23.1.42 Positive fiber flux9880531015C7

Proof. Set $m(x)=\int _ EV\, \mathrm dA_ E$ and $e(x)=\int _ E|V_ w|^2\, \mathrm dA_ E$. Periodic integration by parts gives

The mean of every fiber derivative vanishes. Expanding $\Delta _ V$ yields

Thus $f(x)=\log (1+\mathrm e^ x)+m(x)-e(x)/2$ is convex.

Both $m$ and $e$ are smooth global functions on the projective base: local fiber coordinates change by holomorphic translations, preserving fiber Haar area, the vertical derivative and the Dirichlet integral. They are $U(2)$-invariant. Smoothness and radial invariance at $\xi =0$ give $m',e'\to 0$ at the negative end. At the positive end use a normal coordinate $z_0/z_1$ to the hyperplane $\{ z_0=0\} $ on the base. The invariant base functions are constant along this hyperplane and smooth across it; their $x$ derivatives tend to zero. This does not require that the pointwise potential be constant on its elliptic fibers at that end. Consequently $f'(-\infty )=0$ and $f'(+\infty )=1$. Integrating the nonnegative $f''$ proves the flux identity.

For each fixed $w$, $R_ V\geq 0$ implies $V_{xx}\geq -1$ on all of $\mathbb R$. Hence for every real $h$,

Taking $h=V_ x(x,w)$ and using $V\leq 0$ proves the derivative bound. Finally $\mu \leq 1$ proves the bound for $A_ V$. $\square$

Proof. The compact group $G=U(3)\times (\mathbb R/\mathbb Z)$ acts transitively and holomorphically on $X$ by unitary transformations and positive scalar multiplication. It preserves $\omega $. Choose a nonnegative smooth central approximate identity $\rho _\epsilon $ on $G$, supported near its identity, and put

Convolution on a compact homogeneous space is smooth. Averaging preserves the current inequality and the bounds, and centrality makes convolution commute with the $U(2)$ action. Transporting the bounded tests defining capacity shows that $\operatorname {Cap}_\omega $ is $G$-invariant.

Suppose $L>0$. By quasi-continuity, choose an open set $O$ of arbitrarily small capacity such that $u$ is continuous on $X\setminus O$. When $g$ is sufficiently close to the identity, $u(gp)$ and $u(p)$ are uniformly close whenever both points lie outside $O$. Thus, outside $O$ and for small kernel support,

Markov's inequality, followed by capacity invariance, gives

where the supremum runs over $v\in \mathrm{PSH}(X,\omega )$. Letting the capacity of $O$ tend to zero proves the claim. If $L=0$, $u=0$ and no approximation is needed. $\square$

Proof. For bounded $u\leq 0$, apply lemma 23.1.46. The bounded capacity continuity of products and quasi-continuous coefficients in section 5.27, section 5.29 passes the total estimate (23.1.45) to $u$. For the tail at $T>0$, use continuous depth cutoffs majorizing $\mathbf1_{\{ -u>T\} }$ and supported in $\{ -u>T-\delta \} $, where $0<\delta <T$. The same product continuity gives a bound by $12Q_{T-\delta }(W)$. Continuity of $(1+\sqrt{2t})W(t)$ gives $Q_{T-\delta }(W)\to Q_ T(W)$. For $T=0$, use the total estimate.

For unbounded $u$, the canonical truncations $u^ K=\max (u,-K)$ remain bounded and invariant. On $\{ u>-K\} $ their measures agree with the finite-level restrictions defining $\mu _ u$. These restrictions increase to $\mu _ u$, so monotone convergence proves (23.1.48).

Splitting at depth $T$ and rescaling $u^ T$ into a capacity test gives, for every Borel set $B$,

Indeed $\omega +\mathrm{dd}^{\mathrm{c}}u^ T\leq (1+T)(\omega +\mathrm{dd}^{\mathrm{c}}(u^ T/(1+T)))$, and the latter potential has range of length at most one. If $\sqrt t\, W(t)\to 0$, then $Q_ T(W)\to 0$. Taking first $T$ large and then the capacity small proves uniform absolute continuity.

Suppose $u_ j\to u$ in capacity in $\mathcal R$. For fixed $K$, the truncations $u_ j^ K$ converge in capacity to $u^ K$, and their weighted bounded products converge weakly. Outside $\{ u\leq -K\} $, locality identifies $\nu _{W,u^ K}$ with $\nu _{W,u}$. Both measures give this exceptional set mass at most $12Q_{K-1}(W)$ when $K>1$, by their tail estimates; this also includes any truncation mass on $\{ u=-K\} $. Hence

uniformly over $u\in \mathcal R$. Letting $K\to +\infty $ removes the truncations and proves the claimed continuity. $\square$