Theorem 23.1.56F1398FE81D8D

Proof. First select the correction with additional local control. Put

Then $t(x_{\rm pre})=\eta ^3$, $x_{\rm pre}\to \infty $, and $b-x_{\rm pre}=O(\sqrt{\eta L})\to 0$. For a fixed large $K$ the preceding defect estimates give $d_\eta \leq (K/2)\eta $ on $[0,x_{\rm pre}]$: on the linear portion $q^2t\leq 4L\eta ^3$, and the exponential error is $O(\exp (-1/(C\kappa \eta ^2L)))=o(\eta )$. On the initial portion use the previously proved $C\eta $ bound. The integral of $d_\eta $ on $[x_{\rm pre},b]$ tends to zero by the same integrated estimates and this interval's length.

Choose a fixed left cutoff, zero for $x\leq -1/2$ and one for $x\geq -1/4$, and a translated right cutoff, one for $x\leq x_{\rm pre}$ and zero for $x\geq x_{\rm pre}+1/4$. Let $r^0_\eta $ be $K\eta $ times their product. It strictly exceeds $d_\eta $ on $[0,x_{\rm pre}]$. The continuous function $e_\eta =\max (d_\eta -r^0_\eta ,0)$ is supported in $[x_{\rm pre},b]$. Add a smooth nonnegative majorant $s_\eta >e_\eta $ on that interval, supported in $(x_{\rm pre}-1/4,b+1/2)$, with $\int s_\eta \leq \int e_\eta +\eta $. Set $r_\eta =r^0_\eta +s_\eta $. Then

After shrinking the fixed $c$, this is bounded by a fixed $m_0$ within all the positivity tolerances above. The conclusions of theorem 23.1.52 therefore hold for this choice, with fixed $A,B,D$. On every fixed compact interval $I\Subset (0,\infty )$, for sufficiently small $\eta $ we have $r_\eta =K\eta $ and exactly

Here $b_0,b_1$ depend only on the fixed left cutoff. Globally $0\leq -g_\eta (x)\leq m_0(x+2)$ for $x\geq 0$.

Let $G(w)=-\int _0^\infty (k_ r(w)-1)\, \mathrm dr$ for $w\ne 0$ on $E$. Gaussian integration gives the uniform bound

Indeed the nearest Gaussian has integral bounded by $C+2\log ^+(1/\operatorname {dist}_ E(w,0)^2)$ on $(0,1]$; the other lattice terms have bounded integral and large times decay. The same bound applies after the cutoff. On a fixed $I\Subset (0,\infty )$, $\vartheta =1$ for small $\eta $ and $t_\eta (x)\leq \exp (-c/(2\eta ^3))$. Every fixed $x$ derivative of $t_\eta $ has this bound times a polynomial in $\eta ^{-1}$. On compact fiber sets away from zero,

To verify derivatives, subtract $G$ from $G_ t=F_\eta (t,\cdot )/\eta $: $G_ t-G=\int _0^ t(k_ r-1)\, \mathrm dr=-t+O(\exp (-d^2/(Ct)))$ when $d$ stays positive. Differentiated Gaussian errors have only polynomial factors in $t^{-1}$, and repeated $x$ differentiation preserves their decay. Together with (23.1.57) and $0\leq \sup V_\eta \leq C\eta $, this gives uniform $C^\ell $ bounds for $u_\eta $ on each such fixed compact product set, for all sufficiently small $\eta $. The threshold may depend on the set and derivative order.

Fix $a_ j=2^{-j-2}$, $j\geq 1$, so $\sum a_ j=1/4$. Choose scales $\eta _ j\downarrow 0$ recursively and write $u_ j=u_{\eta _ j}$, $K_ j=K_{\eta _ j}$. Require the minimum $x$ on $K_ j$ to exceed every earlier $x_{*,i}$ by two. Choose a relatively compact product neighborhood $O_ j$ of $K_ j$, with positive $x$ interval and fiber closure away from zero, on which the earlier $u_ i$ are constant. The $K_ j$ are pairwise disjoint. With $E_ j=\sum _{i<j}a_ i\| u_ i\| _\infty $, impose

For each earlier $O_ i$, require the preceding uniform derivative bounds to hold for $u_ j$ through order $j$. All are finite constraints at stage $j$, achievable by making $\eta _ j$ smaller. They ensure that the future series $\sum _{k>i}a_ ku_ k$ converges in $C^\infty $ on compact subsets of $O_ i$: for every fixed derivative order the sufficiently late terms have a uniform bound there, and $\sum a_ k<\infty $.

These constraints also give $\sum _{k>j}a_ k\eta _ k\leq a_ j\eta _ j$. On $K_ j$ the coordinate and distance estimates are

Using (23.1.58), the global slope bound on $g_ k$, and the supremum constants, the future depth is at most

The last term is at most a fixed multiple of $a_ j\eta _ j^{-2}$ since $\eta _ j\leq a_ j$. Including the earlier and current terms, there is a fixed $C_*>0$, independent of the scales, such that

Fix this constant before the scale choice. Since $\sqrt T W(T)$ is unbounded, at each stage choose arbitrarily large $T_ j$ such that

and so small a scale that all finite preceding constraints hold. Such values occur arbitrarily far out because $W$ is finite on compact intervals. Thus the recursive choice is possible.

The smooth nonpositive partial sums $v_ N=\sum _{j\leq N}a_ ju_ j$ decrease, and

By theorem 5.25.2, $\int _ X(-v_ N)\, \mathrm dV\leq C\sum _ j a_ j<+\infty $. Thus corollary 5.25.3 gives a $\omega $-plurisubharmonic limit $v$ with $\omega +\mathrm{dd}^{\mathrm{c}}v\geq 3\omega /4$. Its supremum is finite and nonpositive.

On $O_ j$ the earlier terms are constant and the future series converges smoothly. Thus $v$ is smooth there and $\omega +\mathrm{dd}^{\mathrm{c}}v\geq a_ j(\omega +\mathrm{dd}^{\mathrm{c}}u_ j)$ pointwise. The determinant inequality for positive Hermitian matrices gives on $K_ j$

On these sets the extended product is the smooth product: on a compact neighborhood of $K_ j$ all sufficiently deep canonical truncations equal $v$. Set $U=v-\sup _ Xv$. Since $\sup v\leq 0$, $-U\leq -v$, and its local currents are unchanged. Monotonicity of $W$ and (23.1.59) give

The disjoint $K_ j$ prove infinite total mass for this single potential. They lie in its smooth locus, so no mass is assigned to poles. $\square$