# Positive Interpolation Operators with Exponential-Type Weights

Authors: Hee Sun Jung; Ryozi Sakai
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We consider positive operators on the real lineℝ with property of interpolation, and we show the weighted Lp-convergence of the operators. We will construct an analogical operator of one which is studied by Knopfmacher (1986). Furthermore, we treat the Shepard-type interpolatory operator (cf. Xie et al. (1998)).


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## Body

## 1. Introduction

In this paper, we consider two interpolatory positive operators. Forγ>1 and -∞<xn,n<⋯<x1,n<∞, we construct an operator
(1)ℱn,γ[f](x)=∑k=1nh(xk,n)wγ-2(xk,n)λknf(xk,n)|Kn(x,xk,n)|γ∑k=1nh(xk,n)wγ-2(xk,n)λkn|Kn(x,xk,n)|γ.
The details will be stated later, and the result is written in Section 2. Knopfmacher [1] studied the positive operator
(2)Fn,γ[f](x)=∑k=1nλknf(xk,n)|Kn(x,xk,n)|γ∑k=1nλkn|Kn(x,xk,n)|γ,
and for 1<γ⩽2, he obtained a certain weighted-convergence theorem on the compact interval I⊂ℝ=(-∞,∞). The operators (1) and (2) have the property of Hermite-Fejér interpolation, that is,
(3)Hn[f](xk,n)=f(xk,n),Hn[f]′(xk,n)=0,k=1,2,…,n.
We also treat the interpolatory positive operator of Shepard-type. Let us define Sn,λ(f;x) for f∈C(ℝ) by
(4)Sn,λ(f;x)∶=∑j=1nf(xj,n)Φn(λ-1)/2(xj,n)|x-xj,n|-λ∑j=1nΦn(λ-1)/2(xj,n)|x-xj,n|-λ,λ⩾1,x∈ℝ.
The operator Sn,λ(f;x) is linear and positive, furthermore it interpolates f(x) at the zeros {xi,n}i=1n. In fact, we see that
(5)Sn,λ(f;xk,n)=limx≠xk,n,x→xk,n(∑j≠k‍f(xj,n)Φn(λ-1)2f(xk,n)Φn(λ-1)/2(xk,n)+∑j≠k‍f(xj,n)Φn(λ-1)/2(xj,n)×|x-xj,n|-λ|x-xk,n|λ∑j≠k‍Φn(λ-1)2)×(∑j≠k‍Φn(λ-1)2Φn(λ-1)/2(xk,n)+∑j≠k‍Φn(λ-1)/2(xj,n)×|x-xj,n|-λ|x-xk,n|λ∑j≠k‍Φn(λ-1)2)-1=f(xk,n),k=1,2,…,n.The related theorem is written in Section4.First we need the following definition from [2]. We say that f:ℝ→[0,∞) is quasi-increasing (quasi-decreasing) if there exists C>0 such that f(x)⩽Cf(y)(f(x)⩾Cf(y)), 0<x<y.Definition 1.
LetQ:ℝ→[0,∞) be an even function and satisfying the following properties.(a)
Q′(x) is continuous in ℝ, with Q(0)=0.(b)
Q′′(x) exists and is positive in ℝ∖{0}.(c)
limx→∞Q(x)=∞.(d)
The function(6)T(x)∶=xQ′(x)Q(x),x≠0
is quasi-increasing in(0,∞), with
(7)T(x)⩾Λ>1,x∈ℝ+∖{0}.
(e)
There existsC1>0 such that
(8)Q′′(x)|Q′(x)|⩽C1|Q′(x)|Q(x),a.e.x∈ℝ∖{0}.
Then, we writew(x)=exp(-Q(x))∈ℱ(C2). If there also exist a compact subinterval J(∋0) of ℝ and C2>0 such that
(9)Q′′(x)|Q′(x)|⩾C2|Q′(x)|Q(x),a.e.x∈ℝ∖J,then we write w(x)=exp(-Q(x))∈ℱ(C2+).Example 2.
There are some typical examples ofQ(x) satisfying w=exp(-Q)∈ℱ(C2+). (1)
IfT(x) is bounded, then the weight w=exp(-Q) is the so-called the Freud-type weight. Then the typical Freud-type example would be
(10)Q(x)=|x|α,α>1.(2)
IfT(x) is unbounded, then the weight w=exp(-Q) is called the Erdös-type weight. Erdös-type examples w=exp(-Q)∈ℱ(C2+) are as follows.
(a)
(see [2, Example 1.2], [3, Theorem 3.1]) For α>1,l=1,2,3,….
(11)Q(x)=Ql,α(x)=expl(|x|α)-expl(0),where(12)expl(x)=exp(exp(exp⋯expx)⋯)(l-times).More precisely, we define forα+m>1, m⩾0, l⩾1 and α⩾0,
(13)Ql,α,m(x)∶=|x|m(expl(|x|α)-α*expl(0)),
where α*=0 if α=0, otherwise α*=1 (but, note that Ql,0,m gives a Freud-type weight).
(b)
(see [3, Theorem 3.5]) For α>1, put Qα(x):=(1+|x|)|x|α-1,α>1.We construct the orthonormal polynomialspn(x)=pn(w2,x) of degree n for w2(x), that is,
(14)∫-∞∞‍pn(w2,x)pm(w2,x)w2(x)dx=δmn(Kroneckerdelta).
Let fw∈Lp(ℝ). The Fourier-type series of f is defined by
(15)f~(x)∶=∑k=0∞‍ak(w2,f)pk(w2,x),ak(w2,f)∶=∫-∞∞‍f(t)pk(w2,t)w2(t)dt.
We denote the partial sum of f~(x) by
(16)sn(f,x)∶=sn(w2,f,x)∶=∑k=0n-1‍ak(w2,f)pk(w2,x).
If we use the Christoffel-Darboux formula, then we obtain
(17)sn(f,x)=∫-∞∞‍Kn(x,t)f(t)w2(t)dt.
Here,
(18)Kn(x,t)∶=∑k=0n-1‍pk(x)pk(t)=γn-1γnpn(x)pn-1(t)-pn(t)pn-1(x)x-t,
where pn(x)=γnxn+⋯. The polynomials of degree ⩽n are denoted by 𝒫n. We define the Christoffel numbers λn(w;x) by
(19)λn(w;x)∶=infP∈𝒫n-1∫-∞∞‍|Pw|2(t)dt|P(x)|2,
then we have
(20)λn(w;x)=1Kn(x,x)=1∑k=0n-1pk2(w2,x).
We denote the zeros of the orthonormal polynomial pn(w2,x) by xn,n<xn-1,n<⋯<x1,n. Then we define the Christoffel numbers λk,n,k=1,2,…,n such as λk,n:=λn(w,xk,n).


## 2. Preliminaries and Theorems

We need the Mhaskar-Rakhmanov-Saff numberax;
(21)x=2π∫01‍axtQ′(axt)(1-t2)1/2dt,x>0.
We define
(22)φu(x)={auu1-(|x|/a2u)1-(|x|/au)+δu,|x|⩽au;φu(au),au<|x|,(23)Φn(x)={1-|x|an+δn,|x|⩽an;δn,an<|x|,
where
(24)δu={uT(au)}-2/3u>0.
Moreover, we define a function ψn(x) for γ>1 and x∈ℝ(25)ψn(x)={an2-γφnγ-1(x),1<γ<2;φn(x)logan,γ=2;φn(x),γ>2,(26)ψn*:={an2-γϕnγ-1,1<γ<2;ϕnlogn,γ=2;ϕn,2<γ,
where ϕn:=max{an/n,anδn}. For the Freud-type weight w we suppose to hold ψn*→0 as n→∞. If w∈ℱ(C2+) is the Erdös-type weight, then it always holds. So for the Freud-type weight we need to limit slightly the weights.To state our main result, we assume some conditions forh(x) as follows.(1)
h(x) is even, positive, and quasi-decreasing on [0,∞).(2)
h(xk,n)~h(xk+1,n) for k=1,2,…,n and n=1,2,…(3)
h(x)Φn-γ/4(x) is bounded on ℝ for n=1,2,….Let{xj,n}j=1n be the zeros of the orthonormal polynomial pn(w2,x). Then we define the operator ℱn,γ[f](x) by (1) with γ>1, h(x), and for each f∈C(ℝ) we define a pointwise modulus of continuity ωx(f;t)=sup{y;|x-y|⩽t,y∈ℝ}|f(x)-f(y)|. When f∈C(ℝ) is uniformly continuous on ℝ, we set
(27)Ω(f;t)=supx∈ℝωx(f;t).Then our first theorem is as follows.Theorem 3.
Letw∈ℱ(C2+), and let ψn*→0 as n→∞. Let γ>1. Then we have the following.(a)
Forxn,n⩽x⩽x1,n,
(28)|ℱn,γ[f](x)-f(x)|⩽Cωx(f;ψn(x))h-1(x)Φn-γ/4(x),
and for |x|>x1,n(29)|ℱn,γ[f](x)-f(x)|⩽Cωx(f;ψn(x))|x|ψn-1(x).(b)
Let0<p⩽∞ and w* be an integrable function satisfying the following condition:
(30)∥w*(x)h-1(x)Φn-γ/4(x)∥Lp([xn,n,x1,n])+∥xψn-1(x)w*(x)∥Lp(|x|⩾x1,n)<∞.Then one has forf(x) being uniformly continuous and bounded on ℝ(31)∥w*{ℱn,γ[f]-f}∥Lp(ℝ)=O(1)Ω(f;ψn*),whereψn* are defined in (26).We prepare some lemmas for the proof of the theorem.Lemma 4.
Letw=exp(-Q)∈ℱ(C2).(1)
(see [2, Lemma 3.5 (3.27)–(3.29)]) For fixed L>0 and uniformly for t>0,
(32)aLt~at,T(aLt)~T(at),Q(j)(aLt)~Q(j)(at),j=0,1.Moreover,(33)T(aLt)~T(at).(2)
(see [2, Lemma 3.4 (3.18),(3.17), Lemma 3.8 (3.42)])
(34)Q(at)~tT(at),Q′(at)~tT(at)at,and forx∈[0,an/2],
(35)Q′(x)~nan(xan)Λ-1,whereΛ>1 is defined in Definition 1(d).
(3)
(see [2, Lemma 3.11 (a), (b)]) Given fixed 0<α, one has uniformly for t>0,
(36)|1-aαtat|~1T(at).
(4)
(see [2, Lemma 3.7(3.38)]) For some 0<ε⩽2, and for large enough t,
(37)T(at)⩽t2-ε.
(5)
(see [2, Lemma 3.8 (a)]) For x∈[0,at),
(38)Q′(x)⩽Ctat11-(x/at).Lemma 5 ([4, Theorem 2.7]).
There existsC>0 such that
(39)supx∈ℝ|pn(x)w(x)Φn1/4(x)|⩽Can-1/2.Lemma 6.
Letw(x)=exp(-Q(x))∈ℱ(C2+). (1)
Letxj,n be the zero of pn(x). Then for n⩾1 and 1⩽j⩽n-1,
(40)xj,n-xj+1,n~φn(xj,n),(41)φn(xj,n)~φn(xj+1,n).(2)
Forn⩾1 and 1⩽j⩽n-1,
(42)Φn(xj,n)~Φn(xj+1,n).Proof.
(1) This follows from [2, Corollary 13.4, Theorem 5.7 (b)].
(2) Recall the definition ofΦn(x) in (21). We have
(43)Φn(xj,n)=1-|xj,n|an+δn=1-|xj+1,n|an+δn-xj,n-xj+1,nan~1-|xj+1,n|an+δn-φn(xj,n)an=1-|xj+1,n|an+δn-1n1-|xj,n/a2n|1-|xj,n|/an.
Hence, if |xk,n|,|xk+1,n|⩽an/2, then we see
(44)Φn(xj,n)~1-|xj+1,n|an+δn-1n1-|xj,n|an+δn.
Here we see
(45)Φn(xj,n)+CnΦn(xj,n)=Φn(xj,n){Φn(xj,n)+Cn}~Φn(xj,n),
because of Φn(xj,n)⩾C/T(an)>1/n1-ε>(1/n)(ε>0). (see Lemma 4 (3), (4)). Therefore, we have
(46)Φn(xj,n)~Φn(xj+1,n).
Let an/2<|xj,n|. Then we see
(47)1n1-|xj,n/a2n|1-|xj,n|/an+δn⩽C1n(nT(an))1/3T(an)~δn.
Therefore we see
(48)Φn(xj,n)~1-|xj+1,n|an+δn=Φn(xj+1,n).Lemma 7 ([2, Theorem 13.3(13.9)]).
Ifx∈[xk+1,n,xk,n], then
(49)(lk,nw)(x)w-1(xk,n)+(lk+1,nw)(x)w-1(xk+,n)~1.Lemma 8.
Letw=exp(-Q)∈ℱ(C2+). Then the following results hold.
(a) For |x|⩽an(1+δn),
(50)Kn(x,x)=∑k=0n-1‍pk2(w2,x)=λn,2-1(w;x)~φn-1(x)w-2(x).
(b) For x∈ℝ(51)Kn(x,x)⩽Cφn-1(x)w-2(x).Proof.
From [2, Theorem 9.3], we have the following.
(1) Uniformly forn⩾1 and |x|⩽an(1+ηn), we have
(52)λn(w;x)~φn(x)w2(x).
(2) Moreover, uniformly forn⩾1 and x∈ℝ,
(53)λn(w;x)⩾Cφn(x)w2(x).
Since Kn(x,x)=1/λn(w;x), we have the following results.


## 3. Proof of Theorem3

To estimate the difference|ℱn,γ[f](x)-f(x)|, we split ∑k=1n‍ into two parts.To prove the theorem we start the estimation of the denominator for the operatorℱn,γ. We will need it in Step 4.Step 1.
Let(54)Hn,γ(x)∶=∑k=1n‍h(xk,n)wγ-2(xk,n)λkn|Kn(x,xk,n)|γ.
Then we have the following.Lemma 9.
There existsC>0 such that uniformly, for x∈[xn,n,x1,n],
(55)Hn,γ(x)⩾Ch(x)φn1-γ(x)w-γ(x).Proof.
By Lemma7, if x∈[xk+1,n,xk,n], xk+1,n⩾0, k⩾1, then
(56)(lk,nw)(x)w-1(xk,n)+(lk+1,nw)(x)w-1(xk+1,n)~1.
Since lk,n(x)=λk,nKn(x,xk,n), we see
(57)λk,nw(x)w(xk,n)Kn(x,xk,n)+λk+1,nw(x)w(xk+1,n)Kn(x,xk+1,n)~1,
and this implies that
(58)0<C⩽(λknw(x)w(xk,n)|Kn(x,xk,n)|+λk+1,nw(x)w(xk+1,n)|Kn(x,xk+1,n)|)=(λknw(x)w2(xk,n)w(xk,n)|Kn(x,xk,n)|+λk+1,nw(x)w2(xk+1,n)w(xk+1,n)|Kn(x,xk+1,n)|)~λknw(x)w2(xk,n)(w(xk,n)|Kn(x,xk,n)|+w(xk+1,n)|Kn(x,xk+1,n)|)⩽Cλknw(x)w2(xk,n)(|Kn(x,xk,n)w(xk,n)|γ+|Kn(x,xk+1,n)w(xk+1,n)|γ)1/γ.
Therefore, from (41) and (52) we can obtain
(59)φn1-γ(x)w-γ(x)~(λknw2(xk,n))1-γw-γ(x)⩽Cλknw2(xk,n)(|Kn(x,xk,n)w(xk,n)|γ+|Kn(x,xk+1,n)w(xk+1,n)|γ).
Using the fact h(xk,n)~h(x)~h(xk+1,n) (see the definition of h(x)), we have by (41) and (52)
(60)Hn,γ(x)⩾h(xk,n)λk,nwγ-2(xk,n)|Kn(x,xk,n)|γ+h(xk+1,n)λk+1,nwγ-2(xk+1,n)|Kn(x,xk+1,n)|γ⩾Ch(x)λk,nw2(xk,n)(|Kn(x,xk,n)w(xk,n)|γ+|Kn(x,xk+1,n)w(xk+1,n)|γ)⩾Ch(x)φn1-γ(x)w-γ(x).
In another case, that is, when xk+1,n<0, we also have the same result.Step 2.
Let|x-xk,n|⩽φn(x). Let f(x) be uniformly continuous and bounded on ℝ, and let γ>1. Then we have
(61)|f(x)-f(xk,n)|⩽ωx(f;φn(x)).
Now, let
(62)∑1‍∶=1Hn,γ(x)∑|x-xk,n|⩽φn(x)‍h(xk,n)wγ-2(xk,n)λkn×|f(x)-f(xk,n)||Kn(x,xk,n)|γ.
We have the following estimation.Lemma 10.
Forx∈ℝ,
(63)∑1‍⩽ωx(f;φn(x)).Proof.
By (61),
(64)∑1‍⩽ωx(f;φn(x))1Hn,γ(x)×∑|x-xk,n|⩽φn(x)‍h(xk,n)wγ-2(xk,n)λkn|Kn(x,xk,n)|γ⩽ωx(f;φn(x)),
because we know from the definition of Hn,γ(x) in (54) that
(65)∑|x-xk,n|⩽φn(x)‍h(xk,n)wγ-2(xk,n)λkn|Kn(x,xk,n)|γ⩽Hn,γ(x).Step 3.
Next, we estimate∑|x-xk,n|>φn(x)‍. Let |x-xk,n|>φn(x) and let |x-xm,n|=min{|x-xk,n|,k=1,2,…,n}. To do so, we prepare the following. By Lemma 6,
(66)|Kn(x,xk,n)|⩽Can|pn(x)pn-1(xk,n)||x-xk,n|⩽Cw-1(x)w-1(xk,n)Φn-1/4(x)×Φn-1/4(xk,n)1|x-xk,n|.
From the property of the modulus of continuity we have, for|x-xk,n|>φn(xk,n),
(67)|f(x)-f(xk,n)|⩽C(|x-xk,n|ψn-1(x)+1)ωx(f;ψn(x)),
where ψn(x) is defined in (25) as ψn(x)→0 uniformly in ℝ as n→∞.We have the following estimate.Lemma 11.
For anyx∈ℝ,
(68)Bn,k(x)∶=1Hn,γ(x)h(xk,n)wγ-2(xk,n)λk,n|Kn(x,xk,n)|γ.
Then
(69)∑|x-xk,n|>φn(x)‍Bn,k(x)⩽1.Proof.
(70)
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(71)∑|x-xk,n|>φn(x)‍h(xk,n)wγ-2(xk,n)λkn|Kn(x,xk,n)|γ⩽Hn,γ(x).Step 4.
Let|x-xk,n|>φn(x). Using the result of Step 1, we have the following estimate.Lemma 12.
For anyx∈ℝ, one sets
(72)Cn,k(x)∶=1Hn,γ(x)h(xk,n)wγ-2(xk,n)λk,n×|x-xk,n||Kn(x,xk,n)|γ.
Then for x∈[xn,n,x1,n],
(73)∑|x-xk,n|>φn(x)‍Cn,k(x)⩽Ch-1(x)Φn-γ/4(x)ψn(x),
and for |x|>x1,n,
(74)∑|x-xk,n|>φn(x)‍Cn,k(x)⩽2|x|.Proof.
First, letx∈[xn,n,x1,n]. Then using (52), (66), and Lemma 9, we have
(75)Cn,k(x)⩽Ch(xk,n)φn(xk,n)Φn-γ/4×(x)Φn-γ/4(xk,n)1|x-xk,n|γ-1×h-1(x)φnγ-1(x).
From the fact that h(x)Φn-γ/4(x) is bounded (recall the definition of h(x)), we can continue as
(76)Cn,k(x)⩽Ch-1(x)Φn-γ/4(x)φnγ-1(x)φn(xk,n)|x-xk,n|γ-1.
Then by (25) and (40),
(77)∑|x-xk,n|>φn(x)‍Cn,k(x)⩽C∑|x-xk,n|>φn(x)‍φn(xk,n)|x-xk,n|γ-1h-1(x)Φn-γ/4(x)φnγ-1(x)⩽Ch-1(x)Φn-γ/4(x)φnγ-1(x)×{an2-γ,1<γ<2;logan,γ=2;φn2-γ(x),γ>2,⩽Ch-1(x)Φn-γ/4(x)ψn(x).
Next, suppose x>x1,n. Then since
(78)Cn,k(x)⩽2|x|1Hn,γ(x)h(xk,n)wγ-2(xk,n)λk,n×|Kn(x,xk,n)|γ,
we have from Lemma 11,
(79)∑|x-xk,n|>φn(x)‍Cn,k(x)⩽2|x|.Step 5.
Using (67) and Lemmas 11 and 12, we can estimate the part ∑|x-xk,n|>φn(x)‍ as follows:
(80)∑2‍∶=(∑|x-xk,n|>φn(x)‍h(xk,n)wγ-2(xk,n)λkn×|f(x)-f(xk,n)||Kn(x,xk,n)|γ∑|x-xk,n|>φn(x)‍)×(∑k=1n‍h(xk,n)wγ-2(xk,n)λkn|Kn(x,xk,n)|γ)-1.
Then forx∈[xn,n,x1,n](81)∑2‍⩽Cωx(f;ψn(x))×(∑|x-xk,n|>φn(x)‍Bn,k(x)+ψn-1(x)×∑|x-xk,n|>φn(x)‍Cn,k(x))⩽Cωx(f;ψn(x))(1+h-1(x)Φn-γ/4(x))⩽Cωx(f;ψn(x))h-1(x)Φn-γ/4(x),
and for |x|⩾x1,n,
(82)∑2‍⩽Cωx(f;ψn(x))×(∑|x-xk,n|>φn(x)‍Bn,k(x)+ψn-1(x)×∑|x-xk,n|>φn(x)‍Cn,k(x))⩽Cωx(f;ψn(x))(1+|x|ψn-1(x))⩽Cωx(f;ψn(x))|x|ψn-1(x).
Therefore, with Lemma 10 we have the following result.Lemma 13.
Forxn,n⩽x⩽x1,n,
(83)|ℱn,γ[f](x)-f(x)|⩽Cωx(f;ψn(x))h-1(x)Φn-γ/4(x)
and for |x|>x1,n(84)|ℱn,γ[f](x)-f(x)|⩽Cωx(f;ψn(x))|x|ψn-1(x).Proof of Theorem3.
(a) follows from Lemma13. We will show (b). Let 0<p⩽∞. Then since we know that φ(x)⩽Cϕn and so ψn(x)⩽Cψn* for all x∈ℝ, we have
(85)∥w*(ℱn,γ[f]-f)∥Lp(ℝ)=O(1)∥w*(∑1‍+∑2‍)∥Lp(ℝ)=O(1)Ω(f;ψn*).Example 14.
Leth(x)=Φγ/4(x) and
(86)w*(x)=Φγ/2(x)(1+|x|)β+1,βp>1,
where
(87)Φ(x)∶=1(1+Q(x))2/3T(x).
Then the condition (30) is satisfied.


## 4. Shepard-Type Operator

Let us define the positive interpolatory operator (4) for f∈C(ℝ) and the zeros {xj,n}j=1n of the orthonormal polynomial pn(w2,x).Let(88)Φ(x)∶=1(1+Q(x))2/3T(x).Lemma 15 ([5, Lemma 3.3]).
Forx∈ℝ, one has
(89)Φ(x)⩽CΦn(x),n⩾1.Assumption 1.
We suppose that, for eachε>0,
(90)T(an)⩽C(ε)nε,n=1,2,3,…,
where C(ε) is a constant depending only on ε.Remark 16.
Letw=exp(-Q)∈ℱ(C2+), and let us define
(91)ν∶=limsupx→∞Q′′(x)/Q′(x)Q′(x)/Q(x),μ∶=liminfx→∞Q′′(x)/Q′(x)Q′(x)/Q(x).
If ν=μ, then we say that the weight w is regular. The regular weights satisfy the condition (90) (see [6, Corollry 5.5]). All weights in Example 2 are regular weights.Lemma 17 ([3, Theorem 1.6]).
Letw=exp(-Q)∈ℱ(C2+), and let an be defined by (21). Then there exists C>0 such that for every n>0(92)an⩽Cn1/Λ,
where Λ>1 is defined in Definition 1 (d). In particular, for the weight wα one has Λ=α. Furthermore, if w is an Erdös-type, then for any η>0, there exists C(η)>0 such that, for every n>0,
(93)an⩽C(η)nη.For each3/2<λ<3 let us set
(94)μn={anTλ/3(an)n1-λ/3,2<λ<3;anTλ/3(an)lognn1-λ/3,λ=2;anTλ/3(an)n(2λ-3)/3,32<λ<2.Our second theorem is as follows.Theorem 18.
Letf∈C(ℝ) be uniformly continuous on ℝ and let 3/2<λ<3. Assume U(x) is a nonnegative and decreasing function with U(x)⩽CΦ(λ-1)/2(x). Then one has for the Erdös-type weights,
(95)∥U(x)(Sn,λ(f;x)-f(x))∥L∞(ℝ)⩽CΩ(f;μn),
where μn is defined in (94).For the Freud weights we have the following. ForΛ>3, let us set (3/2)(1+(1/Λ))<λ<3(1-(1/Λ)) and
(96)μn,Λ={1n1-λ/3-1/Λ,2<λ<3(1-1Λ);1n1/3-1/Λ,λ=2;1n(2λ-3)/3-1/Λ,32(1+1Λ)<λ<2
(note (92) and (94)).Corollary 19.
LetΛ>3, where Λ is defined in Definition 1 (d), and let (3/2)(1+(1/Λ))<λ<3(1-(1/Λ)). Then, for the Freud-type weights, (95) holds with μn,Λ. In particular, when w(x)=exp(-|x|α), one can take Λ=α.Remark 20.
For the Freud-type weights we seelimn→∞μn,Λ=0. If we assume (90), then for the Erdös-type weights, from Lemma 17 (93), we also have limn→∞μn=0.Proof of Theorem18.
Let3/2<λ<3. We see that
(97)Sn,λ(f;x)-f(x)=∑j=1n{f(xj,n)-f(x)}Φn(λ-1)/2(xj,n)|x-xj,n|-λ∑j=1nΦn(λ-1)/2(xj,n)|x-xj,n|-λ.
Let (xm+1,n+xm,n)/2<x⩽xm,n or (xm,n+xm-1,n)/2<x⩽xm,n. Then, we see
(98)|f(xm,n)-f(x)|⩽ωx(f;|xm,n-x|)⩽Cωx(f;φn(x))⩽Cω(f;μn),
where μn is defined in (94). If j≠m, then we have
(99)|f(xj,n)-f(x)|⩽ωx(f;|x-xj,n|)⩽(|x-xj,n|μn-1+1)Ω(f;μn).
Let
(100)∑1‍∶=∑j≠mΦn(λ-1)/2(xj,n)|x-xj,n|-λ∑j=1nΦnλ-1/2(xj,n)|x-xj,n|-λ,∑2‍∶=∑j≠mΦn(λ-1)/2(xj,n)|x-xj,n|-(λ-1)∑j=1nΦnλ-1/2(xj,n)|x-xj,n|-λ.
Then we see that 0<∑1‍⩽1. Now, we will estimate ∑2‍. We see that
(101)1|x-xj,n|~(∑j⩽s⩽m+1orm-1⩽s⩽j‍φn(xs,n))-1~nan(∑j⩽s⩽m+1orm-1⩽s⩽j‍1-|xs,n|/a2n(1-|xs,n|/an+δn)1/2)-1⩾nan(nT(an))-1/3(∑j⩽s⩽m+1orm-1⩽s⩽j‍(1-|xs,n|/a2n))-1⩾n2/3anT1/3(an)1|m-j|.
Hence we have
(102)∑j=1n‍Φn(λ-1)/2(xj,n)|x-xj,n|-λ⩾(n2/3anT1/3(an))λ∑j≠m‍Φn(λ-1)/2(xj,n)1|m-j|λ⩾(n2/3anT1/3(an))λ∑|xj,n|⩽an/2j≠m‍Φn(λ-1)/2(xj,n)1|m-j|λ⩾C(n2/3anT1/3(an))λ∑|xj,n|⩽an/2j≠m‍1|m-j|λ⩾C(n2/3anT1/3(an))λ{1,λ>1;logn,λ=1.
Using for 1⩽j⩽n(103)1-|xj,n|a2n⩾C(1-|xj,n|an+δn),
we see that
(104)1|x-xj,n|~(∑j⩽s⩽m+1orm-1⩽s⩽j‍φn(xs,n))-1~nan(∑j⩽s⩽m+1orm-1⩽s⩽j‍1-|xs,n|/a2n(1-|xs,n|/an+δn)1/2)-1⩽Cnan(∑j⩽s⩽m+1orm-1⩽s⩽j‍(1-|xs,n|/an+δn)1/2)-1⩽Cnan(Φn-1/2(x)+Φn-1/2(xj,n))1|m-j|.
Therefore, we have
(105)∑j≠m‍U(x)Φn(λ-1)/2(xj,n)|x-xj,n|-(λ-1)⩽C(nan)λ-1∑j≠m‍(Φn(λ-1)/2U(x)Φn-(λ-1)/2(x)×Φn(λ-1)/2(xj,n)+U(x))×1|m-j|λ-1⩽C(nan)λ-1∑j≠m‍1|m-j|λ-1⩽C(nan)λ-1{1,2<λ;logn,λ=2;n2-λ,1⩽λ<2.
Then, with (102) we see
(106)|U(x)∑2‍(x)|=|∑j≠mU(x)Φn(λ-1)/2(xj,n)|x-xj,n|-(λ-1)∑j=1nΦn(λ-1)/2(xj,n)|x-xj,n|-λ|⩽C(anT1/3(an)n2/3)λ(nan)λ-1×{1,2<λ;logn,λ=2;n2-λ,1<λ<2;n2-λlogn,λ=1,⩽CanTλ/3(an)n1-λ/3{1,2<λ;logn,λ=2;n2-λ,1<λ<2;n2-λlogn,λ=1.
Hence, using μn in (94), we have that, for 3/2<λ<3,
(107)|U(x)∑2‍(x)|⩽Cμn.
Consequently, with 0<∑1‍⩽1 we have
(108)U(x)|Sn,λ(f;x)-f(x)|⩽CΩ(f;μn).



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*Source: 421328-2013-04-11.xml*