14.24.12 Concentrating smooth test potentialsE597C8D6FBF4

Proof. Choose finitely many coordinate balls $B_ j=\{ |z_ j|<R_ j\} $, with closures in their charts, such that the concentric smaller balls $\{ |z_ j|<r_ j\} $ cover $X$. On $B_ j$, let $\psi _ j$ solve the Dirichlet problem

The density of $g\omega ^ n$ with respect to coordinate volume is smooth and strictly positive. The classical smooth Dirichlet theorem therefore applies, since $C_ j(|z_ j|^2-R_ j^2)$ is a smooth strict subsolution with zero boundary values for sufficiently large $C_ j>0$. Thus $\psi _ j$ is smooth and strictly plurisubharmonic on $\overline B_ j$, and the maximum principle gives $\psi _ j\leq 0$.

We need a lower bound for $\psi _ j$ independent of $\sup _ Xg$. Write the coordinate equation as $\det (\partial _ a\bar\partial _ b\psi _ j)=D_ j$, and put $M_ j=-\min _{B_ j}\psi _ j$. Let $x_ j$ be a minimum point. For each real slope $p$ with $|p|<M_ j/(2R_ j)$, the function $\psi _ j-p\cdot (x-x_ j)$ attains its minimum at an interior point, because its value at $x_ j$ is $-M_ j$, while its boundary values are larger than $-M_ j$. At such a contact point the real Hessian $S$ of $\psi _ j$ is nonnegative, and the gradient of $\psi _ j$ maps the contact set onto the ball of these slopes. The area formula and the matrix inequality

therefore give $M_ j\leq C(n,R_ j)\| D_ j\| _{L^2(B_ j)}^{1/n}$. To prove the matrix inequality, let $J$ be the standard complex structure on the real coordinate space and average $S$ with $J^\mathsf TSJ$. Concavity of $\log \det $, and the fact that the eigenvalues of this $J$-invariant average occur in pairs, give the displayed factor; singular matrices are handled by approximation. Since the ratio of $\omega ^ n$ to coordinate volume is fixed and bounded, $\| D_ j\| _2$ is controlled by $B_0$. Consequently

where $M_0$ depends only on $X,\omega ,B_0$ and the chosen finite cover.

Next we glue the local solutions. Put $\beta _ j=\mathrm{dd}^{\mathrm{c}}|z_ j|^2$. Choose $a_ j>0$ such that $a_ j\beta _ j\leq \omega /2$ on $B_ j$, and let $t_ j=(R_ j^2+r_ j^2)/2$. Fix $\tau >0$ so small that $\tau M_0<a_ j(R_ j^2-r_ j^2)/4$ for every $j$. The smooth function

is bounded below by a fixed positive constant on the smaller ball, negative on a collar of $\partial B_ j$, and satisfies $\omega _{s_ j}\geq \omega /2+\tau \mathrm{dd}^{\mathrm{c}}\psi _ j$. Choose a smooth convex function $\phi _ j:\mathbb R\to \mathbb R$ approximating $s\mapsto \max (s,0)$, equal to zero below a small negative threshold and to $s$ above the corresponding positive threshold, with $0\leq \phi _ j'\leq 1$ and $\phi _ j''\geq 0$; the thresholds are chosen smaller in absolute value than both margins of $s_ j$. Set $Q_ j=\phi _ j(s_ j)$ on $B_ j$ and extend it by zero. This extension is smooth, and

On the smaller ball, $Q_ j=s_ j$ and $\omega _{Q_ j}\geq \tau \mathrm{dd}^{\mathrm{c}}\psi _ j$. If $N$ is the number of balls, put $q=N^{-1}\sum _ jQ_ j$. Then $\omega _ q=N^{-1}\sum _ j\omega _{Q_ j}\geq \omega /2$ everywhere. Each point belongs to at least one smaller ball $\{ |z_ j|<r_ j\} $, so positivity of the other summands and (14.24.15) give

Thus (14.24.14) holds with $\varepsilon =(\tau /N)^ n$, independent of $\sup g$ and of all derivatives of $g$. $\square$

of theorem 14.24.1. By corollary 14.19.29, $u,v$ are continuous. In dimension one, $\mathrm{dd}^{\mathrm{c}}w=0$, so the maximum principle makes $w$ constant. Assume $n\geq 2$.

Suppose that $w=u-v$ is nonconstant. Let $a=\min _ Xw$, $K=\{ w=a\} $, and $\mu =F\omega ^ n$. By lemma 14.24.11, $\mu (K)=0$. Choose $B_0>\| 1\| _{L^2(\omega ^ n)}$ and let $\varepsilon >0$ be the constant from proposition 14.24.13. Since $\mu (K)=0$, the compact set $K_0=K\cap \{ F\geq \varepsilon /2\} $ has zero $\omega ^ n$-volume. Fix $A>1$ with $\varepsilon A>\max _ XF$. By outer regularity of the smooth volume measure, choose an open neighborhood $O$ of $K_0$ so small that

Take a smooth $0\leq \chi \leq 1$ equal to one near $K_0$ and supported in $O$, and set $g=1+(A-1)\chi $. If $K_0=\varnothing $, take $\chi =0$. Then $g\geq 1$, $g=A$ on $K_0$, and $\| g\| _{L^2(\omega ^ n)}<B_0$.

By proposition 14.24.13, there is a smooth strictly $\omega $-plurisubharmonic $q$ with $\omega _ q^ n=G\omega ^ n\geq \varepsilon g\omega ^ n$. On $K_0$ we have $G\geq \varepsilon A>F$; on $K\setminus K_0$ we have $F<\varepsilon /2<\varepsilon \leq G$. Thus $G>F$ throughout the compact set $K$. By compactness and continuity, we can choose $0<\kappa <1$ such that $F<\kappa G$ on $K$; put $W=\{ F<\kappa G\} $. The complement of this open neighborhood of $K$ is compact, and $w-a$ has a strictly positive minimum there if it is nonempty. Hence some $\eta >0$ satisfies $\{ w<a+\eta \} \subset W$.

Proposition 14.24.2 asserts that $\mu (\{ w<a+\eta \} \cap \{ F>\kappa G\} )>0$. By the preceding inclusion, however, this set is empty. This contradiction shows that $w$ is constant. If $\sup _ Xu=\sup _ Xv$, the constant is zero. $\square$

Proof. The functions $u,v$ are continuous by corollary 14.19.29, and in dimension one the assertion follows from $\mathrm{dd}^{\mathrm{c}}(u-v)=0$ and the maximum principle. Let $n\geq 2$, suppose that $w=u-v$ is nonconstant, and put $K=\{ w=\min _ Xw\} $. The proof of lemma 14.24.11 applies with $U=X$, since proposition 14.24.5 gives the pushforward identity and local coercivity on every chart. Thus $K$ has zero volume.

Fix $B_0>\| 1\| _{L^2(\omega ^ n)}$ and the corresponding $\varepsilon $ in proposition 14.24.13. Choose $A>1$ with $\varepsilon A>C_0$, and a smooth cutoff function $\chi $, equal to one on a neighborhood $W$ of $K$ and supported in an open set of volume so small that $g=1+(A-1)\chi $ satisfies $\| g\| _2<B_0$. The resulting test potential $q$ has density $G\geq \varepsilon g$. Choose $\kappa <1$ with $C_0<\kappa \varepsilon A$. Then $F<\kappa G$ almost everywhere on $W$. By continuity of $w$, a sufficiently thin set $\{ w<\min _ Xw+\eta \} $ lies inside $W$, which contradicts proposition 14.24.2. Hence $w$ is constant. $\square$