Positive Polynomials and Boundary Interpolation with Finite Blaschke Products

The famous Jones–Ruscheweyh theorem states that n distinct points on the unit circle can be mapped to n arbitrary points on the unit circle by a Blaschke product of degree at most n − 1. In this paper, we provide a new proof using real algebraic techniques. First, the interpolation conditions are rewritten into complex equations. These complex equations are transformed into a system of polynomial equations with real coefﬁcients. This step leads to a “geometric representation” of Blaschke products. Then another set of transformations is applied to reveal some structure of the equations. Finally, the following two fundamental tools are used: a Positivstellensatz by Prestel and Delzell describing positive polynomials on compact semialgebraic sets using Archimedean module of length N . The other tool is a representation of positive polynomials in a speciﬁc form due to Berr and Wörmann. This, combined with a careful calculation of leading terms of occurring polynomials ﬁnishes the proof.


Introduction
Interpolating with Blaschke products has a large literature. In this paper, we focus exclusively on boundary interpolation, that is when we interpolate with Blaschke products and the nodes are on the unit circle. Necessarily, the values are on the unit circle too. The possibility of such interpolation was established in Cantor and Phelps [3]. Jones and Ruscheweyh [15] sharpened that result and showed that any m pairs of data can be interpolated with a Blaschke product of degree at most m − 1.
The outline of the paper is the following. After this introduction, we present a new proof in detail. The proof consists of two main steps: first, we parametrize Blaschke products and transform the boundary Blaschke interpolation problem to interpolation of real values with special real rational functions. After this transformation, we have two sets: one coming from the Blaschke product representation, and another one coming from interpolation data. These sets are subsets of some high dimensional real euclidean space and the assertion of the Jones-Ruscheweyh theorem is equivalent to that their intersection is not empty. We show this by employing tools from real algebraic geometry. In particular, the main tool in this step is a Positivstellensatz of Prestel and Delzell. Finally, we present some technical lemmas and their proofs in Sect. 5. It would be interesting to compare our approach with that of Semmler and Wegert, and also investigate the structure of solutions as a future paper.

First Part of the Proof
In this part, we rephrase our problem completely. First, we apply the Cayley transform on finite Blaschke products. Let us mention that Semmler and Wegert used this approach, and they established a natural description (see [17,Lem. 3]). As we need the exact dependence of the coefficients on the zeros of the Blaschke product, we detail this step. In this way, we transform Blaschke products (of degree at most m, denoted by B ≤m ) to a special subset of real rational functions (denoted by H m ). Here the coordinates represent the coefficients, and we use a natural choice to exclude the ambiguity (caused by multiplying the numerator and the denominator by the same constant). We also transform the interpolation data and this yields a homogeneous set of data (S, solution set). If the number of interpolation pairs (n) and the degree of Blaschke products (≤ m) differ by one only (i.e. n = m + 1), then there is a nice description of the solution set S. Formulating this description (linear parametrization of S) finishes the rephrasing of the problem to real algebraic equations and it is the last step of the first part.

Parametrizing the Blaschke Products
We use the unit disk D := {z ∈ C : |z| < 1} and the unit circle T := {z ∈ C : |z| = 1}. We introduce a parametrization as follows. Let Now we investigate how E m and F m can be used to parametrize Blaschke products. Consider the parametrization mapping where we set for convenience (see (10), (11) and Lemma 1 below). Note that if a → a 0 , |a| < 1, |a 0 | = 1, then (z − a)/(1 − az) → −a 0 locally uniformly in D, and also on D ∪ T. Hence, P m is a continuous mapping (from the euclidean topology to the locally uniform convergence topology), see also Lemma 3. Roughly speaking, P m maps E m , and F m to Blaschke products of degree m and Blaschke products of degree at most m. More precisely,

The Cayley Transform
In this section, we give a description of how the Cayley transform behaves on Blaschke products of degree at most m. We also need the inverse Cayley transform too. We denote the Cayley transform by z = T (u): The set of real rational functions of degree at most m is: for some polynomials P, Q with real coefficients and deg(P), deg(Q) ≤ m. This follows immediately from [17,Lem. 3] (Representation Lemma) and we reprove and investigate it in detail in the next subsection.

Structure of the Coefficients After the Cayley Transform
We also need the structure of the coefficients of P and Q when B is a Blaschke product (|γ | = 1) later, so we detail this calculation. So let B ∈ B m and we write To simplify this, we introduce so we can write Denote the coefficients of C and D by c j and d j respectively: Note that are (holomorphic) polynomials in a 1 , . . . , a m . We also have D(u) = (−1) m C(u) (when u ∈ R), therefore for j = 0, 1, . . . , m. We can express the leading coefficients, c m = m j=1 (1 + a j ) and d m = m j=1 (−1 − a j ). We are going to express P and Q using C, D and (7). First, the leading coefficient We use Lemma 1 (with W = (−1) m d m ) since |γ | = 1 and Lemma 2; Lemmas 1 and 2 can be found in Sect. 5. This way we get that if m is even (11) are polynomials with real coefficients and Let us remark here that P(u) and Q(u) cannot have common zeros. Otherwise, if u ∈ R is such that P(u) = 0 and Q(u) = 0, then C(u) = 0 and D(u) = 0 too. Considering the definitions of C(u) and D(u), there should be a j , a k with |a j | < 1, |a k | < 1 and i(a j − 1)/(a j + 1) = i(a k − 1)/(−a k − 1), a contradiction. This is also in accord with [17,Rem. 1].
Up to now, any branch of the square root can be used (± √ γ ). Now using (4), hence √ γ = δ, we use δ instead below. Continuing the calculation for the coefficients of P(u) = m j=0 p j u j (and using |δ| = 1 again), we have The coefficients of Q(u) = m j=0 q j u j are

Parametrization of the Real Rational Functions
We investigate the parametrization of rational functions. In the previous section, p j 's and q j 's are polynomials which are the coefficients of rational functions depending on the zeros of the Blaschke product. In this paragraph, we use p j 's and q j 's as variables which are the coefficients of rational functions for convenience (and slightly abusing the notation). Afterward, we use them again as polynomials. Let A := R 2m+2 , . . , p 1 , p 0 , q m , . . . , q 1 , q 0 ) ∈ A : q 2 m + · · · + q 2 1 + q 2 0 = 0 .

It is standard that
is a surjective but not bijective mapping, and if p ∈ A 1 ⊂ R 2m+2 , then cp determines the same rational function (c ∈ R\{0}). In other words, the coefficients of the numerator and denominator of a rational function from H m are not uniquely determined (unless some type of normalization is imposed on the denominator and numerator). Therefore we directly define the coefficients where if m is odd, then p j = −2 Re(δd j ), q j = 2 Im(δd j ), and if m is even, then p j = 2 Im(δd j ), q j = 2 Re(δd j ), where we used (12), (13) and (6). We know that where J c = {1, 2, . . . , m}\J ; in particular, if j = 0, then J = ∅ and the sum consists of only one term, hence Finally, we switch to real algebraic language, in particular, we use new variables as follows With these substitutions, we introduce Let U and V be the following polynomials if m is even (19) and where d depends on a 1 , . . . , a m while U and V depend on x 1 , y 1 , . . . , x m , y m but they are connected by (15).

Applying Cayley Transform on the Interpolation Data
Here we consider the interpolation data and transform it with the Cayley transform. Suppose that pairwise distinct z 1 , . . . , z n ∈ C with |z 1 | = . . . = |z n | = 1 are given and w 1 , . . . , w n ∈ C with |w 1 | = . . . = |w n | = 1 are also given. We transform these and consider Note that if w j = −1 for some j, then v j = ∞; also if z j = −1 for some j, then u j = ∞. By appropriate rotations, this can be avoided. To be precise, let ω, χ ∈ C, |ω| = |χ | = 1 such that none of ωw 1 , . . . , ωw n is equal to −1 and none of z 1 χ, . . . , z n χ is equal to −1. Then we find a Blaschke product Then just take and this will interpolate w j at z j : B(z j ) = w j , j = 1, . . . , m. Therefore, we may assume that all u j and v j are finite, i.e. u 1 , . . . , u n , v 1 , . . . , v n ∈ R, u 1 , . . . , u n are pairwise distinct. (21)

Introducing Two Real Sets Coming from Blaschke Products and Interpolation Data
Here we would like to find real polynomials This equation is equivalent to We will return to this condition later. An equivalent form of (22) is Note that this is a homogeneous linear equation in α 0 , β 0 , . . . , α m , β m , so it always has a solution. Let U be the following Vandermonde matrix where . means transpose hence v is a column vector) and introduce for short. Hence, (24) can be written as We assume that So U is a square matrix, U ∈ R n×n , and since u 1 , . . . , u n are pairwise different (see (21)), U is a non-singular matrix. Note that M can be thought of as a mapping from A = R 2m+2 . Therefore, Using the rank-nullity theorem, this implies that Note that S = S(u 1 , . . . , u n , v 1 , . . . , v n ) ⊂ A depends on the interpolation data (i.e. on z 1 , . . . , z n and w 1 , . . . , w n ) and contains all the solutions of (24) while the "geometrical representation" of general Blaschke products, is independent of the interpolation data. We return to the condition (23). It holds because holds for some Blaschke product B ∈ B ≤m . Therefore P and Q cannot have common zeros.
Note that G ≤m ⊂ A 1 , moreover It is straightforward to see that the Jones-Ruscheweyh theorem is equivalent to
To exploit the dimension condition (28), we introduce the matrix A by collecting the coefficients of α 0 , α 1 , . . . , α m , β 0 , β 1 , . . . , β m . Let A ∈ R 2n×n be the matrix for which where t is a column vector. Now we use the dimension condition (28). Hence, it is standard (see e.g. [14, p. 13, 0.4.6 (f)]) that there is an invertible matrix B 1 ∈ R n×n and a set I 1 ⊂ {1, 2, . . . , 2n} where |I 1 | = n such that the rows of AB 1 with indices from I 1 is the identity matrix of size n.
For ease of notation, we do not introduce new variables for B −1 1 t , that is, we assume that the rows of A with indices from I 1 give the identity matrix. For simplicity, we label the set of For convenience, we introduce for = 0, 1, . . . , m. Therefore

A Second Transformation Applied on the Two Sets
Our ultimate goal is to show that the two sets (S and G ≤m ) coming from different "sides" of the problem have non-empty intersection, i.e (29) holds. In this section, we apply a second transformation which is an adapted form of the rational parametrization of the unit circle. This transformation changes the occurring polynomials and reveals a crucial property of the polynomials (see (44), (45), and (46)). We exploit this property with a Positivstellensatz of Delzell and Prestel which is a special description of positive polynomials on compact, semialgebraic sets. Instead of sums of squares of polynomials, it features higher powers of polynomials.

Describing the Second Transformation
We will transform our system of equations by substituting variables and replacing equations so that we can apply a Positivstellensatz.
We rewrite the polynomials U and V 's using substitutions (38) and (39). So we introduce s 1 , s 2 , . . . , s m , r 1 , r 2 , . . . , r m )  For simplicity, we define It is very important that R j , j = 1, 2, . . . , 2m do not have constant terms (because of the first factor on the right of (41)) In other words, using the substitution we can also write that The last two R j 's, namely R 2m+1 and R 2m+2 behave differently. The expressions (19) and (20) for U m and V m show that -if m is odd, then -if m is even, then where the . . . stand for terms that are multiplied with an s j . Taking into account these four lines above and the substitutions (38) and (39), we can write Hence R j 0 has zero constant term and R j 1 has non-zero constant term. These observations will be crucial for the argument later.

Eliminating the Parametrizing Auxiliary Variables
Since there is an identity submatrix within the matrix A (coming from the coefficients of L j 's; see the definition of A, (33), and that of L k 's, (34) and (35)) there exists an invertible matrix B 2 ∈ R 2n×2n such that with ⎛ This is a simple elimination on the left-hand sides (using row operations on A instead of column operations which we used in Sect. 2.7). We transform the right-hand sides accordingly, hence we introduce R 1 , R 2 , . . . , R 2n as The system (47) with condition (48) is equivalent to with conditions (48), that is, when x = (u, s 1 , . . . , s m , r 1 , . . . , r m ) ∈ W . Introduce Using (52), we can express t k 's with u, s 1 , . . . , s m , r 1 , . . . , r m as follows Therefore if x ∈ W 2 , then there is a unique t ∈ R n such that (x, t) ∈ W , because of the following. Note that R j (x) are polynomials independent of t 1 , . . . , t n . Furthermore, 1 − u 2 + u 4 ≥ 3/4 when u ∈ [−1, 1] and S ≥ 3/2 m+2 holds also. Therefore, for all possible x ∈ W , R j (x)/S(x) is continuous and bounded. This implies that W is compact. Hence, the t's coming from a solution are bounded, i.e. there is M 0 > 0 such that for all t ∈ R n such that (x, t) ∈ W , we have |t j | ≤ M 0 , j = 1, 2, . . . , n.

Application of the Positivstellensatz
We are going to use a form of Positivstellensatz which can be found in the book of Prestel and Delzell [16]. Briefly, it is for sums of even powers and for compact (semialgebraic) sets.
First, we introduce our new set of notations. Then we apply the Positivstellensatz to find a solution. This indirect argument features a step-by-step simplification of the representation (61) provided by the Positivstellensatz. As the first step of simplification, we apply a substitution (63) which turns the representation into a univariate identity (68). Then a careful comparison of the leading terms and degrees leads to an even more simplified identity (69). Finally, exploiting the special structure of the equation (comparing (72) and (71)) leads to a contradiction. We remark that in this section we do not use the polynomials c 0 , c 1 , . . . , c m from Sect. 2.3 (and we use c 0 , c 1 as new symbols).
As the next step, we set N := 8.
Recall that x = (u, s 1 , . . . , s m , r 1 , . . . , r m ) and for unifying the notation, we introduce the following: Put i.e. W is a compact, semialgebraic set. Introduce and only if (x, t) is a solution of (47) and also f ≥ 0.
As the next step, we apply a form of Positivstellensatz, more precisely [16,Thm. 7 where σ 0 , σ 1 , . . . , σ N 1 are sums of N -th powers, i.e. they are from Indirectly, assume that f > 0 on W which implies that (61) holds.
As the next step, we apply a substitution for (61) to simplify it. The expression on the right of (61) after substitution Y, where Y : s 1 = . . . = s m = 0, r 1 = . . . = r m = 0, t 1 = . . . = t n = 0 (63) has the following structure. Obviously, . Therefore the right-hand side will have this form: where slightly abusing the notation, we write σ | Y = σ .
As the next step, we collect the results of substitution and simplification. Using (61), (66) and (67), the right-hand side is relatively simple, while regarding the left-hand side, we use (65) and we write To compare the two sides, we need the powers of 1 − u 6 : (1 − u 6 ) 1 = (−1)u 6 + 1, We investigate the degrees and the leading terms in (68). On the left-hand side, deg f 1 = N = 8. The right-hand side is more involved. We remark that deg σ j = N k j for some k j ∈ N and the leading coefficient lc( σ j ) of σ j is positive. We investigate the degrees of σ j (1 − u 6 ) j , j = 0, 1, . . . , 7 modulo N . Consider the following groups of powers: {0, 4} and {1, 5} and {2, 6} and {3, 7}. In each group they have the same degrees mod N and the same signs of leading coefficients, e.g. if we take {1, 5}, then deg σ 1 (1−u 6 ) 1 = N k 1 +6, deg σ 5 (1−u 6 ) 5 = N k 5 +30 = N (k 5 +3)+6 and the signs of the leading coefficients are the same sign lc σ 1 (1 − u 6 ) 1 = sign lc σ 5 (1 − u 6 ) 5 ) = −1. Since the signs are the same, the leading terms in the same group cannot cancel. Also, since the degrees of σ j (1 − u 6 ) j are different modulo N = 8, the leading terms of different groups cannot cancel either. Hence Taking into account that deg( f 1 ) = 8, this can happen only when i.e. σ 1 is a constant. Finally, we obtain that where c 1 ≥ 0.
So we have where N 2 is a positive integer and A j (u) are polynomials. Again, degree considerations show that the A j 's must be constant or linear polynomials, so we can write where λ j > 0, ζ j ∈ R are pairwise different and c 0 ∈ R and N 3 ≤ N 2 .
We finally reach a contradiction by showing that (71) cannot hold. Note that Exploiting this (and that N is even), we write Observe that all coefficients of (u − ζ j ) N + (u + ζ j ) N /2 are non-negative (e.g. the coefficient of u 2 is 28ζ 6 j ). Therefore the coefficient of u 2 is 4 m+1 (−2) < 0 (according to the left-hand side of (71)) and it is N 3 j=1 λ j · 28ζ 6 j ≥ 0 (according to the right-hand side of (71)). This gives a contradiction.
Therefore, we have f > 0. Obviously f ≥ 0. These two imply that f must have a zero on W , (x, t) ∈ W with f (x, t) = 0, so the system (52) and (53) with x ∈ W has a solution. In turn, this implies that W = 0 which leads to the fact that (47) has a solution from W (see (48)), which implies that (31) has a solution under the conditions (32), that is, (29) holds which is equivalent to the assertion of Theorem 1. Therefore the proof is complete.

Some Technical Lemmas
and the numerator and denominator on the right are real, i.e.
In particular, f = √ γ will do. Furthermore, a similar identity also holds: there exists and again, the numerator and the denominator on the right are real. In this case, Proof We use a separate set of notation in this proof. To see the first assertion, write W , γ and f in polar form in this proof: W = re iω , where r > 0, γ = e iα and f = e iϕ . We also have Im f W + γ W = r Im e i(ϕ−ω) + e i(ϕ+ω+α) = r (sin(ϕ − ω) + sin(ϕ + ω + α)) = 2r sin ϕ + α 2 cos ω + α 2 , and similarly for the denominator, So we have to find ϕ for the given α and ω such that sin ϕ + α 2 cos ω + α 2 = 0 and sin ω + α 2 sin ϕ + α 2 = 0.

Lemma 2
Let P, Q be complex polynomials without common zeros. Assume that H (u) = P(u)/Q(u) is a real rational function, i.e. if u ∈ R and H (u) is finite, then H (u) ∈ R. Also assume that the leading coefficients of P and Q are real. Then all the coefficients of P and Q are real.

Proof
We prove it by induction as follows. Write P(u) = au n + P 1 (u) and Q(u) = bu m + Q 1 (u) where deg(P 1 ) < deg(P) and deg(Q 1 ) < deg(Q). By the assumptions, a, b ∈ R, a = 0, b = 0. If n ≥ m > 0, we can write which implies that with P 2 (u) := P(u) − a b u n−m Q(u), P 2 (u)/Q(u) is a real rational function. Note that n 1 := deg(P 2 ) < deg(P) = n. Denote the leading coefficient of P 2 by c, c ∈ C, c = 0. It is standard to see that lim u→∞ u m u n 1 where the left-hand side is real and b on the right is also real. Hence the leading coefficient c of P 2 is real. If m > n > 0, then consider 1/H (u) = Q(u)/P(u) which is again a real rational function.
If m = 0, then H (u) = P(u)/Q(u) is actually a polynomial. Also, Q(u) = Q(0) ∈ R, and hence H (u) is a real polynomial, that is, if u ∈ R, then H (u) ∈ R. It is then standard that the polynomial H (u) must have real coefficients.
Finally, if n = 0, then consider 1/H (u) and this way we reduce this case to the case discussed in the previous paragraph.
We also need the following lemma ( [17, Lem. 2]) Lemma 3 For any sequence (B n ) of Blaschke products of degree m there exist a Blaschke product B of degree k ≤ m and a subsequence of (B n ) which converges to B locally uniformly on a set which contains all points of D ∪ T with the possible exception of at most m − k boundary points.
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