Higher order Hermite-Fejér interpolation polynomials with Laguerre-type weights

Abstract

Let ℝ⁺ = [0, ∞) and R : ℝ⁺ → ℝ⁺ be a continuous function which is the Laguerre-type exponent, and p_{n, ρ}(x), $\rho >-\frac{1}{2}$ be the orthonormal polynomials with the weight w _ρ (x) = x^ρ e^-R(x). For the zeros ${\left\{x_{k,n,\rho }\right\}}_{k=1}^n$ of $p_{n,\rho }\left(x\right)=p_n\left(w_{\rho }^2;x\right)$, we consider the higher order Hermite-Fejér interpolation polynomial L _n (l, m, f; x) based at the zeros ${\left\{x_{k,n,\rho }\right\}}_{k=1}^n$, where 0 ≤ l ≤ m - 1 are positive integers.

2010 Mathematics Subject Classification: 41A10.

1. Introduction and main results

Let ℝ = [-∞, ∞) and ℝ⁺ = [0, ∞). Let R : ℝ⁺ → ℝ⁺ be a continuous, non-negative, and increasing function. Consider the exponential weights w _ρ (x) = x^ρ exp(-R(x)), ρ > -1/2, and then we construct the orthonormal polynomials ${\left\{p_{n,\rho }\left(x\right)\right\}}_{n=0}^{\mathrm{\infty }}$ with the weight w _ρ (x). Then, for the zeros ${\left\{x_{k,n,\rho }\right\}}_{k=1}^n$ of $p_{n,\rho }\left(x\right)=p_n\left(w_{\rho }^2;x\right)$, we obtained various estimations with respect to $p_{n,\rho }^{\left(j\right)}\left(x_{k,n,\rho }\right)$, k = 1, 2, ..., n, j = 1, 2, ..., ν, in [1]. Hence, in this article, we will investigate the higher order Hermite-Fejér interpolation polynomial L _n (l, m, f; x) based at the zeros ${\left\{x_{k,n,\rho }\right\}}_{k=1}^n$, using the results from [1], and we will give a divergent theorem. This article is organized as follows. In Section 1, we introduce some notations, the weight classes ${\mathcal{L}}_2$, ${\tilde{\mathcal{L}}}_{\nu }$ with $\mathcal{L}\left(C^2\right)$, $\mathcal{L}\left(C^2+\right)$, and main results. In Section 2, we will introduce the classes $\mathcal{F}\left(C^2\right)$ and $\mathcal{F}\left(C^2+\right)$, and then, we will obtain some relations of the factors derived from the classes $\mathcal{F}\left(C^2\right)$, $\mathcal{F}\left(C^2+\right)$ and the classes $\mathcal{L}\left(C^2\right)$, $\mathcal{L}\left(C^2+\right)$. Finally, we will prove the main theorems using known results in [1–5], in Section 3.

We say that f : ℝ → ℝ⁺ is quasi-increasing if there exists C > 0 such that f(x) ≤ Cf(y) for 0 < x < y. The notation f(x) ~ g(x) means that there are positive constants C₁, C₂ such that for the relevant range of x, C₁ ≤ f(x)/g(x) ≤ C₂. The similar notation is used for sequences, and sequences of functions. Throughout this article, C, C₁, C₂, ... denote positive constants independent of n, x, t or polynomials P _n (x). The same symbol does not necessarily denote the same constant in different occurrences. We denote the class of polynomials with degree n by ${\mathcal{P}}_n$.

First, we introduce classes of weights. Levin and Lubinsky [5, 6] introduced the class of weights on ℝ⁺ as follows. Let I = [0, d), where 0 < d ≤ ∞.

Definition 1.1.[5, 6] We assume that R : I → [0, ∞) has the following properties: Let Q(t) = R(x) and x = t².

• (a) $\sqrt{x}R\left(x\right)$ is continuous in I, with limit 0 at 0 and R(0) = 0;

• (b) R″(x) exists in (0, d), while Q″ is positive in $\left(0,\sqrt{d}\right)$;

• (c)
  $$\underset{x\to d-}{lim}R\left(x\right)=\mathrm{\infty };$$

• (d) The function

  $$T\left(x\right):=\frac{x{R}^{\prime }\left(x\right)}{R\left(x\right)}$$

is quasi-increasing in (0, d), with

$$T\left(x\right)\ge \Lambda >\frac{1}{2},x\in \left(0,d\right);$$

• (e) There exists C ₁ > 0 such that

  $$\frac{\mid R^{″}\left(x\right)\mid }{R\prime \left(x\right)}\le C_1\frac{R^{\prime }\left(x\right)}{R\left(x\right)},\text{a}.\text{e}.x\in \left(0,d\right).$$

Then, we write $w\in \mathcal{L}\left(C^2\right)$. If there also exist a compact subinterval J* ∋ 0 of $I^{*}=\left(-\sqrt{d},\sqrt{d}\right)$ and C₂ > 0 such that

$$\frac{Q^{″}\left(t\right)}{\mid Q^{\prime }\left(t\right)\mid }\ge C_2\frac{\mid Q^{\prime }\left(t\right)\mid }{Q\left(t\right)},\text{a}.\text{e}.t\in I^{*}\backslash J^{*},$$

then we write $w\in \mathcal{L}\left(C^2+\right)$.

We consider the case d = ∞, that is, the space ℝ⁺ = [0, ∞), and we strengthen Definition 1.1 slightly.

Definition 1.2. We assume that R : ℝ⁺ → ℝ⁺ has the following properties:

• (a) R(x), R'(x) are continuous, positive in ℝ⁺, with R(0) = 0, R'(0) = 0;

• (b) R″(x) > 0 exists in ℝ⁺\{0};

• (c)
  $$\underset{x\to \mathrm{\infty }}{lim}R\left(x\right)=\mathrm{\infty };$$

• (d) The function

  $$T\left(x\right):=\frac{x{R}^{\prime }\left(x\right)}{R\left(x\right)}$$

is quasi-increasing in ℝ⁺\{0}, with

$$T\left(x\right)\ge \Lambda >\frac{1}{2},x\in {ℝ}^{+}\backslash \left\{0\right\};$$

• (e) There exists C ₁ > 0 such that

  $$\frac{R^{″}\left(x\right)}{R^{\prime }\left(x\right)}\le C_1\frac{R^{\prime }\left(x\right)}{R\left(x\right)},\text{a}.\text{e}.x\in {ℝ}^{+}\backslash \left\{0\right\}.$$

There exist a compact subinterval J ∋ 0 of ℝ⁺ and C₂ > 0 such that

$$\frac{R^{″}\left(x\right)}{R^{\prime }\left(x\right)}\ge C_2\frac{R^{\prime }\left(x\right)}{R\left(x\right)},\text{a}.\text{e}.t\in {ℝ}^{+}\backslash J,$$

then we write $w\in {\mathcal{L}}_2$.

To obtain estimations of the coefficients of higher order Hermite-Fejér interpolation polynomial based at the zeros ${\left\{x_{k,n,\rho }\right\}}_{k=1}^n$, we need to focus on a smaller class of weights.

Definition 1.3. Let $w=exp\left(-R\right)\in {\mathcal{L}}_2$ and let ν ≥ 2 be an integer. For the exponent R, we assume the following:

• (a) R ^(j)(x) > 0, for 0 ≤ j ≤ ν and x > 0, and R ^(j)(0) = 0, 0 ≤ j ≤ ν - 1.

• (b) There exist positive constants C _i > 0, i = 1, 2, ..., ν - 1 such that for i = 1, 2, ..., ν - 1

  $$R^{\left(i+1\right)}\left(x\right)\le C_i{R}^{\left(i\right)}\left(x\right)\frac{R^{\prime }\left(x\right)}{R\left(x\right)},\text{a}.\text{e}.x\in {ℝ}^{+}\backslash \left\{0\right\}.$$

• (c) There exist positive constants C, c ₁ > 0 and 0 ≤ δ < 1 such that on x ∈ (0, c ₁)

  $$R^{\left(\nu \right)}\left(x\right)\le C{\left(\frac{1}{x}\right)}^{\delta }.$$

  (1.1)

• (d) There exists c ₂ > 0 such that we have one among the following

(d1) $T\left(x\right)\text{/}\sqrt{x}$ is quasi-increasing on (c₂, ∞),

(d2) R^(ν)(x) is nondecreasing on (c₂, ∞).

Then we write $w\left(x\right)=e^{-R\left(x\right)}\in {\tilde{\mathcal{L}}}_{\nu }$.

Example 1.4.[6, 7] Let ν ≥ 2 be a fixed integer. There are some typical examples satisfying all conditions of Definition 1.3 as follows: Let α > 1, l ≥ 1, where l is an integer. Then we define

$$R_{l,\alpha }\left(x\right)={exp}_l\left(x^{\alpha }\right)-{exp}_l\left(0\right),$$

where exp _l (x) = exp(exp(exp ... exp(x)) ...) is the l-th iterated exponential.

• (1) If α > ν, $w\left(x\right)=e^{-R_{l,\alpha }\left(x\right)}\in {\tilde{\mathcal{L}}}_{\nu }$.

• (2) If α ≤ ν and α is an integer, we define

  $$R_{l,\alpha }^{*}\left(x\right)={exp}_l\left(x^{\alpha }\right)-{exp}_l\left(0\right)-\sum _{j=1}^r\frac{R_{l,\alpha }^{\left(j\right)}\left(0\right)}{j!}{x}^j.$$

Then $w\left(x\right)=e^{-R_{l,\alpha }^{*}\left(x\right)}\in {\tilde{\mathcal{L}}}_{\nu }$.

In the remainder of this article, we consider the classes ${\mathcal{L}}_2$ and ${\tilde{\mathcal{L}}}_{\nu }$; Let $w\in {\mathcal{L}}_2$ or $w\in {\tilde{\mathcal{L}}}_{\nu }\nu \ge 2$. For $\rho >-\frac{1}{2}$, we set w _ρ (x): = x^ρw(x). Then we can construct the orthonormal polynomials $p_{n,\rho }\left(x\right)=p_n\left(w_{\rho }^2;x\right)$ of degree n with respect to $w_{\rho }^2\left(x\right)$. That is,

$${\int }_0^{\mathrm{\infty }}{p}_{n,\rho }\left(u\right)p_{m,\rho }\left(u\right)w_{\rho }^2\left(u\right)du={\delta }_{nm}\left(\text{Kronecker’s delta}\right)n,m=0,1,2,\dots .$$

Let us denote the zeros of p _n,ρ (x) by

$$0<x_{n,n,\rho }<\cdots <x_{2,n,\rho }<x_{1,n,\rho }<\mathrm{\infty }.$$

The Mhaskar-Rahmanov-Saff numbers a _v is defined as follows:

$$v=\frac{1}{\pi }{\int }_0^1\frac{a_{v}t{R}^{\prime }\left(a_{v}t\right)}{\sqrt{t\left(1-t\right)}}\text{d}t,v>0.$$

Let l, m be non-negative integers with 0 ≤ l < m ≤ ν. For f ∈ C^(l)(ℝ), we define the (l, m)-order Hermite-Fejér interpolation polynomials $L_n\left(l,m,f;x\right)\in {\mathcal{P}}_{mn-1}$ as follows: For each k = 1, 2, ..., n,

$$\begin{array}{c}{L}_n^{\left(j\right)}\left(l,m,f;x_{k,n,\rho }\right)=f^{\left(j\right)}\left(x_{k,n,\rho }\right),j=0,1,2,\dots ,l,\\ L_n^{\left(j\right)}\left(l,m,f;x_{k,n,\rho }\right)=0,j=l+1,l+2,\dots ,m-1.\end{array}$$

For each $P\in {\mathcal{P}}_{mn-1}$, we see L _n (m - 1, m, P; x) = P(x). The fundamental polynomials $h_{s,k,n,\rho }\left(m;x\right)\in {\mathcal{P}}_{mn-1}$, k = 1, 2, ..., n, of L _n (l, m, f; x) are defined by

$$h_{s,k,n,\rho }\left(l,m;x\right)=l_{k,n,\rho }^m\left(x\right)\sum _{i=s}^{m-1}{e}_{s,i}\left(l,m,k,n\right){\left(x-x_{k,n,\rho }\right)}^i.$$

(1.2)

Here, l_{k, n, ρ}(x) is a fundamental Lagrange interpolation polynomial of degree n - 1 [[8], p. 23] given by

$$l_{k,n,\rho }\left(x\right)=\frac{p_n\left(w_{\rho }^2;x\right)}{\left(x-x_{k,n,\rho }\right)p_n^{\prime }\left(w_{\rho }^2;x_{k,n,\rho }\right)}$$

and h_s,_{k, n, ρ}(l, m; x) satisfies

$$h_{s,k,n,\rho }^{\left(j\right)}\left(l,m;x_{p,n,\rho }\right)={\delta }_{s,j}{\delta }_{k,p}j,s=0,1,\dots ,m-1,p=1,2,\dots ,n.$$

(1.3)

Then

$$L_n\left(l,m,f;x\right)=\sum _{k=1}^n\sum _{s=0}^l{f}^{\left(s\right)}\left(x_{k,n,\rho }\right)h_{s,k,n,\rho }\left(l,m;x\right).$$

In particular, for f ∈ C(ℝ), we define the m-order Hermite-Fejér interpolation polynomials $L_n\left(m,f;x\right)\in {\mathcal{P}}_{mn-1}$ as the (0, m)-order Hermite-Fejér interpolation polynomials L _n (0, m, f; x). Then we know that

$$L_n\left(m,f;x\right)=\sum _{k=1}^{n}f\left(x_{k,n,\rho }\right)h_{k,n,\rho }\left(m;x\right),$$

where e _i (m, k, n): = e_0,i(0, m, k, n) and

$$h_{k,n,\rho }\left(m;x\right)=l_{k,n,\rho }^m\left(x\right)\sum _{i=0}^{m-1}{e}_i\left(m,k,n\right){\left(x-x_{k,n,\rho }\right)}^i.$$

(1.4)

We often denote l_{k, n}(x): = l_{k, n, ρ}(x), h_{s, k, n}(x): = h_{s, k, n, ρ}(x), and x_{k, n}: = x_{k, n, ρ}if they do not confuse us.

Theorem 1.5. Let$w\left(x\right)=exp\left(-R\left(x\right)\right)\in \mathcal{L}\left(C^2+\right)$and ρ > -1/2.

• (a) For each m ≥ 1 and j = 0, 1, ..., we have

  $$\mid {\left(l_{k,n}^m\right)}^{\left(j\right)}\left(x_{k,n}\right)\mid \le C{\left(\frac{n}{\sqrt{a_{2n}-x_{k,n}}}\right)}^j{x}_{k,n}^{-\frac{j}{2}}.$$

  (1.5)

• (b) For each m ≥ 1 and j = s, ..., m - 1, we have e _{s, s}(l, m, k, n) = 1/s! and

  $$\mid e_{s,j}\left(l,m,k,n\right)\mid \le C{\left(\frac{n}{\sqrt{a_{2n}-x_{k,n}}}\right)}^{j-s}{x}_{k,n}^{-\frac{j-s}{2}}.$$

  (1.6)

We remark ${\mathcal{L}}_2\subset \mathcal{L}\left(C^2+\right)$.

Theorem 1.6. Let$w\left(x\right)=exp\left(-R\left(x\right)\right)\in {\tilde{\mathcal{L}}}_{\nu },\nu \ge 2$and ρ > -1/2. Assume that 1 + 2ρ -δ/2 ≥ 0 for ρ < -1/4 and if T(x) is bounded, then assume that

$$a_n\le C{n}^{2∕\left(1+\nu -\delta \right)},$$

(1.7)

where 0 ≤ δ < 1 is defined in (1.1). Then we have the following:

• (a) If j is odd, then we have for m ≥ 1 and j = 0, 1, ..., ν - 1,

  $$\begin{array}{cc}\hfill \mid {\left(l_{k,n}^m\right)}^{\left(j\right)}\left(x_{k,n}\right)\mid \le & C\left(\frac{T\left(a_n\right)}{\sqrt{a_n{x}_{k,n}}}+R^{\prime }\left(x_{k,n}\right)+\frac{1}{x_{k,n}}\right)\hfill \\ \times {\left(\frac{n}{\sqrt{a_{2n}}-\sqrt{x_{k,n}}}+\frac{T\left(a_n\right)}{\sqrt{a_n}}\right)}^{j-1}{x}_{k,n}^{-\frac{j-1}{2}}.\hfill \end{array}$$

  (1.8)

• (b) If j - s is odd, then we have for m ≥ 1 and 0 ≤ s ≤ j ≤ m - 1,

  $$\begin{array}{cc}\hfill \mid e_{s,j}\left(l,m,k,n\right)\mid \le & C\left(\frac{T\left(a_n\right)}{\sqrt{a_n{x}_{k,n}}}+R^{\prime }\left(x_{k,n}\right)+\frac{1}{x_{k,n}}\right)\hfill \\ \times {\left(\frac{n}{\sqrt{a_{2n}}-\sqrt{x_{k,n}}}+\frac{T\left(a_n\right)}{\sqrt{a_n}}\right)}^{j-s-1}{x}_{k,n}^{-\frac{j-s-1}{2}}.\hfill \end{array}$$

  (1.9)

Theorem 1.7. Let 0 < ε < 1/4. Let$\frac{1}{\epsilon }\frac{a_n}{n^2}\le x_{k,n}\le \epsilon a_n$. Let s be a positive integer with 2 ≤ 2s ≤ ν. Then under the same conditions as the assumptions of Theorem 1.6, there exists μ₁(ε, n) > 0 such that

$$\left|p_{n,\rho }^{\left(2s\right)}\left(x_{k,n}\right)\right|\le C\delta \left(\epsilon ,n\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s-1}\left|p_n^{\prime }\left(x_{k,n}\right)\right|x_{k,n}^{-\frac{\left(2s-1\right)}{2}}$$

and δ (ε, n) → 0 as n → ∞ and ε → 0.

Theorem 1.8. [4, Lemma 10] Let 0 < ε < 1/4. Let$\frac{1}{\epsilon }\frac{a_n}{n^2}\le x_{k,n}\le \epsilon a_n$. Let s be a positive integer with 2 ≤ 2s ≤ ν - 1. Suppose the same conditions as the assumptions of Theorem 1.6. Then

• (a) for 1 ≤ 2s - 1 ≤ ν - 1,

  $$\left|{\left(l_{k,n}^m\right)}^{\left(2s-1\right)}\left(x_{k,n}\right)\right|\le C\delta \left(\epsilon ,n\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s-1}{x}_{k,n}^{-\frac{2s-1}{2}},$$

  (1.10)

where δ(ε, n) is defined in Theorem 1.7.

• (b) there exists β(n, k) with 0 < D ₁ ≤ β(n, k) ≤ D ₂ for absolute constants D ₁, D ₂ such that the following holds:

  $${\left(l_{k,n}^m\right)}^{\left(2s\right)}\left(x_{k,n}\right)={\left(-1\right)}^s{\varphi }_s\left(m\right){\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}\left(1+{\xi }_s\left(m,\epsilon ,x_{k,n},n\right)\right)$$

  (1.11)

and |ξ _s (m, ε, x_{k, n}, n)| → 0 as n → ∞ and ε → 0.

Theorem 1.9. [4, (4.16)], [9]Let 0 < ε < 1/4. Let$\frac{1}{\epsilon }\frac{a_n}{n^2}\le x_{k,n}\le \epsilon a_n$. Let s be a positive integer with 2 ≤ 2s ≤ m - 1. Suppose the same conditions as the assumptions of Theorem 1.6. Then for j = 0, 1, 2, ..., there is a polynomial Ψ _j (x) of degree j such that (-1) ^j ψ _j (-m) > 0 for m = 1, 3, 5, ... and the following relation holds:

$$e_{2s}\left(m,k,n\right)=\frac{{\left(-1\right)}^s}{\left(2s\right)!}{\mathrm{\Psi }}_s\left(-m\right){\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}\left(1+{\eta }_s\left(m,\epsilon ,x_{k,n},n\right)\right)$$

(1.12)

and |η _s (m, ε, x_{k, n}, n)| → 0 as n → ∞ and ε → 0.

Theorem 1.10. Let m be an odd positive integer. Suppose the same conditions as the assumptions of Theorem 1.6. Then there is a function f in C(ℝ⁺) such that for any fixed interval [a, b], a > 0,

$$\underset{n\to \mathrm{\infty }}{\text{lim}\text{sup}}\underset{a\le x\le b}{\text{max}}\left|L_n\left(m,f;x\right)\right|=\mathrm{\infty }.$$

2. Preliminaries

Levin and Lubinsky introduced the classes $\mathcal{L}\left(C^2\right)$ and $\mathcal{L}\left(C^2+\right)$ as analogies of the classes $\mathcal{F}\left(C^2\right)$ and $\mathcal{F}\left(C^2+\right)$ defined on $I^{*}=\left(-\sqrt{d},\sqrt{d}\right)$. They defined the following:

Definition 2.1.[10] We assume that Q : I* → [0, ∞) has the following properties:

• (a) Q(t) is continuous in I*, with Q(0) = 0;

• (b) Q″(t) exists and is positive in I*\{0};

• (c)
  $$\underset{t\to \sqrt{d}-}{\text{lim}}Q\left(t\right)=\mathrm{\infty };$$

• (d) The function

  $$T^{*}\left(t\right):=\frac{t{Q}^{\prime }\left(t\right)}{Q\left(t\right)}$$

is quasi-increasing in $\left(0,\sqrt{d}\right)$, with

$$T^{*}\left(t\right)\ge {\Lambda }^{*}>1,t\in I^{*}\backslash \left\{0\right\};$$

• (e) There exists C ₁ > 0 such that

  $$\frac{Q^{″}\left(t\right)}{\mid Q^{\prime }\left(t\right)\mid }\le C_1\frac{\mid Q^{\prime }\left(t\right)\mid }{Q\left(t\right)},\text{a}.\text{e}.t\in I^{*}\backslash \left\{0\right\}.$$

Then we write $W\in \mathcal{F}\left(C^2\right)$. If there also exist a compact subinterval J* ∋ 0 of I* and C₂ > 0 such that

$$\frac{Q^{″}\left(t\right)}{\mid Q^{\prime }\left(t\right)\mid }\ge C_2\frac{Q^{\prime }\left(t\right)}{\mid Q\left(t\right)\mid },\text{a}.\text{e}.t\in I^{*}\backslash J^{*},$$

then we write $W\in \mathcal{F}\left(C^2+\right)$.

Then we see that $w\in \mathcal{L}\left(C^2\right)\iff W\in \mathcal{F}\left(C^2\right)$ and $w\in \mathcal{L}\left(C^2+\right)\iff W\in \mathcal{F}\left(C^2+\right)$ where W(t) = w(x), x = t², from [6, Lemma 2.2]. In addition, we easily have the following:

Lemma 2.2.[1]Let Q(t) = R(x), x = t². Then we have

$$w\in {\mathcal{L}}_2\Rightarrow W\in \mathcal{F}\left(C^2+\right),$$

where W(t) = w(x); x = t².

On ℝ, we can consider the corresponding class to ${\tilde{\mathcal{L}}}_{\nu }$ as follows:

Definition 2.3.[11] Let $W=exp\left(-Q\right)\in \mathcal{F}\left(C^2+\right)$ and ν ≥ 2 be an integer. Let Q be a continuous and even function on ℝ. For the exponent Q, we assume the following:

• (a) Q ^(j)(x) > 0, for 0 ≤ j ≤ ν and t ∈ ℝ⁺/{0}.

• (b) There exist positive constants C _i > 0 such that for i = 1, 2, ..., ν - 1,

  $$Q^{\left(i+1\right)}\left(t\right)\le C_i{Q}^{\left(i\right)}\left(t\right)\frac{Q^{\prime }\left(t\right)}{Q\left(t\right)},\text{a}.\text{e}.x\in {ℝ}^{+}\backslash \left\{0\right\}.$$

• (c) There exist positive constants C, c ₁ > 0, and 0 ≤ δ* < 1 such that on t ∈ (0, c ₁),

  $$Q^{\left(\nu \right)}\left(t\right)\le C{\left(\frac{1}{t}\right)}^{{\delta }^{*}}.$$

  (2.1)

• (d) There exists c ₂ > 0 such that we have one among the following:

(d1) T* (t)/t is quasi-increasing on (c₂, ∞),

(d2) Q^(ν)(t) is nondecreasing on (c₂, ∞).

Then we write $W\left(t\right)=e^{-Q\left(t\right)}\in {\tilde{\mathcal{F}}}_{\nu }$.

Let $W\in {\tilde{\mathcal{F}}}_{\nu }$, and ν ≥ 2. For ${\rho }^{*}>-\frac{1}{2}$, we set

$$W_{{\rho *}^{}}\left(t\right):=\mid t{\mid }^{\rho *}W\left(t\right).$$

Then we can construct the orthonormal polynomials $P_{{n,\rho *}^{}}\left(t\right)=P_n\left(W_{{\rho *}^{}}^2;t\right)$ of degree n with respect to W_ρ*(t). That is,

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{P}_{{n,\rho *}^{}}\left(v\right)P_{{m,\rho *}^{}}\left(v\right)W_{{\rho *}^{}}^2\left(v\right)\text{d}t={\delta }_{nm},n,m=0,1,2,\dots .$$

Let us denote the zeros of P_{n, ρ*}(t) by

$$-\mathrm{\infty }<t_{nn}<\cdots <t_{2n}<t_{1n}<\mathrm{\infty }.$$

There are many properties of P_{n, ρ*}(t) = P _n (W_ρ*; t) with respect to W_ρ*(t), $W\in {\tilde{\mathcal{F}}}_{\nu },\nu =2,3,\dots$ of Definition 2.3 in [2, 3, 7, 11–13]. They were obtained by transformations from the results in [5, 6]. Jung and Sakai [2, Theorem 3.3 and 3.6] estimate $P_{{n,\rho *}^{}}^{\left(j\right)}\left(t_{k,n}\right)$, k = 1, 2, ..., n, j = 1, 2, ..., ν and Jung and Sakai [1, Theorem 3.2 and 3.3] obtained analogous estimations with respect to $p_{n,\rho }^{\left(j\right)}\left(x_{k,n}\right)$, k = 1, 2, ..., n, j = 1, 2, ..., ν. In this article, we consider $w=exp\left(-R\right)\in {\tilde{\mathcal{L}}}_{\nu }$ and p_{n, ρ}(x) = p _n (w _ρ ; x). In the following, we give the transformation theorems.

Theorem 2.4. [13, Theorem 2.1] Let W(t) = W(x) with x = t². Then the orthonormal polynomials P_{n, ρ*}(t) on ℝ can be entirely reduced to the orthonormal polynomials p _n , _ρ (x) in ℝ⁺as follows: For n = 0, 1, 2, ...,

$$P_{2n,2\rho +\frac{1}{2}}\left(t\right)=p_{n,\rho }\left(x\right)and{P}_{2n+1,2\rho -\frac{1}{2}}\left(t\right)=t{p}_{n,\rho }\left(x\right).$$

In this article, we will use the fact that w _ρ (x) = x^ρ exp(-R(x)) is transformed into W_2ρ+1/2(t) = |t|^2ρ+1/2exp (-Q(t)) as meaning that

$$\begin{array}{cc}\hfill {\int }_0^{\mathrm{\infty }}{p}_{n,\rho }\left(x\right)p_{m,\rho }\left(x\right)w_{\rho }^2\left(x\right)\text{d}x& =2{\int }_0^{\mathrm{\infty }}{p}_{n,\rho }\left(t^2\right)p_{m,\rho }\left(t^2\right)t^{4\rho +1}{W}^2\left(t\right)\text{d}t\hfill \\ ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{P}_{2n,2\rho +1∕2}\left(t\right)P_{2m,2\rho +1∕2}\left(t\right)W_{2\rho +1∕2}^2\left(t\right)\text{d}t.\hfill \end{array}$$

Theorem 2.5. [1, Theorem 2.5] Let Q(t) = R(x), x = t². Then we have

$$w\left(x\right)=exp\left(-R\left(x\right)\right)\in {\tilde{\mathcal{L}}}_{\nu }\Rightarrow W\left(t\right)=exp\left(-Q\left(t\right)\right)\in {\tilde{\mathcal{F}}}_{\nu }.$$

(2.2)

In particular, we have

$$Q^{\left(\nu \right)}\left(t\right)\le C{\left(\frac{1}{t}\right)}^{\delta },$$

where 0 ≤ δ < 1 is defined in (1.1).

For convenience, in the remainder of this article, we set as follows:

$${\rho }^{*}:=2\rho +\frac{1}{2}\text{for}\rho >-\frac{1}{2},p_n\left(x\right):=p_{n,\rho }\left(x\right),P_n\left(t\right):=P_{n,{\rho }^{*}}\left(t\right),$$

(2.3)

and $x_{k,n}=x_{k,n,\rho },t_{kn}=t_{k,n,{\rho }^{*}}$. Then we know that ${\rho }^{*}>-\frac{1}{2}$ and

$$p_n\left(x\right)=P_{2n,{\rho }^{*}}\left(t\right),x=t^2,x_{k,n}=t_{k,2n}^2,t_{k,2n}>0,k=1,2,\dots ,n.$$

(2.4)

In the following, we introduce useful notations:

• (a) The Mhaskar-Rahmanov-Saff numbers a _v and $a_u^{*}$ are defined as the positive roots of the following equations, that is,

  $$v=\frac{1}{\pi }{\int }_0^1{a}_{v}t{R}^{\prime }\left(a_{v}t\right){\{t\left(1-t\right)\}}^{-\frac{1}{2}}dt,v>0$$

and

$$u=\frac{2}{\pi }{\int }_0^1{a}_u^{*}t{Q}^{\prime }\left(a_u^{*}t\right){\left(1-t^2\right)}^{-\frac{1}{2}}dt,u>0.$$

• (b) Let

  $${\eta }_n={\left\{nT\left(a_n\right)\right\}}^{-\frac{2}{3}}\text{and}{\eta }_n^{*}={\left\{n{T}^{*}\left(a_n^{*}\right)\right\}}^{-\frac{2}{3}}.$$

Then we have the following:

Lemma 2.6. [6, (2.5),(2.7),(2.9)]

$$a_n={a_{2n}^{*}}^2,{\eta }_n=4^{2/3}{\eta }_{2n}^{*},T\left(a_n\right)=\frac{1}{2}{T}^{*}\left(a_{2n}^{*}\right).$$

To prove main results, we need some lemmas as follows:

Lemma 2.7. [13, Theorem 2.2, Lemma 3.7] For the minimum positive zero, t_[n/2],n([n/2] is the largest integer ≤ n/2), we have

$$t_{\left[n/2\right],n}~a_n^{*}{n}^{-1},$$

and for the maximum zero t_1nwe have for large enough n,

$$1-\frac{t_{1n}}{a_n^{*}}~{\eta }_n^{*},{\eta }_n^{*}={\left(n{T}^{*}\left(a_n^{*}\right)\right)}^{-\frac{2}{3}}.$$

Moreover, for some constant 0 < ε ≤ 2 we have

$$T^{*}\left(a_n^{*}\right)\le C{n}^{2-\epsilon }.$$

Remark 2.8. (a) Let $W\left(t\right)\in \mathcal{F}\left(C^2+\right)$. Then

(a-1) T(x) is bounded ⇔ T*(t) is bounded.

(a-2) T(x) is unbounded ⇒ a _n ≤ C(η)n^η for any η > 0.

(a-3) T(a _n ) ≤ Cn^2-εfor some constant 0 < ε ≤ 2.

• (b) Let $w\left(x\right)\in {\tilde{\mathcal{L}}}_{\nu }$. Then

(b-1) ρ > -1/2 ⇒ ρ* > -1/2.

(b-2) 1 + 2ρ - δ/2 ≥ 0 for ρ < -1/4 ⇒ 1 + 2ρ* - δ* ≥ 0 for ρ* < 0.

(b-3) $a_n\le C{n}^{2/\left(1+\nu -\delta \right)}\Rightarrow a_n^{*}\le C{n}^{1/\left(1+\nu -{\delta }^{*}\right)}$.

Proof of Remark 2.8. (a) (a-1) and (a-3) are easily proved from Lemma 2.6. From [11, Theorem 1.6], we know the following: When T*(t) is unbounded, for any η > 0 there exists C(η) > 0 such that

$$a_t^{*}\le C\left(\eta \right)t^{\eta },t\ge 1.$$

In addition, since T(x) = T*(t)/2 and $a_n=a_{2n}^{*}{2}^{}$, we know that (a-2).

• (b) Since $w\left(x\right)\in {\tilde{\mathcal{L}}}_{\nu }$, we know that $W\left(t\right)\in {\tilde{\mathcal{F}}}_{\nu }$ and δ* = δ by Theorem 2.5. Then from (2.3) and Lemma 2.6, we have (b-1), (b-2), and (b-3).    □

Lemma 2.9. [1, Lemma 3.6] For j = 1, 2, 3, ..., we have

$$p_n^{\left(j\right)}\left(x\right)=\sum _{i=1}^j{\left(-1\right)}^{j-i}{c}_{j,i}{P}_{2n}^{\left(i\right)}\left(t\right)t^{-2j+i},$$

where c _{j, i} > 0(1 ≤ i ≤ j, j = 1, 2, ...) satisfy the following relations: for k = 1, 2, ...,

$$c_{k+1,1}=\frac{2k-1}{2}{c}_{k,1},c_{k+1,k+1}=\frac{1}{2^{k+1}},c_{1,1}=\frac{1}{2},$$

and for 2 ≤ i ≤ k,

$$c_{k+1,i}=\frac{c_{k,i-1}+\left(2k-i\right)c_{k,i}}{2}.$$

3. Proofs of main results

Our main purpose is to obtain estimations of the coefficients e _{s, i} (l, m, k, n), k = 1, 2, ..., 0 ≤ s ≤ l, s ≤ i ≤ m - 1.

Theorem 3.1. [1, Theorem 1.5] Let$w\left(x\right)=exp\left(-R\left(x\right)\right)\in \mathcal{L}\left(C^2+\right)$and let ρ > -1/2. For each k = 1, 2, ..., n and j = 0, 1, ..., we have

$$\mid p_{n,\rho }^{\left(j\right)}\left(x_{k,n}\right)\mid \le C{\left(\frac{n}{\sqrt{a_{2n}-x_{k,n}}}\right)}^{j-1}{x}_{k,n}^{-\frac{j-1}{2}}\mid p_{n,\rho }^{\prime }\left(x_{k,n}\right)\mid .$$

Proof of Theorem 1.5. (a) From Theorem 3.1 we know that

$$\mid l_{k,n}^{\left(j\right)}\left(x_{k,n}\right)\mid =\left|\frac{p_n^{\left(j+1\right)}\left(x_{k,n}\right)}{\left(j+1\right)p_n^{\prime }\left(x_{k,n}\right)}\right|\le C{\left(\frac{n}{\sqrt{a_{2n}-x_{k,n}}}\right)}^j{x}_{k,n}^{-\frac{j}{2}}.$$

Then, assuming that (a) is true for 1 ≤ m' < m, we have

$$\begin{array}{cc}\hfill \mid {\left(l_{k,n}^m\right)}^{\left(j\right)}\left(x_{k,n}\right)\mid & =\left|\sum _{s=0}^j\left(\begin{array}{c}\hfill j\hfill \\ \hfill s\hfill \end{array}\right){\left(l_{k,n}^{m-1}\right)}^{\left(s\right)}\left(x_{k,n}\right){l_{k,n}}^{\left(j-s\right)}\left(x_{k,n}\right)\right|\hfill \\ \le C\sum _{s=0}^j{\left(\frac{n}{\sqrt{a_{2n}-x_{k,n}}}\right)}^s{x}_{k,n}^{-\frac{s}{2}}{\left(\frac{n}{\sqrt{a_{2n}-x_{k,n}}}\right)}^{j-s}{x}_{k,n}^{-\frac{j-s}{2}}\hfill \\ \le {\left(\frac{n}{\sqrt{a_{2n}-x_{k,n}}}\right)}^j{x}_{k,n}^{-\frac{j}{2}}.\hfill \end{array}$$

Therefore, the result is proved by induction with respect to m.

• (b) From (2) and (3), we know e _{s, s} (l, m, k, n) = 1/s! and the following recurrence relation: for s + 1 ≤ i ≤ m - 1,

  $$e_{s,i}\left(l,m,k,n\right)=-\sum _{p-s}^{i-1}\frac{1}{\left(i-p\right)!}{e}_{s,p}\left(l,m,k,n\right){\left(l_{k,n}\right)}^{\left(i-p\right)}\left(x_{k,n}\right).$$

  (3.5)

Therefore, we have the result by induction on i and (3.5).

Theorem 3.2. [1, Theorem 1.6] Let$w\left(x\right)=exp\left(-R\left(x\right)\right)\in {\tilde{\mathcal{L}}}_{\nu }$and let ρ > -1/2. Suppose the same conditions as the assumptions of Theorem 1.6. For each k = 1, 2, ..., n and j = 1, ..., ν, we have

$$\mid p_{n,\rho }^{\left(j\right)}\left(x_{k,n}\right)\mid \le C{\left(\frac{n}{\sqrt{a_n}-\sqrt{x_{k,n}}}+\frac{T\left(a_n\right)}{\sqrt{a_n}}\right)}^{j-1}{x}_{k,n}^{-\frac{j-1}{2}}\mid p_{n,\rho }^{\prime }\left(x_{k,n}\right)\mid$$

and in particular, if j is even, then we have

$$\begin{array}{cc}\hfill \mid p_{n,\rho }^{\left(j\right)}\left(x_{k,n}\right)\mid \le & C\left(\frac{T\left(a_n\right)}{\sqrt{a_n{x}_{k,n}}}+R^{\prime }\left(x_{k,n}\right)+\frac{1}{x_{k,n}}\right)\hfill \\ \times {\left(\frac{n}{\sqrt{a_n}-\sqrt{x_{k,n}}}+\frac{T\left(a_n\right)}{\sqrt{a_n}}\right)}^{j-2}{x}_{k,n}^{-\frac{j-2}{2}}\mid p_{n,\rho }^{\prime }\left(x_{k,n}\right)\mid .\hfill \end{array}$$

Proof of Theorem 1.6. We use the induction method on m.

• (a) For m = 1, we have the result because of

  $$l_{k,n}^{\left(j\right)}\left(x_{k,n}\right)=\frac{p_n^{\left(j+1\right)}\left(x_{k,n}\right)}{\left(j+1\right)p_n^{\prime }\left(x_{k,n}\right)},j=1,2,3,\dots ,$$

and Theorem 3.2. Now we assume the theorem for 1 ≤ m' < m. Then, we have the following: For 1 ≤ 2s - 1 ≤ ν - 1,

$$\begin{array}{cc}\hfill {\left(l_{k,n}^m\right)}^{\left(2s-1\right)}\left(x_{k,n}\right)=& \sum _{r=0}^s\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill 2r\hfill \end{array}\right){\left(l_{k,n}^{m-1}\right)}^{\left(2r\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r-1\right)}\left(x_{k,n}\right)\hfill \\ +\sum _{r=0}^s\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill 2r+1\hfill \end{array}\right){\left(l_{k,n}^{m-1}\right)}^{\left(2r+1\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r-2\right)}\left(x_{k,n}\right).\hfill \end{array}$$

Since

$$\frac{n}{\sqrt{a_{2n}-x_{k,n}}}\le \frac{n}{\sqrt{a_{2n}}-\sqrt{x_{k,n}}},$$

we have

$$\begin{array}{c}\left|{\left(l_{k,n}^{m-1}\right)}^{\left(2r\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r-1\right)}\left(x_{k,n}\right)\right|\hfill \\ \le C\left(\frac{T\left(a_n\right)}{\sqrt{a_n{x}_{k,n}}}+R^{\prime }\left(x_{k,n}\right)+\frac{1}{x_{k,n}}\right)\hfill \\ \times {\left(\frac{n}{\sqrt{a_{2n}-x_{k,n}}}\right)}^{2r}{\left(\frac{n}{\sqrt{a_{2n}}-\sqrt{x_{k,n}}}+\frac{T\left(a_n\right)}{\sqrt{a_n}}\right)}^{2s-2r-2}{x}_{k,n}^{-s+1}\hfill \\ \le C\left(\frac{T\left(a_n\right)}{\sqrt{a_n{x}_{k,n}}}+R^{\prime }\left(x_{k,n}\right)+\frac{1}{x_{k,n}}\right)\hfill \\ \times {\left(\frac{n}{\sqrt{a_{2n}}-\sqrt{x_{k,n}}}+\frac{T\left(a_n\right)}{\sqrt{a_n}}\right)}^{2s-2}{x}_{k,n}^{-s+1},\hfill \end{array}$$

and similarly

$$\begin{array}{cc}\hfill \left|{\left(l_{k,n}^{m-1}\right)}^{\left(2r+1\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r-2\right)}\left(x_{k,n}\right)\right|\le & C\left(\frac{T\left(a_n\right)}{\sqrt{a_n{x}_{k,n}}}+R^{\prime }\left(x_{k,n}\right)+\frac{1}{x_{k,n}}\right)\hfill \\ \times {\left(\frac{n}{\sqrt{a_{2n}}-\sqrt{x_{k,n}}}+\frac{T\left(a_n\right)}{\sqrt{a_n}}\right)}^{2s-2}{x}_{k,n}^{-s+1}.\hfill \end{array}$$

Therefore, we have

$$\begin{array}{cc}\hfill \left|{\left(l_{k,n}^m\right)}^{\left(2s-1\right)}\left(x_{k,n}\right)\right|\le & C\left(\frac{T\left(a_n\right)}{\sqrt{a_n{x}_{k,n}}}+R^{\prime }\left(x_{k,n}\right)+\frac{1}{x_{k,n}}\right)\hfill \\ \times {\left(\frac{n}{\sqrt{a_{2n}}-\sqrt{x_{k,n}}}+\frac{T\left(a_n\right)}{\sqrt{a_n}}\right)}^{2s-2}{x}_{k,n}^{-s+1}.\hfill \end{array}$$

• (b) To prove the result, we proceed by induction on i. From (1.2) and (1.3) we know e _{s, s} (l, m, k, n) = 1/s! and the following recurrence relation: for s + 1 ≤ i ≤ m - 1,

  $$e_{s,i}\left(l,m,k,n\right)=-\sum _{p=s}^{i-1}\frac{1}{\left(i-p\right)!}{e}_{s,p}\left(l,m,k,n\right){\left(l_{k,n}^m\right)}^{\left(i-p\right)}\left(x_{k,n}\right).$$

  (3.6)

When i - s is odd, we know that

$$\left\{\begin{array}{c}\hfill i-p:\text{odd},\text{if}p-s:\text{even}\hfill \\ \hfill i-p:\text{even},\text{if}p-s:\text{odd}.\hfill \end{array}\right.$$

Then, we have (1.9) from (1.5), (1.8), (3.6), and the assumption of induction on i.   □

Theorem 3.3. [1, Theorem 1.7] Let 0 < ε < 1/4. Let $\frac{1}{\epsilon }\frac{a_n}{n^2}\le x_{k,n}\le \epsilon a_n$and let s be a positive integer with 2 ≤ 2s ≤ ν - 1. Suppose the same conditions as the assumptions of Theorem 1.6. Then there exists β(n, k), 0 < D₁ ≤ β(n, k) ≤ D₂for absolute constants D₁, D₂such that the following equality holds:

$$p_{n,\rho }^{\left(2s+1\right)}\left(x_{k,n}\right)={\left(-1\right)}^s{\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}\left(1+{\rho }_s\left(\epsilon ,x_{k,n},n\right)\right)p_n^{\prime }\left(x_{k,n}\right)x_{k,n}^{-s}$$

and |ρ _s (ε, x _{k, n} , n)| → 0 as n → ∞ and ε → 0.

Lemma 3.4. [3, Theorem 2.5] Let$W\in \mathcal{F}\left(C^2+\right)$and r = 1, 2, ⋯. Then, uniformly for 1 ≤ k ≤ n,

$$\left|\frac{P_n^{(r)}(t_{k,n})}{{P^{\prime }}_n(t_{k,n})}\right|\le C{\left(\frac{n}{\sqrt{a_{2n}^{*}{2}^{}-t_{k,n}^2}}\right)}^{r-1}.$$

Lemma 3.5. [2, Theorem 3.3] Let ρ* > -1/2 and$W\left(x\right)=exp\left(-Q\left(x\right)\right)\in {\tilde{\mathcal{F}}}_{\nu }$, ν ≥ 2. Assume that 1 + 2ρ* - δ* ≥ 0 for ρ* < 0 and if T*(t) is bounded, then assume

$$a_n^{*}\le C{n}^{1/\left(1+\nu -{\delta }^{*}\right)},$$

where 0 ≤ δ* < 1 is defined in (2.1). Let 0 < α < 1/2. Let$\frac{1}{\epsilon }\frac{a_n^{*}}{n}\le \mid t_{kn}\mid \le \epsilon a_n^{*}$and let s be a positive integer with 2 ≤ 2s ≤ ν. Then there exists μ(ε, n) > 0 such that

$$\left|P_n^{\left(2s\right)}\left(t_{k,n}\right)\right|\le C\mu \left(\epsilon ,n\right){\left(\frac{n}{a_n}\right)}^{2s-1}\left|P_n^{\prime }\left(t_{k,n}\right)\right|$$

and μ(ε, n) → 0 as n → ∞ and ε → 0.

Proof of Theorem 1.7. By Lemma 2.9, we have

$$\begin{array}{cc}\hfill \left|p_n^{\left(2s\right)}\left(x_{k,n}\right)\right|& =\left|\sum _{i=1}^{2s}{\left(-1\right)}^{2s-i}{c}_{2s,i}{P}_{2n}^{\left(i\right)}\left(t_{k,n}\right)t_{k,n}^{-4s+i}\right|\hfill \\ \le C\left|c_{2s,2s}{P}_{2n}^{\left(2s\right)}\left(t_{k,n}\right)t_{k,n}^{-2s}\right|+\left|\sum _{i=1}^{2s-1}{\left(-1\right)}^{2s-i}{c}_{2s,i}{P}_{2n}^{\left(i\right)}\left(t_{k,n}\right)t_{k,n}^{-4s+i}\right|.\hfill \end{array}$$

Since, we have by Lemma 3.5,

$$\left|c_{2s,2s}{P}_{2n}^{\left(2s\right)}\left(t_{k,n}\right)t_{k,n}^{-2s}\right|\le C\mu \left(\epsilon ,2n\right){\left(\frac{n}{a_{2n}^{*}}\right)}^{2s-1}\left|P_{2n}^{\prime }\left(t_{k,n}\right)\right|{\left|t_{k,n}\right|}^{-2s}$$

and by Lemma 3.4,

$$\begin{array}{c}\left|\sum _{i=1}^{2s-1}{\left(-1\right)}^{2s-i}{c}_{2s,i}{P}_{2n}^{\left(i\right)}\left(t_{k,n}\right)t_{k,n}^{-4s+i}\right|\hfill \\ \hfill \le & C{\left(\frac{n}{a_{2n}^{*}}\right)}^{2s-1}\left|P_{2n}^{\prime }\left(t_{k,n}\right)\right|{\left|t_{k,n}\right|}^{-2s}\sum _{i=1}^{2s-1}{\left(\frac{n}{a_{2n}^{*}}\mid t_{k,n}\mid \right)}^{-2s+i}\hfill \\ \hfill \le & C\epsilon {\left(\frac{n}{a_{2n}^{*}}\right)}^{2s-1}\left|P_{2n}^{\prime }\left(t_{k,n}\right)\right|{\left|t_{k,n}\right|}^{-2s},\hfill \end{array}$$

we have

$$\begin{array}{cc}\hfill \left|p_n^{\left(2s\right)}\left(x_{k,n}\right)\right|& \le C\delta \left(\epsilon ,n\right){\left(\frac{n}{a_{2n}^{*}}\right)}^{2s-1}\left|P_{2n}^{\prime }\left(t_{k,n}\right)\right|{\left|t_{k,n}\right|}^{-2s}\hfill \\ \le C\delta \left(\epsilon ,n\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s-1}\left|p_n^{\prime }\left(x_{k,n}\right)\right|x_{k,n}^{-\frac{\left(2s-1\right)}{2}},\hfill \end{array}$$

where δ(ε, n) = μ(ε, 2n) + ε.   □

Here we can estimate the coefficients e _i (ν, k, n) of the fundamental polynomials h _kn (ν; x).

For j = 0, 1, ..., define ϕ _j (1): = (2j + 1)^-1 and for k ≥ 2,

$${\phi }_j\left(k\right):=\sum _{r=0}^j\frac{1}{2j-2r+1}\left(\begin{array}{c}\hfill 2j\hfill \\ \hfill 2r\hfill \end{array}\right){\phi }_r\left(k-1\right).$$

(3.7)

Proof of Theorem 1.8. In a manner analogous to the proof of Theorem 1.6 (a), we use mathematical induction with respect to m.

• (a) From Theorem 1.7, we know that for 1 ≤ 2s -1 ≤ ν - 1,

  $$\left|l_{k,n}^{\left(2s-1\right)}\left(x_{k,n}\right)\right|=\left|\frac{p_n^{\left(2s\right)}\left(x_{k,n}\right)}{2s{p}_n^{\prime }\left(x_{k,n}\right)}\right|\le C\delta \left(\epsilon ,n\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s-1}{x}_{k,n}^{-\frac{2s-1}{2}}.$$

From Theorem 1.5, we know that for x _{k, n} ≤ a _n /4,

$$\left|{\left(l_{k,n}^m\right)}^{\left(j\right)}\left(x_{k,n}\right)\right|\le C{\left(\frac{n}{\sqrt{a_n}}\right)}^j{x}_{k,n}^{-\frac{j}{2}}.$$

(3.8)

Then, we have by mathematical induction on m,

$$\begin{array}{cc}\hfill \left|{\left(l_{k,n}^m\right)}^{\left(2s-1\right)}\left(x_{k,n}\right)\right|\le & C\sum _{r=0}^s\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill 2r\hfill \end{array}\right)\left|{\left(l_{k,n}^{m-1}\right)}^{\left(2r\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r-1\right)}\left(x_{k,n}\right)\right|\hfill \\ +\sum _{r=0}^s\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill 2r+1\hfill \end{array}\right)\left|{\left(l_{k,n}^{m-1}\right)}^{\left(2r+1\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r-2\right)}\left(x_{k,n}\right)\right|\hfill \\ \hfill \le & C\delta \left(\epsilon ,n\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s-1}{x}_{k,n}^{-\frac{2s-1}{2}}.\hfill \end{array}$$

• (b) From Theorem 3.3, we know that for 0 ≤ 2s ≤ ν - 1,

  $$\begin{array}{cc}\hfill l_{k,n}^{\left(2s\right)}\left(x_{k,n}\right)& =\frac{p_n^{\left(2s+1\right)}\left(x_{k,n}\right)}{\left(2s+1\right)p_n^{\prime }\left(x_{k,n}\right)}\hfill \\ ={\left(-1\right)}^s{\varphi }_s\left(1\right){\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}\left(1+{\rho }_s\left(\epsilon ,x_{k,n},n\right)\right).\hfill \end{array}$$

  (3.9)

If we let ξ _s (1, ε, x _{k, n} , n) = ρ _s (ε, x _{k, n} , n), then (1.11) holds for m = 1 because |ξ _s (1, ε, x _{k, n} , n)| → 0 as n → ∞ and ε → 0. Now, we split ${\left(l_{k,n}^m\right)}^{\left(2s\right)}\left(x_{k,n}\right)$ into two terms as follows:

$$\begin{array}{cc}\hfill {\left(l_{k,n}^m\right)}^{\left(2s\right)}\left(x_{k,n}\right)=& \sum _{0\le 2r\le 2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\left(l_{k,n}^{m-1}\right)}^{\left(2r\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r\right)}\left(x_{k,n}\right)\hfill \\ +\sum _{1\le 2r-1\le 2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r-1\hfill \end{array}\right){\left(l_{k,n}^{m-1}\right)}^{\left(2r-1\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r+1\right)}\left(x_{k,n}\right).\hfill \end{array}$$

(3.10)

For the second term, we have from (1.10),

$$\left|\sum _{1\le 2r-1\le 2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r-1\hfill \end{array}\right){\left(l_{k,n}^{m-1}\right)}^{\left(2r-1\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r+1\right)}\left(x_{k,n}\right)\right|\le C{\delta }^2\left(\epsilon ,n\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}.$$

(3.11)

For the first term, we let ξ _s (m) = ξ _s (m, ε, x _{k, n} , n) for convenience. Then we know that

$$l_{k,n}^{\left(2s-2r\right)}\left(x_{k,n}\right)={\left(-1\right)}^{s-r}{\varphi }_{s-r}\left(1\right){\beta }^{s-r}\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s-r}{x}_{k,n}^{-\left(s-r\right)}\left(1+{\xi }_{s-r}\left(1\right)\right)$$

and |ξ_s-r(1)| → 0 as n → ∞ and ε → 0. By mathematical induction, we assume for 0 ≤ 2r ≤ 2s;

$${\left(l_{k,n}^{m-1}\right)}^{\left(2r\right)}\left(x_{k,n}\right)={\left(-1\right)}^r{\varphi }_r\left(m-1\right){\beta }^r\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2r}{x}_{k,n}^{-r}\left(1+{\xi }_r\left(m-1\right)\right)$$

and |ξ _r (m - 1)| → 0 as n → ∞ and ε → 0. Then, since

$$\begin{array}{c}\hfill {\left(l_{k,n}^{m-1}\right)}^{\left(2r\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r\right)}\left(x_{k,n}\right)={\left(-1\right)}^s{\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}\\ \hfill \times {\varphi }_r\left(m-1\right){\varphi }_{s-r}\left(1\right)\left(1+{\xi }_r\left(m-1\right)\right)\left(1+{\xi }_{s-r}\left(1\right)\right),\end{array}$$

we have for 0 ≤ 2r ≤ 2s, using the definition of (3.7),

$$\begin{array}{c}\sum _{0\le 2r\le 2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\left(l_{k,n}^{m-1}\right)}^{\left(2r\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r\right)}\left(x_{k,n}\right)\hfill \\ \hfill =& {\left(-1\right)}^s{\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}\hfill \\ \times \sum _{0\le 2r\le 2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\varphi }_r\left(m-1\right){\varphi }_{s-r}\left(1\right)\left(1+{\xi }_r\left(m-1\right)\right)\left(1+{\xi }_{s-r}\left(1\right)\right)\hfill \\ \hfill =& {\left(-1\right)}^s{\varphi }_s\left(m\right){\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}+{\left(-1\right)}^s{\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}\hfill \\ \times \sum _{0\le 2r\le 2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\varphi }_r\left(m-1\right){\varphi }_{s-r}\left(1\right)\left({\xi }_r\left(m-1\right)+{\xi }_{s-r}\left(1\right)+{\xi }_r\left(m-1\right){\xi }_{s-r}\left(1\right)\right).\hfill \end{array}$$

Here, we consider (3.10). If we let

$$\begin{array}{c}{\xi }_s\left(m,\epsilon ,x_{k,n},n\right)={\xi }_s\left(m\right)=\\ \sum _{0\le 2r\le 2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right)\frac{{\varphi }_r\left(m-1\right){\varphi }_{s-r}\left(1\right)}{{\varphi }_s\left(m\right)}\left({\xi }_r\left(m-1\right)+{\xi }_{s-r}\left(1\right)+{\xi }_r\left(m-1\right){\xi }_{s-r}\left(1\right)\right)\\ +\sum _{1\le 2r-1\le 2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r-1\hfill \end{array}\right)\frac{{\left(l_{k,n}^{m-1}\right)}^{\left(2r-1\right)}\left(x_{k,n}\right)l_{k,n}^{\left(2s-2r+1\right)}\left(x_{k,n}\right)}{{\left(-1\right)}^s{\varphi }_s\left(m\right){\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}},\end{array}$$

then we have

$$\begin{array}{cc}\hfill {\xi }_s\left(m\right)& \le \sum _{0\le 2r\le 2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right)\frac{{\varphi }_r\left(m-1\right){\varphi }_{s-r}\left(1\right)}{{\varphi }_s\left(m\right)}\left({\xi }_r\left(m-1\right)+{\xi }_{s-r}\left(1\right)+{\xi }_r\left(m-1\right){\xi }_{s-r}\left(1\right)\right)\hfill \\ +C\frac{{\delta }^2\left(\epsilon ,n\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}}{{\left(-1\right)}^s{\varphi }_s\left(m\right){\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}}\text{by the definition of }\left(3.11\right)\hfill \\ \le \sum _{0\le 2r\le 2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right)\frac{{\varphi }_r\left(m-1\right){\varphi }_{s-r}\left(1\right)}{{\varphi }_s\left(m\right)}\left({\xi }_r\left(m-1\right)+{\xi }_{s-r}\left(1\right)+{\xi }_r\left(m-1\right){\xi }_{s-r}\left(1\right)\right)\hfill \\ +C^{\prime }{\delta }^2\left(\epsilon ,n\right).\hfill \end{array}$$

Then, we know that (1.11) holds and |ξ _i (j)| → 0 as n → ∞ and ε → 0, using mathematical induction on m. Therefore, we have the result.

We rewrite the relation (3.7) in the form for ν = 1, 2, 3 ...,

$${\varphi }_0\left(\nu \right):=1$$

and for j = 1, 2, 3 ..., ν = 2, 3, 4, ...,

$${\varphi }_j\left(\nu \right)-{\varphi }_j\left(\nu -1\right)=\frac{1}{2j+1}\sum _{r=0}^{j-1}\left(\begin{array}{c}\hfill 2j+1\hfill \\ \hfill 2r\hfill \end{array}\right){\varphi }_r\left(\nu -1\right).$$

Now, for every j we will introduce an auxiliary polynomial determined by ${\left\{{\mathrm{\Psi }}_j\left(y\right)\right\}}_{j=1}^{\mathrm{\infty }}$ as the following lemma:

Lemma 3.6. [4, Lemma 11] (i) For j = 0, 1, 2 ..., there exists a unique polynomial Ψ _j (y) of degree j such that

$${\Psi }_j\left(\nu \right)={\varphi }_j\left(\nu \right),\nu =1,2,3,\dots .$$

• (ii) Ψ₀(y) = 1 and Ψ _j (0) = 0, j = 1, 2, ....

Since Ψ _j (y) is a polynomial of degree j, we can replace ϕ _j (ν) in (3.7) with Ψ _j (y), that is,

$${\mathrm{\Psi }}_j\left(y\right)=\sum _{r=0}^j\frac{1}{2j-2r+1}\left(\begin{array}{c}\hfill 2j\hfill \\ \hfill 2r\hfill \end{array}\right){\mathrm{\Psi }}_r\left(y-1\right),j=0,1,2,...,$$

for an arbitrary y and j = 0, 1, 2, .... We use the notation F _kn (x, y) = (l _{k, n} (x)) ^y which coincides with $l_{k,n}^y\left(x\right)$ if y is an integer. Since l _{k, n} (x _{k, n} ) = 1, we have F _kn (x, t) > 0 for x in a neighborhood of x_{k, n}and an arbitrary real number t.

We can show that (∂/∂x) ^j F _kn (x _{k, n} , y) is a polynomial of degree at most j with respect to y for j = 0, 1, 2, ..., where (∂/∂x) ^j F _kn (x _{k, n} , y) is the j th partial derivative of F _kn (x, y) with respect to x at (x _{k, n} , y) [14, p. 199]. We prove these facts by induction on j. For j = 0 it is trivial. Suppose that it holds for j ≥ 0. To simplify the notation, let F(x) = F _kn (x, y) and l(x) = l _{k, n} (x) for a fixed y. Then F'(x)l(x) = yl'(x)F(x). By Leibniz's rule, we easily see that

$$F^{\left(j+1\right)}\left(x_{k,n}\right)=-\sum _{s=0}^{j-1}\left(\begin{array}{c}\hfill j\hfill \\ \hfill s\hfill \end{array}\right)F^{\left(s+1\right)}\left(x_{k,n}\right)l^{\left(j-s\right)}\left(x_{k,n}\right)+y\sum _{s=0}^j\left(\begin{array}{c}\hfill j\hfill \\ \hfill s\hfill \end{array}\right)l^{\left(s+1\right)}\left(x_{k,n}\right)F^{\left(j-s\right)}\left(x_{k,n}\right),$$

which shows that F^(j+1)(x _{k, n} ) is a polynomial of degree at most j + 1 with respect to y. Let $P_{kn}^{\left[j\right]}\left(y\right)$, j = 0, 1, 2, ... be defined by

$${\left(\partial ∕\partial x\right)}^{2j}{F}_{kn}\left(x_{k,n},y\right)={\left(-1\right)}^j{\beta }^j\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2j}{x}_{k,n}^{-j}{\mathrm{\Psi }}_j\left(y\right)+P_{kn}^{\left[j\right]}\left(y\right).$$

(3.12)

Then $P_{kn}^{\left[j\right]}\left(y\right)$ is a polynomial of degree at most 2j.

By Theorem 1.8 (1.11), we have the following:

Lemma 3.7. [4, Lemma 12] Let j = 0, 1, 2, ..., and M be a positive constant. Let 0 < ε < 1/4, $\frac{1}{\epsilon }\frac{a_n}{n^2}\le x_{k,n}\le \epsilon a_n$, and |y| ≤ M. Then

• (a) there exists κ _j (ε, x _{k, n} , n) > 0 such that

  $$\left|{\left(\partial ∕\partial y\right)}^s{P}_{kn}^{\left[j\right]}\left(y\right)\right|\le C{\kappa }_j\left(\epsilon ,x_{k,n},n\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2j}{x}_{k,n}^{-j},s=0,1$$

  (3.13)

and κ _j (ε, x _{k, n} , n) → 0 as n → ∞ and ε → 0.

• (b) there exists γ _j (ε, n) > 0 such that

  $$\left|{\left(\partial ∕\partial x\right)}^{2j+1}{F}_{kn}\left(x_{k,n},y\right)\right|\le C{\gamma }_j\left(\epsilon ,n\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2j+1}{x}_{k,n}^{-\frac{2j+1}{2}}$$

  (3.14)

and γ _j (ε, n) → 0 as n → ∞ and ε → 0.

Lemma 3.8. [4, Lemma 13] If y < 0, then for j = 0, 1, 2 ...,

$${\left(-1\right)}^j{\mathrm{\Psi }}_j\left(y\right)>0.$$

Lemma 3.9. For positive integers s and m with 1 ≤ m ≤ ν,

$$\sum _{r=0}^s\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\mathrm{\Psi }}_r\left(-m\right){\phi }_{s-r}\left(m\right)=0.$$

Proof. If we let $C_s\left(y\right)={\sum }_{r=0}^s\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\mathrm{\Psi }}_r\left(-y\right){\mathrm{\Psi }}_{s-r}\left(y\right)$, then it suffices to show that C _s (m) = 0. For every s,

$$\begin{array}{cc}\hfill 0& ={\left(l_{k,n}^{-m+m}\right)}^{2s}\left(x_{k,n}\right)=\sum _{i=0}^{2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill i\hfill \end{array}\right){\left(l_{k,n}^{-m}\right)}^{\left(i\right)}\left(x_{k,n}\right){\left(l_{k,n}^m\right)}^{\left(2s-i\right)}\left(x_{k,n}\right)\hfill \\ =\sum _{r=0}^s\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\left(\partial ∕\partial x\right)}^{2r}{F}_{kn}\left(x_{k,n},-m\right){\left(l_{k,n}^m\right)}^{\left(2s-2r\right)}\left(x_{k,n}\right)\hfill \\ +\sum _{r=0}^{s-1}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r+1\hfill \end{array}\right){\left(\partial ∕\partial x\right)}^{2r+1}{F}_{kn}\left(x_{k,n},-m\right){\left(l_{k,n}^m\right)}^{\left(2s-2r-1\right)}\left(x_{k,n}\right).\hfill \end{array}$$

By (1.11), (3.12) and (3.13), we see that the first sum ${\sum }_{r=0}^s$ has the form of

$$\sum _{r=0}^s={\left(-1\right)}^s{\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}\left(\sum _{r=0}^s\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\mathrm{\Psi }}_r\left(-m\right){\varphi }_{s-r}\left(m\right)+{\tilde{\eta }}_s\left(-m,\epsilon ,x_{k,n},n\right)\right).$$

Then, since

$$\begin{array}{cc}\hfill {\tilde{\eta }}_s\left(-m,\epsilon ,x_{k,n},n\right)& =\sum _{r=0}^s\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\mathrm{\Psi }}_r\left(-m\right){\varphi }_{s-r}\left(m\right){\xi }_{s-r}\left(m,\epsilon ,x_{k,n},n\right)\hfill \\ +\sum _{r=0}^s\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\left(-1\right)}^{-r}{\beta }^{-r}\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{-2r}{x}_{k,n}^r\hfill \\ \times {\varphi }_{s-r}\left(m\right)P_{kn}^{\left[r\right]}\left(-m\right)\left(1+{\xi }_{s-r}\left(m,\epsilon ,x_{k,n},n\right)\right),\hfill \end{array}$$

we know that $\mid {\tilde{\eta }}_s\left(-m,\epsilon ,x_{k,n},n\right)\mid \to 0$ as n → ∞ and ε → 0 (see (3.12)). By (3.14) and (3.8), the second sum ${\sum }_{r=0}^{s-1}$ is bounded by $C{\left(\frac{n}{\sqrt{a_n}}\right)}^{2s+1}{x}_{k,n}^{-\frac{2s+1}{2}}{\sum }_{r=0}^{2s-1}{\gamma }_r\left(\epsilon ,n\right)$, and we know that ${\sum }_{r=0}^{2s-1}{\gamma }_r\left(\epsilon ,n\right)\to 0$ as n → ∞ and ε → 0. Therefore, we obtain the following result: for every s,

$$0=\sum _{r=0}^s\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\mathrm{\Psi }}_r\left(-m\right){\mathrm{\Psi }}_{s-r}\left(m\right).$$

Theorem 1.9 is important to show a divergence theorem with respect to L _n (m, f; x), where m is an odd integer.

Proof of Theorem 1.9. We prove (1.12) by induction on s. Since e₀(m, k, n) = 1 and Ψ₀(y) = 1, (1.12) holds for s = 0. From (3.6) we write e_2s(m, k, n) in the form of

$$\begin{array}{cc}\hfill e_{2s}\left(m,k,n\right)& =-\sum _{r=0}^{s-1}\frac{1}{\left(2s-2r\right)!}{e}_{2r}\left(m,k,n\right){\left(l_{k,n}^m\right)}^{\left(2s-2r\right)}\left(x_{k,n}\right)\hfill \\ -\sum _{r=1}^s\frac{1}{\left(2s-2r+1\right)!}{e}_{2r-1}\left(m,k,n\right){\left(l_{k,n}^m\right)}^{\left(2s-2r+1\right)}\left(x_{k,n}\right)\hfill \\ =:I+II.\hfill \end{array}$$

From (1.6), we know that for x _{k, n} ≤ a _n /4,

$$\mid e_j\left(l,m,k,n\right)\mid \le C{\left(\frac{n}{\sqrt{a_n}}\right)}^j{x}_{k,n}^{-\frac{j}{2}}.$$

(3.15)

Then, by (1.10) and (3.15), |II| is bounded by $C{\sum }_{r=1}^s\delta \left(\epsilon ,n\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}$. For 0 ≤ i < s, we suppose (1.12). Then, we have for I,

$$\begin{array}{cc}\hfill -\sum _{r=0}^{s-1}& =\frac{{\left(-1\right)}^{s+1}}{\left(2s\right)!}{\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}\hfill \\ \times \sum _{r=0}^{s-1}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\mathrm{\Psi }}_r\left(-m\right){\varphi }_{s-r}\left(m\right)\left(1+{\eta }_r\right)\left(1+{\xi }_{s-r}\right),\hfill \end{array}$$

where ξ _s-r : = ξ _s-r (m, ε, x _{k, n} , n) and η _r : = η _r (m, ε, x _{k, n} , n) which are defined in (1.11) and (1.12). Then, using Lemma 3.9 and ϕ₀ (m) = 1, we have the following form:

$$e_{2s}\left(m,k,n\right)=\frac{{\left(-1\right)}^s}{\left(2s\right)!}{\mathrm{\Psi }}_s\left(-m\right){\beta }^s\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{2s}{x}_{k,n}^{-s}\left(1+{\eta }_s\left(m,\epsilon ,x_{k,n},n\right)\right).$$

Here, since

$$\begin{array}{cc}\hfill {\eta }_s\left(m,\epsilon ,x_{k,n},n\right)& =\sum _{r=0}^{s-1}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill 2r\hfill \end{array}\right){\mathrm{\Psi }}_r\left(-m\right){\varphi }_{s-r}\left(m\right)\left({\eta }_r+{\xi }_{s-r}+{\eta }_r{\xi }_{s-r}\right)\hfill \\ +{\left(-1\right)}^s{\beta }^{-s}\left(2n,k\right){\left(\frac{n}{\sqrt{a_n}}\right)}^{-2s}{x}_{k,n}^s\frac{\left(2s\right)!}{{\mathrm{\Psi }}_s\left(-m\right)}II,\hfill \end{array}$$

we see that |η _s (m, ε, x _{k, n} , n)| → 0 as n → ∞ and ε → 0 (recall above estimation of |II|). Therefore, we proved the result.

Lemma 3.10. [5, Theorem 1.3] Let$\rho >-\frac{1}{2}$and w(x) (C²+). There exists n₀such that uniformly for n ≥ n₀, we have the following:

• (a) For 1 ≤ j ≤ n,

  $$\mid p_{n,\rho }^{\prime }\left(x_{j,n}\right)\mid w_{\rho }\left(x_{j,n}\right)\widetilde{{\phi }_n}{\left(x_{j,n}\right)}^{-1}{\left[x_{j,n}\left(a_n-x_{j,n}\right)\right]}^{-1∕4}.$$

  (3.16)

• (b) For j ≤ n - 1 and x ∈ [x _j+1,n, x _j,n ],

  $$\begin{array}{c}\mid p_{n,\rho }\left(x\right)\mid w\left(x\right){\left(x+\frac{a_n}{n^2}\right)}^{\rho }\\ ~min\left\{\mid x-x_{j,n}\mid ,\mid x-x_{j+1,n}\mid \right\}{\phi }_n{\left(x_{j,n}\right)}^{-1}{\left[x_{j,n}\left(a_n-x_{j,n}\right)\right]}^{-1∕4}.\end{array}$$

  (3.17)

• (c) For 0 < a ≤ x _{k, n} ≤ b < ∞,

  $$\mid p_{n,\rho }^{\prime }\left(x_{k,n}\right)\mid w_{\rho }\left(x_{k,n}\right)~\frac{n}{a_n^{3∕4}}.$$

  (3.18)

• (d) For 0 < a ≤ x _k+1,n, x _{k, n} ≤ b < ∞ and x ∈ [(x _k+1,n+ x _{k, n} )/2 x _{k, n}+ x _k-1,n)/2],

  $$\mid p_{n,\rho }\left(x\right)\mid w\left(x\right){\left(x+\frac{a_n}{n^2}\right)}^{\rho }~\frac{1}{a_n^{1∕4}}.$$

  (3.19)

Moreover, for 0 < a ≤ x ≤ b < ∞, there exists a constant C > 0 such that

$$\mid p_{n,\rho }\left(x\right)\mid w\left(x\right){\left(x+\frac{a_n}{n^2}\right)}^{\rho }\le C\frac{1}{a_n^{1∕4}}.$$

(3.20)

• (e) Uniformly for n ≥ 1 and 1 ≤ j < n,

  $$x_{j,n}-x_{j+1,n}~{\phi }_n\left(x_{j,n}\right).$$

  (3.21)

• (f) Let Λ be defined in Definition 1.2 (d). There exists C > 0 such that for n ≥ 1,

  $$a_n\le C{n}^{1∕\Lambda }.$$

Proof. (a) and (b) follow from [5, Theorem 1.3]. (e) follows from [5, Theorem 1.4]. We need to prove (c), (d), and (f).

• (c) For 0 < a ≤ x _{k, n} ≤ b < ∞, we have (2.11);

  $${\phi }_n\left(x_{k,n}\right)~\frac{\sqrt{a_n}}{n},$$

so applying (a), we have the result.

• (d) Let 0 < a ≤ x _{k+1, n} < x _{k, n} ≤ b < ∞. We take a constant δ > 0 as

  $$\text{min}\left\{\mid x-x_{k,n}\mid ,\mid x-x_{k+1,n}\mid \right\}=\delta \frac{\sqrt{a_n}}{n}.$$

Then, by (b) we have

$$\mid p_{n,\rho }\left(x\right)\mid w\left(x\right){\left(x+\frac{a_n}{n^2}\right)}^{\rho }~\delta \frac{\sqrt{a_n}}{n}\frac{n}{a_n^{3∕4}}=\delta \frac{1}{a_n^{1∕4}}.$$

Moreover, by [6, Theorem 1.2] the second inequality holds.

• (f) We see

  $$\frac{R^{\prime }\left(x\right)}{R\left(x\right)}=\frac{T\left(x\right)}{x}\ge \frac{\Lambda }{x},$$

so that by an integration, R(x) ≥ R(1) x^Λ for x ≥ 1, and hence we have

$$R^{\prime }\left(x\right)\ge \Lambda R\left(1\right)x^{\Lambda -1}\left(x\ge 1\right).$$

Since lim_n→∞a _n = ∞, we can choose n₀ such that a _n ≥ 2 for all n ≥ n₀. Then for some C₁ > 0,

$$\begin{array}{cc}\hfill n& =\frac{2}{\pi }{\int }_0^1\frac{a_{n}t{R}^{\prime }\left(a_{n}t\right)}{\sqrt{t\left(1-t\right)}}\text{d}t\ge \frac{2}{\pi }{\int }_0^1\frac{a_{n}t\Lambda R\left(1\right){\left(a_{n}t\right)}^{\Lambda -1}}{\sqrt{t\left(1-t\right)}}\text{d}t\hfill \\ \hfill \ge & a_n^{\Lambda }\frac{2\Lambda R\left(1\right)}{\pi }{\int }_{1∕2}^1\frac{t^{\Lambda }}{\sqrt{t\left(1-t\right)}}\text{d}t=:C_1^{\Lambda }{a}_n^{\Lambda }.\hfill \end{array}$$

Hence, we have the result.

Lemma 3.11. Let the function h _kn (m; x) be defined by (1.4) and let 0 < c < a < b < d < ∞. Then we have

$$\underset{a\le x\le b}{max}\sum _{c\le x_{k,n}\le d}\left|l_{k,n}^m\left(x\right)\sum _{i=0}^{m-2}{e}_i\left(m,k,n\right){\left(x-x_{k,n}\right)}^i\right|\le C.$$

Proof. Let c ≤ x_k+1,n< x _{k, n} ≤ d. Then by (3.21), we see $\left|x_{k,n}-x_{k+1,n}\right|~\sqrt{a_n}∕n$.

Now, choose α, β > 0 satisfying for all x_k+1,n, x _{k, n} ∈ [c, d],

$$\alpha \frac{\sqrt{a_n}}{n}\le \left|x_{k,n}-x_{k+1,n}\right|\le \beta \frac{\sqrt{a_n}}{n}.$$

(3.22)

Let x ∈ [a, b] and |x - _xj(x),n| = min {|x - x _{k, n} }|; x _{k, n} ∈ [a, b]}, _xj(c)+1,n< c ≤ _xj(c),n, and _xj(d),n≤ d < _xj(d)-1,n. Moreover, we take a non-negative integer j _k satisfying for each x _{k, n} ∈ [a, b] and k ≠ j(x),

$$\left(j_k+\frac{1}{2}\right)\alpha \frac{\sqrt{a_n}}{n}\le \left|x-x_{k,n}\right|\le \left(j_k+1\right)\beta \frac{\sqrt{a_n}}{n}.$$

(3.23)

Then we have

$$\begin{array}{c}\underset{a\le x\le b}{max}\sum _{c\le x_{k,n}\le d}\left|l_{k,n}^m\left(x\right)\sum _{i=0}^{m-2}{e}_i\left(m,k,n\right){\left(x-x_{k,n}\right)}^i\right|\\ \le \underset{a\le x\le b}{max}\sum _{i=0}^{m-2}\sum _{c\le x_{k,n}\le d}\left|{\left(\frac{p_n\left(w_{\rho }^2;x\right)}{\left(x-x_{k,n}\right)p_n^{\prime }\left(w_{\rho }^2;x_{k,n}\right)}\right)}^m{e}_i\left(m,k,n\right){\left(x-x_{k,n}\right)}^i\right|\\ \le \underset{a\le x\le b}{max}\sum _{i=0}^{m-2}\left[{\left|\frac{p_n\left(w_{\rho }^2;x\right)}{\left(x-x_{j\left(x\right),n}\right)p_n^{\prime }\left(w_{\rho }^2;x_{j\left(x\right),n}\right)}\right|}^m\mid e_i\left(m,k,n\right){\left(x-x_{j\left(x\right),n}\right)}^i\mid \right.\\ +\sum _{\underset{x_{k,n}\ne x_{j\left(x\right),n}}{c\le x_{k,n}\le d}}\left.\left|{\left(\frac{p_n\left(w_{\rho }^2;x\right)}{\left(x-x_{k,n}\right)p_n^{\prime }\left(w_{\rho }^2;x_{k,n}\right)}\right)}^m{e}_i\left(m,k,n\right){\left(x-x_{k,n}\right)}^i\right|\right].\end{array}$$

Here, by (3.16) and (3.17) we see

$${\left|\frac{p_n\left(w_{\rho }^2;x\right)}{\left(x-x_{j\left(x\right),n}\right)p_n^{\prime }\left(w_{\rho }^2;x_{j\left(x\right),n}\right)}\right|}^m\le C$$

and from (3.20), we have |_xj(x),n| ≤ b + 1, and so by (1.6) we have

$$\mid e_i\left(m,j\left(x\right),n\right){\left(x-x_{j\left(x\right),n}\right)}^i\mid \le C{\left(\frac{n}{\sqrt{a_{2n}-x_{j\left(x\right),n}}}\right)}^i{x}_{j\left(x\right),n}^{-\frac{i}{2}}{\left(\frac{\sqrt{a_n}}{n}\right)}^i\le C.$$

Consequently, we have

$${\left|\frac{p_n\left(w_{\rho }^2;x\right)}{\left(x-x_{j\left(x\right),n}\right)p_n^{\prime }\left(w_{\rho }^2;x_{j\left(x\right),n}\right)}\right|}^m\mid e_i\left(m,j\left(x\right),n\right){\left(x-x_{j\left(x\right)n}\right)}^i\mid \le C.$$

Similarly, for c ≤ x _{k, n} ≤ d with x _{k, n} ≠ _xj(x),n, we have by (3.18) and (3.20),

$${\left|\frac{p_n\left(w_{\rho }^2;x\right)}{p_n^{\prime }\left(w_{\rho }^2;x_{k,n}\right)}\right|}^m\le {\left(\frac{\sqrt{a_n}}{n}\right)}^m$$

and by (1.6) and (3.23),

$$\mid e_i\left(m,k,n\right){\left(x-x_{k,n}\right)}^{i-m}\mid \le C{\left(\frac{n}{\sqrt{a_n}}\right)}^i{\left(\frac{n}{\sqrt{a_n}}\frac{1}{\left(j_k+1∕2\right)\alpha }\right)}^{m-i}.$$

Therefore, we have for 0 ≤ i ≤ m - 2,

$$\begin{array}{c}\sum _{\underset{x_{k,n}\ne x_{j\left(x\right),n}}{c\le x_{k,n}\le d}}\left|{\left(\frac{p_n\left(w_{\rho }^2;x\right)}{\left(x-x_{k,n}\right)p_n^{\prime }\left(w_{\rho }^2;x_{k,n}\right)}\right)}^m{e}_i\left(m,k,n\right){\left(x-x_{k,n}\right)}^i\right|\\ \le C\sum _{\underset{x_{k,n}\ne x_{j\left(x\right),n}}{c\le x_{k,n}\le d}}{\left(\frac{1}{\left(j_k+1∕2\right)\alpha }\right)}^2\le C.\end{array}$$

Therefore,

$$\underset{a\le x\le b}{max}\sum _{c\le x_{k,n}\le d}\left|l_{k,n}^m\left(x\right)\sum _{i=0}^{m-2}{e}_i\left(m,k,n\right){\left(x-x_{k,n}\right)}^i\right|\le C.$$

Proof of Theorem 1.10. We use Theorem 1.9 and Lemma 3.11. We find a lower bound for the Lebesgue constants ${\lambda }_n\left(m,\left[a,b\right]\right)=\underset{a\le x\le b}{max}{\sum }_{k=1}^n\left|h_{kn}\left(m;x\right)\right|$ with a positive odd order m and a given interval [a, b], 0 < a < b < ∞. By the expression (1.4) we have

$$\begin{array}{cc}\hfill \left|h_{kn}\left(m;x\right)\right|& \ge \mid l_{k,n}^m\left(x\right)e_{m-1}\left(m,k,n\right){\left(x-x_{k,n}\right)}^{m-1}\mid \hfill \\ -\left|l_{k,n}^m\left(x\right)\sum _{i=0}^{m-2}{e}_i\left(m,k,n\right){\left(x-x_{k,n}\right)}^i\right|.\hfill \end{array}$$

Let c = a/2, d = b + (b - a) and

$$\begin{array}{cc}\hfill {\lambda }_n\left(m,\left[a,b\right]\right)& \ge \underset{a\le x\le b}{max}\sum _{c\le x_{k,n}\le d}\left|l_{k,n}^m\left(x\right)e_{m-1}\left(m,k,n\right){\left(x-x_{k,n}\right)}^{m-1}\right|\hfill \\ -\underset{a\le x\le b}{max}\sum _{c\le x_{k,n}\le d}\left|l_{k,n}^m\left(x\right)\sum _{i=0}^{m-2}{e}_i\left(m,k,n\right){\left(x-x_{k,n}\right)}^i\right|\hfill \\ =\underset{a\le x\le b}{max}{F}_n\left(x\right)-\underset{a\le x\le b}{max}{G}_n\left(x\right).\hfill \end{array}$$

It follows from Lemma 3.11 that max_{a ≤ x ≤ b}G _n (x) ≤ C with C independent of n. Therefore, it is enough to show that max_{a ≤ x ≤ b}F _n (x) ≥ C log (1 + n). We consider α, β and j _k defined in (3.22) and (3.23). Let K (x; [c, d]) be the set of numbers defined as

$$K\left(x;\left[c,d\right]\right)=\left\{j_k;\left(j_k+1∕2\right)\alpha \frac{\sqrt{a_n}}{n}\le \left|x-x_{k,n}\right|\le \left(j_k+1\right)\beta \frac{\sqrt{a_n}}{n},x_{k,n}\in \left[c,d\right],k\ne j\left(x\right)\right\},$$

where j _k is a non-negative integer. Then, there exist γ > 0 and C > 0 such that

$$C{n}^{\gamma }\le max\left\{j_k\in K\left(x;\left[c,d\right]\right)\right\},$$

that is, we see

$$\left\{0,1,2,\dots ,\left[C{n}^{\gamma }\right]\right\}\subset K\left(x;\left[c,d\right]\right).$$

(3.24)

In fact, from Lemma 3.10 (f), we see a _n ≤ c₁n^1/Λ, Λ > 1/2. By (3.22) and (3.23), we see 0 ∈ K(x; [c, d]) and

$$\left(\text{max}\left\{j_k\in K\left(x;\left[c,d\right]\right)\right\}+1\right)\beta \frac{\sqrt{a_n}}{n}\ge b-a.$$

Hence, we have

$$\text{max}\left\{j_k\in K\left(x;\left[c,d\right]\right)\right\}+1\ge \frac{b-a}{\beta }\frac{n}{\sqrt{a_n}}\ge \frac{b-a}{\beta }\frac{n}{\sqrt{a_n}}\frac{1}{\sqrt{c_1}}{n}^{1-1∕\left(2\Lambda \right)}=:C{n}^{\gamma }.$$

So, we have (3.24). Now, we take an interval [x_l+1,n, x_{l, n}] ⊂ [a, b], and we put

$$x^{*}:=\left(x_{\ell +1,n}+x_{\ell ,n}\right)∕2.$$

By Lemma 3.10 (d), we have

$$\left|\left(p_n{w}_{\rho }\right)\left(x^{*}\right)\right|~\frac{1}{a_n^{1∕4}}.$$

From Lemma 3.10 (e), we know for c ≤ x _{k, n} ≤ d

$$\left|x^{*}-x_{k,n}\right|\ge \frac{\alpha }{2}\frac{\sqrt{a_n}}{n}.$$

(3.25)

By Lemma 3.10 (c), we have for c ≤ x _{k, n} ≤ d

$$\left|p_n^{\prime }\left(x_{k,n}\right)\right|w_{\rho }\left(x_{k,n}\right)~\frac{n}{a_n^{3∕4}}.$$

Then, we have

$$\begin{array}{cc}\hfill \mid l_{k,n}\left(x^{*}\right)\mid & =\frac{\mid p_n\left(x^{*}\right)\mid w\left(x^{*}\right){\left(x^{*}+\frac{a_n}{n^2}\right)}^{\rho }}{\mid \left(x^{*}-x_{k,n}\right)p_n^{\prime }\left(x_{k,n}\right)\mid w_{\rho }\left(x_{k,n}\right)}\frac{w_{\rho }\left(x_{k,n}\right)}{w\left(x^{*}\right){\left(x^{*}+\frac{a_n}{n^2}\right)}^{\rho }}\hfill \\ \ge C\frac{\sqrt{a_n}}{n}\frac{1}{\left|x^{*}-x_{k,n}\right|}.\hfill \end{array}$$

Here, we used the following facts:

$${\left(x^{*}+\frac{a_n}{n}\right)}^{\rho }~x^{*\rho }$$

and

$$\frac{w_{\rho }\left(x_{k,n}\right)}{w_{\rho }\left(x^{*}\right)}~1,x^{*},x_{k,n}\in \left[a,b\right].$$

Now we use Theorem 1.9, that is, for c ≤ x _{k, n} ≤ d we have

$$e_{m-1}\left(m,k,n\right)~{\left(\frac{n}{\sqrt{a_n}}\right)}^{m-1}.$$

Therefore, with (3.25) and (3.26), we have

$$\begin{array}{cc}\hfill F_n\left(x^{*}\right)& \ge \sum _{j_k\in K\left(x^{*};\left[c,d\right]\right)}\left|l_{k,n}^m\left(x^{*}\right)e_{m-1}\left(m,k,n\right){\left(x^{*}-x_{k,n}\right)}^{m-1}\right|\hfill \\ \ge C\sum _{j_k\in K\left(x^{*};\left[c,d\right]\right)}{\left(\frac{\sqrt{a_n}}{n}\frac{1}{\left|x^{*}-x_{k,n}\right|}\right)}^m{\left(\frac{n}{\sqrt{a_n}}\right)}^{m-1}{\left|x^{*}-x_{k,n}\right|}^{m-1}\hfill \\ =C\sum _{j_k\in K\left(x^{*};\left[c,d\right]\right)}{\left(\frac{\sqrt{a_n}}{n}\right)}^m\frac{1}{\left|x^{*}-x_{k,n}\right|}{\left(\frac{n}{\sqrt{a_n}}\right)}^{m-1}\hfill \\ \ge C\sum _{j_k\in K\left(x^{*};\left[c,d\right]\right)}{\left(\frac{\sqrt{a_n}}{n}\right)}^m\frac{1}{\left(j_k+1\right)\beta \left(\sqrt{a_n}∕n\right)}{\left(\frac{n}{\sqrt{a_n}}\right)}^{m-1}\hfill \\ \ge C\left(\beta \right)\sum _{j_k\in K\left(x^{*};\left[c,d\right]\right)}\frac{1}{j_k+1}\ge C\sum _{0\le j\le n^{\gamma }}\frac{1}{j+1}\ge Clogn.\hfill \end{array}$$

Consequently, the theorem is complete.   □