for the complex Monge–Amp` ere operator

POLONICI MATHEMATICI LXV.1 (1996)

Some sufficient conditions for solvability of the Dirichlet problem

for the complex Monge–Amp` ere operator

by S lawomir Ko lodziej (Krak´ ow)

Abstract. We find a bounded solution of the non-homogeneous Monge–Amp`ere equa- tion under very weak assumptions on its right hand side.

Introduction. In this paper we are interested in solving, under possibly weak assumptions on the measure dµ, the following Dirichlet problem for the complex Monge–Amp`ere equation in a given strictly pseudoconvex domain Ω ⊂ C

:

(∗)

u ∈ PSH ∩ L

(Ω), (dd

u)

= dµ,

z

lim

u(z

) = φ(z), z ∈ ∂Ω, φ ∈ C(∂Ω),

where d = ∂ + ∂, d

= i(∂ − ∂) and so dd

= 2πi∂∂. It has been shown by E. Bedford and B. A. Taylor [BT1] that the wedge product (dd

u)

= dd

u ∧ . . . ∧ dd

u is well defined for plurisubharmonic (psh), locally bounded functions u, and that (∗) is solvable for measures having continuous densities with respect to the Lebesgue measure (here denoted by dλ). The equation has attracted attention of a number of authors; we refer to [B] for a more detailed account. In particular, it is known that continuous solutions exist if dµ = f dλ, where f ∈ L

(Ω, dλ) (U. Cegrell–L. Persson [CP]), but for f ∈ L

(Ω, dλ) this is not necessarily true [CS]. In Theorem 3 below we show that if f ∈ L

(Ω, dλ), p > 1, then there exists a continuous solution of (∗). This is the answer to the question posed in [CS] and [P] (see also [B], [BL]). For the case of rotation invariant measures in a ball a solution was given in [P].

The result can be extended from L

, p > 1, to some Orlicz spaces as shown

1991 Mathematics Subject Classification: Primary 32F07; Secondary 32F05.

Key words and phrases: plurisubharmonic function, complex Monge–Amp`ere operator.

Research partially supported by KBN Grant 2 P03A 058 09.

[11]

in Theorem 4. To prove it we use an a priori estimate for the kuk

norm of a solution of (∗) if dµ satisfies a certain integral condition (Theorem 1).

E. Bedford [B] conjectured that some such estimate is possible. It is shown that the integral condition cannot be substantially weakened. Combining Theorem 1 with the results of [KO] we solve the Dirichlet problem (∗) for a large family of measures dµ.

I am indebted to my colleagues Z. B locki and G. Lewicki for valuable suggestions and U. Cegrell (Ume˚ a University) for pointing out an error in a previous version of the paper. The referee’s remarks simplified the proof of Theorem 1.

Preliminaries. Here we present some notions and results which are used in the paper. The background material can be found in [B], [K], [S]. Ω will denote throughout a strictly pseudoconvex domain in C

. For a compact subset K ⊂ Ω we define the relative extremal function and the relative capacity [BT2] (see also [B], [K]) by the formulas

u

(z) = sup{u(z) : u ∈ PSH ∩ L

, u < 0 in Ω, u ≤ −1 on K}, cap(K, Ω) = sup n

K

(dd

u)

: u ∈ PSH(Ω), −1 ≤ u < 0 o . By [BT2],

cap(K, Ω) =

\

K

(dd

u

)

=

\

Ω

(dd

u

)

,

where u

:= lim

u

(z). If u

= u

we say that K is regular. For an open subset U ⊂ Ω the relative capacity is defined by

cap(U, Ω) = sup{cap(K, Ω) : K ⊂ U, K compact}.

Another extremal function (of logarithmic growth) and an associated capac- ity were introduced by J. Siciak (see [S], [AT], [B], [K]):

L

(z) = sup{u(z) : u ∈ PSH(C

),

u(z) < log(1 + |z|) + O(1), u ≤ 0 on K}, T

(K) := exp(− sup{L

(z) : |z| ≤ R})

for a compact set K ⊂ C

and a given R > 0. We extend the definition of T

to open sets in the same way as the definition of cap above.

Important inequalities between cap and T were proved by H. Alexander and B. A. Taylor [AT]. If B := B(0, R) and K ⊂ B(0, r), r < R, is compact, then

exp(−A(r)(cap(K, B))

) ≤ T

(K) ≤ exp(−2π(cap(K, B))

).

The main tool in pluripotential theory is the following Comparison Prin-

ciple of Bedford and Taylor [BT2]:

Comparison Principle. If u, v ∈ PSH∩ L

(Ω) and lim inf

(u(z)

− v(z)) ≥ 0, then

\

{u<v}

(dd

v)

≤

\

{u<v}

(dd

u)

.

Due to the same authors and presented here in a simplified version, sufficient for our applications, is

Convergence Theorem [BT2]. If u

∈ PSH ∩ L

(Ω), j = 1, 2, . . . , and u

↑u a.e. in Ω or u

↓u with u ∈ PSH ∩ L

(Ω) then

(dd

u

)

→ (dd

u)

in the sense of currents.

An a priori estimate. We begin with proving an a priori estimate for the L

norm of a solution to the Dirichlet problem (∗) when dµ is assumed to satisfy a certain integral condition.

Theorem 1. Let Ω be a strictly pseudoconvex domain in C

and let µ be a Borel measure in Ω such that

T

Ω

dµ ≤ 1. Consider an increasing function h : R → (1, ∞) satisfying

∞\

1

(yh

(y))

dy < ∞.

If µ satisfies the integral condition (∗∗)

\

Ω

|v|

h(|v|) dµ ≤ A whenever

v ∈ PSH(Ω) ∩ C(Ω), v = 0 on ∂Ω,

\

Ω

(dd

v)

≤ 1,

then the norm kuk

of a solution of the Dirichlet problem (∗) is bounded by a constant B = B(h, A) which does not depend on µ.

P r o o f. It is no restriction to assume that φ = 0 in (∗): the general case will follow by the Comparison Principle [BT2]. Let u be a solution of (∗).

For s < 0 denote by U

the open set {u < s} and put

a(s) := cap(U

, Ω) = cap(U

), b(s) := µ(U

).

Our proof rests on the following two propositions.

Proposition 1. b(s) ≤ Aa(s)h

([a(s)]

).

Proposition 2. t

a(s) ≤ b(s + t) if t > 0 and s + t < 0.

P r o o f o f P r o p o s i t i o n 1. Consider v = (ra(s))

u

, where K ⊂ U

is a compact regular set with cap(K) = ra(s) (r < 1). Then

T

(dd

v)

= 1 and so the integral condition (∗∗) applies, giving A ≥

\

Ω

|v|

h(|v|) dµ ≥

\

K

|v|

h(|v|) dµ = (ra(s))

h([ra(s)]

)µ(K), which is just the desired estimate as r → 1 (and so µ(K) → b(s)).

P r o o f o f P r o p o s i t i o n 2. We apply the Comparison Principle [BT2] to the pair of functions u

and v := (rt)

(u − s − t), where K, r are defined as above. Note that K ⊂ {v < u

} ⊂ U

. Hence

ra(s) =

\

{v<uK}

(dd

u

)

≤ (rt)

\

{v<uK}

(dd

u)

≤ (rt)

µ(U

) = (rt)

b(s + t).

The proposition follows if we let r → 1.

E n d o f t h e p r o o f o f T h e o r e m 1. Fix s

so that a = a(s

) 6= 0.

We need to find a lower bound for s

. To this end we first define an increasing sequence s

, s

, . . . , s

by

s

:= sup{s : a(s) ≤ lim

t→sj−1+

ea(t)}.

Then

t→s

lim

a(t) ≤ lim

t→sj−1+

ea(t) and a(s

) ≥ ea(s

).

We continue this process till

(1) 1 ≤ a(s

).

For fixed s and s

such that a(s) ≤ ea(s

) and t := s − s

we have by the above two propositions

a(s

) ≤ t

b(s) ≤ At

a(s)h

([a(s)]

)

= Aet

a(s

)h

([a(s)]

).

Hence

t ≤ (Ae)

h

(a(s))

where h

(x) := h

(x

). Letting s → s

− and s

→ s

+ we thus get t

:= s

− s

≤ (Ae)

h

(a(s

)).

Since the function h

(x) := h

(e

) = h

(e

) is increasing we can

further estimate

N −1

X

j=0

t

≤ (Ae)

N −1

X

j=0

h

(log a(s

)) (2)

≤ (Ae)



X

j=0

log a(sj+2)

\

log a(sj)

h

(x) dx + 2h

(log a(s

)) 

≤ 2(Ae)



−∞

h

(x) dx + h

(∞)  . By our hypothesis on h, we have h

(∞) ≤ 1 and

0

\

−∞

h

(x) dx =

0

\

−∞

h

(e

) dx

= n

∞

\

1

h

(y)y

dy =: nc(h) < ∞.

These remarks combined with (2) give s

− s

=

N −1

X

j=0

t

≤ 2(Ae)

(nc(h) + 1) =: c.

This means that for s

≥ s

+ c we have a(s

) > 1 (see (1)). So fixing s

= s

+ c + 1 we conclude that s

≥ 0 because otherwise, by applying Proposition 2, we would get a contradiction with the assumptions:

µ(U

) > 1.

Thus s

≥ −c − 1 =: B. The proof is complete.

R e m a r k. The hypothesis that µ satisfies (∗∗) can be replaced by µ(K) ≤ A cap(K)h

((cap(K))

)

for any K ⊂ Ω compact and regular. The above proof still works.

It turns out that the integral condition (∗∗) is not far from being sharp.

From [BL, Corollary 2.2] (see also [D, Th. 2.2]) it follows that any bounded solution of (∗) satisfies (∗∗) with h ≡ 1 and A = n!kuk

T

Ω

dµ. However, if we let h ≡ 1 then (∗∗) ceases to be a sufficient condition for boundedness of u (when n > 1). This can be seen by considering radial psh functions in a ball B = B(0, R). In that case we have a characterization of bounded solutions of (∗) given in [P] (see also [M]). A radial psh function u is bounded if and only if

(3)

R

\

0

r

F

(r) dr < ∞, where F (r) =

T

B(0,r)

(dd

u)

.

It is easy to see that for the rotation invariant measure dµ = (dd

u)

the integral in (∗∗) assumes its maximal value for v(z) = (2π)

log |z|. Suppose that

(4) (2π)

\

B

|v|

dµ =

R\

0

|log r|

F

(r) dr < ∞.

Via integration by parts this is equivalent to

R

\

0

|log r|

r

F (r) dr < ∞.

Write F (r) = |log r|

g

(r). Then (4) takes the form

R\

0

[|log r|rg(r)]

dr < ∞, whereas (3) now says

R

\

0

[|log r|rg

(r)]

dr < ∞.

Taking g such that the former inequality is satisfied but the latter is not, e.g. g(r) = (log |log(r)|)

, we arrive at the desired conclusion.

Coupling Theorem 1 above with Theorem 1 from [KO] we obtain a fairly general class of measures for which the Dirichlet problem (∗) is solvable. For the definition of a measure locally dominated by capacity which we need in the statement of the next theorem we refer to [KO]. Essentially we require from such a measure (say µ) that there exists c > 0 such that given two concentric balls B

:= B(a, r) ⊂ B

:= B(a, 2r) ⊂ Ω and a compact subset E ⊂ B

, the following estimate holds:

µ(E) ≤ cµ(B

) cap(E, B

).

(The actual definition is a bit less restrictive.)

Theorem 2. If a measure µ in Ω is locally dominated by capacity and satisfies the condition (∗∗) from Theorem 1 with h such that

h(ax) ≤ bh(x), x > 0,

for some a > 1 and b > 1, then there exists a solution of (∗).

P r o o f. For a while we assume that µ has compact support in Ω. Define a regularizing sequence of measures µ

by fixing a radial non-negative function ω ∈ C

(B) with

T

ω dλ = 1 (here B is the unit ball in C

) and setting

µ

= ω

∗ µ, where ω

(z) = t

ω(z/t), t > 0.

By Theorem 1 and Remark following it, it is enough to find t

> 0 and A > 0 such that for any compact set K ⊂ Ω,

(ι) µ

(K) ≤ A cap(K, Ω)h

((cap(K, Ω))

), t < t

.

Proposition 3. If E ⋐ Ω is regular then for any d > 1 there exists t

such that

cap(K

, Ω) ≤ d cap(K, Ω), |y| < t

, where K ⊂ E is regular and K

:= {x : x − y ∈ K}.

P r o o f. For K ⊂ E define w

:= u

(x + y), where u

is the extremal function of K

. For any c such that 0 < c < 1/2 define Ω

= {u

< −c}.

By continuity of u

one can fix t

> 0 such that if |y| ≤ t

and x ∈ Ω

then x + y ∈ Ω. Therefore

g(x) :=  max(w

− c, (1 + 2c)u

)(x), x ∈ Ω

, (1 + 2c)u

(x), x 6∈ Ω

,

is a well defined plurisubharmonic function in Ω. Since K ⊂ E and w

= −1 on K one concludes that g = w

− c in a neighbourhood of K. Hence

cap(K, Ω) ≥ (1 + 2c)

\

K

(dd

g)

= (1 + 2c)

\

K

(dd

w

)

= (1 + 2c)

\

Ky

(dd

u

)

= (1 + 2c)

cap(K

, Ω).

Thus the proposition is proved.

To complete the proof of Theorem 2 let us fix a set E and a positive number t

such that the above proposition holds with E := S

t<t0

supp µ

⋐ Ω and d = a

. By the assumptions there exists A

> 0 such that µ(K) ≤ A

cap(K)h

((cap(K))

).

Hence for t < t

we have by Proposition 3 and the extra assumption on h, µ

(K) ≤ sup

|y|<t

µ(K

) ≤ A

sup

|y|<t

cap(K

)h

((cap(K

))

)

≤ A

d cap(K)h

((d cap(K))

)

≤ A

db

cap(K)h

((cap(K))

).

Setting A := A

a

b

we verify this way that µ

satisfies (ι) for t < t

, with the constant A independent of t. Thus by Theorem 1 the family of solutions of (∗) for µ

, t < t

, is uniformly bounded. So one can apply [KO, Th. 1] to get the conclusion.

To verify the statement for an arbitrary measure µ note that by the

above argument the solutions exist for χ

dµ, where χ

is a non-decreasing

sequence of smooth cut-off functions with χ

↑1 in Ω. Moreover, the L

norms of those solutions are uniformly bounded by a constant depending only on A. Hence the result follows by applying the monotone convergence theorem of [BT2].

Solutions for measures having densities in L

, p > 1. In Theorem 3 we are going to prove that for dµ = f dλ, f ∈ L

(Ω), p > 1, the Dirichlet problem (∗) has a continuous solution. To this end we shall use the following

Lemma 1. Suppose v ∈ PSH(Ω) ∩ C(Ω), v=0 on ∂Ω and

T

(dd

v)

=1.

Then the Lebesgue measure λ(U

) of the set {v < s} is bounded from above by c exp(−2π|s|), where c does not depend on v.

P r o o f. The proof is a variation of the proof of Proposition 2 of [KO].

First we shall estimate cap(U

) = cap(U

, Ω) applying the Comparison Prin- ciple [BT2]. For t > 1 and a regular compact set K ⊂ U

we have

cap(K) =

\

K

(dd

u

)

=

\

{−ts⁻¹v<uK}

(dd

u

)

≤ t

s

\

Ω

(dd

v)

≤ t

s

. Hence

(5) cap(U

) ≤ |s|

.

Write (z

, z

) ∈ C × C

and set U

(z

) := {z

∈ C : (z

, z

) ∈ U

}. Let V

(resp. V ) be the extremal function of logarithmic growth of U

(z

) (resp.

U

). Then (see [TS])

λ(U

(z

)) ≤ C

T

(U

(z

)),

where λ denotes the Lebesgue measure in C, C

is an independent constant and

T

(U

(z

)) := exp(− sup

|z1|<R

V

), with R chosen so that Ω ⊂ B(0, R). Thus

λ(U

) =

\

λ(U

(z

)) dλ(z

) ≤ C

\

T

(U

(z

)) dλ(z

) (6)

= C

\

exp(− sup

|z1|<R

V (z

, z

)) dλ(z

).

A simple argument using a result of Alexander [A] shows that the right hand side of (6) is dominated by

C

exp(− sup

|z|<R

V (z)) = C

T

(U

)

(see [KO] for details). Finally, we apply an inequality between the capacities cap and T proved in [AT] to obtain

λ(U

) ≤ C

exp[−2π(cap(U

, B(0, R)))

] ≤ C

exp[−2π(cap(U

, Ω))

].

Hence by (5) we get

λ(U

) ≤ C

exp(−2π|s|), which was to be proved.

Corollary. If v ∈ PSH(Ω) ∩ C(Ω), v = 0 on ∂Ω and

T

Ω

(dd

v)

≤ 1, then kvk

≤ c(p).

P r o o f. By the lemma,

\

|v|

dλ ≤

\

Ω

dλ +

∞

X

s=1

\

{−s−1<v<−s}

|v|

dλ ≤ c

∞

X

s=1

(s + 1)

e

=: c(p) < ∞.

Now we are in a position to prove

Theorem 3. If f ∈ L

(Ω, dλ), p > 1, f ≥ 0 then the Dirichlet problem (∗) has a continuous solution for dµ = f dλ.

P r o o f. Set f

:= min(f, j). Let u

be the continuous solution of (dd

u)

= f

dλ,

z

lim

u(z

) = φ(z), z ∈ ∂Ω

(see [C], [CP]). Then by the convergence theorem of [BT2], u = lim u

is the desired solution provided u

is uniformly bounded. This is the case if the integral condition (∗∗) in Theorem 1 is satisfied for dµ = f dλ and some suitable h. Let us verify this condition for h(x) = max(1, x). By H¨older’s inequality we have

\

|v|

h(|v|)f dλ =

\

{v≥−1}

+

\

{v<−1}

≤ kf k

+ 

|v|

dλ 

kf k

, where p

+ q

= 1. Since by the Corollary above,

\

|v|

dλ ≤ c(q(n + 1)),

one can apply Theorem 1 to conclude that u = lim u

is bounded.

Now, if u

solves (dd

u)

= |f

− f

| dλ, u = 0 on ∂Ω, then by the Comparison Principle and the above argument,

ku

− u

k ≤ −u

≤ c

kf

− f

k

. So u

is uniformly convergent and u is continuous.

The last result readily extends to cover densities belonging to some Orlicz spaces. As an example (which can be refined yet) we give the following

Theorem 4. Let L

(Ω, dλ) denote the Orlicz space corresponding to

ϕ(t) = |t|(log(1 + |t|))

h(log(1 + |t|)) with h satisfying the hypothesis of

Theorem 1. If f ∈ L

(Ω, dλ) then (∗) is solvable with dµ = f dλ.

P r o o f. As in the preceding proof, it is enough to verify the condi- tion (∗∗). We apply Young’s inequality for the function g(log(1 + r)) = (log(1 + r))

h(log(1 + r)) and its inverse. Then

g(|v(x)|)f (x) ≤

f (x)

\

0

g(log(1 + r)) dr +

g(|v(x)|)

\

0

[exp(g

(t)) − 1] dt

≤ f (x)g(log(1 + f (x))) +

|v(x)|

\

0

e

g

(s) ds

≤ kf k

+ g(|v(x)|)e

.

When integrated over Ω, the right hand side remains bounded since by the lemma,

\

Ω

g(|v(x)|)e

dx ≤ c

∞

X

s=1

e

g(s + 1) < ∞.

Thus the result follows from Theorem 1.

Example. If ϕ(t) = |t|(log(1 + |t|))

(log(log(1 + |t|)))

, m > n, then Theorem 4 applies. On the other hand, if ϕ(t) = |t|(log(1 + |t|))

, m < n, it is no longer true; a suitable counterexample is given in [P].

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Institute of Mathematics Jagiellonian University Reymonta 4

30-059 Krak´ ow, Poland E-mail: kolodzie@im.uj.edu.pl

Re¸ cu par la R´ edaction le 10.9.1994

R´ evis´ e le 20.1.1995