On Linear Combinations of Two Orthogonal Polynomial Sequences on the Unit Circle
© C. Suárez. 2010
Received: 1 August 2009
Accepted: 5 March 2010
Published: 8 March 2010
Let be a monic orthogonal polynomial sequence on the unit circle. We define recursively a new sequence of polynomials by the following linear combination: , , . In this paper, we give necessary and sufficient conditions in order to make be an orthogonal polynomial sequence too. Moreover, we obtain an explicit representation for the Verblunsky coefficients and in terms of and . Finally, we show the relation between their corresponding Carathéodory functions and their associated linear functionals.
1. Notation and Preliminary Results
Along this paper, we will use the following notations. We denote by the linear space of Laurent polynomials with complex coefficients and by the dual algebraic space of . Let be the space of complex polynomials.
Let . Denoting by , , we say that
is Hermitian if , ;
- (ii)is regular or quasidefinite (positive definite) if the principal minors of the moment matrix are nonsingular (positive), that is,(1.1)
In any case we denote , with .
The sequence is said to be the sequence of the moments associated with .
Furthermore, if is a positive definite linear functional then a finite nontrivial positive Borel measure supported on the unit circle exists such that
In the sequel, we denote by the monic orthogonal polynomial sequence (MOPS) associated with .
For simplicity, along this paper we also assume that is normalized (i.e., ). It is well known that the regularity of is a necessary and sufficient condition for the existence of a sequence of orthogonal polynomials on the unit circle. On the other hand, the polynomials satisfy the so-called Szegö recurrence relations
where is the reversed polynomial of , .
This sequence verifies the following properties:
To the linear functional we can associate a formal series as follows:
In the positive definite case, is called the Carathéodory function associated with . In this case, can be written as
The measure can be reconstructed from by means of the inversion formula. The aim of this paper is the analysis of the following problem. Given an MOPS on the unit circle , orthogonal with respect to a linear functional , to find necessary and sufficient conditions in order to make a sequence of monic polynomials defined by
an MOPS with respect to a linear functional . Further more, to find the relation between the linear functionals and and their corresponding Carathéodory functions.
where and are fixed functions. As a next step, to find the relation between the functionals.
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, in  there 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 of and . Section 4 is devoted to find the functional relation between and . Finally, Section 5 contains the rational relation between the corresponding Carathéodory functions.
2. Necessary Conditions
Let be a monic orthogonal polynomial sequence and let be a monic polynomial sequence. We assume that there exist sequences of complex numbers and such that the following relation holds:
Also, we assume and and , with and .
In this section, we find some necessary conditions in order to make the sequence defined recursively from by relation (2.1) an MOPS.
With this aim, we define the complex numbers and as follows:
The following proposition justifies this choice.
Taking we have (2.4). Observe that, this is as same as (2.1) for .
Identifying the coefficients of degree , then (2.5) holds. Therefore, we can rewrite (2.11) as (2.6).
On the other hand, applying the -operator in (2.1) we have . Substituting in (2.6) and using (2.1), we obtain (2.7).
In the sequel, we denote by the linear regular functional associated with and by the linear regular functional associated with . Besides, we denote by the real number such that with . Therefore, .
Under the same conditions as in Proposition 2.1, the following assertions hold:
If , then and , for all ,
If and , for all , then , for all ,
Assume and , for all , then if and only if , for all . Moreover, for all .
- (i)We eliminate using equalities (2.2)–(2.4) and (2.3)–(2.5). By doing this, we get(2.12)
Taking , we obtain .
If , then and thus . Wherefrom . Now, using (2.1) we have .
Assume that there exits such that . From (2.1), written for , it holds that and then , and thus successively.
The result follows from (2.6) and the above item.
On the other hand, applying the -operator in (2.6), we obtain for .
The situation is the trivial case, that is, , for all . For this reason, in the sequel, it will be excluded.
The next result will be used later.
Since that , then (2.13) follows.
We obtain (2.14) changing by .
For all such that , we define the following complex number:
This number plays a very important role in the solution of our problem.
is not null, since together with Corollary 2.2(iii). Wherefrom, it has a unique solution for .
By solving this, we get (2.17).
In the sequel, we denote by the sequence of the kernels corresponding to . For the sequence we keep the same notations as in Section 1.
Assume that and are two MOPS that verify (2.1) with and , for all . Also assume , for all . Under these conditions, then the following assertions hold
, for all
- (ii)There exist two complex numbers and with such that(2.20)(2.21)
Here, the initial parameters and are given by and ,
- (iii)The sequences and are connected by the following formulas:(2.22)(2.23)
Item (i) follows immediately from (2.17). Indeed, if we take , we obtain .
Let us proceed with the proof of (ii). Inserting
Given that and have not common roots, then , for all .
Using (2.28) we obtain
This complex constant is denoted in the statement by . The property is a consequence of Corollary 2.2(iii). On the other hand, the explicit expressions of and follow from (2.7) for and , respectively.
This completes the proof of (ii) because the complex number exits by the symmetry of the problem.
Finally, we show (iii). Using again (2.17), we have
Substituting this relation in (2.31) and using the recurrences of the kernels (1.9) and (1.10), (2.22) holds.
In order to state the converse we need the following assertions.
Using again the fact that and have no common roots and , it follows that the coefficients in the last relation are zero and this implies (2.33) and (2.34).
Let us proceed with (2.35). From (2.33) and (2.34), for all , we have
since is a constant.
and it is true according to (2.42).
3. Some Solutions
We state a necessary and sufficient condition in terms of the data .
and with , , . Then, is a MOPS different from if and only if the following formulas hold:
, , where is defined by (2.16),
- (iii)there exist two complex numbers such that(3.2)
It only remains to establish the sufficient condition.
We first show that (3.6) implies (3.1)
Therefore, the result follows immediately from (3.5).
In order to obtain , we take in (3.6)
Now, we show that the sequence given by (3.6) satisfies (1.4) with , then it is a MOPS.
We will apply (1.12) as well as , since
If we show that the coefficient of is null and the coefficient of is , then (1.4) is true.
At first, we compute the coefficient of .
and this is equal to zero from (3.3).
We can obtain the coefficient of by observation of (3.13). It is easy to see that this coefficient is . By virtue of (3.9) it is equal to , and then the required result follows.
Finally, the condition follows from (3.11) by using the same method as in the proof of (2.35).
Observe that condition (i) together with (3.6) gives .
We obtain and from (2.22) and (2.23) for , respectively. The items (iii) and (iv) are straightforward from (2.20) and (2.21).
Now, we are going to express the Verblunsky coefficients for the solutions in terms of and . We remember that to give a MOPS on the unit circle is equivalent to know the sequence of complex numbers with .
Let and be two sequences of complex numbers such that
and , for all ,
and , for all , where is given by (3.15) and ,
and , for all , where is given by (3.14),
, for all .
Then, the only MOPS solutions of (3.1), such that for all , verify
Moreover, the sequences and are connected by
In order to obtain and we use the hypothesis (iv) as well as (2.33)–(3.16) and (2.34)–(3.17), respectively. The conditions , follow from (ii) and (iii), respectively.
Applying in (3.6), it holds that
This completes the proof because of the symmetry of the problem.
Notice that the restrictions given for and in the previous theorem ensure that the sequences generated by and are MPOS, but they do not ensure that and fulfill (3.1). In fact, other similar conditions to (2.42) seem to be necessary in order to obtain a characterization of the Verblunsky coefficients in terms of and .
4. Linear Functionals
In this section we establish the relation between the regular functionals associated with the sequences and .
wherefrom the result follows because is a basis in .
If the equality is trivial by definition of .
If , the left-hand side in (4.1) is
where the last equality follows from (2.34).
We compute the right-hand side using(2.31)
Finally, we see that the equality is true due to (3.14) and (3.15).
The opposite question has been proved in . That is, if and are regular functionals related by (4.1) with , then the corresponding orthogonal polynomials satisfy (2.1).
5. Carathéodory's Functions
In this section we obtain the relation between the Carathéodory functions associated with the sequences and . We denote by the sequence of the moments corresponding to , that is, , for all .
Putting and we find (5.1).
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