The uniqueness easily follows from the comparison principle as we explain in proposition 4.3 below. We are going to prove the existence by a fixed point method.

Fix $\psi\in{\mathcal{E}}^{1}(X,\omega)$ such that $\int_{X}\psi d\mu=0$, and let us consider the equation

$$MA(\psi)\hskip 56.9055pt(\omega+dd^{c}\varphi)^{n}=e^{t\psi-c_{\psi}}\mu,$$

where the constant $c_{\psi}:=\log[\int_{X}e^{t\psi}d\mu]$ is chosen so that

$$1=\int_{X}(\omega+dd^{c}\varphi)^{n}=e^{-c_{\psi}}\int_{X}e^{t\psi}d\mu.$$

Observe that $\mu_{\psi}:=e^{t\psi-c_{\psi}}\mu$ satisfies condition ${\mathcal{H}}(\alpha,A_{\psi},\omega)$, where $A_{\psi}=\exp(t\sup_{X}\psi-c_{\psi})$. It follows therefore from Theorem 2.1 that there exists a unique continuous function $\varphi\in PSH(X,\omega)$ solution to $MA(\psi)$ and normalized by $\int_{X}\varphi d\mu=0$. We use here this linear normalization rather than the non-linear $\sup$-normalization: they are comparable thanks to proposition 2.7 in [GZ 1], which shows that

$$-M_{\mu}\leq\int_{X}ud\mu-\sup_{X}u\leq 0,$$

for all functions $u\in PSH(X,\omega)$ and for some uniform constant $M_{\mu}>0$. Since $\int_{X}\psi d\mu=0$, we infer

(6)

$$0\leq{\mathcal{E}}_{\omega}(\varphi):=\int_{X}|\varphi|\omega_{\varphi}^{n}=e^{-c_{\psi}}\int_{X}|\varphi|e^{t\psi}d\mu\leq 2M_{\mu}e^{tM_{\mu}},$$

by observing that $c_{\psi}\geq 0$ since $t\geq 0$, and

$$\int_{X}|\varphi|d\mu\leq\int_{X}|\varphi-\sup_{X}\varphi|d\mu+\sup_{X}\varphi\leq 2M_{\mu},$$

since $\int_{X}\varphi d\mu=0$.

The important fact here is that the energy ${\mathcal{E}}_{\omega}(\varphi)$ of $\varphi$ is bounded from above by a constant $M_{0}:=2M_{\mu}e^{tM_{\mu}}$ which is independent of $\psi$. We have thus defined an operator

$$T:\psi\in{\mathcal{C}}_{M}\mapsto\varphi\in{\mathcal{C}}_{M_{0}}$$

which associates to $\psi\in{\mathcal{C}}_{M}$ the unique solution $\varphi\in{\mathcal{C}}_{M_{0}}$ to $MA(\psi)$, where

$${\mathcal{C}}_{M}:=\left\{\psi\in{\mathcal{E}}^{1}(X,\omega)\,/\,\int_{X}\psi d\mu=0\text{ and }{\mathcal{E}}_{\omega}(\psi)\leq M\right\}.$$

It follows from proposition 3.2.3 in [GZ 2] that ${\mathcal{E}}^{1}(X,\omega)$ is convex. So is the subset of functions $\psi\in{\mathcal{E}}^{1}(X,\omega)$ such that $\int_{X}\psi d\mu=0$. The set ${\mathcal{C}}_{M}$ is not convex, but it is relatively compact in $L^{1}(X)$ and its closed convex hull $\hat{{\mathcal{C}}}_{M}$ is contained in ${\mathcal{C}}_{\kappa_{n}M}$ for some uniform constant $\kappa_{n}$ which only depends on the dimension of $X$: this follows from easy computations (see lemma 7.2 and the proof of proposition 3.2 in [GZ 2]). Therefore $T$ maps the compact convex set $\hat{{\mathcal{C}}}_{M}$ into itself if $M$ is large enough.

We claim that $T$ is continuous. Let $(\psi_{j})\in{\mathcal{C}}_{M}^{\mathbb{N}}$ be a sequence of functions which converges in $L^{1}(X)$ towards $\psi\in{\mathcal{C}}_{M}$. We need to show that $\varphi_{j}:=T(\psi_{j})$ converges in $L^{1}(X)$ towards $T(\psi)$. Since the set $\{u\in PSH(X,\omega)\,/$ $\,\int_{X}ud\mu=0\}$ is relatively compact in $L^{1}(X)$ (see proposition 2.7 in [GZ 1]), we can assume – relabelling if necessary – that $(\varphi_{j})$ converges in $L^{1}(X)$ towards a function $\varphi\in PSH(X,\omega)$. We show in lemma 4.2 below that $(\varphi_{j})$ converges in $L^{1}(\mu)$ towards $\varphi$. In particular $\int_{X}\varphi d\mu=0$ and, passing to a subsequence if necessary, we can assume that $e^{t\psi_{j}(x)}\rightarrow e^{t\psi(x)}$ for $\mu$ almost every point $x$. Set

$$\hat{\varphi}_{j}:=\left(\sup_{l\geq j}\varphi_{l}\right)^{*}\;\;\text{ and }\;\;\check{\psi}_{j}:=\inf_{l\geq j}\psi_{l}.$$

Observe that $(\hat{\varphi}_{j})$ decreases towards $\varphi$, while $(e^{t\check{\psi}_{j}})$ increases towards $e^{t\psi}$ at $\mu$ almost every point. The energy of $\hat{\varphi}_{j}$ is controlled by that of $\varphi_{j}$ since $\hat{\varphi}_{j}\geq\varphi_{j}$ (see lemma 7.2 in [GZ 2]), and ${\mathcal{E}}_{\omega}(\varphi_{j})\leq M_{0}$ by (5), therefore $\varphi\in{\mathcal{E}}^{1}(X,\omega)$ and $(\omega+dd^{c}\hat{\varphi}_{j})^{n}\rightarrow(\omega+dd^{c}\varphi)^{n}$. It follows from an inequality of J.-P.Demailly [Dem 1] that

$$(\omega+dd^{c}\hat{\varphi}_{j})^{n}\geq e^{t\check{\psi}_{j}-\hat{c}_{j}}\mu,$$

where $\hat{c}_{j}:=\sup_{l\geq j}c_{\psi_{l}}$. Observe that $\hat{c}_{j}\rightarrow c_{\psi}$, thus

$$(\omega+dd^{c}\varphi)^{n}\geq e^{t\psi-c_{\psi}}\mu.$$

Since these are two probability measures, there is actually equality hence $\varphi=T(\psi)$: this shows that $T$ is continuous.

We can now invoke Schauder fixed point theorem, which yields a fixed point $\varphi=T(\varphi),\varphi\in{\mathcal{C}}_{M}$. The function $\varphi$ is automatically continuous (by Theorem 2.1, since $e^{t\varphi-c_{\varphi}}\mu$ satisfies ${\mathcal{H}}(\alpha,A^{\prime},\omega))$, hence $\Phi:=\varphi-t^{-1}c_{\varphi}$ is the solution we were looking for. ∎

We first show that $(\varphi_{j})$ converges to $\varphi$ in $L^{1}(\mu)$. Observe that the sequence $(\varphi_{j})$ is uniformly bounded: this follows from Theorem 2.1 since $(\omega+dd^{c}\varphi_{j})^{n}$ satisfies ${\mathcal{H}}(\alpha,A_{j},\omega)$, where $A_{j}=e^{t\sup_{X}\psi_{j}-c_{\psi_{j}}}A\leq e^{tM_{\mu}}A$ is bounded from above. It follows then from standard arguments that $\int_{X}\varphi_{j}d\mu\rightarrow\int_{X}\varphi d\mu$ (see e.g. the proof of lemma 5.2 in [Ce]).

Fix $\varepsilon>0$ and let $G$ be an open set of $X$ such that $\varphi$ is continuous on $X\setminus G$ and $Cap_{\omega}(G)\leq\varepsilon$ (see corollary 3.8 in [GZ 1]). By Hartogs’ lemma, $\varphi_{j}\leq\varphi+\varepsilon$ on the compact set $X\setminus G$, if $j\geq j_{\varepsilon}$. Observe that

$$\int_{X\setminus G}|\varphi-\varphi_{j}|d\mu\leq 2\varepsilon+\int_{X\setminus G}[\varphi-\varphi_{j}]d\mu\leq 3\varepsilon,$$

if $j\geq j^{\prime}_{\varepsilon}$. O the other hand since $\mu$ satisfies ${\mathcal{H}}(\alpha,A,\omega)$, we get

$$\int_{G}|\varphi-\varphi_{j}|d\mu\leq 2M\mu(G)\leq 2MA\varepsilon^{1+\alpha},$$

where $M=\sup_{j}||\varphi_{j}||_{L^{\infty}(X)}$. This shows that $||\varphi-\varphi_{j}||_{L^{1}(\mu)}\rightarrow 0$.

The proof for $(e^{t\psi_{j}})$ is similar: it suffices to note that the functions $u_{j}:=e^{t\psi_{j}-t\sup_{X}\psi_{j}}$ are $\omega$-psh and uniformly bounded. One can then apply the rest of the argument. ∎

It follows from the comparison principle (see [GZ 2]) that

$$\int_{(\varphi<\psi)}(\omega+dd^{c}\psi)^{n}\leq\int_{(\varphi<\psi)}(\omega+dd^{c}\varphi)^{n}=\int_{(\varphi<\psi)}e^{t(\varphi-\psi)}(\omega+dd^{c}\psi)^{n}.$$

Since $e^{t(\varphi-\psi)}<1$ on $(\varphi<\psi)$, we infer $\varphi\geq\psi$ for $\nu$ almost every point, where $\nu=(\omega+dd^{c}\psi)^{n}$. Reversing the roles of $\varphi,\psi$ yields $\varphi=\psi$ for $\nu$ almost every point. Therefore $(\omega+dd^{c}\varphi)^{n}=(\omega+dd^{c}\psi)^{n}$, hence $\varphi-\psi=c$ is constant by Theorem 3.4 in [GZ 2]. Finally $c=0$ since $e^{tc}=1$ and $t>0$. ∎

Set $\psi_{j}:=\max(\psi,-j)\in PSH(X,\omega)\cap L^{\infty}(X)$. Observe that the probability measures $(\omega+dd^{c}\psi_{j})^{n}$ converge in $X\setminus E$ towards the measure $\nu=e^{t\psi}\mu$. Since $\nu(X)=\nu(X\setminus E)=1$, it follows that $(\omega+dd^{c}\varphi_{j})^{n}$ converges to $\nu$ on all of $X$.

Fix $\varepsilon>0$ and set $v_{\varepsilon}:=(\psi+\varepsilon v)/(1+\varepsilon)\in PSH(X,\omega)$, where $v\in PSH(X,\omega)$, $v\leq 0$, is such that $e^{v}$ is continuous and $(v=-\infty)=E$. It follows from lemma 2.2 that for all $s>0$,

$$Cap_{\omega}(\psi_{j}<-s-1)\leq\int_{(\psi_{j}<-s)}(\omega+dd^{c}\psi_{j})^{n}\leq\int_{(v_{\varepsilon}\leq-s/(1+\varepsilon))}(\omega+dd^{c}\psi_{j})^{n}.$$

Observe that $e^{v_{\varepsilon}}$ is continuous on $X$, hence the sublevel sets $(v_{\varepsilon}\leq c)$ are compact. We infer, letting $j\rightarrow+\infty$,

$$Cap_{\omega}(\psi<-s-1)\leq\int_{(v_{\varepsilon}\leq-s/(1+\varepsilon))}e^{t\psi}d\mu.$$

Letting $\varepsilon$ go to zero and using that $\mu(X)=1$ yields

$$Cap_{\omega}(\psi<-s-1)\leq\int_{(\psi<-s)}e^{t\psi}d\mu\leq e^{-s}.$$

Therefore the capacity of the sublevel sets of $\psi$ decreases fast as $s\rightarrow+\infty$, hence by lemma 6.2 in [GZ 2] we get $\psi\in{\mathcal{E}}^{1}(X,\omega)$. Since $e^{-t\varphi}(\omega+dd^{c}\varphi)^{n}\equiv e^{-t\psi}(\omega+dd^{c}\psi)^{n}$, it follows from proposition 4.3 that $\varphi\equiv\psi$. ∎

The proof of Theorem 3.5 applies here almost verbatim. ∎