The smooth recovery theorem extends to a bounded solution whose complex Hessian is continuous and everywhere positive. In this case one can vary the potential in both directions, and the resulting densities force every supporting probability to annihilate the same elliptic operator. The Sobolev range theorem then gives uniqueness.
Theorem 11.34.11.F4FDB7BC8A03 Let $n\geq 2$, let $\theta =\omega $ be a Hermitian form, and let $\varphi =0$. Suppose $F\in C(X)$ and a bounded solution $u\in \mathrm{PSH}(X,\omega )$ of (11.33.1) has distributional complex Hessian represented by a continuous form
\[ \sigma =\omega +\mathrm{dd}^{\mathrm{c}}u>0 \]
everywhere on $X$. Then $\mathcal A_0(F)$ consists of one probability $\nu _ F$. It satisfies
\begin{align*} \int _ X L_\sigma h\, \mathrm d\nu _ F& =0\quad (h\in C^\infty (X)),\\ DB_0(F)[H]& =-\int _ XH\, \mathrm d\nu _ F\quad (H\in C(X)). \end{align*}
The derivative is taken in the Gâteaux sense.
Proof.
The continuous trace of $\mathrm{dd}^{\mathrm{c}}u$ is the Euclidean Laplacian of $u$, up to the fixed normalization. Local constant-coefficient elliptic regularity gives $u\in W^{2,p}$ for every finite $p$. Indeed, subtract the Newtonian potential of a cutoff of this trace. Its second high-frequency derivatives are order-zero multipliers, and the low-frequency kernels are locally bounded; the remaining distribution is harmonic. Sobolev embedding gives a $C^{1,\alpha }$ representative for every $\alpha <1$, which agrees with the quasi-psh representative.
On a coordinate ball choose smooth $\rho $ with $\mathrm{dd}^{\mathrm{c}}\rho \geq \omega $. The radial regularizations of the bounded psh function $u+\rho $ decrease to it, and their complex Hessians converge uniformly on smaller balls. Continuity of bounded products and the finite smooth-background expansion identify $(\omega +\mathrm{dd}^{\mathrm{c}}u)^ n$ with $\sigma ^ n$. For smooth real $h$, the same argument applies to $u+th$, which is admissible for all sufficiently small signed $t$ by uniform positivity. Define
\[ F_ t=F+\log \frac{(\sigma +t\mathrm{dd}^{\mathrm{c}}h)^ n}{\sigma ^ n}. \]
Its equation has the same normalization scalar as the equation for $u$, so $B_0(F_ t)=B_0(F)$. The supporting inequality gives $\int _ X(F_ t-F)\, \mathrm d\nu \leq 0$ for each $\nu \in \mathcal A_0(F)$. Since $F_ t-F=tL_\sigma h+O(t^2)$ uniformly, using both signs yields stationarity.
Choose $p=4n+1$ and $q=p/(p-1)=1+1/(4n)$. By corollary 12.9.3, every supporting probability has an $L^ q$ density, since $q<n/(n-1)$. Smooth density in $W^{2,p}$ extends its stationarity to every $W^{2,p}$ test. By lemma 11.24.1, all these densities lie in a one-dimensional space. Their integrals equal one, so they coincide. Nonemptiness is theorem 11.34.2, and the derivative formula follows from proposition 11.34.5.
$\square$
Pome
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