Lemma 5.42.1096444124D09C

Proof. Step 1: the classical expansion. Apply the classical Chern-connection regularization expansion to the closed current $\mathrm{dd}^{\mathrm{c}}a$ and the smooth lower bound $-\theta $. It gives monotonicity and logarithmic convexity: for a constant $K_0>0$, the function $\lambda (x,t)=t\partial _ t(R_ ta(x)+K_0t^2)$ is nonnegative and increasing in $t$. Instead of inferring a rate from the unspecified modulus in that theorem, we estimate the error terms of the expansion directly, using the smoothness of $\theta $.

Choose the auxiliary curvature form to be $A\omega $, with $A$ large enough that the Chern curvature plus $A\omega \otimes \operatorname {Id}$ is nonnegative on every decomposable tensor. Equivalently, after absorbing the normalization constants, the Chern curvature coefficients $R_{j\bar k\ell \bar m}$ in a unitary frame satisfy

for all tangent vectors $\xi ,\zeta $. Such an $A$ exists by compactness. Pairing this inequality with the nonnegative eigenvalues of a positive complex Hessian bounds its curvature contraction from below by $-A$ times its trace, with no further error term.

Step 2: a frozen quadratic correction. Fix $x\in X$ and a complex radius variable $w\ne 0$, and put $t=|w|$. Write $\xi \in T_ x^{1,0}X$ and $\tau \in \mathbb C$ for the base and radius components of a tangent vector. In a coordinate chart of uniform size about $x$, write $\beta =\mathrm{dd}^{\mathrm{c}}|z|^2$ and choose a quadratic function $q_{x,t}$ with

where the constant $B$ is chosen so large that $a+q_{x,t}$ is plurisubharmonic on a ball of radius of order $t$ containing the support of the averages at $x$ and at nearby points. This is possible by the bound $\| \theta (y)-\theta (x)\| \leq C|y-x|$ on such a ball. For the Hessian calculation at $(x,w)$, the quadratic function is held fixed; its coefficients and all needed derivatives are uniformly bounded.

Step 3: the error terms. By normalized compactness, $\| a\| _{L^1}$ is bounded. On each fixed larger chart, add one fixed strictly plurisubharmonic quadratic function $Q$ to $a$. For the trace measure $\mu =\mathrm{dd}^{\mathrm{c}}(a+Q)\wedge \beta ^{n-1}$, the local Lelong mass inequality and a bound for the mass of a fixed larger ball give $\mu (B(x,t))\leq Ct^{2n-2}$. The coefficients of $\partial \bar\partial (a+q_{x,t})$ are controlled in total variation by those of $\mathrm{dd}^{\mathrm{c}}(a+Q)$ plus a uniformly bounded smooth form. Write $|\partial \bar\partial (a+q_{x,t})|$ for the sum of the total variations of its coefficient measures. The rescaled averaging kernels are supported in $B(x,Ct)$ and bounded by a constant times $t^{-2n}$, so their integrals against this measure are bounded by

Thus the terms of the expansion with a coefficient $O(t^3)$ contribute $O(t)$, and the remainder of the expansion of order two is $O(t)$ as well. The mixed and vertical contributions are bounded by $C(|\xi ||\tau |+|\tau |^2)$. By the choice of $A$, the curvature term is bounded below by $-A\lambda (x,t)|\xi |^2$; here the radial term of the same expansion identifies the relevant mass with $\lambda $ up to $O(t^2)$. Subtracting the regularization of $q_{x,t}$ restores $-\theta (x)$: its base Hessian differs from $\mathrm{dd}^{\mathrm{c}}q_{x,t}$ by $O(t)$, its mixed and vertical Hessians are bounded, and its radial logarithmic derivative is $O(t^2)$. The extra term $Bt\beta $ also costs only $O(t)$. For a real $(1,1)$-form $\alpha $, write $\alpha [v]^2$ for its associated Hermitian quadratic form evaluated on $v$. Consequently, for a fixed constant $C_1>0$, the complex Hessian of $(x,w)\mapsto R_{|w|}a(x)$ satisfies

All constants are independent of $x,t,c,\delta $ and uniform for normalized $a$. In particular, setting $\xi =0$ in the same expansion gives a uniform lower bound for the vertical Hessian. Adding a fixed multiple of $|w|^2$ makes the function subharmonic in $w$, which gives the logarithmic convexity and monotonicity above with a uniform constant $K_0$.

Step 4: the minimum. Take $K\geq K_0$ and add $K(t^2+t)$ to the regularization. The positive vertical Hessian of $K|w|$ absorbs the mixed term by the inequality

and it absorbs the remaining bounded vertical error for small $t$. The logarithmic derivative of the corrected function is at least $\lambda $. The Kiselman minimum argument with an upper endpoint, as in the proof of lemma 5.42.4, therefore gives the last inequality in (5.42.11), now with error $C\delta $. On compact subsets of $\Omega $, local boundedness of $a$ makes the term $-c\log (t/\delta )$ dominate uniformly as $t\downarrow 0$, so that the minimum is attained away from zero. As before, minimizers at the upper endpoint are handled without differentiating their radius.

Step 5: the remaining bounds. The other bounds in (5.42.11) follow from the monotonicity of $R_ ta+Kt^2$ and from evaluation at $t=\delta $. In particular, the transform is bounded above near the analytic set $X\setminus \Omega $. The removable singularity theorem therefore extends it across this set, and local freezing and passage to the limit preserve the exact smooth lower bound for its Hessian. Its lower bound $a-K(\delta ^2+\delta )$ also extends, since quasi-plurisubharmonic functions are determined by their values off an analytic set.

The adjoint computation in the proof of lemma 5.42.2 gives $R_ t^*1=1+O(t^2)$. It only needs $a\in L^1$, since

Together with $R_ ta-a\geq -Kt^2$, this proves the stated $L^1$ estimate. Finally, near $K_1$ the smoothing kernel only samples a fixed larger compact subset of $\Omega $, on which $a$ is bounded. The kernel derivative estimate in the proof of lemma 5.42.2 then gives the bound $C_{K_1}/t$. $\square$