On Linear Combinations of Two Orthogonal Polynomial Sequences on the Unit Circle

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doi:10.1155/2010/406231

Research Article

On Linear Combinations of Two Orthogonal

Polynomial Sequences on the Unit Circle

C. Su ´arez

Departamento de Matem´atica Aplicada I, E.T.S.I.I., Universidad de Vigo, Campus Lagoas-Marcosende, 36200 Vigo, Spain

Correspondence should be addressed to C. Su´arez,[email protected] Received 1 August 2009; Revised 1 December 2009; Accepted 5 March 2010 Academic Editor: Panayiotis Siafarikas

Copyrightq2010 C. Su´arez. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Let{Φn}be a monic orthogonal polynomial sequence on the unit circle. We define recursively

a new sequence{Ψn}of polynomials by the following linear combination:Ψnz pnΨn−1z

Φnz qnΦn−1z,pn, qn∈C,pnqn/0. In this paper, we give necessary and suﬃcient conditions

in order to make{Ψn}be an orthogonal polynomial sequence too. Moreover, we obtain an explicit

representation for the Verblunsky coeﬃcients{Φn0}and{Ψn0}in terms ofpnandqn. Finally,

we show the relation between their corresponding Carath´eodory functions and their associated linear functionals.

1. Notation and Preliminary Results

We recall some definitions and general results about orthogonal polynomials on the unit circle

OPUC. They can be seen in1–3.

Along this paper, we will use the following notations. We denote byΛ span{zk_{, k}_∈ Z} the linear space of Laurent polynomials with complex coeﬃcients and by Λ the dual algebraic space ofΛ. LetPspan{zk_{, k}_∈_N}_{be the space of complex polynomials.}

Definition 1.1. Letu∈Λ. Denoting byunuzn,n∈Z, we say that iuis Hermitian if for alln≥0,u₋nun;

iiuis regular or quasidefinitepositive definiteif the principal minors of the moment matrix are nonsingularpositive, that is,

∀n≥0, Δndetuzi−j

i0···n;j0···n/0 >0. 1.1

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The sequence{un}is said to be the sequence of the moments associated withu. Furthermore, ifuis a positive definite linear functional then a finite nontrivial positive Borel measureμsupported on the unit circle exists such that

uPz 1

2π

₂_π

0

Peiθdμ, P ∈Λ. 1.2

Definition 1.2. Let{Φnz}₀∞be a complex polynomial sequence with degΦnz n. We say

that{Φnz}₀∞is a sequence of orthogonal polynomialsOPSswith respect to the linear and Hermitian functionaluif

∀n, m≥0, u

ΦnzΦm

1

z

enδnm withen/0. 1.3

In the sequel, we denote by{Φn}the monic orthogonal polynomial sequenceMOPS associated withu.

For simplicity, along this paper we also assume thatuis normalizedi.e.,u0 1. It is well known that the regularity ofuis a necessary and suﬃcient condition for the existence of a sequence of orthogonal polynomials on the unit circle. On the other hand, the polynomials

Φnsatisfy the so-calledSzeg¨o recurrence relations

∀n≥1, Φnz zΦn₋1z Φn0Φ∗n−1z, 1.4

∀n≥1, Φ∗_nz Φ∗_n₋₁z Φn0zΦn−1z, 1.5

∀n≥1, Φnz 1− |Φn0|2zΦn−1z Φn0Φ∗nz, 1.6

∀n≥1, Φ∗_nz 1− |Φn0|2Φ∗_n₋₁z Φn0Φnz, 1.7

whereΦ∗_nz zn_Φn₁_/z_{is the reversed polynomial of}_Φn_z_,_n_≥_0.

Definition 1.3. Given an MOPS{Φn}, the sequence of kernels of parametery ∈Cassociated

with the linear functionaluis defined by

∀n≥0, Kn

z, y n

j0 Φjy

ej Φj

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This sequence verifies the following properties:

∀n≥0, Kn

z, y 1 en

⎛

⎝Φ∗nzΦ∗n

y−zyΦnzΦny

1−zy

⎞

⎠_, _1.9

∀n≥0, Kn

z, y 1 en1

⎛

⎝Φ∗n1zΦ∗n1

y−Φn1zΦn1

y

1−zy

⎞

⎠_, _1.10

∀n≥0, Kn

z, yKn

y, z Φ∗_nz enKnz,0, 1.11

∀n≥1, Kn

z, yzyKn−1

z, yΦ ∗ ny en Φ

∗

nz. 1.12

To the linear functionaluwe can associate a formal seriesFuas follows:

Fuz 12 ∞

n1

unzn. 1.13

In the positive definite case,Fuis called the Carath´eodory function associated withu. In this

case,Fucan be written as

Fuz 1

2π

0

eiθ_z

eiθ−_zdμθ, |z|<1. 1.14

The measuredμcan be reconstructed fromFuby means of the inversion formula. The aim of

this paper is the analysis of the following problem. Given an MOPS on the unit circle{Φn}, orthogonal with respect to a linear functionalu, to find necessary and suﬃcient conditions in order to make a sequence of monic polynomials{Ψn}defined by

∀n≥1, Φnz qnΦn−1z Ψnz pnΨn−1z withpnqn/0 1.15

an MOPS with respect to a linear functionalL. Further more, to find the relation between the linear functionalsLanduand their corresponding Carath´eodory functions.

Many authors have dealt with this kind of problems. In the constructive theory of orthogonal polynomials they have been called inverse problems. Concretely, an inverse problem for linear functionals can be stated as follows. Given two sequences of monic polynomials {Φn} and {Ψn}, to find necessary and suﬃcient conditions in order to make

{Ψn}an MOPS when{Φn}is a MOPS and they are related by

FΦn, . . . ,Φn−l GΨn, . . . ,Ψn−k, 1.16

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For instance, this subject has been treated in 4–6 in the context of the theory of orthogonal polynomials on the real line. For orthogonal polynomials with respect to measures supported on the unit circle, in7there have been relevant results.

The structure of this paper is the following. In Section 2 we give the necessary conditions in order to be sure that the problem 1.15 admits a nontrivial solution. In

Section 3, we prove a suﬃcient condition and we obtain the explicit solution in terms ofpn

andqn.Section 4is devoted to find the functional relation betweenLandu. Finally,Section 5

contains the rational relation between the corresponding Carath´eodory functions.

2. Necessary Conditions

Let {Φn}_n_≥₀ be a monic orthogonal polynomial sequence and let {Ψn}_n_≥₀ be a monic polynomial sequence. We assume that there exist sequences of complex numbers{qn}_n_≥₂and

{pn}_n_≥₂such that the following relation holds:

∀n≥2, Φnz qnΦn₋1z Ψnz pnΨn−1z. 2.1

Also, we assumeΦ0z Ψ0z 1 andΦ1z z−q1andΨ1z z−p1, with|q1|/1 and |p1|/1.

In this section, we find some necessary conditions in order to make the sequence{Ψn} defined recursively from{Φn}by relation2.1an MOPS.

With this aim, we define the complex numbersNn1andDn1as follows:

∀n≥1, Nn1 Φn10 qn1Φn0, 2.2

∀n≥1, Dn1qn1−qn Φn0Φn10. 2.3

The following proposition justifies this choice.

Proposition 2.1. Let{Ψn}_n_≥₀be the monic sequence given as in2.1. If{Ψn}_n_≥₀is an MOPS, then the following relations hold:

∀n≥1, Nn1 Ψn10 pn1Ψn0, 2.4

∀n≥1, Dn1pn1−pn Ψn0Ψn10. 2.5

Moreover,

∀n≥1, zDn1Φn−1z−Ψn−1z Nn1

Ψ∗

n−1z−Φ∗n−1z

, 2.6

∀n≥1, Dn2

qnΦn₋1z−pnΨn−1z

Nn2

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Proof. From1.4together with the definition ofΨn2.1, we have

∀n≥1, Ψn1z Φn1z qn1Φnz−pn1Ψnz zqn1

Φnz Φn10Φ∗nz−pn1Ψnz

zΨnz−zqnΦn−1zzpnΨn−1zqn1ΦnzΦn10Φ∗nz−pn1Ψnz. 2.8

Since{Ψn}is a MOPS, using1.4we have

Ψn10Ψ∗nz −zqnΦn−1z zpnΨn−1z qn1Φnz Φn10Φ∗nz−pn1Ψnz. 2.9

That is,

pn1Ψnz Ψn10Ψ∗nz−zpnΨn−1z qn1Φnz Φn10Φ∗nz−zqnΦn−1z.

2.10

Using1.4and1.5in both sequences, we get

∀n≥1, pn1−pn Ψn10Ψn0

zΨn₋1z

Ψn10 pn1Ψn0

Ψ∗ n−1z

qn1−qn Φn10Φn0

zΦn₋1z

Φn10 qn1Φn0

Φ∗ n−1z.

2.11

Takingz0 we have2.4. Observe that, this is as same as2.1forz0.

Identifying the coeﬃcients of degreen, then 2.5 holds. Therefore, we can rewrite

2.11as2.6.

On the other hand, applying the∗-operator in2.1we have Ψ∗_nz pnzΨ∗_n₋₁z Φ∗

nz qnzΦ∗_n₋₁z. Substituting in2.6and using2.1, we obtain2.7.

In the sequel, we denote byuthe linear regular functional associated with{Φn}and byLthe linear regular functional associated with{Ψn}. Besides, we denote byEn the real

number such thatEnLΨnzz−nwithE01. Therefore,En/En−11− |Ψn0|2.

Corollary 2.2. Under the same conditions as inProposition 2.1, the following assertions hold:

iIfp1q1, thenpnqnandΦnz Ψnz, for alln≥1,

iiIfp1/q1andpnqn/0, for alln≥2, thenΦnz/ Ψnz, for alln≥1,

iiiAssumep1/q1 andpnqn/0, for alln≥ 2, thenNn1/0 if and only ifDn1/0, for all

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Proof. iWe eliminateΨn10using equalities2.2–2.4and2.3–2.5. By doing this, we get

∀n≥1, qn1−qn−

pn1

En En−1 −

pn

Φn10

Ψn0−Φn0qn1Φn0Ψn0. 2.12

Takingn1, we obtainq2−q1−p2E1−p1 Φ20q1−p1 q2q1p1.

Ifp1q1, thenq2−p2|q1|2q2−p2and thusq2−p2e10. Wherefromq2p2. Now, using2.1we haveΨ2z Φ2z.

Proceeding in the same way forn2 we obtainq3−p3e2/e1 0, henceq3 p3and Ψ3z Φ3z, and thus successively.

iiAssume that there exitsn0≥2 such thatΦn0 Ψn0. From2.1, written fornn0, it holds thatqn0pn0and thenΦn0−1 Ψn0−1, and thus successively.

Hence,Ψ1 Φ1, in contradiction with the hypothesis. iiiThe result follows from2.6and the above item.

On the other hand, applying the∗n-operator in2.6, we obtain|Nn1||Dn1|forn≥2.

Remark 2.3. The situationp1 q1 is the trivial case, that is,Ψn Φn, for alln ≥ 1. For this

reason, in the sequel, it will be excluded.

The next result will be used later.

Lemma 2.4. Under the same conditions as inProposition 2.1 together withp1/q1, the following

assertions hold:

∀n≥2, uΨnz −1n1pn· · ·p2

q1−p1

, 2.13

∀n≥2, LΦnz −1n1qn· · ·q2

p1−q1

. 2.14

Proof. Using2.1we obtain

∀n≥2, uΨn −pnuΨn−1. 2.15

Since thatuΨ1z uz−p1 q1−p1, then2.13follows. We obtain2.14changingubyL.

For alln≥1 such thatDn1/0, we define the following complex number:

Tn1

Nn1

Dn1

Φn10 qn1Φn0

qn1−qn Φn10Φn0

. 2.16

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Proposition 2.5. Assume that{Φn}and{Ψn}are two MOPSs that verify 2.1with p1/q1 and

pnqn/0, for alln ≥2. If moreoverDn2/0for alln≥ 1, then the following relation linkingΨnz

andΦnzholds:

∀n≥1,

z−pn1Tn2 pn1Tn3

Ψnz

z− qn1Tn2 pn1Tn3

Φnz Tn2 pn1

pn1−qn1

Φ∗ nz.

2.17

Proof. From2.6and2.7, for alln≥1, we have the system

pn1Dn3Ψnz pn1Nn3Ψ∗nz qn1Dn3Φnz qn1Nn3Φ∗nz,

zDn2Ψnz Nn2Ψ∗nz zDn2Φnz Nn2Φ∗nz.

2.18

The corresponding determinant

∀n≥1,

pn1Dn3 pn1Nn3

zDn2 Nn2

2.19

is not null, sinceDn2/0 together withCorollary 2.2iii. Wherefrom, it has a unique solution forΨn.

By solving this, we get2.17.

In the sequel, we denote by Knz, ythe sequence of the kernels corresponding to

{Ψn}. For the sequence{Φn}we keep the same notations as inSection 1.

Proposition 2.6. Assume that{Ψn} and {Φn} are two MOPS that verify2.1with p1/q1 and

pnqn/0, for alln≥2. Also assumeDn2/0, for alln≥1. Under these conditions, then the following

assertions hold

ipn/qn, for alln≥2,

iiThere exist two complex numbersαandβwith|α||β|1such that

∀n≥2, α pn pn

Tn1

Tn2, 2.20

∀n≥2, β qn qn

Tn1

Tn2

. 2.21

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iiiThe sequences{Φn}and{Ψn}are connected by the following formulas:

Let us proceed with the proof ofii. Inserting

Ψnz

Using the recurrences1.6and1.7in the right-hand side we deduce

∀n≥1, By doing this, we obtain

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where

Combining these relations, we deduce

∀n≥2, pn consequence ofCorollary 2.2iii. On the other hand, the explicit expressions ofT3 and T4 follow from2.7forn1 andn2, respectively.

This completes the proof ofiibecause the complex numberβexits by the symmetry of the problem.

Substituting this relation in2.31and using the recurrences of the kernels1.9and1.10,

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In order to state the converse we need the following assertions.

Proposition 2.7. Under the hypothesis ofProposition 2.6,

∀n≥2, Φn0 Tn1Tn2qn, 2.33

Equating this formula with2.25, previously written forΨn, and applying1.4we get

∀n≥2,

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On the other hand, substituting in Nn1 Ψn10 pn1Ψn0 Φn10 qn1Φn0the

Finally, we use2.42in order to calculate the right-hand side

∀n≥2, −pn

Corollary 2.8. Under the hypothesis ofProposition 2.6,

∀n≥1,

Proof. From1.6it follows that the formula in the statement is equivalent to

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Now, writtenTn2/Tn3in terms ofαas in2.20, the previous relation becomes

qn2−pn2

αpn2 pn2

pn2

qn1−pn1

_p

n1

pn1

αqn1

pn2−qn2

, 2.48

and it is true according to2.42.

3. Some Solutions

We state a necessary and suﬃcient condition in terms of the data{Φn}.

Theorem 3.1. Let{Φn}_n_≥₀ be a MOPS such thatΦ1z z−q1,q1 ∈Cand|q1|/1. Also assume

Dn2/0, for alln≥1. We define recursively a sequence{Ψn}n≥0of monic polynomials by the relations

∀n≥2, Φnz qnΦn₋1z Ψnz pnΨn−1z, pn, qn∈C, pnqn/0, 3.1

andΨ1z z−p1withp1∈C,|p1|/1,p1/q1. Then,{Ψnz}is a MOPS diﬀerent from{Φnz}

if and only if the following formulas hold:

iFor alln≥2, pn/qn,

iiFor alln≥2,|Tn1|1, whereTnis defined by2.16, iiithere exist two complex numbersα, βsuch that

∀n≥2, α pn pn

Tn1

Tn2

, β qn qn

Tn1

Tn2

, 3.2

iv

∀n≥1, α

pn1−qn1

pn1 Φn

α

qn1−pn1

pn1 Φ

∗

nαTn2, 3.3

v

∀n≥2, Φn0 Tn1Tn2qn, 3.4

vi

∀n≥1,

pn2−qn2

pn2

en1

en

qn1−pn1

Φn1α

Φnα . 3.5

Moreover, the sequences{Φn}and{Ψn}are connected by

∀n≥1, Ψnz Φnz en Φnα

pn1−qn1

pn1

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Proof. It only remains to establish the suﬃcient condition.

The task is now to obtain that the expression in the brackets is null. Using 1.8, this expression becomes

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We will apply1.12as well asK∗_nz, α αn_K_n_{z, α}_{, since}_|_α_|₁

Corollary 3.2. Under the same conditions as in the previous theorem, the following relations hold

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ii

Now, we are going to express the Verblunsky coeﬃcients for the solutions in terms of{pn}and{qn}. We remember that to give a MOPS{Φn}on the unit circle is equivalent to

Moreover, the sequences{Φnz}and{Ψnz}are connected by

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Proof. In order to obtainΦn0andΨn0we use the hypothesisivas well as2.33–3.16

and2.34–3.17, respectively. The conditions|Φn0|,|Ψn0|/1, follow fromiiandiii, respectively.

Applyinguin3.6, it holds that

∀n≥1, uΨnz en Φnα

pn1−qn1

pn1

. 3.20

Combining with2.13, we get

en −1n1pn· · ·p2

q1−p1

pn1

pn1−qn1

Φnα. 3.21

Again from2.22we get

∀n≥2, Ψnz Φnz −1n−1p2· · ·pn

q1−p1

Kn−1z, α. 3.22

This completes the proof because of the symmetry of the problem.

Remark 3.4. Notice that the restrictions given forpn andqn in the previous theorem ensure

that the sequences generated byΦn0and Ψn0 are MPOS, but they do not ensure that

ΦnzandΨnzfulfill3.1. In fact, other similar conditions to2.42seem to be necessary in order to obtain a characterization of the Verblunsky coeﬃcients in terms ofpnandqn.

4. Linear Functionals

In this section we establish the relation between the regular functionals associated with the sequences{Φn}and{Ψn}.

Proposition 4.1. LetuandLbe the regular functionals normalized byu1 u0 1andL1

v01associated with{Φn}and{Ψn}, respectively. Then, the following relation holds

λz−βL z−αu, where λ Φ1α

Ψ1

β. 4.1

Proof. We will show that

∀n≥0, λz−βLΨn z−αuΨn, 4.2

wherefrom the result follows because{Ψn}is a basis inP. Ifn0 the equality is trivial by definition ofλ. Ifn≥1, the left-hand side in4.1is

λLz−βΨnz−λΨn10En−λ

Tn2pn1Tn3

En, 4.3

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We compute the right-hand side using2.31

∀n≥2, uz−αΨnz −Φn10en Tn2

pn1

pn1−qn1

en. 4.4

In the same way, by virtue of 2.33, the right-hand side is equal to −qn1/pn1Tn2

pn1Tn3en.Therefore, it only remains to check the equalityλEn qn1/pn1en. In order to

do this we take the conjugate in2.35, obtaining

∀n≥2, En en

qn1

pn1

p2

q2

E1

e1

. 4.5

Finally, we see that the equalityλ α−q1/β−p1 q2/p2e1/E1is true due to3.14 and3.15.

Remark 4.2. The opposite question has been proved in 8. That is, if uand L are regular

functionals related by4.1with|α||β|1, then the corresponding orthogonal polynomials satisfy2.1.

5. Carath ´eodory’s Functions

In this section we obtain the relation between the Carath´eodory functions associated with the sequences{Φn}and{Ψn}. We denote by{vn}the sequence of the moments corresponding to L, that is,Lzn _v

n, for alln≥0.

Proposition 5.1. Let Fu and FL be the Carath´eodory functions associated with {Φn} and {Ψn},

respectively. Then,FLis the following rational transformation ofFu:

FLz 1/λz−αFuz zα−

zβ

z−β . 5.1

Proof. Indeed, from1.13FLz v02

_∞

n1vkzk, thus

1−zβF_Lz 1−zβv02v1z2 ∞

n1

vn1−βvn

zn1_. _5.2

Using4.1, it holds thatλvn1−βvn un1−αun. Therefore,

∞

n1

vn1−βvn

zn1 1 λ

∞

n1

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where from

F_Lz

1/λ1−zαFu−1zα

1zβv0

1−zβ

. 5.4

Puttingαβλ/λandv01,we find5.1.

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