Indeterminate moment problems
Christian Berg
University of Copenhagen, Denmark
Hilbert spaces of entire functions and their applications,
B¦dlewo
May 2226, 2017
Based on joint work with
Ryszard Szwarc (Wrocªaw)
Christian Berg
Indeterminate moment problems
Overview
1. Basic facts about indeterminate Hamburger moment
problems
2. Order and type, logarithmic order and type of indeterminate
moment problems
3. Determination of the order as exponent of convergence
E(bn )
4. Symmetric indeterminate moment problems
5. Multi-zeta values
6. Birth and death processes
7. Valent's conjectures
Christian Berg
Indeterminate moment problems
The basic data of a Hamburger moment problem
A Hamburger moment problem is characterized either by the
moment sequence (sn ) or by two real sequences (an ), (bn ) with
bn > 0 entering in a symmetric Jacobi matrix


a0 b0 0 0 · · ·
b0 a1 b1 0 · · ·


 0 b a b · · ·
1
2
2

J=


 0 0 b2 a3 . . . 


..
..
..
..
..
.
.
.
.
.
These sequences determine the orthonormal polynomials Pn , n ≥ 0
via the three-term recurrence relation
xPn (x ) = bn Pn+1 (x ) + an Pn (x ) + bn−1 Pn−1 (x ), n ≥ 0,
(3trl)
with the initial conditions P−1 (x ) = 0, P0 (x ) = 1.
The polynomials of the second kind (Qn (x )) satisfy (3trl) with
Q−1 (x ) = −1, Q0 (x ) = 0, b−1 = 1.
Christian Berg
Indeterminate moment problems
Characterizations of indeterminacy
The following is a classical result
Theorem
Given the data (sn ), (an ), (bn ) the following conditions are
equivalent:
P∞
2
2
(i)
n=0 Pn (0) + Qn (0) < ∞,
P∞
2 1/2
(ii) P (z ) =
< ∞, z ∈ C.
n=0 |Pn (z )|
(iii)
J has deciency indices (1, 1) in `2 .
If (i)(iii) hold (the indeterminate case), then
P∞
2 1/2
Q (z ) =
< ∞ for z ∈ C, and the series for
n=0 |Qn (z )|
P , Q are uniformly convergent on compact subsets of C.
In the following we P
will only discuss such indeterminate cases,
hence by Carleman (1/bn ) < ∞.
Christian Berg
Indeterminate moment problems
Nevanlinna matrices 1
The following polynomials
An (z ) = z
n −1
X
k=
Qk (0)Qk (z ),
0
Bn (z ) = −1 + z
Cn (z ) = 1 + z
Dn (z ) = z
n−1
X
k=
n−1
X
k=
n−1
X
k=
Qk (0)Pk (z ),
0
Pk (0)Qk (z ),
0
Pk (0)Pk (z ).
0
tend to entire functions denoted A, B , C , D, when n tends to
innity. They are needed in the Nevanlinna parametrization of all
the solutions to the indeterminate moment problem.
Christian Berg
Indeterminate moment problems
Nevanlinna matrices 2
Marcel Riesz (1923): A, . . . , D have minimal exponential type.
Berg and Pedersen (1994): A, . . . , D , P , Q have the same order ρ,
type τ and Phragmén-Lindelöf indicator called order, type and
indicator of the moment problem.
By M. Riesz: ρ ≤ 1 and if ρ = 1, then τ = 0.
Berg and Pedersen (2005): If ρ = 0 then A, . . . , D , P , Q have the
same logarithmic order ρ[1] and logarithmic type τ [1] called
logarithmic order and logarithmic type of the moment problem.
The logarithmic order and type of a function f (of order 0) is
dened in terms of its maximum modulus Mf as
ρf = inf {α > 0 | Mf (r ) ≤as r (log r ) }
α
[1]
[1]
τf
[1]
ρ
= inf {c > 0 | Mf (r ) ≤as r c (log r ) }.
f
Christian Berg
Indeterminate moment problems
Nevanlinna matrices 3
The rst explicit examples of Nevanlinna matrices were calculated
around 1994:
Ismail-Masson: Order 0, logarithmic order 1. The functions are
related to theta-functions.
Berg-Valent: Order 1/4. The functions are related to trigonometric
functions of order 4.
Main Question: Is it possible to calculate the order of the
momentproblem directly from the recurrence coecients or the
moments?
Christian Berg
Indeterminate moment problems
Some preliminary results
Proposition
Consider an indeterminate moment problem corresponding to
sequences (an ), (bn ) from the three-term recurrence relation. If
another moment problem is given in terms of (ãn ), (b̃n ), and if
ãn = an , b̃n = bn for n ≥ n0 , then the second problem is also
indeterminate and the two problems have the same order, type and
Phragmén-Lindelöf indicator function.
Proposition
Consider an indeterminate Hamburger moment problem of order ρ,
type τ and indicator h corresponding to the sequences (an ), (bn )
from the three-term recurrence relation. For c > 0 the moment
problem corresponding to the sequences (can ), (cbn ) is also
indeterminate with order ρ(c ) = ρ, type τ (c ) = τ /c ρ and indicator
h(c )(θ) = h(θ)/c ρ .
Christian Berg
Indeterminate moment problems
The Liv²ic function
In 1939 Liv²ic considered the function
F (z ) =
∞
X
z 2n
n= s n
0
2
which is entire of minimal exponential type for any indeterminate
√
problem because lim n/ 2 s2n = 0. This holds by Carleman's
criterion giving that
∞
X
√
1/ 2 s2n < ∞.
n
n
√
n=
0
Moreover,
s2n is increasing for n ≥ 1.
Liv²ic proved that ρF ≤ ρ, where ρ is the order of the moment
problem.
We shall discuss the Question: Is ρF = ρ?
2n
Christian Berg
Indeterminate moment problems
The Liv²ic function 2
Berg-Szwarc, Adv. Math. 2014: The answer is yes under certain
"regularity" conditions.
Pruckner-Romanov-Woracek, ArXiv, Dec. 2015 Examples where
ρF < ρ.
The results of Berg-Szwarc build on renements/extensions of work
by Berezanski (1956), where (bn ) is assumed log-concave, i.e.,
bn2 ≥ bn−1 bn+1 .
Denition: We say that (bn ) is regular if (bn ) is either eventually
log-convex, i.e., bn ≤ bn− bn+ for n ≥ n , or eventually
log-concave, i.e. bn ≥ bn− bn+Pfor n ≥ n .
Examples of regular (bn ) with (1/bn ) < ∞:
α
bn = nα , bn = n logα n, α > 1, bn = an , a > 1, α > 0.
2
2
1
1
1
1
Christian Berg
0
0
Indeterminate moment problems
Results from Berg-Szwarc: Adv. Math. 2014
Denition: For a sequence (zn ) of complex numbers for which
|zn | → ∞, we introduce the exponent of convergence
(
)
∞
X
1
E(zn ) = inf α > 0 |
<∞ ,
α
n=n∗ |zn |
where n∗ ∈ N is such that |zn | > 0 for n ≥ n∗ .
Theorem
Assume that (bn ) is regular and that
∞
X
1 + |an |
p
< ∞.
n=1 bn bn−1
Then the order of the moment problem is given by ρ = E(bn ).
If the order is ρ = 0, then the logarithmic order of the moment
problem is ρ[1] = E(log bn ).
Christian Berg
Indeterminate moment problems
Some other relevant entire functions
Write
Pn (x ) =
n
X
k=
bk ,n x k ,
0
hence bn,n = 1/(b0 b1 · · · bn−1 ). Consider the entire functions
H (z ) =
∞
X
n=
b n ,n z n ,
G (z ) =
0
∞
X
zn
in addition to the Livsic functions F (z ) =
.
n=
z n /s n .
0
P
bnn
2
2
Theorem
Under the same conditions as before
ρ = ρF = ρG = ρH .
If ρ = 0, then
[1]
[1]
[1]
ρ[1] = ρF = ρG = ρH .
Christian Berg
Indeterminate moment problems
Some remarks
1. Consider the symmetric case : an ≡ 0 and assume (bn ) regular:
Then
∞
∞
X
X
1
1 + |an |
p
< ∞ ⇐⇒
< ∞( Carleman condition)
n=0 bn
n=1 bn bn−1
so in this case the Carleman condition is equivalent to
indeterminacy.
2. For an arbitrary indeterminate problem dene
ck =
∞
X
n=k
!1/2
bk2,n
,
Φ(z ) =
∞
X
k=
ck z k .
0
The order and type of Φ is always equal to the order and type of
the moment problem.
Christian Berg
Indeterminate moment problems
Symmetric indeterminate moment problems
Theorem (Berg-Szwarc, Math. Sbornik 2017)
Consider a symmetric indeterminate Hamburger moment problem
vn = P22n (0),
un = Q22n−1 (0),
n ≥ 1.
Assume that
(un ) ∈ `α , (vn ) ∈ `β , where 0 < α, β ≤ 1.
Assume also that there exist constants C , D > 0 such that
(i)
(ii)
(iii)
β/α
un ≤ Cvn
, n ≥ 1.
vj ≤ Dvi for i ≤ j.
Let γ be the harmonic mean of α and β , i.e.,
γ −1 =
1 −1
α + β −1 .
2
Then the order ρ of the moment problem is ≤ γ .
Christian Berg
Indeterminate moment problems
Proof: Product representation of the Nevanlinna matrix
Using ideas of Barry Simon, the Nevanlinna matrix can be given in
the symmetric case as
#
"Y
∞
A(z ) B (z )
0 −1
=
(I − zVn )(I + zUn )
.
C (z ) D (z )
1 z
n=
1
where the product expands to the left and
0 0
0 un
Vn =
, Un =
.
vn 0
0 0
We observe that Um Un = Vm Vn = 0 and
0
0
un vm 0
Un Vm =
, Vn Um =
.
0
0
0 vn um
We want to analyse the innite product and its power series
∞
∞
Y
X
m11 (z ) m12 (z )
M (z ) =
:=
(I −zVn )(I +zUn ) = I +
Mn z n .
m21 (z ) m22 (z )
n=
n=
1
Christian Berg
Indeterminate moment problems
1
Further analysis
We nd
M1 =
−
P∞
0
P∞
k = vk
1
k = uk
0
1
and for n ≥ 1
M2n = (−1)n
X
≤k1 ≤k2 <...<k2n−1 ≤k2n
Vk2 Uk2
1
· · · Vk2 Uk1
Uk2 Vk2
1
· · · Uk2 Vk1 ,
n
n−
1
+ (−1)n
X
≤k1 <k2 ≤...≤k2n−1 <k2n
n
n−
1
because we shall consider all possible choices of 2n parentheses,
where we select the term containing z. Since the product of two
consecutive U 0 s or V 0 s gives zero, we have to alternate between
U 0 s and V 0 s, and this also determines the inequalities between the
indices. We see that M2n is a diagonal matrix. There is a similar
expression for M2n+1 showing that the diagonal elements vanish.
Christian Berg
Indeterminate moment problems
The function
C
from the Nevanlinna matrix
Let us focus on the lower right corner of M2n . This matrix entry is
equal to
X
a2n = (−1)n
uk1 vk2 · · · uk2 −1 vk2 ,
≤k1 ≤k2 <...<k2n−1 ≤k2n
n
n
1
Furthermore,
M (z ) =
A(z )z − B (z ) A(z )
C (z )z − D (z ) C (z )
Proposition
Consider a symmetric indeterminate Hamburger moment problem
and let vn = P22n (0), un = Q22n−1 (0). The order, type and
Phragmén-Lindelöf indicator of the moment problem is equal to the
order, type and Phragmén-Lindelöf indicator of the function
m22 (z ) = 1 +
∞
X
n
a2n z 2n = C (z ),
Christian Berg =1 Indeterminate moment problems
Multi-zeta values
For real s > 1 and n ≥ 1 dene the special multi-zeta value
X
γn (s ) =
(k1 k2 . . . k2n−1 k2n )−s .
≤k1 ≤k2 <...<k2n−1 ≤k2n
1
Observe that the inequalities between the indices kj are alternating
between ≤ and <: k2j −1 ≤ k2j < k2j +1 ≤ k2j +2 .
We shall consider the entire function
∞
X
Gs (z ) = 1 +
γn (s )z n .
n=
1
We shall se that ρG = 1/(2s ). Call its type Ts = τG .
The canonical product
s
Ps (z ) =
s
∞ Y
n=
1
z 1+ s ,
n
1 < s, z ∈ C
has order ρP = 1/s and type τP = π/ sin(π/s ), cf. Boas.
s
s
Christian Berg
Indeterminate moment problems
Multi-zeta values 2
Choosing k1 = k2 = l1 , k3 = k4 = l2 , . . . k2n−1 = k2n = ln we get
X
γn (s ) >
(l1 l2 . . . ln )−2s ,
≤l1 <l2 <...<ln
1
which is an ordinary multi-zeta value, showing that for r > 0
∞ ∞
∞ Y
X
Y
r2
r 2
1 + 2s < 1 +
γn (s )r 2n <
1+ s
n
n
n=
n=
1
1
i.e.,
n=
1
P2s (r 2 ) < Gs (r 2 ) < Ps (r )2 .
The rst and third function have order 1/s and so does Gs (z 2 ).
Therefore ρG = 1/(2s ). Since the type of Gs (z ) and Gs (z 2 ) are
the same called Ts , we get
s
π
2π
π
≤ Ts ≤
=
.
sin(π/(2s ))
sin(π/s )
sin(π/(2s ) cos(π/(2s ))
Christian Berg
Indeterminate moment problems
Symmetric indeterminate moment problems II
Theorem (Berg-Szwarc, Math. Sbornik 2017)
Consider a symmetric indeterminate Hamburger moment problem
vn = P22n (0) = c1 n−1/β (1+δn ),
un = Q22n−1 (0) = c2 n−1/α (1+εn ),
where c1 , c2 > 0, 0 < α, β < 1, |δn |, |εn | ≤ K /n for some constant
K > 0 independent of n. Let γ be the harmonic mean of α, β .
Then the order ρ and type τ of the moment problem is the same as
the order and type of the function
∞
X
n=
γn (1/γ)(c1 c2 )n z 2n = G1/γ (c1 c2 z 2 )
1
hence
ρ = γ,
τ = (c1 c2 )γ/2 T1/γ ,
where T1/γ is the type of G1/γ .
Christian Berg
Indeterminate moment problems
Birth and death processes with polynomial rates
Birth and death processes lead to Stieltjes moment problems
according to a theory developed by Karlin and McGregor. The
theory depends on two sequences (λn ), (µn ) of birth and death
rates with the constraints λn , µn+1 > 0, n ≥ 0 and µ0 = 0.
The recurrence coecients (an ), (bn ) of the corresponding Stieltjes
moment problem are given by
p
an = λn + µn , bn = λn µn+1 , n ≥ 0.
In a series of papers Valent and his co-authors studied Stieltjes
moment problems coming from birth and death processes with
polynomial rates of degree p:.
Christian Berg
Indeterminate moment problems
Valent's polynomial rates of degree
and
p
λn = (pn + e1 ) . . . (pn + ep ),
n ≥ 0,
µn = (pn + d1 ) . . . (pn + dp ),
n ≥ 0,
where it is assumed that
0 < e1 ≤ e2 ≤ . . . ≤ ep ,
−p < d1 ≤ d2 ≤ . . . ≤ dp ,
d1 . . . dp = 0,
so one has λn , µn+1 > 0 for n ≥ 0 and µ0 = 0.
By introducing the quantities
E = e1 + · · · + ep ,
D = d1 + · · · + dp ,
Valent proved that the Stieltjes problem is indeterminate if and
only if
E −D
1<
< p − 1.
p
Christian Berg
Indeterminate moment problems
Valent's conjectures
Assume p > 2: The indeterminate cases.
In the special cases p = 3, 4 and for special values of
ej , dj , j = 1, . . . , p, it has been possible to calculate the Nevanlinna
matrices using elliptic functions. On the basis of these calculations
Valent (1998) formulated the following conjecture:
Valent's conjecture For the indeterminate Stieltjes moment
problem with indeterminate rates as above (hence p ≥ 3), the
order, type and Phragmén-Lindelöf indicator function are given as
Z 1
du
, h(θ) = τ cos((θ−π)/p ), θ ∈ [0, 2π].
ρ = 1/p , τ =
(1 − u p )2/p
0
(1)
Recently, Romanov (TAMS 2017) has proved Valent's conjecture
concerning order using the powerful theory of canonical systems.
Christian Berg
Indeterminate moment problems
The symmetric problem corresponding to Valent's rates
The sequence (bn ) in this case is
q
p
b2n = λn = (pn + e1 ) . . . (pn + ep )
b2n+1 =
√
µn+1
P2n (0) = (−1)n
q
= (p (n + 1) + d1 ) . . . (p (n + 1) + dp ).
b0 b2 . . . b2n−2
,
b1 b3 . . . b2n−1
Q2n−1 (0) = (−1)n
b1 b3 . . . b2n−3
b0 b2 . . . b2n−2
Then one nds
vn = P2n (0) = c1 n
2
−[p −(E −D )/p ]
(1 + δn ),
c1 :=
p
Y
Γ(1 + dj /p )
Γ(ej /p )
j=
1
un = Q2n−1 (0) = c2 n
2
−(E −D )/p
(1+εn ),
c2 := p
−p
p
Y
j=
1
Christian Berg
Γ(ej /p )
Γ(1 + dj /p )
Indeterminate moment problems
Application of the theorems of Berg-Szwarc
We have
β = (p − (E − D )/p )−1 ,
α = p /(E − D ),
c1 c2 = p −p
The harmonic mean of α and β is γ = 2/p.
The symmetric problem corresponding to the Valent rates:
Order ρ = 2/p, type τ = (1/p )Tp/2
The Stieltjes problem corresponding to the Valent rates:
Order ρ = 1/p, type τ = (1/p )Tp/2 .
From the inequalities for the type Ts of Gs we get
π
π
≤τ ≤
.
p sin(π/p )
p sin(π/p ) cos(π/p )
We have not been able to prove as conjectured by Valent
Z 1
du
τ=
(1 − u p )2/p
0
However, the value of the integral lies between the bounds above.
The conjecture about type is asymptotically correct as p → ∞.
Christian Berg
Indeterminate moment problems