15.36 Radial selection for integrable densitiesC10FB3AB7407

Proof. Choose a smooth nondecreasing concave function $\chi $ on $[0,+\infty )$, equal to the identity on $[0,1]$ and constant on $[2,+\infty )$, and put $q_ j=j\chi (q_\alpha /j)$. Concavity gives $q_ j\uparrow q_\alpha $ and

The functions $q_ j$ are smooth and positive on $Y$: near $o$ they are constant, and elsewhere the density is smooth. Let $w_ j$ be the continuous solution of $\mathrm{MA}_\kappa (w_ j)=\mathrm e^{\lambda w_ j}q_ j\kappa ^ k$ from theorem 14.19.3. It is the greatest subsolution by theorem 15.35.2 and bounded uniqueness. Since $u_\alpha \leq 0$ and $\mathrm{MA}^{\mathrm{ext}}_\kappa (u_\alpha )=q_\alpha \kappa ^ k$, it is a subsolution for every $q_ j$, whence $u_\alpha \leq w_ j$. Density comparison gives $w_ j\downarrow w$.

The pullback $W_ j=\lambda ^{-1}\log b+w_ j$ is a bounded solution on $X$, by the product expansion in theorem 15.30.3. For the smooth density $q_ j\mu $ it is therefore also the greatest subsolution. Every subsolution for $q_\alpha \mu $ is a subsolution for $q_ j\mu $, so $U_{\lambda ,\alpha }\leq W_ j$.

Monotone convergence in capacity and theorem 15.34.2 on $Y$ identify the equation for $w\geq u_\alpha $. Moreover, Stokes gives $\int _ Y\mathrm e^{\lambda w_ j}q_ j\kappa ^ k=1$. The integrands are dominated by $\mathrm e^{\lambda \sup _ Yw_1}q_\alpha $, so dominated convergence gives the same identity for $w$. Its pullback is a product solution, hence lies below the greatest solution on $X$. Together with $U_{\lambda ,\alpha }\leq \lim _ jW_ j$, this proves separation and all measure assertions.

Uniqueness makes each $w_ j$ invariant under the unitary action on the affine coordinates $z\in \mathbb C^ k$, and the limit has the same invariance. A radial $\kappa $-plurisubharmonic function has a convex nondecreasing profile $\rho +w$. Symmetric convolution of this convex profile on compact $s$ intervals, decreasing as the scale decreases, and bounded-product continuity give its radial measure $\mathrm d((\rho +w)'^ k)$, as in lemma 15.30.1. The bound $w\geq u_\alpha $, whose growth is $o(|s|)$ at $-\infty $, forces the initial derivative to vanish. The positive density makes the flux continuous and positive at finite $s$, yielding (15.36.3). Its final value is the mass one already proved. The profile is consequently smooth on finite $s$ intervals. $\square$

Proof. Choose $[\alpha _-,\alpha _+]\Subset (0,1)$ containing all late parameters. The density formula in lemma 15.30.1 gives, uniformly for $\delta $ in this interval and $t=-s$ large,

Away from $o$ the densities are jointly smooth, including on the hyperplane at infinity. Thus one integrable density $g$ dominates all $q_\delta $, and dominated convergence gives $q_{\alpha _ j}\to q_\alpha $ in $L^1$.

For every normalized $\kappa $-plurisubharmonic $v$, theorem 5.25.2 gives $\int _ Y(-v)\kappa ^ k\leq C_ k$. Absolute continuity of the integral of $g$ therefore implies

Taking $T=\lambda ^{-1/2}$ shows that

uniformly in $\delta $ and $v$. Write $v_ j=v_{\lambda _ j,\alpha _ j}$ and $a_ j=a_{\lambda _ j,\alpha _ j}$. The exact mass identity of the lemma is

It gives $\mathrm e^{\lambda _ ja_ j}\to 1$, hence $\lambda _ ja_ j\to 0$. The actual densities $\mathrm e^{\lambda _ ja_ j}\mathrm e^{\lambda _ jv_ j}q_{\alpha _ j}$ converge in $L^1$ to $q_\alpha $ and are bounded by $2g$ for all late $j$.

Their cumulative radial fluxes $Q_ j$ consequently converge uniformly to the flux $Q_\alpha $ of $u_\alpha $. By (15.36.3), $v_ j'\to u_\alpha '$ uniformly on compact $s$ intervals. The normalized $L^1$ bound controls one value of each profile on a fixed annulus; thus every sequence has a subsequence converging uniformly on compact $s$ intervals to $u_\alpha +d$ for some finite constant $d$.

Near infinity the common bound for the densities is smooth and bounded. Hence $1-Q_ j(s)\leq C\mathrm e^{-s}$ and $|v_ j'(s)|\leq C\mathrm e^{-s}$ for large $s$. Integrating gives uniform convergence up to the hyperplane at infinity. At the other end, monotonicity of $\rho +v_ j$ gives, for $s<s_0$,

Choose $s_0$ so negative that the right side is negative for all late members of the subsequence. The supremum zero is then attained outside this inner ball. Uniform convergence on its compact complement gives $\sup _ Y(u_\alpha +d)=0$, so $d=0$. Normalized compactness identifies every $L^1$ cluster limit with $u_\alpha $; this proves convergence of the entire family and its local uniform convergence off $o$.

For capacity convergence, fix $T>0$. The sublevel $\{ u_\alpha <-T\} $ is a neighborhood of $o$, and its complement is a compact set on which convergence is uniform. Its capacity is at most $C/T$ by (5.29.4). Let $j\to \infty $ and then $T\to \infty $. The same argument on $X$ uses the normalized pullback of $u_\alpha $ and proves convergence in Hermitian capacity there; comparison sets omit pole sets. Pullback also preserves the $L^1$ and local uniform limits. The product expansion proves total-variation convergence of the measures, and theorem 15.30.3 supplies the exact limiting mass and envelope class. $\square$

Proof. As $t=-s\to \infty $, the radial profile of lemma 15.30.1 satisfies

Thus the first moment is finite precisely when $(k+1)\alpha <k$. In that case, write $x=1-\psi $. Its cumulative flux is

Integration by parts gives (15.36.9); the pole boundary term vanishes since $(k+1)\alpha -k<0$.

Suppose $\alpha <k/(k+1)$ and choose the parameter interval in the preceding proof with $\alpha _+<k/(k+1)$. Its dominating density $g$ satisfies $(-u_{\alpha _+})g\in L^1$. Write $w_ j=w_{\lambda _ j,\alpha _ j}=v_ j+a_ j$. Since $w_ j\geq u_{\alpha _ j}\geq u_{\alpha _+}$, the mass identity and $\mathrm e^ r\geq 1+r$ give $\int _ Yw_ jq_{\alpha _ j}\kappa ^ k\leq 0$. On a fixed annulus $A$ away from $o$, the normalized smooth-volume $L^1$ bound gives

On the complement, use $w_ j\geq u_{\alpha _+}$ and the integrable lower bound. The annulus has uniformly positive measure, so $0\leq a_ j\leq C_*$. Along any subsequence with $a_ j\to a$, local uniform convergence gives $w_ j\to u_\alpha +a$ off $o$. In the exact identity

the negative part is dominated by $(-u_{\alpha _+})g$ and the positive part by $C_*\mathrm e^{C_*}g$. Dominated convergence gives $M_\alpha +a=0$. Every subsequence therefore has the asserted scalar limit. Adding these convergent constants to the normalized potentials gives all three claimed topologies.

If $\alpha \geq k/(k+1)$, use instead

The integrands are nonnegative and tend off $o$ to $(-u_\alpha )q_\alpha $. Fatou's lemma forces $a_ j\to +\infty $. The preceding theorem gives $\lambda _ ja_ j\to 0$; local uniform convergence of $v_ j$ gives the stated divergence of $w_ j$. This includes sequences approaching the moment boundary from below. $\square$

Proof. The product formula reduces the inequality to $\mathrm{MA}^{\mathrm{ext}}_\kappa (r)\geq q_\alpha \kappa ^ k$. Closed Stokes gives each bounded truncation mass one, so the extended measure of $r$ has mass at most one. The inequality therefore gives equality and full mass. Closed-class uniqueness yields $r=u_\alpha +d$. In the finite-moment range its mean is $M_\alpha +d$, which is nonpositive exactly when $d\leq -M_\alpha $. $\square$