15.31 Singular measures on products with a curveE01ACE366291

Proof. The Poincaré–Lelong formula gives $\mathrm{dd}^{\mathrm{c}}_ xG(x,y)=\delta _ y-\kappa $. After a rotation sending $y$ to $[1:0]$, the variable $t=d_{\mathrm{ch}}(x,y)^2$ is uniformly distributed on $[0,1]$ for the measure $\kappa $. Thus

The continuous potentials $\int \max (G(x,y),-L)\, \mathrm d\mu (y)$ are $\kappa $-plurisubharmonic by lemma 5.24.1 and averaging, and decrease to $Q_\mu $. Tonelli's theorem gives its integrability and mean zero; the decreasing-limit property gives its canonical values. Distributional Fubini proves the curvature identity.

In degree one the extended measure of $p$ is

The pole set of $Q_\mu $ is polar, so non-polarity gives its asserted measure. For any $p$, distributional Stokes gives the full current $\kappa +\mathrm{dd}^{\mathrm{c}}p$ mass one, and its extended measure has mass at most one. Domination of the probability $\mu $ therefore forces both measures to equal $\mu $. The distribution $p-Q_\mu $ is harmonic on the compact connected curve, hence constant, and canonical equality gives the assertion everywhere. $\square$

Proof. Continuity gives the uniform kernel bound

To prove non-polarity, take a nontrivial subharmonic function on a coordinate disk. Cut off its positive Hessian measure on a slightly larger disk. The logarithmic potential of this finite measure has the same Hessian on the smaller disk, so their difference is harmonic there. This is the local Riesz decomposition: it also follows directly from the distributional fundamental solution and harmonic regularity. On compact subdisks the kernel $-\log |z-a|^2$ differs from $K(z,a)$ by a bounded smooth term. Symmetry of $K$, Tonelli's theorem and (15.31.6) make the logarithmic potential integrable against $\mu $; its harmonic remainder is bounded. The pole set consequently has $\mu $-mass zero. A countable disk cover proves the claim for polar Borel sets.

The group $\mathrm{SU}(2)$ acts on $Y$ by Fubini–Study isometries. Average with smooth nonnegative conjugation-invariant kernels of mass one supported near the identity. This gives smooth $Q_\epsilon $, probability measures $\mu _\epsilon $, and

Smooth Stokes gives $E(Q_\epsilon ,Q_\epsilon )=-\int Q_\epsilon \, \mathrm d\mu _\epsilon \leq \| Q\| _\infty $. Poincaré's inequality and weak Hilbert compactness show that $Q\in H^1$. Its distributional curvature identity gives (15.31.5) first for smooth $h$. Apply this identity to $\sqrt{h^2+\epsilon ^2}$, whose gradient has magnitude at most $|\mathrm dh|$. Letting $\epsilon \downarrow 0$ yields

Completion defines the trace and extends the identity.

These group averages converge strongly in $H^1$: rotations act continuously and isometrically on that space, and the averaging kernels approach the identity. They also converge pointwise to every canonical quasi-subharmonic function. Indeed their induced averages about a point are radial by conjugation invariance. In a coordinate centered at that point they are positive averages of circular means; after a smooth quadratic correction, the submean inequality and upper semicontinuity give the required limit. The correction error tends to zero, and upper semicontinuity gives the limit also at a pole. For a $\kappa $-plurisubharmonic $H^1$ function, strong convergence and the trace bound give $L^1(\mu )$ convergence, while a subsequence converges $\mu $-almost everywhere to the trace. Non-polarity makes the canonical value finite there and identifies the two. Finally, the Sobolev chain rule and smooth approximation show that Lipschitz compositions are continuous in $H^1$; the corresponding $L^1(\mu )$ limits prove the composition assertion. $\square$

Proof. Variational existence. On the real Hilbert space

consider the extended functional

It is finite at zero. Jensen's inequality and (15.31.5) give

so it is coercive and bounded below. Strong $H^1$ convergence gives $L^1(\mu )$ convergence of traces. A subsequence realizing the lower limit converges almost everywhere for $\mu $, and Fatou's lemma proves strong lower semicontinuity of the exponential term. Jensen excludes a zero integral or a logarithm tending to $-\infty $. Hölder's inequality makes the logarithmic term convex, while the Dirichlet term is strictly convex on $\mathcal H$. Separation of a closed convex epigraph makes $J_\lambda $ weakly lower semicontinuous. Weak compactness therefore gives a unique minimizer $q_\lambda $, with

Variation in a smooth mean-zero direction $h$ is justified by its boundedness, which dominates the differentiated integrand by a constant times the integrable exponential. Put $\rho _\lambda =\mathrm e^{\lambda \operatorname {Tr}_\mu q_\lambda }/Z_\lambda $. The Euler equation is

The second identity follows for arbitrary smooth tests by subtracting their $\kappa $-mean; $\int \rho _\lambda \, \mathrm d\mu =1$ fixes the constant term. The minimizer has a canonical $\kappa $-plurisubharmonic representative. Its mean-zero Green representation and lemma 15.31.4 give

The kernel bound then gives $q_\lambda \geq 1-(\mathrm e^\lambda /Z_\lambda )A_\mu $.

To obtain continuity, the continuous functions $\int \min (K(x,y),L)\, \mathrm d\mu (y)$ increase to the continuous function $1-Q(x)$. Dini's theorem gives

Multiplication by the bounded density $\rho _\lambda $ preserves this estimate. Its truncated Green potentials converge uniformly, so $q_\lambda $ is continuous. Now

solves (15.31.9) with precisely the displayed constant.

Uniqueness and the barriers. Every bounded $\kappa $-plurisubharmonic function is in $H^1$, by the same smoothing and energy bound used above after subtracting its $\kappa $-mean. For two bounded solutions $w_1,w_2$, their right-hand sides have bounded densities relative to $\mu $. The trace inequality extends distributional Stokes to their difference, giving

The difference is constant; the unit exponential integrals force that constant to be zero.

Set $L_0=Q-\sup Q\leq 0$, $R_0=Q-\inf Q\geq 0$, and $z=(L_0-w_\lambda )_+$. The $H^1$ chain rule, canonical trace and extended Stokes give

Indeed $z>0$ implies $w_\lambda <L_0\leq 0$. Thus $z$ is constant. If it were a positive constant, continuity would give $w_\lambda =L_0-z\leq -z$ everywhere, contradicting its unit exponential integral. This proves the lower barrier. Testing $(w_\lambda -R_0)_+$ gives the upper barrier with the same argument and the reversed exponential difference. In particular $\| w_\lambda \| _\infty \leq \operatorname {osc}Q$.

Selection. Write $q_\lambda =w_\lambda -\int _ Yw_\lambda \kappa $, consistent with the variational normalization. The barriers give $\mathrm e^{\lambda w_\lambda }=1+O(\lambda )$ uniformly, and the Green identity gives

The last kernel integral is bounded by $1+A_\mu $. The $q_\lambda $ are uniformly bounded, so Taylor's formula gives

This proves the whole-family estimate. Finally, lemma 15.31.3 identifies every critical subsolution as $Q+C$. Its $\mu $-mean is at most zero exactly when $C\leq -\int Q\, \mathrm d\mu $, proving the order characterization. $\square$

Proof. A nontrivial local plurisubharmonic function on a product chart has a nontrivial subharmonic restriction to the curve for smooth-volume-almost every point of $X_0$, by local integrability and Fubini's theorem. Non-polarity of $\mu $ and a second application of Fubini give its pole set product measure zero. A countable cover proves that $\nu $ is non-pluripolar.

For bounded $p\in \mathrm{PSH}(Y,\kappa )$, bounded multilinearity and lemma 5.28.4 give

Only one curve factor survives in the expansion. For arbitrary $p$, apply this to $p_ j=\max (p,-j)$, restrict to the pullback of $\{ p>-j\} $ and increase $j$. The defining extended limit gives the same formula with $\mathrm{MA}^{\mathrm{ext}}_\kappa (p)$ on the right. Lemma 15.31.3 now proves the equation, finite mass and separated uniqueness. Boundedness and continuity follow from the same constant-difference identity.

If $v_-(h_0)=0$, pull back bounded potentials whose $h_0$-masses tend to zero. Their $H$-masses are $N$ times those masses, because $\int _ Y\kappa =1$, so $v_-(H)=0$. A closed $h_0$ would give every bounded potential mass $V_0>0$ by Stokes. Hence $h_0$ is not closed; restricting $H$ to a factor slice shows that $H$ is not closed either. $\square$

Proof. Lemma 15.31.4 gives non-polarity, and theorem 15.31.8 constructs $w_\lambda $. Substituting it in (15.31.14) proves the exponential equation and its finite mass. Supremum normalization gives the displayed $c_\lambda $, and the uniform curve estimate gives both limits. For any bounded $H$-plurisubharmonic $p$ and $t\geq 1$, the constant $a_ t=t\inf _ Xp$ is an $H$-plurisubharmonic competitor below $tp$. Hence $P_ H(tp)$ is genuine for every $t$, which proves the envelope-class assertion.

Greatestness. Write $s=U_\lambda $. If $b$ is a bounded exponential subsolution, then on $\{ s<b-\epsilon \} $,

The bounded strict domination theorem, theorem 14.19.10, gives $s\geq b-\epsilon $. Letting $\epsilon \downarrow 0$ proves $s\geq b$, and applying the argument in both directions proves bounded uniqueness.

For an arbitrary exponential subsolution $r$, fix $A>0$ and put $p=s-A$, $W=\max (r,p)$. These maxima are $H$-plurisubharmonic by lemma 5.24.1, and $W$ is bounded. Choose $j$ with $-j<\inf _ Xp$ and set $r_ j=\max (r,-j)$; then $W=\max (r_ j,p)$. Plurifine locality on $\{ r_ j>p\} $ and lemma 5.28.3, applied to $p\leq W$ on their contact set, give

On the first set $r_ j=r>-j$, so strict-superlevel locality identifies the measure with $\mathrm{MA}^{\mathrm{ext}}_ H(r)$. On the second, $W=p$ and $\mathrm{MA}_ H(p)=\mathrm e^{\lambda s}\nu \geq \mathrm e^{\lambda p}\nu $. Thus $W$ is a bounded exponential subsolution, and $r\leq W\leq s$. This uses bounded products before every restriction and applies also when the measure of $r$ has infinite mass.

The maximal coefficient. Suppose $\mathrm{MA}^{\mathrm{ext}}_ H(r)\geq a\nu $ with $a>0$, and put $B=\sup _ Xr$. For every $\lambda >0$ the potential $r+\lambda ^{-1}\log a-B$ is an exponential subsolution, since its exponential is at most $a$. Greatestness gives

Evaluate at a fixed finite point of $r$, multiply by $\lambda $, and let $\lambda \downarrow 0$. The uniform barriers for $w_\lambda $ give $a\leq c$. The equation for $u$ attains $c$ in each stated class. $\square$

Proof. Since $K=-\log d_{\mathrm{ch}}^2$, layer-cake integration gives

Consequently the continuous truncated-kernel potentials converge uniformly to $Q_\mu $. Weighted arclength on the circle has the ball bound with $\delta =1$. On that circle the kernel is $1+\log \sin ^2((\vartheta -\varphi )/2)$, whose first cosine coefficient is $-2$. Integrating its $L^1$ Fourier series against $1+\epsilon \cos \varphi $ gives

This proves nonconstancy, and the support has smooth curve volume zero. Its product with $X_0$ also has smooth volume zero. $\square$