POLONICI MATHEMATICI LXVII.1 (1997)
An energy estimate for the complex Monge–Amp` ere operator
by Urban Cegrell and Leif Persson (Ume˚ a)
Abstract. We prove an energy estimate for the complex Monge–Amp`ere operator, and a comparison theorem for the corresponding capacity and energy. The results are pluricomplex counterparts to results in classical potential theory.
Introduction. Recall that in classical potential theory, a positive mea- sure µ is said to have finite energy if
\
−G
Ω(x, y) dµ(x) dµ(y) < ∞,
where G
Ωis the Green function for the domain Ω. It is shown that
\
−G
Ω(x, y) dµ(x) dν(y)
defines an inner product on the linear space of measures spanned by the measures of finite energy. In particular, we have the Cauchy–Schwarz in- equality
\−G
Ωdµ dν
≤
\−G
Ωdµ dµ
1/2\−G
Ωdν dν
1/2.
In this paper, we prove the following analogue of this inequality for the complex Monge–Amp`ere operator:
Theorem 1.1. Let Ω be a domain in C
n, n ≥ 2. Suppose u, v ∈ PSH ∩ L
∞(Ω) with lim
z→ξu(z) = lim
z→ξv(z) = 0, ∀ξ ∈ ∂Ω. If p ≥ 1, 0 ≤ j ≤ n, then
\
(−u)
p(dd
cu)
j∧ (dd
cv)
n−j≤ D
p,j\(−u)
p(dd
cu)
n(p+j)/(n+p)\(−v)
p(dd
cv)
n(n−j)/(n+p)1991 Mathematics Subject Classification: Primary 32F07; Secondary 31C10.
Key words and phrases : capacity, complex Monge–Amp`ere operator, energy estimate, plurisubharmonic function.
[95]
where D
p,j= p
(p+j)(n−j)/(p−1)for p > 1 and D
p,j= exp(1 + j)(n − j) for p = 1.
For the classical notation of energy and Green potentials we refer to Landkof [6], and for the pluripotential theory to the survey article by Bed- ford [1].
2. Proof of the theorem. In order to be able to integrate by parts, we first assume that
(2.1)
\
Ω
((dd
cu)
n+ (dd
cv)
n) < ∞.
Then for the mixed terms we have
\
Ω
(dd
cu)
j∧ (dd
cv)
n−j≤
\
Ω
(dd
c(u + v))
n< ∞, 0 ≤ j ≤ n,
where the last inequality is obtained from the comparison principle and the assumption above (cf. [5]). For let µ = (dd
c(u + v))
nand choose 1 < α < 2 such that µ{u = αv} = 0. Then µΩ = µ{(1+α)u/α < u+v}+µ{(1+α)v <
u+v}, and thus µΩ ≤ 3
nT
Ω
((dd
cu)
n+(dd
cv)
n) by the comparison principle, which proves the boundedness of the mixed terms.
Since d
cu ∧ (dd
cu)
j−1∧ (dd
cv)
n−jis a positive measure on {u = −ε}
(cf. [4]), we have 0 ≤
\
{u=−ε}
(−v)
pd
cu ∧ (dd
cu)
j−1∧ (dd
cv)
n−j≤ sup{(−v(z))
p| u(z) = −ε} ·
\
Ω
(dd
cu)
j∧ (dd
cv)
n−j→ 0, ε ց 0.
Therefore, we can integrate by parts in this case. Define x
j= log
\
(−u)
p(dd
cu)
j∧ (dd
cv)
n−j, y
j= log
\
(−v)
p(dd
cv)
j∧ (dd
cu)
n−j. Then integration by parts and H¨older’s inequality give
\
(−u)
p(dd
cu)
j∧(dd
cv)
n−j= −
\
dv ∧ d
c(−u)
p∧ (dd
cu)
j∧ (dd
cv)
n−j−1=
\
vdd
c(−u)
p∧ (dd
cu)
j∧ (dd
cv)
n−j−1= p(p − 1)
\
v(−u)
p−2du ∧ d
cu ∧ (dd
cu)
j∧ (dd
cv)
n−j−1+ p
\
(−v)(−u)
p−1(dd
cu)
j+1∧ (dd
cv)
n−j−1≤ p
\
(−v)(−u)
p−1(dd
cu)
j+1∧ (dd
cv)
n−j−1≤ p
\
(−v)
p(dd
cu)
j+1∧ (dd
cv)
n−j−11/p× p
\
(−u)
p(dd
cu)
j+1∧ (dd
cv)
n−j−1(p−1)/p. Taking logarithms, we get
x
j≤ p − 1
p x
j+1+ 1
p y
n−j−1+ log p and
y
j≤ p − 1
p y
j+1+ 1
p x
n−j−1+ log p.
In matrix notation,
(2.2) S
x
0y
0.. . x
ny
n
≤ log p
1 .. . 1
where S is the 2n × (2n + 2) matrix
S =
1 0
1−pp0 0 · · · 0 0 −
1p0 0
0 1 0
1−pp0 · · · 0 −
1p0 0 0
0 0 1 0
1−pp· · · −
1p0 0 0 0
.. . .. . .. . .. . .. . . .. .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . . .. .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . . .. .. . .. . .. . .. . .. .
0 0 −
p10 0 · · · 1 0
1−pp0 0
0 −
p10 0 0 · · · 0 1 0
1−pp0
−
1p0 0 0 0 · · · 0 0 1 0
1−pp
.
Let A denote the left 2n×2n submatrix of S. We will find that A is invertible and that A
−1has nonnegative elements. So multiplication of the system (2.2) with A
−1will preserve the inequality and give a reduced row-echelon form.
To this end consider the system of equations
A
x
0y
0.. . x
n−1y
n−1
=
c
0d
0.. . c
n−1d
n−1
.
A calculation shows that then x
j= n − j
(p − 1)(p + n)
j−1
X
k=0
(k + 1)c
k(2.3)
+ p + j
(p − 1)(p + n)
n−1
X
k=j
(p − 1 + n − k)c
k+ n − j
(p − 1)(p + n)
n−1
X
k=n−j
(p − 1 + n − k)d
k+ p + j
(p − 1)(p + n)
n−j−1
X
k=0
(k + 1)d
k,
and similarly for y
j. This shows that A
−1exists and has nonnegative ele- ments. It follows from (2.3) that
(2.4) A
−1S =
I 0 0 · · · 0 0 A
00 I 0 · · · 0 0 A
1.. . .. . .. . . .. ... ... .. . 0 0 0 · · · I 0 A
n−20 0 0 · · · 0 I A
n−1
,
where I is the 2 × 2 identity matrix and A
j= −
p+j p+n
n−j p+n n−j p+n
p+j p+n
! . Then (2.2) implies that
(2.5) A
−1S
x
0y
0.. . x
ny
n
≤ log p A
−1
1 .. . 1
.
To compute the right hand side of (2.5), we have to find
(2.6) A
−1
1 .. . 1
=
x
′0y
′0.. . x
′n−1y
′n−1
.
Thus we put c
k= d
k= 1 in (2.3) and get
(2.7) x
′j= y
′j= (p + j)(n − j) p − 1 . We substitute (2.7) and (2.6) in (2.5) and obtain
(2.8)
x
j− p + j
p + n x
n− n − j
p + n y
n≤ (p + j)(n − j) p − 1 log p, y
j− n − j
p + n x
n− p + j
p + n y
n≤ (p + j)(n − j) p − 1 log p.
This concludes the proof for the case p > 1 and the extra assumption (2.1).
Since the integrals are continuous in p, and since
p→1
lim log p p − 1 = 1,
the inequality also holds for p = 1. To complete the proof of the theorem, we have to remove the assumption (2.1). We can assume that
\
((−u)
p(dd
cu)
n+ (−v)
p(dd
cv)
n) < ∞,
otherwise there is nothing to prove. Let ε > 0 be given an let u
rdenote the usual regularization
u
r(z) =
\
u(z − rξ)φ(ξ) dV (ξ),
where V is the Lebesgue measure on C
n, and φ is a fixed radial, nonn- egative, smooth and compactly supported function in the unit ball of C
nwith
T
φ dV = 1. Let ω ⋐ Ω be a strictly pseudoconvex domain containing {u < −ε/4}. Then u
r∈ PSH(ω) ∩ C
∞(ω) if r < d(ω,
cΩ), and we define
u
ωr,ε=
u
rif u
r< −ε, εh
ω{ur<−ε}
if u
r≥ −ε, where h
ωEis the relative extremal function
(2.9) h
ωE(z) = sup{φ(z) | φ ∈ PSH(ω), φ ≤ 0, φ|
E≤ −1}
with respect to ω. By Sard’s theorem, the boundary of {u
r< −ε} is a smooth manifold for all ε outside a set of Lebesgue measure zero. We consider only those ε’s. Then lim
{ur≤−ε}∋ξ→zh
ω{ur<−ε}
(ξ) = −1 for all z ∈ {u
r< −ε}, so u
ωr,εis plurisubharmonic on ω. Now,
\
ω
(−u
ωr,ε)
p(dd
cu
ωr,ε)
n=
\
{ur<−ε}
. . . +
\
{ur≥−ε}
. . .
≤
\
K
(−u
r)
p(dd
cu
r)
n+ ε
p\
{ur=−ε}
(dd
cu
ωr,ε)
nfor all compact sets K in ω containing {u < −ε}. Furthermore,
\
ω
(dd
cu
ωr,ε)
n=
\
ω
(dd
cεh
ω{ur<−ε})
n=
\
{ur=−ε}
(dd
cεh
ω{ur<−ε})
n≤
\
{u<(ε/4)hω{ur <−ε}−ε/4}
(dd
cεh
ω{ur<−ε})
n= 4
n\
{u<(ε/4)hω{ur <−ε}−ε/4}
dd
cε
4 h
ω{ur<−ε}− ε 4
n≤ 4
n\
{u<−ε/4}
(dd
cu)
nby the comparison principle. Combining these two inequalities, we get
\
ω
(−u
ωr,ε)
p(dd
cu
ωr,ε)
n≤
\
K
(−u
r)
p(dd
cu
r)
n+ ε
p\
{u<−ε/4}
(dd
cu)
n. We now let r ց 0; then u
ωr,εdecreases to
u
ωε=
u if u < −ε, εh
ω{u<−ε}if u ≥ −ε, and
\
ω
(−u
ωε)
p(dd
cu
ωε)
n≤
\
K
(−u)
p(dd
cu)
n+ ε
p\
{u<ε/4}
(dd
cu)
nso if we let ω and K increase to Ω, then u
ωεdecreases to u
Ωεand
\
(−u
Ωε)
p(dd
cu
Ωε)
n≤
\
(−u)
p(dd
cu)
n+ ε
p\
{u<ε/4}
(dd
cu)
n. If we now let ε ց 0 then
ε→0
lim
\
(−u
Ωε)
p(dd
cu
Ωε)
n≤
\
(−u)
p(dd
cu)
nand similarly for v. Also, by semicontinuity we have
lim inf
ε→0
\
(−u
Ωε)
p(dd
cu
Ωε)
j∧ (dd
cu
Ωε)
n−j≥
\
(−u)
p(dd
cu)
j∧ (dd
cv)
n−j. We have already proved the inequalities for u
Ωεand v
Ωεso the above inequal- ities complete the proof of the theorem.
R e m a r k. The theorem can be generalized to more than two functions.
Also, it can be proved that D
1,j= 1 (see [7]).
3. An application. Let Ω be a strictly pseudoconvex set in C
n, n ≥ 2, and denote by P the class of bounded plurisubharmonic functions φ on Ω such that lim
z→ξφ(z) = 0, ∀ξ ∈ ∂Ω and
T
Ω
(dd
cφ)
n< ∞. In analogy with
the notation of capacity and energy in classical potential theory, we consider the pluricomplex capacity, defined by Bedford and Taylor in [2],
d(F ) = sup n
\F
(dd
cu)
nu ∈ P, −1 ≤ u ≤ 0 o , and the pluricomplex energy,
I(F ) = inf n
\−u(dd
cu)
nu ∈ P,
\
F
(dd
cu)
n≥ 1 o , of a compact subset F of Ω. If
T
F
(dd
cu)
n= 0, ∀u ∈ P , we say that F has infinite energy ; this happens exactly when F is pluripolar.
Theorem 3.1. Suppose that F is not pluripolar. Then (3.1) D
−(n+1)/n1,0≤ d(F )
1/nI(F ) ≤ 1.
P r o o f. Let ψ = h
∗F/d(F )
1/n∈ P , where h
∗Fdenotes the smallest upper semicontinuous majorant of the relative extremal function h
F= h
ΩFdefined by (2.9). Then supp(dd
cψ)
n⊂ F and
T
F
(dd
cψ)
n= 1 by [2]. Therefore, I(F ) ≤ 1
d(E)
\
− h
∗Fd(F )
1/n(dd
ch
∗F)
n= 1 d(F )
1/nsince h
∗F= −1 on F outside a pluripolar set. This proves the last inequality in (3.1).
To prove the first inequality we use Theorem 1.1. If u ∈ P with
T
F
(dd
cu)
n≥ 1, then 1 ≤
\
−h
F(dd
cu)
n≤ D
1,0\−h
F(dd
ch
F)
n1/(n+1)\−u(dd
cu)
nn/(n+1)= D
1,0d(F )
1/(n+1)\−u(dd
cu)
nn/(n+1)so
D
−(n+1)/n1,0≤ d(F )
1/n\
