Open Access

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

Journal of Inequalities and Applications20112011:122

https://doi.org/10.1186/1029-242X-2011-122

Received: 28 July 2011

Accepted: 25 November 2011

Published: 25 November 2011

Abstract

Let + = [0, ∞) and R : ++ be a continuous function which is the Laguerre-type exponent, and pn, ρ(x), ρ > - 1 2 be the orthonormal polynomials with the weight w ρ (x) = x ρ e-R(x). For the zeros { x k , n , ρ } k = 1 n of p n , ρ ( x ) = p n ( w ρ 2 ; x ) , we consider the higher order Hermite-Fejér interpolation polynomial L n (l, m, f; x) based at the zeros { x k , n , ρ } k = 1 n , where 0 ≤ lm - 1 are positive integers.

2010 Mathematics Subject Classification: 41A10.

Keywords

Laguerre-type weightsorthonormal polynomialshigher order Hermite-Fejér interpolation polynomials

1. Introduction and main results

Let = [-∞, ∞) and + = [0, ∞). Let 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 { p n , ρ ( x ) } n = 0 with the weight w ρ (x). Then, for the zeros { x k , n , ρ } k = 1 n of p n , ρ ( x ) = p n ( w ρ 2 ; x ) , we obtained various estimations with respect to p n , ρ ( j ) ( x k , n , ρ ) , 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 { x k , n , ρ } 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 L 2 , L ̃ ν with L ( C 2 ) , L ( C 2 + ) , and main results. In Section 2, we will introduce the classes F ( C 2 ) and F ( C 2 + ) , and then, we will obtain some relations of the factors derived from the classes F ( C 2 ) , F ( C 2 + ) and the classes L ( C 2 ) , L ( C 2 + ) . Finally, we will prove the main theorems using known results in [15], in Section 3.

We say that f : + 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, C1f(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 P n .

First, we introduce classes of weights. Levin and Lubinsky [5, 6] introduced the class of weights on + 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.
  1. (a)

    x R ( x ) is continuous in I, with limit 0 at 0 and R(0) = 0;

     
  2. (b)

    R″(x) exists in (0, d), while Q″ is positive in ( 0 , d ) ;

     
  3. (c)
    lim x d - R ( x ) = ;
     
  4. (d)
    The function
    T ( x ) : = x R ( x ) R ( x )
     
is quasi-increasing in (0, d), with
T ( x ) Λ > 1 2 , x ( 0 , d ) ;
  1. (e)
    There exists C 1 > 0 such that
    R ( x ) R ( x ) C 1 R ( x ) R ( x ) , a . e . x ( 0 , d ) .
     
Then, we write w L ( C 2 ) . If there also exist a compact subinterval J* 0 of I * = ( - d , d ) and C2 > 0 such that
Q ( t ) Q ( t ) C 2 Q ( t ) Q ( t ) , a . e . t I * \ J * ,

then we write w L ( C 2 + ) .

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

Definition 1.2. We assume that R : ++ has the following properties:
  1. (a)

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

     
  2. (b)

    R″(x) > 0 exists in +\{0};

     
  3. (c)
    lim x R ( x ) = ;
     
  4. (d)
    The function
    T ( x ) : = x R ( x ) R ( x )
     
is quasi-increasing in +\{0}, with
T ( x ) Λ > 1 2 , x + \ { 0 } ;
  1. (e)
    There exists C 1 > 0 such that
    R ( x ) R ( x ) C 1 R ( x ) R ( x ) , a . e . x + \ { 0 } .
     
There exist a compact subinterval J 0 of + and C2 > 0 such that
R ( x ) R ( x ) C 2 R ( x ) R ( x ) , a . e . t + \ J ,

then we write w L 2 .

To obtain estimations of the coefficients of higher order Hermite-Fejér interpolation polynomial based at the zeros { x k , n , ρ } k = 1 n , we need to focus on a smaller class of weights.

Definition 1.3. Let w = exp ( - R ) L 2 and let ν ≥ 2 be an integer. For the exponent R, we assume the following:
  1. (a)

    R (j)(x) > 0, for 0 ≤ jν and x > 0, and R (j)(0) = 0, 0 ≤ jν - 1.

     
  2. (b)
    There exist positive constants C i > 0, i = 1, 2, ..., ν - 1 such that for i = 1, 2, ..., ν - 1
    R ( i + 1 ) ( x ) C i R ( i ) ( x ) R ( x ) R ( x ) , a . e . x + \ { 0 } .
     
  3. (c)
    There exist positive constants C, c 1 > 0 and 0 ≤ δ < 1 such that on x (0, c 1)
    R ( ν ) ( x ) C 1 x δ .
    (1.1)
     
  4. (d)

    There exists c 2 > 0 such that we have one among the following

     

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

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

Then we write w ( x ) = e - R ( x ) L ̃ ν .

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 , α ( x ) = exp l ( x α ) - exp l ( 0 ) ,
where exp l (x) = exp(exp(exp ... exp(x)) ...) is the l-th iterated exponential.
  1. (1)

    If α > ν, w ( x ) = e - R l , α ( x ) L ̃ ν .

     
  2. (2)
    If αν and α is an integer, we define
    R l , α * ( x ) = exp l ( x α ) - exp l ( 0 ) - j = 1 r R l , α ( j ) ( 0 ) j ! x j .
     

Then w ( x ) = e - R l , α * ( x ) L ̃ ν .

In the remainder of this article, we consider the classes L 2 and L ̃ ν ; Let w L 2 or w L ̃ ν ν 2 . For ρ > - 1 2 , we set w ρ (x): = x ρ w(x). Then we can construct the orthonormal polynomials p n , ρ ( x ) = p n ( w ρ 2 ; x ) of degree n with respect to w ρ 2 ( x ) . That is,
0 p n , ρ ( u ) p m , ρ ( u ) w ρ 2 ( u ) d u = δ n m ( Kronecker’s delta ) n , m = 0 , 1 , 2 , .
Let us denote the zeros of p n,ρ (x) by
0 < x n , n , ρ < < x 2 , n , ρ < x 1 , n , ρ < .
The Mhaskar-Rahmanov-Saff numbers a v is defined as follows:
v = 1 π 0 1 a v t R ( a v t ) t ( 1 - t ) d t , v > 0 .
Let l, m be non-negative integers with 0 ≤ l < mν. For f C(l)(), we define the (l, m)-order Hermite-Fejér interpolation polynomials L n ( l , m , f ; x ) P m n - 1 as follows: For each k = 1, 2, ..., n,
L n ( j ) ( l , m , f ; x k , n , ρ ) = f ( j ) ( x k , n , ρ ) , j = 0 , 1 , 2 , , l , L n ( j ) ( l , m , f ; x k , n , ρ ) = 0 , j = l + 1 , l + 2 , , m - 1 .
For each P P m n - 1 , we see L n (m - 1, m, P; x) = P(x). The fundamental polynomials h s , k , n , ρ ( m ; x ) P m n - 1 , k = 1, 2, ..., n, of L n (l, m, f; x) are defined by
h s , k , n , ρ ( l , m ; x ) = l k , n , ρ m ( x ) i = s m - 1 e s , i ( l , m , k , n ) ( x - x k , n , ρ ) i .
(1.2)
Here, lk, n, ρ(x) is a fundamental Lagrange interpolation polynomial of degree n - 1 [[8], p. 23] given by
l k , n , ρ ( x ) = p n ( w ρ 2 ; x ) ( x - x k , n , ρ ) p n ( w ρ 2 ; x k , n , ρ )
and hs,k, n, ρ(l, m; x) satisfies
h s , k , n , ρ ( j ) ( l , m ; x p , n , ρ ) = δ s , j δ k , p j , s = 0 , 1 , , m - 1 , p = 1 , 2 , , n .
(1.3)
Then
L n ( l , m , f ; x ) = k = 1 n s = 0 l f ( s ) ( x k , n , ρ ) h s , k , n , ρ ( l , m ; x ) .
In particular, for f C(), we define the m-order Hermite-Fejér interpolation polynomials L n ( m , f ; x ) P m n - 1 as the (0, m)-order Hermite-Fejér interpolation polynomials L n (0, m, f; x). Then we know that
L n ( m , f ; x ) = k = 1 n f ( x k , n , ρ ) h k , n , ρ ( m ; x ) ,
where e i (m, k, n): = e0,i(0, m, k, n) and
h k , n , ρ ( m ; x ) = l k , n , ρ m ( x ) i = 0 m - 1 e i ( m , k , n ) ( x - x k , n , ρ ) i .
(1.4)

We often denote lk, n(x): = lk, n, ρ(x), hs, k, n(x): = hs, k, n, ρ(x), and xk, n: = xk, n, ρif they do not confuse us.

Theorem 1.5. Let w ( x ) = exp ( - R ( x ) ) L ( C 2 + ) and ρ > -1/2.
  1. (a)
    For each m ≥ 1 and j = 0, 1, ..., we have
    ( l k , n m ) ( j ) ( x k , n ) C n a 2 n - x k , n j x k , n - j 2 .
    (1.5)
     
  2. (b)
    For each m ≥ 1 and j = s, ..., m - 1, we have e s, s(l, m, k, n) = 1/s! and
    e s , j ( l , m , k , n ) C n a 2 n - x k , n j - s x k , n - j - s 2 .
    (1.6)
     

We remark L 2 L ( C 2 + ) .

Theorem 1.6. Let w ( x ) = exp ( - R ( x ) ) L ̃ ν , ν 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 C n 2 ( 1 + ν - δ ) ,
(1.7)
where 0 ≤ δ < 1 is defined in (1.1). Then we have the following:
  1. (a)
    If j is odd, then we have for m ≥ 1 and j = 0, 1, ..., ν - 1,
    ( l k , n m ) ( j ) ( x k , n ) C T ( a n ) a n x k , n + R ( x k , n ) + 1 x k , n × n a 2 n - x k , n + T ( a n ) a n j - 1 x k , n - j - 1 2 .
    (1.8)
     
  2. (b)
    If j - s is odd, then we have for m ≥ 1 and 0 ≤ sjm - 1,
    e s , j ( l , m , k , n ) C T ( a n ) a n x k , n + R ( x k , n ) + 1 x k , n × n a 2 n - x k , n + T ( a n ) a n j - s - 1 x k , n - j - s - 1 2 .
    (1.9)
     
Theorem 1.7. Let 0 < ε < 1/4. Let 1 ε a n n 2 x k , n ε a n . Let s be a positive integer with 2 ≤ 2sν. Then under the same conditions as the assumptions of Theorem 1.6, there exists μ1(ε, n) > 0 such that
p n , ρ ( 2 s ) ( x k , n ) C δ ( ε , n ) n a n 2 s - 1 p n ( x k , n ) x k , n - ( 2 s - 1 ) 2

and δ (ε, n) → 0 as n → ∞ and ε → 0.

Theorem 1.8. [4, Lemma 10] Let 0 < ε < 1/4. Let 1 ε a n n 2 x k , n ε a n . Let s be a positive integer with 2 ≤ 2sν - 1. Suppose the same conditions as the assumptions of Theorem 1.6. Then
  1. (a)
    for 1 ≤ 2s - 1 ≤ ν - 1,
    ( l k , n m ) ( 2 s - 1 ) ( x k , n ) C δ ( ε , n ) n a n 2 s - 1 x k , n - 2 s - 1 2 ,
    (1.10)
     
where δ(ε, n) is defined in Theorem 1.7.
  1. (b)
    there exists β(n, k) with 0 < D 1β(n, k) ≤ D 2 for absolute constants D 1, D 2 such that the following holds:
    ( l k , n m ) ( 2 s ) ( x k , n ) = ( - 1 ) s ϕ s ( m ) β s ( 2 n , k ) n a n 2 s x k , n - s ( 1 + ξ s ( m , ε , x k , n , n ) )
    (1.11)
     

and |ξ s (m, ε, xk, n, n)| → 0 as n → ∞ and ε → 0.

Theorem 1.9. [4, (4.16)], [9]Let 0 < ε < 1/4. Let 1 ε a n n 2 x k , n ε a n . Let s be a positive integer with 2 ≤ 2sm - 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 2 s ( m , k , n ) = ( - 1 ) s ( 2 s ) ! Ψ s ( - m ) β s ( 2 n , k ) n a n 2 s x k , n - s 1 + η s ( m , ε , x k , n , n )
(1.12)

and |η s (m, ε, xk, 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(+) such that for any fixed interval [a, b], a > 0,
lim sup n max a x b | L n ( m , f ; x ) | = .

2. Preliminaries

Levin and Lubinsky introduced the classes L ( C 2 ) and L ( C 2 + ) as analogies of the classes F ( C 2 ) and F ( C 2 + ) defined on I * = ( - d , d ) . They defined the following:

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

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

     
  2. (b)

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

     
  3. (c)
    lim t d - Q ( t ) = ;
     
  4. (d)
    The function
    T * ( t ) : = t Q ( t ) Q ( t )
     
is quasi-increasing in ( 0 , d ) , with
T * ( t ) Λ * > 1 , t I * \ { 0 } ;
  1. (e)
    There exists C 1 > 0 such that
    Q ( t ) Q ( t ) C 1 Q ( t ) Q ( t ) , a . e . t I * \ { 0 } .
     
Then we write W F ( C 2 ) . If there also exist a compact subinterval J* 0 of I* and C2 > 0 such that
Q ( t ) Q ( t ) C 2 Q ( t ) Q ( t ) , a . e . t I * \ J * ,

then we write W F ( C 2 + ) .

Then we see that w L ( C 2 ) W F ( C 2 ) and w L ( C 2 + ) W F ( C 2 + ) 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 L 2 W F ( C 2 + ) ,

where W(t) = w(x); x = t2.

On , we can consider the corresponding class to L ̃ ν as follows:

Definition 2.3.[11] Let W = exp ( - Q ) F ( C 2 + ) and ν ≥ 2 be an integer. Let Q be a continuous and even function on . For the exponent Q, we assume the following:
  1. (a)

    Q (j)(x) > 0, for 0 ≤ jν and t +/{0}.

     
  2. (b)
    There exist positive constants C i > 0 such that for i = 1, 2, ..., ν - 1,
    Q ( i + 1 ) ( t ) C i Q ( i ) ( t ) Q ( t ) Q ( t ) , a . e . x + \ { 0 } .
     
  3. (c)
    There exist positive constants C, c 1 > 0, and 0 ≤ δ* < 1 such that on t (0, c 1),
    Q ( ν ) ( t ) C 1 t δ * .
    (2.1)
     
  4. (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 ( t ) = e - Q ( t ) F ̃ ν .

Let W F ̃ ν , and ν ≥ 2. For ρ * > - 1 2 , we set
W ρ * ( t ) : = t ρ * W ( t ) .
Then we can construct the orthonormal polynomials P n , ρ * ( t ) = P n ( W ρ * 2 ; t ) of degree n with respect to Wρ*(t). That is,
- P n , ρ * ( v ) P m , ρ * ( v ) W ρ * 2 ( v ) d t = δ n m , n , m = 0 , 1 , 2 , .
Let us denote the zeros of Pn, ρ*(t) by
- < t n n < < t 2 n < t 1 n < .

There are many properties of Pn, ρ*(t) = P n (Wρ*; t) with respect to Wρ*(t), W F ̃ ν , ν = 2 , 3 , of Definition 2.3 in [2, 3, 7, 1113]. They were obtained by transformations from the results in [5, 6]. Jung and Sakai [2, Theorem 3.3 and 3.6] estimate P n , ρ * ( j ) ( t k , n ) , 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 , ρ ( j ) ( x k , n ) , k = 1, 2, ..., n, j = 1, 2, ..., ν. In this article, we consider w = exp ( - R ) L ̃ ν and pn, 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 Pn, ρ*(t) on can be entirely reduced to the orthonormal polynomials p n , ρ (x) in +as follows: For n = 0, 1, 2, ...,
P 2 n , 2 ρ + 1 2 ( t ) = p n , ρ ( x ) a n d P 2 n + 1 , 2 ρ - 1 2 ( t ) = t p n , ρ ( x ) .
In this article, we will use the fact that w ρ (x) = x ρ exp(-R(x)) is transformed into W2ρ+1/2(t) = |t|2ρ+1/2exp (-Q(t)) as meaning that
0 p n , ρ ( x ) p m , ρ ( x ) w ρ 2 ( x ) d x = 2 0 p n , ρ ( t 2 ) p m , ρ ( t 2 ) t 4 ρ + 1 W 2 ( t ) d t = - P 2 n , 2 ρ + 1 2 ( t ) P 2 m , 2 ρ + 1 2 ( t ) W 2 ρ + 1 2 2 ( t ) d t .
Theorem 2.5. [1, Theorem 2.5] Let Q(t) = R(x), x = t2. Then we have
w ( x ) = exp ( - R ( x ) ) L ̃ ν W ( t ) = exp ( - Q ( t ) ) F ̃ ν .
(2.2)
In particular, we have
Q ( ν ) ( t ) C 1 t δ ,

where 0 ≤ δ < 1 is defined in (1.1).

For convenience, in the remainder of this article, we set as follows:
ρ * : = 2 ρ + 1 2 for ρ > - 1 2 , p n ( x ) : = p n , ρ ( x ) , P n ( t ) : = P n , ρ * ( t ) ,
(2.3)
and x k , n = x k , n , ρ , t k n = t k , n , ρ * . Then we know that ρ * > - 1 2 and
p n ( x ) = P 2 n , ρ * ( t ) , x = t 2 , x k , n = t k , 2 n 2 , t k , 2 n > 0 , k = 1 , 2 , , n .
(2.4)
In the following, we introduce useful notations:
  1. (a)
    The Mhaskar-Rahmanov-Saff numbers a v and a u * are defined as the positive roots of the following equations, that is,
    v = 1 π 0 1 a v t R ( a v t ) { t ( 1 - t ) } - 1 2 d t , v > 0
     
and
u = 2 π 0 1 a u * t Q ( a u * t ) ( 1 - t 2 ) - 1 2 d t , u > 0 .
  1. (b)
    Let
    η n = { n T ( a n ) } - 2 3 and η n * = { n T * ( a n * ) } - 2 3 .
     

Then we have the following:

Lemma 2.6. [6, (2.5),(2.7),(2.9)]
a n = a 2 n * 2 , η n = 4 2 3 η 2 n * , T ( a n ) = 1 2 T * ( a 2 n * ) .

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 integern/2), we have
t [ n 2 ] , n ~ a n * n - 1 ,
and for the maximum zero t1nwe have for large enough n,
1 - t 1 n a n * ~ η n * , η n * = ( n T * ( a n * ) ) - 2 3 .
Moreover, for some constant 0 < ε ≤ 2 we have
T * ( a n * ) C n 2 - ε .

Remark 2.8. (a) Let W ( t ) F ( C 2 + ) . 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 ) ≤ Cn2-εfor some constant 0 < ε ≤ 2.
  1. (b)

    Let w ( x ) L ̃ ν . 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 C n 2 ( 1 + ν - δ ) a n * C n 1 ( 1 + ν - δ * ) .

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 * C ( η ) t η , t 1 .
In addition, since T(x) = T*(t)/2 and a n = a 2 n * 2 , we know that (a-2).
  1. (b)

    Since w ( x ) L ̃ ν , we know that W ( t ) F ̃ ν 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 ( j ) ( x ) = i = 1 j ( - 1 ) j - i c j , i P 2 n ( i ) ( t ) t - 2 j + i ,
where c j, i > 0(1 ≤ ij, j = 1, 2, ...) satisfy the following relations: for k = 1, 2, ...,
c k + 1 , 1 = 2 k - 1 2 c k , 1 , c k + 1 , k + 1 = 1 2 k + 1 , c 1 , 1 = 1 2 ,
and for 2 ≤ ik,
c k + 1 , i = c k , i - 1 + ( 2 k - i ) 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 ≤ sl, sim - 1.

Theorem 3.1. [1, Theorem 1.5] Let w ( x ) = exp ( - R ( x ) ) L ( C 2 + ) and let ρ > -1/2. For each k = 1, 2, ..., n and j = 0, 1, ..., we have
p n , ρ ( j ) ( x k , n ) C n a 2 n - x k , n j - 1 x k , n - j - 1 2 p n , ρ ( x k , n ) .
Proof of Theorem 1.5. (a) From Theorem 3.1 we know that
l k , n ( j ) ( x k , n ) = p n ( j + 1 ) ( x k , n ) ( j + 1 ) p n ( x k , n ) C n a 2 n - x k , n j x k , n - j 2 .
Then, assuming that (a) is true for 1 ≤ m' < m, we have
( l k , n m ) ( j ) ( x k , n ) = s = 0 j j s ( l k , n m - 1 ) ( s ) ( x k , n ) l k , n ( j - s ) ( x k , n ) C s = 0 j n a 2 n - x k , n s x k , n - s 2 n a 2 n - x k , n j - s x k , n - j - s 2 n a 2 n - x k , n j x k , n - j 2 .
Therefore, the result is proved by induction with respect to m.
  1. (b)
    From (2) and (3), we know e s, s (l, m, k, n) = 1/s! and the following recurrence relation: for s + 1 ≤ im - 1,
    e s , i ( l , m , k , n ) = - p - s i - 1 1 ( i - p ) ! e s , p ( l , m , k , n ) ( l k , n ) ( i - p ) ( x k , n ) .
    (3.5)
     

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

Theorem 3.2. [1, Theorem 1.6] Let w ( x ) = exp ( - R ( x ) ) L ̃ ν 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
p n , ρ ( j ) ( x k , n ) C n a n - x k , n + T ( a n ) a n j - 1 x k , n - j - 1 2 p n , ρ ( x k , n )
and in particular, if j is even, then we have
p n , ρ ( j ) ( x k , n ) C T ( a n ) a n x k , n + R ( x k , n ) + 1 x k , n × n a n - x k , n + T ( a n ) a n j - 2 x k , n - j - 2 2 p n , ρ ( x k , n ) .
Proof of Theorem 1.6. We use the induction method on m.
  1. (a)
    For m = 1, we have the result because of
    l k , n ( j ) ( x k , n ) = p n ( j + 1 ) ( x k , n ) ( j + 1 ) p n ( x k , n ) , j = 1 , 2 , 3 , ,
     
and Theorem 3.2. Now we assume the theorem for 1 ≤ m' < m. Then, we have the following: For 1 ≤ 2s - 1 ≤ ν - 1,
( l k , n m ) ( 2 s - 1 ) ( x k , n ) = r = 0 s 2 s - 1 2 r ( l k , n m - 1 ) ( 2 r ) ( x k , n ) l k , n ( 2 s - 2 r - 1 ) ( x k , n ) + r = 0 s 2 s - 1 2 r + 1 ( l k , n m - 1 ) ( 2 r + 1 ) ( x k , n ) l k , n ( 2 s - 2 r - 2 ) ( x k , n ) .
Since
n a 2 n - x k , n n a 2 n - x k , n ,
we have
( l k , n m - 1 ) ( 2 r ) ( x k , n ) l k , n ( 2 s - 2 r - 1 ) ( x k , n ) C T ( a n ) a n x k , n + R ( x k , n ) + 1 x k , n × n a 2 n - x k , n 2 r n a 2 n - x k , n + T ( a n ) a n 2 s - 2 r - 2 x k , n - s + 1 C T ( a n ) a n x k , n + R ( x k , n ) + 1 x k , n × n a 2 n - x k , n + T ( a n ) a n 2 s - 2 x k , n - s + 1 ,
and similarly
( l k , n m - 1 ) ( 2 r + 1 ) ( x k , n ) l k , n ( 2 s - 2 r - 2 ) ( x k , n ) C T ( a n ) a n x k , n + R ( x k , n ) + 1 x k , n × n a 2 n - x k , n + T ( a n ) a n 2 s - 2 x k , n - s + 1 .
Therefore, we have
( l k , n m ) ( 2 s - 1 ) ( x k , n ) C T ( a n ) a n x k , n + R ( x k , n ) + 1 x k , n × n a 2 n - x k , n + T ( a n ) a n 2 s - 2 x k , n - s + 1 .
  1. (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 ≤ im - 1,
    e s , i ( l , m , k , n ) = - p = s i - 1 1 ( i - p ) ! e s , p ( l , m , k , n ) ( l k , n m ) ( i - p ) ( x k , n ) .
    (3.6)
     
When i - s is odd, we know that
i - p : odd , if p - s : even i - p : even , if p - s : odd .

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 1 ε a n n 2 x k , n ε a n and let s be a positive integer with 2 ≤ 2sν - 1. Suppose the same conditions as the assumptions of Theorem 1.6. Then there exists β(n, k), 0 < D1β(n, k) ≤ D2for absolute constants D1, D2such that the following equality holds:
p n , ρ ( 2 s + 1 ) ( x k , n ) = ( - 1 ) s β s ( 2 n , k ) n a n 2 s ( 1 + ρ s ( ε , x k , n , n ) ) p n ( x k , n ) 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 F ( C 2 + ) and r = 1, 2, . Then, uniformly for 1 ≤ kn,
| P n ( r ) ( t k , n ) P n ( t k , n ) | C ( n a 2 n * 2 t k , n 2 ) r 1 .
Lemma 3.5. [2, Theorem 3.3] Let ρ* > -1/2 and W ( x ) = exp ( - Q ( x ) ) F ̃ ν , ν ≥ 2. Assume that 1 + 2ρ* - δ* ≥ 0 for ρ* < 0 and if T*(t) is bounded, then assume
a n * C n 1 ( 1 + ν - δ * ) ,
where 0 ≤ δ* < 1 is defined in (2.1). Let 0 < α < 1/2. Let 1 ε a n * n t k n ε a n * and let s be a positive integer with 2 ≤ 2sν. Then there exists μ(ε, n) > 0 such that
P n ( 2 s ) ( t k , n ) C μ ( ε , n ) n a n 2 s - 1 P n ( t k , n )

and μ(ε, n) → 0 as n → ∞ and ε → 0.

Proof of Theorem 1.7. By Lemma 2.9, we have
p n ( 2 s ) ( x k , n ) = i = 1 2 s ( - 1 ) 2 s - i c 2 s , i P 2 n ( i ) ( t k , n ) t k , n - 4 s + i C c 2 s , 2 s P 2 n ( 2 s ) ( t k , n ) t k , n - 2 s + i = 1 2 s - 1 ( - 1 ) 2 s - i c 2 s , i P 2 n ( i ) ( t k , n ) t k , n - 4 s + i .
Since, we have by Lemma 3.5,
c 2 s , 2 s P 2 n ( 2 s ) ( t k , n ) t k , n - 2 s C μ ( ε , 2 n ) n a 2 n * 2 s - 1 P 2 n ( t k , n ) t k , n - 2 s
and by Lemma 3.4,
i = 1 2 s - 1 ( - 1 ) 2 s - i c 2 s , i P 2 n ( i ) ( t k , n ) t k , n - 4 s + i C n a 2 n * 2 s - 1 P 2 n ( t k , n ) t k , n - 2 s i = 1 2 s - 1 n a 2 n * t k , n - 2 s + i C ε n a 2 n * 2 s - 1 P 2 n ( t k , n ) t k , n - 2 s ,
we have
p n ( 2 s ) ( x k , n ) C δ ( ε , n ) n a 2 n * 2 s - 1 P 2 n ( t k , n ) t k , n - 2 s C δ ( ε , n ) n a n 2 s - 1 p n ( x k , n ) x k , n - ( 2 s - 1 ) 2 ,

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,
φ j ( k ) : = r = 0 j 1 2 j - 2 r + 1 2 j 2 r φ r ( k - 1 ) .
(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.
  1. (a)
    From Theorem 1.7, we know that for 1 ≤ 2s -1 ≤ ν - 1,
    l k , n ( 2 s - 1 ) ( x k , n ) = p n ( 2 s ) ( x k , n ) 2 s p n ( x k , n ) C δ ( ε , n ) n a n 2 s - 1 x k , n - 2 s - 1 2 .
     
From Theorem 1.5, we know that for x k, n a n /4,
( l k , n m ) ( j ) ( x k , n ) C n a n j x k , n - j 2 .
(3.8)
Then, we have by mathematical induction on m,
( l k , n m ) ( 2 s - 1 ) ( x k , n ) C r = 0 s 2 s - 1 2 r ( l k , n m - 1 ) ( 2 r ) ( x k , n ) l k , n ( 2 s - 2 r - 1 ) ( x k , n ) + r = 0 s 2 s - 1 2 r + 1 ( l k , n m - 1 ) ( 2 r + 1 ) ( x k , n ) l k , n ( 2 s - 2 r - 2 ) ( x k , n ) C δ ( ε , n ) n a n 2 s - 1 x k , n - 2 s - 1 2 .
  1. (b)
    From Theorem 3.3, we know that for 0 ≤ 2sν - 1,
    l k , n ( 2 s ) ( x k , n ) = p n ( 2 s + 1 ) ( x k , n ) ( 2 s + 1 ) p n ( x k , n ) = ( - 1 ) s ϕ s ( 1 ) β s ( 2 n , k ) n a n 2 s x k , n - s ( 1 + ρ s ( ε , x k , n , n ) ) .
    (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 ( l k , n m ) ( 2 s ) ( x k , n ) into two terms as follows:
( l k , n m ) ( 2 s ) ( x k , n ) = 0 2 r 2 s 2 s 2 r ( l k , n m - 1 ) ( 2 r ) ( x k , n ) l k , n ( 2 s - 2 r ) ( x k , n ) + 1 2 r - 1 2 s 2 s 2 r - 1 ( l k , n m - 1 ) ( 2 r - 1 ) ( x k , n ) l k , n ( 2 s - 2 r + 1 ) ( x k , n ) .
(3.10)
For the second term, we have from (1.10),
1 2 r - 1 2 s 2 s 2 r - 1 ( l k , n m - 1 ) ( 2 r - 1 ) ( x k , n ) l k , n ( 2 s - 2 r + 1 ) ( x k , n ) C δ 2 ( ε , n ) n a n 2 s 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 ( 2 s - 2 r ) ( x k , n ) = ( - 1 ) s - r ϕ s - r ( 1 ) β s - r ( 2 n , k ) n a n 2 s - r x k , n - ( s - r ) ( 1 + ξ s - r ( 1 ) )
and |ξs-r(1)| → 0 as n → ∞ and ε → 0. By mathematical induction, we assume for 0 ≤ 2r ≤ 2s;
( l k , n m - 1 ) ( 2 r ) ( x k , n ) = ( - 1 ) r ϕ r ( m - 1 ) β r ( 2 n , k ) n a n 2 r x k , n - r ( 1 + ξ r ( m - 1 ) )
and |ξ r (m - 1)| → 0 as n → ∞ and ε → 0. Then, since
( l k , n m - 1 ) ( 2 r ) ( x k , n ) l k , n ( 2 s - 2 r ) ( x k , n ) = ( - 1 ) s β s ( 2 n , k ) n a n 2 s x k , n - s × ϕ r ( m - 1 ) ϕ s - r ( 1 ) ( 1 + ξ r ( m - 1 ) ) 1 + ξ s - r ( 1 ) ,
we have for 0 ≤ 2r ≤ 2s, using the definition of (3.7),
0 2 r 2 s 2 s 2 r ( l k , n m - 1 ) ( 2 r ) ( x k , n ) l k , n ( 2 s - 2 r ) ( x k , n ) = ( - 1 ) s β s ( 2 n , k ) n a n 2 s x k , n - s × 0 2 r 2 s 2 s 2 r ϕ r ( m - 1 ) ϕ s - r ( 1 ) ( 1 + ξ r ( m - 1 ) ) ( 1 + ξ s - r ( 1 ) ) = ( - 1 ) s ϕ s ( m ) β s ( 2 n , k ) n a n 2 s x k , n - s + ( - 1 ) s β s ( 2 n , k ) n a n 2 s x k , n - s × 0 2 r 2 s 2 s 2 r ϕ r ( m - 1 ) ϕ s - r ( 1 ) ( ξ r ( m - 1 ) + ξ s - r ( 1 ) + ξ r ( m - 1 ) ξ s - r ( 1 ) ) .
Here, we consider (3.10). If we let
ξ s ( m , ε , x k , n , n ) = ξ s ( m ) = 0 2 r 2 s 2 s 2 r ϕ r ( m - 1 ) ϕ s - r ( 1 ) ϕ s ( m ) ( ξ r ( m - 1 ) + ξ s - r ( 1 ) + ξ r ( m - 1 ) ξ s - r ( 1 ) ) + 1 2 r - 1 2 s 2 s 2 r - 1 ( l k , n m - 1 ) ( 2 r - 1 ) ( x k , n ) l k , n ( 2 s - 2 r + 1 ) ( x k , n ) ( - 1 ) s ϕ s ( m ) β s ( 2 n , k ) n a n 2 s x k , n - s ,
then we have
ξ s ( m ) 0 2 r 2 s 2 s 2 r ϕ r ( m - 1 ) ϕ s - r ( 1 ) ϕ s ( m ) ( ξ r ( m - 1 ) + ξ s - r ( 1 ) + ξ r ( m - 1 ) ξ s - r ( 1 ) ) + C δ 2 ( ε , n ) ( n a n ) 2 s x k , n - s ( - 1 ) s ϕ s ( m ) β s ( 2 n , k ) n a n 2 s x k , n - s by the definition of  ( 3 . 11 ) 0 2 r 2 s 2 s 2 r ϕ r ( m - 1 ) ϕ s - r ( 1 ) ϕ s ( m ) ( ξ r ( m - 1 ) + ξ s - r ( 1 ) + ξ r ( m - 1 ) ξ s - r ( 1 ) ) + C δ 2 ( ε , n ) .

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 ...,
ϕ 0 ( ν ) : = 1
and for j = 1, 2, 3 ..., ν = 2, 3, 4, ...,
ϕ j ( ν ) - ϕ j ( ν - 1 ) = 1 2 j + 1 r = 0 j - 1 2 j + 1 2 r ϕ r ( ν - 1 ) .

Now, for every j we will introduce an auxiliary polynomial determined by { Ψ j ( y ) } j = 1 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
Ψ j ( ν ) = ϕ j ( ν ) , ν = 1 , 2 , 3 , .
  1. (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,
Ψ j ( y ) = r = 0 j 1 2 j - 2 r + 1 2 j 2 r Ψ r ( y - 1 ) , 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 ( x ) 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 xk, nand 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 ( j + 1 ) ( x k , n ) = - s = 0 j - 1 j s F ( s + 1 ) ( x k , n ) l ( j - s ) ( x k , n ) + y s = 0 j j s l ( s + 1 ) ( x k , n ) F ( j - s ) ( x k , n ) ,
which shows that F(j+1)(x k, n ) is a polynomial of degree at most j + 1 with respect to y. Let P k n [ j ] ( y ) , j = 0, 1, 2, ... be defined by
( x ) 2 j F k n ( x k , n , y ) = ( - 1 ) j β j ( 2 n , k ) n a n 2 j x k , n - j Ψ j ( y ) + P k n [ j ] ( y ) .
(3.12)

Then P k n [ j ] ( y ) 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, 1 ε a n n 2 x k , n ε a n , and |y| ≤ M. Then
  1. (a)
    there exists κ j (ε, x k, n , n) > 0 such that
    ( y ) s P k n [ j ] ( y ) C κ j ( ε , x k , n , n ) n a n 2 j x k , n - j , s = 0 , 1
    (3.13)
     
and κ j (ε, x k, n , n) → 0 as n → ∞ and ε → 0.
  1. (b)
    there exists γ j (ε, n) > 0 such that
    ( x ) 2 j + 1 F k n ( x k , n , y ) C γ j ( ε , n ) n a n 2 j + 1 x k , n - 2 j + 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 ...,
( - 1 ) j Ψ j ( y ) > 0 .
Lemma 3.9. For positive integers s and m with 1 ≤ mν,
r = 0 s 2 s 2 r Ψ r ( - m ) φ s - r ( m ) = 0 .
Proof. If we let C s ( y ) = r = 0 s 2 s 2 r Ψ r ( - y ) Ψ s - r ( y ) , then it suffices to show that C s (m) = 0. For every s,
0 = ( l k , n - m + m ) 2 s ( x k , n ) = i = 0 2 s 2 s i ( l k , n - m ) ( i ) ( x k , n ) ( l k , n m ) ( 2 s - i ) ( x k , n ) = r = 0 s 2 s 2 r ( x ) 2 r F k n ( x k , n , - m ) ( l k , n m ) ( 2 s - 2 r ) ( x k , n ) + r = 0 s - 1 2 s 2 r + 1 ( x ) 2 r + 1 F k n ( x k , n , - m ) ( l k , n m ) ( 2 s - 2 r - 1 ) ( x k , n ) .
By (1.11), (3.12) and (3.13), we see that the first sum r = 0 s has the form of
r = 0 s = ( - 1 ) s β s ( 2 n , k ) n a n 2 s x k , n - s r = 0 s 2 s 2 r Ψ r ( - m ) ϕ s - r ( m ) + η ̃ s ( - m , ε , x k , n , n ) .
Then, since
η ̃ s ( - m , ε , x k , n , n ) = r = 0 s 2 s 2 r Ψ r ( - m ) ϕ s - r ( m ) ξ s - r ( m , ε , x k , n , n ) + r = 0 s 2 s 2 r ( - 1 ) - r β - r ( 2 n , k ) n a n - 2 r