2. Monge-Ampère capacity [032S]

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2. Monge-Ampère capacity

We assume in this section that $\omega$ is a Kähler form on $X$ . Let $T$ be a positive closed current of bidegree $(p,p)$ on $X$ , $0\leq p\leq n=\dim_{\mathbb{C}}X$ . It can be thought of as a closed differential form of bidegree $(p,p)$ with measure coefficients whose total variation is controlled by

||T||:=\int_{X}T\wedge\omega^{n-p}.

We refer the reader to chapter 3 of [15] for basic properties of positive currents. Given $\varphi\in PSH(X,\omega)$ we write $\varphi\in L^{1}(T)$ if $\varphi$ is integrable with respect to each (measure) coefficient of $T$ . This is equivalent to $\varphi$ being integrable with respect to the trace measure $T\wedge\omega^{n-p}$ . In this case the current $\varphi T$ is well defined, hence so is

\omega_{\varphi}\wedge T:=\omega\wedge T+dd^{c}(\varphi T).

This is again a positive closed current on $X$ , of bidegree $(p+1,p+1)$ . Indeed positivity is a local property which is stable under taking limits. One can locally regularize $\varphi$ and approximate $\omega_{\varphi}\wedge T$ by the currents $\omega_{\varphi_{\varepsilon}}\wedge T$ which are positive since $\omega_{\varphi_{\varepsilon}}$ are smooth positive forms.

When $\varphi\in PSH(X,\omega)\cap L^{\infty}(X)$ then $\varphi\in L^{1}(T)$ for any positive closed current $T$ of bidegree $(p,p)$ . One can thus inductively define $\omega_{\varphi}^{j}\wedge T$ , $1\leq j\leq n-p$ , for $\varphi\in PSH(X,\omega)\cap L^{\infty}(X)$ . For $T=0$ and $j=n$ one obtains the complex Monge-Ampère operator, $\omega_{\varphi}^{n}$ . It follows from the local theory that the operator $\varphi\mapsto\omega_{\varphi}^{n}$ is continuous under monotone sequences (see [5]). The proof of these continuity properties is simpler in our compact setting. We refer the reader to [24] where this is proved in a more general global context.

Proposition 2.1 (Chern-Levine-Nirenberg inequalities).

Let $T$ be a positive closed current of bidegree $(p,p)$ on $X$ and $\varphi\in PSH(X,\omega)\cap L^{\infty}(X)$ .

Then $||\omega_{\varphi}\wedge T||=||T||$ . Moreover if $\psi\in PSH(X,\omega)\cap L^{1}(T)$ , then $\psi\in L^{1}(T\wedge\omega_{\varphi})$ and

||\psi||_{L^{1}(T\wedge\omega_{\varphi})}\leq||\psi||_{L^{1}(T)}+[2\sup_{X}\psi+\sup_{X}\varphi-\inf_{X}\varphi]||T||.

Proof.

By Stokes theorem, $\int_{X}dd^{c}\varphi\wedge T\wedge\omega^{n-p-1}=0$ , hence

||\omega_{\varphi}\wedge T||:=\int_{X}\omega_{\varphi}\wedge T\wedge\omega^{n-p-1}=\int_{X}T\wedge\omega^{n-p}:=||T||.

Consider now $\psi\in L^{1}(T)$ . Since $T$ has measure coefficients, this simply means that $\psi$ is integrable with respect to the total variation of these measures. Assume first $\psi\leq 0$ , $\varphi\geq 0$ and $\varphi,\psi$ are smooth. Then

||\psi||_{L^{1}(T\wedge\omega_{\varphi})}:=\int_{X}(-\psi)T\wedge\omega_{\varphi}\wedge\omega^{n-p-1}=||\psi||_{L^{1}(T)}+\int_{X}(-\psi)T\wedge dd^{c}\varphi\wedge\omega^{n-p-1}.

Now it follows from Stokes theorem that

\int_{X}(-\psi)T\wedge dd^{c}\varphi\wedge\omega^{n-p-1}=\int_{X}\varphi T\wedge(-dd^{c}\psi)\wedge\omega^{n-p-1}

\leq\int_{X}\varphi T\wedge\omega^{n-p}\leq\sup_{X}\varphi\int_{X}T\wedge\omega^{n-p},

where the forelast inequality follows from $\varphi T\wedge\omega^{n-p}\geq 0$ and $-dd^{c}\psi\leq\omega$ . This yields

||\psi||_{L^{1}(T\wedge\omega_{\varphi})}\leq||\psi||_{L^{1}(T)}+\sup_{X}\varphi||T||.

The general case follows by regularizing $\varphi,\psi$ , observing that $\omega_{\varphi}=\omega_{\varphi^{\prime}}$ where $\varphi^{\prime}=\varphi-\inf_{X}\varphi\geq 0$ , and decomposing $\psi=\psi^{\prime}+\sup_{X}\psi$ with $\psi^{\prime}=\psi-\sup_{X}\psi\leq 0$ . ∎

Remark 2.2.

The fact that the $L^{1}$ -norm of $\psi$ with respect to the probability measure $T\wedge\omega_{\varphi}\wedge\omega^{n-p-1}$ is controlled by its $L^{1}$ -norm with respect to $T\wedge\omega^{n-p}$ is similar to the phenomenon already encountered in example 1.8: one can write

T\wedge\omega_{\varphi}\wedge\omega^{n-p-1}=T\wedge\omega^{n-p}+dd^{c}S,\;S=(\varphi-\inf_{X}\varphi)T\wedge\omega^{n-p-1}\geq 0.

This type of estimates is usually referred to as ”Chern-Levine-Nirenberg inequalities”, in reference to [8] where simpler -but fondamental- $L^{\infty}$ -estimates were established (with $\psi=constant$ ). Estimates involving the $L^{1}$ -norm of $\psi$ were first proved in the local context by Cegrell [6] and Demailly [12].

A straightforward induction yields the following

Corollary 2.3.

Let $\psi,\varphi\in PSH(X,\omega)$ with $0\leq\varphi\leq 1$ . Then

0\leq\int_{X}|\psi|\omega_{\varphi}^{n}\leq\int_{X}|\psi|\omega^{n}+n[1+2\sup_{X}\psi]\int_{X}\omega^{n}.

Following Bedford-Taylor [5] and Kolodziej [29] we introduce the following Monge-Ampère capacity.

Definition 2.4.

Let $K$ be a Borel subset of $X$ . We set

Cap_{\omega}(K):=\sup\left\{\int_{K}\omega_{\varphi}^{n}\,/\,\varphi\in PSH(X,\omega),0\leq\varphi\leq 1\right\}.

Note that this definition only makes sense when the cohomology class $[\omega]$ is big, i.e. when $[\omega]^{n}>0$ , and when it admits locally bounded potentials $\varphi$ . This implies, by a regularization result of Demailly [12], that $[\omega]$ is big and nef. To simplify we actually assume that $\omega$ (hence $[\omega]$ ) is Kähler.

Proposition 2.5.

1) If $K\subset K^{\prime}\subset X$ are Borel subsets then

Vol_{\omega}(K):=\int_{K}\omega^{n}\leq Cap_{\omega}(K)\leq Cap_{\omega}(K^{\prime})\leq Cap_{\omega}(X)=Vol_{\omega}(X).

2) If $K_{j}$ are Borel subsets of $X$ then $Cap_{\omega}(\cup K_{j})\leq\sum Cap_{\omega}(K_{j})$ . Moreover $Cap_{\omega}(\cup K_{j})=\lim Cap_{\omega}(K_{j})$ if $K_{j}\subset K_{j+1}$ .

3) If $\omega_{1}\leq\omega_{2}$ then $Cap_{\omega_{1}}(\cdot)\leq Cap_{\omega_{2}}(\cdot)$ . For all $A\geq 1$ , $Cap_{\omega}(\cdot)\leq Cap_{A\omega}(\cdot)\leq A^{n}Cap_{\omega}(\cdot)$ . In particular if $\omega,\omega^{\prime}$ are two Kähler forms then there exists $C\geq 1$ such that

\frac{1}{C}Cap_{\omega}(\cdot)\leq Cap_{\omega^{\prime}}(\cdot)\leq C\cdot Cap_{\omega}(\cdot).

4) If $f:X\rightarrow X$ is holomorphic then for all Borel subset $K$ of $X$ ,

Cap_{\omega}(f(K))\leq Cap_{f^{*}\omega}(K).

In particular $Cap_{\omega}(f(K))=Cap_{\omega}(K)$ for every $\omega$ -isometry $f$ .

Proof.

That $Vol_{\omega}(\cdot)\leq Cap_{\omega}(\cdot)$ is a straightforward consequence of the definition (since $\omega_{0}=\omega$ ). It then follows from Stokes theorem that $\int_{X}\omega_{\varphi}^{n}=\int_{X}\omega^{n}$ for every $\varphi\in PSH(X,\omega)\cap L^{\infty}(X)$ , thus $Vol_{\omega}(X)=Cap_{\omega}(X)$ .

Property 2) is a straightforward consequence of the definitions.

If $\omega_{1}\leq\omega_{2}$ then $PSH(X,\omega_{1})\subset PSH(X,\omega_{2})$ hence $Cap_{\omega_{1}}(\cdot)\leq Cap_{\omega_{2}}(\cdot)$ . Fix $A\geq 1$ . If $\psi\in PSH(X,A\omega)$ is such that $0\leq\psi\leq 1$ then $\psi/A\in PSH(X,\omega)$ with $0\leq\psi/A\leq 1/A\leq 1$ . Moreover $(A\omega+dd^{c}\psi)^{n}=A^{n}(\omega+dd^{c}(\psi/A))^{n}$ . This shows $Cap_{A\omega}(\cdot)\leq A^{n}Cap_{\omega}(\cdot)$ .

In particular if $\omega,\omega^{\prime}$ are both Kähler then $A^{-1}\omega\leq\omega^{\prime}\leq A\omega$ for some constant $A\geq 1$ , hence $C^{-1}Cap_{\omega}(\cdot)\leq Cap_{\omega^{\prime}}(\cdot)\leq C\cdot Cap_{\omega}(\cdot)$ with $C=A^{n}$ .

It remains to prove 4). It follows from the change of variables formula that if $\varphi\in PSH(X,\omega)$ with $0\leq\varphi\leq 1$ then

\int_{f(K)}\omega_{\varphi}^{n}\leq\int_{K}f^{*}\omega_{\varphi}^{n}=\int_{K}(f^{*}\omega+dd^{c}(\varphi\circ f))^{n}\leq Cap_{f^{*}\omega}(K)

since $\varphi\circ f\in PSH(X,f^{*}\omega)$ with $0\leq\varphi\circ f\leq 1$ . We infer $Cap_{\omega}(f(K))\leq Cap_{f^{*}\omega}(K)$ . When $f$ is a $\omega$ -isometry, i.e. $f\in Aut(X)$ with $f^{*}\omega=\omega$ , then the mapping $\varphi\mapsto\varphi\circ f$ is an isomorphism of $\{u\in PSH(X,\omega)\,/\,0\leq u\leq 1\}$ , whence $Cap_{\omega}(f(K))=Cap_{f^{*}\omega}(K)=Cap_{\omega}(K)$ . ∎

Let $PSH^{-}(X,\omega)$ denote the set of negative $\omega$ -psh functions. A set is said to be $PSH(X,\omega)$ -polar if it is included in the $-\infty$ locus of some function $\psi\in PSH(X,\omega)$ , $\psi\not\equiv-\infty$ . As we shall soon see, the sets of zero Monge-Ampère capacity are precisely the $PSH(X,\omega)$ -polar sets. We start by establishing the following:

Proposition 2.6.

If $P$ is a $PSH(X,\omega)$ -polar set, then $Cap_{\omega}(P)=0$ . More precisely, if $\psi\in PSH^{-}(X,\omega)$ then

Cap_{\omega}(\psi<-t)\leq\frac{1}{t}\left[\int_{X}(-\psi)\omega^{n}+nVol_{\omega}(X)\right],\;\forall t>0.

Proof.

Fix $\varphi\in PSH(X,\omega)$ such that $0\leq\varphi\leq 1$ . Fix $t>0$ and set $K_{t}=\{x\in X\,/\,\psi(x)<-t\}$ . By Chebyshev’s inequality,

\int_{K_{t}}\omega_{\varphi}^{n}\leq\int_{X}(-\psi/t)\omega_{\varphi}^{n}\leq\frac{1}{t}\left[\int_{X}(-\psi)\omega^{n}+nVol_{\omega}(X)\right],

where the last inequality follows from corollary 2.3. Taking supremum over all $\varphi^{\prime}$ s yields the claim. ∎

Observe that the previous proposition says that $Cap_{\omega}^{*}(P)=0$ , where

Cap_{\omega}^{*}(E):=\inf\{Cap_{\omega}(G)\,/G\text{ open with }E\subset G\},

is the outer capacity associate to $Cap_{\omega}$ .

Our aim is now to show that $\omega$ -psh functions are quasicontinuous with respect to $Cap_{\omega}$ (corollary 2.8). We first need to show that decreasing sequences of $\omega$ -psh functions converge ”in capacity”.

Proposition 2.7.

Let $\psi,\psi_{j}\in PSH(X,\omega)\cap L^{\infty}(X)$ such that $(\psi_{j})$ decreases to $\psi$ . Then for each $\delta>0$ ,

Cap_{\omega}(\{\psi_{j}>\psi+\delta\})\rightarrow 0.

Proof.

We can assume w.l.o.g. that $Vol_{\omega}(X)=1$ and $0\leq\psi_{j}-\psi\leq 1$ . Fix $\delta>0$ and $\varphi\in PSH(X,\omega)$ , $0\leq\varphi\leq 1$ . By Chebyshev inequality, it suffices to control $\int_{X}(\psi_{j}-\psi)\omega_{\varphi}^{n}$ uniformly in $\varphi$ . It follows from Stokes theorem that

\int_{X}(\psi_{j}-\psi)\omega_{\varphi}^{n}=\int_{X}(\psi_{j}-\psi)\omega\wedge\omega_{\varphi}^{n-1}-\int_{X}d(\psi_{j}-\psi)\wedge d^{c}\varphi\wedge\omega_{\varphi}^{n-1}.

Now by Cauchy-Schwartz inequality,

\left|\int_{X}df_{j}\wedge d^{c}\varphi\wedge\omega_{\varphi}^{n-1}\right|\leq\left(\int_{X}df_{j}\wedge d^{c}f_{j}\wedge\omega_{\varphi}^{n-1}\right)^{1/2}\cdot\left(\int_{X}d\varphi\wedge d^{c}\varphi\wedge\omega_{\varphi}^{n-1}\right)^{1/2},

where we set $f_{j}:=\psi_{j}-\psi\geq 0$ . Moreover

\int_{X}d\varphi\wedge d^{c}\varphi\wedge\omega_{\varphi}^{n-1}=\int_{X}\varphi(-dd^{c}\varphi)\wedge\omega_{\varphi}^{n-1}\leq\int_{X}\varphi\omega\wedge\omega_{\varphi}^{n-1}\leq 1,

since $\varphi\omega_{\varphi}^{n-1}\geq 0$ , $-dd^{c}\varphi\leq\omega$ and $\varphi\leq 1$ . Similarly

\int_{X}df_{j}\wedge d^{c}f_{j}\wedge\omega_{\varphi}^{n-1}=\int_{X}-f_{j}dd^{c}f_{j}\wedge\omega_{\varphi}^{n-1}\leq\int_{X}f_{j}\omega_{\psi}\wedge\omega_{\varphi}^{n-1}.

Altogether this yields

	$\displaystyle\int_{X}(\psi_{j}-\psi)\omega_{\varphi}^{n}$	$\displaystyle\leq$	$\displaystyle\int_{X}(\psi_{j}-\psi)\omega\wedge\omega_{\varphi}^{n-1}+\left(\int_{X}(\psi_{j}-\psi)\omega_{\psi}\wedge\omega_{\varphi}^{n-1}\right)^{1/2}$
		$\displaystyle\leq$	$\displaystyle\sqrt{2}\left(\int_{X}(\psi_{j}-\psi)(\omega+\omega_{\psi})\wedge\omega_{\varphi}^{n-1}\right)^{1/2},$

where the last inequality follows from the elementary inequalities $0\leq a\leq\sqrt{a}\leq 1$ and $\sqrt{a}+\sqrt{b}\leq\sqrt{2}\sqrt{a+b}$ .

Going on replacing at each step a term $\omega_{\varphi}$ by $\omega+\omega_{\psi}$ , we end up with

\int_{X}(\psi_{j}-\psi)\omega_{\varphi}^{n}\leq 2\left(\int_{X}(\psi_{j}-\psi)(\omega+\omega_{\psi})^{n}\right)^{1/2^{n}}.

The majorant being independent of $\varphi$ and converging to $0$ as $j\rightarrow+\infty$ (by dominated convergence theorem), this completes the proof. ∎

Corollary 2.8 (Quasicontinuity).

Let $\varphi\in PSH(X,\omega)$ . For each $\varepsilon>0$ there exists an open subset $O_{\varepsilon}$ of $X$ such that $Cap_{\omega}(O_{\varepsilon})<\varepsilon$ and $\varphi$ is continuous on $X\setminus O_{\varepsilon}$ .

Proof.

For $t>0$ large enough, the set $O_{1}=\{\varphi<-t\}$ has capacity $<\varepsilon/2$ by proposition 2.6. Working in $X\setminus O_{1}$ we can thus replace $\varphi$ by $\varphi_{t}=\max(\varphi,-t)$ which is bounded on $X$ . Regularizing $\varphi$ (see Appendix), we can find a sequence $\psi_{j}$ of smooth $A\omega$ -psh functions which decrease to $\varphi_{t}$ on $X$ , for some $A\geq 1$ . By proposition 2.7, the set $O_{j}=\{\psi_{k_{j}}>\varphi_{t}+1/j\}$ has capacity $<\varepsilon 2^{-j-1}$ if $k_{j}$ is large enough. Now $\psi_{k_{j}}$ uniformly converges to $\varphi=\varphi_{t}$ on $X\setminus O_{\varepsilon}$ , $O_{\varepsilon}=\cup_{j\geq 1}O_{j}$ , so $\varphi$ is continuous on $X\setminus O_{\varepsilon}$ and $Cap_{\omega}(O_{\varepsilon})\leq\varepsilon$ . ∎

Example 2.9.

The capacity $Cap_{\omega}(\cdot)$ does not distinguish between ”big sets”. Assume indeed there exists an ample divisor $D$ such that $[D]\sim k\omega$ , $k\in\mathbb{N}$ . Then there exists $\varphi\in PSH(X,\omega)$ such that $dd^{c}\varphi=k^{-1}[D]-\omega$ . Note that $\varphi\in{\mathcal{C}}^{\infty}(X\setminus D)$ , $e^{\varphi}\in{\mathcal{C}}^{0}(X)$ and $\{\varphi=-\infty\}=D$ . Replacing $\varphi$ by $\varphi-\sup_{X}\varphi$ if necessary, we may assume $\sup_{X}\varphi=0$ . Consider $\varphi_{c}=\max(\varphi,-c)\in PSH(X,\omega)\cap{\mathcal{C}}^{0}(X)$ . Then $\varphi_{c}\equiv\varphi$ outside some neighborhood $V_{c}=\{\varphi<-c\}$ of $D$ . Since $0\leq 1+\varphi_{1}\leq 1$ and $\omega_{1+\varphi_{1}}=\omega_{\varphi}=0$ in $X\setminus V_{1}$ , we get

Cap_{\omega}(X)=\int_{X}(\omega_{1+\varphi_{1}})^{n}=\int_{V_{1}}(\omega_{1+\varphi_{1}})^{n}\leq Cap_{\omega}(V_{1}),

hence $Cap_{\omega}(V_{1})=Cap_{\omega}(X)$ .

As a concrete example take $X=\mathbb{C}\mathbb{P}^{n}$ and $\omega=\omega_{FS}$ , $D$ being some hyperplane $H_{\infty}$ ”at infinity” ( $k=1$ ). Set $\varphi[z:t]=\log|t|-\frac{1}{2}\log[||z||^{2}+|t|^{2}]$ where $z$ denotes the euclidean coordinates in $\mathbb{C}^{n}=\mathbb{C}\mathbb{P}^{n}\setminus H_{\infty}$ and $H_{\infty}=(t=0)$ . Observe that $\sup_{\mathbb{C}\mathbb{P}^{n}}\varphi=0$ . One then computes

\mathbb{C}\mathbb{P}^{n}\setminus V_{1}=\left\{z\in\mathbb{C}^{n}\,/\,|z|\leq\sqrt{e^{2}-1}\right\}.

Thus the capacity of the complement of any euclidean ball of radius smaller than $\sqrt{e^{2}-1}$ equals $1$ .

The definition of $Cap_{\omega}$ mimics the definition of the relative Monge-Ampère capacity introduced by Bedford and Taylor in [5]. Fix ${\mathcal{U}}=\{{\mathcal{U}}_{\alpha}\}$ a finite covering of $X$ by strictly pseudoconvex open subsets of $X$ , ${\mathcal{U}}_{\alpha}=\{x\in X\,/\,\varrho_{\alpha}(x)<0\}$ , where $\varrho_{\alpha}$ is a strictly psh smooth function defined in a neighborhood of $\overline{{\mathcal{U}}_{\alpha}}$ . Fix $\delta>0$ such that ${\mathcal{U}}^{\delta}=\{{\mathcal{U}}_{\alpha}^{\delta}\}$ is still a covering of $X$ , where ${\mathcal{U}}_{\alpha}^{\delta}=\{x\in X\,/\,\varrho_{\alpha}(x)<-\delta\}$ . For a Borel subset $K$ of $X$ , we set

Cap_{BT}(K):=\sum_{\alpha}Cap_{BT}(K\cap{\mathcal{U}}_{\alpha}^{\delta},{\mathcal{U}}_{\alpha}),

where

Cap_{BT}(E,\Omega):=\sup\left\{\int_{E}(dd^{c}u)^{n}\,/\,u\in PSH(\Omega),\,0\leq u\leq 1\right\}

is the capacity studied by Bedford and Taylor. The next proposition is due to Kolodziej [29]. We include a slightly different proof.

Proposition 2.10.

There exists $C\geq 1$ such that

\frac{1}{C}Cap_{\omega}(\cdot)\leq Cap_{BT}(\cdot)\leq C\cdot Cap_{\omega}(\cdot).

Proof.

Let $E$ be a Borel subset of $X$ . Since $Cap_{\omega}(E\cap{\mathcal{U}}_{\alpha}^{\delta})\leq Cap_{\omega}(E)\leq\sum_{\alpha}Cap_{\omega}(E\cap{\mathcal{U}}_{\alpha}^{\delta})$ , it is sufficient to show that if $\Omega=\{x\in X\,/\,\varrho(x)<0\}$ is a smooth hyperconvex subset of $X$ , then there exists $C\geq 1$ such that for all $E\subset\Omega_{\delta}$ ,

\frac{1}{C}Cap_{\omega}(E)\leq Cap_{BT}(E,\Omega)\leq C\cdot Cap_{\omega}(E),

where $\Omega_{\delta}=\{x\in X\,/\,\varrho(x)<-\delta\}$ .

It is an easy and well known fact in the local theory that the capacities $Cap(\cdot,\Omega)$ and $Cap(\cdot,\Omega^{\prime})$ are comparable when $\Omega^{\prime}\subset\Omega$ (see e.g. theorem 6.5 in [12]). Therefore we can assume (passing to a finer covering if necessary) that $\omega=dd^{c}\psi$ near $\overline{\Omega}$ . Fix $C_{1}>0$ such that $-C_{1}\leq\psi\leq C_{1}$ on $\Omega$ . Fix $\varphi\in PSH(X,\omega)$ such that $0\leq\varphi\leq 1$ on $X$ and set $u=(2C_{1})^{-1}(\varphi+\psi+C_{1})$ . Then $u\in PSH(\Omega)$ and $0\leq u\leq 1$ , hence

\int_{E}\omega_{\varphi}^{n}=(2C_{1})^{n}\int_{E}(dd^{c}u)^{n}\leq(2C_{1})^{n}Cap_{BT}(E,\Omega),

which yields $Cap_{\omega}(E)\leq(2C_{1})^{n}Cap_{BT}(E,\Omega)$ . Observe that we have not used here that $\omega$ is Kähler.

For the reverse inequality we consider $\chi\in{\mathcal{C}}^{\infty}(X)$ such that $\chi\equiv 0$ in $X\setminus\Omega$ and $\chi<0$ in $\Omega$ . Replacing $\chi$ by $\varepsilon\chi$ if necessary, we can assume $\chi\in PSH(X,\omega)$ . This is because $\omega$ is Kähler (and this is the only place where we shall use this crucial assumption). Fix $\varepsilon>0$ so small that $\chi\leq-\varepsilon$ on $\Omega_{\delta}$ . Let now $u\in PSH(\Omega)$ be such that $0\leq u\leq 1$ on $\Omega$ . Consider

\varphi(x)=\left\{\begin{array}[]{cl}\frac{u-\psi+C_{1}}{2+2C_{1}}&\text{ in }\Omega_{\delta}\\ \max\left(\frac{u-\psi+C_{1}}{2+2C_{1}},\frac{2}{\varepsilon}\chi(x)+1\right)&\text{ in }\Omega\setminus\Omega_{\delta}\\ 1&\text{ in }X\setminus\Omega\end{array}\right.

Observe that $0\leq u^{\prime}:=(u-\psi+C_{1})/(2+2C_{1})\leq(1+2C_{1})/(2+2C_{1})<1$ in $\Omega$ . Therefore $\varphi\in PSH(X,\frac{2}{\varepsilon}\omega)$ since $\frac{2}{\varepsilon}\chi(x)+1\leq-1<u^{\prime}$ in $\Omega_{\delta}$ , while $\frac{2}{\varepsilon}\chi(x)+1\equiv 1>u^{\prime}$ on $\partial\Omega$ . Note also that $0\leq\varphi\leq 1$ thus for $E\subset\Omega_{\delta}$ ,

	$\displaystyle\frac{1}{(2+2C_{1})^{n}}\int_{E}(dd^{c}u)^{n}$	$\displaystyle=$	$\displaystyle\int_{E}\left(\frac{\omega}{2+2C_{1}}+dd^{c}\varphi\right)^{n}\leq\int_{E}\left(\frac{2}{\varepsilon}\omega+dd^{c}\varphi\right)^{n}$
		$\displaystyle\leq$	$\displaystyle Cap_{2\omega/\varepsilon}(E)\leq\left(\frac{2}{\varepsilon}\right)^{n}Cap_{\omega}(E)$

hence $Cap_{BT}(E,\Omega)\leq 4^{n}(1+C_{1})^{n}\varepsilon^{-n}Cap_{\omega}(E)$ . ∎

Since locally pluripolar sets are precisely the sets of zero relative capacity [5], we obtain the following

Corollary 2.11.

$Cap_{\omega}(P)=0\Leftrightarrow Cap_{BT}(P)=0\Leftrightarrow P\text{ is locally pluripolar}.$

We shall show later on that locally pluripolar sets are $PSH(X,\omega)$ -polar when $\omega$ is Kähler (see theorem 5.2). Note that we rely here on the local theory to prove that $Cap_{\omega}(P)=0\Rightarrow P$ is locally pluripolar. One reason is that we do not know an equivalent in our global context of the ”relative extremal function” [5].

The second important result we shall borrow from the local theory is a direct consequence of the corresponding result of Bedford and Taylor [4].

Theorem 2.12 (Dirichlet Problem).

Let $\varphi\in PSH(X,\omega)\cap L^{\infty}(X)$ . Let $B$ be a small ball in $X$ . Then there exists $\hat{\varphi}\in PSH(X,\omega)$ such that $\hat{\varphi}=\varphi$ in $X\setminus B$ , $\hat{\varphi}\geq\varphi$ and $(\omega_{\hat{\varphi}})^{n}=0$ in $B$ . Moreover if $\varphi_{1}\leq\varphi_{2}$ then $\hat{\varphi_{1}}\leq\hat{\varphi_{2}}$ .