Theorem 11.45.1623BDF62163E8

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. $\square$