11.45 Selection for logarithmic models on productsA167FA614566

Proof. First take the ordinary envelope on the curve with background $K\omega _1$ and obstacle $\min (\psi +C,0)$. Rotation of $z$ preserves both the obstacle and the admissible family, so the envelope is rotation invariant. With $s=\log |z|^2$ and $F(s)=K\log (1+\mathrm{e}^ s)$, its profile plus $F$ is convex and lies below $F$. The comparison with $\psi +C$ bounds its final slope by $a$. At $s_*=\log R$, one has $F'(s_*)=a$. Above $s_*$, convexity and the final slope bound give

and below $s_*$ the obstacle gives $u\leq 0$. The function in (11.45.2) attains both bounds. Its convex profile, including its logarithmic value at infinity, gives a curve potential, and direct calculation gives $\psi \leq \phi _{a,d}\leq \psi +C_*$. It is therefore a competitor for every $C\geq C_*$, so the curve envelope is exactly $\phi _{a,d}$.

A product competitor restricts to each curve $\mathbb P^1\times \{ y\} $ as either the constant $-\infty $ or a curve competitor. It is thus bounded above by $\phi _{a,d}$. Conversely, the curve model is a product competitor because the vertical form $b\kappa $ is positive. Hence the product envelope is the same function. Ordinary-type invariance of the singularity envelope gives its model property. Moreover,

which proves the smaller-background assertion.

Off $D$, the horizontal curvature of $\phi _{a,d}$ is $K\mathbf1_{\{ |z|^2<R\} }\omega _1$. The first derivatives of its convex profile agree at $R$, so there is no boundary-circle atom. Expanding the product off $D$ gives the displayed measure, and the non-pluripolar product assigns $D$ zero mass. Finally, in terms of $\ell $ the residual extension is

The formulas agree at the moving interface and extend to both endpoints. On compact positive parameter ranges each logarithm is used away from its singular endpoint. Joint continuity on the resulting compact set proves uniform variation. $\square$

Proof. The one-sided ordinary class has the description

Indeed, $u\preceq \psi $ gives $\nu (u,D)\geq d$. Divisorial removal, lemma 5.38.1, shows that $\theta +\mathrm{dd}^{\mathrm{c}}u-d[D]\geq 0$. Thus the locally integrable difference $u-\psi $ has an $\eta $-plurisubharmonic representative $r$. The equality $u=\psi +r$ holds off $D$ by canonical equality and on $D$ because both sides equal $-\infty $. Conversely, adding $\psi $ to $r$ adds the positive current $d[D]$; since $r$ has a finite compact upper bound, the one-sided comparison follows. Off $D$ the currents agree, so finite-level locality identifies their extended products, including at any additional poles of $r$. Both measures give $D$ zero mass, proving the global equality. The bounded ordinary type corresponds exactly to bounded $r$, as in theorem 11.41.1.

The function $g_\sigma =\mathrm{e}^{\sigma \psi }f$ extends continuously by zero on $D$ and is positive elsewhere. Since $\eta >0$, theorem 14.19.3 gives a continuous solution $R_\sigma $ of

For this bounded density, the greatest solution of theorem 15.35.2 is bounded and coincides with $R_\sigma $ by corollary 14.19.12. Thus it dominates every $\eta $-plurisubharmonic exponential subsolution. The correspondence (11.45.6) transfers this order to the entire one-sided ordinary class and gives $s_\sigma =\psi +R_\sigma $. Bounded residual uniqueness gives uniqueness in the ordinary type. The bounded residual equation has finite mass, so the singular equation does too. $\square$

Proof. We construct uniformly bounded residuals on the curve, then use their equation to determine both the normalized limit and the scalar error.

Uniform construction. The kernel has uniformly integrable logarithmic tails and $C_ G=\sup _ x\int |G(x,z)|\omega _1(z)<\infty $. These facts follow by rotation invariance and the radial integral of $|\log \ell +1|$ on $0<\ell <1$; the Poincaré–Lelong formula and $\int _0^1\log \ell \, \mathrm{d}\ell =-1$ give its curvature and mean. Truncating the kernel shows that convolution of bounded densities is continuous.

Temporarily omit subscripts. On continuous mean-zero functions on the curve define

For all late parameters choose fixed bounds $a\leq a_{\max }$, $d\leq d_{\max }$ and $0<h_{\min }\leq H\leq h_{\max }$. Put $B=4a_{\max }h_{\max }/h_{\min }$ and choose $R_0>C_ G(B+a_{\max })$. On the ball $\| w\| _\infty \leq R_0$, for sufficiently small $\sigma $,

Here $\int \ell ^{\sigma d}\omega _1=(1+\sigma d)^{-1}$. Thus $T_\sigma $ maps the mean-zero ball into itself. Since $\int A(w)\omega _1=a$, its density derivative in direction $h$ is

It follows that $\| T_\sigma (w)-T_\sigma (v)\| _\infty \leq 2\sigma B C_ G\| w-v\| _\infty $. The map is a contraction for all sufficiently small exponents, uniformly over the late parameters. Its fixed point $w_ j$ has mean zero and $\| w_ j\| _\infty \leq R_0$.

Write $Z_ j=Z(w_ j)$ and $A_ j=A(w_ j)$. The fixed-point equation gives $\mathrm{dd}^{\mathrm{c}}w_ j=(A_ j-a_ j)\omega _1$, and the potential

has residual form $A_ j\omega _1+b_ j\kappa $, continuous and semipositive. On coordinate balls, mollification of $w_ j$ converges together with its continuous complex Hessian. The local bounded product, expanded after a common strict smooth correction, therefore equals $N b_ j^ kA_ j\, \mathrm{d}W$. Removing $D$ identifies the singular product with the same measure. The identity

shows that (11.45.11) solves the exponential equation in the ordinary type. By lemma 11.45.4, it is $s_ j$.

Residual and normalized limits. Uniform boundedness and dominated convergence give $Z_ j\to m$ and $A_ j\to A_0$ in $L^1(\omega _1)$, with a common bound on $A_ j$. For uniformly bounded densities $F_ j\to 0$ in $L^1$, one has

First letting $j\to \infty $ and then $M\to \infty $ proves $w_ j\to w_0$ uniformly. Although $A_ j=0$ on $D$ and $A_0>0$ there, the $L^1$ convergence suffices for this argument.

Put $S_ j=\sup _ X(\psi _ j+w_ j)$. A point with $\ell =1$ supplies a uniformly bounded competing value. Since $d_ j\geq d_{\min }>0$, the upper bound $d_{\min }\log \ell +R_0$ tends uniformly to $-\infty $ near $D$, so all maximizing points stay in a fixed compact subset of $X\setminus D$. Uniform convergence there gives $S_ j\to S_0$. Hence $v_ j=\psi _ j+w_ j-S_ j$ has the asserted $L^1$ and local uniform limits. For capacity convergence, the complements of $\{ \psi <-T\} $ are compact subsets of $X\setminus D$. Choose a fixed Hermitian form for which $\psi $ is plurisubharmonic and apply (5.29.4); their omitted sublevels have capacity at most $C/T$. Local uniform convergence, followed by $T\to \infty $, proves convergence in capacity, with the common poles omitted.

The measure formula and uniform convergence of $b_ j$ give total-variation convergence to $N b^ kA_0\, \mathrm{d}W=c_0 f\, \mathrm{d}W$. The same bounded product calculation gives this equation for $v_0$. The residual extensions in proposition 11.45.3 show directly that $v_0\sim \phi _{a,d}$. Choose $C$ with $v_0\geq \phi _{a,d}-C$. For every $t\geq 1$, the potential $q_ t=\phi _{a,d}-tC$ satisfies

which proves membership in the relative envelope class by definition 16.28.61.

Scalar centering. The exact formula is $c_ j=a_ j\mathrm{e}^{\sigma _ j S_ j}/Z_ j$. In the difference quotient

the negative part is bounded by $h_{\max }(d_{\max }|\log \ell |+R_0)$ and the positive part by a constant. Dominated convergence gives a limit $\int H(\psi +w_0)\omega _1=mJ$. Since $m_ j\to m>0$,

The identity $c_ j/c_{0,j}=m_ j\mathrm{e}^{\sigma _ j S_ j}/Z_ j$ proves (11.45.10). Subtracting $\sigma _ j^{-1}\log c_{0,j}$ in (11.45.11) leaves $\psi _ j+w_ j-\sigma _ j^{-1}\log (Z_ j/m_ j)$, proving the centered limit in all three topologies. $\square$

Proof. For a separated critical subsolution, (11.45.6) and the product expansion give $\mathrm{MA}^{\mathrm{ext}}_{a\omega _1}(r)\geq A_0\omega _1$. Closed curve truncations have mass $a$, so the non-pluripolar measure has mass at most $a$. Since $\int A_0\omega _1=a$, it must equal $A_0\omega _1$. The positive closed current $a\omega _1+\mathrm{dd}^{\mathrm{c}}r$ also has mass $a$, so it has no remaining measure on the pole set. Consequently $\mathrm{dd}^{\mathrm{c}}r=(A_0-a)\omega _1$ as distributions, and $r=w_0+C$ because harmonic distributions on $\mathbb P^1$ are constant. Supremum zero gives $C=-S_0$, while the mean condition gives $C\leq -J$.

For a general $u\preceq \phi _{a,d}$ with coefficient $c>0$, write $u=\psi +r$ by (11.45.6) and put $B_ r=\sup _ Xr$. Since $\psi \leq 0$, the potential $u+\sigma ^{-1}\log c-B_ r$ is an exponential subsolution for every $\sigma >0$. Greatestness gives

At a point off $D$ where $r$ is finite, subtract $\sigma ^{-1}\log c_0$, multiply by $\sigma $ and use the fixed-data case of theorem 11.45.9. This yields $\log c\leq \log c_0$. The potential $v_0$ attains $c_0$. $\square$

Proof. Let $H_ t$ be the sphere heat operator. On spherical harmonics of degree $\ell $, it acts by $\mathrm{e}^{-t\ell (\ell +1)}$. The harmonic series defines a smooth heat solution for $t>0$; for smooth initial functions, the maximum principle gives positivity and contraction in the uniform norm. Density of smooth functions extends $H_ t$ to a positive, strongly continuous contraction on $C(\mathbb P^1)$. Choose $M\geq \max V$ and define

Then $\| R_ M\| \leq M^{-1}$ and $(M-\Delta )R_ Mg=g$ in distributions. For any $g\in C(\mathbb P^1)$, the map

contracts by at most $1-\min V/M<1$. Its continuous fixed point satisfies $\Delta f=Vf+g$ in distributions, with a continuous right-hand side. The functions $f_\varepsilon =H_\varepsilon f$ are smooth and satisfy

uniformly. Thus $(\Delta -V)f_\varepsilon \to g$ uniformly. Testing against the finite measure $T$ gives $\int g\, \mathrm{d}T=0$ for every continuous $g$, hence $T=0$. $\square$

Proof. We first define the weighted current and variation, then prove invariant-measure uniqueness and apply the supporting inequality on the full space $C(X)$.

The current and operator. Since $A/b$ depends only on the first curve, $\mathrm{d}((A/b)\omega _1)=0$ distributionally; this follows also by uniform smooth approximation of its coefficient. Thus $Q$ is closed, and $\int \rho \Omega ^2=\int A\omega _1/a=1$. The coefficient $\rho $ is smooth and $\Omega $ is continuous, so the product defining $Q$ is well defined. Uniform positivity of $\Omega $ gives $\Omega +t\mathrm{dd}^{\mathrm{c}}h>0$ for sufficiently small signed $t$. The bounded-product calculation used in theorem 11.45.9, now with the smooth perturbation $th$, gives

This proves the two linearizations; mixed components of $\mathrm{dd}^{\mathrm{c}}h$ have zero wedge with the diagonal $\Omega $. Distributional Stokes gives $\int L_ uh\, \mathrm{d}\nu =2\int \mathrm{dd}^{\mathrm{c}}h\wedge Q=0$.

Invariant probabilities. Choose real spherical harmonics $e_ j$ on the second sphere, with $e_0=1$ and $\Delta _2e_ j=-\lambda _ je_ j$. Only the constant mode has eigenvalue zero. For an invariant probability $\mu $, define finite signed measures $T_ j$ on the first sphere by

Testing $L_ u$ on $f(x)e_ j(y)$ gives

For $j=0$, testing on spherical harmonics of the first sphere shows that $T_0=C\omega _1$. Their finite span is uniformly dense, because restrictions of polynomials decompose into harmonics and polynomials are dense on the sphere. The first marginal of $\mu $ is therefore $CA\omega _1$, and probability normalization gives $C=1/a$. For every nonconstant mode, lemma 11.45.15 with $V=\lambda _ jA/b>0$ gives $T_ j=0$. As these are identities of measures, their test functions may then be multiplied by the continuous function $A$. They determine all spherical harmonic moments of $\mu $. Uniform density of harmonics and of finite product tests gives $\mu =(A/a)\omega _1\wedge \omega _2$.

Supporting probabilities and weights. Let $\mu \in \mathcal A_\phi (F)$ and, for smooth real $h$, set

This is continuous, and $u+th$ solves the ordinary-type equation with density $\mathrm{e}^{F_ t}$ and scalar $c_0$. Thus $B_\phi (F_ t)=B_\phi (F)$, by uniqueness of the scalar. Uniform positivity gives $F_ t-F=tL_ uh+O(t^2)$ in the uniform norm. Apply (11.34.1) for both signs of $t$ to obtain $\int L_ uh\, \mathrm{d}\mu =0$. The perturbed density need not depend on the first curve alone: the inequality is the one on all of $C(X)$. Invariant uniqueness and nonemptiness of $\mathcal A_\phi (F)$ from theorem 11.34.2 now prove (11.45.17).

Finally, $\Omega ^2/\mathrm{d}W$ is bounded above and below by positive constants, so a normalized $r$ as in the statement belongs to $L^1(\mathrm{d}W)$ and $r\Omega $ is an order-zero current. Distributional Stokes makes $r\Omega ^2$ invariant. The probability uniqueness just proved gives $r\Omega ^2=\rho \Omega ^2$, hence $r=\rho $ almost everywhere. $\square$

Proof. Existence of the smooth positive weights is theorem 11.4.2, applied to the residual metrics $\Omega _ j$ on the surface. Put $\nu _ j=\rho _ j\Omega _ j^2$ and $L_ jh=2\Omega _ j\wedge \mathrm{dd}^{\mathrm{c}}h/\Omega _ j^2$. Distributional Stokes gives $\int L_ jh\, \mathrm{d}\nu _ j=0$. Uniform convergence to the positive form $\Omega $ supplies a common determinant lower bound and gives $L_ jh\to L_ uh$ uniformly for every fixed smooth $h$. Every weak cluster point of the probabilities $\nu _ j$ is therefore invariant for $L_ u$. By theorem 11.45.16, it equals $\nu $, proving convergence of the whole sequence. The same determinant convergence gives $F_ j\to F$ uniformly.

For a smooth real $(1,1)$ test form $\xi $, the ratios $\Omega _ j\wedge \xi /\Omega _ j^2$ converge uniformly to $\Omega \wedge \xi /\Omega ^2$. Consequently,

Similarly, the positive densities $p_ j=\Omega _ j^2/\mathrm{d}W$ converge uniformly to $p=\Omega ^2/\mathrm{d}W$. For every $g\in C(X)$,

These arguments require no pointwise control of the weights.

For existence, heat-smooth $A$ on the first sphere to positive smooth functions $A_ j$ of the same mean $a$, converging uniformly to $A$. Green inversion gives smooth mean-zero $w_ j(x)$ with $\mathrm{dd}^{\mathrm{c}}w_ j=(A_ j-a)\omega _1$ and $w_ j\to w_0$ uniformly. Their residuals converge uniformly to $\Omega $. Adding $\varepsilon _ j h(x,y)$ for a fixed smooth function depending on both variables and $\varepsilon _ j\to 0$ preserves positivity for all sufficiently small coefficients and supplies nonsymmetric examples. Each $u_ j$ retains the fixed logarithmic divisor, and its non-pluripolar equation has scalar $c_0$ and logarithmic density $F_ j$. $\square$

Proof. The height $t$ is uniform on $[-1,1]$, so $\int A\omega _1=1$ and $\int tA\omega _1=0$. Since $\Omega ^2=2bA\, \mathrm{d}W$, its mass is four and its normalized height moment is

The weight is $1/(2b)$ and $\nu =A\, \mathrm{d}W$. Moreover, $\mathrm{dd}^{\mathrm{c}}\theta =-2t\, \mathrm{d}W$, so the background has no smooth positive Gauduchon potential perturbation.

For the smoothing family, direct differentiation gives

It is positive of mass one and converges weakly to $\delta _ p$. Thus the positive smooth forms are $\widehat\Omega _\delta =a\omega _1+dP_\delta +b\omega _2$. Their classical weights and probabilities are

The weighted forms are closed, so these are the normalized classical adjoints. Their limits follow from the curve measure convergence. In particular, the weighted-current limit contains $d[D]/(2(a+d)b(p))$. These smoothings fill the pole and do not satisfy the fixed-divisor and uniform-residual hypotheses of proposition 11.45.18. $\square$