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

Abstract

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

MSC:41A05.

1 Introduction and preliminaries

Let $\mathbb{R}=(-\mathrm{\infty },\mathrm{\infty })$, and let $Q\in C^1(\mathbb{R}):\mathbb{R}\to {\mathbb{R}}^{+}:=[0,\mathrm{\infty })$ be an even function, and $w(x)=exp(-Q(x))$ be the weight such that ${\int }_0^{\mathrm{\infty }}{x}^n{w}^2(x)dx<\mathrm{\infty }$ for all $n=0,1,2,\dots$ . 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,

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{p}_n(x)p_m(x)w^2(x)dx={\delta }_{mn}(\text{Kronecker’s delta})$$

and

$$p_n(x)={\gamma }_n{x}^n+\cdots ,{\gamma }_n>0.$$

We denote the zeros of $p_n(x)$ 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 $L_n(f;x)$ based at the zeros ${\{x_{k,n}\}}_{k=1}^n$ as follows:

$$L_n(f;x):=\sum _{k=1}^{n}f(x_{k,n})l_{k,n}(x),l_{k,n}(x):=\frac{p_n(x)}{(x-x_{k,n})p_n^{\prime }(x_{k,n})}.$$

A function $f:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ 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 $Q:\mathbb{R}\to {\mathbb{R}}^{+}$ be an even function satisfying the following properties:

• (a) $Q^{\prime }(x)$ is continuous in ℝ, with $Q(0)=0$.

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

• (c) ${lim}_{x\to \mathrm{\infty }}Q(x)=\mathrm{\infty }$.

• (d) The function

  $$T(x):=\frac{x{Q}^{\prime }(x)}{Q(x)},x\ne 0$$

is quasi-increasing in $(0,\mathrm{\infty })$ with

$$T(x)⩾\mathrm{\Lambda }>1,x\in {\mathbb{R}}^{+}\mathrm{\setminus }\{0\}.$$

• (e) There exists $C_1>0$ such that

  $$\frac{Q^{″}(x)}{|Q^{\prime }(x)|}⩽C_1\frac{|Q^{\prime }(x)|}{Q(x)},\text{a.e. }x\in \mathbb{R}\mathrm{\setminus }\{0\}.$$

Then we write $w(x)=exp(-Q(x))\in \mathcal{F}(C^2)$. If there also exist a compact subinterval J (∋0) of ℝ and $C_2>0$ such that

$$\frac{Q^{″}(x)}{|Q^{\prime }(x)|}⩾C_2\frac{|Q^{\prime }(x)|}{Q(x)},\text{a.e. }x\in \mathbb{R}\mathrm{\setminus }J,$$

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

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:

$$Q(x)={|x|}^{\alpha },\alpha >1.$$

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 $w=exp(-Q)$.

• (2) [2, Theorem 3.1] For $\alpha >1$, $l=1,2,3,\dots$

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

where

$${exp}_l(x)=exp(exp(exp\cdots expx)\cdots )(l\text{-times}).$$

More generally, we define for $\alpha +u>1$, $\alpha ⩾0$, $u⩾0$ and $l⩾1$,

$$Q_{l,\alpha ,u}(x):={|x|}^u({exp}_l\left({|x|}^{\alpha }\right)-{\alpha }^{\ast }{exp}_l(0)),$$

where ${\alpha }^{\ast }=0$ if $\alpha =0$, otherwise ${\alpha }^{\ast }=1$. (We note that $Q_{l,0,u}(x)$ gives a Freud-type weight.)

• (3) We define $Q_{\alpha }(x):={(1+|x|)}^{{|x|}^{\alpha }}-1$, $\alpha >1$.

In this paper, we investigate the convergence of the Lagrange interpolation polynomials with respect to the weight $w\in \mathcal{F}(C^2+)$. 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 $Q:\mathbb{R}\to \mathbb{R}$ is even and continuous. $Q^{″}$ exists in $(0,\mathrm{\infty })$, $Q^{(j)}⩾0$, in $(0,\mathrm{\infty })$, $j=0,1,2$, and the function

$$T^{\ast }(x):=1+\frac{x{Q}^{″}(x)}{Q^{\prime }(x)}$$

is increasing in $(0,\mathrm{\infty })$ with

$$\underset{x\to \mathrm{\infty }}{lim}{T}^{\ast }(x)=\mathrm{\infty };T^{\ast }(0+):=\underset{x\to 0+}{lim}{T}^{\ast }(x)>1.$$

(1.1)

Moreover, we assume that for some constants $C_1,C_2,C_3>0$,

$$C_1⩽T^{\ast }(x)/\left(\frac{x{Q}^{\prime }(x)}{Q(x)}\right)⩽C_2,x⩾C_3,$$

and for every $\epsilon >0$,

$$T^{\ast }(x)=O\left(Q{(x)}^{\epsilon }\right),x\to \mathrm{\infty }.$$

(1.2)

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

Damelin and Lubinsky [3] got the following results with the Erdős-type weights $w=exp(-Q)\in \mathcal{E}$.

Theorem A ([3, Theorem 1.3])

Let $w=exp(-Q)\in \mathcal{E}$. Let $L_n(f,x)$ denote the Lagrange interpolation polynomial to f at the zeros of $p_n(w^2,x)$. Let $1<p<\mathrm{\infty }$, $\mathrm{\Delta }\in \mathbb{R}$, $\kappa >0$. Then for

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel (f-L_n(f))w{(1+Q)}^{-\mathrm{\Delta }}\parallel }_{L_p(\mathbb{R})}=0$$

to hold for every continuous function $f:\mathbb{R}\to \mathbb{R}$ satisfying

$$\underset{|x|\to \mathrm{\infty }}{lim}|f(x)w(x){(log|x|)}^{1+\kappa }|=0,$$

it is necessary and sufficient that

$$\mathrm{\Delta }>max\{0,\frac{2}{3}(\frac{1}{4}-\frac{1}{p})\}.$$

Our main purpose in this paper is to give mean and uniform convergence theorems with respect to $\{L_n(f)\}$, $n=1,2,\dots$ , in $L_p$-norm, $1<p⩽\mathrm{\infty }$. The proof for $1<p<\mathrm{\infty }$ 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 $1<p<\mathrm{\infty }$. 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 $f(x)\sim g(x)$ 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 ${\{c_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{d_n\}}_{=1}^{\mathrm{\infty }}$, 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:

$$x=\frac{2}{\pi }{\int }_0^1\frac{a_{x}u{Q}^{\prime }(a_{x}u)}{{(1-u^2)}^{\frac{1}{2}}}du,x>0.$$

The function ${\phi }_u(x)$ is defined as follows:

$${\phi }_u(x)=\{\begin{array}{cc}\frac{a_u}{u}\frac{1-\frac{|x|}{a_{2u}}}{\sqrt{1-\frac{|x|}{a_u}}+{\delta }_u},\hfill & |x|⩽a_u,\hfill \\ {\phi }_u(a_u),\hfill & a_u<|x|,\hfill \end{array}$$

where

$${\delta }_x={(xT(a_x))}^{-\frac{2}{3}},x>0.$$

We define

$$\mathrm{\Phi }(x):=\frac{1}{{(1+Q(x))}^{\frac{2}{3}}T(x)}$$

and

$${\mathrm{\Phi }}_n(x):=max\{{\delta }_n,1-\frac{|x|}{a_n}\}.$$

Here we note that for $0<d⩽|x|$,

$$\mathrm{\Phi }(x)\sim \frac{Q{(x)}^{\frac{1}{3}}}{x{Q}^{\prime }(x)}$$

and we see

$$\mathrm{\Phi }(x)⩽C{\mathrm{\Phi }}_n(x),n⩾1$$

(see Lemma 3.3 below). Moreover, we define

$${\mathrm{\Phi }}^{{(\frac{1}{4}-\frac{1}{p})}^{+}}(x):=\{\begin{array}{cc}1,\hfill & 0<p<4,\hfill \\ {\mathrm{\Phi }}^{\frac{1}{4}-\frac{1}{p}}(x),\hfill & 4⩽p⩽\mathrm{\infty }.\hfill \end{array}$$

Let $1<p<\mathrm{\infty }$. 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 $w=exp(-Q)\in \mathcal{F}(C^2+)$.

• (a) Let $T(x)$ be unbounded. Then for any $\eta >0$, there exists a constant $C(\eta )>0$ such that for $t⩾1$,

  $$a_t⩽C(\eta )t^{\eta }.$$

• (b) Assume

  $$\frac{Q^{″}(x)}{Q^{\prime }(x)}⩽\lambda (b)\frac{Q^{\prime }(x)}{Q(x)},|x|⩾b>0,$$

  (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 $\lambda :=\lambda (b)>1$, then there exists a constant $C(\lambda ,\eta )$ such that for $a_t⩾1$,

$$T(a_t)⩽C(\lambda ,\eta )t^{\frac{2(\eta +\lambda -1)}{\lambda +1}}.$$

(2.2)

If $0<\lambda ⩽1$, then for any $\mu >0$, there exists $C(\lambda ,\mu )$ such that

$$T(a_t)⩽C(\lambda ,\mu )t^{\mu },t⩾1.$$

(2.3)

For a fixed constant $\beta >0$, we define

$$\varphi (x):={(1+x^2)}^{-\beta /2}.$$

(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 $f\in C_0(\mathbb{R})$ mean that $f\in C(\mathbb{R})$ and ${lim}_{|x|\to \mathrm{\infty }}f(x)=0$.

Theorem 2.2 Let $w=exp(-Q)\in \mathcal{F}(C^2+)$, and let $T(x)$ be unbounded. Let $1<p<\mathrm{\infty }$ and $\beta >0$, and let us define ϕ as (2.4), and $\lambda =\lambda (b)⩾1$ as (2.1). We suppose that for $f\in C(\mathbb{R})$,

$${\varphi }^{-1}(x)w(x)f(x)\in C_0(\mathbb{R}),$$

and

$$\mathrm{\Delta }>\frac{9}{4}\frac{\lambda -1}{3\lambda -1}.$$

(2.5)

Then we have

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel (f-L_n(f))w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(\mathbb{R})}=0.$$

We remark that if $w\in \mathcal{F}(C^2+)$ is the Erdős-type weight, then we have $\lambda =\lambda (b)⩾1$ in (2.1). In fact, if $\lambda <1$, then by Lemma 3.9 below, we see that for $x⩾b>0$,

$$T(x)=\frac{x{Q}^{\prime }(x)}{Q(x)}⩽\frac{x}{Q(x)}{Q}^{\prime }(b){\left(\frac{Q(x)}{Q(b)}\right)}^{\lambda }=\frac{Q^{\prime }(b)}{Q{(b)}^{\lambda }}\frac{x}{Q{(x)}^{1-\lambda }}\to 0\text{as }x\to \mathrm{\infty }.$$

This contradicts our assumption for $T(x)$. In Example 1.2, we consider the weight $w_{l,\alpha ,m}=exp(-Q_{l,\alpha ,m})$. In (2.1), we set $Q:=Q_{l,\alpha ,m}$ and $\lambda :=\lambda (b)$. If $w_{l,\alpha ,m}$ is an Erdős-type weight, that is, $T(x):=T_{l,\alpha ,m}(x)$ is unbounded, then it is easy to show

$$\underset{b\to \mathrm{\infty }}{lim}\lambda (b)=1.$$

Therefore, when we give any $\mathrm{\Delta }>0$, there exists a constant b large enough such that

$$\mathrm{\Delta }>\frac{9}{4}\frac{\lambda (b)-1}{3\lambda (b)-1}.$$

Hence, we have the following corollary.

Corollary 2.3 Let $1<p<\mathrm{\infty }$ and $\mathrm{\Delta }>0$. Then for the weight $w_{l,\alpha ,m}=exp(-Q_{l,\alpha ,m})$ ($\alpha >0$), we have

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel (f-L_n(f))w_{l,\alpha ,m}{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(\mathbb{R})}=0.$$

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

Theorem 2.4 Let $w=exp(-Q)\in \mathcal{F}(C^2+)$, and let $T(x)$ be unbounded. For every $f\in C_0(\mathbb{R})$ and $n⩾1$, we have

$${\parallel (f-L_n(f))w{\mathrm{\Phi }}^{3/4}\parallel }_{L_{\mathrm{\infty }}(\mathbb{R})}⩽C{E}_{n-1}(w;f)logn,$$

where

$$E_m(w;f)=\underset{P_m\in {\mathcal{P}}_m}{inf}\underset{x\in \mathbb{R}}{max}|(f(x)-P_m(x))w(x)|,m=0,1,2,\dots .$$

Moreover, if $f^{(r)}$, $r⩾1$, is an integer, then for $n>r+1$ we have

$${\parallel (f-L_n(f))w{\mathrm{\Phi }}^{3/4}\parallel }_{L_{\mathrm{\infty }}(\mathbb{R})}⩽C{\left(\frac{a_n}{n}\right)}^r{E}_{n-r-1}(w;f^{(r)})logn.$$

3 Fundamental lemmas

To prove the theorems we need some lemmas.

Lemma 3.1 Let $w=exp(-Q)\in \mathcal{F}(C^2+)$. Then we have the following.

• (a) [1, Lemma 3.11(a), (b)] Given fixed $0<\alpha$, $\alpha \ne 1$, we have uniformly for $t>0$,

  $$|1-\frac{a_{\alpha t}}{a_t}|\sim \frac{1}{T(a_t)},$$

and we have for $t>0$,

$$|1-\frac{a_t}{a_s}|\sim \frac{1}{T(a_t)}|1-\frac{t}{s}|,\frac{1}{2}⩽\frac{t}{s}⩽2.$$

• (b) [1, Lemma 3.7 (3.38)] For some $0<\epsilon ⩽2$, and for large enough t,

  $$T(a_t)⩽t^{2-\epsilon }.$$

Lemma 3.2 Let $w=exp(-Q)\in \mathcal{F}(C^2+)$. Then we have the following.

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

  $$Q(a_{Lt})\sim Q(a_t)\mathit{\text{and}}{Q}^{\prime }(a_{Lt})\sim Q^{\prime }(a_t).$$

Moreover,

$$a_{Lt}\sim a_t\mathit{\text{and}}T(a_{Lt})\sim T(a_t).$$

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

  $$Q^{\prime }(x)\sim \frac{t\sqrt{T(x)}}{a_t}\mathit{\text{and}}Q(x)\sim \frac{t}{\sqrt{T(x)}}.$$

• (c) [1, Lemma 3.8(a)] For $x\in [0,a_t)$,

  $$Q^{\prime }(x)⩽C\frac{t}{a_t}\frac{1}{\sqrt{1-\frac{x}{a_t}}}.$$

Lemma 3.3 Let $w=exp(-Q)\in \mathcal{F}(C^2+)$. For $x\in \mathbb{R}$, we have

$$\mathrm{\Phi }(x)⩽C{\mathrm{\Phi }}_n(x),n⩾1.$$

Proof Let $x=a_u$, $u⩾1$. By Lemma 3.2(b), we have

$$u\sim Q(a_u)\sqrt{T(a_u)}.$$

So, we have

$${\delta }_u^{-1}\sim Q^{\frac{2}{3}}(a_u)T(a_u)=\frac{a_u{Q}^{\prime }(a_u)}{Q^{\frac{1}{3}}(a_u)}=\frac{x{Q}^{\prime }(x)}{Q^{\frac{1}{3}}(x)}.$$

(3.1)

Now, if $u⩽\frac{n}{2}$, then we have

$$\begin{array}{rcl}1-\frac{a_u}{a_n}& ⩾& 1-\frac{a_{n/2}}{a_n}\sim \frac{1}{T(a_n)}\left(\text{by Lemma 3.1(a)}\right)\\ ⩾& \frac{1}{{(nT(a_n))}^{\frac{2}{3}}}={\delta }_n\left(\text{by Lemma 3.1(b)}\right).\end{array}$$

So, we have

$$\begin{array}{rcl}{\mathrm{\Phi }}_n(x)& =& 1-\frac{a_u}{a_n}⩾1-\frac{a_u}{a_{2u}}\sim \frac{1}{T(a_u)}\left(\text{by Lemma 3.1(a)}\right)\\ ⩾& \frac{1}{{(uT(a_u))}^{\frac{2}{3}}}={\delta }_u\sim \mathrm{\Phi }(x)\left(\text{by Lemma 3.2(b) and (3.1)}\right).\end{array}$$

Let $\frac{n}{2}<u<n$. Then we have

$${\mathrm{\Phi }}_n(x)⩾{\delta }_n\sim {\delta }_u\sim \mathrm{\Phi }(x)\left(\text{by Lemma 3.2(a), (b) and (3.1)}\right).$$

 □

Lemma 3.4 Let $w\in \mathcal{F}(C^2+)$. Then we have the following.

• (a) [1, Theorem 1.19(f)] For the minimum positive zero $x_{[n/2],n}$,

  $$x_{[n/2],n}\sim \frac{a_n}{n},$$

and for the maximum zero $x_{1,n}$,

$$1-\frac{x_{1,n}}{a_n}\sim {\delta }_n.$$

• (b) [1, Theorem 1.19(e)] For $n⩾1$ and $1⩽j⩽n-1$,

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

• (c) [1, p.329, (12.20)] Uniformly for $n⩾1$, $1⩽k⩽n-1$,

  $${\phi }_n(x_{k,n})\sim {\phi }_n(x_{k+1,n}).$$

• (d) Let $max\{|x_{k,n}|,|x_{k+1,n}|\}⩽a_{\alpha n}$, $0<\alpha <1$. Then we have

  $$w(x_{k,n})\sim w(x_{k+1,n})\sim w(x)(x_{k+1,n}⩽x⩽x_{k,n}).$$

So, for given $C>0$ and $|x|⩽a_{\beta n}$, $0<\beta <\alpha$, if $|x-x_{k,n}|⩽C{\phi }_n(x)$, then we have

$$w(x)\sim w(x_{k,n}).$$

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

$$|x_{k,n}-x_{k+1,n}|⩽C{\phi }_n(x_{k,n}).$$

Then we see

$$\begin{array}{rcl}{\phi }_n(x_{k,n})& \sim & \frac{a_n}{n}\frac{1-\frac{|x_{k,n}|}{a_{2n}}}{\sqrt{1-\frac{|x_{k,n}|}{a_n}}}=\frac{a_n}{n}\frac{1-\frac{|x_{k,n}|}{a_n}+|x_{k,n}|\{\frac{1}{a_n}-\frac{1}{a_{2n}}\}}{\sqrt{1-\frac{|x_{k,n}|}{a_n}}}\\ =& \frac{a_n}{n}\frac{1-\frac{|x_{k,n}|}{a_n}+\frac{|x_{k,n}|}{a_n}(1-\frac{a_n}{a_{2n}})}{\sqrt{1-\frac{|x_{k,n}|}{a_n}}}\sim \frac{a_n}{n}\frac{1-\frac{|x_{k,n}|}{a_n}+C\frac{|x_{k,n}|}{a_n}\frac{1}{T(a_n)}}{\sqrt{1-\frac{|x_{k,n}|}{a_n}}}\\ \sim & \frac{a_n}{n}\sqrt{1-\frac{|x_{k,n}|}{a_n}}.\end{array}$$

(3.2)

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

$$\begin{array}{rcl}|Q(x_{k,n})-Q(x_{k+1,n})|& =& |Q^{\prime }(\xi )(x_{k,n}-x_{k+1,n})|⩽C|Q^{\prime }(\xi )|{\phi }_n(x)(x_{k+1,n}⩽\xi ⩽x_{k,n})\\ ⩽& C|Q^{\prime }(x_{k,n})|\frac{a_n}{n}\sqrt{1-\frac{|x_{k,n}|}{a_n}}⩽C\frac{n}{a_n}\frac{1}{\sqrt{1-\frac{|x_{k,n}|}{a_n}}}\frac{a_n}{n}\sqrt{1-\frac{|x_{k,n}|}{a_n}}⩽C.\end{array}$$

Consequently,

$$w(x_{k,n})\sim w(x_{k+1,n})\sim w(x)(x_{k+1,n}⩽x⩽x_{k,n}).$$

Let $|x-x_{k,n}|⩽C{\phi }_n(x)$ 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

$$|x_{k,n}|⩽|x|+C{\phi }_n(x)⩽|x|+C\frac{a_n}{n}\sqrt{1-\frac{|x|}{a_n}},$$

and if we take n large enough, then we have

$$\begin{array}{rcl}\frac{d}{dt}(t+C\frac{a_n}{n}\sqrt{1-\frac{t}{a_n}})& =& 1-C\frac{1}{n}\frac{1}{2\sqrt{1-\frac{t}{a_n}}}⩾1-C\frac{1}{n}\frac{1}{2\sqrt{1-\frac{a_{n/3}}{a_n}}}\\ ⩾& 1-C\frac{\sqrt{T(a_n)}}{2n}⩾1-C\frac{1}{2n^{\epsilon /2}}>0,\end{array}$$

that is, $g(t)=t+C\frac{a_n}{n}\sqrt{1-\frac{t}{a_n}}$ is increasing. So, we see

$$|x_{k,n}|⩽a_{\beta n}+C\frac{a_n}{n}\sqrt{1-\frac{a_{\beta n}}{a_n}}⩽a_{\beta n}+C\frac{a_n}{n}\frac{1}{\sqrt{T(a_n)}}.$$

Therefore, we have

$$\begin{array}{rcl}{a}_{\alpha n}-(a_{\beta n}+C\frac{a_n}{n}\frac{1}{\sqrt{T(a_n)}})& \sim & \frac{a_n}{T(a_n)}-C\frac{a_n}{n}\frac{1}{\sqrt{T(a_n)}}\\ =& \frac{a_n}{T(a_n)}(1-C\frac{\sqrt{T(a_n)}}{n})⩾\frac{a_n}{T(a_n)}(1-C\frac{1}{n^{\epsilon /2}})>0.\end{array}$$

Now, we can show (d). Without loss of generality, we may assume $x\in [x_{j+1,n},x_{j,n}]\subset \{x_{k,n}||x-x_{k,n}|⩽C{\phi }_n(x)\}$. We define

$$x_{k_1,n}:=min\{x_{k,n}||x-x_{k,n}|⩽C{\phi }_n(x)\},x_{k_2,n}:=max\{x_{k,n}||x-x_{k,n}|⩽C{\phi }_n(x)\}.$$

Here we note that $k_1$, $k_2$ are decided depending only on the constant C. Then by former result, we have

$$w(x_{k_1,n})\sim w(x_{k_2,n})\sim w(x)(x_{k_1,n}⩽x⩽x_{k_2,n}).$$

 □

Lemma 3.5 Let $w=exp(-Q)\in \mathcal{F}(C^2+)$. Then we have the following.

• (a) [1, Theorem 1.17] Uniformly for $n⩾1$,

  $$\underset{x\in \mathbb{R}}{sup}|p_n(x)|w(x)|x^2-a_n^2{|}^{\frac{1}{4}}\sim 1.$$

• (b) [1, Theorem 1.19(a)] Uniformly for $n⩾1$ and $1⩽j⩽n$,

  $$|\left(p_n^{\prime }w\right)(x_{j,n})|\sim {\phi }_n^{-1}(x_{j,n})a_n^{-\frac{1}{2}}{(1-\frac{|x_{j,n}|}{a_n})}^{-\frac{1}{4}}.$$

• (c) [1, Theorem 1.19(d)] For $x\in [x_{k+1,n},x_{k,n}]$, if $k⩽n-1$,

  $$|p_n(x)w(x)|\sim min\{|x-x_{k,n}|,|x-x_{k+1,n}|\}{a}_n^{1/2}{\phi }_n{(x)}^{-1}{(1-\frac{|x_{k,n}|}{a_n})}^{-1/4}.$$

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

Let $w\in \mathcal{F}(C^2+)$ and $0<p⩽\mathrm{\infty }$. Then uniformly $n⩾2$,

$${\parallel {\mathrm{\Phi }}^{{(\frac{1}{4}-\frac{1}{p})}^{+}}{p}_{n}w\parallel }_{L_p(\mathbb{R})}⩽C{a}_n^{\frac{1}{p}-\frac{1}{2}}\{\begin{array}{cc}1,\hfill & 0<p<4\mathit{\text{or}}p=\mathrm{\infty };\hfill \\ log(1+n),\hfill & 4⩽p,\hfill \end{array}$$

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

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

Let $f\in L_{p,w}(\mathbb{R})$. The Fourier-type series of f is defined by

$$\tilde{f}(x):=\sum _{k=0}^{\mathrm{\infty }}{a}_k(w^2,f)p_k(w^2,x),a_k(w^2,f):={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)p_k(w^2,t)w^2(t)dt.$$

We denote the partial sum of $\tilde{f}(x)$ by

$$s_n(f,x):=s_n(w^2,f,x):=\sum _{k=0}^{n-1}{a}_k(w^2,f)p_k(w^2,x).$$

The partial sum $s_n(f)$ admits the representation

$$s_n(f,x)=\sum _{j=0}^{n-1}{a}_j{p}_j(x)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)K_n(x,t)w^2(t)dt,$$

where

$$K_n(x,t):=\sum _{j=0}^{n-1}{p}_j(x)p_j(t).$$

The Christoffel-Darboux formula

$$K_n(x,t)=\frac{{\gamma }_{n-1}}{{\gamma }_n}\frac{p_n(x)p_{n-1}(t)-p_{n-1}(x)p_n(t)}{x-t}$$

(3.3)

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

Lemma 3.7 ([6, Lemma 9.2.6])

Let $1<p<\mathrm{\infty }$ and $g\in L_p(\mathbb{R})$. Then for the Hilbert transform

$$H(g,x):=\underset{\epsilon \to 0+}{lim}{\int }_{|x-t|⩾\epsilon }\frac{g(t)}{x-t}dt,x\in \mathbb{R},$$

(3.4)

we have

$${\parallel H(g)\parallel }_{L_p(\mathbb{R})}⩽C{\parallel g\parallel }_{L_p(\mathbb{R})},$$

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

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

Let $w=exp(-Q)\in \mathcal{F}(C^2)$, $1⩽p⩽\mathrm{\infty }$ and $\gamma ⩾0$. Then for any $\epsilon >0$, there exists a polynomial P such that

$${\parallel (f(x)-P(x)){(1+x^2)}^{\gamma }w(x)\parallel }_{L_p(\mathbb{R})}<\epsilon .$$

Lemma 3.9 Let $w\in \mathcal{F}(C^2+)$ 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 $Q^{\prime }(x)/Q(x)=T(x)/x⩾M/x$ for $x⩾x_M$. Hence, we see

$$log\frac{Q(x)}{Q(x_M)}⩾log{\left(\frac{x}{x_M}\right)}^M,x⩾x_M,$$

that is,

$$Q(x)⩾\frac{Q(x_M)}{{(x_M)}^M}{x}^M,x⩾x_M.$$

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 $w\in \mathcal{F}(C^2+)$. 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 $w\in \mathcal{F}(C^2+)$. Let $0<\alpha <\frac{1}{4}$ and

$$\sum _n(x):=\sum _{|x_{k,n}|⩾a_{\alpha n}}|l_{k,n}(x)|w^{-1}(x_{k,n}).$$

Then we have for $|x|⩽a_{\alpha n/2}$ and $|x|⩾a_{2n}$,

$$\sum _n(x)w(x)⩽C.$$

Moreover, for $a_{\alpha n/2}⩽|x|⩽a_{2n}$,

$$\sum _n(x)w(x)⩽C(logn+a_n^{\frac{1}{2}}|p_n(x)w(x)|T^{-\frac{1}{4}}(a_n)).$$

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 $T^{\ast }(x)$ with $T(x)$. □

Lemma 4.2 ([3, Lemma 3.2])

Let $0<\eta <1$. Let $\psi :\mathbb{R}\mapsto (0,\mathrm{\infty })$ be a continuous function with the following property: For $n⩾1$, there exist polynomials $R_n$ of degree ⩽n such that

$$C_1⩽\frac{\psi (t)}{R_n(t)}⩽C_2,|t|⩽a_{4n}.$$

Then for $n⩾n_0$ and $P\in {\mathcal{P}}_n$,

$$\sum _{|x_{k,n}|⩽a_{\eta n}}{\lambda }_{k,n}|P(x_{k,n})|w^{-1}(x_{k,n})\psi (x_{k,n})⩽C{\int }_{-a_{4n}}^{a_{4n}}|P(t)w(t)|\psi (t)dt.$$

Remark 4.3 To prove Lemma 4.7 below, we apply this lemma with $\psi (t)=\varphi (t)={(1+t^2)}^{-\beta /2}$, $\beta >0$. In fact, when ${\varphi }^{\ast }(x)=\varphi (t)$, $t=a_{4n}x$, we can approximate ${\varphi }^{\ast }$ by polynomials $R_n^{\ast }\in {\mathcal{P}}_n$ on $[-1,1]$, that is, for any $\epsilon >0$ there exists $R_n^{\ast }\in {\mathcal{P}}_n$ such that

$$|{\varphi }^{\ast }(x)-R_n^{\ast }(x)|<\epsilon ,x\in [-1,1].$$

Therefore,

$$|\frac{R_n^{\ast }(x)}{{\varphi }^{\ast }(x)}-1|<\frac{\epsilon }{{\varphi }^{\ast }(x)},x\in [-1,1],$$

and so there exist $C_1,C_2>0$ such that

$$C_1⩽1-\frac{\epsilon }{{\varphi }^{\ast }(x)}⩽\left|\frac{R_n^{\ast }(x)}{{\varphi }^{\ast }(x)}\right|<1+\frac{\epsilon }{{\varphi }^{\ast }(x)}⩽C_2,x\in [-1,1].$$

Now, if we set $R_n(t)=R_n^{\ast }(x)$, then we have the result.

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

Let ${\{f_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of measurable functions from $\mathbb{R}\mapsto \mathbb{R}$ such that for $n⩾1$,

$$f_n(x)=0,|x|<a_{\frac{n}{9}};|f_n(x)|w(x)⩽\varphi (x),x\in \mathbb{R}.$$

Then for $1⩽p⩽\mathrm{\infty }$ and $\mathrm{\Delta }>0$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel L_n(f_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(\mathbb{R})}=0.$$

(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

$$|L_n(f_n;x)w(x)|⩽\varphi (a_{\frac{n}{9}})\sum _{|x_{k,n}|⩾a_{\frac{n}{9}}}|l_{k,n}(x)|w^{-1}(x_{k,n})w(x)⩽C_1\varphi (a_{\frac{n}{9}}).$$

So,

$$\begin{array}{rcl}{\parallel L_n(f_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(|x|⩽a_{\frac{n}{18}}\mathrm{o}\mathrm{r}|x|⩾a_{2n})}& ⩽& \varphi (a_{\frac{n}{9}}){\parallel {\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(\mathbb{R})}\\ ⩽& C_2\varphi (a_{\frac{n}{9}})=o(1)\end{array}$$

(4.2)

by Lemma 3.9 (note the definition of $\mathrm{\Phi }(x)$) 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

$$|L_n(f_n;x)w(x)|⩽\varphi (a_{\frac{n}{9}})(logn+a_n^{\frac{1}{2}}|p_n(x)|w(x)T^{-\frac{1}{4}}(a_n)).$$

Also, for this range of x, we see that

$$\mathrm{\Phi }(x)=\frac{1}{{(1+Q(x))}^{\frac{2}{3}}T(x)}\sim \frac{1}{{(1+Q(a_n))}^{\frac{2}{3}}T(a_n)}\sim \frac{T^{\frac{1}{3}}(a_n)}{n^{\frac{2}{3}}T(a_n)}={\delta }_n$$

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

Then since $\mathrm{\Delta }>0$, using Lemma 3.1(a), Lemma 2.1(a), and Lemma 3.6, we have

$$logn{\parallel {\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(a_{\frac{n}{18}}⩽|x|⩽a_{2n})}⩽C{\delta }_n^{\mathrm{\Delta }}{(a_{2n}-a_{\frac{n}{18}})}^{\frac{1}{p}}logn⩽C$$

and

Therefore, we have by (2.4)

$${\parallel L_n(f_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(a_{\frac{n}{18}}⩽|x|⩽a_{2n})}⩽C_4\varphi (a_{\frac{n}{9}})=o(1).$$

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

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

Let $1⩽p⩽\mathrm{\infty }$. Let ${\{g_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of measurable functions from $\mathbb{R}\mapsto \mathbb{R}$ such that for $n⩾1$,

$$g_n(x)=0,|x|⩾a_{\frac{n}{9}};|g_n(x)|w(x)⩽\varphi (x),x\in \mathbb{R}.$$

(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

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel L_n(g_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(|x|⩾a_{\frac{n}{8}})}=0.$$

(4.5)

Proof Using Lemma 3.5(b) and Lemma 3.4(b), we have for $x⩾a_{\frac{n}{8}}$,

$$\begin{array}{rcl}|L_n(g_n;x)|& ⩽& \sum _{|x_{k,n}|⩽a_{\frac{n}{9}}}|l_{k,n}(x)|w^{-1}(x_{k,n})\varphi (x_{k,n})\\ ⩽& C_1{a}_n^{\frac{1}{2}}|p_n(x)|\sum _{|x_{k,n}|⩽a_{\frac{n}{9}}}(x_{k,n}-x_{k+1,n})\frac{{(1-\frac{|x_{k,n}|}{a_n}+{\delta }_n)}^{\frac{1}{4}}}{|x-x_{k,n}|}\varphi (x_{k,n})\\ ⩽& C_2{a}_n^{\frac{1}{2}}|p_n(x)|{\int }_{-a_{\frac{n}{9}}}^{a_{\frac{n}{9}}}\frac{{(1-\frac{|t|}{a_n}+{\delta }_n)}^{\frac{1}{4}}}{|x-t|}\varphi (t)dt.\end{array}$$

(4.6)

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

$$|x-t|\sim |x-x_{k,n}|,t\in [x_{k+1,n},x_{k,n}].$$

(4.7)

Let $|x|⩾a_{\frac{n}{8}}$ and $t\in [x_{k+1,n},x_{k,n}]$. Then

$$|\frac{x-t}{x-x_{k,n}}-1|=\left|\frac{t-x_{k,n}}{x-x_{k,n}}\right|⩽\frac{x_{k,n}-x_{k+1,n}}{|x_{k\pm 2,n}-x_{k,n}|}⩽c<1.$$

Now, we use the fact that $x+C\phi (x)$, $x>0$ is increasing for $0<x⩽a_{n/2}$, and then

$$x_{k,n}+C{\phi }_n(x_{k,n})⩽a_{\frac{n}{9}}+C{\phi }_n(a_{\frac{n}{9}})⩽a_{\frac{n}{8}}⩽x.$$

Here, the second inequality follows from the definition of ${\phi }_n(x)$ and Lemma 3.1(a), (b). Hence, we have (4.7). Now, we use the monotonicity of ${(1-\frac{|x|}{a_n}+{\delta }_n)}^{\frac{1}{4}}\varphi (x)$. From (4.7) there exists $C>0$ such that for $t\in [x_{k+1,n},x_{k,n}]$,

$$\begin{array}{rcl}(x_{k,n}-x_{k+1,n})\frac{{(1-\frac{|x_{k,n}|}{a_n}+{\delta }_n)}^{\frac{1}{4}}}{|x-x_{k,n}|}\varphi (x_{k,n})& ⩽& {\int }_{x_{k+1,n}}^{x_{k,n}}\frac{{(1-\frac{|t|}{a_n}+{\delta }_n)}^{\frac{1}{4}}}{|x-x_{k,n}|}\varphi (t)dt\\ ⩽& \frac{1}{C}{\int }_{x_{k+1,n}}^{x_{k,n}}\frac{{(1-\frac{|t|}{a_n}+{\delta }_n)}^{\frac{1}{4}}}{|x-t|}\varphi (t)dt.\end{array}$$

Hence, (4.6) holds. Next, for $t\in [0,a_{\frac{n}{9}}]$ and $x⩾a_{\frac{n}{8}}$, we know by Lemma 3.1(a),

$$1⩽\frac{a_n-t}{x-t}⩽1+\frac{a_n-a_{\frac{n}{8}}}{a_{\frac{n}{8}}-t}⩽1+\frac{a_n-a_{\frac{n}{8}}}{a_{\frac{n}{8}}-a_{\frac{n}{9}}}⩽1+C\frac{a_{\frac{n}{8}}}{a_{\frac{n}{9}}}\frac{T(a_{\frac{n}{9}})}{T(a_{\frac{n}{8}})}⩽C_3$$

and

$$1-\frac{|t|}{a_n}⩾C_4\frac{1}{T(a_n)}⩾{\delta }_n.$$

So, we have

$$|L_n(g_n;x)|⩽C_6{a}_n^{\frac{1}{4}}|p_n(x)|{\int }_0^{a_{\frac{n}{9}}}{(x-t)}^{-\frac{3}{4}}\varphi (t)dt.$$

Let $t=a_s$, $\frac{n}{9}⩾s⩾1$. Then, since we know for $x⩾a_{\frac{n}{8}}$,

$$x-t=x(1-\frac{t}{x})⩾a_{\frac{n}{8}}(1-\frac{a_s}{a_{\frac{9}{8}s}})⩾C_7\frac{a_n}{T(a_s)},$$

we obtain

$$|L_n(g_n;x)|⩽C_8{a}_n^{-\frac{1}{2}}|p_n(x)|{\int }_0^{a_{\frac{n}{9}}}{T}^{\frac{3}{4}}(t)\varphi (t)dt⩽C_8{a}_n^{\frac{1}{2}}{T}^{\frac{3}{4}}(a_n)|p_n(x)|.$$

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 $\mathrm{\Delta }>\frac{9}{4}\frac{\lambda -1}{3\lambda -1}$,

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel L_n(g_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(|x|⩾a_{\frac{n}{8}})}=0.$$

 □

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

Let $1<p<\mathrm{\infty }$. Let $\sigma :\mathbb{R}\mapsto \mathbb{R}$ be a bounded measurable function. Let $\lambda =\lambda (b)⩾1$ be defined in Lemma 2.1, and then we suppose

$$\mathrm{\Delta }>\{\begin{array}{cc}0,\hfill & 1<p⩽2;\hfill \\ \frac{3}{2}\frac{(\lambda -1)}{3\lambda -1}\frac{p-2}{p},\hfill & 2<p⩽4;\hfill \\ max\{\frac{\lambda -1}{3\lambda -1}\frac{p-1}{p}-\frac{1}{4}\frac{\lambda +1}{3\lambda -1}\frac{p-4}{p},0\},\hfill & 4<p.\hfill \end{array}$$

(4.8)

Then for $1<p<\mathrm{\infty }$ and the partial sum $s_n$ of the Fourier series, we have

$${\parallel s_n\left[\sigma \varphi w^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(|x|⩽a_{\frac{n}{8}})}⩽C{\parallel \sigma \parallel }_{L_{\mathrm{\infty }}(\mathbb{R})}$$

(4.9)

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

Proof We may suppose that ${\parallel \sigma \parallel }_{L_{\mathrm{\infty }}(\mathbb{R})}=1$. By (3.3), (3.4) and Lemma 3.5(a),

$$|s_n\left[\sigma \varphi w^{-1}\right](x)|w(x)⩽a_n^{\frac{1}{2}}{(1-\frac{|x|}{a_n})}^{-\frac{1}{4}}\sum _{j=n-1}^n|H[\sigma \varphi p_{j}w](x)|.$$

(4.10)

Let us choose $l:=l(n)$ such that $2^l⩽\frac{n}{8}⩽2^{l+1}$. Then we know

$$2^{l+3}⩽n⩽2^{l+4}.$$

(4.11)

Define

$${\mathcal{I}}_k=[a_{2^k},a_{2^{k+1}}],1⩽k⩽l+2.$$

For $j=n-1,n$ and $x\in {\mathcal{I}}_k$, we split

$$\begin{array}{rcl}H[\sigma \varphi p_{j}w](x)w(x)& =& ({\int }_{-\mathrm{\infty }}^0+{\int }_0^{a_{2^{k-1}}}+\mathit{P}.\mathit{V}.{\int }_{a_{2^{k-1}}}^{a_{2^{k+2}}}+{\int }_{a_{2^{k+2}}}^{\mathrm{\infty }})\frac{(\sigma \varphi p_{j}w)(t)}{x-t}dt\\ :=& I_1(x)+I_2(x)+I_3(x)+I_4(x).\end{array}$$

(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 $p=1$,

$$\begin{array}{rcl}|I_1(x)|& ⩽& {\int }_0^{\mathrm{\infty }}\frac{|(p_{j}w\varphi )(-t)|}{t+x}⩽C_1{a}_n^{-\frac{1}{2}}{\int }_0^{\frac{a_n}{2}}\frac{\varphi (t)}{t+a_2}dt+C_2{a}_n^{-1}{\int }_{\frac{a_n}{2}}^{\mathrm{\infty }}|p_j(t)|w(t)dt\\ ⩽& C_2(a_n^{-\frac{1}{2}}+a_n^{-1}{a}_n^{1-\frac{1}{2}})⩽C_3{a}_n^{-\frac{1}{2}}.\end{array}$$

(4.13)

Here we have used

$${\int }_0^{\mathrm{\infty }}\frac{\varphi (t)}{1+t}dt<\mathrm{\infty }.$$

(4.14)

By Lemma 3.5(a), and noting $1-x/a_n⩽1-t/a_n$ for $x\in {\mathcal{I}}_k$,

$$\begin{array}{rcl}|I_2(x)|& ⩽& {\int }_0^{a_{2^{k-1}}}\frac{|(p_{j}w\varphi )(t)|}{x-t}dt⩽C_4{a}_n^{-\frac{1}{2}}{\int }_0^{a_{2^{k-1}}}\frac{{(1-\frac{t}{a_n})}^{-\frac{1}{4}}}{x-t}dt\\ ⩽& C_4{a}_n^{-\frac{1}{2}}{(1-\frac{x}{a_n})}^{-\frac{1}{4}}{\int }_0^{a_{2^{k-1}}}\frac{dt}{x-t}\\ =& C_4{a}_n^{-\frac{1}{2}}{(1-\frac{x}{a_n})}^{-\frac{1}{4}}log{(1-\frac{a_{2^{k-1}}}{x})}^{-1}.\end{array}$$

Using

$$1-\frac{a_{2^{k-1}}}{x}⩾1-\frac{a_{2^{k-1}}}{a_{2^k}}⩾C\frac{1}{T(a_{2^k})}⩾C\frac{1}{T(x)},$$

we can see

$$|I_2(x)|⩽C_6{a}_n^{-\frac{1}{2}}{(1-\frac{x}{a_n})}^{-\frac{1}{4}}log\left(\frac{T(x)}{C}\right).$$

(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,

$$\begin{array}{rcl}|I_4(x)|& ⩽& {\int }_{a_{2^{k+2}}}^{2a_{2^{k+2}}}\frac{|(p_{j}w\varphi )(t)|}{t-x}dt+C{\int }_{2a_{2^{k+2}}}^{\mathrm{\infty }}\frac{|(p_{j}w\varphi )(t)|}{t}dt(\text{by }t⩽2(t-x))\\ ⩽& C_7\left(a_n^{-\frac{1}{2}}{\int }_{a_{2^{k+2}}}^{2a_{2^{k+2}}}\right|1-\frac{t}{a_n}{|}^{-\frac{1}{4}}\frac{dt}{t-x}\\ +a_n^{-\frac{1}{2}}{\int }_{2a_{2^{k+2}}}^{max\{2a_{2^{k+2}},\frac{1}{2}{a}_n\}}\frac{\varphi (t)}{t}dt+{\int }_{\frac{1}{2}{a}_n}^{\mathrm{\infty }}\frac{|(p_{j}w)(t)|}{t}dt)\\ ⩽& C_7\left(a_n^{-\frac{1}{2}}{\int }_{a_{2^{k+2}}}^{2a_{2^{k+2}}}\right|1-\frac{t}{a_n}{|}^{-\frac{1}{4}}\frac{dt}{t-x}+C{a}_n^{-\frac{1}{2}}+a_n^{-1}{a}_n^{1-\frac{1}{2}})\\ (\text{by (4.14) and Lemma 3.6 with }p=1)\\ ⩽& C_8{a}_n^{-\frac{1}{2}}[J+1],\end{array}$$

(4.16)

where

$$J:={\int }_{a_{2^{k+2}}}^{2a_{2^{k+2}}}|1-\frac{t}{a_n}{|}^{-\frac{1}{4}}\frac{dt}{t-x}.$$

Since, if

$$|1-\frac{t}{a_n}|⩽\frac{1}{2}(1-\frac{x}{a_n}),$$

then we see

$$|t-x|=a_n|(1-\frac{x}{a_n})-(1-\frac{t}{a_n})|⩾\frac{a_n}{2}(1-\frac{x}{a_n}).$$

Now, we have

$$\begin{array}{lll}J& ⩽& C_9({(1-\frac{x}{a_n})}^{-\frac{1}{4}}\underset{t\in [a_{2^{k+2}},2a_{2^{k+2}}]}{{\int }_{|1-\frac{t}{a_n}|⩾\frac{1}{2}(1-\frac{x}{a_n}),}}\frac{1}{t-x}dt\\ +a_n^{-1}{(1-\frac{x}{a_n})}^{-1}\underset{t\in [a_{2^{k+2}},2a_{2^{k+2}}]}{{\int }_{|1-\frac{t}{a_n}|⩽\frac{1}{2}(1-\frac{x}{a_n}),}}|1-\frac{t}{a_n}{|}^{-\frac{1}{4}}dt)\\ ⩽& C_{10}({(1-\frac{x}{a_n})}^{-\frac{1}{4}}log(1+\frac{a_{2^{k+2}}}{a_{2^{k+2}}-a_{2^{k+1}}})\\ +{(1-\frac{x}{a_n})}^{-1}{\int }_{|1-s|⩽\frac{1}{2}(1-\frac{x}{a_n})}{|1-s|}^{-\frac{1}{4}}ds)\\ ⩽& C_{10}({(1-\frac{x}{a_n})}^{-\frac{1}{4}}log(1+CT(a_{2^{k+2}}))+\frac{4}{3}{\left(\frac{1}{2}(1-\frac{x}{a_n})\right)}^{\frac{3}{4}}{(1-\frac{x}{a_n})}^{-1})\\ ⩽& C_{11}{(1-\frac{x}{a_n})}^{-\frac{1}{4}}log(CT(x)).\end{array}$$

So, from (4.16) we have

$$|I_4(x)|⩽C_{12}{a}_n^{-\frac{1}{2}}{(1-\frac{x}{a_n})}^{-\frac{1}{4}}log(CT(x)).$$

(4.17)

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

$$|I_1+I_2+I_4|⩽C_{13}{a}_n^{-\frac{1}{2}}{(1-\frac{x}{a_n})}^{-\frac{1}{4}}log(CT(x)).$$

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

(4.18)

We must estimate the $L_p$-norm with respect to $I_3$, that is, ${\parallel \mathit{P}.\mathit{V}.{\int }_{a_{2^{k-1}}}^{a_{2^{k+2}}}\frac{(\sigma \varphi p_{j}w)(t)}{x-t}dt\parallel }_{L_p({\mathcal{I}}_k)}$. We use M. Riesz’s theorem on the boundedness of the Hilbert transform from $L_p(\mathbb{R})$ to $L_p(\mathbb{R})$ (Lemma 3.7) to deduce that by Lemma 3.5(a) and the boundedness of $|\sigma \varphi |$,

$$\begin{array}{rcl}{\parallel \mathit{P}.\mathit{V}.{\int }_{a_{2^{k-1}}}^{a_{2^{k+2}}}\frac{(\sigma \varphi p_{j}w)(t)}{x-t}dt\parallel }_{L_p({\mathcal{I}}_k)}& ⩽& C_{15}{\left({\int }_{a_{2^{k-1}}}^{a_{2^{k+2}}}{|(\sigma \varphi p_{j}w)(t)|}^{p}dt\right)}^{\frac{1}{p}}\\ ⩽& C_{16}{a}_n^{-\frac{1}{2}}{(1-\frac{a_{2^{k+2}}}{a_n})}^{-\frac{1}{4}}{(a_{2^{k+2}}-a_{2^{k-1}})}^{\frac{1}{p}}.\end{array}$$

(4.19)

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

(4.20)

Noting (4.11), we see $n⩾2^{l+3}$ for $k⩽l$, so

$$1-\frac{a_{2^{k+1}}}{a_n}⩾1-\frac{a_{2^{k+1}}}{a_{2^{k+3}}}⩾C_{19}\frac{1}{T(a_{2^k})}\text{and}{a}_{2^{k+1}}-a_{2^k}⩽C_{20}\frac{a_{2^k}}{T(a_{2^k})}.$$

On the other hand, using Lemma 3.2(b), we see $\mathrm{\Phi }(a_t)\sim {\delta }_t$. Hence, we have

$$\begin{array}{rcl}{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}(a_{2^k})& \sim & {\delta }_{2^k}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}={\left(\frac{1}{2^{k}T(a_{2^k})}\right)}^{\frac{2}{3}(\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+})}\\ =& \{\begin{array}{cc}{(\frac{1}{2^{k}T(a_{2^k})})}^{\frac{2}{3}\mathrm{\Delta }},\hfill & 0<p⩽4;\hfill \\ {(\frac{1}{2^{k}T(a_{2^k})})}^{\frac{2}{3}(\mathrm{\Delta }+(\frac{1}{4}-\frac{1}{p}))},\hfill & 4<p.\hfill \end{array}\end{array}$$

Hence, from (4.20) we have

From Lemma 2.1 (2.2), we know

$$T^{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p}}(a_{2^k})⩽C_{1}C(\lambda ,\eta ){\left(2^k\right)}^{\frac{2(\eta +\lambda -1)}{\lambda +1}max\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p},0\}},$$

and

$$T^{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}(1-\frac{1}{p})}(a_{2^k})⩽C_{2}C(\lambda ,\eta ){\left(2^k\right)}^{\frac{2(\eta +\lambda -1)}{\lambda +1}max\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}(1-\frac{1}{p}),0\}}.$$

Therefore, we continue with Lemma 2.1(a) as

$$\begin{array}{rc}⩽& C_{20}C(\lambda ,\eta )log(CT(a_{2^{k+1}}))\\ \times \{\begin{array}{cc}{(\frac{1}{2^k})}^{\frac{2}{3}\mathrm{\Delta }-\frac{\eta }{p}-\frac{2(\eta +\lambda -1)}{\lambda +1}max\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p},0\}},\hfill & 1<p⩽4;\hfill \\ {(\frac{1}{2^k})}^{\frac{2}{3}(\mathrm{\Delta }+(\frac{1}{4}-\frac{1}{p}))-\frac{\eta }{p}-\frac{2(\eta +\lambda -1)}{\lambda +1}max\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}(1-\frac{1}{p}),0\}},\hfill & 4<p.\hfill \end{array}\end{array}$$

(4.21)

First, let $1<p⩽4$. Then (4.8), that is,

$$\mathrm{\Delta }>\{\begin{array}{cc}0,\hfill & 1<p⩽2;\hfill \\ \frac{3}{2}\frac{\lambda -1}{3\lambda -1}\frac{p-2}{p},\hfill & 2<p⩽4\hfill \end{array}$$

implies

$$\mathrm{\Delta }>\frac{3}{2}\frac{\lambda -1}{3\lambda -1}\frac{p-2}{p}\text{and}\mathrm{\Delta }>0$$

iff

$$\frac{2}{3}\mathrm{\Delta }-\frac{2(\lambda -1)}{\lambda +1}(-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p})>0\text{and}\mathrm{\Delta }>0$$

iff

$$\frac{2}{3}\mathrm{\Delta }-\frac{2(\lambda -1)}{\lambda +1}max\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p},0\}>0.$$

This means that there exists a positive constant ${\eta }_1>0$ small enough such that

$$A({\eta }_1):=\frac{2}{3}\mathrm{\Delta }-\frac{{\eta }_1}{p}-\frac{2({\eta }_1+\lambda -1)}{\lambda +1}max\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{2}-\frac{1}{p},0\}>0.$$

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

$$\mathrm{\Delta }>\frac{\lambda -1}{3\lambda -1}(1-\frac{1}{p})-\frac{\lambda +1}{3\lambda -1}(\frac{1}{4}-\frac{1}{p})\text{and}\mathrm{\Delta }+\frac{1}{4}-\frac{1}{p}>0$$

iff

$$\frac{2}{3}(\mathrm{\Delta }+(\frac{1}{4}-\frac{1}{p}))-\frac{2(\lambda -1)}{\lambda +1}(-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}(1-\frac{1}{p}))>0$$

and

$$\frac{2}{3}(\mathrm{\Delta }+(\frac{1}{4}-\frac{1}{p}))>0$$

iff

$$\frac{2}{3}(\mathrm{\Delta }+(\frac{1}{4}-\frac{1}{p}))-\frac{2(\lambda -1)}{\lambda +1}max\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}(1-\frac{1}{p}),0\}>0.$$

Similarly to the previous case, this means that there exists a positive constant ${\eta }_2>0$ small enough such that

$$B({\eta }_2):=\frac{2}{3}(\mathrm{\Delta }+(\frac{1}{4}-\frac{1}{p}))-\frac{{\eta }_2}{p}-\frac{2({\eta }_2+\lambda -1)}{\lambda +1}max\{-\frac{2}{3}\mathrm{\Delta }+\frac{1}{3}(1-\frac{1}{p}),0\}>0.$$

Now, we estimate $I_{p,k}$. From (4.21), we have

For $\eta >0$ small enough, we can see $A(\eta )>A({\eta }_1)>0$ and $B(\eta )>B({\eta }_2)>0$. Let $\tau :=min\{A({\eta }_1),B({\eta }_2)\}/2$. Then for small enough $\eta >0$, we have

$$\begin{array}{rcl}{\parallel s_n\left[\sigma \varphi w^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p({\mathcal{I}}_k)}& ⩽& C_{20}C(\lambda ,\eta )log(CT(a_{2^{k+1}})){\left(\frac{1}{2^k}\right)}^{2\tau }\\ ⩽& C_{21}C(\lambda ,\eta ){\left(\frac{1}{2^k}\right)}^{\tau },\end{array}$$

because we see that for all $k>0$,

$$log(CT(a_{2^{k+1}})){\left(\frac{1}{2^k}\right)}^{\tau }<C_{22}.$$

Therefore, under the conditions (4.8) we have

$$\begin{array}{rcl}{\parallel s_n\left[\sigma \varphi w^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(a_2⩽|x|⩽a_{\frac{n}{8}})}^p& ⩽& \sum _{k=1}^l{\parallel s_n\left[\sigma \varphi w^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p({\mathcal{I}}_k)}^p\\ ⩽& C_{21}C(\lambda ,\eta )\sum _{k=1}^l{\left(\frac{1}{2^k}\right)}^{\tau }⩽C_{23}C(\lambda ,\eta ).\end{array}$$

(4.22)

The estimation of

$${\parallel s_n\left[\sigma \varphi w^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(|x|⩽a_2)}^p$$

is similar. In fact, for $x\in [-a_2,a_2]$, we split

$$H[\sigma \varphi p_{j}w](x)=({\int }_{-\mathrm{\infty }}^{-2a_2}+\mathit{P}.\mathit{V}.{\int }_{-2a_2}^{2a_2}+{\int }_{2a_2}^{\mathrm{\infty }})\frac{(\sigma \varphi p_{j}w)(t)}{x-t}dt.$$

Here we see that

$$\begin{array}{rcl}\left|{\int }_{-\mathrm{\infty }}^{-2a_2}\frac{(\sigma \varphi p_{j}w)(t)}{x-t}dt\right|& =& \left|{\int }_{2a_2}^{\mathrm{\infty }}\frac{(\sigma \varphi p_{j}w)(-t)}{x+t}dt\right|⩽\left|{\int }_{2a_2}^{\mathrm{\infty }}\frac{(\sigma \varphi p_{j}w)(-t)}{t-a_2}dt\right|\\ =& \left|{\int }_0^{\mathrm{\infty }}\frac{(\sigma \varphi p_{j}w)(-s-2a_2)}{s+a_2}dt\right|\end{array}$$

and

$$\begin{array}{rcl}\left|{\int }_{2a_2}^{\mathrm{\infty }}\frac{(\sigma \varphi p_{j}w)(t)}{x-t}dt\right|& =& \left|{\int }_{2a_2}^{\mathrm{\infty }}\frac{(\sigma \varphi p_{j}w)(t)}{t-x}dt\right|⩽\left|{\int }_{2a_2}^{\mathrm{\infty }}\frac{(\sigma \varphi p_{j}w)(t)}{t-a_2}dt\right|\\ =& \left|{\int }_0^{\mathrm{\infty }}\frac{(\sigma \varphi p_{j}w)(s+2a_2)}{s+a_2}ds\right|.\end{array}$$

So, we can estimate ${\int }_{-\mathrm{\infty }}^{-2a_2}$ and ${\int }_{2a_2}^{\mathrm{\infty }}$ as we did $I_1$ before (see (4.12)). We can estimate the second integral as follows: By M. Riesz’s theorem,

$${\parallel \mathit{P}.\mathit{V}.{\int }_{-2a_2}^{2a_2}\frac{(\sigma \varphi p_{j}w)(t)}{x-t}dt\parallel }_{L_p(|t|⩽2a_2)}^p⩽C{\int }_{-2a_2}^{2a_2}{|(\sigma \varphi p_{j}w)(t)|}^{p}dt⩽C{a}_n^{-\frac{p}{2}}⩽C.$$

Now, under the assumption (4.8), we can select ${\eta }_0>0$ small enough such that

$$\mathrm{\Delta }>\{\begin{array}{cc}0,\hfill & 1<p⩽2;\hfill \\ \frac{3}{2}\frac{\lambda +{\eta }_0-1}{3\lambda +2{\eta }_0-1}\frac{p-2}{p},\hfill & 2<p⩽4;\hfill \\ max\{\frac{\lambda +{\eta }_0-1}{3\lambda +2{\eta }_0-1}\frac{p-1}{p}-\frac{1}{4}\frac{\lambda +1}{3\lambda +2{\eta }_0-1}\frac{p-4}{p},0\},\hfill & 4<p.\hfill \end{array}$$

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 $L_n(g_n)$ over $[-a_{\alpha n},a_{\alpha n}]$.

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

Let $1<p<\mathrm{\infty }$ and $0<\epsilon <1$. Let $\{g_n\}$ be as in Lemma 4.4, but we exchange (4.3) with

$$|g_n(x)w(x)|⩽\epsilon \varphi (x),x\in \mathbb{R},n⩾1.$$

Then for $1<p<\mathrm{\infty }$,

$$lim\underset{n\to \mathrm{\infty }}{sup}{\parallel L_n(g_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(|x|⩽a_{\frac{n}{8}})}⩽C\epsilon .$$

Proof Let

$${\chi }_n:={\chi }_{[-a_{\frac{n}{8}},a_{\frac{n}{8}}]};h_n:=sign(L_n(g_n)){|L_n(g_n)|}^{p-1}{\chi }_n{w}^{p-2}{\mathrm{\Phi }}^{(\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+})p}$$

and

$${\sigma }_n:=sign{s}_n[h_n].$$

We shall show that

$${\parallel L_n(g_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(|x|⩽a_{\frac{n}{8}})}⩽\epsilon {\parallel s_n\left[{\sigma }_n\varphi w^{-1}\right]w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(|x|⩽a_{\frac{n}{8}})}.$$

(4.23)

Then from Lemma 4.5 we will conclude (4.22). Using orthogonality of $f-s_n[f]$ 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

$$\begin{array}{rc}⩽& C\epsilon {\int }_{\mathbb{R}}|s_n[h_n](x)|\varphi (x)w(x)dx\\ =& C\epsilon {\int }_{\mathbb{R}}{s}_n[h_n](x){\sigma }_n\varphi (x)w^{-1}(x)w^2(x)dx=C\epsilon {\int }_{\mathbb{R}}{h}_n(x)s_n\left[{\sigma }_n\varphi w^{-1}\right](x)w^2(x)dx\\ =& C\epsilon {\int }_{-a_{\frac{n}{8}}}^{a_{\frac{n}{8}}}{h}_n(x)s_n\left[{\sigma }_n\varphi w^{-1}\right](x)w^2(x)dx.\end{array}$$

Using Hölder’s inequality with $q=p/(p-1)$, we continue this as

Cancellation of ${\parallel L_n(g_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(|x|⩽a_{\frac{n}{8}})}^{p-1}$ gives (4.23). □

Proof of Theorem 2.2 In proving the theorem, we split our functions into pieces that vanish inside or outside $[-a_{\frac{n}{9}},a_{\frac{n}{9}}]$. Throughout, we let ${\chi }_S$ denote the characteristic function of a set S. Also, we set for some fixed $\beta >0$,

$$\varphi (x)={(1+x^2)}^{-\beta /2},$$

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

$${\parallel (f-P)w{\varphi }^{-1}\parallel }_{L_{\mathrm{\infty }}(\mathbb{R})}⩽\epsilon$$

(see Lemma 3.8). Then we have

(4.24)

Here we used that

$${\parallel \varphi {\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(\mathbb{R})}<\mathrm{\infty },$$

because $\mathrm{\Delta }>0$ and ${\mathrm{\Phi }}^{-1}$ grows faster than any power of x (see Lemma 3.9). Next, let

$${\chi }_n:=\chi [-a_{\frac{n}{9}},a_{\frac{n}{9}}],$$

and write

$$P-f=(P-f){\chi }_n+(P-f)(1-{\chi }_n)=:g_n+f_n.$$

By Lemma 4.4 we have

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel L_n(f_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{4})}^{+}}\parallel }_{L_p(\mathbb{R})}=0.$$

By Lemma 4.5 we have

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel L_n(g_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{4})}^{+}}\parallel }_{L_p(|x|⩾a_{\frac{n}{8}})}=0,$$

and by Lemma 4.7,

$$lim\underset{n\to \mathrm{\infty }}{sup}{\parallel L_n(g_n)w{\mathrm{\Phi }}^{\mathrm{\Delta }+{(\frac{1}{4}-\frac{1}{p})}^{+}}\parallel }_{L_p(|x|⩽a_{\frac{n}{8}})}⩽C\epsilon .$$

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 $w\in \mathcal{F}(C^2+)$. Let $0<\alpha <\frac{1}{4}$ and

$$\sum _n(x):=\sum _{|x_{k,n}|⩾a_{\alpha n}}|l_{k,n}(x)|w^{-1}(x_{k,n}).$$

Then we have for $x\in \mathbb{R}$,

$$\sum _n(x)w(x){\mathrm{\Phi }}^{1/4}(x)⩽Clogn.$$

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

Lemma 5.2 Let $w\in \mathcal{F}(C^2+)$. Let $0<\alpha <\frac{1}{4}$ and

$${\sum _n}^{\prime }(x):=\sum _{|x_{k,n}|⩽a_{\alpha n}}|l_{k,n}(x)|w^{-1}(x_{k,n}).$$

Then we have

$${\sum _n}^{\prime }(x)w(x)\mathrm{\Phi }{(x)}^{3/4}⩽Clogn.$$

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

$$\begin{array}{rcl}{\sum _n}^{\prime }(x)& =& \sum _{|x_{k,n}|⩽a_{\alpha n}}|l_{k,n}(x)|w^{-1}(x_{k,n})\\ =& \frac{|p_n(x)|}{|x-x_{j_x,n}||P_n^{\prime }(x_{j_x,n})|w(x_{j_x,n})}+\underset{k\ne j_x}{\sum _{|x_{k,n}|⩽a_{\alpha n},}}\frac{|p_n(x)|}{|x-x_{k,n}||P_n^{\prime }(x_{k,n})|w(x_{k,n})}\\ ⩽& Cw{(x)}^{-1}+a_n^{1/2}|p_n(x)|\underset{k\ne j_x}{\sum _{|x_{k,n}|⩽a_{\alpha n},}}\frac{{\phi }_n(x_{k,n})(1-\frac{|x_{k,n}|}{a_n})}{|x-x_{k,n}|}\\ \sim & Cw{(x)}^{-1}+\frac{a_n^{3/2}}{n}|p_n(x)|\underset{k\ne j_x}{\sum _{|x_{k,n}|⩽a_{\alpha n},}}\frac{1-\frac{|x_{k,n}|}{a_{2n}}}{\sqrt{1-\frac{|x_{k,n}|}{a_n}}}{(1-\frac{|x_{k,n}|}{a_n})}^{1/4}\frac{1}{|x-x_{k,n}|}\\ \sim & Cw{(x)}^{-1}+\frac{a_n^{3/2}}{n}|p_n(x)|\underset{k\ne j_x}{\sum _{|x_{k,n}|⩽a_{\alpha n},}}{(1-\frac{|x_{k,n}|}{a_n})}^{3/4}\frac{1}{|x-x_{k,n}|},\end{array}$$

where we used the fact

$$1-\frac{|x_{k,n}|}{a_{2n}}\sim 1-\frac{|x_{k,n}|}{a_n},|x_{k,n}|⩽a_{\alpha n}.$$

So,

$$\begin{array}{rcl}{\sum _n}^{\prime }(x)& ⩽& Cw{(x)}^{-1}+\frac{a_n^{3/2}}{n}|p_n(x)|\underset{k\ne j_x}{\sum _{|x_{k,n}|⩽a_{\alpha n},}}{(1-\frac{|x_{k,n}|}{a_n})}^{3/4}\frac{1}{|x_{j_x,n}-x_{k,n}|}\\ ⩽& Cw{(x)}^{-1}+\frac{a_n^{3/2}}{n}|p_n(x)|\underset{k\ne j_x}{\sum _{|x_{k,n}|⩽a_{\alpha n},}}{(1-\frac{|x_{k,n}|}{a_n})}^{3/4}\frac{1}{{\sum }_{j_x\lessgtr i\lessgtr k}{\phi }_n(x_{i,n})}\\ ⩽& Cw{(x)}^{-1}+a_n^{1/2}|p_n(x)|\underset{k\ne j_x}{\sum _{|x_{k,n}|⩽a_{\alpha n},}}{(1-\frac{|x_{k,n}|}{a_n})}^{3/4}\frac{1}{{\sum }_{j_x\lessgtr i\lessgtr k}\sqrt{1-|x_{i,n}|/a_n}}.\end{array}$$

Therefore we have by Lemma 3.6 with $p=\mathrm{\infty }$,

$$\begin{array}{rcl}{\sum _n}^{\prime }(x)w(x)\mathrm{\Phi }{(x)}^{3/4}& ⩽& C+C{a}_n^{1/2}|p_n(x)|w(x)\mathrm{\Phi }{(x)}^{1/4}\\ \times \underset{k\ne j_x}{\sum _{|x_{k,n}|⩽a_{\alpha n},}}{(1-\frac{|x_{k,n}|}{a_n})}^{3/4}{(1-\frac{|x_{j_x,n}|}{a_n})}^{1/2}\frac{1}{{\sum }_{j_x\lessgtr i\lessgtr k}\sqrt{1-|x_{i,n}|/a_n}}\\ ⩽& C\underset{k\ne j_x}{\sum _{|x_{k,n}|⩽a_{\alpha n},}}\frac{1}{|j_x-k|}\sim logn.\end{array}$$

 □

Lemma 5.3 ([8, Theorem 1])

Let $w\in \mathcal{F}(C^2+)$. Then there exists a constant $C_0>0$ such that for every absolutely continuous function f with $w{f}^{\prime }\in C_0(\mathbb{R})$ (this means $w(x)f^{\prime }(x)\to 0$ as $|x|\to \mathrm{\infty }$) and every $n\in \mathbb{N}$, we have

$$E_n(w;f)⩽C\frac{a_n}{n}{E}_{n-1}(w;f^{\prime }).$$

Proof of Theorem 2.4 There exists $P_{n-1}\in {\mathcal{P}}_n$ such that

$$|(f(x)-P_{n-1}(x))w(x)|⩽2E_{n-1}(w;f).$$

Therefore, by Lemma 5.1 and Lemma 5.2,

Let $w{f}^{(r)}\in C_0(\mathbb{R})$. If we repeatedly use Lemma 5.3, then we have

$$|(f(x)-L_n(f)(x))w(x){\mathrm{\Phi }}^{3/4}(x)|⩽C_r{\left(\frac{a_n}{n}\right)}^r{E}_{n-r-1}(w;f^{(r)})logn.$$

 □