On m\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{m}$$\end{document}-Subharmonic Ordering of Measures

We study an order-relation induced by m-subharmonic functions. We shall consider maximality with respect to this order and a related notion of minimality for certain m-subharmonic functions. This concept is then applied to the problem of convergence of measures in the weak*-topology, in particular Hessian measures.


Introduction
In this paper we study an order-relation between measures on an m-hyperconvex domain Ω in C n . Let μ and ν be measures on Ω. We say that μ is msubharmonically greater than ν if Ω (−ϕ)dμ ≥ Ω (−ϕ)dν, ∀ ϕ ∈ E 0,m (Ω) ∩ C(Ω) and write μ ν, where E 0,m (Ω) is the Cegrell class of negative msubharmonic functions defined in Sect. 2. It is easy to see that the condition μ ≥ ν implies μ ν. But the inverse is not true (see Example 1). We also show that if u, v are functions in the Cegrell class F m (Ω) such that u ≤ v, then their complex Hessian measures are in the relation H m (u) H m (v) (see Proposition 2). But the inverse is not true (see Example 2).
In Sect. 4, we study maximality with respect to the -ordering, and a related notion of minimality for m-subharmonic functions in the class F m (Ω). A finite measure μ on Ω is said to be maximal if for any measure ν on Ω such that ν(Ω) = μ(Ω), the relation ν μ implies that ν = μ. The Dirac measure is a maximal measure. Theorem 9 shows that each finite measure on Ω with compact support is majorized in the -ordering by a maximal measure with the same total mass. A function u ∈ F m (Ω) is said to be minimal if for any Theorem 3 [9,Theorem 1.6.5]. If E is m-polar set, then there exists u ∈ SH − m (C n ) such that E ⊂ {u = −∞}.
Throughout this paper Ω will denote a bounded m-hyperconvex domain in C n . Now we recall the definitions of the Cegrell classes. (2) A function u ∈ SH m (Ω) belongs to E m (Ω) if for each z 0 ∈ Ω, there exists an open neighborhood U ⊂ Ω of z 0 and a decreasing sequence We have the following inclusions Below we present some of the basic properties of the Cegrell classes.
The following lemma explains why the functions in E 0,m (Ω) are sometimes called test functions.
Theorem 5 [2,9].  1 1 ∧ · · · ∧ dd c u m j ∧ β n−m converge to a Radon measure in weak*-topology independent to the choice of sequences {u k j }. We define dd c u 1 ∧ · · · ∧ dd c u m ∧ β n−m to be this limit. Integration by parts formula is true for the function from the Cegrell class F m (Ω).
where T = dd c w 1 ∧ · · · ∧ dd c w m−1 ∧ β n−m and the equality means that if one of the two terms is finite then they are equal.
The following theorem is sometimes called the Cegrell decomposition theorem. Proof. By the proof of [10,Theorem 4.14], we can find a function u ∈ E 1,m (Ω) and 0 ≤ f ∈ L 1 (H m (u)) such that μ = fH m (u) + ν, where ν is charged by an m-polar subset of Ω. The rest of the proof goes verbatim as the proof of [10, Theorem 5.3].

The m-Subharmonic Ordering
Let μ j , μ be measures on Ω. By Theorem 5, we can see that following conditions are equivalent If one of above assertion is satisfied, we say that μ j tends to μ on Ω in the weak*-topology. (2) Assume that {μ j } j is a sequence measures on Ω and sup j μ j (Ω) < ∞, then there exists a subsequence {μ j k } k ⊂ {μ j } j such that μ j k converges to a measure μ in the weak*-topology as k → ∞.

Definition 6.
Let μ and ν be measures on Ω. We write μ ν if and only if And we say that μ is m-subhamonically greater than ν.
Example 1. For a ∈ Ω, let δ a be the Dirac measure at a. Let σ r be the normalized measure on the sphere ∂B(a, r), where r enough small such that Thus δ a σ r , but it is clear that δ a is not greater than σ r even though δ a (Ω) = σ r (Ω) = 1.
The following example shows that the converse implication to the statement given in Proposition 2 is not true.
Let Ω is the unit ball B in C n , n ≥ 2 and define the functions For more details, we can compute (see [12]) where dV is the Lebesgue measure on C n . By [12] one can compute the solution u to the equation The solution is given by Remark 3. The relation defines a partial order on the set of positive Borel measures on Ω. But it is not a total order. To see that consider the Dirac Remark 4. We have thatÊ is closed in Ω. Moreover, if E is relatively compact in Ω, so isÊ.

Proposition 3. Let μ, ν be finite regular measures on
Proof. Put K = suppμ. IfK = Ω then Proposition 3 is clear. Therefore we assume that Ω\K = ∅. Suppose that suppν K . SinceK is closed in Ω, it follows that ν(Ω\K) > 0. By the regularity of ν, we can find a compact set L ∈ Ω\K such that ν(L) > 0. From the definition ofK, for each z ∈ L, there exist a neighborhood U (z) of z and a function ϕ On m-Subharmonic Ordering of Measures Page 7 of 18 5 Proposition 3 is proved by a contradiction.

Maximal Measures and Minimal Functions
We want to study the maximality with respect to the m-subharmonic ordering by using some kind of normalization.
(3) We will show that the condition μ 1 , μ 2 are maximal does not imply the maximality of μ 1 + μ 2 (see Example 5). This implies that the set of maximal measures on Ω is not a convex cone.

Definition 9.
We say that a set K Ω is an interpolation set for SH − m (Ω) if for each f ∈ C(K), f < 0 there exists a function ϕ ∈ SH − m (Ω) such that ϕ = f on K.
For each value c j < 0, we take d j > 0 such that d j ψ j (a j ) = c j . Define ϕ = max(d 1 ψ 1 , . . . , d k ψ k ). Then we have ϕ ∈ SH − m (Ω) and ϕ(a j ) = c j . Thus the finite set {a 1 , . . . , a k } is an interpolation set for SH − m (Ω). And Proposition 4 implies that the measure k j=1 b j δ aj is maximal, where δ aj is the Dirac measure at the point a j and b 1 , . . . , b k are given nonnegative numbers.
We will show that each finite measure with compacted support is majorized by a maximal measure with the same total mass. Proof. For given f ∈ C ∞ 0 (Ω), choose a constant c > 0 so that (±f + cϕ) ∈ SH − m (Ω). Then we have which implies that Ω ±fdμ ≥ Ω ±fdν. So μ = ν.

Proof. Put K = suppμ and
Because μ ∈ M μ , so M μ = ∅. By Proposition 3, suppν ⊂ K for each ν ∈ M μ . Let ρ be the exhaustion function of Ω that is negative, continuous strictly m-subharmonic. We define Since ρ is bounded on K, it follows that A is finite. Let {ν j } j be a sequence in M μ such that Ω (−ρ)dν j → A, as j → ∞. By Remark 1, we may assume that ν j tend to some measure μ 0 in the weak*-topology and μ 0 (Ω) ≤ μ(Ω).

Definition 10. A function u ∈ F m (Ω) is said to be minimal if for any function
Then u is minimal.
To prove this proposition we need the following lemma.
Proof. We use a method from [7]. Using integration by parts, we have is a strictly m-subharmonic exhaustion function of Ω (see [2]). Hence, to prove u = v it is enough to show that (2) Thus, for every couple j, k, j for every couple j, k, j + k = m − 2. Assume that by assumption (4). So (2) is true by taking l = m − 1 in (4).

Proof of Proposition 5. Assume that
From the assumption H m (u) is maximal, we get H m (u) = H m (v). Now Proposition 5 follows from Lemma 1.
Proof. We use a method from [1]. For Proof. We use the same idea as in [5]. We choose a sequence
We will show that each function in F m (Ω) is minorized by a minimal function with the same total Hessian mass.
Then v j ≥ u j , v j ∈ E 0,m (Ω) and v j ↓ u as j → ∞. Theorem . We shall prove that t ∈ S. It is obvious that t ≤ u. Let {K i } be a compact exhaustion sets of Ω and let {t j } be a sequence of continuous functions such that t j ≥ t and t j ↓ t as j → ∞. For each z ∈ K i , choose v z ∈ T such that v z (z) < t j (z) and define the open set U z = {w ∈ Ω : v z (w) < t j (w)}. Take z 1 , . . . , z N ∈ K i such that ∪ N k=1 U z k ⊃ K i . Since T is totally ordered, we may choose v j i to be the smallest of the functions v z1 , . . . , v zN , which implies that v j i < t j on K i . Now let u 1 = v 1 1 and u j be the smallest of the functions {u 1 , . . . , u j−1 , v j j } if j ≥ 2, since T is totally ordered. Then {u j } is a decreasing sequence of functions in T such that u j ≤ v j j < t j on K j . Therefore u j ∈ F m (Ω), H m (u j )(Ω) = H m (u)(Ω) and u j ↓ t, as j → ∞. Proposition 7 implies t ∈ F m (Ω) and H m (t)(Ω) = H m (u)(Ω).