Proposition 17.9.2A6533ED6CC68

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 $R_ s^ u=R_ s^{u-c}\leq R_ s^{v-c'}=R_ s^ v$. $\square$