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

Journal of Inequalities and Applications20122012:237

https://doi.org/10.1186/1029-242X-2012-237

• Received: 10 April 2012
• Accepted: 2 October 2012
• Published:

## Abstract

Let $\mathbb{R}=\left(-\mathrm{\infty },\mathrm{\infty }\right)$, and let $Q\in {C}^{1}\left(\mathbb{R}\right):\mathbb{R}\to {\mathbb{R}}^{+}:=\left[0,\mathrm{\infty }\right)$ be an even function. We consider the exponential-type weights $w\left(x\right)={e}^{-Q\left(x\right)}$, $x\in \mathbb{R}$. In this paper, we obtain a mean and uniform convergence theorem for the Lagrange interpolation polynomials ${L}_{n}\left(f\right)$ in ${L}_{p}$, $1 with the weight w.

MSC:41A05.

## Keywords

• exponential-type weight
• Lagrange interpolation polynomial

## 1 Introduction and preliminaries

Let $\mathbb{R}=\left(-\mathrm{\infty },\mathrm{\infty }\right)$, and let $Q\in {C}^{1}\left(\mathbb{R}\right):\mathbb{R}\to {\mathbb{R}}^{+}:=\left[0,\mathrm{\infty }\right)$ be an even function, and $w\left(x\right)=exp\left(-Q\left(x\right)\right)$ be the weight such that ${\int }_{0}^{\mathrm{\infty }}{x}^{n}{w}^{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx<\mathrm{\infty }$ for all $n=0,1,2,\dots$ . Then we can construct the orthonormal polynomials ${p}_{n}\left(x\right)={p}_{n}\left({w}^{2};x\right)$ of degree n with respect to ${w}^{2}\left(x\right)$. That is,
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{p}_{n}\left(x\right){p}_{m}\left(x\right){w}^{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx={\delta }_{mn}\phantom{\rule{1em}{0ex}}\left(\text{Kronecker’s delta}\right)$
and
${p}_{n}\left(x\right)={\gamma }_{n}{x}^{n}+\cdots ,\phantom{\rule{1em}{0ex}}{\gamma }_{n}>0.$
We denote the zeros of ${p}_{n}\left(x\right)$ by
$-\mathrm{\infty }<{x}_{n,n}<{x}_{n-1,n}<\cdots <{x}_{2,n}<{x}_{1,n}<\mathrm{\infty }.$
We denote the Lagrange interpolation polynomial ${L}_{n}\left(f;x\right)$ based at the zeros ${\left\{{x}_{k,n}\right\}}_{k=1}^{n}$ as follows:
${L}_{n}\left(f;x\right):=\sum _{k=1}^{n}f\left({x}_{k,n}\right){l}_{k,n}\left(x\right),\phantom{\rule{1em}{0ex}}{l}_{k,n}\left(x\right):=\frac{{p}_{n}\left(x\right)}{\left(x-{x}_{k,n}\right){p}_{n}^{\prime }\left({x}_{k,n}\right)}.$

A function $f:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is said to be quasi-increasing if there exists $C>0$ such that $f\left(x\right)⩽Cf\left(y\right)$ for $0.

We are interested in the following subclass of weights from .

Definition 1.1 Let $Q:\mathbb{R}\to {\mathbb{R}}^{+}$ be an even function satisfying the following properties:
1. (a)

${Q}^{\prime }\left(x\right)$ is continuous in , with $Q\left(0\right)=0$.

2. (b)

${Q}^{″}\left(x\right)$ exists and is positive in $\mathbb{R}\mathrm{\setminus }\left\{0\right\}$.

3. (c)

${lim}_{x\to \mathrm{\infty }}Q\left(x\right)=\mathrm{\infty }$.

4. (d)
The function
$T\left(x\right):=\frac{x{Q}^{\prime }\left(x\right)}{Q\left(x\right)},\phantom{\rule{1em}{0ex}}x\ne 0$

is quasi-increasing in $\left(0,\mathrm{\infty }\right)$ with
$T\left(x\right)⩾\mathrm{\Lambda }>1,\phantom{\rule{1em}{0ex}}x\in {\mathbb{R}}^{+}\mathrm{\setminus }\left\{0\right\}.$
1. (e)
There exists ${C}_{1}>0$ such that

Then we write $w\left(x\right)=exp\left(-Q\left(x\right)\right)\in \mathcal{F}\left({C}^{2}\right)$. If there also exist a compact subinterval J (0) of and ${C}_{2}>0$ such that

then we write $w\left(x\right)=exp\left(-Q\left(x\right)\right)\in \mathcal{F}\left({C}^{2}+\right)$.

Example 1.2 (1) If $T\left(x\right)$ is bounded, then the weight $w=exp\left(-Q\right)$ is called the Freud-type weight. The following example is the Freud-type weight:
$Q\left(x\right)={|x|}^{\alpha },\phantom{\rule{1em}{0ex}}\alpha >1.$
If $T\left(x\right)$ is unbounded, then the weight $w=exp\left(-Q\right)$ is called the Erdős-type weight. The following examples give the Erdős-type weights $w=exp\left(-Q\right)$.
1. (2)
[2, Theorem 3.1] For $\alpha >1$, $l=1,2,3,\dots$
$Q\left(x\right)={Q}_{l,\alpha }\left(x\right)={exp}_{l}\left({|x|}^{\alpha }\right)-{exp}_{l}\left(0\right),$

where
${exp}_{l}\left(x\right)=exp\left(exp\left(exp\cdots expx\right)\cdots \right)\phantom{\rule{1em}{0ex}}\left(l\text{-times}\right).$
More generally, we define for $\alpha +u>1$, $\alpha ⩾0$, $u⩾0$ and $l⩾1$,
${Q}_{l,\alpha ,u}\left(x\right):={|x|}^{u}\left({exp}_{l}\left({|x|}^{\alpha }\right)-{\alpha }^{\ast }{exp}_{l}\left(0\right)\right),$
where ${\alpha }^{\ast }=0$ if $\alpha =0$, otherwise ${\alpha }^{\ast }=1$. (We note that ${Q}_{l,0,u}\left(x\right)$ gives a Freud-type weight.)
1. (3)

We define ${Q}_{\alpha }\left(x\right):={\left(1+|x|\right)}^{{|x|}^{\alpha }}-1$, $\alpha >1$.

In this paper, we investigate the convergence of the Lagrange interpolation polynomials with respect to the weight $w\in \mathcal{F}\left({C}^{2}+\right)$. When we consider the Erdős-type weights, the following definition follows from Damelin and Lubinsky .

Definition 1.3 Let $w\left(x\right)=exp\left(-Q\left(x\right)\right)$, where $Q:\mathbb{R}\to \mathbb{R}$ is even and continuous. ${Q}^{″}$ exists in $\left(0,\mathrm{\infty }\right)$, ${Q}^{\left(j\right)}⩾0$, in $\left(0,\mathrm{\infty }\right)$, $j=0,1,2$, and the function
${T}^{\ast }\left(x\right):=1+\frac{x{Q}^{″}\left(x\right)}{{Q}^{\prime }\left(x\right)}$
is increasing in $\left(0,\mathrm{\infty }\right)$ with
$\underset{x\to \mathrm{\infty }}{lim}{T}^{\ast }\left(x\right)=\mathrm{\infty };\phantom{\rule{2em}{0ex}}{T}^{\ast }\left(0+\right):=\underset{x\to 0+}{lim}{T}^{\ast }\left(x\right)>1.$
(1.1)
Moreover, we assume that for some constants ${C}_{1},{C}_{2},{C}_{3}>0$,
${C}_{1}⩽{T}^{\ast }\left(x\right)/\left(\frac{x{Q}^{\prime }\left(x\right)}{Q\left(x\right)}\right)⩽{C}_{2},\phantom{\rule{1em}{0ex}}x⩾{C}_{3},$
and for every $\epsilon >0$,
${T}^{\ast }\left(x\right)=O\left(Q{\left(x\right)}^{\epsilon }\right),\phantom{\rule{1em}{0ex}}x\to \mathrm{\infty }.$
(1.2)

Then we write $w\in \mathcal{E}$.

Damelin and Lubinsky  got the following results with the Erdős-type weights $w=exp\left(-Q\right)\in \mathcal{E}$.

Theorem A ([3, Theorem 1.3])

Let $w=exp\left(-Q\right)\in \mathcal{E}$. Let ${L}_{n}\left(f,x\right)$ denote the Lagrange interpolation polynomial to f at the zeros of ${p}_{n}\left({w}^{2},x\right)$. Let $1, $\mathrm{\Delta }\in \mathbb{R}$, $\kappa >0$. Then for
$\underset{n\to \mathrm{\infty }}{lim}{\parallel \left(f-{L}_{n}\left(f\right)\right)w{\left(1+Q\right)}^{-\mathrm{\Delta }}\parallel }_{{L}_{p}\left(\mathbb{R}\right)}=0$
to hold for every continuous function $f:\mathbb{R}\to \mathbb{R}$ satisfying
$\underset{|x|\to \mathrm{\infty }}{lim}|f\left(x\right)w\left(x\right){\left(log|x|\right)}^{1+\kappa }|=0,$
it is necessary and sufficient that
$\mathrm{\Delta }>max\left\{0,\frac{2}{3}\left(\frac{1}{4}-\frac{1}{p}\right)\right\}.$

Our main purpose in this paper is to give mean and uniform convergence theorems with respect to $\left\{{L}_{n}\left(f\right)\right\}$, $n=1,2,\dots$ , in ${L}_{p}$-norm, $1. The proof for $1 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. Finally, we will prove the theorem for the uniform convergence in Section 5.

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

Throughout $C,{C}_{1},{C}_{2},\dots$ 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:
$x=\frac{2}{\pi }{\int }_{0}^{1}\frac{{a}_{x}u{Q}^{\prime }\left({a}_{x}u\right)}{{\left(1-{u}^{2}\right)}^{\frac{1}{2}}}\phantom{\rule{0.2em}{0ex}}du,\phantom{\rule{1em}{0ex}}x>0.$
The function ${\phi }_{u}\left(x\right)$ is defined as follows:
${\phi }_{u}\left(x\right)=\left\{\begin{array}{cc}\frac{{a}_{u}}{u}\frac{1-\frac{|x|}{{a}_{2u}}}{\sqrt{1-\frac{|x|}{{a}_{u}}}+{\delta }_{u}},\hfill & |x|⩽{a}_{u},\hfill \\ {\phi }_{u}\left({a}_{u}\right),\hfill & {a}_{u}<|x|,\hfill \end{array}$
where
${\delta }_{x}={\left(xT\left({a}_{x}\right)\right)}^{-\frac{2}{3}},\phantom{\rule{1em}{0ex}}x>0.$
We define
$\mathrm{\Phi }\left(x\right):=\frac{1}{{\left(1+Q\left(x\right)\right)}^{\frac{2}{3}}T\left(x\right)}$
and
${\mathrm{\Phi }}_{n}\left(x\right):=max\left\{{\delta }_{n},1-\frac{|x|}{{a}_{n}}\right\}.$
Here we note that for $0,
$\mathrm{\Phi }\left(x\right)\sim \frac{Q{\left(x\right)}^{\frac{1}{3}}}{x{Q}^{\prime }\left(x\right)}$
and we see
$\mathrm{\Phi }\left(x\right)⩽C{\mathrm{\Phi }}_{n}\left(x\right),\phantom{\rule{1em}{0ex}}n⩾1$
(see Lemma 3.3 below). Moreover, we define
${\mathrm{\Phi }}^{{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\left(x\right):=\left\{\begin{array}{cc}1,\hfill & 0

Let $1. We give a convergence theorem as an analogy of Theorem A for ${L}_{n}\left(f\right)$ in ${L}_{p}$-norm. We need to prepare a lemma.

Lemma 2.1 ([4, Theorem 1.6])

Let $w=exp\left(-Q\right)\in \mathcal{F}\left({C}^{2}+\right)$.
1. (a)
Let $T\left(x\right)$ be unbounded. Then for any $\eta >0$, there exists a constant $C\left(\eta \right)>0$ such that for $t⩾1$,
${a}_{t}⩽C\left(\eta \right){t}^{\eta }.$

2. (b)
Assume
$\frac{{Q}^{″}\left(x\right)}{{Q}^{\prime }\left(x\right)}⩽\lambda \left(b\right)\frac{{Q}^{\prime }\left(x\right)}{Q\left(x\right)},\phantom{\rule{1em}{0ex}}|x|⩾b>0,$
(2.1)

where $b>0$ is large enough. Suppose that there exist constants $\eta >0$ and ${C}_{1}>0$ such that ${a}_{t}⩽{C}_{1}{t}^{\eta }$. If $\lambda :=\lambda \left(b\right)>1$, then there exists a constant $C\left(\lambda ,\eta \right)$ such that for ${a}_{t}⩾1$,
$T\left({a}_{t}\right)⩽C\left(\lambda ,\eta \right){t}^{\frac{2\left(\eta +\lambda -1\right)}{\lambda +1}}.$
(2.2)
If $0<\lambda ⩽1$, then for any $\mu >0$, there exists $C\left(\lambda ,\mu \right)$ such that
$T\left({a}_{t}\right)⩽C\left(\lambda ,\mu \right){t}^{\mu },\phantom{\rule{1em}{0ex}}t⩾1.$
(2.3)
For a fixed constant $\beta >0$, we define
$\varphi \left(x\right):={\left(1+{x}^{2}\right)}^{-\beta /2}.$
(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 $f\in {C}_{0}\left(\mathbb{R}\right)$ mean that $f\in C\left(\mathbb{R}\right)$ and ${lim}_{|x|\to \mathrm{\infty }}f\left(x\right)=0$.

Theorem 2.2 Let $w=exp\left(-Q\right)\in \mathcal{F}\left({C}^{2}+\right)$, and let $T\left(x\right)$ be unbounded. Let $1 and $\beta >0$, and let us define ϕ as (2.4), and $\lambda =\lambda \left(b\right)⩾1$ as (2.1). We suppose that for $f\in C\left(\mathbb{R}\right)$,
${\varphi }^{-1}\left(x\right)w\left(x\right)f\left(x\right)\in {C}_{0}\left(\mathbb{R}\right),$
and
$\mathrm{\Delta }>\frac{9}{4}\frac{\lambda -1}{3\lambda -1}.$
(2.5)
Then we have
$\underset{n\to \mathrm{\infty }}{lim}{\parallel \left(f-{L}_{n}\left(f\right)\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(\mathbb{R}\right)}=0.$
We remark that if $w\in \mathcal{F}\left({C}^{2}+\right)$ is the Erdős-type weight, then we have $\lambda =\lambda \left(b\right)⩾1$ in (2.1). In fact, if $\lambda <1$, then by Lemma 3.9 below, we see that for $x⩾b>0$,
This contradicts our assumption for $T\left(x\right)$. In Example 1.2, we consider the weight ${w}_{l,\alpha ,m}=exp\left(-{Q}_{l,\alpha ,m}\right)$. In (2.1), we set $Q:={Q}_{l,\alpha ,m}$ and $\lambda :=\lambda \left(b\right)$. If ${w}_{l,\alpha ,m}$ is an Erdős-type weight, that is, $T\left(x\right):={T}_{l,\alpha ,m}\left(x\right)$ is unbounded, then it is easy to show
$\underset{b\to \mathrm{\infty }}{lim}\lambda \left(b\right)=1.$
Therefore, when we give any $\mathrm{\Delta }>0$, there exists a constant b large enough such that
$\mathrm{\Delta }>\frac{9}{4}\frac{\lambda \left(b\right)-1}{3\lambda \left(b\right)-1}.$

Hence, we have the following corollary.

Corollary 2.3 Let $1 and $\mathrm{\Delta }>0$. Then for the weight ${w}_{l,\alpha ,m}=exp\left(-{Q}_{l,\alpha ,m}\right)$ ($\alpha >0$), we have
$\underset{n\to \mathrm{\infty }}{lim}{\parallel \left(f-{L}_{n}\left(f\right)\right){w}_{l,\alpha ,m}{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(\mathbb{R}\right)}=0.$

We also consider the case of $p=\mathrm{\infty }$.

Theorem 2.4 Let $w=exp\left(-Q\right)\in \mathcal{F}\left({C}^{2}+\right)$, and let $T\left(x\right)$ be unbounded. For every $f\in {C}_{0}\left(\mathbb{R}\right)$ and $n⩾1$, we have
${\parallel \left(f-{L}_{n}\left(f\right)\right)w{\mathrm{\Phi }}^{3/4}\parallel }_{{L}_{\mathrm{\infty }}\left(\mathbb{R}\right)}⩽C{E}_{n-1}\left(w;f\right)logn,$
where
${E}_{m}\left(w;f\right)=\underset{{P}_{m}\in {\mathcal{P}}_{m}}{inf}\underset{x\in \mathbb{R}}{max}|\left(f\left(x\right)-{P}_{m}\left(x\right)\right)w\left(x\right)|,\phantom{\rule{1em}{0ex}}m=0,1,2,\dots .$
Moreover, if ${f}^{\left(r\right)}$, $r⩾1$, is an integer, then for $n>r+1$ we have
${\parallel \left(f-{L}_{n}\left(f\right)\right)w{\mathrm{\Phi }}^{3/4}\parallel }_{{L}_{\mathrm{\infty }}\left(\mathbb{R}\right)}⩽C{\left(\frac{{a}_{n}}{n}\right)}^{r}{E}_{n-r-1}\left(w;{f}^{\left(r\right)}\right)logn.$

## 3 Fundamental lemmas

To prove the theorems we need some lemmas.

Lemma 3.1 Let $w=exp\left(-Q\right)\in \mathcal{F}\left({C}^{2}+\right)$. Then we have the following.
1. (a)
[1, Lemma 3.11(a), (b)] Given fixed $0<\alpha$, $\alpha \ne 1$, we have uniformly for $t>0$,
$|1-\frac{{a}_{\alpha t}}{{a}_{t}}|\sim \frac{1}{T\left({a}_{t}\right)},$

and we have for $t>0$,
$|1-\frac{{a}_{t}}{{a}_{s}}|\sim \frac{1}{T\left({a}_{t}\right)}|1-\frac{t}{s}|,\phantom{\rule{1em}{0ex}}\frac{1}{2}⩽\frac{t}{s}⩽2.$
1. (b)
[1, Lemma 3.7 (3.38)] For some $0<\epsilon ⩽2$, and for large enough t,
$T\left({a}_{t}\right)⩽{t}^{2-\epsilon }.$

Lemma 3.2 Let $w=exp\left(-Q\right)\in \mathcal{F}\left({C}^{2}+\right)$. Then we have the following.
1. (a)
[1, Lemma 3.5(a), (b)] Let $L>0$ be a fixed constant. Uniformly for $t>0$,
$Q\left({a}_{Lt}\right)\sim Q\left({a}_{t}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{Q}^{\prime }\left({a}_{Lt}\right)\sim {Q}^{\prime }\left({a}_{t}\right).$

Moreover,
${a}_{Lt}\sim {a}_{t}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}T\left({a}_{Lt}\right)\sim T\left({a}_{t}\right).$
1. (b)
[1, Lemma 3.4 (3.18), (3.17)] Uniformly for $x>0$ with ${a}_{t}:=x$, $t>0$, we have
${Q}^{\prime }\left(x\right)\sim \frac{t\sqrt{T\left(x\right)}}{{a}_{t}}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}Q\left(x\right)\sim \frac{t}{\sqrt{T\left(x\right)}}.$

2. (c)
[1, Lemma 3.8(a)] For $x\in \left[0,{a}_{t}\right)$,
${Q}^{\prime }\left(x\right)⩽C\frac{t}{{a}_{t}}\frac{1}{\sqrt{1-\frac{x}{{a}_{t}}}}.$

Lemma 3.3 Let $w=exp\left(-Q\right)\in \mathcal{F}\left({C}^{2}+\right)$. For $x\in \mathbb{R}$, we have
$\mathrm{\Phi }\left(x\right)⩽C{\mathrm{\Phi }}_{n}\left(x\right),\phantom{\rule{1em}{0ex}}n⩾1.$
Proof Let $x={a}_{u}$, $u⩾1$. By Lemma 3.2(b), we have
$u\sim Q\left({a}_{u}\right)\sqrt{T\left({a}_{u}\right)}.$
So, we have
${\delta }_{u}^{-1}\sim {Q}^{\frac{2}{3}}\left({a}_{u}\right)T\left({a}_{u}\right)=\frac{{a}_{u}{Q}^{\prime }\left({a}_{u}\right)}{{Q}^{\frac{1}{3}}\left({a}_{u}\right)}=\frac{x{Q}^{\prime }\left(x\right)}{{Q}^{\frac{1}{3}}\left(x\right)}.$
(3.1)
Now, if $u⩽\frac{n}{2}$, then we have
$\begin{array}{rcl}1-\frac{{a}_{u}}{{a}_{n}}& ⩾& 1-\frac{{a}_{n/2}}{{a}_{n}}\sim \frac{1}{T\left({a}_{n}\right)}\phantom{\rule{1em}{0ex}}\left(\text{by Lemma 3.1(a)}\right)\\ ⩾& \frac{1}{{\left(nT\left({a}_{n}\right)\right)}^{\frac{2}{3}}}={\delta }_{n}\phantom{\rule{1em}{0ex}}\left(\text{by Lemma 3.1(b)}\right).\end{array}$
So, we have
$\begin{array}{rcl}{\mathrm{\Phi }}_{n}\left(x\right)& =& 1-\frac{{a}_{u}}{{a}_{n}}⩾1-\frac{{a}_{u}}{{a}_{2u}}\sim \frac{1}{T\left({a}_{u}\right)}\phantom{\rule{1em}{0ex}}\left(\text{by Lemma 3.1(a)}\right)\\ ⩾& \frac{1}{{\left(uT\left({a}_{u}\right)\right)}^{\frac{2}{3}}}={\delta }_{u}\sim \mathrm{\Phi }\left(x\right)\phantom{\rule{1em}{0ex}}\left(\text{by Lemma 3.2(b) and (3.1)}\right).\end{array}$
Let $\frac{n}{2}. Then we have
${\mathrm{\Phi }}_{n}\left(x\right)⩾{\delta }_{n}\sim {\delta }_{u}\sim \mathrm{\Phi }\left(x\right)\phantom{\rule{1em}{0ex}}\left(\text{by Lemma 3.2(a), (b) and (3.1)}\right).$

□

Lemma 3.4 Let $w\in \mathcal{F}\left({C}^{2}+\right)$. Then we have the following.
1. (a)
[1, Theorem 1.19(f)] For the minimum positive zero ${x}_{\left[n/2\right],n}$,
${x}_{\left[n/2\right],n}\sim \frac{{a}_{n}}{n},$

and for the maximum zero ${x}_{1,n}$,
$1-\frac{{x}_{1,n}}{{a}_{n}}\sim {\delta }_{n}.$
1. (b)
[1, Theorem 1.19(e)] For $n⩾1$ and $1⩽j⩽n-1$,
${x}_{j,n}-{x}_{j+1,n}\sim {\phi }_{n}\left({x}_{j,n}\right).$

2. (c)
[1, p.329, (12.20)] Uniformly for $n⩾1$, $1⩽k⩽n-1$,
${\phi }_{n}\left({x}_{k,n}\right)\sim {\phi }_{n}\left({x}_{k+1,n}\right).$

3. (d)
Let $max\left\{|{x}_{k,n}|,|{x}_{k+1,n}|\right\}⩽{a}_{\alpha n}$, $0<\alpha <1$. Then we have
$w\left({x}_{k,n}\right)\sim w\left({x}_{k+1,n}\right)\sim w\left(x\right)\phantom{\rule{1em}{0ex}}\left({x}_{k+1,n}⩽x⩽{x}_{k,n}\right).$

So, for given $C>0$ and $|x|⩽{a}_{\beta n}$, $0<\beta <\alpha$, if $|x-{x}_{k,n}|⩽C{\phi }_{n}\left(x\right)$, then we have
$w\left(x\right)\sim w\left({x}_{k,n}\right).$
Proof (d) Let $max\left\{|{x}_{k,n}|,|{x}_{k+1,n}|\right\}=|{x}_{k,n}|$ (for the case of $max\left\{|{x}_{k,n}|,|{x}_{k+1,n}|\right\}=|{x}_{k+1,n}|$, we also have the result similarly). By (b) there exists a constant $C>0$ such that
$|{x}_{k,n}-{x}_{k+1,n}|⩽C{\phi }_{n}\left({x}_{k,n}\right).$
Then we see
$\begin{array}{rcl}{\phi }_{n}\left({x}_{k,n}\right)& \sim & \frac{{a}_{n}}{n}\frac{1-\frac{|{x}_{k,n}|}{{a}_{2n}}}{\sqrt{1-\frac{|{x}_{k,n}|}{{a}_{n}}}}=\frac{{a}_{n}}{n}\frac{1-\frac{|{x}_{k,n}|}{{a}_{n}}+|{x}_{k,n}|\left\{\frac{1}{{a}_{n}}-\frac{1}{{a}_{2n}}\right\}}{\sqrt{1-\frac{|{x}_{k,n}|}{{a}_{n}}}}\\ =& \frac{{a}_{n}}{n}\frac{1-\frac{|{x}_{k,n}|}{{a}_{n}}+\frac{|{x}_{k,n}|}{{a}_{n}}\left(1-\frac{{a}_{n}}{{a}_{2n}}\right)}{\sqrt{1-\frac{|{x}_{k,n}|}{{a}_{n}}}}\sim \frac{{a}_{n}}{n}\frac{1-\frac{|{x}_{k,n}|}{{a}_{n}}+C\frac{|{x}_{k,n}|}{{a}_{n}}\frac{1}{T\left({a}_{n}\right)}}{\sqrt{1-\frac{|{x}_{k,n}|}{{a}_{n}}}}\\ \sim & \frac{{a}_{n}}{n}\sqrt{1-\frac{|{x}_{k,n}|}{{a}_{n}}}.\end{array}$
(3.2)
Therefore, from (3.2) and Lemma 3.2(c), we have
$\begin{array}{rcl}|Q\left({x}_{k,n}\right)-Q\left({x}_{k+1,n}\right)|& =& |{Q}^{\prime }\left(\xi \right)\left({x}_{k,n}-{x}_{k+1,n}\right)|⩽C|{Q}^{\prime }\left(\xi \right)|{\phi }_{n}\left(x\right)\phantom{\rule{1em}{0ex}}\left({x}_{k+1,n}⩽\xi ⩽{x}_{k,n}\right)\\ ⩽& C|{Q}^{\prime }\left({x}_{k,n}\right)|\frac{{a}_{n}}{n}\sqrt{1-\frac{|{x}_{k,n}|}{{a}_{n}}}⩽C\frac{n}{{a}_{n}}\frac{1}{\sqrt{1-\frac{|{x}_{k,n}|}{{a}_{n}}}}\frac{{a}_{n}}{n}\sqrt{1-\frac{|{x}_{k,n}|}{{a}_{n}}}⩽C.\end{array}$
Consequently,
$w\left({x}_{k,n}\right)\sim w\left({x}_{k+1,n}\right)\sim w\left(x\right)\phantom{\rule{1em}{0ex}}\left({x}_{k+1,n}⩽x⩽{x}_{k,n}\right).$
Let $|x-{x}_{k,n}|⩽C{\phi }_{n}\left(x\right)$ and $|x|⩽{a}_{\beta n}$. Then we see that there exists ${n}_{0}>0$ such that $|{x}_{k,n}|⩽{a}_{\alpha n}$, $n⩾{n}_{0}$. In fact, we can show it as follows. We use Lemma 3.1(a) and (b). For $|x|⩽{a}_{\beta n}$, we see
$|{x}_{k,n}|⩽|x|+C{\phi }_{n}\left(x\right)⩽|x|+C\frac{{a}_{n}}{n}\sqrt{1-\frac{|x|}{{a}_{n}}},$
and if we take n large enough, then we have
$\begin{array}{rcl}\frac{d}{dt}\left(t+C\frac{{a}_{n}}{n}\sqrt{1-\frac{t}{{a}_{n}}}\right)& =& 1-C\frac{1}{n}\frac{1}{2\sqrt{1-\frac{t}{{a}_{n}}}}⩾1-C\frac{1}{n}\frac{1}{2\sqrt{1-\frac{{a}_{n/3}}{{a}_{n}}}}\\ ⩾& 1-C\frac{\sqrt{T\left({a}_{n}\right)}}{2n}⩾1-C\frac{1}{2{n}^{\epsilon /2}}>0,\end{array}$
that is, $g\left(t\right)=t+C\frac{{a}_{n}}{n}\sqrt{1-\frac{t}{{a}_{n}}}$ is increasing. So, we see
$|{x}_{k,n}|⩽{a}_{\beta n}+C\frac{{a}_{n}}{n}\sqrt{1-\frac{{a}_{\beta n}}{{a}_{n}}}⩽{a}_{\beta n}+C\frac{{a}_{n}}{n}\frac{1}{\sqrt{T\left({a}_{n}\right)}}.$
Therefore, we have
$\begin{array}{rcl}{a}_{\alpha n}-\left({a}_{\beta n}+C\frac{{a}_{n}}{n}\frac{1}{\sqrt{T\left({a}_{n}\right)}}\right)& \sim & \frac{{a}_{n}}{T\left({a}_{n}\right)}-C\frac{{a}_{n}}{n}\frac{1}{\sqrt{T\left({a}_{n}\right)}}\\ =& \frac{{a}_{n}}{T\left({a}_{n}\right)}\left(1-C\frac{\sqrt{T\left({a}_{n}\right)}}{n}\right)⩾\frac{{a}_{n}}{T\left({a}_{n}\right)}\left(1-C\frac{1}{{n}^{\epsilon /2}}\right)>0.\end{array}$
Now, we can show (d). Without loss of generality, we may assume $x\in \left[{x}_{j+1,n},{x}_{j,n}\right]\subset \left\{{x}_{k,n}||x-{x}_{k,n}|⩽C{\phi }_{n}\left(x\right)\right\}$. We define
${x}_{{k}_{1},n}:=min\left\{{x}_{k,n}||x-{x}_{k,n}|⩽C{\phi }_{n}\left(x\right)\right\},\phantom{\rule{2em}{0ex}}{x}_{{k}_{2},n}:=max\left\{{x}_{k,n}||x-{x}_{k,n}|⩽C{\phi }_{n}\left(x\right)\right\}.$
Here we note that ${k}_{1}$, ${k}_{2}$ are decided depending only on the constant C. Then by former result, we have
$w\left({x}_{{k}_{1},n}\right)\sim w\left({x}_{{k}_{2},n}\right)\sim w\left(x\right)\phantom{\rule{1em}{0ex}}\left({x}_{{k}_{1},n}⩽x⩽{x}_{{k}_{2},n}\right).$

□

Lemma 3.5 Let $w=exp\left(-Q\right)\in \mathcal{F}\left({C}^{2}+\right)$. Then we have the following.
1. (a)
[1, Theorem 1.17] Uniformly for $n⩾1$,
$\underset{x\in \mathbb{R}}{sup}|{p}_{n}\left(x\right)|w\left(x\right)|{x}^{2}-{a}_{n}^{2}{|}^{\frac{1}{4}}\sim 1.$

2. (b)
[1, Theorem 1.19(a)] Uniformly for $n⩾1$ and $1⩽j⩽n$,
$|\left({p}_{n}^{\prime }w\right)\left({x}_{j,n}\right)|\sim {\phi }_{n}^{-1}\left({x}_{j,n}\right){a}_{n}^{-\frac{1}{2}}{\left(1-\frac{|{x}_{j,n}|}{{a}_{n}}\right)}^{-\frac{1}{4}}.$

3. (c)
[1, Theorem 1.19(d)] For $x\in \left[{x}_{k+1,n},{x}_{k,n}\right]$, if $k⩽n-1$,
$|{p}_{n}\left(x\right)w\left(x\right)|\sim min\left\{|x-{x}_{k,n}|,|x-{x}_{k+1,n}|\right\}{a}_{n}^{1/2}{\phi }_{n}{\left(x\right)}^{-1}{\left(1-\frac{|{x}_{k,n}|}{{a}_{n}}\right)}^{-1/4}.$

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

Let $w\in \mathcal{F}\left({C}^{2}+\right)$ and $0. Then uniformly $n⩾2$,
${\parallel {\mathrm{\Phi }}^{{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}{p}_{n}w\parallel }_{{L}_{p}\left(\mathbb{R}\right)}⩽C{a}_{n}^{\frac{1}{p}-\frac{1}{2}}\left\{\begin{array}{cc}1,\hfill & 0

where ${x}^{+}=0$ if $x⩽0$, ${x}^{+}=x$ if $x>0$.

Proof From Lemma 3.3, we know $\mathrm{\Phi }\left(x\right)⩽{\mathrm{\Phi }}_{n}\left(x\right)$, then in [5, Theorem 2.7] we only exchange ${\mathrm{\Phi }}_{n}$ with Φ. □

Let $f\in {L}_{p,w}\left(\mathbb{R}\right)$. The Fourier-type series of f is defined by
$\stackrel{˜}{f}\left(x\right):=\sum _{k=0}^{\mathrm{\infty }}{a}_{k}\left({w}^{2},f\right){p}_{k}\left({w}^{2},x\right),\phantom{\rule{1em}{0ex}}{a}_{k}\left({w}^{2},f\right):={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(t\right){p}_{k}\left({w}^{2},t\right){w}^{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$
We denote the partial sum of $\stackrel{˜}{f}\left(x\right)$ by
${s}_{n}\left(f,x\right):={s}_{n}\left({w}^{2},f,x\right):=\sum _{k=0}^{n-1}{a}_{k}\left({w}^{2},f\right){p}_{k}\left({w}^{2},x\right).$
The partial sum ${s}_{n}\left(f\right)$ admits the representation
${s}_{n}\left(f,x\right)=\sum _{j=0}^{n-1}{a}_{j}{p}_{j}\left(x\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(t\right){K}_{n}\left(x,t\right){w}^{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,$
where
${K}_{n}\left(x,t\right):=\sum _{j=0}^{n-1}{p}_{j}\left(x\right){p}_{j}\left(t\right).$
The Christoffel-Darboux formula
${K}_{n}\left(x,t\right)=\frac{{\gamma }_{n-1}}{{\gamma }_{n}}\frac{{p}_{n}\left(x\right){p}_{n-1}\left(t\right)-{p}_{n-1}\left(x\right){p}_{n}\left(t\right)}{x-t}$
(3.3)

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

Lemma 3.7 ([6, Lemma 9.2.6])

Let $1 and $g\in {L}_{p}\left(\mathbb{R}\right)$. Then for the Hilbert transform
$H\left(g,x\right):=\underset{\epsilon \to 0+}{lim}{\int }_{|x-t|⩾\epsilon }\frac{g\left(t\right)}{x-t}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}x\in \mathbb{R},$
(3.4)
we have
${\parallel H\left(g\right)\parallel }_{{L}_{p}\left(\mathbb{R}\right)}⩽C{\parallel g\parallel }_{{L}_{p}\left(\mathbb{R}\right)},$

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

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

Let $w=exp\left(-Q\right)\in \mathcal{F}\left({C}^{2}\right)$, $1⩽p⩽\mathrm{\infty }$ and $\gamma ⩾0$. Then for any $\epsilon >0$, there exists a polynomial P such that
${\parallel \left(f\left(x\right)-P\left(x\right)\right){\left(1+{x}^{2}\right)}^{\gamma }w\left(x\right)\parallel }_{{L}_{p}\left(\mathbb{R}\right)}<\epsilon .$
Lemma 3.9 Let $w\in \mathcal{F}\left({C}^{2}+\right)$ be an Erdős-type weight, that is, $T\left(x\right)$ is unbounded. Then for any $M>1$, there exist ${x}_{M}>0$ and ${C}_{M}>0$ such that
$Q\left(x\right)⩾{C}_{M}{x}^{M},\phantom{\rule{1em}{0ex}}x⩾{x}_{M}.$
Proof For every $M>1$, there exists ${x}_{M}>0$ such that $T\left(x\right)⩾M$ for $x⩾{x}_{M}$, so that ${Q}^{\prime }\left(x\right)/Q\left(x\right)=T\left(x\right)/x⩾M/x$ for $x⩾{x}_{M}$. Hence, we see
$log\frac{Q\left(x\right)}{Q\left({x}_{M}\right)}⩾log{\left(\frac{x}{{x}_{M}}\right)}^{M},\phantom{\rule{1em}{0ex}}x⩾{x}_{M},$
that is,
$Q\left(x\right)⩾\frac{Q\left({x}_{M}\right)}{{\left({x}_{M}\right)}^{M}}{x}^{M},\phantom{\rule{1em}{0ex}}x⩾{x}_{M}.$

Let us put ${C}_{M}:=Q\left({x}_{M}\right)/{\left({x}_{M}\right)}^{M}$. □

## 4 Proof of Theorem 2.2 by Damelin and Lubinsky methods

In this section, we assume $w\in \mathcal{F}\left({C}^{2}+\right)$. To prove the theorem we need some lemmas, and we will use the Damelin and Lubinsky methods of .

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

Let $w\in \mathcal{F}\left({C}^{2}+\right)$. Let $0<\alpha <\frac{1}{4}$ and
$\sum _{n}\left(x\right):=\sum _{|{x}_{k,n}|⩾{a}_{\alpha n}}|{l}_{k,n}\left(x\right)|{w}^{-1}\left({x}_{k,n}\right).$
Then we have for $|x|⩽{a}_{\alpha n/2}$ and $|x|⩾{a}_{2n}$,
$\sum _{n}\left(x\right)w\left(x\right)⩽C.$
Moreover, for ${a}_{\alpha n/2}⩽|x|⩽{a}_{2n}$,
$\sum _{n}\left(x\right)w\left(x\right)⩽C\left(logn+{a}_{n}^{\frac{1}{2}}|{p}_{n}\left(x\right)w\left(x\right)|{T}^{-\frac{1}{4}}\left({a}_{n}\right)\right).$

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 $T\left(x\right)$. 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 ${T}^{\ast }\left(x\right)$ with $T\left(x\right)$. □

Lemma 4.2 ([3, Lemma 3.2])

Let $0<\eta <1$. Let $\psi :\mathbb{R}↦\left(0,\mathrm{\infty }\right)$ be a continuous function with the following property: For $n⩾1$, there exist polynomials ${R}_{n}$ of degree n such that
${C}_{1}⩽\frac{\psi \left(t\right)}{{R}_{n}\left(t\right)}⩽{C}_{2},\phantom{\rule{1em}{0ex}}|t|⩽{a}_{4n}.$
Then for $n⩾{n}_{0}$ and $P\in {\mathcal{P}}_{n}$,
$\sum _{|{x}_{k,n}|⩽{a}_{\eta n}}{\lambda }_{k,n}|P\left({x}_{k,n}\right)|{w}^{-1}\left({x}_{k,n}\right)\psi \left({x}_{k,n}\right)⩽C{\int }_{-{a}_{4n}}^{{a}_{4n}}|P\left(t\right)w\left(t\right)|\psi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$
Remark 4.3 To prove Lemma 4.7 below, we apply this lemma with $\psi \left(t\right)=\varphi \left(t\right)={\left(1+{t}^{2}\right)}^{-\beta /2}$, $\beta >0$. In fact, when ${\varphi }^{\ast }\left(x\right)=\varphi \left(t\right)$, $t={a}_{4n}x$, we can approximate ${\varphi }^{\ast }$ by polynomials ${R}_{n}^{\ast }\in {\mathcal{P}}_{n}$ on $\left[-1,1\right]$, that is, for any $\epsilon >0$ there exists ${R}_{n}^{\ast }\in {\mathcal{P}}_{n}$ such that
$|{\varphi }^{\ast }\left(x\right)-{R}_{n}^{\ast }\left(x\right)|<\epsilon ,\phantom{\rule{1em}{0ex}}x\in \left[-1,1\right].$
Therefore,
$|\frac{{R}_{n}^{\ast }\left(x\right)}{{\varphi }^{\ast }\left(x\right)}-1|<\frac{\epsilon }{{\varphi }^{\ast }\left(x\right)},\phantom{\rule{1em}{0ex}}x\in \left[-1,1\right],$
and so there exist ${C}_{1},{C}_{2}>0$ such that
${C}_{1}⩽1-\frac{\epsilon }{{\varphi }^{\ast }\left(x\right)}⩽|\frac{{R}_{n}^{\ast }\left(x\right)}{{\varphi }^{\ast }\left(x\right)}|<1+\frac{\epsilon }{{\varphi }^{\ast }\left(x\right)}⩽{C}_{2},\phantom{\rule{1em}{0ex}}x\in \left[-1,1\right].$

Now, if we set ${R}_{n}\left(t\right)={R}_{n}^{\ast }\left(x\right)$, then we have the result.

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

Let ${\left\{{f}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of measurable functions from $\mathbb{R}↦\mathbb{R}$ such that for $n⩾1$,
${f}_{n}\left(x\right)=0,\phantom{\rule{1em}{0ex}}|x|<{a}_{\frac{n}{9}};\phantom{\rule{2em}{0ex}}|{f}_{n}\left(x\right)|w\left(x\right)⩽\varphi \left(x\right),\phantom{\rule{1em}{0ex}}x\in \mathbb{R}.$
Then for $1⩽p⩽\mathrm{\infty }$ and $\mathrm{\Delta }>0$, we have
$\underset{n\to \mathrm{\infty }}{lim}{\parallel {L}_{n}\left({f}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(\mathbb{R}\right)}=0.$
(4.1)
Proof Let $|x|⩽{a}_{\frac{n}{18}}$ or $|x|⩾{a}_{2n}$. We use the first inequality of Lemma 4.1 with $\alpha =\frac{1}{9}$, then from the assumption with respect to ${f}_{n}$, we see that
$|{L}_{n}\left({f}_{n};x\right)w\left(x\right)|⩽\varphi \left({a}_{\frac{n}{9}}\right)\sum _{|{x}_{k,n}|⩾{a}_{\frac{n}{9}}}|{l}_{k,n}\left(x\right)|{w}^{-1}\left({x}_{k,n}\right)w\left(x\right)⩽{C}_{1}\varphi \left({a}_{\frac{n}{9}}\right).$
So,
$\begin{array}{rcl}{\parallel {L}_{n}\left({f}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩽{a}_{\frac{n}{18}}\phantom{\rule{0.25em}{0ex}}\mathrm{o}\mathrm{r}\phantom{\rule{0.25em}{0ex}}|x|⩾{a}_{2n}\right)}& ⩽& \varphi \left({a}_{\frac{n}{9}}\right){\parallel {\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(\mathbb{R}\right)}\\ ⩽& {C}_{2}\varphi \left({a}_{\frac{n}{9}}\right)=o\left(1\right)\end{array}$
(4.2)
by Lemma 3.9 (note the definition of $\mathrm{\Phi }\left(x\right)$) and the definition of ϕ in (2.4). Next, we let ${a}_{\frac{n}{18}}⩽|x|⩽{a}_{2n}$. From the second inequality in Lemma 4.1, we see that
$|{L}_{n}\left({f}_{n};x\right)w\left(x\right)|⩽\varphi \left({a}_{\frac{n}{9}}\right)\left(logn+{a}_{n}^{\frac{1}{2}}|{p}_{n}\left(x\right)|w\left(x\right){T}^{-\frac{1}{4}}\left({a}_{n}\right)\right).$
Also, for this range of x, we see that
$\mathrm{\Phi }\left(x\right)=\frac{1}{{\left(1+Q\left(x\right)\right)}^{\frac{2}{3}}T\left(x\right)}\sim \frac{1}{{\left(1+Q\left({a}_{n}\right)\right)}^{\frac{2}{3}}T\left({a}_{n}\right)}\sim \frac{{T}^{\frac{1}{3}}\left({a}_{n}\right)}{{n}^{\frac{2}{3}}T\left({a}_{n}\right)}={\delta }_{n}$
Then since $\mathrm{\Delta }>0$, using Lemma 3.1(a), Lemma 2.1(a), and Lemma 3.6, we have
$logn{\parallel {\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left({a}_{\frac{n}{18}}⩽|x|⩽{a}_{2n}\right)}⩽C{\delta }_{n}^{\mathrm{\Delta }}{\left({a}_{2n}-{a}_{\frac{n}{18}}\right)}^{\frac{1}{p}}logn⩽C$
Therefore, we have by (2.4)
${\parallel {L}_{n}\left({f}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left({a}_{\frac{n}{18}}⩽|x|⩽{a}_{2n}\right)}⩽{C}_{4}\varphi \left({a}_{\frac{n}{9}}\right)=o\left(1\right).$

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

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

Let $1⩽p⩽\mathrm{\infty }$. Let ${\left\{{g}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of measurable functions from $\mathbb{R}↦\mathbb{R}$ such that for $n⩾1$,
${g}_{n}\left(x\right)=0,\phantom{\rule{1em}{0ex}}|x|⩾{a}_{\frac{n}{9}};\phantom{\rule{2em}{0ex}}|{g}_{n}\left(x\right)|w\left(x\right)⩽\varphi \left(x\right),\phantom{\rule{1em}{0ex}}x\in \mathbb{R}.$
(4.3)
Let us suppose
$\mathrm{\Delta }>\frac{9}{4}\frac{\lambda -1}{3\lambda -1},$
(4.4)
where $\lambda ⩾1$ is defined in Lemma 2.1. Then for $1⩽p⩽\mathrm{\infty }$, we have
$\underset{n\to \mathrm{\infty }}{lim}{\parallel {L}_{n}\left({g}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩾{a}_{\frac{n}{8}}\right)}=0.$
(4.5)
Proof Using Lemma 3.5(b) and Lemma 3.4(b), we have for $x⩾{a}_{\frac{n}{8}}$,
$\begin{array}{rcl}|{L}_{n}\left({g}_{n};x\right)|& ⩽& \sum _{|{x}_{k,n}|⩽{a}_{\frac{n}{9}}}|{l}_{k,n}\left(x\right)|{w}^{-1}\left({x}_{k,n}\right)\varphi \left({x}_{k,n}\right)\\ ⩽& {C}_{1}{a}_{n}^{\frac{1}{2}}|{p}_{n}\left(x\right)|\sum _{|{x}_{k,n}|⩽{a}_{\frac{n}{9}}}\left({x}_{k,n}-{x}_{k+1,n}\right)\frac{{\left(1-\frac{|{x}_{k,n}|}{{a}_{n}}+{\delta }_{n}\right)}^{\frac{1}{4}}}{|x-{x}_{k,n}|}\varphi \left({x}_{k,n}\right)\\ ⩽& {C}_{2}{a}_{n}^{\frac{1}{2}}|{p}_{n}\left(x\right)|{\int }_{-{a}_{\frac{n}{9}}}^{{a}_{\frac{n}{9}}}\frac{{\left(1-\frac{|t|}{{a}_{n}}+{\delta }_{n}\right)}^{\frac{1}{4}}}{|x-t|}\varphi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
(4.6)
Equation (4.6) is shown as follows: First, we see
$|x-t|\sim |x-{x}_{k,n}|,\phantom{\rule{1em}{0ex}}t\in \left[{x}_{k+1,n},{x}_{k,n}\right].$
(4.7)
Let $|x|⩾{a}_{\frac{n}{8}}$ and $t\in \left[{x}_{k+1,n},{x}_{k,n}\right]$. Then
$|\frac{x-t}{x-{x}_{k,n}}-1|=|\frac{t-{x}_{k,n}}{x-{x}_{k,n}}|⩽\frac{{x}_{k,n}-{x}_{k+1,n}}{|{x}_{k±2,n}-{x}_{k,n}|}⩽c<1.$
Now, we use the fact that $x+C\phi \left(x\right)$, $x>0$ is increasing for $0, and then
${x}_{k,n}+C{\phi }_{n}\left({x}_{k,n}\right)⩽{a}_{\frac{n}{9}}+C{\phi }_{n}\left({a}_{\frac{n}{9}}\right)⩽{a}_{\frac{n}{8}}⩽x.$
Here, the second inequality follows from the definition of ${\phi }_{n}\left(x\right)$ and Lemma 3.1(a), (b). Hence, we have (4.7). Now, we use the monotonicity of ${\left(1-\frac{|x|}{{a}_{n}}+{\delta }_{n}\right)}^{\frac{1}{4}}\varphi \left(x\right)$. From (4.7) there exists $C>0$ such that for $t\in \left[{x}_{k+1,n},{x}_{k,n}\right]$,
$\begin{array}{rcl}\left({x}_{k,n}-{x}_{k+1,n}\right)\frac{{\left(1-\frac{|{x}_{k,n}|}{{a}_{n}}+{\delta }_{n}\right)}^{\frac{1}{4}}}{|x-{x}_{k,n}|}\varphi \left({x}_{k,n}\right)& ⩽& {\int }_{{x}_{k+1,n}}^{{x}_{k,n}}\frac{{\left(1-\frac{|t|}{{a}_{n}}+{\delta }_{n}\right)}^{\frac{1}{4}}}{|x-{x}_{k,n}|}\varphi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\\ ⩽& \frac{1}{C}{\int }_{{x}_{k+1,n}}^{{x}_{k,n}}\frac{{\left(1-\frac{|t|}{{a}_{n}}+{\delta }_{n}\right)}^{\frac{1}{4}}}{|x-t|}\varphi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
Hence, (4.6) holds. Next, for $t\in \left[0,{a}_{\frac{n}{9}}\right]$ and $x⩾{a}_{\frac{n}{8}}$, we know by Lemma 3.1(a),
$1⩽\frac{{a}_{n}-t}{x-t}⩽1+\frac{{a}_{n}-{a}_{\frac{n}{8}}}{{a}_{\frac{n}{8}}-t}⩽1+\frac{{a}_{n}-{a}_{\frac{n}{8}}}{{a}_{\frac{n}{8}}-{a}_{\frac{n}{9}}}⩽1+C\frac{{a}_{\frac{n}{8}}}{{a}_{\frac{n}{9}}}\frac{T\left({a}_{\frac{n}{9}}\right)}{T\left({a}_{\frac{n}{8}}\right)}⩽{C}_{3}$
and
$1-\frac{|t|}{{a}_{n}}⩾{C}_{4}\frac{1}{T\left({a}_{n}\right)}⩾{\delta }_{n}.$
So, we have
$|{L}_{n}\left({g}_{n};x\right)|⩽{C}_{6}{a}_{n}^{\frac{1}{4}}|{p}_{n}\left(x\right)|{\int }_{0}^{{a}_{\frac{n}{9}}}{\left(x-t\right)}^{-\frac{3}{4}}\varphi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$
Let $t={a}_{s}$, $\frac{n}{9}⩾s⩾1$. Then, since we know for $x⩾{a}_{\frac{n}{8}}$,
$x-t=x\left(1-\frac{t}{x}\right)⩾{a}_{\frac{n}{8}}\left(1-\frac{{a}_{s}}{{a}_{\frac{9}{8}s}}\right)⩾{C}_{7}\frac{{a}_{n}}{T\left({a}_{s}\right)},$
we obtain
$|{L}_{n}\left({g}_{n};x\right)|⩽{C}_{8}{a}_{n}^{-\frac{1}{2}}|{p}_{n}\left(x\right)|{\int }_{0}^{{a}_{\frac{n}{9}}}{T}^{\frac{3}{4}}\left(t\right)\varphi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt⩽{C}_{8}{a}_{n}^{\frac{1}{2}}{T}^{\frac{3}{4}}\left({a}_{n}\right)|{p}_{n}\left(x\right)|.$
Hence, if $1⩽\lambda$, 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 $\eta >0$ can be taken small enough. Then we have (4.5), that is, for $\mathrm{\Delta }>\frac{9}{4}\frac{\lambda -1}{3\lambda -1}$,
$\underset{n\to \mathrm{\infty }}{lim}{\parallel {L}_{n}\left({g}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩾{a}_{\frac{n}{8}}\right)}=0.$

□

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

Let $1. Let $\sigma :\mathbb{R}↦\mathbb{R}$ be a bounded measurable function. Let $\lambda =\lambda \left(b\right)⩾1$ be defined in Lemma 2.1, and then we suppose
$\mathrm{\Delta }>\left\{\begin{array}{cc}0,\hfill & 1
(4.8)
Then for $1 and the partial sum ${s}_{n}$ of the Fourier series, we have
${\parallel {s}_{n}\left[\sigma \varphi {w}^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩽{a}_{\frac{n}{8}}\right)}⩽C{\parallel \sigma \parallel }_{{L}_{\mathrm{\infty }}\left(\mathbb{R}\right)}$
(4.9)

for $n⩾1$. Here C is independent of σ and n.

Proof We may suppose that ${\parallel \sigma \parallel }_{{L}_{\mathrm{\infty }}\left(\mathbb{R}\right)}=1$. By (3.3), (3.4) and Lemma 3.5(a),
$|{s}_{n}\left[\sigma \varphi {w}^{-1}\right]\left(x\right)|w\left(x\right)⩽{a}_{n}^{\frac{1}{2}}{\left(1-\frac{|x|}{{a}_{n}}\right)}^{-\frac{1}{4}}\sum _{j=n-1}^{n}|H\left[\sigma \varphi {p}_{j}w\right]\left(x\right)|.$
(4.10)
Let us choose $l:=l\left(n\right)$ such that ${2}^{l}⩽\frac{n}{8}⩽{2}^{l+1}$. Then we know
${2}^{l+3}⩽n⩽{2}^{l+4}.$
(4.11)
Define
${\mathcal{I}}_{k}=\left[{a}_{{2}^{k}},{a}_{{2}^{k+1}}\right],\phantom{\rule{1em}{0ex}}1⩽k⩽l+2.$
For $j=n-1,n$ and $x\in {\mathcal{I}}_{k}$, we split
$\begin{array}{rcl}H\left[\sigma \varphi {p}_{j}w\right]\left(x\right)w\left(x\right)& =& \left({\int }_{-\mathrm{\infty }}^{0}+{\int }_{0}^{{a}_{{2}^{k-1}}}+\phantom{\rule{0.5em}{0ex}}\mathit{P}.\mathit{V}.{\int }_{{a}_{{2}^{k-1}}}^{{a}_{{2}^{k+2}}}+{\int }_{{a}_{{2}^{k+2}}}^{\mathrm{\infty }}\right)\frac{\left(\sigma \varphi {p}_{j}w\right)\left(t\right)}{x-t}\phantom{\rule{0.2em}{0ex}}dt\\ :=& {I}_{1}\left(x\right)+{I}_{2}\left(x\right)+{I}_{3}\left(x\right)+{I}_{4}\left(x\right).\end{array}$
(4.12)
Here $\mathit{P}.\mathit{V}.$ stands for the principal value. First, we give the estimations of ${I}_{1}$ and ${I}_{2}$ for $x\in {\mathcal{I}}_{k}$. Let $x\in {\mathcal{I}}_{k}$. Then we have by Lemma 3.5(a) and Lemma 3.6 with $p=1$,
$\begin{array}{rcl}|{I}_{1}\left(x\right)|& ⩽& {\int }_{0}^{\mathrm{\infty }}\frac{|\left({p}_{j}w\varphi \right)\left(-t\right)|}{t+x}⩽{C}_{1}{a}_{n}^{-\frac{1}{2}}{\int }_{0}^{\frac{{a}_{n}}{2}}\frac{\varphi \left(t\right)}{t+{a}_{2}}\phantom{\rule{0.2em}{0ex}}dt+{C}_{2}{a}_{n}^{-1}{\int }_{\frac{{a}_{n}}{2}}^{\mathrm{\infty }}|{p}_{j}\left(t\right)|w\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\\ ⩽& {C}_{2}\left({a}_{n}^{-\frac{1}{2}}+{a}_{n}^{-1}{a}_{n}^{1-\frac{1}{2}}\right)⩽{C}_{3}{a}_{n}^{-\frac{1}{2}}.\end{array}$
(4.13)
Here we have used
${\int }_{0}^{\mathrm{\infty }}\frac{\varphi \left(t\right)}{1+t}\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty }.$
(4.14)
By Lemma 3.5(a), and noting $1-x/{a}_{n}⩽1-t/{a}_{n}$ for $x\in {\mathcal{I}}_{k}$,
$\begin{array}{rcl}|{I}_{2}\left(x\right)|& ⩽& {\int }_{0}^{{a}_{{2}^{k-1}}}\frac{|\left({p}_{j}w\varphi \right)\left(t\right)|}{x-t}\phantom{\rule{0.2em}{0ex}}dt⩽{C}_{4}{a}_{n}^{-\frac{1}{2}}{\int }_{0}^{{a}_{{2}^{k-1}}}\frac{{\left(1-\frac{t}{{a}_{n}}\right)}^{-\frac{1}{4}}}{x-t}\phantom{\rule{0.2em}{0ex}}dt\\ ⩽& {C}_{4}{a}_{n}^{-\frac{1}{2}}{\left(1-\frac{x}{{a}_{n}}\right)}^{-\frac{1}{4}}{\int }_{0}^{{a}_{{2}^{k-1}}}\frac{dt}{x-t}\\ =& {C}_{4}{a}_{n}^{-\frac{1}{2}}{\left(1-\frac{x}{{a}_{n}}\right)}^{-\frac{1}{4}}log{\left(1-\frac{{a}_{{2}^{k-1}}}{x}\right)}^{-1}.\end{array}$
Using
$1-\frac{{a}_{{2}^{k-1}}}{x}⩾1-\frac{{a}_{{2}^{k-1}}}{{a}_{{2}^{k}}}⩾C\frac{1}{T\left({a}_{{2}^{k}}\right)}⩾C\frac{1}{T\left(x\right)},$
we can see
$|{I}_{2}\left(x\right)|⩽{C}_{6}{a}_{n}^{-\frac{1}{2}}{\left(1-\frac{x}{{a}_{n}}\right)}^{-\frac{1}{4}}log\left(\frac{T\left(x\right)}{C}\right).$
(4.15)
Next, we give an estimation of ${I}_{4}$ for $x\in {\mathcal{I}}_{k}$. Let $x\in {\mathcal{I}}_{k}$. From Lemma 3.5(a) again,
(4.16)
where
$J:={\int }_{{a}_{{2}^{k+2}}}^{2{a}_{{2}^{k+2}}}|1-\frac{t}{{a}_{n}}{|}^{-\frac{1}{4}}\frac{dt}{t-x}.$
Since, if
$|1-\frac{t}{{a}_{n}}|⩽\frac{1}{2}\left(1-\frac{x}{{a}_{n}}\right),$
then we see
$|t-x|={a}_{n}|\left(1-\frac{x}{{a}_{n}}\right)-\left(1-\frac{t}{{a}_{n}}\right)|⩾\frac{{a}_{n}}{2}\left(1-\frac{x}{{a}_{n}}\right).$
Now, we have
$\begin{array}{lll}J& ⩽& {C}_{9}\left({\left(1-\frac{x}{{a}_{n}}\right)}^{-\frac{1}{4}}\underset{t\in \left[{a}_{{2}^{k+2}},2{a}_{{2}^{k+2}}\right]}{{\int }_{|1-\frac{t}{{a}_{n}}|⩾\frac{1}{2}\left(1-\frac{x}{{a}_{n}}\right),}}\frac{1}{t-x}\phantom{\rule{0.2em}{0ex}}dt\\ +{a}_{n}^{-1}{\left(1-\frac{x}{{a}_{n}}\right)}^{-1}\underset{t\in \left[{a}_{{2}^{k+2}},2{a}_{{2}^{k+2}}\right]}{{\int }_{|1-\frac{t}{{a}_{n}}|⩽\frac{1}{2}\left(1-\frac{x}{{a}_{n}}\right),}}|1-\frac{t}{{a}_{n}}{|}^{-\frac{1}{4}}\phantom{\rule{0.2em}{0ex}}dt\right)\\ ⩽& {C}_{10}\left({\left(1-\frac{x}{{a}_{n}}\right)}^{-\frac{1}{4}}log\left(1+\frac{{a}_{{2}^{k+2}}}{{a}_{{2}^{k+2}}-{a}_{{2}^{k+1}}}\right)\\ +{\left(1-\frac{x}{{a}_{n}}\right)}^{-1}{\int }_{|1-s|⩽\frac{1}{2}\left(1-\frac{x}{{a}_{n}}\right)}{|1-s|}^{-\frac{1}{4}}\phantom{\rule{0.2em}{0ex}}ds\right)\\ ⩽& {C}_{10}\left({\left(1-\frac{x}{{a}_{n}}\right)}^{-\frac{1}{4}}log\left(1+CT\left({a}_{{2}^{k+2}}\right)\right)+\frac{4}{3}{\left(\frac{1}{2}\left(1-\frac{x}{{a}_{n}}\right)\right)}^{\frac{3}{4}}{\left(1-\frac{x}{{a}_{n}}\right)}^{-1}\right)\\ ⩽& {C}_{11}{\left(1-\frac{x}{{a}_{n}}\right)}^{-\frac{1}{4}}log\left(CT\left(x\right)\right).\end{array}$
So, from (4.16) we have
$|{I}_{4}\left(x\right)|⩽{C}_{12}{a}_{n}^{-\frac{1}{2}}{\left(1-\frac{x}{{a}_{n}}\right)}^{-\frac{1}{4}}log\left(CT\left(x\right)\right).$
(4.17)
Therefore, from (4.13), (4.15) and (4.17), we have
$|{I}_{1}+{I}_{2}+{I}_{4}|⩽{C}_{13}{a}_{n}^{-\frac{1}{2}}{\left(1-\frac{x}{{a}_{n}}\right)}^{-\frac{1}{4}}log\left(CT\left(x\right)\right).$
We must estimate the ${L}_{p}$-norm with respect to ${I}_{3}$, that is, ${\parallel \mathit{P}.\mathit{V}.{\int }_{{a}_{{2}^{k-1}}}^{{a}_{{2}^{k+2}}}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(t\right)}{x-t}\phantom{\rule{0.2em}{0ex}}dt\parallel }_{{L}_{p}\left({\mathcal{I}}_{k}\right)}$. We use M. Riesz’s theorem on the boundedness of the Hilbert transform from ${L}_{p}\left(\mathbb{R}\right)$ to ${L}_{p}\left(\mathbb{R}\right)$ (Lemma 3.7) to deduce that by Lemma 3.5(a) and the boundedness of $|\sigma \varphi |$,
$\begin{array}{rcl}{\parallel \mathit{P}.\mathit{V}.{\int }_{{a}_{{2}^{k-1}}}^{{a}_{{2}^{k+2}}}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(t\right)}{x-t}\phantom{\rule{0.2em}{0ex}}dt\parallel }_{{L}_{p}\left({\mathcal{I}}_{k}\right)}& ⩽& {C}_{15}{\left({\int }_{{a}_{{2}^{k-1}}}^{{a}_{{2}^{k+2}}}{|\left(\sigma \varphi {p}_{j}w\right)\left(t\right)|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}\\ ⩽& {C}_{16}{a}_{n}^{-\frac{1}{2}}{\left(1-\frac{{a}_{{2}^{k+2}}}{{a}_{n}}\right)}^{-\frac{1}{4}}{\left({a}_{{2}^{k+2}}-{a}_{{2}^{k-1}}\right)}^{\frac{1}{p}}.\end{array}$
(4.19)
Noting (4.11), we see $n⩾{2}^{l+3}$ for $k⩽l$, so
$1-\frac{{a}_{{2}^{k+1}}}{{a}_{n}}⩾1-\frac{{a}_{{2}^{k+1}}}{{a}_{{2}^{k+3}}}⩾{C}_{19}\frac{1}{T\left({a}_{{2}^{k}}\right)}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{a}_{{2}^{k+1}}-{a}_{{2}^{k}}⩽{C}_{20}\frac{{a}_{{2}^{k}}}{T\left({a}_{{2}^{k}}\right)}.$
On the other hand, using Lemma 3.2(b), we see $\mathrm{\Phi }\left({a}_{t}\right)\sim {\delta }_{t}$. Hence, we have
$\begin{array}{rcl}{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\left({a}_{{2}^{k}}\right)& \sim & {\delta }_{{2}^{k}}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}={\left(\frac{1}{{2}^{k}T\left({a}_{{2}^{k}}\right)}\right)}^{\frac{2}{3}\left(\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}\right)}\\ =& \left\{\begin{array}{cc}{\left(\frac{1}{{2}^{k}T\left({a}_{{2}^{k}}\right)}\right)}^{\frac{2}{3}\mathrm{\Delta }},\hfill & 0
From Lemma 2.1 (2.2), we know
${T}^{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p}}\left({a}_{{2}^{k}}\right)⩽{C}_{1}C\left(\lambda ,\eta \right){\left({2}^{k}\right)}^{\frac{2\left(\eta +\lambda -1\right)}{\lambda +1}max\left\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p},0\right\}},$
and
${T}^{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}\left(1-\frac{1}{p}\right)}\left({a}_{{2}^{k}}\right)⩽{C}_{2}C\left(\lambda ,\eta \right){\left({2}^{k}\right)}^{\frac{2\left(\eta +\lambda -1\right)}{\lambda +1}max\left\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}\left(1-\frac{1}{p}\right),0\right\}}.$
Therefore, we continue with Lemma 2.1(a) as
$\begin{array}{rc}⩽& {C}_{20}C\left(\lambda ,\eta \right)log\left(CT\left({a}_{{2}^{k+1}}\right)\right)\\ ×\left\{\begin{array}{cc}{\left(\frac{1}{{2}^{k}}\right)}^{\frac{2}{3}\mathrm{\Delta }-\frac{\eta }{p}-\frac{2\left(\eta +\lambda -1\right)}{\lambda +1}max\left\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p},0\right\}},\hfill & 1
(4.21)
First, let $1. Then (4.8), that is,
$\mathrm{\Delta }>\left\{\begin{array}{cc}0,\hfill & 1
implies
$\mathrm{\Delta }>\frac{3}{2}\frac{\lambda -1}{3\lambda -1}\frac{p-2}{p}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{\Delta }>0$
iff
$\frac{2}{3}\mathrm{\Delta }-\frac{2\left(\lambda -1\right)}{\lambda +1}\left(-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p}\right)>0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{\Delta }>0$
iff
$\frac{2}{3}\mathrm{\Delta }-\frac{2\left(\lambda -1\right)}{\lambda +1}max\left\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p},0\right\}>0.$
This means that there exists a positive constant ${\eta }_{1}>0$ small enough such that
$A\left({\eta }_{1}\right):=\frac{2}{3}\mathrm{\Delta }-\frac{{\eta }_{1}}{p}-\frac{2\left({\eta }_{1}+\lambda -1\right)}{\lambda +1}max\left\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p},0\right\}>0.$
Now, let $p>4$. Then (4.8), that is,
$\mathrm{\Delta }>\frac{\lambda -1}{3\lambda -1}\frac{p-1}{p}-\frac{1}{4}\frac{\lambda +1}{3\lambda -1}\frac{p-4}{p}$
implies
$\mathrm{\Delta }>\frac{\lambda -1}{3\lambda -1}\left(1-\frac{1}{p}\right)-\frac{\lambda +1}{3\lambda -1}\left(\frac{1}{4}-\frac{1}{p}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{\Delta }+\frac{1}{4}-\frac{1}{p}>0$
iff
$\frac{2}{3}\left(\mathrm{\Delta }+\left(\frac{1}{4}-\frac{1}{p}\right)\right)-\frac{2\left(\lambda -1\right)}{\lambda +1}\left(-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}\left(1-\frac{1}{p}\right)\right)>0$
and
$\frac{2}{3}\left(\mathrm{\Delta }+\left(\frac{1}{4}-\frac{1}{p}\right)\right)>0$
iff
$\frac{2}{3}\left(\mathrm{\Delta }+\left(\frac{1}{4}-\frac{1}{p}\right)\right)-\frac{2\left(\lambda -1\right)}{\lambda +1}max\left\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}\left(1-\frac{1}{p}\right),0\right\}>0.$
Similarly to the previous case, this means that there exists a positive constant ${\eta }_{2}>0$ small enough such that
$B\left({\eta }_{2}\right):=\frac{2}{3}\left(\mathrm{\Delta }+\left(\frac{1}{4}-\frac{1}{p}\right)\right)-\frac{{\eta }_{2}}{p}-\frac{2\left({\eta }_{2}+\lambda -1\right)}{\lambda +1}max\left\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}\left(1-\frac{1}{p}\right),0\right\}>0.$
For $\eta >0$ small enough, we can see $A\left(\eta \right)>A\left({\eta }_{1}\right)>0$ and $B\left(\eta \right)>B\left({\eta }_{2}\right)>0$. Let $\tau :=min\left\{A\left({\eta }_{1}\right),B\left({\eta }_{2}\right)\right\}/2$. Then for small enough $\eta >0$, we have
$\begin{array}{rcl}{\parallel {s}_{n}\left[\sigma \varphi {w}^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left({\mathcal{I}}_{k}\right)}& ⩽& {C}_{20}C\left(\lambda ,\eta \right)log\left(CT\left({a}_{{2}^{k+1}}\right)\right){\left(\frac{1}{{2}^{k}}\right)}^{2\tau }\\ ⩽& {C}_{21}C\left(\lambda ,\eta \right){\left(\frac{1}{{2}^{k}}\right)}^{\tau },\end{array}$
because we see that for all $k>0$,
$log\left(CT\left({a}_{{2}^{k+1}}\right)\right){\left(\frac{1}{{2}^{k}}\right)}^{\tau }<{C}_{22}.$
Therefore, under the conditions (4.8) we have
$\begin{array}{rcl}{\parallel {s}_{n}\left[\sigma \varphi {w}^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left({a}_{2}⩽|x|⩽{a}_{\frac{n}{8}}\right)}^{p}& ⩽& \sum _{k=1}^{l}{\parallel {s}_{n}\left[\sigma \varphi {w}^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left({\mathcal{I}}_{k}\right)}^{p}\\ ⩽& {C}_{21}C\left(\lambda ,\eta \right)\sum _{k=1}^{l}{\left(\frac{1}{{2}^{k}}\right)}^{\tau }⩽{C}_{23}C\left(\lambda ,\eta \right).\end{array}$
(4.22)
The estimation of
${\parallel {s}_{n}\left[\sigma \varphi {w}^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩽{a}_{2}\right)}^{p}$
is similar. In fact, for $x\in \left[-{a}_{2},{a}_{2}\right]$, we split
$H\left[\sigma \varphi {p}_{j}w\right]\left(x\right)=\left({\int }_{-\mathrm{\infty }}^{-2{a}_{2}}+\phantom{\rule{0.5em}{0ex}}\mathit{P}.\mathit{V}.{\int }_{-2{a}_{2}}^{2{a}_{2}}+{\int }_{2{a}_{2}}^{\mathrm{\infty }}\right)\frac{\left(\sigma \varphi {p}_{j}w\right)\left(t\right)}{x-t}\phantom{\rule{0.2em}{0ex}}dt.$
Here we see that
$\begin{array}{rcl}|{\int }_{-\mathrm{\infty }}^{-2{a}_{2}}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(t\right)}{x-t}\phantom{\rule{0.2em}{0ex}}dt|& =& |{\int }_{2{a}_{2}}^{\mathrm{\infty }}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(-t\right)}{x+t}\phantom{\rule{0.2em}{0ex}}dt|⩽|{\int }_{2{a}_{2}}^{\mathrm{\infty }}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(-t\right)}{t-{a}_{2}}\phantom{\rule{0.2em}{0ex}}dt|\\ =& |{\int }_{0}^{\mathrm{\infty }}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(-s-2{a}_{2}\right)}{s+{a}_{2}}\phantom{\rule{0.2em}{0ex}}dt|\end{array}$
and
$\begin{array}{rcl}|{\int }_{2{a}_{2}}^{\mathrm{\infty }}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(t\right)}{x-t}\phantom{\rule{0.2em}{0ex}}dt|& =& |{\int }_{2{a}_{2}}^{\mathrm{\infty }}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(t\right)}{t-x}\phantom{\rule{0.2em}{0ex}}dt|⩽|{\int }_{2{a}_{2}}^{\mathrm{\infty }}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(t\right)}{t-{a}_{2}}\phantom{\rule{0.2em}{0ex}}dt|\\ =& |{\int }_{0}^{\mathrm{\infty }}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(s+2{a}_{2}\right)}{s+{a}_{2}}\phantom{\rule{0.2em}{0ex}}ds|.\end{array}$
So, we can estimate ${\int }_{-\mathrm{\infty }}^{-2{a}_{2}}$ and ${\int }_{2{a}_{2}}^{\mathrm{\infty }}$ as we did ${I}_{1}$ before (see (4.12)). We can estimate the second integral as follows: By M. Riesz’s theorem,
${\parallel \mathit{P}.\mathit{V}.{\int }_{-2{a}_{2}}^{2{a}_{2}}\frac{\left(\sigma \varphi {p}_{j}w\right)\left(t\right)}{x-t}\phantom{\rule{0.2em}{0ex}}dt\parallel }_{{L}_{p}\left(|t|⩽2{a}_{2}\right)}^{p}⩽C{\int }_{-2{a}_{2}}^{2{a}_{2}}{|\left(\sigma \varphi {p}_{j}w\right)\left(t\right)|}^{p}\phantom{\rule{0.2em}{0ex}}dt⩽C{a}_{n}^{-\frac{p}{2}}⩽C.$
Now, under the assumption (4.8), we can select ${\eta }_{0}>0$ small enough such that
$\mathrm{\Delta }>\left\{\begin{array}{cc}0,\hfill & 1

Consequently, from (4.22) with ${\eta }_{0}$ we have the result (4.9). □

Let $0<\alpha <1$, then for ${g}_{n}$ in Lemma 4.5 we estimate ${L}_{n}\left({g}_{n}\right)$ over $\left[-{a}_{\alpha n},{a}_{\alpha n}\right]$.

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

Let $1 and $0<\epsilon <1$. Let $\left\{{g}_{n}\right\}$ be as in Lemma 4.4, but we exchange (4.3) with
$|{g}_{n}\left(x\right)w\left(x\right)|⩽\epsilon \varphi \left(x\right),\phantom{\rule{1em}{0ex}}x\in \mathbb{R},n⩾1.$
Then for $1,
$lim\underset{n\to \mathrm{\infty }}{sup}{\parallel {L}_{n}\left({g}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩽{a}_{\frac{n}{8}}\right)}⩽C\epsilon .$
Proof Let
${\chi }_{n}:={\chi }_{\left[-{a}_{\frac{n}{8}},{a}_{\frac{n}{8}}\right]};\phantom{\rule{2em}{0ex}}{h}_{n}:=sign\left({L}_{n}\left({g}_{n}\right)\right){|{L}_{n}\left({g}_{n}\right)|}^{p-1}{\chi }_{n}{w}^{p-2}{\mathrm{\Phi }}^{\left(\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}\right)p}$
and
${\sigma }_{n}:=sign{s}_{n}\left[{h}_{n}\right].$
We shall show that
${\parallel {L}_{n}\left({g}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩽{a}_{\frac{n}{8}}\right)}⩽\epsilon {\parallel {s}_{n}\left[{\sigma }_{n}\varphi {w}^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩽{a}_{\frac{n}{8}}\right)}.$
(4.23)
Then from Lemma 4.5 we will conclude (4.22). Using orthogonality of $f-{s}_{n}\left[f\right]$ to ${\mathcal{P}}_{n-1}$, and the Gauss quadrature formula, we see that
Here, if we use Lemma 4.2 with $\psi =\varphi$, we continue as
$\begin{array}{rc}⩽& C\epsilon {\int }_{\mathbb{R}}|{s}_{n}\left[{h}_{n}\right]\left(x\right)|\varphi \left(x\right)w\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& C\epsilon {\int }_{\mathbb{R}}{s}_{n}\left[{h}_{n}\right]\left(x\right){\sigma }_{n}\varphi \left(x\right){w}^{-1}\left(x\right){w}^{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx=C\epsilon {\int }_{\mathbb{R}}{h}_{n}\left(x\right){s}_{n}\left[{\sigma }_{n}\varphi {w}^{-1}\right]\left(x\right){w}^{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& C\epsilon {\int }_{-{a}_{\frac{n}{8}}}^{{a}_{\frac{n}{8}}}{h}_{n}\left(x\right){s}_{n}\left[{\sigma }_{n}\varphi {w}^{-1}\right]\left(x\right){w}^{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
Using Hölder’s inequality with $q=p/\left(p-1\right)$, we continue this as

Cancellation of ${\parallel {L}_{n}\left({g}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩽{a}_{\frac{n}{8}}\right)}^{p-1}$ gives (4.23). □

Proof of Theorem 2.2 In proving the theorem, we split our functions into pieces that vanish inside or outside $\left[-{a}_{\frac{n}{9}},{a}_{\frac{n}{9}}\right]$. Throughout, we let ${\chi }_{S}$ denote the characteristic function of a set S. Also, we set for some fixed $\beta >0$,
$\varphi \left(x\right)={\left(1+{x}^{2}\right)}^{-\beta /2},$
and suppose (2.5). We note that (2.5) means (4.8). Let $0<\epsilon <1$. We can choose a polynomial P such that
${\parallel \left(f-P\right)w{\varphi }^{-1}\parallel }_{{L}_{\mathrm{\infty }}\left(\mathbb{R}\right)}⩽\epsilon$
Here we used that
${\parallel \varphi {\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(\mathbb{R}\right)}<\mathrm{\infty },$
because $\mathrm{\Delta }>0$ and ${\mathrm{\Phi }}^{-1}$ grows faster than any power of x (see Lemma 3.9). Next, let
${\chi }_{n}:=\chi \left[-{a}_{\frac{n}{9}},{a}_{\frac{n}{9}}\right],$
and write
$P-f=\left(P-f\right){\chi }_{n}+\left(P-f\right)\left(1-{\chi }_{n}\right)=:{g}_{n}+{f}_{n}.$
By Lemma 4.4 we have
$\underset{n\to \mathrm{\infty }}{lim}{\parallel {L}_{n}\left({f}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{4}\right)}^{+}}\parallel }_{{L}_{p}\left(\mathbb{R}\right)}=0.$
By Lemma 4.5 we have
$\underset{n\to \mathrm{\infty }}{lim}{\parallel {L}_{n}\left({g}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{4}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩾{a}_{\frac{n}{8}}\right)}=0,$
and by Lemma 4.7,
$lim\underset{n\to \mathrm{\infty }}{sup}{\parallel {L}_{n}\left({g}_{n}\right)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{\left(\frac{1}{4}-\frac{1}{p}\right)}^{+}}\parallel }_{{L}_{p}\left(|x|⩽{a}_{\frac{n}{8}}\right)}⩽C\epsilon .$

Here we take $\epsilon >0$ as $\epsilon \to 0$, then with (4.24) we have the result. □

## 5 Proof of Theorem 2.4

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

Let $w\in \mathcal{F}\left({C}^{2}+\right)$. Let $0<\alpha <\frac{1}{4}$ and
$\sum _{n}\left(x\right):=\sum _{|{x}_{k,n}|⩾{a}_{\alpha n}}|{l}_{k,n}\left(x\right)|{w}^{-1}\left({x}_{k,n}\right).$
Then we have for $x\in \mathbb{R}$,
$\sum _{n}\left(x\right)w\left(x\right){\mathrm{\Phi }}^{1/4}\left(x\right)⩽Clogn.$

Proof From Lemma 4.1 and Lemma 3.6 with $p=\mathrm{\infty }$, we have the result easily. □

Lemma 5.2 Let $w\in \mathcal{F}\left({C}^{2}+\right)$. Let $0<\alpha <\frac{1}{4}$ and
${\sum _{n}}^{\prime }\left(x\right):=\sum _{|{x}_{k,n}|⩽{a}_{\alpha n}}|{l}_{k,n}\left(x\right)|{w}^{-1}\left({x}_{k,n}\right).$
Then we have
${\sum _{n}}^{\prime }\left(x\right)w\left(x\right)\mathrm{\Phi }{\left(x\right)}^{3/4}⩽Clogn.$
Proof By Lemma 3.5(c), Lemma 3.4(d) and Lemma 3.5(b),
$\begin{array}{rcl}{\sum _{n}}^{\prime }\left(x\right)& =& \sum _{|{x}_{k,n}|⩽{a}_{\alpha n}}|{l}_{k,n}\left(x\right)|{w}^{-1}\left({x}_{k,n}\right)\\ =& \frac{|{p}_{n}\left(x\right)|}{|x-{x}_{{j}_{x},n}||{P}_{n}^{\prime }\left({x}_{{j}_{x},n}\right)|w\left({x}_{{j}_{x},n}\right)}+\underset{k\ne {j}_{x}}{\sum _{|{x}_{k,n}|⩽{a}_{\alpha n},}}\frac{|{p}_{n}\left(x\right)|}{|x-{x}_{k,n}||{P}_{n}^{\prime }\left({x}_{k,n}\right)|w\left({x}_{k,n}\right)}\\ ⩽& Cw{\left(x\right)}^{-1}+{a}_{n}^{1/2}|{p}_{n}\left(x\right)|\underset{k\ne {j}_{x}}{\sum _{|{x}_{k,n}|⩽{a}_{\alpha n},}}\frac{{\phi }_{n}\left({x}_{k,n}\right)\left(1-\frac{|{x}_{k,n}|}{{a}_{n}}\right)}{|x-{x}_{k,n}|}\\ \sim & Cw{\left(x\right)}^{-1}+\frac{{a}_{n}^{3/2}}{n}|{p}_{n}\left(x\right)|\underset{k\ne {j}_{x}}{\sum _{|{x}_{k,n}|⩽{a}_{\alpha n},}}\frac{1-\frac{|{x}_{k,n}|}{{a}_{2n}}}{\sqrt{1-\frac{|{x}_{k,n}|}{{a}_{n}}}}{\left(1-\frac{|{x}_{k,n}|}{{a}_{n}}\right)}^{1/4}\frac{1}{|x-{x}_{k,n}|}\\ \sim & Cw{\left(x\right)}^{-1}+\frac{{a}_{n}^{3/2}}{n}|{p}_{n}\left(x\right)|\underset{k\ne {j}_{x}}{\sum _{|{x}_{k,n}|⩽{a}_{\alpha n},}}{\left(1-\frac{|{x}_{k,n}|}{{a}_{n}}\right)}^{3/4}\frac{1}{|x-{x}_{k,n}|},\end{array}$
where we used the fact
$1-\frac{|{x}_{k,n}|}{{a}_{2n}}\sim 1-\frac{|{x}_{k,n}|}{{a}_{n}},\phantom{\rule{1em}{0ex}}|{x}_{k,n}|⩽{a}_{\alpha n}.$
So,
$\begin{array}{rcl}{\sum _{n}}^{\prime }\left(x\right)& ⩽& Cw{\left(x\right)}^{-1}+\frac{{a}_{n}^{3/2}}{n}|{p}_{n}\left(x\right)|\underset{k\ne {j}_{x}}{\sum _{|{x}_{k,n}|⩽{a}_{\alpha n},}}{\left(1-\frac{|{x}_{k,n}|}{{a}_{n}}\right)}^{3/4}\frac{1}{|{x}_{{j}_{x},n}-{x}_{k,n}|}\\ ⩽& Cw{\left(x\right)}^{-1}+\frac{{a}_{n}^{3/2}}{n}|{p}_{n}\left(x\right)|\underset{k\ne {j}_{x}}{\sum _{|{x}_{k,n}|⩽{a}_{\alpha n},}}{\left(1-\frac{|{x}_{k,n}|}{{a}_{n}}\right)}^{3/4}\frac{1}{{\sum }_{{j}_{x}\lessgtr i\lessgtr k}{\phi }_{n}\left({x}_{i,n}\right)}\\ ⩽& Cw{\left(x\right)}^{-1}+{a}_{n}^{1/2}|{p}_{n}\left(x\right)|\underset{k\ne {j}_{x}}{\sum _{|{x}_{k,n}|⩽{a}_{\alpha n},}}{\left(1-\frac{|{x}_{k,n}|}{{a}_{n}}\right)}^{3/4}\frac{1}{{\sum }_{{j}_{x}\lessgtr i\lessgtr k}\sqrt{1-|{x}_{i,n}|/{a}_{n}}}.\end{array}$
Therefore we have by Lemma 3.6 with $p=\mathrm{\infty }$,
$\begin{array}{rcl}{\sum _{n}}^{\prime }\left(x\right)w\left(x\right)\mathrm{\Phi }{\left(x\right)}^{3/4}& ⩽& C+C{a}_{n}^{1/2}|{p}_{n}\left(x\right)|w\left(x\right)\mathrm{\Phi }{\left(x\right)}^{1/4}\\ ×\underset{k\ne {j}_{x}}{\sum _{|{x}_{k,n}|⩽{a}_{\alpha n},}}{\left(1-\frac{|{x}_{k,n}|}{{a}_{n}}\right)}^{3/4}{\left(1-\frac{|{x}_{{j}_{x},n}|}{{a}_{n}}\right)}^{1/2}\frac{1}{{\sum }_{{j}_{x}\lessgtr i\lessgtr k}\sqrt{1-|{x}_{i,n}|/{a}_{n}}}\\ ⩽& C\underset{k\ne {j}_{x}}{\sum _{|{x}_{k,n}|⩽{a}_{\alpha n},}}\frac{1}{|{j}_{x}-k|}\sim logn.\end{array}$

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Lemma 5.3 ([8, Theorem 1])

Let $w\in \mathcal{F}\left({C}^{2}+\right)$. Then there exists a constant ${C}_{0}>0$ such that for every absolutely continuous function f with $w{f}^{\prime }\in {C}_{0}\left(\mathbb{R}\right)$ (this means $w\left(x\right){f}^{\prime }\left(x\right)\to 0$ as $|x|\to \mathrm{\infty }$) and every $n\in \mathbb{N}$, we have
${E}_{n}\left(w;f\right)⩽C\frac{{a}_{n}}{n}{E}_{n-1}\left(w;{f}^{\prime }\right).$
Proof of Theorem 2.4 There exists ${P}_{n-1}\in {\mathcal{P}}_{n}$ such that
$|\left(f\left(x\right)-{P}_{n-1}\left(x\right)\right)w\left(x\right)|⩽2{E}_{n-1}\left(w;f\right).$
Let $w{f}^{\left(r\right)}\in {C}_{0}\left(\mathbb{R}\right)$. If we repeatedly use Lemma 5.3, then we have
$|\left(f\left(x\right)-{L}_{n}\left(f\right)\left(x\right)\right)w\left(x\right){\mathrm{\Phi }}^{3/4}\left(x\right)|⩽{C}_{r}{\left(\frac{{a}_{n}}{n}\right)}^{r}{E}_{n-r-1}\left(w;{f}^{\left(r\right)}\right)logn.$

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## Declarations

### Acknowledgements

The authors thank the referees for many kind suggestions and comments.

## Authors’ Affiliations

(1)
Department of Mathematics Education, Sungkyunkwan University, Seoul, 110-745, Republic of Korea
(2)
Department of Mathematics, Meijo University, Nagoya 468-8502, Japan

## References 