We then introduce paraorthogonal polynomials of the second kind and prove that zeros of paraorthogonal polynomials of the first and second kind are intertwined. Finally, we present two examples, one on ∂D and one on R, so that adding a point mass will generate non-exponential perturbations to the recurrence coefficients. Uvarov [51] and Nevai [38] were unaware of the work of Jost-Kohn and of each other and discovered the formulas for adding point masses for orthogonal polynomials on the real line.
Then one of the following is true: i) Since αn → L6= 0, this measure has the same essential spectrum as the measure dµ0 with Verblunsky coefficients αn(dµ0) ≡ L, which is supported. ii) Case (1) is a special case of Corollary 24.3 of [49], in which Simon proved that varying the weight of an isolated pure point in the gap will result in an exponentially small perturbation of αn(dµ). iii) By (2c), adding a pure point to the gap will preserve the bounded variational property of (αn)n. Let Γβ be the union of open arcs that form the interior of the bonds that form ess supp(dµβ). Therefore, two paraorthogonal polynomials of the same degree are linearly independent if and only if all their zeros are different.
Para-orthogonal polynomials of the second kind arise from orthogonal polynomials of the second kind, namely ψk(z), which are orthogonal polynomials. The existence of the measure is guaranteed by Verblunsky's theorem which states that for any given sequence of complex numbers in D, there corresponds a measure on the unit circle with coefficients such as Verblunsky.
Proof of Theorem 2.0.2
The proof can be performed similarly, except that after (4.1.2) all representations ehn will be replaced by mehn+1. Since τ and λ are distinct zeros of hn, we can apply a similar argument as in Lemma 4.1.2 to (z−τ)(z−λ)hn(z) and obtain the following.
Proof of Theorem 2.0.3
Proof of Lemma 2.0.1
The proof is essentially the same as the one of Lemma 4.1.1, except for a few differences. The L2 norm here refers to the one taken with respect to ν and hn is replaced by sn.
Proof of Theorem 2.0.4
Second, note that kΦn+1k is independent of j so it can be omitted from the summation. Suppose we can prove that φ∗n(ζ) tends to a nonzero limitLasn tends to infinity, then 1/Kn =O(1/n). As a result, if we add another pure point dµ1, we can use an argument similar to the one above and the point mass formula (2.0.11) to prove that αn(dν) is the sum of αn(dµ0) plus two tails. terms and an error term.
In general, if we have a measure dµm as defined in (2.0.20), then we add one pure point after another and use the point mass formula (2.0.11) inductively. Therefore, we will be able to express αn(dµm) as the sum of αn(dµ0) plus m tail terms, and an error term. By an argument similar to the one above, we see that tj,n is O(1/n) and zjtj,n −tj,n−1 is small.
Obviously the 'smallness' must be determined by rigorous calculations that we will present in the proof. Nevertheless, these observations led us to introduce the concept of generalized bounded variation Wm, and from this we were able to deduce that limn→∞ϕ∗n(z, dµm) exists. The technique used in this proof is a generalization of the technique used in proving Theorem 2.0.10.
The above calculations also show that the sum on the right-hand side of (6.1.16) is absolutely convergent as n → ∞, and the convergence is uniform on any compact subset of ∂D\{ζ1, ζ2,.
Proof of Theorem 2.0.9
We considered the induction step µme defined in (2.0.20) as the measure produced by adding the pure point dµm−1 as follows. Consequently, we can use a similar argument as in the base case and derive this. By evaluating successive Verblunsky coefficients in the same way as in the basic case, we prove that dµm ∈Wm+1(1, z1, z2, . . . , zm).
In this section we consider measures with asymptotically periodic Verblunsky coefficients for p-type bounded variation (this term was first introduced in [23]), i.e. given a periodic sequence βn of period p,. It is well known that any measure satisfying (7.0.2) has the same essential spectrum as dµβ (the measure of periodic Verblunsky coefficients (βn)n) which is supported on a finite number of bands. Applying Weyl's theorem to the CMV matrix (see Theorem 4.3.5 of [46]), αn → L implies that dµ has the same essential spectrum as the target dµ0 with Verblunsky coefficients αn(dµ0)≡L(the target dµ0 is known to be associated with Geronimus polynomials).
In other words, there is a gap in the spectrum in which there is at most one pure point. Note that αn≡L can be seen as a periodic sequence of period one, in fact there is a more general result on gaps in the spectrum for measures with periodic Verblunsky coefficients. As a result, in both cases we consider, there are gaps in the spectrum, and when z ∈∂D is in one of these open gaps, we have |T rTp(z)|>2.
Tools
The Stolz–Ces` aro Theorem
Kooman’s Theorem
Outline of the Proof
To understand the asymptotics of ϕn(z) and ϕ∗n(z), we will study the behavior of the product in (7.3.4). First note that An(z) is hyperbolic for large enough n when z ∈ GL, so An(z) has distinct eigenvalues. This allows us to apply Kooman's result (see Theorem 7.2.2) to the productAn· · ·A0 to prove that there exists an integer N (which only depends on how fast An(z)→ A ∞(z)) such that there exist w≡w(N)∈C2, matrices Gj(z) and diagonal matricesDj(z) such that:.
In the former case, we will prove that changing the weight of an existing point mass will lead to a perturbation that is exponentially small.
Proof of Theorem 2.0.11
By constructing the function D, DAn is diagonal under this new basis, so a diagonal matrix exists. Since U is analytic on S, there exist on some compact subset of S constants η1, η2 >0 such that. First we want to show that given any >0 there exists an integer J such that |rj|< for all j ≥J.
By assumption, we are free to choose any M, so we choose an integer M such that 1/M <. Moreover, log is analytic close to 1, so in the -neighborhood of 1 there exists a constant E such that. We decide as follows: we note that y±(ζ) is continuous with respect to ζ, so if.
Otherwise, there must be some ζ1 in the gap, such that |λ1(ζ1)| = 1, which contradicts the hyperbolicity of A∞(ζ) in the gap. Now that we have successfully separated the real and imaginary parts of L(∆∞(ζ) +L), we can prove this with a direct calculation. From now on we will write all error terms in the order O(|αn−αn−1|) as and.
We will see later that the introduction of these objects will not affect the result of our calculation. Now we will prove that each of the sums on the right-hand side of (7.4.97) is summable.
Proof of Theorem 2.0.12
Using a similar argument as in Section 7.4 for the family Bn(ζ), we can show that there exists a nonzero vector w and an integer N such that. Finally, by a similar argument as in the proof of Theorem 2.0.11, we could prove that for every fixed j (∆kp+j(ζ))k is a bounded variation.
Proof of Theorem 2.0.13
Proof of Corollary 2.0.1
Moreover, since ζ is not a pure point of dµ, Θn(ζ) is a strictly increasing series tending to +∞, so we can apply the Stolz-Ces`aro theorem and conclude that ζn∆n(ζ) =ζnPn( ζ )/Θn(ζ) → −2L. We will use the Stolz-Ces`aro theorem again to prove that the limit in (7.7.2) exists and is finite.
Szeg˝ o condition and bounded variation
According to the Killip–Simon theorem [27], condition (7.8.3) implies that such a measure does not satisfy the quasi-Szeg˝ condition, i.e. We now use the inverse Szeg˝ mapping (see Chapter 13 of [47]) on dγy to form the probability measure µy on ∂D. Furthermore, since γy(x) is symmetric, each of the two components on the right-hand side of (7.8.13) is symmetric along the imaginary axis. dµy can be viewed as a duplicate of the probability measure. 7.8.16).
Therefore, to any monotonic sequence an → 1 and any 0 < y < 2 there corresponds a family αn(dνy) of bounded variation that converges to −ay <0.
The Szeg˝ o Mapping
The measure dγ0 is purely absolutely continuous and symmetrically supported on [−2,2], with no pure points outside [−2,2]. Also, dµy is symmetric along both the x and y axes due to the symmetry of dγy and the Szeg˝o map. Next, we add a clean point at z = 1 to dµy to form the measure dµ˜y and calculate the perturbed Verblunsky coefficients αn(d˜µy).
As the final step, we show that for some constants Cx0, Dx0 (both dependent on x0), such that.
The Proof
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