Proposition 14.11.160FD5A56EE10

Proof. We first bound the scalar. At a maximum of $u$, the first derivatives of $u$ vanish and $\mathrm{dd}^{\mathrm{c}}u\leq 0$, so $\Gamma _ u\leq \gamma _0^{n-1}$; at a minimum the reverse inequality holds. The determinant is monotone with respect to the order of positive matrices, and $(\eta _ u^ n/\gamma ^ n)^{n-1}$ is the determinant of the coefficient matrix of $\Gamma _ u$ relative to $\gamma $ (see (14.13.1) below). Evaluating the equation at these two points gives

This proves the scalar bound.

Consider the fixed real linear elliptic operator

Its principal part is $\mathrm{dd}^{\mathrm{c}}u\wedge \gamma ^{n-1}/\gamma ^ n$, its remaining term is of first order, and it has no zeroth-order term. Since $\Gamma _ u$ is positive, $\gamma \wedge \Gamma _ u\geq 0$, and therefore $Lu\geq -C_0$ for a fixed constant $C_0$.

By the Fredholm and maximum-principle argument in the proof of theorem 11.4.2, the operator $L$ has an adjoint probability measure $\nu $ with smooth strictly positive density, that is, $\int _ XLv\, \mathrm{d}\nu =0$ for every smooth $v$. Since $Lv\leq 0$ at a maximum of $v$, its normalized Green kernel can be written

Here the kernel is smooth off the diagonal and locally integrable. To obtain this representation, construct the inverse of $L$ on functions of zero $\nu $-mean by the Fredholm alternative. With the minus sign above, the local fundamental solution has a positive leading singularity, so compactness gives a lower bound for the kernel. Adding a constant to $K$ makes it nonnegative and does not change the formula, since $\int _ XLv\, \mathrm{d}\nu =0$. The local estimates for the Green kernel give the uniform integral bound.

Apply the formula to $u$ and use $Lu\geq -C_0$. At a maximum point, where $u=0$, we obtain

Because $u\leq 0$ and $\nu $ has a fixed smooth positive density, this is a uniform $L^1$ bound for $u$ with respect to any fixed smooth volume form.

Now let $M=\min _ Xu$, and choose holomorphic coordinates on a ball centered at a minimum point. By compactness, the coordinate radius can be fixed, with uniform bounds for the metric coefficients and their derivatives; rescale the ball to the unit ball $B\subset \mathbb R^{2n}$. Write $x$ for the real coordinates, $D$ and $D^2$ for the Euclidean gradient and Hessian, $\mathrm{d}x$ for Euclidean volume, and $I$ for an identity matrix of the appropriate size. Fix a small number $\varepsilon >0$, depending only on those bounds, and set

For every real vector $p$ with $|p|<\varepsilon /2$, the function $v(x)-p\cdot x$ attains its minimum at an interior point, because its boundary values are larger than $M+\varepsilon /2$, whereas its value at the center is $M$. At such a point,

Let $A\subset B$ be the set of all points satisfying $|Dv|\leq \varepsilon /2$, $D^2v\geq 0$, and $v\leq M+\varepsilon /2$. Since $u\geq M$, this set lies in $\{ |x|\leq 1/\sqrt2\} $. It is closed there, hence compact in $B$, and by the preceding paragraph its image under the gradient map $Dv$ contains the ball $B_{\varepsilon /2}$ of radius $\varepsilon /2$. The area formula for the smooth gradient map therefore gives

The Hessian is nonnegative on $A$, so no absolute value is needed in the determinant.

We now bound this determinant pointwise on $A$. There,

The complex Hessian of $u$ has the corresponding lower bound. Wedging with the fixed positive form $\gamma ^{n-2}$ and bounding the first-order term of $\Gamma _ u$ by the gradient estimate, we find that $\Gamma _ u\geq \gamma _0^{n-1}-C\varepsilon \gamma ^{n-1}$. Let $Q_ u$ be the coefficient matrix of $\Gamma _ u$ in a fixed local frame of $(n-1,0)$-forms. Since $\gamma _0^{n-1}$ is strictly positive, choosing $\varepsilon $ small gives

with a uniform $c>0$. The equation and the bounds for $F+b$ already proved bound $\det Q_ u$ from above, because the determinant of this matrix is, up to a fixed positive factor depending on the frame, the $(n-1)$st power of the density of $\eta _ u^ n$. All eigenvalues of $Q_ u$ are therefore bounded above as well: their product is bounded, and each of them is at least $c$.

In particular, $\gamma \wedge \Gamma _ u/\gamma ^ n$ is uniformly bounded on $A$. Subtracting its fixed background term and its bounded first-order term, we obtain an upper bound for

This is a fixed positive multiple of the $\gamma $-trace of the complex Hessian of $u$, which differs from that of $v$ by a bounded term. The complex Hessian of $v$ is nonnegative on $A$, since the real Hessian is nonnegative there, so comparison of metrics bounds its Euclidean trace from above. The real trace of $D^2v$ is four times this complex trace. It follows that

Inequality (14.11.2) now gives $|A|\geq c_2>0$, where $c_2$ is independent of $u$. On $A$ we also have $u\leq v\leq M+\varepsilon /2$. Consequently the uniform $L^1$ bound gives

This bounds $-M$ and completes the proof. $\square$