Proposition 15.18.46.29094CECD90F There is $\alpha _*<1$, depending only on $n$ and the exponent $N$ in proposition 15.18.35, with the following property. Let $B_0$ be a Euclidean ball, fix a smooth real $(1,1)$-form $\vartheta $ and a smooth function $F>0$ near $\overline{B_0}$, and let $u_ j$ be smooth exact solutions of $(\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)^ n=F\beta ^ n$ with $\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j>0$. If
\[ \sup _ j\lVert u_ j\rVert _{C^{1,\alpha }(B_0)}<\infty \qquad \text{for some }\alpha _*<\alpha <1, \]
then $\operatorname {tr}_\beta (\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)$ is uniformly bounded on every smaller ball.
Proof.
Fix a center for concentric balls $B_ r\Subset B_0$, put $s_ j=\operatorname {tr}_\beta (\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j)$, and set
\[ M_ j(r)=1+\sup _{B_ r}s_ j. \]
For concentric balls $B_ r\Subset B_ R\Subset B_0$, $d=R-r\leq 1$, proposition 15.18.35 gives
7B1658CB459D
15.18.47
\begin{equation} |\nabla s_ j|\leq Cd^{-1}M_ j(R)^ N \quad \text{on }B_{r+d/2}. \end{equation}
Fix $p\in B_ r$ and let $\chi _ h$ be a nonnegative mollifier of radius $h\leq d/4$. Since
\[ s_ j=\operatorname {tr}_\beta \vartheta +\Delta _\beta u_ j, \]
integration by parts, after subtracting the constant vector $\nabla u_ j(p)$, and the uniform $C^{1,\alpha }$ bound give
\[ (s_ j*\chi _ h)(p)\leq C(1+h^{\alpha -1}). \]
Equation (15.18.47) gives
\[ |s_ j(p)-(s_ j*\chi _ h)(p)| \leq Cd^{-1}M_ j(R)^ Nh. \]
Choose
\[ h=\frac d4M_ j(R)^{-N/(2-\alpha )}. \]
Taking the supremum over $p\in B_ r$ and absorbing fixed powers of $d^{-1}$, we obtain
66EE1806AFB7
15.18.48
\begin{equation} M_ j(r)\leq Cd^{-1}M_ j(R)^ q, \qquad q=\frac{N(1-\alpha )}{2-\alpha }. \end{equation}
Choose $\alpha _*$ so close to one that $q<1$ for $\alpha >\alpha _*$. Choose $0<r_0<R_0$ with $B_{R_0}\Subset B_0$. Let $r_ k\uparrow R_0$ with $r_{k+1}-r_ k=(R_0-r_0)2^{-k-1}$. Iterating (15.18.48) gives
\[ M_ j(r_0) \leq \prod _{k=0}^{m-1}\left[C(r_{k+1}-r_ k)^{-1}\right]^{q^ k} M_ j(r_ m)^{q^ m}. \]
The infinite product converges because $\sum q^ k(1+k)<\infty $. For each fixed $j$, the final factor tends to one as $m\to \infty $. The resulting bound is independent of $j$.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.