Proposition 8.3.3.B863BB49A505 Fix a compact complex manifold $X$ of dimension $n$, a smooth Hermitian form $\omega $, a smooth real $(1,1)$-form $\theta $, an integer $2\leq m\leq n$, and constants $M,A,B>0$. There is a uniform bound for $\sup _ X|\partial u|_\omega $ over all smooth real $u$ such that $\theta _ u=\theta +\mathrm{dd}^{\mathrm{c}}u\in \Gamma _ m(\omega )$ and
\[ |u|\leq M,\qquad \frac{\theta _ u^ m\wedge \omega ^{n-m}}{\omega ^ n}\leq B, \qquad \operatorname {tr}_\omega \theta _ u \leq A\bigl (1+\sup _ X|\partial u|_\omega ^2\bigr ). \]
The bound requires no derivatives of the Hessian density.
Proof.
In local holomorphic coordinates write $\omega =\mathrm i g_{i\bar j}\, \mathrm dz_ i\wedge \mathrm d\bar z_ j$ and $\theta _ u=\mathrm i(b_{i\bar j}+\kappa u_{i\bar j})\, \mathrm dz_ i\wedge \mathrm d\bar z_ j$, where the fixed constant $\kappa >0$ accounts for the convention for $\mathrm{dd}^{\mathrm{c}}$. For eigenvalues $\lambda \in \Gamma _ m$, $m\geq 2$, one has
\[ \sum _ i\lambda _ i^2 =\left(\sum _ i\lambda _ i\right)^2-2S_2(\lambda ) <\left(\sum _ i\lambda _ i\right)^2. \]
Thus the assumed trace estimate bounds every eigenvalue in absolute value by $A(1+\sup |\partial u|^2)$.
Suppose contrary to the claim that a sequence $u_ j$ has maxima $R_ j=\sup |\partial u_ j|_\omega \to \infty $, attained at $x_ j$. Pass to $x_ j\to x_\infty $ and work in a fixed chart, changing coordinates linearly so that $g(x_\infty )=I$. If $y_ j$ is the coordinate of $x_ j$, set
\begin{align*} v_ j(z)& =u_ j(y_ j+z/R_ j),\qquad g_ j(z)=g(y_ j+z/R_ j),\\ E_ j(z)& =R_ j^{-2}b(y_ j+z/R_ j)+\kappa H(v_ j)(z). \end{align*}
These are defined on balls with radii tending to infinity. On each fixed ball $g_ j\to I$, the smooth background term tends to zero, and $E_ j$ is uniformly bounded by the eigenvalue estimate. In particular $H(v_ j)$ and its Euclidean trace are uniformly bounded. Interior estimates for the fixed Euclidean Laplacian give local uniform $W^{2,q}$ bounds for every finite $q$. A diagonal subsequence converges in $C^1$ on compact subsets to a bounded globally Lipschitz $v$, with $|\partial v(0)|=1$.
The Hermitian matrices $g_ j^{-1/2}E_ jg_ j^{-1/2}$ belong to the interior of $C_{n,m}$ and differ from $\kappa H(v_ j)$ by a term tending uniformly to zero on fixed balls. Pairing with any fixed matrix in the dual cone and passing in distributions proves that $v$ is admissible. Its equation is
\[ \frac{S_ m(g_ j^{-1/2}E_ jg_ j^{-1/2})}{\binom nm} =R_ j^{-2m}\frac{\theta _{u_ j}^ m\wedge \omega ^{n-m}}{\omega ^ n} (y_ j+z/R_ j)\leq BR_ j^{-2m}. \]
We show maximality by lower tests, rather than by an unproved convergence assertion for variable-background Hessian measures. If a smooth lower test $\psi $ at $z_0$ had $H(\psi )(z_0)$ in the interior, choose a small $\delta >0$ so that $H(\psi )-\delta I$ remains there near $z_0$. Strictifying the contact by subtracting $\delta |z-z_0|^2$ gives interior minimum points $z_ j\to z_0$ for $v_ j$ minus this test. At those points
\[ E_ j\geq R_ j^{-2}b+\kappa (H(\psi )-\delta I). \]
The matrix on the right, measured with $g_ j$, is uniformly inside the cone. Monotonicity of $S_ m$ under nonnegative addition gives a fixed positive lower bound for the left side of the preceding equation. This contradicts $BR_ j^{-2m}\to 0$. Hence $v$ is maximal. The Liouville theorem makes it constant, contradicting $|\partial v(0)|=1$.
$\square$
Pome
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