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Pluripolar sets and the subextension inCegrell's classesPham Hoang Hiep aa Department of Mathematics, University of Education, Cau Giay,Hanoi, VietnamPublished online: 22 Sep 2010.

To cite this article: Pham Hoang Hiep (2008) Pluripolar sets and the subextension in Cegrell'sclasses, Complex Variables and Elliptic Equations: An International Journal, 53:7, 675-684, DOI:10.1080/17476930801966893

To link to this article: http://dx.doi.org/10.1080/17476930801966893

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Complex Variables and Elliptic Equations

Vol. 53, No. 7, July 2008, 675–684

Pluripolar sets and the subextension in Cegrell’s classes

Pham Hoang Hiep*

Department of Mathematics, University of Education, Cau Giay, Hanoi, Vietnam

Communicated by R.P. Gilbert

(Received 4 July 2006; final version received 3 February 2008)

In this article, we prove that if E is a complete pluripolar set in �, thenE¼ {’¼�1} for some ’2F1(�). Moreover, we study the subextension inCegrell’s class Ep.

Keywords: complex Monge-Ampere operator; plurisubharmonic function;

AMS Subject Classifications: primary 32W20; secondary 32U05

1. Introduction

Let � be a bounded hyperconvex domain in Cn. Denote by PSH(�) the plurisubharmonic

(psh) functions on �. The complex Monge-Ampere operator (dd c)n is well-defined over the

class of locally bounded psh functions, according to the fundamental work of Bedford and

Taylor in [1,2]. Recently, Cegrell introduced in [3,4] new classes of psh functions on which

the complex Monge-Ampere can be defined and enjoy important properties. In particular,

this operator is continuous with respect to monotone sequences. The aim of the present

note is to study pluripolar sets and the subextension in Cegrell’s classes. The subextension

was studied in [5–8]. First, in Section 3, we prove that for every complete pluripolar set E

in � there exists a function ’2F1 (�) such that E¼ {’¼�1} . Next, in Section 4, we

show that if u2PSH�(�) such that u2Ep(�), p40, then for every bounded hyperconvex

domain ~�, satisfying � � ~�, there exists ~u 2 Epð ~�Þ such that ~u� u on �,

(ddc ~u)n� 1�(ddcu)n and ep( ~u)� ep(u).

2. Preliminaries

In this section, we recall some elements of pluripotential theory that will be used

throughout this article. All this can be found in [2–4,9].

(1) Unless otherwise specified, � will be a bounded hyperconvex domain in Cn

meaning that there exists a negative exhaustive psh function for �.

*Email: phhiep_vn@yahoo.com

ISSN 1747–6933 print/ISSN 1747–6941 online

� 2008 Taylor & Francis

DOI: 10.1080/17476930801966893

http://www.informaworld.com

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(2) Let � be a bounded domain in Cn. The Cn-capacity in the sense of Bedford and

Taylor on � is the set function given by

CnðEÞ ¼ CnðE,�Þ ¼ sup

ZE

ðdd cuÞn : u 2 PSHð�Þ, �1 � u � 0

� �

for every Borel set E in �. It is known [2] that

CnðEÞ ¼

Z�

ðdd ch�E,�Þn:

where h�E,� is the relative extremal psh function for E (relative to �) defined as the smallest

upper semicontinuous majorant of hE,�

hE,�ðzÞ ¼ sup uðzÞ : u 2 PSHð�Þ : �1 � u � 0, u � �1 on E� �

:

(3) The following classes of psh functions were introduced by Cegrell in [3,4]

E0ð�Þ ¼ ’ 2 PSH \ L1ð�Þ : limz!@�

’ ðzÞ ¼ 0,

Z�

ðdd c’Þn 5 þ1� �

,

D0ð�Þ ¼ f’ 2 E0ð�Þ : suppðddc’Þn �� �g,

Fð�Þ ¼ ’ 2 PSHð�Þ : 9E0ð�Þ 3 ’j & ’, supj�1

Z�

ðdd c’jÞn 5 þ1

( ),

Epð�Þ ¼ ’ 2 PSHð�Þ : 9E0ð�Þ 3 ’j & ’, supj�1

Z�

ð�’jÞpðdd c’jÞ

n < þ1

( ),

F pð�Þ ¼ Epð�Þ \ Fð�Þ,

where p40. By Holder’s inequality it follows that

F pð�Þ � F qð�Þ 8 05 q5 p:

We set

F1ð�Þ ¼\p>0

F pð�Þ:

(4) For each u2Ep(�) we set

epðuÞ ¼

Z�

ð�uÞpðdd cuÞn:

3. Pluripolar sets in the class F1In [4] Cegrell has proved that for every pluripolar set E in � we can find u2F 1(�) such

that u��1 on E. According to the ideas given in [3] we prove the following

THEOREM 3.1 Let E be a complete pluripolar set in �. Then, there exists a function

’2F1 (�) such that (E¼ {’¼�1}.

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We need some lemmas

LEMMA 3.2 For every u2PSH(�), there is v2PSH�(�) such that

fz 2 � : uðzÞ ¼ �1g ¼ fz 2 � : vðzÞ ¼ �1g:

Proof Choose ’2PSH�(�)\L1(�), limz!@� ’ðzÞ ¼ 0, ’ 6� 0. We set �j ¼ f’ < �1=j2g,

then �j��� for j� 1 and {�j} is an increasing exhaustion sequence of �. We set

mj ¼ maxðsup��j

u, 0Þ, uj ¼1

j2ð jþmjÞðu�mjÞ:

The following assertions are straightforward:

uj 2 PSHð�Þ, uj � 0 on �j, uj � �1

j2on fu � �jg,ðaÞ

uj � �1 on X \�j where X ¼ fz 2 � : uðzÞ ¼ �1g:ðbÞ

It follows from (a) that the function

vj ¼maxðuj �

1

j 2,’Þ on �j

’ on �n�j

8<:

belongs to PSH�(�)\L1(�). Now we set v ¼P1

j¼1 vj. It is clear that v2PSH�(�)(possibly��1). All that remains is to check {z2� : u(z)¼�1}¼ {z2� : v(z)¼�1}.To see this, let z02X. Then there is j0 such that z02X\�j 8 j� j0. It follows that

vðz0Þ ¼Xj0j¼1

vjðz0Þ þX1

j¼j0þ1

vjðz0Þ

¼Xj0j¼1

vjðz0Þ þX1

j¼j0þ1

’ðz0Þ ¼ �1

because ’(z0)50. Finally, if z02� nX then there is j0 such that z02�jn{u4�j} 8 j� j0.We have

vðz0Þ ¼Xj0j¼1

vjðz0Þ þX1

j¼j0þ1

vjðz0Þ

� j0’ðz0Þ þX1

j¼j0þ1

ujðz0Þ �1

j2

� �

� j0’ðz0Þ þX1

j¼j0þ1

�2

j24 �1:

From the ideas given in [10, Corollary 2.6], we have the following

LEMMA 3.3 Let u2E0 and k¼ 0, 1, . . . . Then we have

ekðuÞ � ekðuþ vÞ � ekðuÞ þ AXnj¼1

Z�

ðdd cvÞn� �j=n

8 v 2 E0, �1 � v � 0:

where A depends on u, k, n.

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Proof We may assume that �1� u� 0. We have

ekðuþ vÞ ¼

Z�

ð�u� vÞkðdd cuÞn þXnj¼1

n

j

!Z�

ð�u� vÞkðdd cuÞn�j ^ ðdd cvÞj

�

Z�

ð�u� vÞkðdd cuÞn þ 2kXnj¼1

n

j

!Z�

ðdd cuÞn�j ^ ðdd cvÞj:

Using Corollary 5.6 in [4] we obtain the following estimates

ekðuþ vÞ �

Z�

ð�uÞkðdd cuÞn þXkm¼1

k

m

!Z�

ð�uÞk�mð�vÞmðdd cuÞn

þ 2kXnj¼1

n

j

! Z�

ðdd cuÞn� �n�j=n Z

�

ðdd cvÞn� �j=n

�

Z�

ð�uÞkðdd cuÞn þXkm¼1

k

m

!Z�

ð�vÞðdd cuÞn

þ 2kXnj¼1

n

j

! Z�

ðdd cuÞn� �n�j=n Z

�

ðdd cvÞn� �j=n

�

Z�

ð�uÞkðdd cuÞn þ 2kZ

�

ð�vÞðdd cuÞn

þ 2kXnj¼1

n

j

! Z�

ðdd cuÞn� �n�j=n Z

�

ðdd cvÞn� �j=n

:

Thus by integrating by parts we obtain

ekðuþ vÞ

�

Z�

ð�uÞkðddcuÞnþ2kZ

�

ð�uÞðddcuÞn�1^ddcvþ2kXnj¼1

n

j

� � Z�

ðddcuÞn� �n�j=n Z

�

ðddcvÞn� �j=n

�

Z�

ð�uÞkðddcuÞnþ2kZ

�

ðddcuÞn�1^ddcvþ2kXnj¼1

n

j

� � Z�

ðddcuÞn� �n�j=n Z

�

ðddcvÞn� �j=n

Again by Corollary 5.6 in [4] we have

ekðuþ vÞ �

Z�

ð�uÞkðdd cuÞn þ 2kZ

�

ðdd cuÞn� �n�1=n Z

�

ðdd cvÞn� �1=n

þ 2kXnj¼1

n

j

! Z�

ðdd cuÞn� �n�j=n Z

�

ðdd cvÞn� �j=n

:

Proof of Theorem 3.1 By Lemma 3.1 we choose a function 2PSH�(�) such that

E¼ { ¼�1}. Since E is Borelian and pluripolar, we have

0 ¼ CnðE Þ ¼ inf CnðU Þ : E � U,U is openg:�

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Thus, there is a sequence of open subsets Uj�� such that E�Uj, 8j� 1 and

limj!1

CnðUjÞ ¼ 0. Let {�j} be an exhaustion sequence of open sets of �. Let hj ¼ h�Uj\�j,�

denote the relative extremal functions of Uj\�j with respect to �. As � is hyperconvex we

have hj2E0, �1� hj� 0, hj��1 on Uj\�j andZ�

ðdd chjÞn¼ CnðUj \�jÞ � CnðUjÞ ! 0 as j!1, ð1Þ

and

hjjUj\�j¼ �1 for j � 1:

For each increasing sequence {aj}j¼ 0, 1, . . .�N considerP1

j¼0 haj 2 PSH�ð�Þ. Obviously,P1j¼0 haj � �1 on E. Indeed, let z02E. Take j0 such that z0 2 �j0 . Since z02Uj\�j for

j� j0 we haveP1

j¼0 hajðz0Þ ¼ �1. We show that there exists a subsequence {aj}�N such

that u ¼P1

j¼0 haj 2T1

k¼1 Fkð�Þ ¼T

p4 0 Fpð�Þ. We claim that there are two sequences

{ak}%1 and {bk} such thatZ�

ð�ha0 � � � � � hakÞmðdd cðha0 þ � � � hak ÞÞ

n 5 bm, 8 0 � m � k: ð2Þ

Indeed, we choose a0¼ 1 and b0 ¼R

�ðddch1Þ

nþ 1. Assume that a0, . . . , ak and b0, . . . , bk

have been chosen. We set uk ¼ ha0 þ � � � þ hak 2 E0. Applying Lemma 3.3 and (1) we can

find akþ1 large enough so thatZ�

ð�uk � hakþ1Þmðdd cðuk þ hakþ1 ÞÞ

nbm, 8 0 � m � k:

We set

bkþ1 ¼

Z�

ð�uk � hakþ1Þkþ1ðdd cðuk þ hakþ1 ÞÞ

nþ 1:

This proves the claim. Consider the sequence uk ¼ ha0 þ � � � þ hak 2 E0. Then

uk& u2PSH�(�). Moreover for every m� 1, by (2) we have

supk�0

Z�

ð�ukÞmðdd cukÞ

n� sup

k�m

Z�

ð�ukÞmðdd cukÞ

n 5 bm5 þ1 8 m � 0:

Thus u2Fm(�), 8m� 0. So u 2 F1ð�Þ ¼T1

m¼0 Fmð�Þ. We set ’¼max(u, ). By

Lemma 3.4 in [3] we have ’2F1(�). Clearly E¼ {z2� : ’(z)¼�1}.

4. The subextension in the class Ep, p40

In this section, for each bounded hyperconvex domain ~�, satisfying � � ~�, and u2Ep(�)

we prove that there exists ~u 2 Epð ~�Þ such that ~u� u on �, (ddc ~u)n� 1�(ddcu)n and

ep( ~u)� ep(u). For each u2PSH�(�) we define the subextension of u to ~� as

~u ¼ supf’ 2 PSH�ð ~�Þ : ’ � u on �g:

In this section, we can state the main results

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THEOREM 4.1 Let �, ~� be such that � � ~�. If u2Ep(�), then there exists ~u 2 Epð ~�Þ such

that ~u� u on �, (dd c ~u) n� 1�(ddcu)n and ep( ~u)� ep(u).

THEOREM 4.2 Let �, ~� and u2PSH�(�) be such that � � ~�, max (u,�1)2Ep(�). Then

there exists ~u 2 PSH�ð ~�Þ such that maxð ~u, �1Þ 2 "pð ~�Þ, ~u� u on � and

(ddc ~u)n� 1�(ddcu)n.

We need some lemmas

LEMMA 4.3 Let � be a domain in Cn and w2C�(�). We set

u ¼ supf’ 2 PSH�ð�Þ : ’ � w on �g:

Then (dd cu)n¼ 0 on {u5w}.

Proof See Corollary 9.2 in [2].

LEMMA 4.4 Let �, ~� be such that � � ~�. For each u2D0\C(�), we set

~u ¼ supf’ 2 PSH�ð ~�Þ : ’ � u on �g:

Then ~u 2 "0ð ~�Þ, (ddc ~u)n¼ 0 on ð ~�n�Þ [ ðf ~u < ug \�Þ and (dd c ~u)n� (dd cu)n on { ~u¼ u}\�.

Proof If u� 0 then ~u� 0. We assume that u =� 0. We choose a ball B(z, r)��� and M40

such that u�MhB(z,r),� on supp(ddcu)n. By the comparison principle we get

u �MhBðz, rÞ,� �MhBðz, rÞ, ~� on �. It implies that ~u �MhBðz, rÞ, ~�. Hence ~u 2 "0ð ~�Þ.By using Lemma 4.3 for w 2 C�ð ~�Þ with

w ¼u on �

0 on ~�n�

�

we have (dd c ~u)n¼ 0 on f ~u5wg ¼ ð ~�n�Þ [ ðf ~u5 ug \�Þ. Let K be a compact set in

{ ~u¼ u}\�. Since K � f ~uþ 1=j4ug, we haveZK

ðdd c ~uÞn ¼ limj!þ1

ZK

ðdd c maxð ~uþ1

j, uÞÞn �

ZK

ðdd cmaxð ~u, uÞÞn

¼

ZK

ðdd cuÞn:

Hence (dd c ~u)n�(ddcu)n on { ~u¼ u}\�.

LEMMA 4.5 Let �, ~� be such that � � ~�. For each u2F (�), we set

~u ¼ supf’ 2 PSH�ð ~�Þ : ’ � u on �g:

Then ~u 2 Fð ~�Þ, (dd c ~u)n� 1�(ddcu)n and (dd c ~u)n¼ 0 on { ~u5u}\�.

Proof If u� 0 then ~u� 0. We assume that u =� 0. By Theorem 2.1 in [4] we can choose

D0\C(�)3 uj& u. We set

~uj ¼ supf’ 2 PSH�ð ~�Þ : ’ � uj on �g:

By Lemma 4.4 we have ~uj 2 E0ð ~�Þ,R

~�ðddc ~ujÞ

n�R

�ðddcujÞ

n and (dd c ~uj)n� 1�(dd

cuj)n.

Hence ~u 2 Fð ~�Þ and 1 ~�n@�ðddc ~uÞn � 1�ðdd

cuÞn. By Lemma 4.4 we get

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ðmaxðuj, �sÞ �maxð ~uj, �sÞÞðddc ~ujÞ

n¼ 0,

for all s40. Letting j!1 we obtain

ðmaxðu, �sÞ �maxð ~u, �sÞÞðdd c ~uÞn ¼ 0,

for all s40. Hence (ddc ~u)n¼ 0 on { ~u5�s5u}\� for all s40. Moreover, since

f ~u < ug \� ¼S

s2Q�f ~u < �sug \� we have (dd c ~u)n¼ 0 on { ~u5u}\�. We only have to

prove that (dd c ~u)n¼ 0 on ~� \ @�. Let K be a compact subset in ~� \ @�. Given "40.

SinceR

�ðdd cuÞn < þ1, we can find X��Y��� such that

R�nXðdd

cuÞn < �. Let U be

open set such that K�U and U\���nY. Let g2C1(Cn) be such that

g ¼ 1 on CnnY, g ¼ 0 on X and 0 � g � 1:

By Lemma 4.4 we haveZK

ðdd c ~uÞn �

ZU

ðdd c ~uÞn � limj!1

ZU

ðdd c ~ujÞn� lim

j!1

ZU\�

ðdd cujÞn� lim

j!1

Z�

gðdd cujÞn

¼ limj!1½

Z�

ðg� 1Þðdd cujÞnþ

Z�

ðdd cujÞn�

¼

Z�

ðg� 1Þðdd cuÞn þ

Z�

ðdd cuÞn

¼

Z�

gðdd cuÞn �

Z�nX

ðdd cuÞn 5 �:

Therefore we obtain (ddc ~u)n(K)¼ 0. From the ideas given in [7], Lemma 2.1, we have the

following

LEMMA 4.6 Let u2Ep(�), then there exists a sequence {uj}�D0(�) such that uj& u and

(ddcuj)n� (dd cu)n.

Proof By Theorem 5.11 in [4] we have (ddcu)n¼ f(dd c’)n with ’2E0(�) and

f 2 L1locððdd

c�ÞnÞ. By Kolodziej’s theorem ([11]) there exists uj2E0(�) such that

ðdd cujÞn¼ minðf, jÞ1�j

ðdd c�Þn,

where �j is a fundamental increasing sequence of �. By Theorem 6.2 in [3] for p� 1 and

Theorem 4.2 in [12] for 05p51 we get uj& u as j!1.

The proof of Theorem 4.1 If u� 0 then ~u� 0. We assume that u =� 0. By Lemma 4.6 we can

choose D0(�)3 uj& u such that (ddcuj)n� (dd cu)n. By Lemma 4.5 we haveZ

~�

ð� ~ujÞpðdd c ~ujÞ

n¼

Zf ~uj¼ujg\�

ð� ~ujÞpðdd c ~ujÞ

n

�

Zf ~uj¼ujg\�

ð�ujÞpðdd cujÞ

n�

Z�

ð�ujÞpðdd cujÞ

n:

Hence

supj�1

epð ~ujÞ � supj�1

epðujÞ

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Therefore ~u 2 Epð ~�Þ and ep( ~u)� ep(u). By Lemma 4.5 we have 1 ~�n@�ðddc ~uÞn � 1�ðdd

cuÞn.We only have to prove that (dd c ~u)n¼ 0 on ~� \ @�. Let K be a compact subset in ~� \ @�.Let �0 be such that K � �0 �� ~�. Given "40. Since

R�ð�uÞpðdd cuÞn < þ1, we can find

X��� such thatR

�nXð�uÞpðdd cuÞn < �. Let U be open set such that K�U���0 and

U\��� nX. By Lemma 4.5 we getZK

ðdd c ~uÞn �

ZU

ðdd c ~uÞn �1

Mp

ZU

ð� ~uÞpðdd c ~uÞn �1

Mplimj!1

ZU

ð� ~ujÞpðdd c ~ujÞ

n

�1

Mplimj!1

ZU\�

ð�ujÞpðdd cujÞ

n

�1

Mp

ZU\�

ð�uÞpðdd cuÞn �1

Mp

Z�nX

ð�uÞpðdd cuÞn 5�

Mp:

where M ¼ �max�0 ~u > 0. Therefore (dd c ~u)n(K)¼ 0.

PROPOSITION 4.7 Let u2E0(�). ThenZ�

�maxð�ð�uÞp, �1Þðdd cuÞn ¼

Z�

ð�maxðu, �1ÞÞpðdd c maxðu, �1ÞÞn

for all p� 0.

Proof By Proposition 4.2 in [13] we haveZfu4�1g

�maxð�ð�uÞp, �1Þðdd cuÞn ¼

Zfu4�1g

ð�uÞpðdd cuÞn

¼

Zfu4�1g

ð�uÞpðdd c maxðu, �1ÞÞn

¼

Zfu4�1g

ð�maxðu, �1ÞÞpðdd c maxðu, �1ÞÞn:

By Stokes’ theorem we haveZfu��1g

�maxð�ð�uÞp, �1Þðdd cuÞn ¼

Zfu��1g

ðdd cuÞn

¼

Z�

ðdd cuÞn �

Zfu4�1g

ðdd cuÞn

¼

Z�

ðdd c maxðu, �1ÞÞn �

Zfu4�1g

ðdd c maxðu, �1ÞÞn

¼

Zfu��1g

ðdd c maxðu, �1ÞÞn

¼

Zfu��1g

ð�maxðu, �1ÞÞpðdd c maxðu, �1ÞÞn:

PROPOSITION 4.8 Let (t)¼max (�(�t)p,�1). Then

(a) E (�)¼ {u2PSH�(�): max (u,�1)2Ep(�)}.(b) E (�)¼ {u2PSH�(�): 9 ’2Ep(�), 9 v2F (�) such that u� ’þ v}.

682 P.H. Hiep

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where we set

" ð�Þ ¼ ’ 2 PSH�ð�Þ : 9 "0ð�Þ 3 ’j & ’, supj�1

Z�

� ð’jÞðddc’jÞ

n < þ1

( ):

Proof (a) Direct application of Proposition 4.7.

(b) Let u2E (�). By Theorem 2.1 in [4] we can choose "0 \ Cð ��Þ 3 uj & u. By

Kolodziej’s theorem ([11]) there exists ’j,vj2E0(�) such that

ðdd c’jÞn¼ 1fuj 4�1gðdd

cujÞn¼ 1fuj 4�1gðdd

c maxðuj, �1ÞÞn

and

ðdd cvjÞn¼ 1fuj��1gðdd

cujÞn

Since

supj�1

Z�

ðdd cvjÞn¼ sup

j�1

Zfuj��1g

� ðujÞðddcujÞ

n� sup

j�1

Z�

� ðujÞðddcujÞ

n < þ1

and [14] we can assume that vj! v2F (�) a.e. By the comparison principle we get

’j�max(uj,�1)�max(u,�1)2Ep(�) and uj� ’jþ vj. Hence u�max(u,�1)þ v.

Proof of Theorem 4.2 We set

~u ¼ supf’ 2 PSH�ð ~�Þ : ’ � u on �g:

By Proposition 4.8 we can find ’2Ep(�), v2F (�) such that u� ’þ v. Since ~u � ~’þ ~v we

have max ( ~u,�1)2Ep(�). By Theorem 4.14 in [15] we can choose F (�)3 uj& u such that

ðdd cujÞn¼ 1�j

ðdd cuÞn where �j is a fundamental increasing sequence of �. By Lemma 4.5

we obtain ðdd c ~uÞn ¼ limj!1ðddc ~ujÞ

n� limj!1 1�ðdd

cujÞn� 1�ðdd

cuÞn.

Acknowledgements

I am grateful to Professor Urban Cegrell for many helpful discussions on subextension that helpedto improve this article during my visit to Mid Sweden University in Sundsvall, Sweden, in 2006.I thank to Dr Nguyen Quang Dieu for some comments on pluripolar sets. The author is alsoindebted to the referee for his useful comments.This work is supported by the National ResearchProgramme for Natural Sciences, Vietnam.

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(2004), pp. 159–179.[5] P. Ahag and R. Czyz, On the Cegrell classes, Math. Zeit. 256 (2007), pp. 243–264.

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