Lemma 15.31.4.F5E3175FA178 Suppose $Q=Q_\mu $ is finite and continuous. Then $\mu $ is non-polar, $Q\in H^1(Y)$, and smooth restriction extends to a continuous linear map
\[ \operatorname {Tr}_\mu :H^1(Y)\longrightarrow L^1(\mu ). \]
For every $h\in H^1(Y)$,
DB0D4C618F78
15.31.5
\begin{equation} \int _ Y\operatorname {Tr}_\mu h\, \mathrm d\mu =\int _ Yh\kappa -E(h,Q). \end{equation}
For $h\in \mathrm{PSH}(Y,\kappa )\cap H^1(Y)$ this trace agrees $\mu $-almost everywhere with its canonical values. The trace commutes with Lipschitz real compositions.
Proof.
Continuity gives the uniform kernel bound
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15.31.6
\begin{equation} A_\mu :=\sup _ x\int _ YK(x,y)\, \mathrm d\mu (y) =1-\inf _ YQ<\infty . \end{equation}
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
\[ Q_\epsilon \longrightarrow Q\ \text{uniformly},\qquad \mathrm{dd}^{\mathrm{c}}Q_\epsilon =\mu _\epsilon -\kappa ,\qquad \int _ YQ_\epsilon \kappa =0. \]
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
\[ \int _ Y|h|\, \mathrm d\mu \leq \| h\| _{L^1(\kappa )}+E(h,h)^{1/2}E(Q,Q)^{1/2} \leq C_\mu \| h\| _{H^1}. \]
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.
$\square$
Pome
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