Theorem 20.3.669EDE5892D63B

Proof. We first remove the logarithmic divisor. If $v\sim r$, then $\nu (v,D)=b$, so lemma 5.38.1 gives $\theta +\mathrm{dd}^{\mathrm{c}}v-b[D]\geq 0$. Taking $q=v-bG_ p$ in $L^1$ and then its quasi-plurisubharmonic representative gives

Indeed the ordinary inequalities hold off $D$ and extend canonically across it. Conversely such a $q$ gives $v=bG_ p+q$ with

Their locally bounded products agree off $D$, and both extended measures give $D$ zero mass. Thus their extended measures agree everywhere.

If $k>0$, choose $L$ with $\epsilon _ L<k/2$. The potential $bG_ p+U_ L$ belongs to the specified type and satisfies

For necessity, average $q$ over the unitary group acting transitively on the second projective line, with its Haar probability measure. The background is invariant. The average has a genuine first-factor representative $q_0\sim U$ and $kw+\mathrm{dd}^{\mathrm{c}}q_0\geq 0$. One can first average in $L^1$ and then take the canonical representative; the curvature inequality and the ordinary bounds pass to it. The positive curve measure has total mass $k$, excluding $k<0$. When $k=0$, $q_0$ is harmonic on the compact curve and therefore constant, contradicting $q_0\sim U$.

It remains to compute the mass for every competitor, before using the average. Choose a closed Kähler form $K=C(w_1+w_2)\geq \beta $ and a fixed $u_0=U_ L$ as above. The closed positive current $S_0=K+\mathrm{dd}^{\mathrm{c}}u_0$ has no coefficient mass on $D$, since its first-factor curvature has no atom at $p$. The product formula gives

For $S=K+\mathrm{dd}^{\mathrm{c}}q$, both $S$ and $S_0$ are closed positive currents in the same Kähler class, with bounded difference of potentials and small unbounded locus $D$. The classical closed-class comparison of non-pluripolar products, applied in both singularity orders in degrees one and two, gives

Here the degree-one non-pluripolar current is the restriction of $S$ off $D$ and is closed. Hence the positive current $S-\langle S\rangle $ has zero trace against $K$, and vanishes. This proves $\langle S\rangle =S$. Equivalently, absence of polar coefficient mass follows from proposition 5.46.3. The closed comparison is used only for $K$.

Put $R=\beta -K$. Bedford–Taylor expansion off $D$, followed by extension by zero, now yields

The first term is finite by closed comparison, the second has finite variation because $S$ has finite trace, and the last is smooth. Since none has coefficient mass on $D$, this is also an identity on $Y$. Distributional Stokes therefore gives

The pairing uses only $q\in L^1$. Averaging preserves it, since $\mathrm{dd}^{\mathrm{c}}\beta =\mathrm{dd}^{\mathrm{c}}a\wedge w_2$. The last measure is positive of total mass $k$, proving the bounds in (20.3.68) for all original competitors. Substituting $q=v-bG_ p$ and $\int G_ p\mathrm{dd}^{\mathrm{c}}a=a(p)-\int aw$ proves (20.3.67).

To approach either bound, choose a point $x$ where $a$ takes the corresponding extreme value, smooth strictly positive measures $\zeta _ j$ of mass $k$ tending weakly to $k\delta _ x$, and smooth $g_ j$ with $\mathrm{dd}^{\mathrm{c}}g_ j=\zeta _ j-kw$. Choose $L_ j$ so that $\epsilon _{L_ j}$ is less than half the minimum density of $\zeta _ j$ relative to $w$ and $|\int U_{L_ j}\mathrm{dd}^{\mathrm{c}}a|<1/j$. Then

Their masses tend to $2ka(x)$ by (20.3.69). $\square$