Proposition 20.3.74E91643D06F3E

Proof. Put $\beta =(c-b)w_1+aw_2$. The representatives of lemma 20.3.65 show that $U$ satisfies (18.27.1); by lemma 18.27.2 so does $\pi ^*U$. The logarithmic divisor is $b\pi ^*D$, whose smooth curvature is $b\pi ^*w_1$, so its residual background is $\pi ^*\beta $. If a strict current existed in the ordinary type of $\pi ^*r$, theorem 18.27.4 would give a smooth $f$ with $\pi ^*\beta +\mathrm{dd}^{\mathrm{c}}f>0$. The exceptional curve $E$ of the last blowup is contracted by the composite map. Restriction to this smooth curve would give $\mathrm{dd}^{\mathrm{c}}(f|_ E)>0$, contrary to its zero integral by Stokes. Only the smooth function $f$ is restricted, so this argument applies also when the original potential has a pole along $E$.

For the one-sided assertion, start with the strict representative $v_0=bG_ p+U_ L$ from theorem 20.3.66. Suppose $v_ j$ is strict with lower bound $\delta _ jh_ j>0$ on the current surface, and let $\pi _{j+1}$ be the next point blowup. For its exceptional divisor $E_{j+1}$, choose a smooth metric on $\mathcal O(E_{j+1})$ with curvature $\eta _{j+1}$ negative along the vertical tangent, and write $\mathrm{dd}^{\mathrm{c}}\ell _{j+1}=[E_{j+1}]-\eta _{j+1}$. For all sufficiently small $t_ j>0$, $\pi _{j+1}^*h_ j-t_ j\eta _{j+1}>0$. To see this, in a blowup chart $(x,z)\mapsto (x,xz)$ the pullback has a uniformly positive horizontal coefficient, mixed coefficient $O(|x|)$ and Schur complement at least $c_1|x|^2$. The perturbation adds a positive vertical term at least $t_ jc_2$, while its error in the Schur complement is bounded by $C(t_ j|x|+t_ j|x|^2+t_ j^2)$. Shrink the exceptional neighborhood and then choose $t_ j$ small; on its compact complement positivity persists under small perturbations. Thus

is strict. These genuine quasi-plurisubharmonic pullbacks are defined even over pole centers. The logarithmic terms are bounded above, so induction gives $v'\leq \pi ^*r+C$. Finally proposition 20.3.73 preserves both volumes in the pulled-back ordinary type. The additional exceptional poles of the one-sided witness are not competitors in that fixed type. $\square$