Unbounded Hankel operators and moment problems

0.16 MB·

by  D. R. Yafaev

UNBOUNDED HANKEL OPERATORS AND MOMENT PROBLEMS
6 D. R. YAFAEV
1
0
Abstract. We find necessary and sufficient conditions for a non-negative Hankel
2
quadraticformtoadmittheclosure. Wealsodescribethedomainofthecorresponding
n
closedform. Thisallowsustodefineunboundednon-negativeHankeloperatorsunder
a
J optimal assumptions on their matrix elements. The results obtained supplement the
9 classical Widom condition for a Hankel operator to be bounded.
2

]
A
1. Main results. Discussion
F
h. 1.1. Hankel operators Q can formally be defined in the space ℓ2(Z ) of sequences
+
t g = (g ,g ,...) by the formula
a 0 1
m
∞
[ (Qg) = q g , n = 0,1,.... (1.1)
n n+m m
X
1 m=0
v
Thus the matrix elements of a Hankel operator depend on the sum of indices only.
2
4 The precise definition of the operator Q requires some accuracy. Let ℓ2(Z+)
D ⊂
0 be the dense set of sequences g = (g ,g ,...) with only a finite number of non-zero
0 1
8 components. If the sequence q = (q ,q ,...) ℓ2(Z ), then for g sequence (1.1)
0 0 1 +
. also belongs to ℓ2(Z ). In this case the operat∈or Q is defined on ,∈anDd it is symmetric
1 + D
0 if q = q . Without any a priori assumptions on q , only the quadratic form
n n n
6
1 q[g,g] = q g g¯ (1.2)
n+m m n
: X
v n,m≥0
i
X is well defined for g .
∈ D
r ThefundamentaltheoremofNehari[6]guaranteesthataHankeloperatorQ(defined,
a
possibly, via its quadratic form (1.2)) is bounded if and only if q are the Fourier
n
coefficients of some bounded function on the unit circle T. The theory of Hankel
operators is a very well developed subject. We refer to the books [7, 8] for basic
information on this theory. However to the best of our knowledge, it was always
assumed that Hankel operators were bounded. The only exception is paper [13] where
Hankel operators were realized as integral operators in the space L2(R ).
+
The goal of this paper is to make first steps in the study of unbounded Hankel
operators. We consider non-negative quadratic forms (1.2) (in particular, we always
2000 Mathematics Subject Classification. Primary 47A05, 47A07; Secondary 47B25, 47B35.
Key words and phrases. Hankel operators, moment problems, Paley-Wiener theorem, Laplace
transform.
1
2 D.R. YAFAEV
assume that q = q ) so that we are tempted to define Q as a self-adjoint operator
n n
corresponding to the quadratic form q[g,g]. Such an operator exists if the form q[g,g]
admits the closure in the space ℓ2(Z ), but as is well known this is not always true.
+
We refer to the book [2] for basic information concerning these notions.
1.2. Below we give necessary and sufficient conditions guaranteeing the existence of
the closure of q[g,g], but previously we discuss the link of Hankel quadratic forms with
the Hamburger moment problem. The following result was obtained in [5].
Theorem 1.1. The condition
q g g¯ 0, g , (1.3)
n+m m n
≥ ∀ ∈ D
X
n,m≥0
is satisfied if and only if there exists a non-negative measure dM(µ) on R such that the
coefficients q admit the representations
n
∞
q = µndM(µ), n = 0,1,.... (1.4)
n Z ∀
−∞
Note that the measure satisfying equations (1.4) is in general not unique (see the
paper [11], for a comprehensive discussion of this phenomenon). Roughly speaking,
the non-uniqueness of solutions of the Hamburger moment problem is due to a very
rapid growth of the coefficients q . Indeed, the famous Stieltjes example shows that
n
the measures
dMθ(µ) = 1R+(µ)µ−lnµ 1+θsin(2πlnµ) dµ, ∀θ ∈ [−1,1],
(cid:0) (cid:1)
solve equations (1.4) with q = √πe(k+1)2/4. On the other hand, if q Rnn! for some
n n
| | ≤
R > 0, then the solution of equations (1.4) for the measure dM(µ) is unique.
1.3. The definition of the Hankel operator requires essentially more restrictive as-
sumptions on the matrix elements q . Let us state our main result.
n
Theorem 1.2. Let assumption (1.3) be satisfied. Then the following conditions are
equivalent:
(i) The form q[g,g] defined on admits the closure in ℓ2(Z ).
+
D
(ii) The matrix elements q 0 as n .
n
→ → ∞
(iii) The measure dM(µ) defined by equations (1.4) satisfies the condition
M(R ( 1,1)) = 0 (1.5)
\ −
(to put it differently, suppM [ 1,1] and M( 1 ) = M( 1 ) = 0).
⊂ − {− } { }
Note that for q ℓ2(Z ) the assertion (i) is obvious because in this case q[g,g] =
+
∈
(Qg,g) where Q is the symmetric operator Q defined on by (1.1). As far as the
D
proof of Theorem 1.2 is concerned, we note that only the implication
(i) = (ii) or (iii)
⇒
is sufficiently non-trivial.
UNBOUNDED HANKEL OPERATORS AND MOMENT PROBLEMS 3
In Section 3 we also give (see Theorem 3.4) an efficient description of the closure of
the form (1.2). In Section 4 we discuss some consequences of our results for moment
problems.
1.4. Theorem 1.2 is to a large extent motivated by the following classical results of
H. Widom.
Theorem 1.3. [12, Theorem 3.1] Let the matrix elements q of the Hankel operator
n
(1.1) be given by the relations
1
q = µndM(µ), n = 0,1,..., M( 1 ) = M( 1 ) = 0, (1.6)
n Z ∀ {− } { }
−1
with some non-negative measure dM(µ). Then the following conditions are equivalent:
(i) The operator Q is bounded.
(ii) q = O(n−1) as n .
n
→ ∞
(iii) M((1 ε,1)) = O(ε) and M(( 1, 1+ε)) = O(ε) as ε 0.
− − − →
Theorem 1.4. [12, Theorem 3.2] Under the same a priori assumptions as in Theo-
rem 1.3 the following conditions are equivalent:
(i) The operator Q is compact.
(ii) q = o(n−1) as n .
n
→ ∞
(iii) M((1 ε,1)) = o(ε) and M(( 1, 1+ε)) = o(ε) as ε 0.
− − − →
Theorems 1.3 and 1.4 give optimal conditions for the Hankel operator Q with matrix
elements (1.6) to be bounded and compact. Roughly speaking, condition (iii) of The-
orem 1.3 means that the measure dM(µ) is “subordinated” to the Lebesgue measure
near the end points 1 and 1 of its support. Similarly, condition (iii) of Theorem 1.4
−
means that the measure dM(µ) is “diluted” compared to the Lebesgue measure near
these end points.
2. Proof of Theorem 1.2
2.1. It is almost obvious that conditions (ii) and (iii) are equivalent. Indeed, if (iii)
is satisfied, then
1−ε 1 −1+ε
q = µndM(µ)+ µndM(µ)+ µndM(µ).
n Z Z Z
−1+ε 1−ε −1
The second and third integrals on the right are bounded by M((1 ε,1)) and
−
M(( 1, 1 + ε)), and hence they tend to zero as ε 0 uniformly in n. The first
− − →
integral is bounded by (1 ε)nM(( 1,1)), and therefore it tends to zero as n
for every ε > 0. Conversely−, if there−exists a set X R ( 1,1) such that M(X)→>∞0,
⊂ \ −
then q M(X), and hence condition (ii) cannot be satisfied.
2n
≥
4 D.R. YAFAEV
It is convenient to reformulate condition (i) in an equivalent form. Let dM be an
arbitrary measure satisfying the condition
∞
µ ndM(µ) < , n = 0,1,..., (2.1)
Z | | ∞ ∀
−∞
and let L2(M) = L2(R;dM) be the space of functions u(µ) with the norm u .
L2(M)
k k
We put
∞
( g)(µ) = g µn (2.2)
n
A
X
n=0
and observe that under assumption (2.1), g L2(M) for all g . Therefore we can
define an auxiliary operator A: ℓ2(Z ) AL2(M∈ ) on domain (A∈)D= by the formula
+
→ D D
Ag = g. In view of equations (1.4) the form q[g,g] defined by relation (1.2) can be
A
written as
∞
q[g,g] = Ag 2 = ( g)(µ) 2dM(µ), g . (2.3)
k kL2(M) Z | A | ∈ D
−∞
This yields the following result.
Lemma 2.1. The form q[g,g] defined on admits the closure in the space ℓ2(Z ) if
+
and only if the operator A: ℓ2(Z ) L2(DM) defined on the same set admits the
+
→ D
closure.
Recall that the operator A admits the closure if and only if its adjoint operator
A∗: L2(M) ℓ2(Z ) is densely defined. So our next goal is to construct A∗. Observe
+
→
that under assumption (2.1) for an arbitrary u L2(M), all the integrals
∈
∞
u(µ)µndM(µ) =: u , n Z , (2.4)
Z n ∈ +
−∞
are absolutely convergent. We denote by L2(M) the set of u L2(M) such that
∗
the sequence u ∞ ℓ2(Z ). D ⊂ ∈
{ n}n=0 ∈ +
Lemma 2.2. Under assumption (2.1) the operator A∗ is given by the equality
∞
(A∗u) = u(µ)µndM(µ), n Z ,
n Z ∈ +
−∞
on the domain (A∗) = .
∗
D D
Proof. Obviously, for all g and all u L2(M), we have
∈ D ∈
∞
(Ag,u) = g u¯ (2.5)
L2(M) n n
X
n=0
where the numbers u are defined by relations (2.4). The right-hand side here equals
n
(g,A∗u) provided u . It follows that (A∗).
∗ ∗
∈ D D ⊂ D
Conversely, if u (A∗), then
∈ D
(Ag,u) = (g,A∗u) g A∗u
| L2(M)| | ℓ2(Z+)| ≤ k kℓ2(Z+)k kℓ2(Z+)
UNBOUNDED HANKEL OPERATORS AND MOMENT PROBLEMS 5
for all g . Therefore it follows from equality (2.5) that
∈ D
∞
g u¯ C(u) g , g .
n n ≤ k kℓ2(Z+) ∀ ∈ D
(cid:12)X (cid:12)
n=0
(cid:12) (cid:12)
Since is dense in ℓ2(Z ), we see that u ∞ ℓ2(Z ), and hence u . Thus
(A∗)D . + { n}n=0 ∈ + ∈ D∗ (cid:3)
∗
D ⊂ D
2.2. Next, we use the following analytical result.
Theorem 2.3. The set is dense in L2(M) if and only if condition (1.5) is satisfied.
∗
D
Proof. Let the set consist of u L2(M) such that suppu [ a,a] for some a < 1.
E ∈ ⊂ −
According to definition (2.4) for u , we have u Can whence . Under
n ∗
∈ E | | ≤ E ⊂ D
assumption (1.5) the set is dense in L2(M) and so is also dense in this space.
∗
E D
Let us prove the converse statement. Suppose that is dense in L2(M). For an
∗
D
arbitrary u L2(M), we put
∈
∞
f(x) = eiµxu(µ)dM(µ). (2.6)
Z
−∞
Obviously, we have
f(x) u M(R). (2.7)
L2(M)
| | ≤ k k
It follows from condition (2.1) that f C∞(R)pand
∈
∞
f(n)(0) = in µnu(µ)dM(µ).
Z
−∞
Assume now that u . Then this sequence is bounded and hence the function
∗
∈ D
∞ f(n)(0)
f(x) = xn
n!
X
n=0
is entire and satisfies the estimate
∞
1
f(z) C z n = Ce|z|, z C. (2.8)
| | ≤ n!| | ∈
X
n=0
Let us now show that
f(z) Ce|Imz|, z C. (2.9)
| | ≤ ∈
Consider, for example the angle argz [0,π/2] and put F(z) = f(z)eiz. Since eiz =
∈ | |
e−Imz, it follows from estimates (2.7) and (2.8) that F(z) Ce|z| for all z and that
| | ≤
the function F(z) is bounded on the rays z = r and z = ir where r 0. Therefore, by
≥
the Phragm´en-Lindelo¨f principle (see, e.g., the book [4]), F(z) is bounded in the whole
angle argz [0,π/2]. This yields estimate (2.9) for f(z) = F(z)e−iz.
∈
According to the Paley-Wiener theorem (see, e.g., Theorem IX.12 of [9]) it follows
from estimate (2.9) that the Fourier transform of f(x) (considered as a distribution in
the Schwartz class ′(R)) is supported by the interval [ 1,1]. Therefore formula (2.6)
S −
6 D.R. YAFAEV
implies that for every u , the distribution u(µ)dM(µ) is also supported by [ 1,1],
∗
∈ D −
that is
∞
ϕ(µ)u(µ)dM(µ) = 0, ϕ C∞(R [ 1,1]). (2.10)
Z ∀ ∈ 0 \ −
−∞
Since is dense in L2(M), we can approximate 1 by functions u in this space.
∗ ∗
D ∈ D
Hence equality (2.10) is true with u(µ) = 1. It follows that
suppM [ 1,1] (2.11)
⊂ −
because ϕ C∞(R [ 1,1]) is arbitrary.
∈ 0 \ −
For the proof of (1.5), it remains to show that M( 1 ) = M( 1 ) = 0. In view of
{− } { }
(2.11) for an arbitrary u L2(M), sequence (2.4) admits the representation
∈
1
u = M( 1 )u(1)+( 1)nM( 1 )u( 1)+ u(µ)µndM (µ) (2.12)
n { } − {− } − Z 0
−1
where M (X) = M(X ( 1,1)) is the restriction of the measure M on the open
0
∩ −
interval ( 1,1). Obviously, for any ε (0,1), we have
− ∈
1 1−ε 1 −1+ε
u(µ)µndM (µ) = u(µ)µndM (µ)+ u(µ)µndM (µ)+ u(µ)µndM (µ).
Z 0 Z 0 Z 0 Z 0
−1 −1+ε 1−ε −1
(2.13)
Applying the Schwarz inequality to each integral on the right, we estimate this expres-
sion by
(1 ε)n M (( 1,1))+ M ((1 ε,1))+ M (( 1, 1+ε)) u .
− 0 − 0 − 0 − − k kL2(M0)
(cid:0) p p p (cid:1)
Since M ((1 ε,1)) 0 and M (( 1, 1+ε)) 0 as ε 0, we see that integral in
0 0
− → − − → →
the left-hand side of (2.13) tends to zero as n . Thus (2.12) implies that
→ ∞
u = M( 1 )u(1)+( 1)nM( 1 )u( 1)+o(1)
n
{ } − {− } −
as n . Therefore if u , or equivalently u ∞ ℓ2(Z ), then necessarily
→ ∞ ∈ D∗ { n}n=0 ∈ +
M( 1 )u(1) = M( 1 )u( 1) = 0. So if M( 1 ) = 0 (or M( 1 ) = 0), then
{ } {− } − { } 6 {− } 6
u(1) = 0 (or u( 1) = 0). It follows that the function u such that u(µ) = 0 for µ = 1
− 6
and u(1) = 1 (or u(µ) = 0 for µ = 1 and u( 1) = 1) cannot be approximated by
functions in . Hence is not d6en−se in L2(M−). (cid:3)
∗ ∗
D D
Remark 2.4. We have used the Phragm´en-Lindelo¨f principle for the proof of estimate
(2.9) only. Actually, relation (2.10) can be directly deduced from estimates (2.7) and
(2.8) using the arguments given in the proof of Theorem 19.3 of the book [10]. However
the intermediary estimate (2.9) makes the proof of (2.10) essentially more transparent.
2.3. Let us come back to the proof of Theorem 1.2. Putting together Lemma 2.2
and Theorem 2.3, we see that the operator A∗ is densely defined and hence A admits
the closure if and only if condition (1.5) is satisfied. In view of Lemma 2.1 this proves
that the conditions (i) and (iii) of Theorem 1.2 are equivalent and thus concludes its
proof.
UNBOUNDED HANKEL OPERATORS AND MOMENT PROBLEMS 7
3. The closure of the Hankel quadratic form
3.1. Let the condition (1.5) be satisfied, and let A¯ be the closure of the operator
A defined by equation (2.2) on domain (A) = . Then the form (2.3) admits the
closure in the space ℓ2(Z ) and the formD D
+
q[g,g] = A¯g 2 (3.1)
k kL2(M)
isclosed ondomain [q] = (A¯). Since A¯ = A∗∗, itremains to describe theset (A∗∗).
Observe that seriDes (2.2)Dconverges for all g ℓ2(Z ) and all µ ( 1,1D). The
+
∈ ∈ −
function ( g)(µ) depends continuously on µ, but only the estimate
A
∞
g µ n (1 µ2)−1/2 g
| n|| | ≤ − k kℓ2(Z+)
X
n=0
holds. So it is of course possible that g L2(M). Now we define the operator A
max
by the formula A g = g on the dAoma6∈in (A ) that consists of all g ℓ2(Z )
max max +
A D ∈
such that g L2(M). Our goal is to show that
A ∈
A∗∗ = A . (3.2)
max
A difficult part in the proof of (3.2) is the inclusion A A∗∗ that is equivalent
max
⊂
to the relation
(A g,u) = (g,A∗u) (3.3)
max L2(M) ℓ2(Z+)
for all g (A ) and all u (A∗) = . In the detailed notation, relation (3.3)
max ∗
∈ D ∈ D D
means that
1 ∞ ∞ 1
g µn u(µ)dM(µ) = g µnu(µ)dM(µ) .
Z n n Z
−1(cid:0)Xn=0 (cid:1) Xn=0 (cid:0) −1 (cid:1)
The problem is that these integrals do not converge absolutely, and so the Fubini
theorem cannot be applied.
3.2. The shortest way to prove (3.2) is to reduce the operator by appropriate
A
unitary transformations to the Laplace transform defined by the relation
∞
( f)(λ) = e−tλf(t)dt. (3.4)
B Z
0
Weconsider itasamapping : L2(R ) L2(R ;dΣ)wherethenon-negativemeasure
+ +
dΣ(λ) on R satisfies the coBndition →
+
∞
(λ+1)−kdΣ(λ) < (3.5)
Z ∞
0
for some k > 0. Let the set D L2(R ) consist of functions compactly supported in
+
R . Iff D,then( f)(λ)isac⊂ontinuousfunctionforallλ 0and( f)(λ) = O(e−cλ)
+
with som∈e c = c(f)B> 0 as λ ; in particular, f L≥2(R ). WBe put Bf = f
+
with (B) = D. → ∞ B ∈ B
D
8 D.R. YAFAEV
It is easy to show (see [13], for details) that the operator B∗ is given by the formula
∞
(B∗v)(t) = e−tλv(λ)dΣ(λ), (3.6)
Z
0
and its domain (B∗) consists of all v L2(R ;dΣ) such that B∗v L2(R ). Ob-
+ +
viously, this conDdition is satisfied if v is∈compactly supported in R . ∈Since the set of
+
such v is dense in L2(R ;dΣ), the operator B∗ is densely defined. Thus B admits the
+
closure and B¯ = B∗∗.
The integral (3.4) converges for all f L2(R ) and λ > 0. The function ( f)(λ) is
+
∈ B
continuous, but of course the estimate
( f)(λ) (2λ)−1/2 f
| B | ≤ k kL2(R+)
does not guarantee that f L2(R ;dΣ). Let us now define the operator B by the
+ max
formula Bf = f on thBe do∈main (B ) that consists of all f L2(R ) such that
max +
f L2(R ;dΣB). We use the folloDwing assertion. ∈
+
B ∈
Lemma 3.1. [13, Theorem 3.9] Let dΣ(λ) be a measure on R such that the condition
+
(3.5) is satisfied for some k > 0. Then
B∗∗ = B . (3.7)
max
3.3. Let us find a relation between the operators A and B. Suppose that the
measures dΣ(λ) and dM(µ) are linked by the equality
2λ 1
dM(µ) = (λ+1/2)−2dΣ(λ), µ = − . (3.8)
2λ+1
Thus if M(( 1,1)) < , then the condition (3.5) holds with k = 2. Let us also set
− ∞
1 2λ 1
(Vu)(λ) = u − . (3.9)
λ+1/2 (cid:16)2λ+1(cid:17)
Obviously, V : L2(( 1,1);dM) L2(R ;dΣ) is a unitary operator. If the measures
+
− →
dM and dΣ are absolutely continuous, that is
dΣ(λ) = σ(λ)dλ, λ > 0, dM(µ) = η(µ)dµ, µ ( 1,1),
∈ −
with some σ L1 (R ) and η L1( 1,1), then relation (3.8) means that
∈ loc + ∈ −
1+µ
η(µ) = σ .
2(1 µ)
(cid:0) − (cid:1)
Recall that the Laguerre polynomial (see the book [3], Chapter 10.12) of degree n is
defined by the formula
n
n!
L (t) = n!−1etdn(e−ttn)/dtn = ( t)m.
n (n m)!(m!)2 −
X
m=0 −
UNBOUNDED HANKEL OPERATORS AND MOMENT PROBLEMS 9
We need the identity (see formula (10.12.32) in [3])
∞ 1 2λ 1 n
L (t)e−(1/2+λ)tdt = − , λ > 1/2. (3.10)
Z n λ+1/2(cid:16)2λ+1(cid:17) −
0
It can be deduced from this fact that the functions L (t)e−t/2, n = 0,1,..., form an
n
orthonormal basis in the space L2(R ), and hence the operator U: l2(Z ) L2(R )
+ + +
→
defined by the formula
∞
(Ug)(t) = g L (t)e−t/2, g = (g ,g ,...), (3.11)
n n 0 1
X
n=0
is unitary.
A link of the operators and is stated in the following assertion.
A B
Lemma 3.2. For all g , the identity holds
∈ D
V g = Ug. (3.12)
A B
Proof. It follows from equalities (3.4), (3.10) and (3.11) that
∞ ∞ ∞ 1 2λ 1 n
( Ug)(λ) = g L (t)e−(1/2+λ)tdt = g − .
B Xn=0 nZ0 n Xn=0 nλ+1/2(cid:16)2λ+1(cid:17)
In view of definitions (2.2), (3.9) this expression equals (V g)(λ). (cid:3)
A
Combining Lemmas 3.1 and 3.2, it is now easy to obtain the following result.
Lemma 3.3. Let dM(µ) be a finite measure on ( 1,1). Then equality (3.2) holds.
−
Proof. Observe that the adjoint of the operator B defined by (3.4) on the set U is
D
still given by formula (3.6). Therefore it follows from (3.12) that A∗V = UB∗ and
hence
VA∗∗ = B∗∗U. (3.13)
Let g ℓ2(Z ) be arbitrary. Approximating it by functions g and using (3.12),
+ n
∈ ∈ D
we see that (V g)(λ) = ( Ug)(λ) for all λ > 0. It follows that
A B
VA = B U. (3.14)
max max
Comparing (3.13) and (3.14), we see that the identities (3.2) and (3.7) are equivalent.
(cid:3)
In view of relation (3.2), formula (3.1) leads to the following result.
Theorem 3.4. Let assumption (1.3) and one of three equivalent conditions (i), (ii) or
(iii) of Theorem 1.2 be satisfied. Let the form q[g,g] be defined on by (1.2), and let
D
be the operator (2.2). Then the closure of q[g,g] is given by the equality
A
1
q[g,g] = ( g)(µ) 2dM(µ) (3.15)
Z | A |
−1
on the set [q] of all g ℓ2(Z ) such that the right-hand side of (3.15) is finite.
+
D ∈
10 D.R. YAFAEV
We note that the non-negative operator Q corresponding to the form (3.15) satisfies
the relations
q[g,h] = (g,Qh), g [q], h (Q) [q],
∀ ∈ D ∀ ∈ D ⊂ D
q[g,g] = Qg 2, g ( Q) = [q],
k k ∀ ∈ D D
p p
but its domain (Q) does not admit an efficient description.
D
4. Moment problems
4.1. Comparing Theorems 1.1 and 1.2, we obtain the following result concerning
moment problems.
Proposition 4.1. A non-negative measure dM(µ) satisfying conditions (1.4) and (1.5)
exists if and only if inequality (1.3) holds and q 0 as n (or, equivalently, the
n
→ → ∞
form (1.2) admits the closure).
Instead of the interval [ 1,1] we can consider an arbitrary finite interval [ a,a].
− −
Our arguments proving the equivalence of conditions (ii) and (iii) in Theorem 1.2 lead
to the following simple assertion.
Proposition 4.2. A non-negative measure dM(µ) satisfying the condition
a
q = µndM(µ), n = 0,1,...,
n Z ∀
−a
exists if and only if inequality (1.3) holds and q = O(an) as n . Moreover,
n
→ ∞
M( a ) = M( a ) = 0 if and only if q = o(an) as n .
n
{− } { } → ∞
4.2. The results obtained above can be combined with the Stieltjes theorem which
states that there exists a non-negative measure dM(µ) satisfying equations (1.4) and
such that suppM [0, ) if and only if inequalities (1.3) and
⊂ ∞
q g g¯ 0, g , (4.1)
n+m+1 m n
≥ ∀ ∈ D
X
n,m≥0
hold.
Let us state analogues of Propositions 4.1 and 4.2.
Proposition 4.3. A non-negative measure dM(µ) satisfying conditions (1.4) and
suppM [0,1], M( 1 ) = 0 exists if and only if inequalities (1.3) and (4.1) hold
⊂ { }
and q 0 as n (or, equivalently, the form q[g,g] admits the closure).
n
→ → ∞
Proposition 4.4. A non-negative measure dM(µ) satisfying the condition
a
q = µndM(µ), n = 0,1,....
n Z ∀
0
exists if and only if inequalities (1.3) and (4.1) hold and q = O(an) as n .
n
→ ∞
Moreover, M( a ) = 0 if and only if q = o(an) as n .
n
{ } → ∞