P}^1$, and let $p_1,p_2$ be the two projections. On

the function $U$ is $p_1^*\theta $-psh. Choose $C>0$ such that

The constant subgeodesic at $\phi -C$ and proposition 17.10.7 give

Fix concentric annuli $A_0\Subset A_1\Subset A$ and a unit-mass Fubini–Study form $\omega _{\mathrm{FS}}$ on $\mathbb {P}^1$. Put

The class represented by $\Theta _ B$ is big. Since $\phi =P_\theta [\phi ]$, we have $\phi \leq 0$. For every $B>0$, $\Phi =P_{\Theta _ B}[\Phi ]$. Indeed, if $\Psi \in \mathrm{PSH}(Y,\Theta _ B)$ satisfies $\Psi \leq 0$ and $\Psi \preceq \Phi $, then on each fibre $X\times \{ z\} $, either $\Psi _ z\equiv -\infty $, or $\Psi _ z\in \mathrm{PSH}(X,\theta )$ is a candidate in the definition of $P_\theta [\phi ]=\phi $. Thus $\Psi _ z\leq \phi $ in both cases. Conversely, $\Phi $ itself is a candidate. Moreover, proposition 13.11.4 6 gives

Choose an $S^1$-invariant smooth function $q$ on $\mathbb {P}^1$ such that

After increasing $B$, we may assume that $B\omega _{\mathrm{FS}}+\mathrm{dd}^{\mathrm{c}}q\geq 0$. Identifying $q$ with $p_2^*q$, define

The gluing lemma corollary 5.14.20 shows that $W\in \mathrm{PSH}(Y,\Theta _ B)$. By construction,

for some $C'>0$. Hence $W\in \mathcal{E}^{\infty }(Y,\Theta _ B;\Phi )$ by (17.10.67).

Set

Define the order-zero current on $\mathbb {P}^1$

The integration-by-parts formula [Xia19b], applied against a smooth test function on $\mathbb {P}^1$, gives

By (16.5.11), (17.10.68), and corollary 13.11.15, on $A_0$ the current $\mathcal{H}_ W$ is represented almost everywhere by

This representative is continuous. Indeed, on every smaller annulus, the convex functions $t\mapsto \varphi _ t(x)-\phi (x)$, considered on the common locus where both values are finite, are uniformly bounded by (17.10.66), hence uniformly Lipschitz. The resulting inequalities between the potentials hold almost everywhere and therefore everywhere by proposition 5.14.6. The monotonicity in 16.5.14 and (16.5.13) then give the local Lipschitz continuity of $t\mapsto E_\theta ^\phi (\varphi _ t)$.

On $X\times A_0$, define the signed Radon measure

This measure is positive. To see this, take a relatively compact product coordinate ball $Q\subset X\times A_0$. Let $a$ be a smooth local potential of $p_1^*\theta $, put $h=U+a\leq 0$ and $h_ k=h\lor (-k)$, and choose a smooth local potential $b$ of $B\beta $ with $|b|\leq C_ Q$. For $p=n,n+1$, set

Plurifine locality and $|b|\leq C_ Q$ give, for $k>C_ Q$,

Thus the two cofinal truncation families have the same limit. By proposition 13.11.4 6 and $\beta ^2=0$,

Passing to the increasing truncation limit yields

By (13.11.17),

since $W_ z\sim \phi $ on every fibre and proposition 16.3.10 applies. On the other hand,

by proposition 13.11.4 6. Together with (17.10.70) and (17.10.71), these identities give on $A_0$

Step 2. We prove that $\mu _ U=0$. Let $D\Subset X\times A$ be a coordinate domain, let $a$ be a smooth local potential of $p_1^*\theta $ near $\overline D$, and set $h=U+a$. If $v$ is a psh function on $D$ with the boundary inequality required in the definition of maximality, max-pasting gives

The function $\widetilde U$ need not be radial. Its $S^1$-symmetrization

is radial and $p_1^*\theta $-psh. The $S^1$-saturation of $D$ is relatively compact in $X\times A$, so $V=U$ near $X\times \partial A$. Hence the family $V_ t(x)=V(x,\mathrm{e}^{-t})$ is a subgeodesic from $\varphi _0$ to $\varphi _1$ by definition 17.10.2. The definition (17.10.21) therefore gives

It follows that $v\leq h$ on $D$, so $h$ is a maximal psh function.

On a product coordinate ball $Q\Subset X\times A_0$, put $O_ k=Q\cap \{ h>-k\} $. By [Proposition 5.9.3, EKF], $h$ is $\mathcal F$-maximal. The function $h_ k=h\lor (-k)$ is locally bounded and agrees with $h$ on the $\mathcal F$-open set $O_ k$, hence is $\mathcal F$-maximal there. Therefore [Theorem 5.10.1, EKF] gives

By (17.10.73), $\mu _ U=0$. Since $A_0\Subset A$ was arbitrary, (17.10.74) shows that $z\mapsto E_\theta ^\phi (\varphi _{-\log |z|})$ is radial and harmonic on $A$. Thus $t\mapsto E_\theta ^\phi (\varphi _ t)$ is affine on $(0,1)$.

Choose $C_0>0$ such that $|\varphi _1-\varphi _0|\leq C_0$ on their common finite locus; equivalently, the two resulting inequalities between the potentials hold everywhere. Affine barriers and proposition 17.10.7 give

By 16.5.14 and (16.5.13), the energy converges to the corresponding endpoint energy as $t\to 0+$ and $t\to 1-$. Hence

when the endpoints belong to $\mathcal E^\infty (X,\theta ;\phi )$.

Step 3. Let now $\varphi _0,\varphi _1\in \mathcal E^1(X,\theta ;\phi )$. By example 17.10.23 and (16.5.13), we may add different constants to the endpoints and assume that $\varphi _0,\varphi _1\leq \phi $. Put

and let $(\varphi _ t^{(k)})_{t\in [0,1]}$ be the corresponding geodesic. The endpoint truncations decrease to $\varphi _ j$, and Perron monotonicity gives

Let $h_ t=\lim _{k\to \infty }\varphi _ t^{(k)}$. The decreasing limit of the complexifications is $p_1^*\theta $-psh and is not identically $-\infty $. Taking the limit in the convex upper bound gives, first almost everywhere and then everywhere by proposition 5.14.6,

Note that

Since $\varphi _ t\to \varphi _0$ in $L^1$ as $t\to 0+$ and $\varphi _0,\varphi _1\in L^1(X)$, it follows that $h_ t\to \varphi _0$ in $L^1$. The same argument at $t=1$ gives $h_ t\to \varphi _1$ in $L^1$. Together with the $p_1^*\theta $-psh complexification constructed above, these endpoint limits show by definition 17.10.2 that $(h_ t)$ is a subgeodesic from $\varphi _0$ to $\varphi _1$. Hence (17.10.21) gives $h_ t\leq \varphi _ t$. Together with (17.10.76), this proves

Applying proposition 16.5.37 at the endpoints and at each interior slice, and then using (17.10.75), gives

This concludes the proof. $\square$

Proof. Put

and let $(g_ t)_{t\in (0,1)}$ be the geodesic from $\varphi _0$ to $\varphi _1$. By proposition 16.5.53, $\varphi _0,\varphi _1\in \mathcal E^1(X,\theta ;\phi )$. The slice families are uniformly bounded from above by proposition 17.10.7; hence $h_ t\in \mathcal E^1(X,\theta ;\phi )$ by proposition 16.5.53. Moreover, proposition 17.10.13 shows that $(h_ t)$ is a subgeodesic from $\varphi _0$ to $\varphi _1$. Thus

For each $t$, the net $(\varphi _ t^ i)_{i\in I}$ is increasing. By proposition 5.14.5, its pointwise supremum agrees with $h_ t$ almost everywhere, and the analogous assertion holds at the endpoints. Therefore proposition 16.5.37, proposition 17.10.65 gives

By (17.10.42), (17.10.78), and 17.10.45,

so $h_ t=g_ t$ for all $t\in (0,1)$. $\square$

Proof. By definition of $d_1$ and proposition 17.10.65, it suffices to show the concavity of

Let $(\eta _ t)_{t\in [0,1]}$ be the geodesic from $\varphi _0\land \psi _0$ to $\varphi _1\land \psi _1$. Then for each $t\in [0,1]$, we have $\eta _ t\leq \varphi _ t\land \psi _ t$. Hence, by the monotonicity of $E_\theta ^\phi $ and proposition 17.10.65,

for each $t\in [0,1]$. Applying the same argument to the restrictions of $(\varphi _ t)_ t$ and $(\psi _ t)_ t$ to any subinterval of $[0,1]$, which are again geodesics by proposition 17.10.24, gives a similar inequality on every subinterval. Hence $t\mapsto E_\theta ^\phi (\varphi _ t\land \psi _ t)$ is concave, and our assertion follows. $\square$

Proof. Symmetry and the triangle inequality follow from the corresponding pointwise properties of $d_1$ on $\mathcal{E}^1(X,\theta ;\phi )$.

We next show that $d_1(\ell ,\ell ')<\infty $ for any $\ell ,\ell '\in \mathcal{R}^1(X,\theta ;\phi )$. Let $C_\ell $ be the constant in proposition 17.10.39, and set $\tilde{\ell }_ t=\ell _ t-C_\ell t$. Then $\tilde{\ell }$ is again a geodesic ray, $\sup _ X\tilde{\ell }_ t=0$, and hence $\tilde{\ell }_ t\leq \phi $ by proposition 16.3.21. Thus, by proposition 17.10.65,

Moreover, by (16.5.13), $d_1(\ell _ t,\tilde{\ell }_ t)=|C_\ell |t\int _ X\theta _{\phi }^ n$. Hence $d_1(\ell _ t,\phi )$ grows at most linearly in $t$. The same argument applies to $\ell '$, and the triangle inequality gives $d_1(\ell ,\ell ')<\infty $.

It remains to prove separation: suppose that $\ell ,\ell '\in \mathcal{R}^1(X,\theta ;\phi )$ and $d_1(\ell ,\ell ')=0$.

Fix $s>0$. Then it follows from proposition 17.10.79 that

Therefore, $\ell _ s=\ell '_ s$ and hence $\ell =\ell '$. $\square$

Proof. By proposition 17.10.65, the energy is linear along finite energy rays. Hence

Since $\ell _ t\leq \ell '_ t$ for all $t\geq 0$, (17.10.42) gives

Letting $t\to \infty $ proves the assertion. $\square$

Proof. For each $t>0$, let $(\ell ''^ t_{s})_{s\in [0,t]}$ be the geodesic from $\phi $ to $\ell _ t\lor \ell '_ t$.

Step 1. We first show that for each fixed $s\geq 0$, $\ell ''^ t_{s}$ is increasing in $t\in [s,\infty )$.

To see this, fix $s\geq 0$ and choose $t'>t\geq s$. We need to show that

Since $(\ell ''^{t'}_ a)_{a\in [0,t]}$ is a geodesic, it suffices to show that $(\ell ''^ t_ a)_{a\in [0,t]}$ is a candidate in the Perron envelope defining the former geodesic. In other words, in verifying (17.10.88), we may assume that either $s=0$ or $s=t$. The case $s=0$ is of course trivial. So it remains to prove the following:

By symmetry, it suffices to prove

But since $(\ell _ a)_{a\in [0,t']}$ is a candidate in the Perron envelope defining $\ell ''^{t'}$, this inequality follows.

Step 2. Next, observe that for a fixed $s\geq 0$, we have

for all $t>s$. The right-hand side is bounded from above by a constant independent of $t\geq s$ by proposition 17.10.39. Let

Then proposition 17.10.77 guarantees that $\ell \lor \ell '\in \mathcal{R}^{\infty }(X,\theta ;\phi )$.

Step 3. We need to show that $\ell \lor \ell '$ defined in this way is indeed the minimal ray lying above $\ell $ and $\ell '$.

First, by Step 1, we have

for any $t\geq s\geq 0$. Therefore,

for all $s\geq 0$. In other words, $\ell \lor \ell '\geq \ell $. Similarly, $\ell \lor \ell '\geq \ell '$.

Next, let $L\in \mathcal{R}^{\infty }(X,\theta ;\phi )$ be a ray lying above both $\ell $ and $\ell '$. Then we have

for all $t\geq 0$. In particular,

for all $t\geq s\geq 0$. It follows that

for all $s\geq 0$.

Step 4. It remains to argue (17.10.87). Since the energy is linear along bounded rays,

Moreover, $\ell ''^ t_1$ increases to $(\ell \lor \ell ')_1$ as $t\to \infty $. By proposition 16.5.37,

Finally, by proposition 17.10.65 applied to $(\ell ''^ t_ s)_{s\in [0,t]}$,

Combining these identities gives (17.10.87). $\square$

Proof. The first inequality is trivial. As for the second, we estimate

By symmetry, we find

By theorem 17.10.48 and 17.10.47, for each $t>0$,

Now (17.10.91) follows. $\square$

Proof. We first show that for each fixed $t\geq 0$, $\ell _ t^{\varphi ,C}$ is increasing in $C\geq t$.

To see this, choose $t\leq C_1<C_2$. We need to show that

Since both sides are geodesics for $t\in [0,C_1]$, it suffices to show that

Now $((V_{\theta }-t)\lor \varphi )_{t\in [0,C_2]}$ is a subgeodesic from $V_{\theta }$ to $(V_{\theta }-C_2)\lor \varphi $ by proposition 17.10.13.⁴ It is therefore a candidate in the Perron envelope defining $\ell ^{\varphi ,C_2}$, and evaluating at $t=C_1$ gives exactly (17.10.95).

From proposition 17.10.7, we know that for any $C>t>0$, we have

while the linear subgeodesic $V_\theta -t$ is a candidate in the same Perron envelope, so $V_\theta -t\leq \ell ^{\varphi ,C}_ t$. Therefore, by proposition 5.14.2,

for all $t\geq 0$. Thanks to proposition 16.5.37, we have

as $C\to \infty $ for all $t\geq 0$. Applying proposition 17.10.77 on each interval $[0,T]$ to the increasing family $(\ell ^{\varphi ,C}_ t)_{t\in [0,T]}$, $C\geq T$, shows that $\ell ^{\varphi }\in \mathcal{R}^{\infty }(X,\theta )$.

It remains to compute the energy of $\ell ^{\varphi }$. We first fix $C\geq t>0$ and compute using proposition 17.10.65:

Letting $C\to \infty $ and applying proposition 16.5.37, we find that

for any $t\geq 0$. It follows that

Using the definition of $E_{\theta }$, in order to obtain (17.10.94), it suffices to show that for each $j=0,\ldots ,n$, we have

For this purpose, for each $C>0$, we decompose $X$ as $\{ \varphi > V_{\theta }-C\} $ and $\{ \varphi \leq V_{\theta }-C\} $. We have

On the other hand,

Observe that for $C>0$, the function $\mathds {1}_{\{ \varphi >V_{\theta }-C\} }C^{-1}(\varphi -V_{\theta })$ is defined quasi-everywhere. Since $\varphi \leq V_\theta $, it takes values in $[-1,0]$ outside a pluripolar set. When $C\to \infty $, these functions converge to $0$ quasi-everywhere, so the first integral above converges to $0$ by dominated convergence. Similarly, the indicators $\mathds {1}_{\{ \varphi >V_\theta -C\} }$ increase to $1$ outside a pluripolar set, hence

Therefore, (17.10.97) follows. $\square$