Let $\theta $ now be any smooth real $(1,1)$-form and let $u\in \mathrm{PSH}(X,\theta )\cap L^\infty (X)$. We use the same averaging operators and the same metric.
We first state the classical input for this background. The classical Chern-connection regularization theorem applies to the closed current $\mathrm{dd}^{\mathrm{c}}u$ with the smooth, possibly nonclosed, lower bound $-\theta $; the curvature of the Chern connection on the tangent bundle is bounded by a fixed multiple of $\omega $. The theorem gives a constant $K_0>0$ such that
\[ R_ tu+K_0t^2\downarrow u. \]
To state its curvature estimate, use local coordinates $z$ on $X$ and a complex radius variable $w$, and put $t=|w|$. Write $\zeta \in T_ z^{1,0}X$ and $\eta \in \mathbb C$ for the base and radius components of a tangent vector, and set $\lambda (z,t)=t\partial _ t(R_ tu(z)+K_0t^2)$. Then $\lambda $ is nonnegative and increasing in $t$, and there are constants $A,C>0$ and a function $\epsilon _0(t)\geq 0$ with $\epsilon _0(t)\downarrow 0$ as $t\downarrow 0$ such that the complex Hessian of $(z,w)\mapsto R_{|w|}u(z)$ is bounded below by the base form $-\theta -A\lambda (z,t)\omega -\epsilon _0(t)\omega $, evaluated on $\zeta $, minus mixed and vertical error terms bounded by $C(|\zeta ||\eta |+|\eta |^2)$. This is the classical regularization estimate for a smooth lower bound.
Lemma 5.42.4.D0B76A103405 For $K$ sufficiently large, there is an increasing function $\epsilon (\delta )\geq 0$ tending to zero at zero such that, for every small $\delta >0$ and every $c>0$, the function
\[ \Phi _{c,\delta } =\inf _{0<t\leq \delta } \{ R_ tu+K(t^2-\delta ^2)+K(t-\delta )-c\log (t/\delta )\} \]
is bounded and quasi-plurisubharmonic and satisfies
\[ \theta +\mathrm{dd}^{\mathrm{c}}\Phi _{c,\delta } \geq -(Ac+\epsilon (\delta ))\omega . \]
The infimum is attained away from zero. Moreover, $R_ tu+Kt^2\downarrow u$ as $t\downarrow 0$.
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
\[ C|\zeta ||\eta | \leq C' t|\zeta |^2+\frac{K}{8t}|\eta |^2, \]
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
\[ -\theta -(A\widetilde\lambda +\epsilon (t))\omega , \]
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 $u$.
$\square$
The classical theorem only gives some modulus $\epsilon $; we do not claim that $\epsilon (\delta )$ is linear in $\delta $. The cancellation and kernel arguments in the proof of lemma 5.42.2 still give
\[ \| R_ tu-u\| _{L^1(\mathrm{d}V)}\leq Ct^2, \qquad \| \nabla R_ tu\| _\infty \leq C/t, \]
because the averaging map is unchanged, $u$ is bounded, and these arguments use only the monotone lower bound $R_ tu-u\geq -Kt^2$.
For continuous potentials, the curvature loss of the averages becomes uniformly small.
Lemma 5.42.5.A340B19CD0CC Let $\alpha $ be a smooth real $(1,1)$-form on $X$, and let $u\in \mathrm{PSH}(X,\alpha )\cap C^0(X)$. There are smooth real functions $u_ j$ and numbers $\varepsilon _ j\downarrow 0$ such that $u_ j\to u$ uniformly and
\[ \alpha +\mathrm{dd}^{\mathrm{c}}u_ j\geq -\varepsilon _ j\omega . \]
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
\[ \alpha +\mathrm{dd}^{\mathrm{c}}R_ tu\geq -(A\lambda (\cdot ,t)+\epsilon _0(t))\omega . \]
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 $A\| \lambda (\cdot ,t_ j)\| _\infty +\epsilon _0(t_ j)$.
$\square$
Pome
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