Convergence in capacity of plurisubharmonic functions with given boundary values

0.14 MB·

by  Nguyen Xuan Hong

CONVERGENCE IN CAPACITY OF PLURISUBHARMONIC
FUNCTIONS WITH GIVEN BOUNDARY VALUES
NGUYEN XUANHONG,NGUYENVAN TRAO ANDTRAN VAN THUY
6
1
Abstract. In this paper, we study the convergence in the capacity of se-
0
quence of plurisubharmonic functions. As an application, we prove stability
2
results for solutions of thecomplex Monge-Amp`ere equations.

r
a
M
1. Introduction

1
1 It is well-known that convergence in the sense of distributions of plurisubhar-
monic functions does not in general imply convergence of their Monge-Amp`ere

]
measures. Therefore, it is important to find conditions on sequences of plurisub-
V
harmonic functions such that the corresponding Monge-Amp`ere measures are
C
convergent in the weak* topology.
.
h Bedford and Taylor [3] introduced and studied in 1982 the C -capacity of
n
t
a Borel sets. Xing [21] proved in 1996 that the complex Monge-Amp`ere opera-
m
tor is continuous under convergence of bounded plurisubharmonic functions in
[ C -capacity. He gave a sufficient condition for the weak convergence of complex
n
2 Monge-Amp`ere mass of bounded plurisubharmonic functions. Later, Xing [22]
v studied in 2008 the convergence in the C -capacity of a sequence of plurisubhar-
n
2 monic functions in the class Fa(Ω). Hiep [15] studied in 2010 the convergence in
5
C -capacity within the class E(Ω). Recently, Cegrell [8] proved in 2012 that if a
1 n
3 sequenceofplurisubharmonicfunctionsisboundedfrombelowbyafunctionfrom
0 the Cegrell class E(Ω) and convergent in C -capacity then the corresponding
n−1
.
1 complex Monge-Amp`ere measures are convergent in the weak* topology.
0
Thepurposeof this paperis to study conditions on a sequenceof plurisubhar-
6
monic functions which are equivalent to convergence in C -capacity. Our main
1 n
: result is the following theorem.
v
i
X Main theorem. Let Ω be a bounded hyperconvex domain in Cn and let f ∈E(Ω),
r w ∈Na(Ω,f) such that (−ρ)(ddcw)n < +∞ for some ρ∈ E (Ω). Assume that
a Ω 0
{uj} ⊂ Na(Ω,f) such thRat uj → u0 a.e. on Ω as j → +∞ and uj ≥ w in Ω for
all j ≥ 0. Then, the following statements are equivalent.
(a) u → u in C -capacity in Ω;
j 0 n
(b) For every a > 0, we have
u u
lim max j,ρ (ddcu )n = max 0,ρ (ddcu )n.
j 0
j→+∞ZΩ (cid:16) a (cid:17) ZΩ (cid:16) a (cid:17)
2010 Mathematics Subject Classification: 32W20.
Keywordsandphrases:Plurisubharmonicfunctions,ComplexMonge-Amp`ereoperator,Con-
vergence in capacity.
1
2 NGUYENXUANHONG,NGUYENVANTRAOANDTRANVANTHUY
(c) For every a > 0, we have
v u
lim max j,ρ −max j,ρ (ddcu )n = 0,
j
j→+∞ZΩh (cid:16) a (cid:17) (cid:16) a (cid:17)i
∗
where v := sup u .
j k≥j k
(cid:0) (cid:1)
The paper is organized as follows. In Section 2 we recall some notions of
pluripotential theory. Section 3 is devoted to the proof of the main theorem. In
Section 4 we apply the main theorem to prove a stability result for the solutions
of certain complex Monge-Amp`ere equations.
2. Preliminaries
Someelements ofpluripotentialtheorythatwillbeusedthroughoutthepaper
can be found in [1]-[22].
Definition 2.1. Let n be a positive integer. A bounded domain Ω in Cn is
called bounded hyperconvex domain if there exists a bounded plurisubharmonic
function ϕ : Ω → (−∞,0) such that the closure of the set {z ∈ Ω : ϕ(z) < c} is
compact in Ω, for every c ∈(−∞,0).
We denote by PSH(Ω) the family of plurisubharmonic functions defined on
Ω and PSH−(Ω) denotes the set of negative plurisubharmonic functions on Ω.
By MPSH(Ω) denotes the set of all maximal plurisubharmonic functions in Ω.
Definition 2.2. Let Ω be a bounded hyperconvex domain in Cn. We say that
a bounded, negative plurisubharmonic function ϕ in Ω belongs to E (Ω) if {ϕ <
0
−ε} ⋐ Ω for all ε > 0 and (ddcϕ)n < +∞.
Ω
Let F(Ω) be the familyRof plurisubharmonic functions ϕ defined on Ω, such
that there exists a decreasing sequence {ϕ } ⊂ E (Ω) that converges pointwise to
j 0
ϕ on Ω as j → +∞ and
sup (ddcϕ )n < +∞.
j
j ZΩ
We denote by E(Ω) the family of plurisubharmonic functions ϕ defined on
Ω such that for every open set G ⋐ Ω there exists a plurisubharmonic function
ψ ∈F(Ω) satisfy ψ = ϕ in G.
Let u∈ E(Ω) and let {Ω } be an increasing sequence of boundedhyperconvex
j
domains such that Ω ⋐ Ω ⋐ Ω and +∞Ω = Ω. Put
j j j=1 j
S
uj := sup{ϕ ∈ PSH−(Ω) :ϕ ≤ u in Ω\Ω }
j
and N(Ω):= {u ∈ E(Ω) :uj ր 0 a.e. in Ω}.
Let K ∈ {F,N,E}. We denote by Ka(Ω) the subclass of K(Ω) such that the
Monge-Amp`ere measure (ddc.)n vanishes on all pluripolar sets of Ω.
Let f ∈ E(Ω) and K ∈ {Fa,Na,Ea,F,N,E}. Then we say that a plurisub-
harmonic function ϕ defined on Ω belongs to K(Ω,f) if there exists a function
ψ ∈K(Ω) such that
ψ+f ≤ ϕ≤ f in Ω.
Now we will show that if u ∈ Na(Ω,f) then the pluripolar part of (ddcu)n is
carried by {f = −∞}.
CONVERGENCE IN CAPACITY 3
Proposition 2.3. Let Ω be a bounded hyperconvex domain in Cn. Assume that
f ∈ E(Ω) and u∈ Na(Ω,f) such that (−ρ)(ddcu)n < +∞ for some ρ∈ E (Ω).
Ω 0
Then R
1 (ddcu)n = 1 (ddcf)n in Ω.
{u=−∞} {f=−∞}
Proof. Let v ∈ Fa(Ω) such that v+f ≤ u≤ f in Ω. By Lemma 4.1 and Lemma
4.12 in [1] we have
1 (ddcf)n ≤ 1 (ddcu)n
{f=−∞} {u=−∞}
≤ 1 (ddc(v+f))n = 1 (ddcf)n.
{v+f=−∞} {f=−∞}
It follows that
1 (ddcu)n = 1 (ddcf)n in Ω.
{u=−∞} {f=−∞}
The proof is complete. (cid:3)
Proposition 2.4. Let Ω be a bounded hyperconvex domain in Cn. Let f ∈ E(Ω)
and u ∈ Na(Ω,f) such that (−ρ)(ddcu)n < +∞ for some ρ ∈ E (Ω). Assume
Ω 0
that v ∈ E(Ω) such that v ≤ fR and (ddcv)n ≥ (ddcu)n in Ω. Then v ≤ u on Ω.
Proof. Sincethemeasure1 (ddcu)n vanishes on all pluripolarsubsets of Ω,
{u>−∞}
by Proposition 4.3 in [19] we get
(ddcmax(u,v))n ≥ 1 (ddcu)n.
{u>−∞}
Hence,
1 (ddcmax(u,v))n ≥ 1 (ddcu)n.
{max(u,v)>−∞} {u>−∞}
Moreover, by the hypotheses and Proposition 2.3 we have
1 (ddcmax(u,v))n = 1 (ddcu)n.
{max(u,v)=−∞} {u=−∞}
Hence, (ddcmax(u,v))n ≥ (ddcu)n in Ω. Therefore, from Theorem 3.6 in [1] it
follows that max(u,v) = u in Ω. Thus, v ≤ u in Ω. The proof is complete. (cid:3)
3. Proof of the main theorem
In order to prove the main theorem, we need the following auxiliary lemmas.
Lemma 3.1. Let Ω be a bounded hyperconvex domain in Cn and let f ∈ E(Ω).
Assume that ρ∈ E (Ω) and u ∈ Na(Ω,f) such that (−ρ)(ddcu)n < +∞. Then
0 Ω
for every v ∈ Ea(Ω,f) and for every ϕ ∈ E0(Ω) withRϕ ≥ ρ, we have
1
(v−u)n(ddcϕ)n + −ϕ(ddcv)n
n!Z Z
{u<v} {u<v}
≤ −ϕ(ddcu)n.
Z
{u<v}
Proof. For j ∈ N∗, put v = max(u,v − 1). Because u ≤ v ≤ f in Ω, we have
j j j
v ∈ Fa(Ω,f). By Lemma 3.5 in [1] we have
j
1
(v −u)n(ddcϕ)n + −ϕ(ddcv )n ≤ −ϕ(ddcu)n.
j j
n!Z Z Z
Ω Ω Ω
By Theorem 4.1 in [19] we have v = v− 1 in {u < v }. Hence,
j j j
1
(v −u)n(ddcϕ)n + −ϕ(ddcv)n
j
n!Z Z
{u<vj} {u<vj}
1
= (v −u)n(ddcϕ)n + −ϕ(ddcv )n
j j
n!Z Z
{u<vj} {u<vj}
4 NGUYENXUANHONG,NGUYENVANTRAOANDTRANVANTHUY
1
≤ (v −u)n(ddcϕ)n + −ϕ(ddcv )n
j j
n!Z Z
Ω {u<vj}
1
≤ (v −u)n(ddcϕ)n + −ϕ(ddcv )n− −ϕ(ddcv )n
j j j
n!Z Z Z
Ω Ω {u=vj}
≤ −ϕ(ddcu)n− −ϕ(ddcv )n.
j
Z Z
Ω {u≥v}
Now, since u= v in {u > v− 1} so by Theorem 4.1 in [19] imply that
j j
(ddcu)n = (ddcv )n in {u≥ v}∩{u > −∞}.
j
Moreover, by Proposition 2.3 we have
1 (ddcu)n = 1 (ddcv )n = 1 (ddcf)n in Ω.
{u=−∞} {vj=−∞} j {f=−∞}
Hence, we obtain that
(ddcu)n = (ddcv )n in {u ≥ v}.
j
Therefore,
1
(v −u)n(ddcϕ)n + −ϕ(ddcv)n
j
n!Z Z
{u<vj} {u<vj}
≤ −ϕ(ddcu)n− −ϕ(ddcu)n
Z Z
Ω {u≥v}
= −ϕ(ddcu)n.
Z
{u<v}
Let j → +∞ we obtain that
1
(v−u)n(ddcϕ)n + −ϕ(ddcv)n ≤ −ϕ(ddcu)n.
n!Z Z Z
{u<v} {u<v} {u<v}
The proof is complete. (cid:3)
Lemma 3.2. Let Ω be a bounded hyperconvex domain in Cn and let {u } ⊂
j
Ea(Ω) such that u ≥ u for every j ≥ 1 and u → u in C -capacity in Ω.
j 1 j 0 n
Assume that {ϕk}, k = 1,2 are sequences of uniformly bounded plurisubharmonic
j
functions in Ω which converges weakly to a plurisubharmonic function ϕk in Ω.
0
Then ϕ1ϕ2(ddcu )n → ϕ1ϕ2(ddcu )n weakly as j → +∞.
j j j 0 0 0
Proof. Withoutloss ofgenerality wecan assumethatu ∈ Fa(Ω)and−1 ≤ ϕk ≤
j j
0 in Ω for all j ≥ 0, k = 1,2. Put
(ϕ1+ϕ2+2)2+4 (ϕ1+2)2 (ϕ2 +2)2
ψ1 = j j , ψ2 = j and ψ3 = j .
j 2 j 2 j 2
It is clear that ψk ∈PSH(Ω), 0 ≤ ψk ≤ 4 and ψk → ψk weakly in Ω as j → +∞,
j j j 0
k = 1,2,3. Since ϕ1ϕ2 = ψ1 −ψ2 −ψ3 in Ω we obtain by Theorem 3.4 in [22]
j j j j j
that
ϕ1ϕ2(ddcu )n = ψ1(ddcu )n −ψ2(ddcu )n−ψ3(ddcu )n
j j j j j j j j j
→ ψ1(ddcu )n −ψ2(ddcu )n−ψ3(ddcu )n = ϕ1ϕ2(ddcu )n
0 0 0 0 0 0 0 0 0
weakly in Ω as j → +∞. The proof is complete. (cid:3)
CONVERGENCE IN CAPACITY 5
Proof of the main theorem. Without loss of generality we can assume that f < 0
and −1≤ ρ≤ 0 in Ω.
(a)⇒(b). Fix a > 0. Put
u
j
ϕ := max ,ρ .
j
a
(cid:16) (cid:17)
Because
0 ≤ sup −ϕ (ddcu )n ≤ −ρ(ddcw)n <+∞,
j j
j ZΩ ZΩ
it remains to prove that there exists a subsequence {u } of sequence {u } such
jk j
that
lim ϕ (ddcu )n = ϕ (ddcu )n.
k→+∞ZΩ jk jk ZΩ 0 0
First we claim that there exists an increasing sequence {j } ⊂N∗ such that
k
u
lim ϕ max 1+ jk,0 (ddcu )n = ϕ (ddcu )n. (3.1)
k→+∞ZΩ jk (cid:16) k (cid:17) jk Z{u0>−∞} 0 0
Indeed,let χ ∈ C∞(Ω) such that 0 ≤ χ ≤ χ ≤ 1 in Ω,{ρ ≤ −1} ⋐ {χ = 1}
k 0 k k+1 k k
and
1
(−ρ)(ddcu )n ≤ .
0
Z k
{χk<1}
Since u → u in C -capacity in Ω as j → +∞, so max(u ,−k) → max(u ,−k)
j 0 n j 0
in C -capacity as j → +∞. By Lemma 3.2 we have
n
u u
ϕ max 1+ j,0 (ddcmax(u ,−k))n → ϕ max 1+ 0,0 (ddcmax(u ,−k))n
j j 0 0
k k
(cid:16) (cid:17) (cid:16) (cid:17)
weakly in Ω as j → +∞. Hence, by Theorem 4.1 in [19] we get
u
lim χ ϕ max 1+ j,0 (ddcu )n
k j j
j→+∞ZΩ (cid:16) k (cid:17)
u
= lim χ ϕ max 1+ j,0 (ddcmax(u ,−k))n
k j j
j→+∞ZΩ (cid:16) k (cid:17)
u
= χ ϕ max 1+ 0,0 (ddcmax(u ,−k))n
k 0 0
ZΩ (cid:16) k (cid:17)
u
= χ ϕ max 1+ 0,0 (ddcu )n.
k 0 0
ZΩ (cid:16) k (cid:17)
Because χ max 1+ u0,0 ր 1 as k → +∞ in Ω, we have
k k {u0>−∞}
(cid:0) (cid:1)
u
lim χ ϕ max 1+ 0,0 (ddcu )n = ϕ (ddcu )n.
k 0 0 0 0
k→+∞ZΩ (cid:16) k (cid:17) Z{u0>−∞}
Therefore, there exists an increasing sequence {j } ⊂ N∗ such that
k
u
lim χ ϕ max 1+ jk,0 (ddcu )n = ϕ (ddcu )n. (3.2)
k→+∞ZΩ k jk (cid:16) k (cid:17) jk Z{u0>−∞} 0 0
6 NGUYENXUANHONG,NGUYENVANTRAOANDTRANVANTHUY
Now, fix k ∈ N∗. By the proof of the theorem in [8] (see (3.1) in [8]) we have
0
u
liminf (1−χ )ϕ max 1+ jk,0 (ddcu )n
k→+∞ZΩ k jk (cid:16) k (cid:17) jk
≥ liminf (1−χ )ϕ (ddcu )n ≥ liminf (1−χ )ρ(ddcu )n
k→+∞ZΩ k jk jk k→+∞ZΩ k0 jk
= lki→m+in∞f(cid:20)ZΩρ(ddcujk)n−ZΩχk0ρ(ddcujk)n(cid:21) (3.3)
= ρ(ddcu )n − χ ρ(ddcu )n
0 k0 0
Z Z
Ω Ω
1
≥ ρ(ddcu )n ≥ − .
0
Z k
{χk0<1} 0
Combining this with (3.2) we arrive at
u
lim ϕ max 1+ jk,0 (ddcu )n
k→+∞ZΩ jk (cid:16) k (cid:17) jk
u
= lim χ ϕ max 1+ jk,0 (ddcu )n
k→+∞ZΩ k jk (cid:16) k (cid:17) jk
= ϕ (ddcu )n.
0 0
Z
{u0>−∞}
This proves the claim.
Themeasure1 ϕ max ujk,−1 (ddcu )n vanishesonall pluripolar
{ujk>−∞} jk (cid:16) k (cid:17) jk
subset of Ω, hence by Lemma 5.14 in [6] there exists h ∈ Fa(Ω) such that
k
u
(ddch )n = 1 ϕ max jk,−1 (ddcu )n.
k {ujk>−∞} jk (cid:16) k (cid:17) jk
Because (ddch )n ≤ (ddcu )n in Ω and the measure (ddch )n vanishes on all
k jk k
pluripolar subset of Ω, from Corollary 3.2 in [1] we have
h ≥ u ≥ w in Ω.
k jk
Weclaimthath → 0inC -capacity inΩ.Indeed,letδ > 0andψ ∈ PSH(Ω)
k n
with −1≤ ψ ≤ 0. By Theorem 3.1 in [1] we have
1
(ddcψ)n ≤ (ddcψ)n ≤ (ddch )n
Z Z δn Z k
{hk<−δ} {hk<δψ} {hk<δψ}
1 u
≤ ϕ max jk,−1 (ddcu )n
δn Z{ujk>−∞} jk (cid:16) k (cid:17) jk
1 u
≤− max jk,ρ (ddcu )n.
δn Z{ujk>−∞} (cid:16) k (cid:17) jk
Therefore, by Lemma 3.3 in [1] and Proposition 2.3 we obtain that
1 u 1
(ddcψ)n ≤ − max jk,ρ (ddcu )n + ρ(ddcu )n
Z{hk<−δ} δn ZΩ (cid:16) k (cid:17) jk δn Z{ujk=−∞} jk
1 w 1
≤ − max ,ρ (ddcw)n + ρ(ddcw)n
δn ZΩ (cid:16)k (cid:17) δn Z{w=−∞}
1 w
= − max ,ρ (ddcw)n.
δn Z{w>−∞} (cid:16)k (cid:17)
CONVERGENCE IN CAPACITY 7
It follows that
1 w
C ({h < −δ}) ≤ − max ,ρ (ddcw)n.
n k δn Z{w>−∞} (cid:16)k (cid:17)
Hence, we get
lim C ({h < −δ}) = 0,
n k
k→+∞
for every δ > 0. Thus, h → 0 in C -capacity in Ω as k → +∞. This proves the
k n
claim, and therefore, by (3.3) and the Theorem in [8] we have
u
0≤ limsup ϕ max jk,−1 (ddcu )n
k→+∞ Z{ujk>−∞} jk (cid:16) k (cid:17) jk
≤ limsup (1−χ )(−ρ)(ddcu )n+limsup χ (ddch )n
Z k0 jk Z k0 k
k→+∞ Ω k→+∞ Ω
1
≤ ,
k
0
for all k ∈ N∗. Thus,
0
u
lim ϕ max jk,−1 (ddcu )n = 0.
k→+∞Z{ujk>−∞} jk (cid:16) k (cid:17) jk
Combining this with (3.1) we arrive at
lim ϕ (ddcu )n = ϕ (ddcu )n.
k→+∞Z{ujk>−∞} jk jk Z{u0>−∞} 0 0
Moreover, by Proposition 2.3, we have
ϕ (ddcu )n = ρ(ddcf)n = ϕ (ddcu )n.
Z jk jk Z Z 0 0
{ujk=−∞} {f=−∞} {u0=−∞}
Hence, we get
lim ϕ (ddcu )n = ϕ (ddcu )n.
k→+∞ZΩ jk jk ZΩ 0 0
(b)⇒(c). Fix a > 0. Since u → u a.e. in Ω as j → +∞ so v ց u as
j 0 j 0
j ր +∞.Hence,v → u inC -capacity inΩ.Therefore,bytheproofof (a)⇒(b)
j 0 n
and Lemma 3.3 in [1], we have
u v
max 0,ρ (ddcu )n = lim max j,ρ (ddcv )n
0 j
ZΩ (cid:16) a (cid:17) j→+∞ZΩ (cid:16) a (cid:17)
v
≥ lim max j,ρ (ddcu )n
j
j→+∞ZΩ (cid:16) a (cid:17)
u
≥ lim max j,ρ (ddcu )n
j
j→+∞ZΩ (cid:16) a (cid:17)
u
= max 0,ρ (ddcu )n.
0
ZΩ (cid:16) a (cid:17)
It follows that
v u
lim max j,ρ (ddcu )n = max 0,ρ (ddcu )n.
j 0
j→+∞ZΩ (cid:16) a (cid:17) ZΩ (cid:16) a (cid:17)
Therefore, we obtain that
v u
lim max j,ρ −max j,ρ (ddcu )n = 0.
j
j→+∞ZΩh (cid:16) a (cid:17) (cid:16) a (cid:17)i
8 NGUYENXUANHONG,NGUYENVANTRAOANDTRANVANTHUY
(c)⇒(a). Because v ց u in Ω as j ր +∞, we get v → u in C -capacity in
j 0 j 0 n
Ω. Hence, it is sufficient to prove that v −u → 0 in C -capacity in Ω. Let K be
j j n
a compact subset of Ω and let ε,δ > 0. Without loss of generality we can assume
that K ⋐ {ρ = −1}. Choose χ ∈ C∞(Ω) and a > b > 1 such that 0 ≤ χ ≤ 1,
0
{ρ ≤ −ε}⋐ {χ = 1}, {χ 6= 0} ⊂ {aρ < −b} and
a w
−max ,ρ (ddcw)n < ε. (3.4)
b Z{w>−∞} (cid:16)a (cid:17)
Let ψ ∈ E (Ω) with ψ ≥ ρ such that
j 0 j
C (K ∩{v −u > 2δ}) < (ddcψ )n+ε. (3.5)
n j j j
Z
K∩{vj−uj>2δ}
Note that u ≤ v in Ω for all j ≥ 1. From the hypotheses we have
j j
0≤ limsup χ(ddcu )n
j
Z
j→+∞ {uj<vj−δ}∩{uj>−b}
1
≤ limsup χ(v −u )(ddcu )n
j j j
δ j→+∞ Z{uj<vj−δ}∩{uj>−b}
a v u
≤ limsup max j,ρ −max j,ρ (ddcu )n
j
δ j→+∞ ZΩh (cid:16) a (cid:17) (cid:16) a (cid:17)i
= 0.
It follow that
lim χ(ddcu )n = 0.
j
j→+∞Z{uj<vj−δ}∩{uj>−b}
By Lemma 3.3 in [1] and Proposition 2.3 we have
limsup χ(ddcu )n
j
Z
j→+∞ {uj<vj−δ}∩{uj>−∞}
= limsup χ(ddcu )n
j
j→+∞ Z{uj<vj−δ}∩{−∞<uj≤−b}
u aρ
≤ limsup −max j, (ddcu )n
j
j→+∞ Z{uj>−∞} (cid:16) b b (cid:17)
a u a u
≤ limsup −max j,ρ (ddcu )n + max j,ρ (ddcu )n
j j
b j→+∞ ZΩ (cid:16) a (cid:17) b Z{uj=−∞} (cid:16) a (cid:17)
a w a w
≤ −max ,ρ (ddcw)n + max ,ρ (ddcw)n
b ZΩ (cid:16)a (cid:17) b Z{w=−∞} (cid:16)a (cid:17)
a w
= −max ,ρ (ddcw)n.
b Z{w>−∞} (cid:16)a (cid:17)
Therefore, by (3.4) we get
limsup χ(ddcu )n < ε. (3.6)
j
j→+∞ Z{uj<vj−δ}∩{uj>−∞}
Now, by Proposition 2.3 and Lemma 3.1 we have
(ddcψ )n ≤ (ddcψ )n
j j
Z Z
K∩{vj−uj>2δ} {uj<vj−2δ}
1
≤ (v −δ−u )n(ddcψ )n
δn Z j j j
{uj<vj−2δ}
CONVERGENCE IN CAPACITY 9
1
≤ (v −δ−u )n(ddcψ )n
δn Z j j j
{uj<vj−δ}
n!
≤ −ψ (ddcu )n.
δn Z j j
{uj<vj−δ}∩{uj>−∞}
Hence, from (3.6) we obtain that
n!
limsup (ddcψ )n ≤ limsup −ψ (ddcu )n
j→+∞ ZK∩{vj−uj>2δ} j δn j→+∞ Z{uj<vj−δ}∩{uj>−∞} j j
n!
≤ limsup −ψ (1−χ)(ddcu )n
δn j→+∞ ZΩ j j
n!
+ limsup χ(ddcu )n
δn Z j
j→+∞ {uj<vj−δ}∩{uj>−∞}
n! n!ε
≤ limsup −ρ(ddcu )n+
δn j→+∞ Z{ρ>−ε} j δn
n! n!ε
≤ limsup −max(ρ,−ε)(ddcu )n +
δn j→+∞ ZΩ j δn
n! n!ε
≤ −max(ρ,−ε)(ddcw)n + .
δn Z δn
Ω
Combining this with (3.5) we get
n! n!
limsupC ({K ∩{v −u > 2δ}) ≤ −max(ρ,−ε)(ddcw)n + +1 ε.
j→+∞ n j j δn ZΩ (cid:18)δn (cid:19)
Let εց 0 we obtain that
lim C ({K ∩{v −u > 2δ}) = 0.
n j j
j→+∞
Thus, v −u → 0 in C -capacity in Ω. The proof is complete. (cid:3)
j j n
4. Application
In this section, we prove a generalization of Cegrell and Kol odziej’s stability
theorem from [9]. First, we need the following.
Lemma 4.1. Let Ω be a bounded hyperconvex domain in Cn and let f ∈ E(Ω),
w ∈ Na(Ω,f) such that (−ρ)(ddcw)n < +∞ for some ρ ∈ E (Ω). Then for
Ω 0
every nonnegative Borel mReasures µ in Ω such that
(ddcf)n ≤ µ ≤ (ddcw)n,
there exists a unique u∈ Na(Ω,f) such that u≥ w and (ddcu)n = µ in Ω.
Proof. The uniqueness imply from Proposition 2.4. From the hypotheses and
Proposition 2.3 we have
1 µ = 1 (ddcf)n in Ω.
{w=−∞} {f=−∞}
Let {Ω } be a sequence of boundedhyperconvex domains such that Ω ⋐ Ω ⋐
j j j+1
Ω and Ω = +∞Ω . Because the measure 1 µ vanishes on all pluripolar
j=1 j {w>−∞}
subsetsofΩ,SapplyingProposition5.1in[14]weseethatthereareuj ∈ Na(Ωj,f)
such that
(ddcu )n = 1 µ+1 (ddcf)n = µ in Ω .
j Ωj∩{w>−∞} Ωj∩{f=−∞} j
10 NGUYENXUANHONG,NGUYENVANTRAOANDTRANVANTHUY
By Proposition 2.4 we have w ≤ u ≤ u ≤ f on Ω . Put u := lim u .
j+1 j j j→+∞ j
Then w ≤ u ≤ f and (ddcu)n = µ in Ω. Moreover, since w ∈ Na(Ω,f), we get
u∈ Na(Ω,f). The proof is complete. (cid:3)
Proposition 4.2. Let Ω be a bounded hyperconvex domain in Cn and let f ∈
E(Ω). Assume that w ∈ Na(Ω,f) such that (−ρ)(ddcw)n < +∞ for some
Ω
ρ ∈ E0(Ω). Then for every sequence of nonneRgative Borel measures {µj} that
converges weakly to a non-negative Borel measure µ in Ω and satisfies
0
(ddcf)n ≤ µ ≤ (ddcw)n for all j ≥ 0,
j
there exist unique u ∈ Na(Ω,f) such that u ≥ w, (ddcu )n = µ and u → u
j j j j j 0
in C -capacity in Ω.
n
Proof. By Lemma 4.1 there exist unique u ∈ Na(Ω,f) such that u ≥ w and
j j
(ddcu )n = µ in Ω. Since u ≥ w, the sequence {u } is compact in L1 (Ω). Let u
j j j j loc
be a cluster point and let {u } be a subsequence of the sequence {u } such that
jk j
∗
u → u a.e. in Ω. Put v := sup u . We claim that
jk k l≥k jl
(cid:0)v (cid:1) u
lim max k,ρ −max jk,ρ (ddcu )n = 0, (4.1)
k→+∞ZΩh (cid:16) a (cid:17) (cid:16) a (cid:17)i jk
for every a > 0. Indeed, let ε > 0. Choose χ ∈ C∞(Ω) such that 0 ≤ χ ≤
0
1 and {χ = 1} ⊂ {ρ < −ε}. By Proposition 2.3 we have that the measure
1 χ(ddcw)n vanishes on all pluripolar subsets of Ω. By Lemma 3.1 in [8]
{f>−∞}
we get
v u
0≤ limsup max k,ρ −max jk,ρ χ(ddcu )n
k→+∞ ZΩh (cid:16) a (cid:17) (cid:16) a (cid:17)i jk
v u
= limsup max k,ρ −max jk,ρ 1 χ(ddcu )n
k→+∞ ZΩh (cid:16) a (cid:17) (cid:16) a (cid:17)i {f>−∞} jk
v u
≤ limsup max k,ρ −max jk,ρ 1 χ(ddcw)n
k→+∞ ZΩh (cid:16) a (cid:17) (cid:16) a (cid:17)i {f>−∞}
u u
= max ,ρ −max ,ρ 1 χ(ddcw)n
ZΩh (cid:16)a (cid:17) (cid:16)a (cid:17)i {f>−∞}
= 0.
It follows that
v u
0≤ limsup max k,ρ −max jk,ρ (ddcu )n
k→+∞ ZΩh (cid:16) a (cid:17) (cid:16) a (cid:17)i jk
v u
≤ limsup max k,ρ −max jk,ρ (1−χ)(ddcu )n
k→+∞ ZΩh (cid:16) a (cid:17) (cid:16) a (cid:17)i jk
v u
≤ limsup −max k,ρ −max jk,ρ (ddcu )n
k→+∞ Z{ρ≥−ε}h (cid:16) a (cid:17) (cid:16) a (cid:17)i jk
≤ 2 −max(ρ,−ε)(ddcw)n.
Z
Ω
Let ε ց 0 we obtain (4.1). This proves the claim, and therefore, by the main
theorem we get u → u in C -capacity in Ω as k → +∞. Hence, by [8] we have
jk n
(ddcu)n = µ in Ω. It is clear that u ∈ Na(Ω,f). From the uniqueness of u we
0 0
get u= u . Thus, u → u a.e. in Ω. It follows that u → u a.e. in Ω. Similarly,
0 jk 0 j 0
we get
v u
lim max j,ρ −max j,ρ (ddcu )n = 0,
j
j→+∞ZΩh (cid:16) a (cid:17) (cid:16) a (cid:17)i