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
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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
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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.
$\square$
Pome
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