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Mean and uniform convergence of Lagrange interpolation with the Erdős-type weights

Abstract

Let R=(,), and let Q C 1 (R):R R + :=[0,) be an even function. We consider the exponential-type weights w(x)= e Q ( x ) , xR. In this paper, we obtain a mean and uniform convergence theorem for the Lagrange interpolation polynomials L n (f) in L p , 1<p with the weight w.

MSC:41A05.

1 Introduction and preliminaries

Let R=(,), and let Q C 1 (R):R R + :=[0,) be an even function, and w(x)=exp(Q(x)) be the weight such that 0 x n w 2 (x)dx< for all n=0,1,2, . Then we can construct the orthonormal polynomials p n (x)= p n ( w 2 ;x) of degree n with respect to w 2 (x). That is,

p n (x) p m (x) w 2 (x)dx= δ m n (Kronecker’s delta)

and

p n (x)= γ n x n +, γ n >0.

We denote the zeros of p n (x) by

< x n , n < x n 1 , n << x 2 , n < x 1 , n <.

We denote the Lagrange interpolation polynomial L n (f;x) based at the zeros { x k , n } k = 1 n as follows:

L n (f;x):= k = 1 n f( x k , n ) l k , n (x), l k , n (x):= p n ( x ) ( x x k , n ) p n ( x k , n ) .

A function f: R + R + is said to be quasi-increasing if there exists C>0 such that f(x)Cf(y) for 0<x<y.

We are interested in the following subclass of weights from [1].

Definition 1.1 Let Q:R R + be an even function satisfying the following properties:

  1. (a)

    Q (x) is continuous in , with Q(0)=0.

  2. (b)

    Q (x) exists and is positive in R{0}.

  3. (c)

    lim x Q(x)=.

  4. (d)

    The function

    T(x):= x Q ( x ) Q ( x ) ,x0

is quasi-increasing in (0,) with

T(x)Λ>1,x R + {0}.
  1. (e)

    There exists C 1 >0 such that

    Q ( x ) | Q ( x ) | C 1 | Q ( x ) | Q ( x ) ,a.e. xR{0}.

Then we write w(x)=exp(Q(x))F( C 2 ). If there also exist a compact subinterval J (0) of and C 2 >0 such that

Q ( x ) | Q ( x ) | C 2 | Q ( x ) | Q ( x ) ,a.e. xRJ,

then we write w(x)=exp(Q(x))F( C 2 +).

Example 1.2 (1) If T(x) is bounded, then the weight w=exp(Q) is called the Freud-type weight. The following example is the Freud-type weight:

Q(x)= | x | α ,α>1.

If T(x) is unbounded, then the weight w=exp(Q) is called the Erdős-type weight. The following examples give the Erdős-type weights w=exp(Q).

  1. (2)

    [2, Theorem 3.1] For α>1, l=1,2,3,

    Q(x)= Q l , α (x)= exp l ( | x | α ) exp l (0),

where

exp l (x)=exp ( exp ( exp exp x ) ) (l-times).

More generally, we define for α+u>1, α0, u0 and l1,

Q l , α , u (x):= | x | u ( exp l ( | x | α ) α exp l ( 0 ) ) ,

where α =0 if α=0, otherwise α =1. (We note that Q l , 0 , u (x) gives a Freud-type weight.)

  1. (3)

    We define Q α (x):= ( 1 + | x | ) | x | α 1, α>1.

In this paper, we investigate the convergence of the Lagrange interpolation polynomials with respect to the weight wF( C 2 +). When we consider the Erdős-type weights, the following definition follows from Damelin and Lubinsky [3].

Definition 1.3 Let w(x)=exp(Q(x)), where Q:RR is even and continuous. Q exists in (0,), Q ( j ) 0, in (0,), j=0,1,2, and the function

T (x):=1+ x Q ( x ) Q ( x )

is increasing in (0,) with

lim x T (x)=; T (0+):= lim x 0 + T (x)>1.
(1.1)

Moreover, we assume that for some constants C 1 , C 2 , C 3 >0,

C 1 T (x)/ ( x Q ( x ) Q ( x ) ) C 2 ,x C 3 ,

and for every ε>0,

T (x)=O ( Q ( x ) ε ) ,x.
(1.2)

Then we write wE.

Damelin and Lubinsky [3] got the following results with the Erdős-type weights w=exp(Q)E.

Theorem A ([3, Theorem 1.3])

Let w=exp(Q)E. Let L n (f,x) denote the Lagrange interpolation polynomial to f at the zeros of p n ( w 2 ,x). Let 1<p<, ΔR, κ>0. Then for

lim n ( f L n ( f ) ) w ( 1 + Q ) Δ L p ( R ) =0

to hold for every continuous function f:RR satisfying

lim | x | |f(x)w(x) ( log | x | ) 1 + κ |=0,

it is necessary and sufficient that

Δ>max { 0 , 2 3 ( 1 4 1 p ) } .

Our main purpose in this paper is to give mean and uniform convergence theorems with respect to { L n (f)}, n=1,2, , in L p -norm, 1<p. The proof for 1<p< will be shown by use of the method of Damelin and Lubinsky. In Section 2, we write the main theorems. In Section 3, we prepare some fundamental lemmas; and in Section 4, we will prove the theorem for 1<p<. Finally, we will prove the theorem for the uniform convergence in Section 5.

For any nonzero real-valued functions f(x) and g(x), we write f(x)g(x) if there exist constants C 1 , C 2 >0 independent of x such that C 1 g(x)f(x) C 2 g(x) for all x. Similarly, for any two sequences of positive numbers { c n } n = 1 and { d n } = 1 , we define c n d n . We denote the class of polynomials of degree at most n by P n .

Throughout C, C 1 , C 2 , denote positive constants independent of n, x, t, and polynomials of degree at most n. The same symbol does not necessarily denote the same constant in different occurrences.

2 Theorems

In the following, we introduce useful notations. Mhaskar-Rakhmanov-Saff numbers (MRS) a x are defined as the positive roots of the following equations:

x= 2 π 0 1 a x u Q ( a x u ) ( 1 u 2 ) 1 2 du,x>0.

The function φ u (x) is defined as follows:

φ u (x)={ a u u 1 | x | a 2 u 1 | x | a u + δ u , | x | a u , φ u ( a u ) , a u < | x | ,

where

δ x = ( x T ( a x ) ) 2 3 ,x>0.

We define

Φ(x):= 1 ( 1 + Q ( x ) ) 2 3 T ( x )

and

Φ n (x):=max { δ n , 1 | x | a n } .

Here we note that for 0<d|x|,

Φ(x) Q ( x ) 1 3 x Q ( x )

and we see

Φ(x)C Φ n (x),n1

(see Lemma 3.3 below). Moreover, we define

Φ ( 1 4 1 p ) + (x):={ 1 , 0 < p < 4 , Φ 1 4 1 p ( x ) , 4 p .

Let 1<p<. We give a convergence theorem as an analogy of Theorem A for L n (f) in L p -norm. We need to prepare a lemma.

Lemma 2.1 ([4, Theorem 1.6])

Let w=exp(Q)F( C 2 +).

  1. (a)

    Let T(x) be unbounded. Then for any η>0, there exists a constant C(η)>0 such that for t1,

    a t C(η) t η .
  2. (b)

    Assume

    Q ( x ) Q ( x ) λ(b) Q ( x ) Q ( x ) ,|x|b>0,
    (2.1)

where b>0 is large enough. Suppose that there exist constants η>0 and C 1 >0 such that a t C 1 t η . If λ:=λ(b)>1, then there exists a constant C(λ,η) such that for a t 1,

T( a t )C(λ,η) t 2 ( η + λ 1 ) λ + 1 .
(2.2)

If 0<λ1, then for any μ>0, there exists C(λ,μ) such that

T( a t )C(λ,μ) t μ ,t1.
(2.3)

For a fixed constant β>0, we define

ϕ(x):= ( 1 + x 2 ) β / 2 .
(2.4)

Using this function, we have the following theorem. We suppose that the weight w is the Erdős-type weight.

Our theorem is as follows. Let f C 0 (R) mean that fC(R) and lim | x | f(x)=0.

Theorem 2.2 Let w=exp(Q)F( C 2 +), and let T(x) be unbounded. Let 1<p< and β>0, and let us define ϕ as (2.4), and λ=λ(b)1 as (2.1). We suppose that for fC(R),

ϕ 1 (x)w(x)f(x) C 0 (R),

and

Δ> 9 4 λ 1 3 λ 1 .
(2.5)

Then we have

lim n ( f L n ( f ) ) w Φ Δ + ( 1 4 1 p ) + L p ( R ) =0.

We remark that if wF( C 2 +) is the Erdős-type weight, then we have λ=λ(b)1 in (2.1). In fact, if λ<1, then by Lemma 3.9 below, we see that for xb>0,

T(x)= x Q ( x ) Q ( x ) x Q ( x ) Q (b) ( Q ( x ) Q ( b ) ) λ = Q ( b ) Q ( b ) λ x Q ( x ) 1 λ 0as x.

This contradicts our assumption for T(x). In Example 1.2, we consider the weight w l , α , m =exp( Q l , α , m ). In (2.1), we set Q:= Q l , α , m and λ:=λ(b). If w l , α , m is an Erdős-type weight, that is, T(x):= T l , α , m (x) is unbounded, then it is easy to show

lim b λ(b)=1.

Therefore, when we give any Δ>0, there exists a constant b large enough such that

Δ> 9 4 λ ( b ) 1 3 λ ( b ) 1 .

Hence, we have the following corollary.

Corollary 2.3 Let 1<p< and Δ>0. Then for the weight w l , α , m =exp( Q l , α , m ) (α>0), we have

lim n ( f L n ( f ) ) w l , α , m Φ Δ + ( 1 4 1 p ) + L p ( R ) =0.

We also consider the case of p=.

Theorem 2.4 Let w=exp(Q)F( C 2 +), and let T(x) be unbounded. For every f C 0 (R) and n1, we have

( f L n ( f ) ) w Φ 3 / 4 L ( R ) C E n 1 (w;f)logn,

where

E m (w;f)= inf P m P m max x R | ( f ( x ) P m ( x ) ) w(x)|,m=0,1,2,.

Moreover, if f ( r ) , r1, is an integer, then for n>r+1 we have

( f L n ( f ) ) w Φ 3 / 4 L ( R ) C ( a n n ) r E n r 1 ( w ; f ( r ) ) logn.

3 Fundamental lemmas

To prove the theorems we need some lemmas.

Lemma 3.1 Let w=exp(Q)F( C 2 +). Then we have the following.

  1. (a)

    [1, Lemma 3.11(a), (b)] Given fixed 0<α, α1, we have uniformly for t>0,

    |1 a α t a t | 1 T ( a t ) ,

and we have for t>0,

|1 a t a s | 1 T ( a t ) |1 t s |, 1 2 t s 2.
  1. (b)

    [1, Lemma 3.7 (3.38)] For some 0<ε2, and for large enough t,

    T( a t ) t 2 ε .

Lemma 3.2 Let w=exp(Q)F( C 2 +). Then we have the following.

  1. (a)

    [1, Lemma 3.5(a), (b)] Let L>0 be a fixed constant. Uniformly for t>0,

    Q( a L t )Q( a t )and Q ( a L t ) Q ( a t ).

Moreover,

a L t a t andT( a L t )T( a t ).
  1. (b)

    [1, Lemma 3.4 (3.18), (3.17)] Uniformly for x>0 with a t :=x, t>0, we have

    Q (x) t T ( x ) a t andQ(x) t T ( x ) .
  2. (c)

    [1, Lemma 3.8(a)] For x[0, a t ),

    Q (x)C t a t 1 1 x a t .

Lemma 3.3 Let w=exp(Q)F( C 2 +). For xR, we have

Φ(x)C Φ n (x),n1.

Proof Let x= a u , u1. By Lemma 3.2(b), we have

uQ( a u ) T ( a u ) .

So, we have

δ u 1 Q 2 3 ( a u )T( a u )= a u Q ( a u ) Q 1 3 ( a u ) = x Q ( x ) Q 1 3 ( x ) .
(3.1)

Now, if u n 2 , then we have

1 a u a n 1 a n / 2 a n 1 T ( a n ) ( by Lemma 3.1(a) ) 1 ( n T ( a n ) ) 2 3 = δ n ( by Lemma 3.1(b) ) .

So, we have

Φ n ( x ) = 1 a u a n 1 a u a 2 u 1 T ( a u ) ( by Lemma 3.1(a) ) 1 ( u T ( a u ) ) 2 3 = δ u Φ ( x ) ( by Lemma 3.2(b) and (3.1) ) .

Let n 2 <u<n. Then we have

Φ n (x) δ n δ u Φ(x) ( by Lemma 3.2(a), (b) and (3.1) ) .

 □

Lemma 3.4 Let wF( C 2 +). Then we have the following.

  1. (a)

    [1, Theorem 1.19(f)] For the minimum positive zero x [ n / 2 ] , n ,

    x [ n / 2 ] , n a n n ,

and for the maximum zero x 1 , n ,

1 x 1 , n a n δ n .
  1. (b)

    [1, Theorem 1.19(e)] For n1 and 1jn1,

    x j , n x j + 1 , n φ n ( x j , n ).
  2. (c)

    [1, p.329, (12.20)] Uniformly for n1, 1kn1,

    φ n ( x k , n ) φ n ( x k + 1 , n ).
  3. (d)

    Let max{| x k , n |,| x k + 1 , n |} a α n , 0<α<1. Then we have

    w( x k , n )w( x k + 1 , n )w(x)( x k + 1 , n x x k , n ).

So, for given C>0 and |x| a β n , 0<β<α, if |x x k , n |C φ n (x), then we have

w(x)w( x k , n ).

Proof (d) Let max{| x k , n |,| x k + 1 , n |}=| x k , n | (for the case of max{| x k , n |,| x k + 1 , n |}=| x k + 1 , n |, we also have the result similarly). By (b) there exists a constant C>0 such that

| x k , n x k + 1 , n |C φ n ( x k , n ).

Then we see

φ n ( x k , n ) a n n 1 | x k , n | a 2 n 1 | x k , n | a n = a n n 1 | x k , n | a n + | x k , n | { 1 a n 1 a 2 n } 1 | x k , n | a n = a n n 1 | x k , n | a n + | x k , n | a n ( 1 a n a 2 n ) 1 | x k , n | a n a n n 1 | x k , n | a n + C | x k , n | a n 1 T ( a n ) 1 | x k , n | a n a n n 1 | x k , n | a n .
(3.2)

Therefore, from (3.2) and Lemma 3.2(c), we have

| Q ( x k , n ) Q ( x k + 1 , n ) | = | Q ( ξ ) ( x k , n x k + 1 , n ) | C | Q ( ξ ) | φ n ( x ) ( x k + 1 , n ξ x k , n ) C | Q ( x k , n ) | a n n 1 | x k , n | a n C n a n 1 1 | x k , n | a n a n n 1 | x k , n | a n C .

Consequently,

w( x k , n )w( x k + 1 , n )w(x)( x k + 1 , n x x k , n ).

Let |x x k , n |C φ n (x) and |x| a β n . Then we see that there exists n 0 >0 such that | x k , n | a α n , n n 0 . In fact, we can show it as follows. We use Lemma 3.1(a) and (b). For |x| a β n , we see

| x k , n ||x|+C φ n (x)|x|+C a n n 1 | x | a n ,

and if we take n large enough, then we have

d d t ( t + C a n n 1 t a n ) = 1 C 1 n 1 2 1 t a n 1 C 1 n 1 2 1 a n / 3 a n 1 C T ( a n ) 2 n 1 C 1 2 n ε / 2 > 0 ,

that is, g(t)=t+C a n n 1 t a n is increasing. So, we see

| x k , n | a β n +C a n n 1 a β n a n a β n +C a n n 1 T ( a n ) .

Therefore, we have

a α n ( a β n + C a n n 1 T ( a n ) ) a n T ( a n ) C a n n 1 T ( a n ) = a n T ( a n ) ( 1 C T ( a n ) n ) a n T ( a n ) ( 1 C 1 n ε / 2 ) > 0 .

Now, we can show (d). Without loss of generality, we may assume x[ x j + 1 , n , x j , n ]{ x k , n ||x x k , n |C φ n (x)}. We define

x k 1 , n :=min { x k , n | | x x k , n | C φ n ( x ) } , x k 2 , n :=max { x k , n | | x x k , n | C φ n ( x ) } .

Here we note that k 1 , k 2 are decided depending only on the constant C. Then by former result, we have

w( x k 1 , n )w( x k 2 , n )w(x)( x k 1 , n x x k 2 , n ).

 □

Lemma 3.5 Let w=exp(Q)F( C 2 +). Then we have the following.

  1. (a)

    [1, Theorem 1.17] Uniformly for n1,

    sup x R | p n (x)|w(x)| x 2 a n 2 | 1 4 1.
  2. (b)

    [1, Theorem 1.19(a)] Uniformly for n1 and 1jn,

    | ( p n w ) ( x j , n )| φ n 1 ( x j , n ) a n 1 2 ( 1 | x j , n | a n ) 1 4 .
  3. (c)

    [1, Theorem 1.19(d)] For x[ x k + 1 , n , x k , n ], if kn1,

    | p n ( x ) w ( x ) | min { | x x k , n | , | x x k + 1 , n | } a n 1 / 2 φ n ( x ) 1 ( 1 | x k , n | a n ) 1 / 4 .

Lemma 3.6 (cf. [5, Theorem 2.7])

Let wF( C 2 +) and 0<p. Then uniformly n2,

Φ ( 1 4 1 p ) + p n w L p ( R ) C a n 1 p 1 2 { 1 , 0 < p < 4 or p = ; log ( 1 + n ) , 4 p ,

where x + =0 if x0, x + =x if x>0.

Proof From Lemma 3.3, we know Φ(x) Φ n (x), then in [5, Theorem 2.7] we only exchange Φ n with Φ. □

Let f L p , w (R). The Fourier-type series of f is defined by

f ˜ (x):= k = 0 a k ( w 2 , f ) p k ( w 2 , x ) , a k ( w 2 , f ) := f(t) p k ( w 2 , t ) w 2 (t)dt.

We denote the partial sum of f ˜ (x) by

s n (f,x):= s n ( w 2 , f , x ) := k = 0 n 1 a k ( w 2 , f ) p k ( w 2 , x ) .

The partial sum s n (f) admits the representation

s n (f,x)= j = 0 n 1 a j p j (x)= f(t) K n (x,t) w 2 (t)dt,

where

K n (x,t):= j = 0 n 1 p j (x) p j (t).

The Christoffel-Darboux formula

K n (x,t)= γ n 1 γ n p n ( x ) p n 1 ( t ) p n 1 ( x ) p n ( t ) x t
(3.3)

is well known (see [6, Theorem 1.1.4]).

Lemma 3.7 ([6, Lemma 9.2.6])

Let 1<p< and g L p (R). Then for the Hilbert transform

H(g,x):= lim ε 0 + | x t | ε g ( t ) x t dt,xR,
(3.4)

we have

H ( g ) L p ( R ) C g L p ( R ) ,

where C>0 is a constant depending upon p only.

Lemma 3.8 (see [7, Theorem 1.4, Theorem 1.6])

Let w=exp(Q)F( C 2 ), 1p and γ0. Then for any ε>0, there exists a polynomial P such that

( f ( x ) P ( x ) ) ( 1 + x 2 ) γ w ( x ) L p ( R ) <ε.

Lemma 3.9 Let wF( C 2 +) be an Erdős-type weight, that is, T(x) is unbounded. Then for any M>1, there exist x M >0 and C M >0 such that

Q(x) C M x M ,x x M .

Proof For every M>1, there exists x M >0 such that T(x)M for x x M , so that Q (x)/Q(x)=T(x)/xM/x for x x M . Hence, we see

log Q ( x ) Q ( x M ) log ( x x M ) M ,x x M ,

that is,

Q(x) Q ( x M ) ( x M ) M x M ,x x M .

Let us put C M :=Q( x M )/ ( x M ) M . □

4 Proof of Theorem 2.2 by Damelin and Lubinsky methods

In this section, we assume wF( C 2 +). To prove the theorem we need some lemmas, and we will use the Damelin and Lubinsky methods of [3].

Lemma 4.1 (cf. [3, Lemma 3.1])

Let wF( C 2 +). Let 0<α< 1 4 and

n (x):= | x k , n | a α n | l k , n ( x ) | w 1 ( x k , n ).

Then we have for |x| a α n / 2 and |x| a 2 n ,

n (x)w(x)C.

Moreover, for a α n / 2 |x| a 2 n ,

n (x)w(x)C ( log n + a n 1 2 | p n ( x ) w ( x ) | T 1 4 ( a n ) ) .

Proof The proof of [3, Lemma 3.1] holds without the condition (1.2) and the second condition in (1.1) and under the assumption of the quasi-increasingness of T(x). The conditions in Definition 1.1 contain all the conditions in Definition 1.3 except for (1.2) and the second condition in (1.1). We see that in [3, Lemma 3.1] we can replace T (x) with T(x). □

Lemma 4.2 ([3, Lemma 3.2])

Let 0<η<1. Let ψ:R(0,) be a continuous function with the following property: For n1, there exist polynomials R n of degree n such that

C 1 ψ ( t ) R n ( t ) C 2 ,|t| a 4 n .

Then for n n 0 and P P n ,

| x k , n | a η n λ k , n | P ( x k , n ) | w 1 ( x k , n )ψ( x k , n )C a 4 n a 4 n | P ( t ) w ( t ) | ψ(t)dt.

Remark 4.3 To prove Lemma 4.7 below, we apply this lemma with ψ(t)=ϕ(t)= ( 1 + t 2 ) β / 2 , β>0. In fact, when ϕ (x)=ϕ(t), t= a 4 n x, we can approximate ϕ by polynomials R n P n on [1,1], that is, for any ε>0 there exists R n P n such that

| ϕ (x) R n (x)|<ε,x[1,1].

Therefore,

| R n ( x ) ϕ ( x ) 1|< ε ϕ ( x ) ,x[1,1],

and so there exist C 1 , C 2 >0 such that

C 1 1 ε ϕ ( x ) | R n ( x ) ϕ ( x ) |<1+ ε ϕ ( x ) C 2 ,x[1,1].

Now, if we set R n (t)= R n (x), then we have the result.

Lemma 4.4 (cf. [3, Lemma 4.1])

Let { f n } n = 1 be a sequence of measurable functions from RR such that for n1,

f n (x)=0,|x|< a n 9 ; | f n ( x ) | w(x)ϕ(x),xR.

Then for 1p and Δ>0, we have

lim n L n ( f n ) w Φ Δ + ( 1 4 1 p ) + L p ( R ) =0.
(4.1)

Proof Let |x| a n 18 or |x| a 2 n . We use the first inequality of Lemma 4.1 with α= 1 9 , then from the assumption with respect to f n , we see that

| L n ( f n ; x ) w ( x ) | ϕ( a n 9 ) | x k , n | a n 9 | l k , n ( x ) | w 1 ( x k , n )w(x) C 1 ϕ( a n 9 ).

So,

L n ( f n ) w Φ Δ + ( 1 4 1 p ) + L p ( | x | a n 18 o r | x | a 2 n ) ϕ ( a n 9 ) Φ Δ + ( 1 4 1 p ) + L p ( R ) C 2 ϕ ( a n 9 ) = o ( 1 )
(4.2)

by Lemma 3.9 (note the definition of Φ(x)) and the definition of ϕ in (2.4). Next, we let a n 18 |x| a 2 n . From the second inequality in Lemma 4.1, we see that

| L n ( f n ; x ) w ( x ) | ϕ( a n 9 ) ( log n + a n 1 2 | p n ( x ) | w ( x ) T 1 4 ( a n ) ) .

Also, for this range of x, we see that

Φ(x)= 1 ( 1 + Q ( x ) ) 2 3 T ( x ) 1 ( 1 + Q ( a n ) ) 2 3 T ( a n ) T 1 3 ( a n ) n 2 3 T ( a n ) = δ n

by Lemma 3.2(b). So, for n large enough,

Then since Δ>0, using Lemma 3.1(a), Lemma 2.1(a), and Lemma 3.6, we have

logn Φ Δ + ( 1 4 1 p ) + L p ( a n 18 | x | a 2 n ) C δ n Δ ( a 2 n a n 18 ) 1 p lognC

and

Therefore, we have by (2.4)

L n ( f n ) w Φ Δ + ( 1 4 1 p ) + L p ( a n 18 | x | a 2 n ) C 4 ϕ( a n 9 )=o(1).

Consequently, with (4.2) we have (4.1). □

Lemma 4.5 (cf. [3, Lemma 4.2])

Let 1p. Let { g n } n = 1 be a sequence of measurable functions from RR such that for n1,

g n (x)=0,|x| a n 9 ; | g n ( x ) | w(x)ϕ(x),xR.
(4.3)

Let us suppose

Δ> 9 4 λ 1 3 λ 1 ,
(4.4)

where λ1 is defined in Lemma 2.1. Then for 1p, we have

lim n L n ( g n ) w Φ Δ + ( 1 4 1 p ) + L p ( | x | a n 8 ) =0.
(4.5)

Proof Using Lemma 3.5(b) and Lemma 3.4(b), we have for x a n 8 ,

| L n ( g n ; x ) | | x k , n | a n 9 | l k , n ( x ) | w 1 ( x k , n ) ϕ ( x k , n ) C 1 a n 1 2 | p n ( x ) | | x k , n | a n 9 ( x k , n x k + 1 , n ) ( 1 | x k , n | a n + δ n ) 1 4 | x x k , n | ϕ ( x k , n ) C 2 a n 1 2 | p n ( x ) | a n 9 a n 9 ( 1 | t | a n + δ n ) 1 4 | x t | ϕ ( t ) d t .
(4.6)

Equation (4.6) is shown as follows: First, we see

|xt||x x k , n |,t[ x k + 1 , n , x k , n ].
(4.7)

Let |x| a n 8 and t[ x k + 1 , n , x k , n ]. Then

| x t x x k , n 1|=| t x k , n x x k , n | x k , n x k + 1 , n | x k ± 2 , n x k , n | c<1.

Now, we use the fact that x+Cφ(x), x>0 is increasing for 0<x a n / 2 , and then

x k , n +C φ n ( x k , n ) a n 9 +C φ n ( a n 9 ) a n 8 x.

Here, the second inequality follows from the definition of φ n (x) and Lemma 3.1(a), (b). Hence, we have (4.7). Now, we use the monotonicity of ( 1 | x | a n + δ n ) 1 4 ϕ(x). From (4.7) there exists C>0 such that for t[ x k + 1 , n , x k , n ],

( x k , n x k + 1 , n ) ( 1 | x k , n | a n + δ n ) 1 4 | x x k , n | ϕ ( x k , n ) x k + 1 , n x k , n ( 1 | t | a n + δ n ) 1 4 | x x k , n | ϕ ( t ) d t 1 C x k + 1 , n x k , n ( 1 | t | a n + δ n ) 1 4 | x t | ϕ ( t ) d t .

Hence, (4.6) holds. Next, for t[0, a n 9 ] and x a n 8 , we know by Lemma 3.1(a),

1 a n t x t 1+ a n a n 8 a n 8 t 1+ a n a n 8 a n 8 a n 9 1+C a n 8 a n 9 T ( a n 9 ) T ( a n 8 ) C 3

and

1 | t | a n C 4 1 T ( a n ) δ n .

So, we have

| L n ( g n ; x ) | C 6 a n 1 4 | p n ( x ) | 0 a n 9 ( x t ) 3 4 ϕ(t)dt.

Let t= a s , n 9 s1. Then, since we know for x a n 8 ,

xt=x ( 1 t x ) a n 8 ( 1 a s a 9 8 s ) C 7 a n T ( a s ) ,

we obtain

| L n ( g n ; x ) | C 8 a n 1 2 | p n ( x ) | 0 a n 9 T 3 4 (t)ϕ(t)dt C 8 a n 1 2 T 3 4 ( a n ) | p n ( x ) | .

Hence, if 1λ, then using Lemma 3.6, (3.1) and (2.2), we have

Here, we may consider that above estimations hold under the condition (4.4), because that η>0 can be taken small enough. Then we have (4.5), that is, for Δ> 9 4 λ 1 3 λ 1 ,

lim n L n ( g n ) w Φ Δ + ( 1 4 1 p ) + L p ( | x | a n 8 ) =0.

 □

Lemma 4.6 (cf. [3, Lemma 4.3])

Let 1<p<. Let σ:RR be a bounded measurable function. Let λ=λ(b)1 be defined in Lemma 2.1, and then we suppose

Δ>{ 0 , 1 < p 2 ; 3 2 ( λ 1 ) 3 λ 1 p 2 p , 2 < p 4 ; max { λ 1 3 λ 1 p 1 p 1 4 λ + 1 3 λ 1 p 4 p , 0 } , 4 < p .
(4.8)

Then for 1<p< and the partial sum s n of the Fourier series, we have

s n [ σ ϕ w 1 ] w Φ Δ + ( 1 4 1 p ) + L p ( | x | a n 8 ) C σ L ( R )
(4.9)

for n1. Here C is independent of σ and n.

Proof We may suppose that σ L ( R ) =1. By (3.3), (3.4) and Lemma 3.5(a),

| s n [ σ ϕ w 1 ] (x)|w(x) a n 1 2 ( 1 | x | a n ) 1 4 j = n 1 n |H[σϕ p j w](x)|.
(4.10)

Let us choose l:=l(n) such that 2 l n 8 2 l + 1 . Then we know

2 l + 3 n 2 l + 4 .
(4.11)

Define

I k =[ a 2 k , a 2 k + 1 ],1kl+2.

For j=n1,n and x I k , we split

H [ σ ϕ p j w ] ( x ) w ( x ) = ( 0 + 0 a 2 k 1 + P . V . a 2 k 1 a 2 k + 2 + a 2 k + 2 ) ( σ ϕ p j w ) ( t ) x t d t : = I 1 ( x ) + I 2 ( x ) + I 3 ( x ) + I 4 ( x ) .
(4.12)

Here P . V . stands for the principal value. First, we give the estimations of I 1 and I 2 for x I k . Let x I k . Then we have by Lemma 3.5(a) and Lemma 3.6 with p=1,

| I 1 ( x ) | 0 | ( p j w ϕ ) ( t ) | t + x C 1 a n 1 2 0 a n 2 ϕ ( t ) t + a 2 d t + C 2 a n 1 a n 2 | p j ( t ) | w ( t ) d t C 2 ( a n 1 2 + a n 1 a n 1 1 2 ) C 3 a n 1 2 .
(4.13)

Here we have used

0 ϕ ( t ) 1 + t dt<.
(4.14)

By Lemma 3.5(a), and noting 1x/ a n 1t/ a n for x I k ,

| I 2 ( x ) | 0 a 2 k 1 | ( p j w ϕ ) ( t ) | x t d t C 4 a n 1 2 0 a 2 k 1 ( 1 t a n ) 1 4 x t d t C 4 a n 1 2 ( 1 x a n ) 1 4 0 a 2 k 1 d t x t = C 4 a n 1 2 ( 1 x a n ) 1 4 log ( 1 a 2 k 1 x ) 1 .

Using

1 a 2 k 1 x 1 a 2 k 1 a 2 k C 1 T ( a 2 k ) C 1 T ( x ) ,

we can see

| I 2 ( x ) | C 6 a n 1 2 ( 1 x a n ) 1 4 log ( T ( x ) C ) .
(4.15)

Next, we give an estimation of I 4 for x I k . Let x I k . From Lemma 3.5(a) again,

| I 4 ( x ) | a 2 k + 2 2 a 2 k + 2 | ( p j w ϕ ) ( t ) | t x d t + C 2 a 2 k + 2 | ( p j w ϕ ) ( t ) | t d t ( by  t 2 ( t x ) ) C 7 ( a n 1 2 a 2 k + 2 2 a 2 k + 2 | 1 t a n | 1 4 d t t x + a n 1 2 2 a 2 k + 2 max { 2 a 2 k + 2 , 1 2 a n } ϕ ( t ) t d t + 1 2 a n | ( p j w ) ( t ) | t d t ) C 7 ( a n 1 2 a 2 k + 2 2 a 2 k + 2 | 1 t a n | 1 4 d t t x + C a n 1 2 + a n 1 a n 1 1 2 ) ( by (4.14) and Lemma 3.6 with  p = 1 ) C 8 a n 1 2 [ J + 1 ] ,
(4.16)

where

J:= a 2 k + 2 2 a 2 k + 2 |1 t a n | 1 4 d t t x .

Since, if

|1 t a n | 1 2 ( 1 x a n ) ,

then we see

|tx|= a n | ( 1 x a n ) ( 1 t a n ) | a n 2 ( 1 x a n ) .

Now, we have

J C 9 ( ( 1 x a n ) 1 4 | 1 t a n | 1 2 ( 1 x a n ) , t [ a 2 k + 2 , 2 a 2 k + 2 ] 1 t x d t + a n 1 ( 1 x a n ) 1 | 1 t a n | 1 2 ( 1 x a n ) , t [ a 2 k + 2 , 2 a 2 k + 2 ] | 1 t a n | 1 4 d t ) C 10 ( ( 1 x a n ) 1 4 log ( 1 + a 2 k + 2 a 2 k + 2 a 2 k + 1 ) + ( 1 x a n ) 1 | 1 s | 1 2 ( 1 x a n ) | 1 s | 1 4 d s ) C 10 ( ( 1 x a n ) 1 4 log ( 1 + C T ( a 2 k + 2 ) ) + 4 3 ( 1 2 ( 1 x a n ) ) 3 4 ( 1 x a n ) 1 ) C 11 ( 1 x a n ) 1 4 log ( C T ( x ) ) .

So, from (4.16) we have

| I 4 ( x ) | C 12 a n 1 2 ( 1 x a n ) 1 4 log ( C T ( x ) ) .
(4.17)

Therefore, from (4.13), (4.15) and (4.17), we have

| I 1 + I 2 + I 4 | C 13 a n 1 2 ( 1 x a n ) 1 4 log ( C T ( x ) ) .

Hence, with (4.10), (4.12) we have

(4.18)

We must estimate the L p -norm with respect to I 3 , that is, P . V . a 2 k 1 a 2 k + 2 ( σ ϕ p j w ) ( t ) x t d t L p ( I k ) . We use M. Riesz’s theorem on the boundedness of the Hilbert transform from L p (R) to L p (R) (Lemma 3.7) to deduce that by Lemma 3.5(a) and the boundedness of |σϕ|,

P . V . a 2 k 1 a 2 k + 2 ( σ ϕ p j w ) ( t ) x t d t L p ( I k ) C 15 ( a 2 k 1 a 2 k + 2 | ( σ ϕ p j w ) ( t ) | p d t ) 1 p C 16 a n 1 2 ( 1 a 2 k + 2 a n ) 1 4 ( a 2 k + 2 a 2 k 1 ) 1 p .
(4.19)

So, by (4.18) and (4.19) we conclude

(4.20)

Noting (4.11), we see n 2 l + 3 for kl, so

1 a 2 k + 1 a n 1 a 2 k + 1 a 2 k + 3 C 19 1 T ( a 2 k ) and a 2 k + 1 a 2 k C 20 a 2 k T ( a 2 k ) .

On the other hand, using Lemma 3.2(b), we see Φ( a t ) δ t . Hence, we have

Φ Δ + ( 1 4 1 p ) + ( a 2 k ) δ 2 k Δ + ( 1 4 1 p ) + = ( 1 2 k T ( a 2 k ) ) 2 3 ( Δ + ( 1 4 1 p ) + ) = { ( 1 2 k T ( a 2 k ) ) 2 3 Δ , 0 < p 4 ; ( 1 2 k T ( a 2 k ) ) 2 3 ( Δ + ( 1 4 1 p ) ) , 4 < p .

Hence, from (4.20) we have

From Lemma 2.1 (2.2), we know

T 2 3 Δ + 1 2 1 p ( a 2 k ) C 1 C(λ,η) ( 2 k ) 2 ( η + λ 1 ) λ + 1 max { 2 3 Δ + 1 2 1 p , 0 } ,

and

T 2 3 Δ + 1 3 ( 1 1 p ) ( a 2 k ) C 2 C(λ,η) ( 2 k ) 2 ( η + λ 1 ) λ + 1 max { 2 3 Δ + 1 3 ( 1 1 p ) , 0 } .

Therefore, we continue with Lemma 2.1(a) as

C 20 C ( λ , η ) log ( C T ( a 2 k + 1 ) ) × { ( 1 2 k ) 2 3 Δ η p 2 ( η + λ 1 ) λ + 1 max { 2 3 Δ + 1 2 1 p , 0 } , 1 < p 4 ; ( 1 2 k ) 2 3 ( Δ + ( 1 4 1 p ) ) η p 2 ( η + λ 1 ) λ + 1 max { 2 3 Δ + 1 3 ( 1 1 p ) , 0 } , 4 < p .
(4.21)

First, let 1<p4. Then (4.8), that is,

Δ>{ 0 , 1 < p 2 ; 3 2 λ 1 3 λ 1 p 2 p , 2 < p 4

implies

Δ> 3 2 λ 1 3 λ 1 p 2 p andΔ>0

iff

2 3 Δ 2 ( λ 1 ) λ + 1 ( 2 3 Δ + 1 2 1 p ) >0andΔ>0

iff

2 3 Δ 2 ( λ 1 ) λ + 1 max { 2 3 Δ + 1 2 1 p , 0 } >0.

This means that there exists a positive constant η 1 >0 small enough such that

A( η 1 ):= 2 3 Δ η 1 p 2 ( η 1 + λ 1 ) λ + 1 max { 2 3 Δ + 1 2 1 p , 0 } >0.

Now, let p>4. Then (4.8), that is,

Δ> λ 1 3 λ 1 p 1 p 1 4 λ + 1 3 λ 1 p 4 p

implies

Δ> λ 1 3 λ 1 ( 1 1 p ) λ + 1 3 λ 1 ( 1 4 1 p ) andΔ+ 1 4 1 p >0

iff

2 3 ( Δ + ( 1 4 1 p ) ) 2 ( λ 1 ) λ + 1 ( 2 3 Δ + 1 3 ( 1 1 p ) ) >0

and

2 3 ( Δ + ( 1 4 1 p ) ) >0

iff

2 3 ( Δ + ( 1 4 1 p ) ) 2 ( λ 1 ) λ + 1 max { 2 3 Δ + 1 3 ( 1 1 p ) , 0 } >0.

Similarly to the previous case, this means that there exists a positive constant η 2 >0 small enough such that

B( η 2 ):= 2 3 ( Δ + ( 1 4 1 p ) ) η 2 p 2 ( η 2 + λ 1 ) λ + 1 max { 2 3 Δ + 1 3 ( 1 1 p ) , 0 } >0.

Now, we estimate I p , k . From (4.21), we have

For η>0 small enough, we can see A(η)>A( η 1 )>0 and B(η)>B( η 2 )>0. Let τ:=min{A( η 1 ),B( η 2 )}/2. Then for small enough η>0, we have

s n [ σ ϕ w 1 ] w Φ Δ + ( 1 4 1 p ) + L p ( I k ) C 20 C ( λ , η ) log ( C T ( a 2 k + 1 ) ) ( 1 2 k ) 2 τ C 21 C ( λ , η ) ( 1 2 k ) τ ,

because we see that for all k>0,

log ( C T ( a 2 k + 1 ) ) ( 1 2 k ) τ < C 22 .

Therefore, under the conditions (4.8) we have

s n [ σ ϕ w 1 ] w Φ Δ + ( 1 4 1 p ) + L p ( a 2 | x | a n 8 ) p k = 1 l s n [ σ ϕ w 1 ] w Φ Δ + ( 1 4 1 p ) + L p ( I k ) p C 21 C ( λ , η ) k = 1 l ( 1 2 k ) τ C 23 C ( λ , η ) .
(4.22)

The estimation of

s n [ σ ϕ w 1 ] w Φ Δ + ( 1 4 1 p ) + L p ( | x | a 2 ) p

is similar. In fact, for x[ a 2 , a 2 ], we split

H[σϕ p j w](x)= ( 2 a 2 + P . V . 2 a 2 2 a 2 + 2 a 2 ) ( σ ϕ p j w ) ( t ) x t dt.

Here we see that

| 2 a 2 ( σ ϕ p j w ) ( t ) x t d t | = | 2 a 2 ( σ ϕ p j w ) ( t ) x + t d t | | 2 a 2 ( σ ϕ p j w ) ( t ) t a 2 d t | = | 0 ( σ ϕ p j w ) ( s 2 a 2 ) s + a 2 d t |

and

| 2 a 2 ( σ ϕ p j w ) ( t ) x t d t | = | 2 a 2 ( σ ϕ p j w ) ( t ) t x d t | | 2 a 2 ( σ ϕ p j w ) ( t ) t a 2 d t | = | 0 ( σ ϕ p j w ) ( s + 2 a 2 ) s + a 2 d s | .

So, we can estimate 2 a 2 and 2 a 2 as we did I 1 before (see (4.12)). We can estimate the second integral as follows: By M. Riesz’s theorem,

P . V . 2 a 2 2 a 2 ( σ ϕ p j w ) ( t ) x t d t L p ( | t | 2 a 2 ) p C 2 a 2 2 a 2 | ( σ ϕ p j w ) ( t ) | p dtC a n p 2 C.

Now, under the assumption (4.8), we can select η 0 >0 small enough such that

Δ>{ 0 , 1 < p 2 ; 3 2 λ + η 0 1 3 λ + 2 η 0 1 p 2 p , 2 < p 4 ; max { λ + η 0 1 3 λ + 2 η 0 1 p 1 p 1 4 λ + 1 3 λ + 2 η 0 1 p 4 p , 0 } , 4 < p .

Consequently, from (4.22) with η 0 we have the result (4.9). □

Let 0<α<1, then for g n in Lemma 4.5 we estimate L n ( g n ) over [ a α n , a α n ].

Lemma 4.7 (cf. [3, Lemma 4.4])

Let 1<p< and 0<ε<1. Let { g n } be as in Lemma 4.4, but we exchange (4.3) with

| g n ( x ) w ( x ) | εϕ(x),xR,n1.

Then for 1<p<,

lim sup n L n ( g n ) w Φ Δ + ( 1 4 1 p ) + L p ( | x | a n 8 ) Cε.

Proof Let

χ n := χ [ a n 8 , a n 8 ] ; h n :=sign ( L n ( g n ) ) | L n ( g n ) | p 1 χ n w p 2 Φ ( Δ + ( 1 4 1 p ) + ) p

and

σ n :=sign s n [ h n ].

We shall show that

L n ( g n ) w Φ Δ + ( 1 4 1 p ) + L p ( | x | a n 8 ) ε s n [ σ n ϕ w 1 ] w Φ Δ + ( 1 4 1 p ) + L p ( | x | a n 8 ) .
(4.23)

Then from Lemma 4.5 we will conclude (4.22). Using orthogonality of f s n [f] to P n 1 , and the Gauss quadrature formula, we see that

Here, if we use Lemma 4.2 with ψ=ϕ, we continue as

C ε R | s n [ h n ] ( x ) | ϕ ( x ) w ( x ) d x = C ε R s n [ h n ] ( x ) σ n ϕ ( x ) w 1 ( x ) w 2 ( x ) d x = C ε R h n ( x ) s n [ σ n ϕ w 1 ] ( x ) w 2 ( x ) d x = C ε a n 8 a n 8 h n ( x ) s n [ σ n ϕ w 1 ] ( x ) w 2 ( x ) d x .

Using Hölder’s inequality with q=p/(p1), we continue this as

Cancellation of L n ( g n ) w Φ Δ + ( 1 4 1 p ) + L p ( | x | a n 8 ) p 1 gives (4.23). □

Proof of Theorem 2.2 In proving the theorem, we split our functions into pieces that vanish inside or outside [ a n 9 , a n 9 ]. Throughout, we let χ S denote the characteristic function of a set S. Also, we set for some fixed β>0,

ϕ(x)= ( 1 + x 2 ) β / 2 ,

and suppose (2.5). We note that (2.5) means (4.8). Let 0<ε<1. We can choose a polynomial P such that

( f P ) w ϕ 1 L ( R ) ε

(see Lemma 3.8). Then we have

(4.24)

Here we used that

ϕ Φ Δ + ( 1 4 1 p ) + L p ( R ) <,

because Δ>0 and Φ 1 grows faster than any power of x (see Lemma 3.9). Next, let

χ n :=χ[ a n 9 , a n 9 ],

and write

Pf=(Pf) χ n +(Pf)(1 χ n )=: g n + f n .

By Lemma 4.4 we have

lim n L n ( f n ) w Φ Δ + ( 1 4 1 4 ) + L p ( R ) =0.

By Lemma 4.5 we have

lim n L n ( g n ) w Φ Δ + ( 1 4 1 4 ) + L p ( | x | a n 8 ) =0,

and by Lemma 4.7,

lim sup n L n ( g n ) w Φ Δ + ( 1 4 1 p ) + L p ( | x | a n 8 ) Cε.

Here we take ε>0 as ε0, then with (4.24) we have the result. □

5 Proof of Theorem 2.4

Lemma 5.1 (cf. [3, Lemma 3.1])

Let wF( C 2 +). Let 0<α< 1 4 and

n (x):= | x k , n | a α n | l k , n ( x ) | w 1 ( x k , n ).

Then we have for xR,

n (x)w(x) Φ 1 / 4 (x)Clogn.

Proof From Lemma 4.1 and Lemma 3.6 with p=, we have the result easily. □

Lemma 5.2 Let wF( C 2 +). Let 0<α< 1 4 and

n (x):= | x k , n | a α n | l k , n ( x ) | w 1 ( x k , n ).

Then we have

n (x)w(x)Φ ( x ) 3 / 4 Clogn.

Proof By Lemma 3.5(c), Lemma 3.4(d) and Lemma 3.5(b),

n ( x ) = | x k , n | a α n | l k , n ( x ) | w 1 ( x k , n ) = | p n ( x ) | | x x j x , n | | P n ( x j x , n ) | w ( x j x , n ) + | x k , n | a α n , k j x | p n ( x ) | | x x k , n | | P n ( x k , n ) | w ( x k , n ) C w ( x ) 1 + a n 1 / 2 | p n ( x ) | | x k , n | a α n , k j x φ n ( x k , n ) ( 1 | x k , n | a n ) | x x k , n | C w ( x ) 1 + a n 3 / 2 n | p n ( x ) | | x k , n | a α n , k j x 1 | x k , n | a 2 n 1 | x k , n | a n ( 1 | x k , n | a n ) 1 / 4 1 | x x k , n | C w ( x ) 1 + a n 3 / 2 n | p n ( x ) | | x k , n | a α n , k j x ( 1 | x k , n | a n ) 3 / 4 1 | x x k , n | ,

where we used the fact

1 | x k , n | a 2 n 1 | x k , n | a n ,| x k , n | a α n .

So,

n ( x ) C w ( x ) 1 + a n 3 / 2 n | p n ( x ) | | x k , n | a α n , k j x ( 1 | x k , n | a n ) 3 / 4 1 | x j x , n x k , n | C w ( x ) 1 + a n 3 / 2 n | p n ( x ) | | x k , n | a α n , k j x ( 1 | x k , n | a n ) 3 / 4 1 j x i k φ n ( x i , n ) C w ( x ) 1 + a n 1 / 2 | p n ( x ) | | x k , n | a α n , k j x ( 1 | x k , n | a n ) 3 / 4 1 j x i k 1 | x i , n | / a n .

Therefore we have by Lemma 3.6 with p=,

n ( x ) w ( x ) Φ ( x ) 3 / 4 C + C a n 1 / 2 | p n ( x ) | w ( x ) Φ ( x ) 1 / 4 × | x k , n | a α n , k j x ( 1 | x k , n | a n ) 3 / 4 ( 1 | x j x , n | a n ) 1 / 2 1 j x i k 1 | x i , n | / a n C | x k , n | a α n , k j x 1 | j x k | log n .

 □

Lemma 5.3 ([8, Theorem 1])

Let wF( C 2 +). Then there exists a constant C 0 >0 such that for every absolutely continuous function f with w f C 0 (R) (this means w(x) f (x)0 as |x|) and every nN, we have

E n (w;f)C a n n E n 1 ( w ; f ) .

Proof of Theorem 2.4 There exists P n 1 P n such that

| ( f ( x ) P n 1 ( x ) ) w(x)|2 E n 1 (w;f).

Therefore, by Lemma 5.1 and Lemma 5.2,

Let w f ( r ) C 0 (R). If we repeatedly use Lemma 5.3, then we have

| ( f ( x ) L n ( f ) ( x ) ) w(x) Φ 3 / 4 (x)| C r ( a n n ) r E n r 1 ( w ; f ( r ) ) logn.

 □

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Jung, H.S., Sakai, R. Mean and uniform convergence of Lagrange interpolation with the Erdős-type weights. J Inequal Appl 2012, 237 (2012). https://doi.org/10.1186/1029-242X-2012-237

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