Holomorphic functions associated with indeterminate ...nalag.cs.kuleuven.be/papers/ade/growth2/slides.pdf · Holomorphic functions associated with indeterminate rational moment problems

Holomorphic functions associated with indeterminaterational moment problems

Adhemar Bultheel1, Erik Hendriksen2, Olav Njastad3

1Dept. Computer Science, KU Leuven2Nieuwkoop, The Netherlands

3Mathematics, Univ. Trondheim

Puerto de la Cruz, Tenerife, January 2014

This is dedicated to the memory of Pablo Gonzalez Vera.

http://nalag.cs.kuleuven.be/papers/ade/growth2

Bultheel, Hendriksen, Njastad (KU Leuven)Indeterminate rational moment problem Tenerife January 2014 1 / 15

Survey

Rational Hamburger moment problem ∞⇒ 1/αkSolution µ ⇔ Nevanlinna functions Ωµ(z)

All solutions µ⇔ Ωµ(z) = A(z)g(z)+B(z)C(z)g(z)+D(z) , g ∈ N

Asymptotic behaviour of A,B,C,D near singularities (αk) is thesame


Classical Hamburger moment problem

Definition (Hamburger MP)

Given M HPD functional on polynomials P by M[tk ] = ck , k = 0, 1, ...Find pos. measure µ on R such that

∫tkdµ(t) = ck , k = 0, 1, ...

Set of all solutions = M.MP is (in)determinate if solution is (not) unique.

Definition (Nevanlinna function)

N = f : f ∈ H(U) & f : U→ U = U ∪ R.Example Sµ(z) =

∫ dµ(t)t−z for µ ∈M.

Expand (t − z)−1 to think of Sµ as a moment generating fct:

Sµ(z) ∼ −1

z

∞∑k=0

ckzk.





∫tkdµ(t) = ck , k = 0, 1, ...






Sµ(z) ∼ −1

z

∞∑k=0

ckzk.





∫tkdµ(t) = ck , k = 0, 1, ...






Sµ(z) ∼ −1

z

∞∑k=0

ckzk.



Theorem (Nevanlinna)

There is a 1-1 relation between N and M given by

Sµ(z) = − a(z)f (z)− c(z)

b(z)f (z)− d(z), f ∈ N

with a, b, c , d entire functions.

Theorem (M. Riesz)

If F ∈ a, b, c , d then for any ε > 0

|F (z)| ≤ M(ε) expε|z |.

(controls growth as z →∞)



Theorem (Nevanlinna)

There is a 1-1 relation between N and M given by

Sµ(z) = − a(z)f (z)− c(z)

b(z)f (z)− d(z), f ∈ N

with a, b, c , d entire functions.

Theorem (M. Riesz)

If F ∈ a, b, c , d then for any ε > 0

|F (z)| ≤ M(ε) expε|z |.

(controls growth as z →∞)


Rational moment problem

rational case

How to generalize this to the case of rational moments?I.e. when polynomials P is replaced by rationals L

L =Pπ∞

, with π∞(z) =∏α∈A

(1− z

α)

Here a finite set Γ of different α’s in R \ 0 = (R ∪ ∞) \ 0.But each α ∈ Γ is has infinite multiplicity. Then L · L = LIf Γ = ∞ then L = P. Here for notational reason α 6=∞.

This presentation

We shall look in particular at the growth of the a, b, c , d near thesingularities α ∈ Γ.



rational case


L =Pπ∞


(1− z

α)


This presentation




rational case


L =Pπ∞


(1− z

α)


This presentation




rational case


L =Pπ∞


(1− z

α)


This presentation




rational case


L =Pπ∞


(1− z

α)


This presentation




rational case


L =Pπ∞


(1− z

α)


This presentation



Orthogonal rational functions

wlog: α0 =∞, α1, . . . , αq︸︷︷︸Γ

, αq+1 = α1, . . . , α2q = αq︸︷︷︸Γ

, . . .

π0 = 1, πn =∏n

k=1(1− zαk

), bn(z) = zn

πn(z) , n = 1, 2, . . .

Ln = spanbk(z), k = 0, ...n =

pn(z)

πn(z), pn ∈ Pn

Definition (Rational moment problem)

Given M HPD functional on L by ck = M[bk ], k = 0, 1, ....Find pos. measure µ such that ck =

∫bk(t)dµ(t), k = 0, 1, ...


D(t, z) =1 + tz

t − z⇒ Ωµ(z) =

∫D(t, z)dµ(t) ∈ N



wlog: α0 =∞, α1, . . . , αq︸︷︷︸Γ

, αq+1 = α1, . . . , α2q = αq︸︷︷︸Γ

, . . .

π0 = 1, πn =∏n

k=1(1− zαk

), bn(z) = zn

πn(z) , n = 1, 2, . . .

Ln = spanbk(z), k = 0, ...n =

pn(z)

πn(z), pn ∈ Pn



∫bk(t)dµ(t), k = 0, 1, ...


D(t, z) =1 + tz

t − z⇒ Ωµ(z) =




wlog: α0 =∞, α1, . . . , αq︸︷︷︸Γ

, αq+1 = α1, . . . , α2q = αq︸︷︷︸Γ

, . . .

π0 = 1, πn =∏n

k=1(1− zαk

), bn(z) = zn

πn(z) , n = 1, 2, . . .

Ln = spanbk(z), k = 0, ...n =

pn(z)

πn(z), pn ∈ Pn



∫bk(t)dµ(t), k = 0, 1, ...


D(t, z) =1 + tz

t − z⇒ Ωµ(z) =




wlog: α0 =∞, α1, . . . , αq︸︷︷︸Γ

, αq+1 = α1, . . . , α2q = αq︸︷︷︸Γ

, . . .

π0 = 1, πn =∏n

k=1(1− zαk

), bn(z) = zn

πn(z) , n = 1, 2, . . .

Ln = spanbk(z), k = 0, ...n =

pn(z)

πn(z), pn ∈ Pn



∫bk(t)dµ(t), k = 0, 1, ...


D(t, z) =1 + tz

t − z⇒ Ωµ(z) =




〈f , g〉 = M[f (z)g(z)] =∫f (t)g(t)dµ(t), µ ∈M

⇒ ORF: ϕn ∈ Ln \ Ln−1, 〈ϕk , ϕl〉 = δk,l

functions of second kind: ψ0(z) = −z and

ψn(z) =

∫D(t, z)[ϕn(t)− ϕn(z)]dµ(t), n ≥ 1, µ ∈M.

... quadrature formulas, Christoffel-Darboux, (long story) ...

Theorem

RMP indeterminate (determinate) iff

∞∑n=0

|ϕn(z)|2 (and∞∑n=0

|ψn(z)|2)

converge (diverge) for all C \ R.






ψn(z) =



Theorem


∞∑n=0

|ϕn(z)|2 (and∞∑n=0

|ψn(z)|2)







ψn(z) =



Theorem


∞∑n=0

|ϕn(z)|2 (and∞∑n=0

|ψn(z)|2)







ψn(z) =



Theorem


∞∑n=0

|ϕn(z)|2 (and∞∑n=0

|ψn(z)|2)



Rational Nevanlinna parametrization

Lemma

There exists x0 ∈ R \ (Γ ∪ 0) not ‘exceptional’ such that[An(z) Cn(z)Bn(z) Dn(z)

]= (x0 − z)

n−1∑k=0

Mk(z) ∈ L2×2n

M0 =

[−1 D(z , x0)

−D(z , x0) 1

], Mk =

[ψk(z)ϕk(z)

][ψk(x0) ϕk(x0)], k ≥ 1

and the limit n→∞ exists locally uniformly in C \ Γ giving A,B,C ,Dholomorphic with a simple pole at ∞ and essential singularities at thepoints of Γ = α1, . . . , αq.

•∫dµ(t) <∞⇒ simple pole at ∞

• Γ-points repeated ∞ times ⇒ essential singularities



Lemma


]= (x0 − z)

n−1∑k=0

Mk(z) ∈ L2×2n

M0 =

[−1 D(z , x0)

−D(z , x0) 1

], Mk =

[ψk(z)ϕk(z)

][ψk(x0) ϕk(x0)], k ≥ 1






Lemma


]= (x0 − z)

n−1∑k=0

Mk(z) ∈ L2×2n

M0 =

[−1 D(z , x0)

−D(z , x0) 1

], Mk =

[ψk(z)ϕk(z)

][ψk(x0) ϕk(x0)], k ≥ 1






Theorem (Nevanlinna parametrization)

Ωµ(z) = −A(z)g(z)− C (z)

B(z)g(z)− D(z)

is a 1-1 relation between all g ∈ N and all µ ∈M.

Theorem (Rational Riesz theorem)

If F ∈ A,B,C ,D then for any ε > 0 near a singularity α ∈ Γ

|F (z)| ≤ M(ε) exp

ε

|z − α|

∀z ∈ V ′α (a pointed disk centered at α excluding all other α’s)



Theorem (Nevanlinna parametrization)

Ωµ(z) = −A(z)g(z)− C (z)

B(z)g(z)− D(z)

is a 1-1 relation between all g ∈ N and all µ ∈M.

Theorem (Rational Riesz theorem)

If F ∈ A,B,C ,D then for any ε > 0 near a singularity α ∈ Γ

|F (z)| ≤ M(ε) exp

ε

|z − α|

∀z ∈ V ′α (a pointed disk centered at α excluding all other α’s)


Order and type

Definition (order and type at α ∈ Γ)

Mα(F , r) = max|z−α|=r

|F (z)|,

then order is defined by

ρα(F ) = infλ : Mα(F , r) ≤ expr−λ, r small

and type is defined by

σα(F ) = infs : Mα(F , r) ≤ expsr−ρα(F ), r small

Theorem

For F ∈ A,B,C ,D and α ∈ Γ

either (1) ρα(F ) < 1 or (2) ρα(F ) = 1 and σα(F ) = 0.


Order and type

Definition (order and type at α ∈ Γ)

Mα(F , r) = max|z−α|=r

|F (z)|,

then order is defined by

ρα(F ) = infλ : Mα(F , r) ≤ expr−λ, r small

and type is defined by

σα(F ) = infs : Mα(F , r) ≤ expsr−ρα(F ), r small

Theorem

For F ∈ A,B,C ,D and α ∈ Γ

either (1) ρα(F ) < 1 or (2) ρα(F ) = 1 and σα(F ) = 0.


Two solutions

Define 2 solutions: µ0 and µ∞ corresponding to g =∞ and g = 0

A(z)

B(z)= −Ωµ∞(z) and

C (z)

D(z)= −Ωµ0(z), for z ∈ C \ Γ

suppµ∞ = Γ ∪ poles of A/B; zeros of A and B simple and interlacesuppµ0 = Γ ∪ poles of C/D; zeros of C and D simple and interlace

thussuppµ∞ = Γ ∪ zeros of B & suppµ0 = Γ ∪ zeros of D

But also the zeros of B and D interlace.

Hence µ∞ and µ0 have common support Γ butall other points in the support interlace.


Two solutions


A(z)


C (z)

D(z)= −Ωµ0(z), for z ∈ C \ Γ






Two solutions


A(z)


C (z)

D(z)= −Ωµ0(z), for z ∈ C \ Γ






Two solutions


A(z)


C (z)

D(z)= −Ωµ0(z), for z ∈ C \ Γ






Two solutions


A(z)


C (z)

D(z)= −Ωµ0(z), for z ∈ C \ Γ






Accumulation points

Each α ∈ Γ is an accumulation point in suppµ∞ and in suppµ0.

Define zFj : j = 1, 2, ... zeros of F and

zFα,j : j = 1, 2, ... (disjoint) subsequence that converges to α ∈ Γ.

The latter exist for F ∈ B,D becausezBj ⊂ suppµ∞ and zDj ⊂ suppµ0

but also for F ∈ A,C by interlacing property.

How do the zFα,j converge to α ∈ Γ?


Accumulation points








Accumulation points








Accumulation points








Accumulation points

Definition (cvg exp and genus at α ∈ Γ)

order zFα,j such that |zFα,j − α| non-increasingthen convergence exponent is defined by

τα(F ) = inft ∈ R :∑∞

j=1 |zFα,j − α|t <∞and genus is defined by

κα(F ) = maxt ∈ Z :∑∞

j=1 |zFα,j − α|t =∞

Theorem

For α ∈ Γ: τα(A) = τα(B) = τα(C ) = τα(D)and κα(A) = κα(B) = κα(C ) = κα(D).


Accumulation points

Definition (cvg exp and genus at α ∈ Γ)

order zFα,j such that |zFα,j − α| non-increasingthen convergence exponent is defined by

τα(F ) = inft ∈ R :∑∞

j=1 |zFα,j − α|t <∞and genus is defined by

κα(F ) = maxt ∈ Z :∑∞

j=1 |zFα,j − α|t =∞

Theorem

For α ∈ Γ: τα(A) = τα(B) = τα(C ) = τα(D)and κα(A) = κα(B) = κα(C ) = κα(D).


Factorization

Theorem

F ∈ A,B,C ,D then F (z) = R(z)∏α∈Γ Fα(z), Fα(z) = Pα(z)Qα(z).

R(z) rational with zeros and poles in Γ.

Qα ∈ H(C \ α) without zeros: Qα = eq( 1z−α

), ∂q ≤ 1

Pα ∈ H(C \ α) is canonical Hadamard productI.e., catches all the zeros zFα,j.

Pα(z) =∞∏j=1

(1−zFα,j − αz − α

)exp

γα(F )∑k=1

1

k

(zFα,j − αz − α

)k

γα(F ) = bρα(F )c here γα(F ) ∈ 0, 1


Equality of orders

For α ∈ Γ:

ρα(F ) completely defined by Fα: ρα(F ) = ρα(Fα) = τα(Fα)

Eventually leads to

Theorem (Equality of orders)

For α ∈ Γ it holds that

ρα(A) = ρα(B) = ρα(C ) = ρα(D)


Equality of orders

For α ∈ Γ:

ρα(F ) completely defined by Fα: ρα(F ) = ρα(Fα) = τα(Fα)

Eventually leads to

Theorem (Equality of orders)

For α ∈ Γ it holds that

ρα(A) = ρα(B) = ρα(C ) = ρα(D)


Holomorphic functions associated with indeterminate ...nalag.cs.kuleuven.be/papers/ade/growth2/slides.pdf · Holomorphic functions associated with indeterminate rational moment problems

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