1. Introduction

In a paper Xie Siqing[6] considered vertical projections of the zeros of (1–x²) , n even or (1 ± x) , n odd; on unit circle and established regularity of (0, 1, …, r – 2, r) – interpolation under certain restrictions on parameters α and β. In another paper[7], considering a particular case for r = 3 and assuming the nodes as :

(1.1)

to be the vertical projections on unit circle of the zeros of (1–x²)P_n(x), where P_n(x) stands for n^th Legendre polynomial having the zeros x_k = cos θ_k. k = 1(1)n, such that 1 > x₁ > x₂ > … > x_n> –1, Xie Siqing claimed regularity, explicit representation and convergence of (0, 1, 3)–interpolation viz. R_n(z) of degree ≤ 6n + 3 on the nodes (1.1) satisfying the conditions :

(1.2)

where α_pk,p = 0,1,3 are arbitrary given complex numbers. The explicit forms of these polynomials are very complicated as such in the last set of conditions (1.2) instead of prescribing , we prescribe , k = 1(1)2n, which give simpler forms of the polynomials.

On the suggestion of P. Turán, J. Balázs[1] for the first time investigated weighted (0,2)–interpolation taking nodes on the real line in[–1.1]. Later on L. Szili[5], K.K. Mathur and R.B. Saxena[3] extended the study of weighted (0,2) and weighted (0,1,3) –interpolations respectively on infinite interval (–∞, ∞).

Weighted lacunary interpolatory problems on unit circle have not been considered on the literature so far, so the object of this paper is to obtain regulatory, explicit representation in a simpler form and a quantitative estimate leading to a convergence theorem in the case of weighted (0,1,3)–interpolation choosing special nodes on unit circles. The authors[4] have also made similar study in the case of weighted (0,2)–interpolation on unit circle.

2. Preliminaries

Let

(2.1)

and

(2.2)

where P_n stands for n^th Legendre Polynomial and

In the well known differential equation :

(2.3)

taking , we get

(2.4)

Further owing to (2.1), (2.4) reduces to

(2.5)

Also for k = 1(1)n, we have

(2.6)

(2.7)

and for k = 1(1)2n, we have

(2.8)

The fundamental functions of Lagrange interpolation based on the nodes as zeros of W(z) and R(z) respectively are given by

(2.9)

and

(2.10)

With the help of (2.2) and (2.9), we define J_k as

(2.11)

(2.12)

(2.13)

For –1 ≤ x ≤ 1, we have

(2.14)

(2.15)

(2.16)

(2.17)

Let x_k's be the zeros of P_n(x), then

(2.18)

(2.19)

For –1 < x < 1, we have

(2.20)

where c is a constant and

3. Regularity Ahd Explicit Representation of Interpolatory Polynomials

Let z_k's be given by (1.1). We seek to determine the regularity of the polynomials R_n(z) of degree ≤ 6n+3, satisfying the conditions :

(3.1)

where α_0k, α_1k and β_3k’s are given arbitrary complex numbers. We call such polynomials as weighted (0,1,3)–interpolation on unit circle.

Theorem 1 : In (3.1) if α_0k = α_1k = 0 for k = 0(1)2n + 1 and β_3k = 0 for k = 1(1)2n, then R_n(z)=0.

Proof: Since R_n(z) is a polynomial of degree at most 6n+3, we have R_n(z)=W(z)R(z)q(z), where W(z) and R(z) are given by (2.1) and (2.2) respectively and q(z) is a polynomial of degree at most 2n+1.

Using the conditions (3.1), we get q’ (z_k) = 0. Therefore, we may suppose q(z)=aJ₀(z)+b, where a and b are constants and J₀(z) is given in (2.11). Now, applying ,we get q(z)=0,leading to R_n(z)=0, which completes the proof of the theorem.

We write R_n(z) satisfying conditions (3.1) as

(3.2)

where A_jk (j = 0,1,3) are unique polynomials each of degree ≤ 6n + 3 determined by the following requirements :

(3.3)

(3.4)

(3.5)

Theorem 2 : Let R(z), W(z) and J_k(z) be given by (2.1), (2.2) and (2.11), then under conditions (3.5), A_3k(z) are given

(3.6)

where

(3.7)

(3.8)

and

(3.9)

Theorem 3 : Let L_k(z) and H_k(z) be given by (2.10) and (2.12), then under the conditions (3.4), A_1k(z) is given by :

(3.10)

where

(3.11)

(3.12)

(3.13)

Theorem 4 : Let L_1k(z), S_k(z), A_3k(z) and A_1k(z) be given by (2.9), (2.13), (3.6) and (3.10), then under the conditions (3.3), A_0k(z) is given by :

(3.14)

where

(3.15)

(3.16)

(3.17)

(3.18)

k = 1(1)2n

Proof of Theorem 2: Let A_3k (z) be in (3.6). Obviously A_3k(z_j) = 0 for j = 0(1)2n+1 and A_3k’ (z_j) = 0 for j = 1(1)2n. From ; j, k = 1(1)2n, we have _. For, result is obvious and for j=k, we get given in (3.7).

Now, from the conditions , we get and as given in (3.8) and (3.9),which proves the theorem.

One can prove theorems 3 and 4 owing to conditions (3.4) and (3.5) respectively, so we omit the details of proof.

4. Estimation of Fundamental Polynomials

Lemma 1 : Let L_1k(z) be given by (2.9), then

(4.1)

where c is a constant independent of n and z.

Proof : Let z = x + iy and |z| = 1, then from (2.6) and (2.9), for 0 ≤ arg z < π and k = 1(1)n, one can see that

(4.2)

Also,

(4.3)

Similarly, for π ≤ arg z < 2π and k = 1(1)n

(4.4)

Therefore, for a fixed z = x + iy, |z| = 1 and |z| = 1 and –1 < x < 1, we have

(4.5)

Now,

Using (2.16), (2.18) and (2.19), we get

(4.6)

Further for –1 < x < 1, we have

(4.7)

(4.5) with the help of (4.6) and (4.7) proves the lemma.

Lemma 2 : Let J₀(z) be given by (2.11), then

(4.8)

and

(4.9)

where K_n is defined by (2.2).

Proof : To prove (4.8), we see that

because of (2.16).

Again

using (2.17), we get the result.

Lemma 3 : Let A_3k(z) be given by (3.6), then

(4.10)

Proof : It is sufficient if we prove the lemma to be true for |z| = 1. Let z = eⁱα (0 ≤ α < 2π). From (3.6), we see that

(4.11)

because from (3.8), lemma 2 and (3.9), one can see that

From (3.7) and (2.6), we have

(4.12)

From (4.11), (4.12) and

(4.13)

owing to (2.1), (2.2) and (2.14), we have

From (2.18) and (2.19), we have

which proves (4.10) owing to lemma 1.

Lemma 4 : Let H_k(z) be given by (2.12), then

Proof : The estimate of |H_k(z)|, for k = 0, 2n+1 follow from (2.12) and (4.8), so we omit the details.

For estimate of , from (2.10), we have

, for k = 1(1)2n.

On further differentiating and rearranging terms, we get

.

From (2.12), we obtain

(4.15)

A little computation yields

(4.16)

(4.17)

and

(4.18)

From (4.15) – (4.18) follows the lemma.

Lemma 5 : For |z| ≤ 1, we have

(4.19)

where A_1k(z) is given by (3.10).

Proof : Considering (3.10), we have

(4.20)

.

From maximal principle, we have

(4.21)

From (3.11), (3.12) and Lemma 2,

(4.22)

Now, for the estimation of , we have

Using (4.13), (4.14) and (4.21), we get

(4.23)

Similarly, one can have

(4.24)

Further, we have

(4.25)

From (4.22), (4.13) and (4.14), we get

(4.26)

From (3.13), we have

.

Using this estimate and lemma 3, we get

(4.27)

Thus, (4.20) owing to (4.23)–(4.27) complete the proof of the lemma.

Lemma 6 : Let S_k(z) be given by (2.13), then

(4.28)

Proof : Let ,

(4.29)

and

(4.30)

then

(4.31)

where S_k(z) is given by (2.13).

Using (4.29) and Lemma 4, one can see that

(4.32)

Similarly, we have

(4.33)

Finally, the lemma follows from (4.31) – (4.33).

Lemma 7 : For |z| ≤ 1, we have

(4.34)

The proof of the lemma is similar to that of Lemma 5, so we omit the details.

5. Quantitative Estimate and Convergence Problem

In this section, we shall prove the following :

Theorem 4 : Let f(z) be continuous in the region |z| ≤ 1 and analytic in |z| < 1, then the sequence {R_n} defined by

(5.1)

satisfies the relation :

(5.2)

where c is independent of n and z, ω₂(f,δ) is the modulus of smoothness of f(z).

Remark : Let f(z) be continuous in |z| ≤ 1 and f '∈Lip. α, α > ½, then the sequence {R_n} converges uniformly to f(z) in |z| ≤ 1, which follows from (5.2) and

.

To prove theorem 4, we need the following well known Jackson’s theorem :

Let f(z) be continuous in the region |z| ≤ 1 and analytic in |z| < 1, then there exists a polynomial F_n(z) of degree ≤ 2n – 2 such that

(5.3)

where z = eⁱθ, 0 < θ ≤ 2π.

To prove our theorem, we shall also require

(5.4)

which is an easy consequence of an inequality of ), Kis[[2],[32]] viz.

, m = 0, 1, …. and

Proof of the main Theorem 4 : Since R_n(z) given by (3.2) is a uniquely determined polynomial of degree ≤ 6n + 3, therefore, the polynomial F_n(z) satisfying (5.3) and (5.4) can be expressed as

Then

Taking z = eⁱθ, 0 < θ ≤ 2π and using (5.3), (5.4), Lemma 3, Lemma 5 and Lemma 7, we get (5.2) which completes the proof of the theorem.