5.42 Chern-connection regularization001730C9ED64

Proof. Let $z$ be local coordinates on $X$ and $\xi \in \mathbb C^ n$ the fiber variable with respect to a smooth orthonormal frame of the tangent bundle. Write $\rho (|\xi |^2)$ for the fixed radial averaging kernel, normalized with respect to Lebesgue measure $\mathrm{d}\xi $ and supported in $|\xi |\leq 1$. In these coordinates the averaging map $F$ gives

The map $F$ is smooth in these variables, and its fiber differential at zero is an isomorphism. Let $R_ t^*$ be the formal adjoint of $R_ t$ with respect to $\mathrm{d}V$, so that $\int _ X(R_ ta)b\, \mathrm{d}V=\int _ Xa\, R_ t^*b\, \mathrm{d}V$. We first claim that

uniformly. To see this, insert a smooth partition of unity in the variable $z$, each term being compactly supported in a chart on which the frame is defined. For fixed $\xi $, change variables $x=F(z,t\xi )$ on a neighborhood of the support of a term. The inverse map and its Jacobian have uniform Taylor expansions in $t$. The term of order zero in the transformed density is the chosen partition function at $x$, and the term of order one is linear in $\xi $, so it integrates to zero against the even radial kernel. The remaining density is bounded by $Ct^2\mathrm{d}V(x)$, since the second derivatives of all maps, partition functions and volume coefficients are bounded on these finitely many compact charts. Summing over the partition functions proves the claim.

Consequently,

By lemma 5.42.1, $R_ tu-u\geq -Kt^2$. A real function $a\geq -Kt^2$ satisfies $|a|\leq a+2Kt^2$; applying this to $a=R_ tu-u$ and integrating gives the first estimate.

For the derivative bound, change variables in the fiber to write $R_ tu(z)=\int _ Xu(x)k_ t(z,x)\, \mathrm{d}V(x)$. In local coordinates the smooth kernel $k_ t$ is supported where $|x-z|\leq Ct$ and satisfies

Indeed, the inverse fiber map and its Jacobian have uniformly bounded derivatives, and differentiating the rescaled radial argument introduces at most one factor $t^{-1}$ in addition to $t^{-2n}$. Since the kernel is smooth with compact support, there are no boundary terms. Integration over a ball of volume at most $Ct^{2n}$ gives $\int |\nabla _ zk_ t|\, \mathrm{d}V\leq C/t$, and multiplication by the uniform bound for $u$ proves the assertion. $\square$

Proof. Take $K\geq K_0$ and put $H(z,t)=R_ tu(z)+Kt^2+Kt$. As a function of $\log t$, $H$ is increasing and strictly convex, because $R_ tu+K_0t^2$ has nonnegative increasing logarithmic derivative and the added terms have positive second logarithmic derivative. Its logarithmic derivative $\widetilde\lambda =t\partial _ tH$ is at least $\lambda $.

The complex Hessian of $K|w|$ has a positive vertical term proportional to $K/|w|$. For sufficiently small $|w|=t$, this term absorbs the vertical error and, by the inequality

also the mixed error. After increasing the constants and replacing $\lambda $ by the larger $\widetilde\lambda $, we find that the complex Hessian of $(z,w)\mapsto H(z,|w|)$ is bounded below by the base form

where $\epsilon $ is an increasing majorant of $\epsilon _0(t)+C't$.

Since $u$ is bounded, the term $-c\log (t/\delta )$ tends uniformly to $+\infty $ as $t\downarrow 0$, so the minimum occurs on a compact interval of radii away from zero. Strict convexity in $\log t$ makes the minimizing radius unique. At an interior minimum $\widetilde\lambda =c$, while at the upper endpoint $t=\delta $ one has $\widetilde\lambda \leq c$. No differentiability of the minimizing radius is needed.

In more detail, localize near a point of $X$ and fix $a>0$. By continuity of the minimizers and of $\widetilde\lambda $, there is an interval of radii containing the minimizers at all nearby points on which $\widetilde\lambda \leq c+a$. If the minimum is attained at the upper endpoint, keep the upper endpoint $\delta $; an open interval approaching it has the same infimum. Because of the uniform logarithmic term, the lower endpoint can be chosen below all minimizers, and since $\widetilde\lambda $ is increasing, the bound $\widetilde\lambda \leq c+a$ holds throughout the chosen interval. The local Kiselman minimum principle for functions depending only on $|w|$, applied after a local quadratic correction, now gives the lower curvature bound with error $A(c+a)+\epsilon (\delta )$, up to an arbitrarily small error from freezing the smooth base forms. Letting these errors and $a$ tend to zero proves the asserted current inequality and the quasi-plurisubharmonicity. Finally, $K\geq K_0$ preserves the monotone convergence of $R_ tu+Kt^2$ to $u$. $\square$

Proof. Use the Chern averaging operators and the classical estimate stated before lemma 5.42.4, with $\alpha $ in place of $\theta $. For some $K_0,A>0$ and $\epsilon _0(t)\to 0$, put $\lambda (x,t)=t\partial _ t(R_ tu(x)+K_0t^2)$. The estimate with vanishing radius component gives

The function $\lambda $ is nonnegative, continuous for $t>0$, and increasing in $t$. For fixed $x$, the convex function $s\mapsto R_{\mathrm{e}^ s}u(x)+K_0\mathrm{e}^{2s}$ has the finite limit $u(x)$ as $s\to -\infty $. Its nonnegative derivative must therefore tend to zero, since a positive limit of the derivative would make the function tend to $-\infty $. Thus $\lambda (x,t)\to 0$ as $t\to 0$, and for any sequence $t_ j\downarrow 0$, Dini's theorem gives $\| \lambda (\cdot ,t_ j)\| _\infty \to 0$.

The averaging kernels have mass one and only sample points at distance at most $Ct$ from the base point. Uniform continuity of $u$ therefore gives $\| R_ tu-u\| _\infty \to 0$. Set $u_ j=R_{t_ j}u$, and let $\varepsilon _ j\to 0$ be a decreasing positive majorant of $A\| \lambda (\cdot ,t_ j)\| _\infty +\epsilon _0(t_ j)$. $\square$

Proof. Use the Chern averages $R_ tu$ in lemma 5.42.5. For a sequence $t_ j\downarrow 0$ there are $\varepsilon _ j\downarrow 0$ such that

We also have a uniform Hölder bound. In a coordinate chart with a smooth orthonormal tangent frame, the averaging formula is

On a smaller chart, smoothness of $F$ bounds $\operatorname {dist}(F(x,t\xi ),F(y,t\xi ))$ by $C\operatorname {dist}(x,y)$, uniformly for small $t$ and $|\xi |\leq 1$. Integrating the Hölder inequality for $u$ gives the same bound for $R_ tu$. A finite cover and the common supremum bound handle pairs outside a common smaller chart.

Set $u_ j=R_{t_ j}u/(1+\varepsilon _ j)$. Then

These functions converge uniformly with a common Hölder bound. Subtracting their suprema gives the asserted normalization when $\sup _ Xu=0$. $\square$

Proof. First we check a common lower curvature bound for the averages. The base part of the classical Chern curvature estimate stated before lemma 5.42.4 gives

Its constants are fixed under the stated bounded range. In the local kernel form used in lemma 5.42.2,

with uniformly bounded first derivatives and compact support in its rescaled second variable. Differentiating with respect to $t$ at fixed $x,y$ and changing variables shows that

Thus $|t\partial _ tR_ t\varphi |\leq CM$, so $\lambda $ is uniformly bounded and the displayed curvature estimate supplies one fixed lower bound $\mathrm{dd}^{\mathrm{c}}R_ t\varphi \geq -C\omega $. Both $R_ t\varphi $ and $\varphi $ also lie in $[-M,0]$.

Apply corollary 5.29.9 to this pair, whose common lower bound has just been established. By lemma 5.42.2, its smooth-volume $L^1$ error is at most $Ct^2$. Domination by $A\omega _ u^ n$ therefore gives

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 $C_{K_1}/t$. $\square$