15.18.13 A polynomial boundary trace estimate94BAFACA36D1

Proof. We follow the classical boundary estimates for the Dirichlet problem on a ball and keep track of the dependence on $L=1+\lVert g\rVert _{C^4}$. Barriers bound the first derivatives, and differentiation along rotations bounds the mixed normal–tangential second derivatives on the boundary. A lower bound for the tangential block of $T$ then bounds the double normal derivative through the equation, and lemma 15.18.8 reduces the interior bound to the boundary bound.

Step 1: first derivatives. Fix a sufficiently large constant $K>0$, depending only on the fixed data. Extend $g$ to $G$ on $\overline B$ with $\lVert G\rVert _{C^4}\leq CL$. For $A=CL$ large enough,

Proposition 15.18.3 gives $\underline u\leq u$. If $h$ solves $\Delta _\beta h=-C_\vartheta $ with $h=g$ on $\partial B$ and $C_\vartheta \geq \sup \operatorname {tr}_\beta \vartheta $, then $u\leq h$. Boundary differentiation of these barriers gives

Let $H=\operatorname {tr}_ T\beta $, $\mathcal L=\operatorname {tr}_ T\mathrm{dd}^{\mathrm{c}}$, and $w=u-\underline u\geq 0$, the linearized operator and the barrier function used below. Taking $K$ fixed and sufficiently large gives

Let $E=\sum _ k(z_ k\partial _{z_ k}+\bar z_ k\partial _{\bar z_ k})$ be the real radial field. Write $\Theta $ for the coefficient matrix of $\vartheta $. Differentiating $\log \det T=\log F$ and commuting $E$ with complex second derivatives yields

Hence $|\mathcal L(Eu)|\leq C(1+H)$. The same estimate holds for $Zu$ when $Z$ is an infinitesimal unitary rotation. Applying the minimum principle to $CL+C'w\pm Yu$, with $Y=E$ or $Z$, gives

Step 2: mixed second derivatives on the boundary. For a rotation $Z$, the function $R_ Z=Z(u-G)$ vanishes on $\partial B$ and satisfies $|\mathcal LR_ Z|\leq CL(1+H)$. The barriers $CLw\pm R_ Z$ and (15.18.17) imply $|\partial _\nu R_ Z|\leq CL^2$ on the boundary. Rotations span the real tangent space of $\partial B$, so all real mixed normal–tangential second derivatives are $O(L^2)$. Pure tangential second derivatives are $O(L)$ by differentiating $u=g$ twice and using (15.18.16). Thus all complex Hessian entries except the complex normal–normal entry are $O(L^2)$.

Step 3: the double normal derivative. It remains to bound the tangential block of $T$ from below. For $n\geq 2$, let

Choose a minimizing pair $(p,\xi )$ and extend $\xi $ to a smooth unit tangential field near $p$, with uniformly bounded derivatives. On the sphere, any smooth function vanishing there satisfies

Define

Then $P\geq 0$ on the nearby spherical boundary and $P(p)=0$. The boundary first-derivative bound also gives $\lambda _0\leq CL$ by the tangential identity above. Consequently

On a fixed cap $D=B\cap B_ r(p)$ consider

Choose $C_2$ so that $\Psi \geq 0$ on the interior spherical part of $\partial D$, and then $C_1$ so that $\mathcal L\Psi \leq 0$ by (15.18.17). The minimum principle yields $\Psi \geq 0$. At $p$ the normal derivative of $|z-p|^2$ vanishes, and therefore

Equations (15.18.16)–(15.18.18) now give $D^2u(\nu ,\nu )(p)\leq CL^2$, hence $\operatorname {tr}_\beta T(p)\leq CL^2$. Positivity and $\det _\beta T=F$ imply

At any boundary point write

in tangential–normal blocks, with $B_0$ an $(n-1)\times (n-1)$ Hermitian matrix, $b\in \mathbb C^{n-1}$ and $d\in \mathbb R$. We have $|b|\leq CL^2$ and $B_0\geq \lambda _0I$. The determinant equation and the Schur complement give

which is polynomial in $L$. For $n=1$, the trace is $F$ and this argument is unnecessary.

Step 4: the interior. Choose a fixed quadratic $q$ with $\vartheta +\mathrm{dd}^{\mathrm{c}}q\geq \varepsilon \beta $. Apply lemma 15.18.8 with one fixed $A$. At an interior maximum of its auxiliary function the trace is uniformly bounded, while at a boundary maximum the preceding estimate applies. Since the remaining terms are uniformly bounded, (15.18.15) follows. $\square$