OF THE L-MOMENT PROBLEM
LUMINIT¸ A LEMNETE-NINULESCU and ANCA ZL ˘ATESCU
We give a necessary and sufficient condition for a complex sequence{yα,β}α,β∈N
to be aL-complex moment sequence for a real,measurable functionhdefined on the closed unit disc in the complex planeC. We also study the support of the functionhunder additional conditions satisfied by the sequence {yα,β}α,β∈N. AMS 2000 Subject Classification: Primary 47A57, 44A60; Secondary 15A57.
Key words: L-moment problem, positive bounded borelian measure, linear func- tional, the support of a function, the support of a measure.
1. INTRODUCTION
This note is devoted to theL-moment problem. TheL-moment problem consists of characterising the sequence of moments an = R
Rtnf(t)dt, n ∈ N of a real measurable function f (with prescribed support) which satisfies a condition such is 0≤f ≤La.e. dt. TheL-moment problem was formulated and completely solved by Akhiezer and Krein [2] in the thirties. The interest for the moments of a bounded measurable function on the real axis goes back to A.A. Markov in the last part of the nineteenth century. It was M.G. Krein who studied again this field with new methods at that time of function the- ory and functional analysis. The aim of the present paper is to study a two dimensional complex L-moment problem. The L-complex moment problem consists of characterising a double complex sequence{yα,β}α,β∈N to represent the moments of a measurable real function h defined on the closed unit disc of the complex plane, which satisfies a condition of boundness 0 ≤h≤L for a positive constant L > 0. We formulate and solve this L-complex moment problem in the following theorem.
REV. ROUMAINE MATH. PURES APPL.,55(2010),3, 197–204
2. EXISTENCE OF A SOLUTION OF A L-MOMENT PROBLEM
Theorem 1. Let D1 ={z∈C,|z| ≤1} be the closed unit disc endowed with a planar positive bounded Borel measure µ. Let{yα,β}αβ∈Nbe a sequence of complex numbers. The following assertions are equivalent.
(a) There exists a positive constant M > 0 such that for all sequences {ξα}α∈N⊂Cwith finitely many nonzero terms and for any k∈N we have
0≤X
α,β
X
0≤θ≤k
(−1)θCkθyα+θ,β+θξαξβ ≤MX
α,β
X
0≤θ≤k
(−1)θCkθcα+θ,β+θξαξβ, where {cα,β}α,β∈N stands for R
D1zαzβdµ(z)∀α, β ∈N.
(b)There exists a real-valued nonnegative, bounded, measurable function h∈L1(D1, µ) such that
yα,β= Z
D1
zαzβh(z)dµ(z) ∀α, β ∈N.
Proof. (a)⇒ (b). Let Pe=
P(z, z) = X
(α,β)∈H
aα,βzαzβ, H finite⊂Z2+, aα,β ∈C
be the C-vector space of polynomial complex function in z, z considered as functions from C to C.
We define the C-linear functional L : Pe → C, L(zαzβ) = yα,β for all (α, β)∈Z2+ and extend it by linearity to Pe.
Let Pan = n
P : C → C, P(z) = P
n∈H⊂N
ξnzn, ξn ∈ C, H finiteo be the analytical polynomial functions. Let P ∈Pan,P(z) = P
n∈H⊂N
ξnzn be an arbitrary polynomial; for this polynomial we have
|P(z)|2(1− |z|2)k= X
α,β∈H⊂N
X
0≤θ≤k
(−1)θCkθzα+θzβ+θξαξβ. From the above definition of L, we obtain
L(P(z)|2(1− |z|2)k) =L
X
α,β∈H⊂N
X
0≤θ≤k
(−1)θCkθzα+θzβ+θξαξβ
=
= X
α,β∈H
X
0≤θ≤k
(−1)θCkθyα+θ,β+θξαξβ.
From assumption (a) the following inequalities hold 0≤X
α,β
X
0≤θ≤k
(−1)θCkθyα+θ,β+θξαξβ =L(|P(z)|2(1− |z|2)k)≤
≤MX
α,β
X
0≤θ≤k
(−1)θCkθcα+θ,β+θξαξβ =M Z
D1
|P(z)|2(1− |z|2)kdµ(z) for all analytical polynomials P ∈Pan and any k∈ N.
We denote by qm(z, z) = X
j∈H⊂N
αj|P(z)|2(1− |z|2)kj, H finite, kj∈N, αj ≥0, P∈Pan, m∈N the polynomials of this type and by R[x, y] the real vector space of polynomials inx, ywith real coefficients. From results in [5], [13], every elementf ∈R[x, y]
can be represented as f(x, y) = qr(z, z)−qm(z, z), with qr, qm as above. We apply these assertions in [5], [13] to decompose an arbitrary polynomialf ∈Pe. We havef(z, z) =f1(z, z) + if2(z, z) with f1, f2 polynomials inz, z with real coefficients. If we write fj(z, z) = fj1(x, y) + ifj2(x, y), j = 1,2, fjk ∈ R[x, y], k = 1,2, from the above mentioned results,we have a decomposition f11(x, y) =q1(z, z)−q2(z, z),f12(x, y) =q3(z, z)−q4(z, z). The same is true of f21 and f22. We have then
(1) |L(f)|=|L(f1) + iL(f2)| ≤ |L(f1)|+|L(f2)|.
Because of (a) we have 0≤L(qj) ≤MR
D1qjdµ =MR
D1|qj|dµ that is
|L(qj)| ≤MR
D1|qj|dµ for allj ∈1,8. From the previous decomposition of f we obtain
(2) |L(f1)| ≤ |L(q1−q2)|+|L(q3−q4)| ≤2M Z
D1
|f1|dµ,
(3) |L(f2)| ≤ |L(q5−q6)|+|L(q7−q8)| ≤2M Z
D1
|f2|dµ.
It follows from (1), (2) and (3) that
(4) |L(f)| ≤4M
Z
D1
|f|dµ ∀f ∈P .e The applicationF :C(D1)→C, F(f) = 4MR
D1|f|dµis a seminorm on C(D1); by Hahn-Banach theorem we can extendLto the space of continuous complex-valued functions on D1 preserving the inequality (4). Let Le denote this extension to C(D1).
The functional Le is positive, i.e., L(fe ) ≥ 0 for any f ∈ CR(D1) with f(z)≥0 for any z∈D1.Indeed, the R-vector space CR(D1) ={f :D1→ R, f continuous onD1}is the closure in the uniform convergence topology on the
compact D1 of Re={P ∈Pe with P(z, z)∈R} and by Proposition 3 [4], any polynomial P ∈Pe withP(z, z)≥0 onD1 can be uniformly approximated on D1 with polynomials of the formqj. But the first inequality of (a) states that L(qj)≥0, which finishes the proof of positivity of L.e
If we consider the uniform normkfk= sup
z∈D1
|f(z)|on C(D1), we obtain
(5) |eL(f)| ≤4M
Z
D1
|f|dµ≤4MkfkD1µ(D1).
This means that Le is a continuous positive functional on C(D1). By Riesz representation theorem, there exists a positive Radon measure dσ on D1, such thatL(fe ) =R
D1fdσ, for any f ∈C(D1), in particular L(zαzβ) =
Z
D1
zαzβdσ(z) =yα,β
for anyα, β ∈N.We prove that the measure dσ is absolutely continuous with respect to the measure dµ. Let E ⊂D1 be a measurable set withµ(E) = 0 and XE the characteristic function ofE. We approximate XE pointwise and monotonically by a bounded sequenceϕkinCR+(D1). For this sequence, using Lebesgue’s dominated convergence theorem we have
0≤ Z
D1
ϕkdσ ≤4M Z
D1
ϕkdµ≤4M Z
D1
lim
k ϕkdµ= 4M Z
D1
XEdµ= 4M µ(E).
Hence, if µ(E) = 0 we also have σ(E) = 0. In this case, there exists h ∈ L1(D1, µ) such that dσ =hdµ.
Because R
D1hϕdµ ≥0 for any ϕ ∈ CR+(D1) we have h ≥ 0 on D1. On the other hand, we observe using also (5) that
0≤ Z
D1
hϕdµ≤4M Z
D1
ϕdµ⇔ Z
D1
ϕ(4M−h)dµ≥0, ∀ϕ∈CR+(D1).
We have then 0≤h≤4M, soh verifies (b).
Conversely, (b)⇒ (a).
If h is a real-valued nonnegative,bounded function, h ∈ L1(D1, µ), 0≤ h≤L, for which (b) holds, we have
0≤X
α,β
X
0≤θ≤k
(−1)θCkθyα+θ,β+θξαξβ =
=X
α,β
X
0≤θ≤k
(−1)θCkθ Z
D1
zα+θzβ+θh(z)dµ(z)ξαξβ =
= Z
D1
(1− |z|2)k|P(z)|2h(z)dµ(z)≤L Z
D1
(1− |z|2)k|P(z)|2dµ
=
=LX
α,β
X
0≤θ≤k
(−1)θCkθcα+θ,β+θξαξβ for a positive constant L with 0 ≤h ≤L and cα,β = R
D1zαzβdµ(z), that is assertion (a).
Additional conditions on the sequences {yα,β}(α,β)∈ Z2
+, {cα,β}(α,β)∈Z2 +, from the above theorem can give us more informations about the support of the function h.
Let us consider the relations A and B below.
A. There exist numberss, S ∈R with 0< s < S <1 such that
(j) X
0≤θ≤k
X
(α,β)⊂Z+2
[S(−1)θCkθyα+θ,β+θ+ (−1)θ+1Ckθyα+θ+1,β+θ+1]ξαξβ = 0 and
(j0) X
0≤θ≤k
X
(α,β)⊂Z+2
[(−1)θCkθyα+θ+1,β+θ+1+s(−1)θ+1Ckθyα+θ,β+θ]ξαξβ = 0 for allk∈N and all sequences{ξα}α∈N⊂C of complex numbers with finitely many nonzero terms.
B. There also exists sequences{ξα}α∈N,{ηα}α∈N such that (jj) X
0≤θ≤k
X
(α,β)⊂Z2+
[S(−1)θCkθcα+θ,β+θ+ (−1)θ+1Ckθcα+θ+1,β+θ+1]ξαξβ 6= 0 and
(jj0) X
0≤θ≤k
X
(α,β)⊂Z2+
[(−1)θCkθcα+θ+1,β+θ+1+s(−1)θ+1Ckθcα+θ,β+θ]ξαξβ 6= 0,
where cα,β =R
D1zαzβdµ(z) for all α, β∈N.
Corollary 1. If the above relations A. and B. hold, then supp [h] ⊂ {z,√
s≤ |z| ≤√ S}.
Proof. Condition (j) from A. means that X
0≤θ≤k
X
(α,β)⊂Z2+
[S(−1)θCkθyα+θ,β+θ+ (−1)θ+1Ckθyα+θ+1,β+θ+1]ξαξβ =
= Z
D1
(S− |z|2)(1− |z|2)k|P(z)|2h(z)dµ(z) = 0 for all P(z) = P
α∈H
ξαzα, H finite ⊂ N and for all k ∈ N. Under these as- sumptions, we can apply Proposition 3 in [4]. This proposition can be stated as follows.
Proposition 3. LetDmthe open unit polydisk in the complex spaceCm, and let A(Dm×Dm) denote the space of functions continuous on the closed polydisk Dm×Dm and analytical on Dm×Dm, equipped with the topology of uniform convergence on compacta. Let M ={f ∈A(Dm×Dm) :f(z, z)≥0 for allzinDm}. IfF denotes the convex hull of{f ∈A(Dm×Dm) :f(z, w) = p(z)Qm
i=1(1−ziwi)kip(w), where ki are non-negative integers and p is a m- variable complex polynomial}, thenF =M, whereF denotes the closure ofF in the topology of uniform convergence on compacta.
From this result we haveR
D1(S− |z|2)(1− |z|2)kP(z, z)h(z)dµ(z) = 0 for all P ∈P withP(z, z)≥0 for anyz∈D1.
Consequently, R
D1(S − |z|2)f(z)h(z)dµ(z) = 0 for any f : D1 → R, f continuous on D1; this means that supp [hdµ]⊂ {z,|z| ≤√
S}.
Similarly, from (j0) we have supp [hdµ]⊂ {z,|z| ≥ √
s}. Condition (jj) from B. ensures that supp [dµ]∈ {z,/ √
s≤ |z| ≤√
S}. From (j), (j0) and (jj) it result that supp [h]⊂ {z,√
s≤ |z| ≤√ S}.
Corollary 2. If X
0≤θ≤k
X
(α,β)⊂Z+2
[(−1)θCkθyα+θ,β+θ+ (−1)θ+1Ckθyα+θ+1,β+θ+1]ξαξβ = 0
for all k∈N and all sequences {ξα}α∈N⊂Cof complex numbers with finitely many nonzero terms, then supp [hdµ]⊂ {z,|z|= 1}.
Proof. The condition in Corollary 2 means that Z
D1
(1− |z|2)k+1|P(z)|2h(z)dµ(z) = 0 for any P(z) = P
α∈H
ξαzα,H finite ⊂N and anyk∈N.
As in the proof of Corollary 1, this means that supp [hdµ]⊂ {z,|z|2 = 1}.
Corollary 3. If X
0≤θ≤k
X
(α,β)⊂Z2+
[(−1)θCkθyα+θ+1,β+θ+ (−1)θ+1Ckθyα+θ,β+θ+1]ξαξβ = 0
for all k∈N and all sequences of complex numbers {ξα}α∈N⊂Cwith finitely many nonzero terms, then supp [hdµ]⊂[−1,1]⊂R.
Proof. The condition in Corollary 3 means that Z
D1
(z−z)(1− |z|2)k|P(z)|2h(z)dµ(z) = 0
for all P(z) = P
α∈H
ξαzα, H finite ⊂N and for all k ∈ N. As in the proof of Corollary 1, this imply supp [hdµ]⊂[−1,1].
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Received 27 February 2008 “Politehnica” University of Bucharest Department of Mathematics
Splaiul Independentei 313 060042 Bucharest, Romania luminita [email protected]
