In the auxiliary function of lemma 15.18.8, the term $-A(u_ j-q)$ penalizes points where $u_ j-q$ is large. A quadratic gap of $u-q$ along the nonregular set therefore suppresses the contribution of the boundary of a neighborhood of that set. Taking $A$ proportional to $\log j$ makes this suppression beat the polynomial growth of the boundary trace, and near the point of minimal $u-q$ the trace then grows more slowly than every positive power of $j$. Interpolation with the polynomial approximation rate gives uniform $C^{1,\alpha }$ bounds.
Distances and norms below are Euclidean. By a smooth exact solution for $(\vartheta ,F)$ we mean a smooth function $u_ j$ with $\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j>0$ and $(\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)^ n=F\beta ^ n$.
Proposition 15.18.25.F9CDEC120FA0 For $R>0$, let $B=B_ R(0)$, let $\vartheta $ and $F>0$ be smooth near $\overline B$, and let $u\in C(\overline B)$. Suppose $q$ is smooth and $\vartheta +\mathrm{dd}^{\mathrm{c}}q\geq \varepsilon \beta $ for some $\varepsilon >0$. Put $v=u-q$ and $m=v(0)$, and fix $\kappa >0$. Let $\mathcal S\subset \overline B$ be closed, contain $0$, and satisfy
87542664E1D7
15.18.26
\begin{equation} v(0)=m, \qquad v(z)\geq m+\kappa |z|^2\quad (z\in \mathcal S). \end{equation}
Assume that smooth exact solutions $u_ j$, $j\geq 2$, satisfy, for fixed constants $a>0$, $b\geq 0$ and $C>0$,
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15.18.27
\begin{align} \lVert u_ j-u\rVert _\infty & \leq Cj^{-a}, \\ \sup _{\overline B}\operatorname {tr}_\beta (\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)& \leq Cj^ b, \tag{15.18.28}\\ \sup _ j\sup _ K\operatorname {tr}_\beta (\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)& <\infty \quad \text{for every }K\Subset \overline B\setminus \mathcal S. \tag{15.18.29} \end{align}
Then for every $\gamma >0$ there are a neighborhood $U_\gamma $ of $0$ and $C_\gamma $ such that
EBBDF08FE74B
15.18.30
\begin{equation} \operatorname {tr}_\beta (\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)\leq C_\gamma j^\gamma \quad \text{on }U_\gamma . \end{equation}
For every $0<\alpha <1$, there is also a neighborhood $U$ of $0$ on which $u_ j\to u$ in $C^{1,\alpha }$. In particular the approximants are uniformly bounded in that norm and $u\in C^{1,\alpha }(U)$.
Proof.
Step 1: the trace bound (15.18.30). Write $e_ j=Cj^{-a}$, fix $d=\kappa R^2/2$, and choose $D>(b+1)/d$. Choose $\eta >0$ so that $2D\eta <\gamma /2$. For small $\delta >0$, the neighborhood
\[ W_\delta =\{ z\in B:\operatorname {dist}(z,\mathcal S)<\delta \} \]
contains $0$, has $v\geq m-\eta $ on its closure, and satisfies $v\geq m+d$ on $\partial B\cap \overline W_\delta $ with the fixed $d$ above. These properties follow from (15.18.26), continuity, and compactness. The remaining part of the boundary of $W_\delta $ is a compact subset of $\overline B\setminus \mathcal S$, where (15.18.29) applies.
Set $A_ j=D\log j$. Apply the maximum principle on $\overline W_\delta $ to
\[ Q_ j=\log \operatorname {tr}_\beta (\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)-A_ j(u_ j-q)+\mathrm{e}^{-(u_ j-q)}. \]
At an interior maximum, lemma 15.18.8 gives $\log \operatorname {tr}_\beta (\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)\leq C+\frac12\log A_ j$. On the part of $\partial W_\delta $ inside $B$, which stays away from $\mathcal S$, the trace has a bound independent of $j$. On the outer part $\partial B\cap \overline W_\delta $, (15.18.28) and the gap $d$ give
\[ Q_ j\leq C+b\log j-A_ j(m+d-e_ j), \]
and the choice $Dd>b+1$ makes this smaller than the bound below for large $j$. Hence throughout $W_\delta $,
\[ Q_ j\leq C_\delta +\tfrac 12\log A_ j-A_ j(m-\eta -e_ j). \]
On a small ball about $0$ where $v\leq m+\eta $, this implies
\[ \log \operatorname {tr}_\beta (\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j) \leq C_\delta +\tfrac 12\log A_ j+A_ j(2\eta +2e_ j). \]
By the choice of $\eta $, the logarithmic factor and $A_ je_ j$ are absorbed into the remaining power. This proves (15.18.30).
Step 2: $C^{1,\alpha }$ convergence. On a smaller ball, the real Laplacian of $u_ j$ is bounded by $Cj^\gamma $. Interior Poisson estimates give a $W^{2,q_0}$ bound for every finite $q_0$, and Sobolev embedding with $q_0$ large then gives, for every $\sigma <1$,
\[ \lVert u_ j\rVert _{W^{2,q_0}}+\lVert u_ j\rVert _{C^{1,\sigma }} \leq Cj^\gamma . \]
For $d_ k=u_{2^{k+1}}-u_{2^ k}$,
\[ \lVert d_ k\rVert _\infty \leq C2^{-ka}, \qquad \lVert d_ k\rVert _{C^{1,\sigma }}\leq C2^{k\gamma }. \]
If $\alpha <\sigma $ and $t=(1+\alpha )/(1+\sigma )$, Hölder interpolation gives
\[ \lVert d_ k\rVert _{C^{1,\alpha }} \leq C2^{k[-a(1-t)+\gamma t]}. \]
Since $\gamma $ may be arbitrarily small, the exponent can be made negative. The dyadic series converges in $C^{1,\alpha }$ to its uniform limit $u$. If $2^ k\leq j<2^{k+1}$, interpolation applied to $u_ j-u_{2^ k}$ gives the same decaying bound. Hence the full sequence converges to $u$ in $C^{1,\alpha }$ on a smaller ball.
$\square$
Proposition 15.18.31.45A010B2A5AF Suppose $u_ j$ solve the same smooth equation as in proposition 15.18.25, satisfy (15.18.27)–(15.18.28), and $u$ is smooth with $\vartheta +\mathrm{dd}^{\mathrm{c}}u>0$ on an open set $O\subset B$. Then, for every $K\Subset O$,
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15.18.32
\begin{equation} \sup _ j\sup _ K\operatorname {tr}_\beta (\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)<\infty . \end{equation}
Proof.
Where $u$ is smooth and strictly $\vartheta $-plurisubharmonic, it provides its own quadratic separation, so the argument of proposition 15.18.25 applies around every point of $O$. A first application gives trace growth of order $\sqrt{\log j}$; the resulting gradient convergence then allows the variant of lemma 15.18.8 with a gradient bound, which removes this growth.
For $p$ in a fixed compact subset of $O$, take a uniformly sized ball $B_ r(p)\Subset O$ and set
\[ q_ p(z)=u(z)-\kappa |z-p|^2 \]
with $\kappa >0$ small, so $\vartheta +\mathrm{dd}^{\mathrm{c}}q_ p\geq \varepsilon \beta $. If $e_ j=Cj^{-a}$ and $v_{j,p}=u_ j-q_ p$, then
\[ v_{j,p}\geq -e_ j, \quad v_{j,p}(p)\leq e_ j, \quad v_{j,p}\geq \kappa r^2-e_ j\text{ on }\partial B_ r(p). \]
With $A_ j=D\log j$ and $D\kappa r^2>b+1$, the same maximum argument as above gives
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15.18.33
\begin{equation} \operatorname {tr}_\beta (\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)(p)\leq C\sqrt{\log j}. \end{equation}
The constants are uniform on compact subsets of $O$.
Put $T_ j=\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j$ and $w_ j=u_ j-u$. On slightly larger compact sets, (15.18.33) bounds $|\Delta w_ j|$ by $C\sqrt{\log j}$. Poisson estimates and Sobolev embedding give
\[ \lVert w_ j\rVert _{C^{1,\sigma }}\leq C_\sigma \sqrt{\log j}, \qquad \sigma <1. \]
Interpolating with $\lVert w_ j\rVert _\infty \leq Cj^{-a}$ shows $\nabla w_ j\to 0$ uniformly on smaller compact sets. The gradients of $v_{j,p}$ are therefore uniformly bounded. Repeat the maximum argument with $\log \operatorname {tr}_\beta T_ j-A_ jv_{j,p}$, now using the variant of lemma 15.18.8 with a gradient bound. Its interior trace bound is now independent of $A_ j$, while the boundary term is suppressed as before. Evaluation at $p$ gives
\[ \operatorname {tr}_\beta T_ j(p)\leq C\exp (2A_ je_ j)\leq C, \]
proving (15.18.32).
$\square$
Pome
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