Proposition 17.10.65.0C8741A27906 Let $(\varphi _ t)_{t\in [a,b]}$ be a geodesic in $\mathcal{E}^1(X,\theta ;\phi )$. Then $t\mapsto E_{\theta }^{\phi }(\varphi _ t)$ is a linear function on $[a,b]$.
Proposition 17.10.650C8741A27906
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
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17.10.66
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
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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,
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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 $\mathbb {P}^1$
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The integration-by-parts formula [Xia19b], applied against a smooth test function on $\mathbb {P}^1$, gives
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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
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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 $\beta ^2=0$,
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Passing to the increasing truncation limit yields
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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 $A_0$
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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
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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,
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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. $\square$
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