14.16 The gradient bound by rescaling40669ABC11F8

Proof. We use the matrices $Q_ u$ of (14.13.1). Their background term $Q_0$ is fixed, and $Z$ is linear in the first derivatives of $u/(2\pi )$.

Suppose that the gradient bound fails. Then there are solutions $(u_ j,b_ j)$ with prescribed functions $F_ j$ such that

After passing to a subsequence, the points $x_ j$ lie in one fixed coordinate neighborhood. Translate the coordinates so that $x_ j$ becomes the origin, and apply a uniformly controlled complex linear change of coordinates so that the matrix of $\gamma $ at $x_ j$ is the identity; a coordinate ball of fixed radius remains available around every $x_ j$. On balls whose radii tend to infinity define

in these coordinates. The coefficient matrices $g_ j(z)$ of $\gamma $ at the points $x_ j+z/A_ j$ converge smoothly to the identity on every fixed ball. In this notation

The assumed estimate (14.16.2) gives a uniform bound for all complex Hessian entries of $v_ j$ on every fixed ball, since two derivatives contribute the factor $A_ j^{-2}$. Hence the Euclidean Laplacians of the $v_ j$ are uniformly bounded, and the interior estimates for the Laplacian give local $W^{2,p}$ bounds for every finite $p$. Choosing $p>2n$ and using Sobolev embedding and compactness, we extract a diagonal subsequence converging in $C^1$ on compact subsets to a function $v$ on $\mathbb C^ n$. This function is bounded and globally Lipschitz, and

Now divide the matrix in (14.13.1) by $A_ j^2$ and evaluate it at the scaled points. The background term becomes $O(A_ j^{-2})$, and the first-order term becomes $O(A_ j^{-1})$, since one derivative of $u_ j$ is $A_ j$ times a derivative of $v_ j$. Since the complex Hessians of $v_ j$ are uniformly bounded and $g_ j\to I$, we obtain on any fixed ball

To be precise, we compute in the unitary frames obtained from the positive square roots $S_ j=g_ j^{-1/2}$. In these frames the principal matrix is $P_ n(S_ jH(v_ j)S_ j/(2\pi ))$, which differs from $P_ n(H(v_ j)/(2\pi ))$ by an error tending to zero uniformly; no derivatives of the frames are taken, and the determinant identity is exact in unitary frames. Since the matrices on the left of (14.16.3) are positive and the positive factor $1/(2\pi )$ does not change the cone, passing to the limit in the sense of distributions gives $P_ n(H(v))\geq 0$.

It remains to check maximality. Suppose that a smooth function $h$ touches $v$ from below at $z_0$ and $P_ n(H(h)(z_0))>0$. Subtract $\delta |z-z_0|^2$ with $\delta >0$ small; on a small ball this makes the contact strict and preserves strict positivity of the matrix. Local uniform convergence gives points $z_ j\to z_0$ at which $v_ j$ minus this perturbed test function has an interior local minimum. At such a minimum,

Because $P_ n$ preserves nonnegative matrices, (14.16.3) now implies $A_ j^{-2}Q_{u_ j}(z_ j)\geq cI$ for a fixed $c>0$ and all sufficiently large $j$. The determinant of this matrix is then at least $c^ n$, which contradicts the limit zero in (14.16.3). Thus $v$ is maximal for the cone of section 14.15.

By theorem 14.15.2, $v$ is constant, which contradicts $|\partial v(0)|=1$. This proves the uniform gradient bound, and substitution in (14.16.2) gives the uniform complex Hessian bound. $\square$