17.9 Extremal slopes and long Perron segments66FE6F0ACABE

Proof. The function $v-u$ is finite almost everywhere, and $v\preceq u$ gives $v-u\leq C'$ for some constant $C'$; hence $C$ is finite. The inequality $v\leq u+C$ holds almost everywhere, hence everywhere, as recalled in section 16.11. Thus the supremum of $v-u$ over the common finite locus is at most $C$; it is at least $C$ because the common finite locus has full measure. The same argument applies to an infimum, and to the complement of any pluripolar set, which still has full measure. The same remarks apply to $G_ t-u$.

By theorem 17.7.8, $G_ t\leq (1-t)u+tv\leq u+tC$. The affine curve $v-C(1-t)$ is a competitor, since $v-C\leq u$, so $G_ t\geq v-(1-t)C$. On the common finite locus $v-u=(v-G_ t)+(G_ t-u)$, and taking essential suprema gives $C\leq (1-t)C+\operatorname *{ess\, sup}_ X(G_ t-u)$. Together with the first bound, this proves the equality for the supremum.

Now let $u\sim v$. The affine curve $u+ct$ is a competitor, since $u+c\leq v$, so $G_ t-u\geq ct$. Applying the equality for the supremum to the reversed segment, $G_ t(u,v)=G_{1-t}(v,u)$, gives $\operatorname *{ess\, sup}_ X(G_ t-v)=(1-t)\operatorname *{ess\, sup}_ X(u-v)=-(1-t)c$, that is, $\operatorname *{ess\, inf}_ X(v-G_ t)=(1-t)c$. Taking essential infima in $v-u=(v-G_ t)+(G_ t-u)$ gives $c\geq (1-t)c+\operatorname *{ess\, inf}_ X(G_ t-u)$, which proves the opposite bound. $\square$

Proof. Since $\phi -T\leq u_ T\leq \phi $, the potential $u_ T$ has bounded difference from $\phi $; in particular $u_ T\in \mathcal E(X,\theta ,\phi )$. By theorem 17.7.8, the Perron segment from $\phi $ to $u_ T$ is a weak geodesic. We regard it as a curve on $[0,T]$, so that its slice at time $s$ is $L_ s^{u,T}$.

Bounds. For $0\leq s\leq T$ put $q_ s=\max \{ u,\phi -s\} $. After the affine change of parameter, the complexification of the curve $(q_ s)_{0\leq s\leq T}$ is $(x,\zeta )\mapsto \max \{ u(x),\phi (x)-T\operatorname{Re}\zeta \} $. This is the maximum of two $p^*\theta $-psh functions, because $\operatorname{Re}\zeta $ is pluriharmonic, and hence it is $p^*\theta $-psh (proposition 5.9.5, applied after adding smooth local potentials of $p^*\theta $, which exist by lemma 5.4.2). Moreover $q_0=\phi $, since $u\leq \phi $, and $q_ T=u_ T$. As $|q_ s-q_ t|\leq |s-t|$ on the finite locus of $\phi $, the curve tends to $\phi $ and $u_ T$ in $L^1$ at the two ends. Thus $(q_ s)$ is a subgeodesic with endpoints $\phi $ and $u_ T$, and by the definition of the Perron segment, $L_ s^{u,T}\geq q_ s\geq \phi -s$. On the other hand, theorem 17.7.8 gives $L_ s^{u,T}\leq (1-s/T)\phi +(s/T)u_ T\leq \phi $.

Monotonicity in $T$. Let $T'>T$. The same subgeodesic, now on $[0,T']$, shows that $L_ T^{u,T'}\geq q_ T=u_ T$. By the last assertion of theorem 17.7.8, the restriction of the segment of length $T'$ to $[0,T]$ is the weak geodesic from $\phi $ to $L_ T^{u,T'}$. Since weak geodesics increase with their endpoints (proposition 17.7.9(1)), this restriction dominates the weak geodesic from $\phi $ to $u_ T$. Hence $L_ s^{u,T'}\geq L_ s^{u,T}$ for $0\leq s\leq T$.

The limit. Fix $s$. The family $(L_ s^{u,T})_{T>s}$ is increasing, bounded above by $\phi $ and bounded below by $\phi -s$. By theorem 5.10.2, applied after adding smooth local potentials of $\theta $, its regularized supremum $R_ s^ u$ is $\theta $-psh, and $\phi -s\leq R_ s^ u\leq \phi $. Since the family is increasing, the supremum may be taken over $T\geq A$ for any $A>s$ without changing $R_ s^ u$. For $s=0$, all the potentials $L_0^{u,T}$ and $R_0^ u$ equal $\phi $.

Comparison. Suppose that $v\leq \phi $ and $u\leq v+C$ with $C\geq 0$. Then

By the monotonicity of weak geodesics in their endpoints and the formula for constant shifts of the endpoints, proposition 17.7.9(1) and (2), we have $G_ t(\phi ,u_ T)\leq G_ t(\phi ,v_ T+C)=G_ t(\phi ,v_ T)+tC$ for $0\leq t\leq 1$. Taking $t=s/T$ gives the bound $L_ s^{u,T}\leq L_ s^{v,T}+Cs/T$. Now fix $s$ and $A>s$. For $T\geq A$ we get $L_ s^{u,T}\leq R_ s^ v+Cs/A$, and taking the regularized supremum over $T\geq A$ gives

Letting $A\to \infty $ yields $R_ s^ u\leq R_ s^ v$.

If $u\sim v$, then $u\leq v+C$ and $v\leq u+C$ for some $C\geq 0$, and the comparison applied in both directions gives $R_ s^ u=R_ s^ v$. If $u\preceq \phi $ is arbitrary, any two constants $c,c'$ with $u-c\leq \phi $ and $u-c'\leq \phi $ give potentials $u-c\sim u-c'$, so $R_ s^{u-c}=R_ s^{u-c'}$. Finally, if $u\preceq v\preceq \phi $, choose constants $c,c'$ with $u-c\leq \phi $ and $v-c'\leq \phi $. Then $u-c\leq v-c'+C$ for some $C\geq 0$, and the comparison gives $R_ s^ u=R_ s^{u-c}\leq R_ s^{v-c'}=R_ s^ v$. $\square$