2. Continuous solutions [02DF]

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2. Continuous solutions

The goal of this section is to prove the following result:

Theorem 2.1.

Let $\mu$ be a probability measure on $X$ which satisfies condition ${\mathcal{H}}(\alpha,A,\omega)$ . Then there exists a unique continuous function $\varphi\in PSH(X,\omega)$ such that

\mu=(\omega+dd^{c}\varphi)^{n}\;\text{ and }\;\sup_{X}\varphi=-1.

Moreover $||\varphi||_{L^{\infty}(X)}\leq C$ , where $C$ only depends on $\alpha,A$ and $\omega$ .

It follows from proposition 1.4 that we already know the existence of a unique solution $\varphi\in{\mathcal{E}}^{1}(X,\omega)$ to this Monge-Ampère equation. We need to show it is continuous. The following result is the key to everything to follow.

Lemma 2.2.

Let $\varphi,\psi\in{\mathcal{E}}^{1}(X,\omega)$ be two negative functions. Then for all $s>0$ and $0\leq t\leq 1$ ,

t^{n}Cap_{\omega}(\varphi-\psi<-s-t)\leq\int_{(\varphi-\psi<-s-t\psi)}\omega_{\varphi}^{n}.

Proof.

Fix $u\in PSH(X,\omega)$ with $0\leq u\leq 1$ . For $\delta>0$ we set $t=\delta/(1+\delta)$ . Observe that $0\leq t\leq 1$ and

\{\varphi-\psi<-s-t\}\subset\{\varphi<\frac{\psi+\delta u}{1+\delta}-s-t\}\subset\{\varphi-\psi<-s-t\psi\}.

Set $\tilde{\varphi}:=(\psi+\delta u)/(1+\delta)-s-t\in PSH(X,\omega)$ . Observe that

t^{n}\int_{(\varphi-\psi<-s-t)}\omega_{u}^{n}\leq\int_{(\varphi-\psi<-s-t)}\left[\frac{1}{1+\delta}\omega_{\psi}+\frac{\delta}{1+\delta}\omega_{u}\right]^{n}\leq\int_{(\varphi<\tilde{\varphi})}[\omega+dd^{c}\tilde{\varphi}]^{n}.

It follows from the comparison principle in class ${\mathcal{E}}^{1}(X,\omega)$ , that

\int_{(\varphi<\tilde{\varphi})}[\omega+dd^{c}\tilde{\varphi}]^{n}\leq\int_{(\varphi<\tilde{\varphi})}[\omega+dd^{c}\varphi]^{n}\leq\int_{(\varphi-\psi<-s-t\psi)}[\omega+dd^{c}\varphi]^{n}.

Taking the supremum over all u’s yields the desired result. ∎

We will also need the following elementary observation:

Lemma 2.3.

Let $f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ be a decreasing right-continuous function such that $\lim_{+\infty}f=0$ . Assume there exists $\alpha,B>0$ such that $f$ satisfies

H(\alpha,B)\hskip 28.45274pttf(s+t)\leq B[f(s)]^{1+\alpha},\;\forall s>0,\,\forall 0\leq t\leq 1.

Then there exists $S_{\infty}=S_{\infty}(\alpha,B)\in\mathbb{R}^{+}$ such that $f(s)=0$ for all $s\geq S_{\infty}$ .

Proof.

Fix $s_{0}>0$ large enough so that $f(s_{0})^{\alpha}<1/2B$ . We define a sequence $(s_{j})\in\mathbb{R}_{+}^{\mathbb{N}}$ by induction in the following way. If $f(s_{0})=0$ we stop here, otherwise we set

s_{1}:=\sup\left\{s>s_{0}\,/\,f(s)>\frac{1}{2}f(s_{0})\right\}.

Observe that $s_{1}\leq 1+s_{0}$ thanks to $H(\alpha,B)$ and by definition of $s_{0}$ .

Since $f$ is right-continuous we get $f(s_{1})\leq f(s_{0})/2$ . If $f(s_{1})=0$ we stop here, otherwise we go on by induction, setting

s_{j+1}:=\sup\left\{s>s_{j}\,/\,f(s)>\frac{1}{2}f(s_{j})\right\}.

At each step $f(s_{j+1})\leq f(s_{j})/2$ and $s_{j+1}\leq 1+s_{j}$ . However the sequence $(s_{j})$ does not grow too fast. It follows indeed from $H(\alpha,B)$ that if $s\in]s_{j},s_{j+1}[$ ,

(s-s_{j})f(s)\leq Bf(s_{j})^{1+\alpha}\leq 2Bf(s)f(s_{j})^{\alpha},

since $f(s_{j})/2\leq f(s)\leq f(s_{j})$ on the interval $[s_{j},s_{j+1}]$ . We infer

s_{j+1}-s_{j}\leq 2Bf(s_{j})^{\alpha}\leq 2B2^{-j\alpha}f(s_{0})^{\alpha}\leq 2^{-j\alpha}.

Thus the sequence $(s_{j})$ is bounded from above, with limit

S_{\infty}=s_{0}+\sum_{j\geq 0}(s_{j+1}-s_{j})\leq s_{0}+\frac{2Bf(s_{0})^{\alpha}}{1-2^{-\alpha}}\leq s_{0}+\frac{1}{1-2^{-\alpha}}.

∎

Remarks 2.4.

Observe that the starting time $s_{0}(f,\alpha,B)$ is invariant under dilatation $f\mapsto\lambda f$ , which transforms $B$ into $B/\lambda^{\alpha}$ . Note also that if $f(0)^{\alpha}<1/2B$ , then we can take $s_{0}=0$ , hence we get in this case

S_{\infty}\leq\frac{2B}{1-2^{-\alpha}}[f(0)]^{\alpha}.

To see how previous lemmas can be used, we first prove that the unique solution $\varphi\in{\mathcal{E}}^{1}(X,\omega)$ given by proposition 1.4 is bounded.

A uniform bound on the solution. Let $\varphi\in{\mathcal{E}}^{1}(X,\omega)$ be the unique function such that $\mu=(\omega+dd^{c}\varphi)^{n}$ and $\sup_{X}\varphi=-1$ . Set

f(s):=[Cap_{\omega}(\varphi<-s)]^{1/n}.

Observe that $f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ is a right-continuous decreasing function with $\lim_{+\infty}f=0$ . Since $\mu=\omega_{\varphi}^{n}$ satisfies ${\mathcal{H}}(\alpha,A,\omega)$ , it follows from lemma 2.2 applied to the function $\psi\equiv 0$ that $f$ satisfies $H(\alpha,B)$ with $B=A^{1/n}$ .

It follows from propositions 2.7 and 3.6 in [GZ 1] that $f(s)\leq C_{1}/s^{1/n}$ for some constant $C_{1}$ which only depends on $\omega$ . We can thus take a starting time $s_{0}=2^{n/\alpha}C_{1}^{n}A^{1/\alpha}$ (see the proof of lemma 2.3) and get $f(s)=0$ for $s\geq S_{\infty}:=s_{0}+(1-2^{-\alpha})^{-1}$ : this shows that the sets $(\varphi<-s)$ are empty if $s\geq S_{\infty}$ , hence

(2)

||\varphi||_{L^{\infty}(X)}\leq 2^{n/\alpha}C_{1}^{n}A^{1/\alpha}+\frac{1}{1-2^{-\alpha}}.

We are now going to use a refinement of the previous reasoning in order to show that $\varphi$ is actually continuous.

Proposition 2.5.

Let $\varphi,\psi\in{\mathcal{E}}^{1}(X,\omega)$ be two negative functions and fix $\varepsilon>0$ . Assume $\omega_{\varphi}^{n}=\mu$ satisfies ${\mathcal{H}}(\alpha,A,\omega)$ and $\psi$ is bounded. There exists $C=C(\alpha,A,\omega,||\psi||_{L^{\infty}(X)})>0$ such that

\sup_{X}(\psi-\varphi)\leq\varepsilon+C\left[Cap_{\omega}(\varphi-\psi<-\varepsilon)\right]^{\alpha/n}.

This inequality can be interpreted as follows. Assume $\psi$ is also a solution to a Monge-Ampère equation $\omega_{\psi}^{n}=\mu^{\prime}$ , where $\mu^{\prime}$ also satisfies ${\mathcal{H}}(\alpha,A,\omega)$ . Then we can interchange the roles of $\varphi$ and $\psi$ and get an upper-bound on $||\varphi-\psi||_{L^{\infty}(X)}$ . The proposition then tells us that if $\varphi$ and $\psi$ are close in capacity, they are close in ${\mathcal{C}}^{0}$ -norm.

Proof.

Set $M:=||\psi||_{L^{\infty}(X)}$ . Observe that when $t\geq 0$ , $(\varphi-\psi<-s-t\psi)\subset(\varphi-\psi<-s+tM)$ , hence it follows from lemma 2.2 that for all $s>0$ , $0\leq t\leq 1$ ,

(3)

t^{n}Cap_{\omega}(\varphi-\psi<-s-t)\leq(1+M)^{n}\int_{(\varphi-\psi<-s)}\omega_{\varphi}^{n},

using the obvious substitutions $s\mapsto s-Mt$ , $t\mapsto(1+M)t$ . Since $\mu=\omega_{\varphi}^{n}$ satisfies ${\mathcal{H}}(\alpha,A,\omega)$ , we infer

t^{n}Cap_{\omega}(\varphi-\psi<-s-t)\leq A(1+M)^{n}Cap_{\omega}(\varphi-\psi<-s)^{1+\alpha}.

Consider

f(s):=\left[Cap_{\omega}(\varphi-\psi<-s-\varepsilon)\right]^{1/n},\,s>0.

Then $f$ satisfies the condition $H(\alpha,B)$ of lemma 2.3 with $B=(1+M)A^{1/n}$ . Assume $f(0)=\left[Cap_{\omega}(\varphi-\psi<-\varepsilon)\right]^{1/n}<\frac{1}{(2B)^{1/\alpha}}$ . It follows in this case from Remarks 2.4 that $f(s)=0$ for $s\geq S_{\infty}$ , where

S_{\infty}\leq\frac{2B}{1-2^{-\alpha}}\left[Cap_{\omega}(\varphi-\psi<-\varepsilon)\right]^{\alpha/n}

Therefore the sets $\{\varphi-\psi<-s-\varepsilon\}$ are empty for $s>S_{\infty}$ , hence

\sup_{X}(\psi-\varphi)\leq\varepsilon+S_{\infty}\leq\varepsilon+\frac{2(1+M)A^{1/n}}{1-2^{-\alpha}}\left[Cap_{\omega}(\varphi-\psi<-\varepsilon)\right]^{\alpha/n}.

Thus we can take here $C\geq 2B/(1-2^{-\alpha})$ .

If $f(0)=[Cap_{\omega}(\varphi-\psi<-\varepsilon)]^{1/n}\geq(2B)^{-1/\alpha}$ , then

\sup_{X}(\psi-\varphi)\leq-\inf_{X}\varphi\leq C_{2}(\alpha,A,\omega),

by (2), hence it suffices to take $C\geq 2BC_{2}$ to conclude. ∎

Proof of theorem 2.1 Let $\varphi\in{\mathcal{E}}^{1}(X,\omega)$ be the unique solution to the normalized Monge-Ampère equation $\mu=(\omega+dd^{c}\varphi)^{n}$ , $\sup_{X}\varphi=-1$ (see proposition 1.4). It follows from (2) that $\varphi$ is bounded.

Since $\omega$ is semi-positive and big, it follows from a result of J.P.Demailly (see [BK] for an elementary proof) that one can approximate any $\omega$ -psh function by a decreasing sequence of smooth $\omega$ -psh functions. Let $\varphi_{j}$ be such an approximating sequence for $\varphi$ . Since $\sup_{X}\varphi=-1$ , we can assume $\varphi_{j}\leq 0$ . Since $\varphi$ is bounded and $\varphi_{j}\geq\varphi$ , the functions $\varphi_{j}$ are uniformly bounded on $X$ .

Observe that $\varphi_{j}$ converges towards $\varphi$ in capacity (see proposition 3.7 in [GZ 1]), hence $\lim Cap_{\omega}(\varphi-\varphi_{j}<-\varepsilon)=0$ , for all $\varepsilon>0$ . It follows therefore from proposition 2.5 (applied with $\psi=\varphi_{j}$ ) that for all $\varepsilon>0$ ,

\lim_{j\rightarrow+\infty}||\varphi-\varphi_{j}||_{L^{\infty}(X)}=\lim_{j\rightarrow+\infty}\sup_{X}(\varphi_{j}-\varphi)\leq\varepsilon.

Thus $(\varphi_{j})$ converges uniformly towards $\varphi$ , hence $\varphi$ is continuous. $\Box$