14.24 Bounded uniqueness and connectednessE0DB623401C5

Proof. Choose $\delta \in (0,1]$ with $\omega _ q\geq \delta \omega $. For $0<t<1$, put $b_ t=(1-t)v+tq$, $m_ t=\min _ X(u-b_ t)$, and $E_ s=\{ u<b_ t+m_ t+s\} $. Then $\omega _{b_ t}\geq t\delta \omega $. Write $\mu =F\omega ^ n$, $d=\kappa ^{-1/n}-1$, and $\gamma =nd$. By (5.28.9),

On $L=\{ F\leq \kappa G\} $ this is at least $(1+\gamma t)\mu $, since $G^{1/n}\geq (1+d)F^{1/n}$ there, and on $R=\{ F>\kappa G\} $ it is at least $(1-nt)\mu $. Choose $s>0$ so small that the reverse comparison factor in (14.19.7), with $\varepsilon =t\delta $, is at most $1+\gamma t/2$. Modified comparison on $E_ s$, with $x=\mu (E_ s\cap L)$ and $y=\mu (E_ s\cap R)$, gives

The open set $E_ s$ is nonempty by the definition of $m_ t$. As $\omega _{b_ t}^ n\geq (t\delta )^ n\omega ^ n$, (14.19.7) forces $\mu (E_ s)>0$, hence $y>0$.

Let $a=\min _ X(u-v)$ and $\Lambda =\| v-q\| _\infty $. Since $u-b_ t=u-v+t(v-q)$, we have $|m_ t-a|\leq t\Lambda $ and $E_ s\subset \{ u-v<a+2t\Lambda +s\} $. Given $\eta $, choose $t$ and then $s$ with $2t\Lambda +s<\eta $. Since $y=\mu (E_ s\cap R)>0$, this proves (14.24.3). Interchanging $u$ and $v$ gives the assertion for the near-maximum set. $\square$

Proof. Locally, expand $T$ into powers of the smooth form $\omega $ multiplied by closed Bedford–Taylor products of $\mathrm{dd}^{\mathrm{c}}u$ and $\mathrm{dd}^{\mathrm{c}}v$. The operator $\mathrm{dd}^{\mathrm{c}}$ then acts only on the powers of $\omega $, and the local mass bounds in section 5.27 show that $\mathrm{dd}^{\mathrm{c}}T$ is a finite signed measure.

We first construct $E$ and prove (14.24.6). On a coordinate ball, convolve $u$, $v$ and the coefficients of $\omega $ with a common nonnegative radial mollifier. Write the resulting smooth positive forms as $A_\rho =\omega _\rho +\mathrm{dd}^{\mathrm{c}}u_\rho $ and $B_\rho =\omega _\rho +\mathrm{dd}^{\mathrm{c}}v_\rho $, and let $T_\rho $ be their mixed sum. On a slightly larger ball, add the same fixed multiple of $|z|^2$ to each potential to obtain plurisubharmonic functions. Their radial regularizations decrease to the originals, so Bedford–Taylor continuity applies to all closed Hessian products. Since $u,v$ are continuous, $u_\rho ,v_\rho ,w_\rho $ converge uniformly on compact subsets.

For a closed smooth product $S_\rho $ appearing in $T_\rho $, the identity

has a distributional limit: both Hessian products on the right converge weakly with uniformly bounded local variation, and their weights converge uniformly. The same holds after multiplication by the convergent smooth factors coming from $\omega _\rho $. Thus the positive measures $E_\rho =\mathrm{d} w_\rho \wedge \mathrm d^{\mathrm c}w_\rho \wedge T_\rho $ converge to a finite positive measure $E$. By (14.24.9), this limit does not depend on the regularization, so the local limits agree on overlapping charts.

For smooth regularizations, the chain rule gives

The last current tends to zero, because $\mathrm{dd}^{\mathrm{c}}w_\rho \wedge T_\rho =A_\rho ^ n-B_\rho ^ n$ tends weakly to $A^ n-B^ n=0$ with uniformly bounded variation, and $\chi '(w_\rho )$ converges uniformly. Passing to the limit and integrating by parts on the compact manifold $X$ gives (14.24.6). A partition of unity justifies this passage without a global regularization; the terms containing derivatives of the cutoff functions cancel after summation.

Let $\sigma =w_*(\mathrm{dd}^{\mathrm{c}}T)$ and $\nu =w_*E$. Equation (14.24.6) says that $\nu ''=\sigma $ as distributions on $\mathbb R$. Stokes' theorem and $\mathrm{dd}^{\mathrm{c}}w\wedge T=0$ give

Consequently $k(t)=\int (s-t)_+\, \mathrm{d}\sigma (s)$ is globally Lipschitz and vanishes outside $[a,b]$. It satisfies $k''=\sigma $, so the compactly supported distribution $\nu -k(t)\, \mathrm{d} t$ has second derivative zero and therefore vanishes. Positivity of $\nu $ and continuity of $k$ imply $k\geq 0$. This proves (14.24.7), and the assertion about level sets follows because $\nu $ is absolutely continuous.

We now prove (14.24.8). On a coordinate neighborhood of $\overline V$, set $\beta =\mathrm{dd}^{\mathrm{c}}|z|^2$ and choose $c_ V>0$ with $F\omega ^ n\geq c_ V\beta ^ n$. Let $A_{\mathrm{ac}}$ denote the absolutely continuous part of the current $A$, whose coefficients are locally integrable functions. Applying lemma 5.28.10 with the constant function $g=c_ V^{1/n}$ and differentiating the coefficient measures, we obtain $\det _\beta A_{\mathrm{ac}}\geq c_ V$ almost everywhere. The local Sobolev regularity in section 5.27 gives $w_\rho \to w$ strongly in $W^{1,2}_{\mathrm{loc}}$, so a subsequence of the gradients converges almost everywhere. Moreover, $A_\rho \to A_{\mathrm{ac}}$ almost everywhere, by differentiation of the coefficient measures.

Since $T_\rho $ contains the term $A_\rho ^{n-1}$ and all its terms are positive, Fatou's lemma, applied to $E_\rho $, yields the measure inequality

The cofactor matrix of $A_{\mathrm{ac}}$ has smallest eigenvalue at least $\det A_{\mathrm{ac}}/\operatorname {tr}A_{\mathrm{ac}}\geq c_ V/\operatorname {tr}A_{\mathrm{ac}}$, where traces and determinants are taken with respect to $\beta $. The trace is locally integrable because $A$ has locally finite mass. The chain rule for Sobolev functions, (14.24.10) and the Cauchy–Schwarz inequality give

The last factor is finite and is absorbed into $C_ V$. The values of $p'$ at the finitely many corners of $p$ are immaterial, because $E$ charges no level set of $w$. $\square$

Proof. Put $a=\min _ Xw<b=\max _ Xw$, and let $k$ be the function of proposition 14.24.5. It is Lipschitz and nonnegative, and $k(a)=0$. Choose $\epsilon _ j\downarrow 0$ with $a+\epsilon _ j<b$. For $\delta >0$, set

If $L$ is a Lipschitz constant for $k$, then $k(t)\leq L(t-a)$ on this interval, so $J_ j(\delta )\to \infty $ as $\delta \downarrow 0$. Choose $\delta _ j>0$ with $J_ j=J_ j(\delta _ j)\geq j$. Let $p_ j=0$ on $(-\infty ,a]$, $p_ j=1$ on $[a+\epsilon _ j,\infty )$, and

The pushforward identity (14.24.7) gives

Every point of $U$ has a coordinate neighborhood $V$ as in proposition 14.24.5, since $F$ is continuous. By (14.24.8), therefore, $\mathrm{d}(p_ j(w))\to 0$ in $L^1_{\mathrm{loc}}(U)$. The bounded functions $p_ j(w)$ converge pointwise and in $L^1_{\mathrm{loc}}$ to $\mathbf1_{\{ w>a\} }$, whose distributional gradient therefore vanishes on $U$. Since $U$ is connected, this indicator function is almost everywhere constant on $U$. As $w$ is nonconstant, $\{ w>a\} $ is a nonempty open set, and its intersection with the dense open set $U$ is nonempty and open; hence the constant is one. Thus $\{ w=a\} \cap U$ has zero volume, and since $F=0$ off $U$, the first assertion follows. Applying the argument to $-w$ gives the assertion for the maximum set. $\square$

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. $\square$