17.7 Subgeodesics and Perron segments4C4B950F1CFF

Proof. (1) Suppose that $g(t_0)=-\infty $ for some $t_0\in I$, and let $t\in I\setminus \{ t_0\} $. Choose $s\in I$ such that $t$ lies strictly between $t_0$ and $s$, and write $t=(1-\lambda )t_0+\lambda s$ with $0<\lambda <1$. Since $g(s)<\infty $, convexity gives $g(t)\leq (1-\lambda )g(t_0)+\lambda g(s)=-\infty $. Hence either $g\equiv -\infty $ or $g$ is finite, and a finite convex function on $I$ is continuous by example 3.4.15.

(2) For $a<b<c$ in $I$ we have $\sigma (a,b)\leq \sigma (a,c)\leq \sigma (b,c)$, as shown in example 3.4.15; this gives the monotonicity of $\sigma $. Now let $t\in I$. For $r<t<r'$ in $I$ we have $\sigma (r,t)\leq \sigma (t,r')$, so $c=\sup _{r<t}\sigma (r,t)$ is finite and $c\leq \sigma (t,r')$ for every $r'>t$. These two inequalities say that $g(s)\geq g(t)+c(s-t)$ for $s<t$ and for $s>t$. The affine function $s\mapsto g(t)+c(s-t)$ is therefore dominated by $g$ and agrees with $g$ at $t$, which proves the remaining assertions.

(3) Fix $s\in (0,1)$. For $0<t<s$ we have $g(t)=g(s)-(s-t)\sigma (t,s)$. By (2), $\sigma (t,s)$ decreases as $t$ decreases, so it has a limit $\sigma _0\in [-\infty ,\infty )$ as $t\downarrow 0$. If $\sigma _0=-\infty $, then $g(t)\to +\infty $ as $t\downarrow 0$, which contradicts the upper bound. Hence $\sigma _0$ is finite and $g(t)\to g(s)-s\sigma _0$. The value at $0$ of the affine function that agrees with $g$ at $t$ and $s$ is $g(s)-s\sigma (t,s)$; it increases as $t$ decreases and tends to $g(s)-s\sigma _0$.

(4) Let $\delta \leq a<b\leq 1-\delta $. For $b<r<1$, (2) gives $\sigma (a,b)\leq \sigma (b,r)\leq 2C/(r-b)$, and letting $r\uparrow 1$ we get $\sigma (a,b)\leq 2C/(1-b)\leq 2C/\delta $. Similarly, for $0<r<a$ we have $\sigma (a,b)\geq \sigma (r,a)\geq -2C/(a-r)$, and letting $r\downarrow 0$ we get $\sigma (a,b)\geq -2C/a\geq -2C/\delta $. $\square$

Proof. Let $D\subset X$ be a connected coordinate chart, identified with a domain in $\mathbb C^ n$, on which $\theta =\mathrm{dd}^{\mathrm{c}}h$ for a smooth function $h$; by lemma 6.14.2, such charts, for instance small coordinate balls, cover $X$. Then $V+h\circ p$ is plurisubharmonic on the domain $D\times S\subset \mathbb C^{n+1}$.

(1) Fix $x\in D$. The complex line $\{ x\} \times \mathbb C$ meets $D\times S$ in $\{ x\} \times S$, so by definition 5.8.1 the function $\zeta \mapsto V(x,\zeta )+h(x)$ is subharmonic on $S$, possibly identically $-\infty $; hence so is $w(\zeta )=g(\operatorname{Re}\zeta )$, where $g(t)=v_ t(x)$. If $w\equiv -\infty $, then $g\equiv -\infty $ is convex. Otherwise $w$ is locally integrable on $S$ by proposition 3.4.8, and $\Delta w\geq 0$ by theorem 3.4.12. For $0<\delta <\frac12$, the mollification $w_\delta =w*\rho _\delta $ is smooth on $S_\delta =\{ \delta <\operatorname{Re}\zeta <1-\delta \} $, and $\Delta w_\delta =(\Delta w)_\delta \geq 0$ there, by proposition 2.5.3(1),(3). Since $w$ is invariant under imaginary translations, so is $w_\delta $; thus $w_\delta (\zeta )=g_\delta (\operatorname{Re}\zeta )$ for the smooth function $g_\delta (t)=w_\delta (t)$ on $(\delta ,1-\delta )$, and $g_\delta ''=\Delta w_\delta \geq 0$, so $g_\delta $ is convex by the mean value theorem. By (3.4.13), $w_\delta $ decreases to $w$ as $\delta \downarrow 0$, so $g_\delta $ decreases to $g$ on $(0,1)$. For $a,b\in (0,1)$ and $0<\lambda <1$, letting $\delta \downarrow 0$ in

shows that $g$ is convex. The rest of (1) is lemma 17.7.2(1).

(2) The plurisubharmonic function $V+h\circ p$ is locally integrable on $D\times S$ (theorem 5.9.1). Let $K\subset D$ be compact and $[a,b]\subset (0,1)$. Since $V$ does not depend on $\operatorname{Im}\zeta $, Tonelli's theorem on $K\times \{ \zeta :a\leq \operatorname{Re}\zeta \leq b,\ 0\leq \operatorname{Im}\zeta \leq 1\} $ gives

Hence, for almost every $x\in K$, the value $v_ t(x)$ is finite for almost every $t\in [a,b]$, and by (1) the function $t\mapsto v_ t(x)$ is then finite on $(0,1)$. Since $X$ is covered by countably many such compact sets $K$, this proves (2).

(3) Fix $t\in (0,1)$. The restriction of $V+h\circ p$ to $D\times \{ t\} $ is the function $v_ t+h$ on $D$. It is upper semicontinuous, and, by definition 5.8.1 applied to the complex lines contained in $\mathbb C^ n\times \{ t\} $, its restriction to every connected component of the intersection of $D$ with a complex line is subharmonic or identically $-\infty $. By (2), $v_ t$ is finite almost everywhere on $D$, so $v_ t+h$ is not identically $-\infty $, and $v_ t+h\in \mathrm{PSH}(D)$. Since such charts cover $X$, $v_ t\in \mathrm{PSH}(X,\theta )$. $\square$

Proof. By lemma 17.7.3, for every $x\in X$ the function $t\mapsto u_ t(x)$ is convex on $(0,1)$, and either identically $-\infty $ or finite and continuous; it is finite for almost every $x$; and every slice $u_ t$ is $\theta $-psh. Since a sequence converging in $L^1(X)$ has a subsequence converging almost everywhere, we can choose sequences $t_ j\downarrow 0$ and $s_ j\uparrow 1$ with $u_{t_ j}\to u_0$ and $u_{s_ j}\to u_1$ almost everywhere. For fixed $t$ and large $j$, convexity on $[t_ j,s_ j]$ gives

and letting $j\to \infty $ gives $u_ t\leq (1-t)u_0+tu_1$ almost everywhere. Both sides are $\theta $-psh, so the inequality holds everywhere.

Let $D$ be a connected coordinate chart of $X$ on which $\theta =\mathrm{dd}^{\mathrm{c}}h$ with $h$ smooth. The function $U+h\circ p$ is plurisubharmonic on $D\times S=D\times ((0,1)+\mathrm i\mathbb R)$ and invariant under imaginary translations, so corollary 5.12.2, with $V=(0,1)$, shows that $w+h=\inf _{0<t<1}(u_ t+h)$ is plurisubharmonic or identically $-\infty $ on $D$. For almost every $x$, the function $t\mapsto u_ t(x)$ is finite and convex, hence bounded below on $(0,1)$ by lemma 17.7.2(2). Thus $w$ is finite almost everywhere, the alternative $-\infty $ is excluded on every chart, and $w$ is $\theta $-psh. Since $w\leq u_ t$ for all $t$ and $u_ t\to u_0$, $u_ t\to u_1$ in $L^1(X)$, we get $w\leq u_0$ and $w\leq u_1$ almost everywhere, hence everywhere. Thus $w$ is a competitor for the obstacle $\min \{ u_0,u_1\} $, and $w\leq P_\theta (u_0,u_1)$.

The bounds $w\leq u_ t\leq \max \{ u_0,u_1\} $ hold for all $t$, with integrable outer functions. For almost every $x$ the function $t\mapsto u_ t(x)$ is finite and continuous on $(0,1)$, so dominated convergence gives continuity of $t\mapsto u_ t$ from $(0,1)$ to $L^1(X)$; continuity at the endpoints is part of the definition. The image of $[0,1]$ under a continuous map is compact. Now fix $x\in X$. If $t\mapsto u_ t(x)$ is identically $-\infty $, its limits at $0$ and $1$ are $-\infty $. Otherwise it is finite and convex and bounded above by $\max \{ u_0(x),u_1(x)\} <\infty $, so by lemma 17.7.2(3), applied to $t\mapsto u_ t(x)$ and to $t\mapsto u_{1-t}(x)$, it has finite limits at $0$ and $1$. By the displayed bound these limits are at most $u_0(x)$ and $u_1(x)$. Since the pointwise limits exist everywhere and $u_{t_ j}\to u_0$, $u_{s_ j}\to u_1$ almost everywhere, they equal $u_0$ and $u_1$ almost everywhere.

It remains to prove equality quasi-everywhere; we treat the endpoint $0$, and the endpoint $1$ follows by reversing the parameter. Fix $s\in (0,1)$ and a sequence $t_ j\downarrow 0$ with $t_ j<s/2$. Consider the value at $t=0$ of the chord of $t\mapsto u_ t$ through the parameters $t_ j$ and $s$:

Where $u_ s$ is finite, the function $t\mapsto u_ t(x)$ is finite and convex, so by lemma 17.7.2(3) $a_ j$ increases with $j$ and tends to $\lim _{t\to 0}u_ t$; this limit equals $u_0$ almost everywhere and is at most $u_0$ everywhere. The functions

are combinations of $u_{t_ j}$ and $u_ s$ with positive coefficients of sum two, hence $2\theta $-psh, since $2\theta +\mathrm{dd}^{\mathrm{c}}(\alpha u_{t_ j}+\beta u_ s)=\alpha \theta _{u_{t_ j}}+\beta \theta _{u_ s}$ for $\alpha ,\beta \geq 0$ with $\alpha +\beta =2$. They increase with $j$ and are bounded above by $u_0+u_ s$. Their pointwise limit is $\lim _{t\to 0}u_ t+u_ s$, which equals $u_0+u_ s$ almost everywhere; hence their regularized upper envelope, a $2\theta $-psh function by theorem 5.10.2 which agrees almost everywhere with the pointwise limit, is $u_0+u_ s$ itself. By corollary 5.47.11(2), applied in charts after adding smooth local potentials of $2\theta $, the pointwise limit $\lim _{t\to 0}u_ t+u_ s$ equals $u_0+u_ s$ quasi-everywhere. Removing the pole set of $u_ s$ proves the assertion at $0$. $\square$

Proof. Let $0\leq a<b\leq 1$. The restriction of $(u_ t)$ to $[a,b]$, reparametrized as $t\mapsto u_{a+(b-a)t}$, has the complexification $U\circ F$, where $F(x,\zeta )=(x,a+(b-a)\zeta )$ is a holomorphic map of $X\times S$ into itself with $p\circ F=p$. In a chart $D$ of $X$ on which $\theta =\mathrm{dd}^{\mathrm{c}}h$, we have $(U+h\circ p)\circ F=U\circ F+h\circ p$, which is plurisubharmonic on $D\times S$ by proposition 5.9.3; it is not identically $-\infty $, because its slices are the functions $u_{a+(b-a)t}+h$, which are not identically $-\infty $ by lemma 17.7.3. Hence $U\circ F$ is $p^*\theta $-psh, and the restriction is a subgeodesic with endpoint potentials $u_ a$ and $u_ b$, by the $L^1$ continuity of proposition 17.7.4. An affine function $c_0+c_1\operatorname{Re}\zeta =\operatorname{Re}(c_0+c_1\zeta )$ is pluriharmonic, so adding it preserves $p^*\theta $-plurisubharmonicity, and the change of background replaces $U$ by $U-g\circ p$ and $p^*\theta $ by $p^*\theta +\mathrm{dd}^{\mathrm{c}}(g\circ p)$; this proves the first assertion.

For the replacement, let $U$ be the complexification of the given subgeodesic, and $V$ that of the new one on $W=X\times \{ a<\operatorname{Re}\zeta <b\} $. We verify the boundary condition of proposition 5.9.5 along each seam $\operatorname{Re}\zeta =a$ or $\operatorname{Re}\zeta =b$ lying in $S$; consider $a>0$. Let $(x_ k,\zeta _ k)\in W$ converge to $(x,\zeta )$ with $\operatorname{Re}\zeta =a$, and put $t_ k=\operatorname{Re}\zeta _ k$. Denote the slices of the new subgeodesic by $v_ t$, $a<t<b$; they converge to $u_ a$ in $L^1(X)$ as $t\downarrow a$.

We show that $\varlimsup _ kV(x_ k,\zeta _ k)\leq U(x,\zeta )=u_ a(x)$. Since $v_{t_ k}\to u_ a$ in $L^1(X)$, lemma 17.4.3 shows that the tail envelopes $w_ l=(\sup _{k\geq l}v_{t_ k})^*$ are $\theta $-psh and decrease pointwise to $u_ a$. For $k\geq l$ we have $V(x_ k,\zeta _ k)=v_{t_ k}(x_ k)\leq w_ l(x_ k)$, and $w_ l$ is upper semicontinuous, so $\varlimsup _ kV(x_ k,\zeta _ k)\leq w_ l(x)$ for every $l$. Letting $l\to \infty $ proves the claim. A seam $\operatorname{Re}\zeta =b$ with $b<1$ is treated in the same way, with $t_ k\uparrow b$. Hence, by proposition 5.9.5 applied in charts after adding smooth local potentials of $p^*\theta $, the function equal to $\max \{ U,V\} =V$ on $W$ and to $U$ elsewhere is $p^*\theta $-psh. Its endpoint limits at $t=0$ and $t=1$ are those of the given subgeodesic, or those of the new one if $a=0$ or $b=1$; in either case they are $u_0$ and $u_1$. $\square$

Proof. Denote by $U^\alpha $ the complexification of $(u_ t^\alpha )$. By proposition 17.7.4, $u_ t^\alpha \leq (1-t)u_0^\alpha +tu_1^\alpha $, so the functions $U^\alpha $ are uniformly bounded above. Their regularized upper envelope $U$ is therefore $p^*\theta $-psh (theorem 5.10.2, applied in charts after adding smooth local potentials of $p^*\theta $), and it is invariant under imaginary translations, because the family is and these translations are homeomorphisms. Thus $U$ is the complexification of the family of its slices $U_ t$, which by lemma 17.7.3 are $\theta $-psh and have convex fibers $t\mapsto U_ t(x)$.

We next reduce to a countable subfamily. Cover $X$ by finitely many coordinate balls $D_\beta $ on which $\theta $ has smooth potentials $h_\beta $ (lemma 6.14.2). By theorem 5.10.2, applied to the plurisubharmonic functions $U^\alpha +h_\beta \circ p$ on $D_\beta \times S$ and to the functions $u_0^\alpha +h_\beta $ and $u_1^\alpha +h_\beta $ on $D_\beta $, there are countable sets $J_\beta \subset A$ such that, on $D_\beta \times S$ and on $D_\beta $ respectively, the regularized suprema over $\alpha \in J_\beta $ of $U^\alpha $, of $u_0^\alpha $ and of $u_1^\alpha $ equal $U$, $v_0$ and $v_1$. The countable set $J=\bigcup _\beta J_\beta $ then satisfies $(\sup _{\alpha \in J}U^\alpha )^*=U$, $(\sup _{\alpha \in J}u_0^\alpha )^*=v_0$ and $(\sup _{\alpha \in J}u_1^\alpha )^*=v_1$, since on each $D_\beta \times S$, respectively $D_\beta $, these regularized suprema lie between the regularized suprema over $J_\beta $ and over $A$.

Let $v_ t=(\sup _\alpha u_ t^\alpha )^*$; we show that $U_ t=v_ t$. Since $U_ t$ is upper semicontinuous on $X$ and dominates every $u_ t^\alpha $, we have $v_ t\leq U_ t$. For $a<t<b$, the inequality $u_ t^\alpha \leq \frac{b-t}{b-a}u_ a^\alpha +\frac{t-a}{b-a}u_ b^\alpha \leq \frac{b-t}{b-a}v_ a+\frac{t-a}{b-a}v_ b$ and upper semicontinuity of the right side give $v_ t\leq \frac{b-t}{b-a}v_ a+\frac{t-a}{b-a}v_ b$. Thus $t\mapsto v_ t(x)$ is convex for every $x$. In the same way, the bound $u_ t^\alpha \leq (1-t)u_0^\alpha +tu_1^\alpha $ gives $v_ t\leq (1-t)v_0+tv_1$. Fix one member $\alpha _0$ of the family and let $w_0=\inf _ tu_ t^{\alpha _0}$, a $\theta $-psh potential by proposition 17.7.4. Since $w_0\leq v_ t\leq U_ t$ and $U$ is bounded above, all $v_ t(x)$ and $U_ t(x)$ are finite on the set where $w_0$ is finite, which has full measure.

The function $F_ J=\sup _{\alpha \in J}U^\alpha $ is Borel, being a countable supremum of upper semicontinuous functions, and by theorem 5.10.2 it agrees with its regularization $U$ outside a Lebesgue null set. Both $F_ J$ and $U$ are invariant under imaginary translations, so this exceptional set is $\{ (x,\zeta ):(x,\operatorname{Re}\zeta )\in N\} $ for a Borel set $N\subset X\times (0,1)$, which is null by Fubini's theorem. Applying Fubini's theorem to $N$, we find that for almost every $x$ we have $U_ t(x)=\sup _{\alpha \in J}u_ t^\alpha (x)\leq v_ t(x)$, hence $U_ t(x)=v_ t(x)$, for almost every $t\in (0,1)$. If moreover $w_0(x)$ is finite, both sides are finite convex functions of $t$, hence continuous (lemma 17.7.2(1)), so they agree for every $t\in (0,1)$. Thus, for each $t$, $U_ t=v_ t$ almost everywhere, and both are $\theta $-psh, so $U_ t=v_ t$ everywhere.

For the endpoints, note first that, since $U_ t=v_ t\leq (1-t)v_0+tv_1\leq \max \{ v_0,v_1\} $, lemma 17.7.2(1),(3) show that $\lim _{t\to 0}U_ t(x)$ exists at every $x$, as an element of $[-\infty ,\infty )$. For $\alpha \in J$ we have $U_ t\geq u_ t^\alpha $, so by proposition 17.7.4 $\lim _{t\to 0}U_ t\geq \lim _{t\to 0}u_ t^\alpha =u_0^\alpha $ outside a pluripolar set. Outside a countable union of pluripolar sets this holds for all $\alpha \in J$, so $\lim _{t\to 0}U_ t\geq \sup _{\alpha \in J}u_0^\alpha =v_0$ quasi-everywhere, by corollary 5.47.11(1); countable unions of pluripolar sets are pluripolar by theorem 5.47.8. The bound $U_ t=v_ t\leq (1-t)v_0+tv_1$ gives the reverse inequality. Hence $U_ t\to v_0$ almost everywhere as $t\to 0$. Since $w_0\leq U_ t\leq \max \{ v_0,v_1\} $, dominated convergence gives $U_ t\to v_0$ in $L^1(X)$. The endpoint $t=1$ is treated in the same way, after reversing the parameter. $\square$

Proof. The rooftop $w=P_\theta (u_0,u_1)$ belongs to $\mathcal E(X,\theta ,\phi )$, by proposition 17.4.7. The constant curve equal to $w$ is a competitor. By lemma 17.7.6, the Perron segment is therefore a subgeodesic, with endpoint potentials $a_0\leq u_0$ and $a_1\leq u_1$, and $w\leq G_ t\leq (1-t)a_0+ta_1\leq (1-t)u_0+tu_1$. Each slice is at most $\phi +C$ for some constant $C$ and at least $w$, so mass monotonicity puts it in $\mathcal E(X,\theta ,\phi )$.

We show that $a_0=u_0$. For $k\geq 1$ the rooftop $q_ k=P_\theta (u_0,u_1+k)$ is not identically $-\infty $, since it dominates $w$, and the affine curve $q_ k-kt$ is a competitor: its value at $t=0$ is at most $u_0$, and its value at $t=1$ is $q_ k-k\leq u_1$. Hence $G_ t\geq q_ k-kt$ for $0<t<1$, and letting $t\to 0$ in $L^1(X)$ gives $a_0\geq q_ k$ almost everywhere, hence everywhere, for every $k$. By the rooftop representation of proposition 16.25.2 and corollary 17.4.8,

Since $a_0$ is upper semicontinuous and dominates every $q_ k$, it dominates $(\sup _ kq_ k)^*=u_0$; hence $a_0=u_0$. Exchanging the roles of the endpoints, with the competitors $P_\theta (u_0+k,u_1)-k(1-t)$, gives $a_1=u_1$.

Let $[a,b]\subset [0,1]$. The restriction of the weak geodesic to $[a,b]$ is a subgeodesic between the slices $G_ a,G_ b\in \mathcal E(X,\theta ,\phi )$. By the part already proved, the weak geodesic between $G_ a$ and $G_ b$ exists and dominates this restriction. By lemma 17.7.5, replacing the restriction by this weak geodesic gives a subgeodesic on $[0,1]$ with endpoints $u_0,u_1$. It is a competitor for the original Perron segment, so its slices are at most $G_ t$; hence the two coincide on $[a,b]$.

Finally, if $u_0,u_1\in \mathcal E^1(X,\theta ,\phi )$, then $w\in \mathcal E^1(X,\theta ,\phi )$ by proposition 17.4.7, and $w\leq G_ t\preceq \phi $ gives $G_ t\in \mathcal E^1(X,\theta ,\phi )$ by corollary 17.3.17. If $u_0,u_1\in \mathcal E^\infty (X,\theta ,\phi )$, then $\phi -C$ is a common minorant for some $C$, so $G_ t\geq w\geq \phi -C$, while the convex upper bound gives $G_ t\leq \phi +C$. $\square$

Proof. Assertion (1) holds because enlarging the endpoints enlarges the family of competitors, and (2) because adding the affine function $(1-t)c_0+tc_1$, respectively reversing the parameter, is a bijection between the families of competitors.

For (3), the geodesics $G^ k=G(u_0^ k,u_1^ k)$ decrease with $k$ by (1) and dominate $G=G(u_0,u_1)$. Their complexifications decrease to a function which dominates the complexification of $G$ and so is not identically $-\infty $ on any chart; by theorem 5.10.1, applied in charts after adding smooth local potentials of $p^*\theta $, it is $p^*\theta $-psh. It is invariant under imaginary translations, and its slices $\widetilde G_ t$ are the pointwise limits of $G^ k_ t$. For every $k$ we have $G_ t\leq \widetilde G_ t\leq G^ k_ t\leq (1-t)u_0^ k+tu_1^ k$, so $G_ t\leq \widetilde G_ t\leq (1-t)u_0+tu_1$. As $t\to 0$, both outer functions converge to $u_0$ in $L^1(X)$, hence so does $\widetilde G_ t$; similarly at $t=1$. Thus $(\widetilde G_ t)$ is a competitor for the Perron segment between $u_0$ and $u_1$, so $\widetilde G\leq G$, and therefore $\widetilde G=G$.

For (4), let $u_1\leq u_0$ and $0<r<s<1$, and put $\delta =(s-r)/(1-r)$. The reparametrized curve $t\mapsto G_{\delta +(1-\delta )t}$ is a subgeodesic by lemma 17.7.5. Its value at $t=0$ is $G_\delta \leq (1-\delta )u_0+\delta u_1\leq u_0$, and its value at $t=1$ is $u_1$. It is therefore a competitor for the segment between $u_0$ and $u_1$, and at $t=r$, where $\delta +(1-\delta )r=s$, it gives $G_ s\leq G_ r$. For the endpoints, $G_ t\leq (1-t)u_0+tu_1\leq u_0$, and the constant curve $u_1$ is a competitor, so $G_ t\geq u_1$.

For (5), note that $u_1\leq u_0+C$ and $u_0\leq u_1+C$ everywhere, because inequalities between quasi-psh functions that hold quasi-everywhere hold everywhere (section 16.11). Hence the affine curves $u_0-Ct$ and $u_1-C(1-t)$ are competitors, so that $G_ t\geq \max \{ u_0-Ct,u_1-C(1-t)\} $. On the other hand $G_ t-u_0\leq t(u_1-u_0)\leq Ct$ and $G_ t-u_1\leq (1-t)(u_0-u_1)\leq C(1-t)$ by the upper bound of theorem 17.7.8. For (6), the upper bound $u_0+tc$ is itself an affine competitor. $\square$

Proof. For each $s$ the potential $u_1-s$ belongs to $\mathcal E(X,\theta ,\phi )$, so the rooftop $P_\theta (u_0,u_1-s)$ is not identically $-\infty $, by proposition 17.4.7, and the affine curve $P_\theta (u_0,u_1-s)+ts$ is a competitor: its value at $t=0$ is at most $u_0$ and its value at $t=1$ is at most $u_1$. Hence the right side is at most $G_ t$.

Conversely, let $(v_ t)$ be a competitor and $s\in \mathbb R$. By corollary 5.12.2, applied with the affine function $\ell (t)=ts$ in charts after adding smooth local potentials of $p^*\theta $, the function $q_ s=\inf _{0<t<1}(v_ t-ts)$ is $\theta $-psh or identically $-\infty $ on each chart. By lemma 17.7.3, for almost every $x$ the function $t\mapsto v_ t(x)$ is finite and convex, hence bounded below on $(0,1)$ by an affine function (lemma 17.7.2(2)), so $q_ s$ is finite almost everywhere and therefore $\theta $-psh. Letting $t\to 0$ and $t\to 1$ in $q_ s\leq v_ t-ts$, and using the $L^1$ convergence of $v_ t$ to its endpoint potentials, which are at most $u_0$ and $u_1$, we get $q_ s\leq u_0$ and $q_ s\leq u_1-s$ almost everywhere, hence everywhere, and therefore $q_ s\leq P_\theta (u_0,u_1-s)$.

Now fix $x$ and $t_0\in (0,1)$. By lemma 17.7.3(1), the fiber $t\mapsto v_ t(x)$ is either identically $-\infty $, in which case $v_{t_0}(x)=-\infty $, or finite. In the second case lemma 17.7.2(2) gives $c\in \mathbb R$ with $v_ t(x)\geq v_{t_0}(x)+c(t-t_0)$ for all $t$, so that $q_ c(x)=\inf _ t(v_ t(x)-ct)=v_{t_0}(x)-ct_0$, and hence $v_{t_0}(x)=q_ c(x)+ct_0\leq \sup _ s(q_ s(x)+t_0s)$. In either case $v_{t_0}(x)\leq \sup _ s\{ P_\theta (u_0,u_1-s)(x)+t_0s\} $, since $q_ s\leq P_\theta (u_0,u_1-s)$. Taking the regularized supremum over all competitors and using lemma 17.7.6 proves the formula. The same lemma, applied to the affine competitors, gives the last assertion. $\square$