# Local saturation of a positive linear convolution operator

## Abstract

Let $\left\{{H}_{n}\left(t\right)\right\}$ be a sequence of non-negative, even, and continuous functions on . In this paper, we consider a convolution operator ${J}_{n}\left(f;x\right)={\int }_{0}^{\mathrm{\infty }}f\left(t\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt$, $f\in {L}_{p}\left({\mathbb{R}}^{+}\right)$, and then investigate the local saturation of ${J}_{n}\left(f;x\right)$.

MSC:44A35.

## 1 Introduction and theorems

Through this paper we let $\left\{{H}_{n}\left(t\right)\right\}$, $n=1,2,\dots$ , be a sequence of non-negative, even, and continuous functions on $\mathbb{R}:=\left(-\mathrm{\infty },\mathrm{\infty }\right)$, and there exist $M,N>0$, and $T>0$ such that $\left\{{H}_{n}\left(t\right)\right\}$ satisfied uniformly

$\underset{|t|⩾T}{sup}{H}_{n}\left(t\right)⩽M,\phantom{\rule{1em}{0ex}}n⩾N,$
(1.1)
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{H}_{n}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=1,\phantom{\rule{1em}{0ex}}n=1,2,\dots ,$
(1.2)
(1.3)

and there exist two positive constants α, β with $2\beta \ge \alpha >3$ such that uniformly for n,

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|t|}^{\alpha }{H}_{n}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\le C{\mu }_{n}^{\beta }.$
(1.4)

As an example for ${H}_{n}\left(t\right)$ we give the following:

Let

${H}_{n}\left(t\right)=\frac{n}{\sqrt{\pi }}{e}^{-{\left(nt\right)}^{2}}.$

Then it satisfies (1.2),

Let us denote ${\mathbb{R}}^{+}:=\left[0,\mathrm{\infty }\right)$. In what follows we assume that ${H}_{n}\left(t\right)$ satisfies the conditions (1.1)-(1.4). Using ${H}_{n}$, we define the convolution operators for $f\in {L}_{p}\left({\mathbb{R}}^{+}\right)$,

${J}_{n}\left(f;x\right):={\int }_{0}^{\mathrm{\infty }}f\left(t\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}n=1,2,\dots .$
(1.5)

Swetits and Wood  studied the operators on a finite interval $\left[0,r\right]$;

${K}_{n}\left(f;x\right):={\int }_{0}^{r}f\left(t\right){H}_{n}^{\ast }\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}f\in {L}_{p}\left(\left[0,r\right]\right),n=1,2,\dots ,$

where ${H}_{n}^{\ast }$ is defined on $\left[-r,r\right]$ as ${H}_{n}$, and then they gave a local saturation theorem. Furthermore, there is a rich bibliography concerning the convergence of positive linear operators on $\left[0,\mathrm{\infty }\right)$ (e.g. see  and the references cited therein).

In this paper we extend  to the infinite interval ${\mathbb{R}}^{+}$. Then we use a similar methods as . For $1 and $c⩾0$, we define ${L}_{p}^{2}\left(\left[c,\mathrm{\infty }\right)\right)$ as the space of those functions f such that $f\in {L}_{p}\left({\mathbb{R}}^{+}\right)$ and ${f}^{\prime }$ is a locally absolutely continuous function on $\left[c,\mathrm{\infty }\right)$, with ${f}^{\prime }\in {L}_{p}\left(\left[c,\mathrm{\infty }\right)\right)$ and ${f}^{″}\in {L}_{p}\left(\left[c,\mathrm{\infty }\right)\right)$. Let ${C}_{c}^{2}\left({\mathbb{R}}^{+}\right)$ be the space of continuous, compactly supported and continuously second differentiable functions on ${\mathbb{R}}^{+}$. Furthermore, the total variation of a real-valued function f defined on an interval $\left[a,b\right]\subset \mathbb{R}$ is the quantity

${V}_{a}^{b}\left(f\right)=\underset{P\in \mathcal{P}}{sup}\sum _{i=0}^{{n}_{P}-1}|f\left({x}_{i+1}\right)-f\left({x}_{i}\right)|.$

Here the supremum is taken over the set $\mathcal{P}$ of all partitions $P=\left\{{x}_{0},{x}_{1},\dots ,{x}_{{n}_{P}}\right\}$ of the interval considered. If $F:\left[b,\mathrm{\infty }\right)\to \mathbb{R}$ and $x\in \left[b,\mathrm{\infty }\right)$, we define

${T}_{F}\left(x\right)=sup\left\{\sum _{1}^{n}|F\left({x}_{j}\right)-F\left({x}_{j-1}\right)|:b={x}_{0}<{x}_{1}<\cdots <{x}_{n}=x,n\in \mathbb{N}\right\}.$

If ${T}_{F}\left(\mathrm{\infty }\right):={lim}_{x\to \mathrm{\infty }}{T}_{F}\left(x\right)$ is finite, we say that F is of bounded variation on $\left[b,\mathrm{\infty }\right)$, and ${T}_{F}\left(\mathrm{\infty }\right)$ is called the total variation of F on $\left[b,\mathrm{\infty }\right)$. We define $BV\left[b,\mathrm{\infty }\right)$ to be the set of all functions on $\left[b,\mathrm{\infty }\right)$ whose total variation on $\left[b,\mathrm{\infty }\right)$ is finite.

Then we first give the pointwise convergence theorem.

Theorem 1.1 (cf. )

Let $f\in {C}_{c}^{2}\left({\mathbb{R}}^{+}\right)$, and let $x\in \left[a,\mathrm{\infty }\right)$, $a>0$. Then we have a pointwise convergence as follows:

$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\mu }_{n}}\left({J}_{n}\left(f;x\right)-f\left(x\right)\right)=\frac{1}{2}{f}^{″}\left(x\right).$
(1.6)

Equation (1.6) holds uniformly on $\left[a,\mathrm{\infty }\right)$.

Then the following is a direct convergence theorem.

Theorem 1.2 Let $0.

1. (i)

If $1, then we have for $f\in {L}_{p}^{2}\left(\left[b,\mathrm{\infty }\right)\right)$

(1.7)
2. (ii)

If $p=1$, then we have for ${f}^{\prime }\in BV\left(\left[b,\mathrm{\infty }\right)\right)$ with $f\in {L}_{1}\left(\left[0,\mathrm{\infty }\right)\right)$

(1.8)
3. (iii)

Let $1⩽p<\mathrm{\infty }$ and f be linear on $\left[b,\mathrm{\infty }\right)$. Then we have

(1.9)

Finally, we give an inverse theorem as follows.

Theorem 1.3 Let $0. Let $f\in {L}_{p}\left({\mathbb{R}}^{+}\right)$ and ${f}^{\prime }\in {L}_{p}\left({\mathbb{R}}^{+}\right)$.

1. (i)

For $1, the condition (1.7) implies $f\in {L}_{p}^{2}\left(\left[a,\mathrm{\infty }\right)\right)$.

2. (ii)

For $p=1$, the condition (1.8) implies ${f}^{\prime }\in BV\left(\left[a,\mathrm{\infty }\right)\right)$.

3. (iii)

For $1⩽p<\mathrm{\infty }$, the condition (1.9) implies that f is linear on $\left[a,\mathrm{\infty }\right)$.

This paper is organized as follows. In Section 2, we will give some fundamental lemmas in order to prove the main results and we will prove the results in Section 3.

## 2 Fundamental lemmas

Throughout this paper $C,{C}_{1},{C}_{2},\dots$ denote positive constants independent of n, x, t or function $f\left(x\right)$. The same symbol does not necessarily denote the same constant in different occurrences.

To prove the theorems we need some lemmas.

Lemma 2.1 Let $\delta >0$ and $\ell =0,1,2$. Then for $2\beta \ge \alpha >3$ defined in (1.4)

${\int }_{|u|⩾\delta }{|u|}^{\ell }{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\le C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha -\ell }}.$
(2.1)

Proof Let $\ell =0,1,2$. Then we have from (1.4)

$\begin{array}{rcl}{\int }_{|u|⩾\delta }{|u|}^{\ell }{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du& =& {\delta }^{\ell }{\int }_{|u|⩾\delta }{\left(\frac{|u|}{\delta }\right)}^{\ell }{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\\ ⩽& {\delta }^{\ell }{\int }_{|u|⩾\delta }{\left(\frac{|u|}{\delta }\right)}^{\alpha }{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du⩽{\delta }^{\ell -\alpha }{\int }_{|u|⩾\delta }{|u|}^{\alpha }{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\\ =& {\delta }^{\ell -\alpha }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|u|}^{\alpha }{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\le C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha -\ell }}.\end{array}$

□

Lemma 2.2 Let ${e}_{0}\left(x\right):=1$, ${e}_{1}\left(x\right):=x$ and let δ be a positive constant. Then for $n=1,2,\dots$ , we have

$|{J}_{n}\left(\left(\cdot -x\right);x\right)|⩽C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha -1}},\phantom{\rule{1em}{0ex}}0<\delta ⩽x$
(2.2)

and

$|{J}_{n}\left({\left(\cdot -x\right)}^{2};x\right)|⩽{\mu }_{n},\phantom{\rule{1em}{0ex}}x⩾0.$
(2.3)

Moreover, we have for $0<\delta ⩽x$

$|{J}_{n}\left({e}_{0};x\right)-{e}_{0}|⩽C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}|{J}_{n}\left({e}_{1};x\right)-{e}_{1}|⩽C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha -1}}.$
(2.4)

Here, α and β are defined in (1.4).

Proof For $0<\delta ⩽x$, we have by (2.1)

For (2.3), we have from (1.3)

${J}_{n}\left({\left(\cdot -x\right)}^{2};x\right)={\int }_{0}^{\mathrm{\infty }}{\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt={\int }_{-x}^{\mathrm{\infty }}{y}^{2}{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy⩽{\mu }_{n}.$

Since we know for $0<\delta ⩽x$

${J}_{n}\left({e}_{0};x\right)={\int }_{0}^{\mathrm{\infty }}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt={\int }_{-x}^{\mathrm{\infty }}{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy=1-{\int }_{-\mathrm{\infty }}^{-x}{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy=1-{\int }_{x}^{\mathrm{\infty }}{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy,$

and from (2.1)

$|{\int }_{x}^{\mathrm{\infty }}{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy|⩽C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }},$
(2.5)

we have for $0<\delta ⩽x$,

$|{J}_{n}\left({e}_{0};x\right)-{e}_{0}|⩽C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }}.$

Next, we give an estimate for ${e}_{1}$. Since

$\begin{array}{rcl}{J}_{n}\left({e}_{1};x\right)& =& {\int }_{0}^{\mathrm{\infty }}t{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt={\int }_{-x}^{\mathrm{\infty }}\left(x+y\right){H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy\\ =& x\left(1-{\int }_{-\mathrm{\infty }}^{-x}{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy\right)+{\int }_{x}^{\mathrm{\infty }}y{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy\\ =& x-x{\int }_{x}^{\mathrm{\infty }}{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy+{\int }_{x}^{\mathrm{\infty }}y{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy\end{array}$

and by (2.1), for $0<\delta ⩽x$,

$|x{\int }_{x}^{\mathrm{\infty }}{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy|+|{\int }_{x}^{\mathrm{\infty }}y{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy|⩽2{\int }_{x}^{\mathrm{\infty }}y{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy⩽C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha -1}},$

we have for $0<\delta ⩽x$,

$|{J}_{n}\left({e}_{1};x\right)-{e}_{1}|⩽C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha -1}}.$

□

Let $b>0$ and then we define

$\chi \left(\left[0,b\right];t\right):=\left\{\begin{array}{cc}1,\hfill & t\in \left[0,b\right],\hfill \\ 0,\hfill & t\notin \left[0,b\right].\hfill \end{array}$

Lemma 2.3 Let $0, $\delta :=a-b$ and let $1\le p<\mathrm{\infty }$. If $f\in {L}_{p}\left({\mathbb{R}}^{+}\right)$, then

${\parallel {J}_{n}\left(\chi \left(\left[0,b\right]\right)f\right)\parallel }_{{L}_{p}\left[a,\mathrm{\infty }\right)}\le C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }}{\parallel f\parallel }_{{L}_{p}\left({\mathbb{R}}^{+}\right)}.$

Proof Let $p=1$. Then we have

$\begin{array}{rcl}{\parallel {J}_{n}\left(\chi \left(\left[0,b\right]\right)f\right)\parallel }_{{L}_{1}\left(\left[a,\mathrm{\infty }\right)\right)}& =& {\int }_{a}^{\mathrm{\infty }}|{\int }_{0}^{\mathrm{\infty }}\chi \left(\left[0,b\right];t\right)f\left(t\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt|\phantom{\rule{0.2em}{0ex}}dx\\ \le & {\int }_{a}^{\mathrm{\infty }}{\int }_{0}^{b}|f\left(t\right)|{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\\ \le & {\int }_{0}^{b}|f\left(t\right)|{\int }_{a}^{\mathrm{\infty }}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \le & \left(\underset{t\in \left[0,b\right]}{sup}{\int }_{a}^{\mathrm{\infty }}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\right){\parallel f\parallel }_{{L}_{1}\left({\mathbb{R}}^{+}\right)}.\end{array}$

For $t\in \left[0,b\right]$ and $x\ge a$, we see $|t-x|\ge a-b=:\delta >0$. From (1.4), we have the following:

$\begin{array}{rcl}\underset{t\in \left[0,b\right]}{sup}{\int }_{a}^{\mathrm{\infty }}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx& \le & \underset{t\in \left[0,b\right]}{sup}{\int }_{a}^{\mathrm{\infty }}{\left(\frac{|t-x|}{\delta }\right)}^{\alpha }{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\\ \le & \frac{1}{{\delta }^{\alpha }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|t|}^{\alpha }{H}_{n}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\le C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }}.\end{array}$
(2.6)

Hence we have

${\parallel {J}_{n}\left(\chi \left(\left[0,b\right]\right)f\right)\parallel }_{{L}_{1}\left[a,\mathrm{\infty }\right)}\le C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }}{\parallel f\parallel }_{{L}_{1}\left({\mathbb{R}}^{+}\right)}.$

Let $1, $0 and $p+q=pq$. By Hölder’s inequality,

$\begin{array}{rcl}|{J}_{n}\left(\chi \left(\left[0,b\right]\right)f,x\right)|& =& |{\int }_{0}^{\mathrm{\infty }}\chi \left(\left[0,b\right];t\right)f\left(t\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt|\\ \le & C{\left({\int }_{0}^{\mathrm{\infty }}\chi \left(\left[0,b\right];t\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\right)}^{1/q}\\ ×{\left({\int }_{0}^{\mathrm{\infty }}\chi \left(\left[0,b\right];t\right)|f\left(t\right){|}^{p}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\right)}^{1/p}.\end{array}$

Here, from (2.1) we have

$\begin{array}{rcl}{\int }_{0}^{\mathrm{\infty }}\chi \left(\left[0,b\right];t\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt& =& {\int }_{0}^{b}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt={\int }_{x-b}^{x}{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy\\ \le & {\int }_{\delta }^{\mathrm{\infty }}{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy\le C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }}.\end{array}$
(2.7)

Hence,

$\begin{array}{c}{\parallel {J}_{n}\left(\chi \left(\left[0,b\right]\right)f\right)\parallel }_{{L}_{p}\left[a,\mathrm{\infty }\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\le C{\left(\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }}\right)}^{1/q}{\left({\int }_{a}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\chi \left(\left[0,b\right];t\right)|f\left(t\right){|}^{p}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}.\hfill \end{array}$
(2.8)

Also, we see by (2.6)

$\begin{array}{c}{\int }_{a}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\chi \left(\left[0,b\right];t\right)|f\left(t\right){|}^{p}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\hfill \\ \phantom{\rule{1em}{0ex}}\le {\int }_{0}^{\mathrm{\infty }}{\int }_{a}^{\mathrm{\infty }}\chi \left(\left[0,b\right];t\right)|f\left(t\right){|}^{p}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{t\in \left[0,b\right]}{sup}{\int }_{a}^{\mathrm{\infty }}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx{\int }_{0}^{\mathrm{\infty }}|f\left(t\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt\le C{\parallel f\parallel }_{{L}_{p}\left({\mathbb{R}}^{+}\right)}^{p}\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }}.\hfill \end{array}$

Thus, by (2.8) we conclude. □

## 3 Proof of theorems

Proof of Theorem 1.1 For $x\in \left[a,\mathrm{\infty }\right)$ and $t⩾0$, we set

$f\left(t\right)=f\left(x\right)+{f}^{\prime }\left(x\right)\left(t-x\right)+\frac{{f}^{″}\left(\eta \right)}{2}{\left(t-x\right)}^{2},\phantom{\rule{1em}{0ex}}x\lessgtr \eta \lessgtr t.$

Then we see

$\begin{array}{c}{J}_{n}\left(f;x\right)-f\left(x\right)-\frac{1}{2}{f}^{″}\left(x\right){\mu }_{n}\hfill \\ \phantom{\rule{1em}{0ex}}=f\left(x\right)\left({J}_{n}\left({e}_{0};x\right)-{e}_{0}\right)+{f}^{\prime }\left(x\right){\int }_{0}^{\mathrm{\infty }}\left(t-x\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{2}\left[{\int }_{0}^{\mathrm{\infty }}{f}^{″}\left(\eta \right){\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt-{f}^{″}\left(x\right){\mu }_{n}\right],\phantom{\rule{1em}{0ex}}x\lessgtr \eta \lessgtr t\hfill \\ \phantom{\rule{1em}{0ex}}:={J}_{1}+{J}_{2}+{J}_{3}.\hfill \end{array}$

For ${J}_{1}$, we have from (2.4)

$|{J}_{1}|=|f\left(x\right)|O\left(\frac{{\mu }_{n}^{\beta }}{{a}^{\alpha }}\right),$

and for ${J}_{2}$, we have from (2.2)

$|{J}_{2}|=|{f}^{\prime }\left(x\right){J}_{n}\left(\left(\cdot -x\right);x\right)|=|{f}^{\prime }\left(x\right)|O\left(\frac{{\mu }_{n}^{\beta }}{{a}^{\alpha -1}}\right).$

Now, we will estimate ${J}_{3}$. We have

$\begin{array}{c}{\int }_{0}^{\mathrm{\infty }}{f}^{″}\left(\eta \right){\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt-{f}^{″}\left(x\right){\mu }_{n}\hfill \\ \phantom{\rule{1em}{0ex}}={\int }_{0}^{\mathrm{\infty }}\left({f}^{″}\left(\eta \right)-{f}^{″}\left(x\right)\right){\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt-{f}^{″}\left(x\right){\int }_{-\mathrm{\infty }}^{0}{\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt.\hfill \end{array}$

For the second term, we have by (2.1)

$|{f}^{″}\left(x\right){\int }_{-\mathrm{\infty }}^{0}{\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt|\le |{f}^{″}\left(x\right)|{\int }_{a}^{\mathrm{\infty }}{u}^{2}{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\le C|{f}^{″}\left(x\right)|\frac{{\mu }_{n}^{\beta }}{{a}^{\alpha -2}}.$

For a given $\epsilon >0$, there exists a positive constant ${\delta }_{1}>0$ (depending only on ε) such that for $|x-\eta |<{\delta }_{1}$

$|{f}^{″}\left(x\right)-{f}^{″}\left(\eta \right)|<\epsilon .$
(3.1)

For the first term, we have using (1.3)

$\begin{array}{c}|{\int }_{|x-t|⩽{\delta }_{1}}\left({f}^{″}\left(\eta \right)-{f}^{″}\left(x\right)\right){\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt|\hfill \\ \phantom{\rule{1em}{0ex}}⩽\epsilon {\int }_{|u|⩽{\delta }_{1}}{u}^{2}{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\le \epsilon {\mu }_{n},\hfill \end{array}$

and by (2.1)

$\begin{array}{c}|{\int }_{|x-t|⩾{\delta }_{1}}\left({f}^{″}\left(\eta \right)-{f}^{″}\left(x\right)\right){\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt|\hfill \\ \phantom{\rule{1em}{0ex}}⩽2{\parallel {f}^{″}\parallel }_{{L}_{\mathrm{\infty }}\left(\mathbb{R}\right)}{\int }_{|u|⩾{\delta }_{1}}{u}^{2}{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du⩽{\parallel {f}^{″}\parallel }_{{L}_{\mathrm{\infty }}\left(\mathbb{R}\right)}O\left(\frac{{\mu }_{n}^{\beta }}{{\delta }_{1}^{\alpha -2}}\right).\hfill \end{array}$

Then we have for some positive constants ${C}_{1}$, ${C}_{2}$, and ${C}_{3}$

$\begin{array}{rcl}|{J}_{n}\left(f;x\right)-f\left(x\right)-\frac{1}{2}{f}^{″}\left(x\right){\mu }_{n}|& \le & {C}_{1}|f\left(x\right)|\frac{{\mu }_{n}^{\beta }}{{a}^{\alpha }}+{C}_{2}|{f}^{\prime }\left(x\right)|\frac{{\mu }_{n}^{\beta }}{{a}^{\alpha -1}}\\ +\epsilon {\mu }_{n}+{C}_{3}{\parallel {f}^{″}\parallel }_{{L}_{\mathrm{\infty }}\left(\mathbb{R}\right)}\frac{{\mu }_{n}^{\beta }}{{\delta }_{1}^{\alpha -2}}.\end{array}$

Therefore, we have for an arbitrary $\epsilon >0$

$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\mu }_{n}}|{J}_{n}\left(f;x\right)-f\left(x\right)-\frac{1}{2}{f}^{″}\left(x\right){\mu }_{n}|\le \epsilon .$

Thus, (1.6) is proved. Moreover, noting (3.1), we see that (1.6) holds uniformly on $\left[a,\mathrm{\infty }\right)$. □

Proof of Theorem 1.2 Let $\delta :=a-b>0$. (i) We start under the condition $1\le p<\mathrm{\infty }$ for the convenience of considering (ii). On the way of the proof we switch over to the assumption with $1. Let $1 and let $\chi :=\chi \left(\left[0,b\right]\right)$ and let ${\chi }_{1}\left(t\right)=1-\chi \left(t\right)$. Since

$f\left(t\right)=\left(\chi f\right)\left(t\right)+\left({\chi }_{1}f\right)\left(t\right),\phantom{\rule{2em}{0ex}}f\left(x\right)=\left({\chi }_{1}f\right)\left(x\right),\phantom{\rule{1em}{0ex}}x\in \left[a,\mathrm{\infty }\right),$

we have

${\parallel {J}_{n}\left(f\right)-f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}\le {\parallel {J}_{n}\left(\chi f\right)\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}+{\parallel {J}_{n}\left({\chi }_{1}f\right)-{\chi }_{1}f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}=:{K}_{1}+{K}_{2}.$
(3.2)

By Lemma 2.3,

${K}_{1}={\parallel {J}_{n}\left(\chi f\right)\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}\le C{\parallel f\parallel }_{{L}_{p}\left({\mathbb{R}}^{+}\right)}\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }}.$
(3.3)

We estimate ${K}_{2}$. Let $f\in {L}_{p}^{2}\left(\left[a,\mathrm{\infty }\right)\right)$. Then

$\begin{array}{rcl}J& :=& {\parallel {J}_{n}\left(\left({\chi }_{1}f\right)\left(t\right)-\left({\chi }_{1}f\right)\left(x\right),x\right)\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}\\ =& {\left({\int }_{a}^{\mathrm{\infty }}|{\int }_{0}^{\mathrm{\infty }}\left({\chi }_{1}\left(t\right)f\left(t\right)-f\left(x\right)\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt{|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ \le & {\left({\int }_{a}^{\mathrm{\infty }}|\left({\int }_{0}^{b}+{\int }_{b}^{\mathrm{\infty }}\right)\left({\chi }_{1}\left(t\right)f\left(t\right)-f\left(x\right)\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt{|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ \le & {\left({\int }_{a}^{\mathrm{\infty }}|f\left(x\right){|}^{p}|{\int }_{0}^{b}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt{|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ +{\left({\int }_{a}^{\mathrm{\infty }}|{\int }_{b}^{\mathrm{\infty }}\left(f\left(t\right)-f\left(x\right)\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt{|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ =:& {I}_{1}+{I}_{2}.\end{array}$
(3.4)

Here, for the first term, using

${\int }_{0}^{b}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\le C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }},$

which is shown in (2.7), we have

${I}_{1}={\left({\int }_{a}^{\mathrm{\infty }}|f\left(x\right){|}^{p}|{\int }_{0}^{b}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt{|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\le C{\parallel f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha }}.$
(3.5)

From this we suppose $1. By

$f\left(t\right)-f\left(x\right)={f}^{\prime }\left(x\right)\left(t-x\right)+{\int }_{x}^{t}\left(t-u\right){f}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du,$

we have

$\begin{array}{rcl}{I}_{2}& =& {\left({\int }_{a}^{\mathrm{\infty }}|{\int }_{b}^{\mathrm{\infty }}\left(f\left(t\right)-f\left(x\right)\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt{|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ \le & {\left({\int }_{a}^{\mathrm{\infty }}|{\int }_{b}^{\mathrm{\infty }}{f}^{\prime }\left(x\right)\left(t-x\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt{|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ +{\left({\int }_{a}^{\mathrm{\infty }}|{\int }_{b}^{\mathrm{\infty }}{\int }_{x}^{t}\left(t-u\right){f}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt{|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ =:& {I}_{2,1}+{I}_{2,2}.\end{array}$

Here we note that $y{H}_{n}\left(y\right)$ is an odd function. Then we have by (2.1)

$\begin{array}{rcl}|{\int }_{b}^{\mathrm{\infty }}\left(t-x\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt|& =& |{\int }_{b-x}^{\mathrm{\infty }}y{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy|={\int }_{x-b}^{\mathrm{\infty }}y{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy\\ \le & {\int }_{y\ge \delta }y{H}_{n}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy\le C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha -1}}\end{array}$

and the first term ${I}_{2,1}$ is estimated as

$\begin{array}{rcl}{I}_{2,1}& =& {\left({\int }_{a}^{\mathrm{\infty }}{\left(|{f}^{\prime }\left(x\right)||{\int }_{b}^{\mathrm{\infty }}\left(t-x\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt|\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ \le & C\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha -1}}{\left({\int }_{a}^{\mathrm{\infty }}{|{f}^{\prime }\left(x\right)|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\le C{\parallel {f}^{\prime }\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}\frac{{\mu }_{n}^{\beta }}{{\delta }^{\alpha -1}}.\end{array}$
(3.6)

Now we set

$\theta \left({f}^{″},x\right):=\underset{b\le t,t\ne x}{sup}\frac{1}{t-x}{\int }_{x}^{t}|{f}^{″}\left(u\right)|\phantom{\rule{0.2em}{0ex}}du,\phantom{\rule{1em}{0ex}}a\le x,$

and denote the Hardy-Littlewood majorant of ${f}^{″}$ at x. Since ${f}^{″}\in {L}_{p}\left(\left[b,\mathrm{\infty }\right)\right)$ and $1, we have

${\parallel \theta \left({f}^{″},x\right)\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}\le {A}_{p}{\parallel {f}^{″}\parallel }_{{L}_{p}\left(\left[b,\mathrm{\infty }\right)\right)},$
(3.7)

where ${A}_{p}>0$ depend only on p [, Theorem 1, p.201]. Since

$|{\int }_{x}^{t}\left(t-u\right){f}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du|\le ±{\int }_{x}^{t}|t-u||{f}^{″}\left(u\right)|\phantom{\rule{0.2em}{0ex}}du\le {\left(t-x\right)}^{2}\frac{1}{t-x}{\int }_{x}^{t}|{f}^{″}\left(u\right)|\phantom{\rule{0.2em}{0ex}}du,$

where $±{\int }_{x}^{t}du\ge 0$, we have by (3.7) and (1.3)

$\begin{array}{rcl}{I}_{2,2}& \le & {\left({\int }_{a}^{\mathrm{\infty }}|{\int }_{b}^{\mathrm{\infty }}\left({\left(t-x\right)}^{2}\frac{1}{t-x}{\int }_{x}^{t}|{f}^{″}\left(u\right)|\phantom{\rule{0.2em}{0ex}}du\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt{|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ \le & {\left({\int }_{a}^{\mathrm{\infty }}|\theta \left({f}^{″};x\right){|}^{p}|{\int }_{b}^{\mathrm{\infty }}{\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt{|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ \le & {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{u}^{2}{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du{\left({\int }_{a}^{\mathrm{\infty }}|\theta \left({f}^{″};x\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{1/p}\\ \le & {A}_{p}{\mu }_{n}{\parallel {f}^{″}\parallel }_{{L}_{p}\left(\left[b,\mathrm{\infty }\right)\right)}.\end{array}$
(3.8)

Hence, from (3.6) and (3.8), we have for some positive constant $C>0$

${I}_{2}\le {I}_{2,1}+{I}_{2,2}\le C\left({\parallel {f}^{\prime }\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}+{\parallel {f}^{″}\parallel }_{{L}_{p}\left(\left[b,\mathrm{\infty }\right)\right)}\right){\mu }_{n}.$
(3.9)

So, from (3.5) and (3.9),

$J\le {I}_{1}+{I}_{2}\le C\left({\parallel f\parallel }_{{L}_{p}\left({\mathbb{R}}^{+}\right)}+{\parallel {f}^{\prime }\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}+{\parallel {f}^{″}\parallel }_{{L}_{p}\left(\left[b,\mathrm{\infty }\right)\right)}\right){\mu }_{n}.$
(3.10)

Now, we see by (3.10) and (2.4)

$\begin{array}{rcl}{K}_{2}& =& {\parallel {J}_{n}\left({\chi }_{1}f\right)-\left({\chi }_{1}f\right)\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}\\ \le & {\parallel {J}_{n}\left(\left({\chi }_{1}f\right)\left(t\right)-\left({\chi }_{1}f\right)\left(x\right),x\right)\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}+{\parallel \left({\chi }_{1}f\right)\left(x\right)\left({J}_{n}\left({e}_{0},x\right)-{e}_{0}\right)\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}\\ =& J+{\parallel \left({\chi }_{1}f\right)\left(x\right)\left({J}_{n}\left({e}_{0},x\right)-{e}_{0}\right)\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}\\ \le & C\left({\parallel f\parallel }_{{L}_{p}\left({\mathbb{R}}^{+}\right)}+{\parallel {f}^{\prime }\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}+{\parallel {f}^{″}\parallel }_{{L}_{p}\left(\left[b,\mathrm{\infty }\right)\right)}\right){\mu }_{n}\\ +{\parallel {J}_{n}\left({e}_{0},x\right)-{e}_{0}\parallel }_{{L}_{\mathrm{\infty }}\left(\left[a,\mathrm{\infty }\right)\right)}{\parallel {\chi }_{1}f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}\\ \le & {C}_{1}\left({\parallel f\parallel }_{{L}_{p}\left({\mathbb{R}}^{+}\right)}+{\parallel {f}^{\prime }\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}+{\parallel {f}^{″}\parallel }_{{L}_{p}\left(\left[b,\mathrm{\infty }\right)\right)}\right){\mu }_{n}=O\left({\mu }_{n}\right).\end{array}$

Consequently, with (3.2) and (3.3) we conclude (i).

(ii) Let $p=1$ and ${f}^{\prime }\in BV\left[b,\mathrm{\infty }\right)$. Then we have for $x,t\in \left[b,\mathrm{\infty }\right)$,

$f\left(t\right)-f\left(x\right)={f}^{\prime }\left(x\right)\left(t-x\right)+{\int }_{x}^{t}\left(t-u\right)\phantom{\rule{0.2em}{0ex}}d{f}^{\prime }\left(u\right).$
(3.11)

On the proof of (i) we recall the part which we assumed as $p=1$. From (3.3) and (3.5) we may only estimate

${I}_{2}^{\prime }:={\int }_{a}^{\mathrm{\infty }}|{\int }_{b}^{\mathrm{\infty }}\left(f\left(t\right)-f\left(x\right)\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx|$
(3.12)

(see (3.4)). By (3.11) we see

$\begin{array}{rcl}{I}_{2}^{\prime }& \le & |{\int }_{a}^{\mathrm{\infty }}{\int }_{b}^{\mathrm{\infty }}{f}^{\prime }\left(x\right)\left(t-x\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx|\\ +|{\int }_{a}^{\mathrm{\infty }}{\int }_{b}^{\mathrm{\infty }}{\int }_{x}^{t}\left(t-u\right)\phantom{\rule{0.2em}{0ex}}d{f}^{\prime }\left(u\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx|\\ =:& {I}_{2,1}^{\prime }+{I}_{2,2}^{\prime }.\end{array}$
(3.13)

Now, we see that

$\begin{array}{rcl}{I}_{2,1}^{\prime }& =& |{\int }_{a}^{\mathrm{\infty }}{\int }_{b}^{\mathrm{\infty }}{f}^{\prime }\left(x\right)\left(t-x\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx|\\ =& |{\int }_{a}^{\mathrm{\infty }}{f}^{\prime }\left(x\right){\int }_{x-b}^{\mathrm{\infty }}u{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dx|\\ =& |{\int }_{a-b}^{\mathrm{\infty }}u{H}_{n}\left(u\right){\int }_{a}^{u+b}{f}^{\prime }\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}du|\\ \le & \underset{a\le x<\mathrm{\infty }}{sup}|{f}^{\prime }\left(x\right)||{\int }_{\delta }^{\mathrm{\infty }}\left(u-\delta \right)u{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du|\\ \le & \underset{\left[b,\mathrm{\infty }\right)}{sup}|{f}^{\prime }\left(x\right)|O\left({\mu }_{n}\right)\end{array}$

by (1.3). We estimate ${I}_{2,2}^{\prime }$. We fix an arbitrary $\eta >0$. Then we have by means of the substitution $u=y+x$ with a new variable y

$\begin{array}{rcl}{I}_{2,2}^{\prime }& =& |{\int }_{a}^{\mathrm{\infty }}{\int }_{b}^{\mathrm{\infty }}{\int }_{x}^{t}\left(t-u\right)\phantom{\rule{0.2em}{0ex}}d{f}^{\prime }\left(u\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx|\\ =& |{\int }_{a}^{\mathrm{\infty }}{\int }_{b}^{\mathrm{\infty }}{\int }_{0}^{t-x}\left(t-x-y\right)\phantom{\rule{0.2em}{0ex}}d{f}^{\prime }\left(x+y\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx|\\ \le & 2{\int }_{a}^{\mathrm{\infty }}{\int }_{b}^{\mathrm{\infty }}{\int }_{0}^{|t-x|}|t-x||d{f}^{\prime }\left(x+y\right)|{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{1em}{0ex}}\because |t-x-y|\le 2|t-x|\\ \le & 2{\int }_{a}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}{\int }_{j\eta \le |t-x|\le \left(j+1\right)\eta }{\int }_{0}^{\left(j+1\right)\eta }|d{f}^{\prime }\left(x+y\right)||t-x|{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\\ \le & 2\sum _{j=0}^{\mathrm{\infty }}{\int }_{j\eta \le |u|\le \left(j+1\right)\eta }|u|{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du{\int }_{a}^{\mathrm{\infty }}{\int }_{0}^{\left(j+1\right)\eta }|d{f}^{\prime }\left(x+y\right)|\phantom{\rule{0.2em}{0ex}}dx.\end{array}$

Then

$\begin{array}{c}{\int }_{a}^{\mathrm{\infty }}{\int }_{0}^{\left(j+1\right)\eta }|d{f}^{\prime }\left(x+y\right)|\phantom{\rule{0.2em}{0ex}}dx\hfill \\ \phantom{\rule{1em}{0ex}}={\int }_{a}^{\mathrm{\infty }}{\int }_{x}^{x+\left(j+1\right)\eta }|d{f}^{\prime }\left(v\right)|\phantom{\rule{0.2em}{0ex}}dx\hfill \\ \phantom{\rule{1em}{0ex}}={\int }_{a\le v\le a+\left(j+1\right)\eta }{\int }_{a\le x\le v}dx|d{f}^{\prime }\left(v\right)|+{\int }_{a+\left(j+1\right)\eta \le v<\mathrm{\infty }}{\int }_{v-\left(j+1\right)\eta \le x\le v}dx|d{f}^{\prime }\left(v\right)|\hfill \\ \phantom{\rule{1em}{0ex}}:={A}_{1}+{A}_{2}.\hfill \end{array}$

Since we can see

${A}_{1}={\int }_{a\le v\le a+\left(j+1\right)\eta }{\int }_{a\le x\le v}dx|d{f}^{\prime }\left(v\right)|={\int }_{a\le v\le a+\left(j+1\right)\eta }\left(v-a\right)|d{f}^{\prime }\left(v\right)|\le \left(j+1\right)\eta BV\left({f}^{\prime };{\mathbb{R}}^{+}\right)$

and

${A}_{2}={\int }_{a+\left(j+1\right)\eta \le v<\mathrm{\infty }}{\int }_{v-\left(j+1\right)\eta \le x\le v}\phantom{\rule{0.2em}{0ex}}dx|d{f}^{\prime }\left(v\right)|\le \left(j+1\right)\eta BV\left({f}^{\prime };{\mathbb{R}}^{+}\right),$

we have

${\int }_{a}^{\mathrm{\infty }}{\int }_{0}^{\left(j+1\right)\eta }|d{f}^{\prime }\left(x+y\right)|\phantom{\rule{0.2em}{0ex}}dx\le 2\left(j+1\right)\eta BV\left({f}^{\prime };{\mathbb{R}}^{+}\right).$

Now, we estimate ${\int }_{j\eta \le |u|\le \left(j+1\right)\eta }|u|{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du$ for a non-negative integer j. Let $j=0$. Then from (1.2) and (1.3), we have

${\int }_{0\le |u|\le \eta }|u|{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\le {\left({\int }_{0\le |u|\le \eta }{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\right)}^{1/2}{\left({\int }_{0\le |u|\le \eta }{u}^{2}{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\right)}^{1/2}=O\left({\mu }_{n}^{1/2}\right).$

Let $j\ge 1$. Then by (2.1) we have

${\int }_{j\eta \le |u|\le \left(j+1\right)\eta }|u|{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\le C\frac{{\mu }_{n}^{\beta }}{{\left(j\eta \right)}^{\alpha -1}}.$

Therefore, we have

$\begin{array}{rcl}{I}_{2,2}^{\prime }& \le & {\int }_{0\le |u|\le \eta }|u|{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du{\int }_{a}^{\mathrm{\infty }}{\int }_{0}^{\eta }|d{f}^{\prime }\left(x+y\right)|\phantom{\rule{0.2em}{0ex}}dx\\ +\sum _{j=1}^{\mathrm{\infty }}{\int }_{j\eta \le |u|\le \left(j+1\right)\eta }|u|{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du{\int }_{a}^{\mathrm{\infty }}{\int }_{0}^{\left(j+1\right)\eta }|d{f}^{\prime }\left(x+y\right)|\phantom{\rule{0.2em}{0ex}}dx\\ \le & O\left({\mu }_{n}^{1/2}\right)\eta BV\left({f}^{\prime };{\mathbb{R}}^{+}\right)+\sum _{j=1}^{\mathrm{\infty }}\frac{1}{{\left(j\eta \right)}^{\alpha -1}}O\left({\mu }_{n}^{\beta }\right)\left(j+1\right)\eta BV\left({f}^{\prime };{\mathbb{R}}^{+}\right)\\ =& O\left({\mu }_{n}^{1/2}\right)\eta BV\left({f}^{\prime };{\mathbb{R}}^{+}\right)+O\left({\mu }_{n}^{\beta }\right)\frac{1}{{\eta }^{\alpha -2}}BV\left({f}^{\prime };{\mathbb{R}}^{+}\right)\phantom{\rule{1em}{0ex}}\left(\because \alpha >3\right).\end{array}$

If we let $\eta ={\mu }_{n}^{1/2}$, then we have ${I}_{2,2}^{\prime }=O\left({\mu }_{n}\right)BV\left({f}^{\prime };{\mathbb{R}}^{+}\right)$, because $\beta -\left(\alpha -2\right)/2\ge 1$. Consequently, (ii) is proved.

(iii) It follows from (2.4). Consequently, for a linear function f on $\left[b,\mathrm{\infty }\right)$

${\parallel {J}_{n}\left(f;x\right)-f\left(x\right)\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}=O\left({\mu }_{n}^{\beta }\right)=o\left({\mu }_{n}\right).$

□

Proof of Theorem 1.3 (i), (ii) hold as follows: Let $f\in {L}_{p}\left({\mathbb{R}}^{+}\right)$. First, we choose $\psi \in {C}^{2}\left({\mathbb{R}}^{+}\right)$ with ${\psi }^{″}\in {L}_{q}\left(\left[a,\mathrm{\infty }\right)\right)$ (for any $1\le q\le \mathrm{\infty }$) such that

$\psi \left(a\right)={\psi }^{\prime }\left(a\right)={\psi }^{″}\left(a\right)=0,\phantom{\rule{2em}{0ex}}\underset{r\to \mathrm{\infty }}{lim}{\psi }^{\left(i\right)}\left(r\right)=0,\phantom{\rule{1em}{0ex}}i=0,1,2$

and

We use the bilinear functional;

${A}_{n}\left(f,\psi \right)=\frac{1}{{\mu }_{n}}{\int }_{0}^{\mathrm{\infty }}\left({J}_{n}\left(f,x\right)-f\left(x\right)\right)\psi \left(x\right)\phantom{\rule{0.2em}{0ex}}dx.$

We will show that for fixed ψ, ${A}_{n}\left(\cdot ,\psi \right)$ is uniformly bounded on ${L}_{p}\left({\mathbb{R}}^{+}\right)$. We see

${\int }_{0}^{\mathrm{\infty }}{J}_{n}\left(f,x\right)\psi \left(x\right)\phantom{\rule{0.2em}{0ex}}dx={\int }_{0}^{\mathrm{\infty }}\psi \left(x\right){\int }_{0}^{\mathrm{\infty }}f\left(t\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx.$

For $x,t\in \left[0,\mathrm{\infty }\right)$ we can write

$\psi \left(x\right)=\psi \left(t\right)+{\psi }^{\prime }\left(t\right)\left(x-t\right)+{\int }_{x}^{t}\left(x-u\right){\psi }^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du.$

Hence,

$\begin{array}{c}{\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}f\left(t\right)\psi \left(x\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}={\int }_{0}^{\mathrm{\infty }}f\left(t\right)\psi \left(t\right){\int }_{0}^{\mathrm{\infty }}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{2em}{0ex}}+{\int }_{0}^{\mathrm{\infty }}f\left(t\right){\psi }^{\prime }\left(t\right){\int }_{0}^{\mathrm{\infty }}\left(x-t\right){H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{2em}{0ex}}+{\int }_{0}^{\mathrm{\infty }}f\left(t\right){\int }_{0}^{\mathrm{\infty }}{\int }_{x}^{t}\left(x-u\right){\psi }^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}={I}_{1}^{″}+{I}_{2}^{″}+{I}_{3}^{″}.\hfill \end{array}$

From (1.2)

${I}_{1}^{″}={\int }_{0}^{\mathrm{\infty }}f\left(t\right)\psi \left(t\right){\int }_{-\mathrm{\infty }}^{t}{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dt={\int }_{0}^{\mathrm{\infty }}f\left(t\right)\psi \left(t\right)\left(1-{\int }_{t}^{\mathrm{\infty }}{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\right)\phantom{\rule{0.2em}{0ex}}dt.$

Since $\psi \left(t\right)=0$ for $t\notin \left[a,\mathrm{\infty }\right)$ (so, we may take $t\ge a$), by (2.1) we have

$\begin{array}{c}|{\int }_{0}^{\mathrm{\infty }}f\left(t\right)\psi \left(t\right){\int }_{t}^{\mathrm{\infty }}{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dt|\hfill \\ \phantom{\rule{1em}{0ex}}\le |{\int }_{0}^{\mathrm{\infty }}f\left(t\right)\psi \left(t\right){\int }_{t}^{\mathrm{\infty }}{\left(\frac{u}{t}\right)}^{\alpha }{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dt|\phantom{\rule{1em}{0ex}}\because {\left(u/t\right)}^{\alpha }\ge 1\hfill \\ \phantom{\rule{1em}{0ex}}\le O\left({\mu }_{n}^{\beta }\right){\int }_{a}^{\mathrm{\infty }}|f\left(t\right)\psi \left(t\right)|\frac{1}{{t}^{\alpha }}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\text{by (1.4)}\hfill \\ \phantom{\rule{1em}{0ex}}\le O\left({\mu }_{n}^{\beta }\right)\underset{a\le t<\mathrm{\infty }}{sup}|\psi \left(t\right)|{\int }_{a}^{\mathrm{\infty }}|f\left(t\right)|\frac{1}{{t}^{\alpha }}\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}\le O\left({\mu }_{n}^{\beta }\right)\underset{a\le t<\mathrm{\infty }}{sup}|\psi \left(t\right)|{\parallel f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}{\parallel {t}^{-\alpha }\parallel }_{{L}_{q}\left(\left[a,\mathrm{\infty }\right)\right)},\hfill \end{array}$

where $1/p+1/q=1$. Thus, we have

${I}_{1}^{″}={\int }_{0}^{\mathrm{\infty }}f\left(t\right)\psi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt+O\left({\mu }_{n}^{\beta }\right){\parallel f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}.$

We estimate ${I}_{2}^{″}$. We may $a\le t$, because $\psi \left(t\right)=0$ for $t\notin \left[a,\mathrm{\infty }\right)$. Noting (1.4) and $\alpha <3$,

$\begin{array}{rcl}|{I}_{2}^{″}|& =& {\int }_{a}^{\mathrm{\infty }}|f\left(t\right){\psi }^{\prime }\left(t\right)|{\int }_{t}^{\mathrm{\infty }}u{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dt\\ \le & {\int }_{a}^{\mathrm{\infty }}|f\left(t\right){\psi }^{\prime }\left(t\right)|{\int }_{t}^{\mathrm{\infty }}\frac{{u}^{\alpha }}{{t}^{\alpha -1}}{H}_{n}\left(u\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dt\\ =& O\left({\mu }_{n}^{\beta }\right){\int }_{a}^{\mathrm{\infty }}|f\left(t\right){\psi }^{\prime }\left(t\right)|\frac{1}{{t}^{\alpha -1}}\phantom{\rule{0.2em}{0ex}}dt\\ =& O\left({\mu }_{n}^{\beta }\right){\parallel {\psi }^{\prime }\parallel }_{{L}_{\mathrm{\infty }}\left(\left[a,\mathrm{\infty }\right)\right)}{\parallel f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}{\parallel {t}^{1-\alpha }\parallel }_{{L}_{q}\left(\left[a,\mathrm{\infty }\right)\right)}\\ =& O\left({\mu }_{n}^{\beta }\right){\parallel f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}.\end{array}$

Finally, we estimate ${I}_{3}^{″}$. Let $1 and $1/p+1/q=1$. As the estimation for ${I}_{2,2}$, we have by (1.3)

$\begin{array}{rcl}|{I}_{3}^{″}|& =& |{\int }_{0}^{\mathrm{\infty }}f\left(t\right){\int }_{0}^{\mathrm{\infty }}{\int }_{x}^{t}\left(x-u\right){\psi }^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt|\\ =& |{\int }_{0}^{\mathrm{\infty }}f\left(t\right){\int }_{0}^{\mathrm{\infty }}\frac{1}{t-x}{\int }_{x}^{t}{\psi }^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du{\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt|\\ \le & {\int }_{a}^{\mathrm{\infty }}|f\left(t\right)|\theta \left({\psi }^{″},t\right){\int }_{0}^{\mathrm{\infty }}{\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ =& O\left({\mu }_{n}\right){\int }_{a}^{\mathrm{\infty }}|f\left(t\right)|\theta \left({\psi }^{″},t\right)\phantom{\rule{0.2em}{0ex}}dt\le O\left({\mu }_{n}\right){\parallel f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}{\parallel {\psi }^{″}\parallel }_{{L}_{q}\left(\left[a,\mathrm{\infty }\right)\right)}\\ =& O\left({\mu }_{n}\right){\parallel f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}.\end{array}$

We estimate ${I}_{3}^{″}$ for $p=1$. There exists η between x and t such that

$\begin{array}{rcl}|{I}_{3}^{″}|& \le & |{\int }_{0}^{\mathrm{\infty }}f\left(t\right){\int }_{0}^{\mathrm{\infty }}{\psi }^{″}\left(\eta \right){\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt|\\ \le & \underset{a\le t<\mathrm{\infty }}{sup}|{\psi }^{″}\left(t\right)|{\int }_{a}^{\mathrm{\infty }}|f\left(t\right)|{\int }_{0}^{\mathrm{\infty }}{\left(t-x\right)}^{2}{H}_{n}\left(t-x\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \le & O\left({\mu }_{n}\right)\underset{a\le t<\mathrm{\infty }}{sup}|{\psi }^{″}\left(t\right)|{\parallel f\parallel }_{{L}_{1}\left(\left[a,\mathrm{\infty }\right)\right)}=O\left({\mu }_{n}\right){\parallel f\parallel }_{{L}_{1}\left(\left[a,\mathrm{\infty }\right)\right)}.\end{array}$

Here we used (1.3). Consequently, it follows that $|{A}_{n}\left(f,\psi \right)|$ is uniformly bounded on ${L}_{p}\left({\mathbb{R}}^{+}\right)$. Next, from Theorem 1.1 we see that for $f\in {C}_{c}^{2}\left({\mathbb{R}}^{+}\right)$,

$\underset{n\to \mathrm{\infty }}{lim}{A}_{n}\left(f,\psi \right)=\frac{1}{2}{\int }_{a}^{\mathrm{\infty }}{f}^{″}\left(x\right)\psi \left(x\right)\phantom{\rule{0.2em}{0ex}}dx=\frac{1}{2}{\int }_{a}^{\mathrm{\infty }}f\left(x\right){\psi }^{″}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.$
(3.14)

Since $\left\{{A}_{n}\left(\cdot ,\psi \right)\right\}$ is uniformly bounded on ${L}_{p}\left({\mathbb{R}}^{+}\right)$, and ${C}_{c}^{2}\left({\mathbb{R}}^{+}\right)$ is dense in ${L}_{p}\left({\mathbb{R}}^{+}\right)$, (3.14) yields

$\underset{n\to \mathrm{\infty }}{lim}{A}_{n}\left(f,\psi \right)=\frac{1}{2}{\int }_{a}^{\mathrm{\infty }}f\left(x\right){\psi }^{″}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx$
(3.15)

for any $f\in {L}_{p}\left({\mathbb{R}}^{+}\right)$. Now, for any fixed $f\in {L}_{p}\left({\mathbb{R}}^{+}\right)$, we consider the sequence of linear functional $\left\{{A}_{n}\left(f,\cdot \right)\right\}$. Since ${\parallel {J}_{n}\left(f\right)-f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}=O\left({\mu }_{n}\right)$, $n\to \mathrm{\infty }$, there exist $h\in {L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)$ ($p>1$) and $h\in BV\left[a,\mathrm{\infty }\right)$ ($p=1$) and a subsequence $\left\{{A}_{{n}_{j}}\left(f,\cdot \right)\right\}$ such that

$\underset{j\to \mathrm{\infty }}{lim}{A}_{{n}_{j}}\left(f,\psi \right)=\left\{\begin{array}{cc}{\int }_{0}^{\mathrm{\infty }}h\left(x\right)\psi \left(x\right)\phantom{\rule{0.2em}{0ex}}dx,\hfill & p>1,\hfill \\ {\int }_{0}^{\mathrm{\infty }}\psi \left(x\right)\phantom{\rule{0.2em}{0ex}}dh\left(x\right),\hfill & p=1.\hfill \end{array}$
(3.16)

From (3.15) and (3.16) we obtain

$\frac{1}{2}{\int }_{0}^{\mathrm{\infty }}f\left(x\right){\psi }^{″}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx=\left\{\begin{array}{cc}{\int }_{0}^{\mathrm{\infty }}h\left(x\right)\psi \left(x\right)\phantom{\rule{0.2em}{0ex}}dx,\hfill & p>1,\hfill \\ {\int }_{0}^{\mathrm{\infty }}\psi \left(x\right)\phantom{\rule{0.2em}{0ex}}dh\left(x\right),\hfill & p=1.\hfill \end{array}$
(3.17)

A particular solution to (3.17) is

$\frac{1}{2}f\left(x\right)=\left\{\begin{array}{cc}{\int }_{\eta }^{x}{\int }_{\eta }^{\xi }h\left(\mu \right)\phantom{\rule{0.2em}{0ex}}d\mu \phantom{\rule{0.2em}{0ex}}d\xi ,\hfill & p>1,\hfill \\ {\int }_{\eta }^{x}{\int }_{\eta }^{\xi }dh\left(\mu \right)\phantom{\rule{0.2em}{0ex}}d\xi ,\hfill & p=1.\hfill \end{array}$

The homogeneous problem

${\int }_{a}^{\mathrm{\infty }}f\left(x\right){\psi }^{″}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx=0$

has the general solution $f\left(x\right)={C}_{1}x+{C}_{2}$ for $a\le x<\mathrm{\infty }$, since we can take $\psi \in {C}^{2}\left(\left[a,\mathrm{\infty }\right)\right)$ arbitrarily as ${\psi }^{″}\in {L}_{q}\left(\left[a,\mathrm{\infty }\right)\right)$, $\psi \left(a\right)={\psi }^{\prime }\left(a\right)=0$, ${lim}_{r\to \mathrm{\infty }}{\psi }^{\left(i\right)}\left(r\right)=0$, $i=0,1$. Hence, if $1, $f\in {L}_{p}^{2}\left(\left[a,\mathrm{\infty }\right)\right)$, and if $p=1$ then ${f}^{\prime }\in BV\left(\left[a,\mathrm{\infty }\right)\right)$. Hence, (i) and (ii) hold. We will show (iii). Now, if

${\parallel {J}_{n}\left(f\right)-f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)}=o\left({\mu }_{n}\right),\phantom{\rule{1em}{0ex}}n\to \mathrm{\infty },$

then

$\begin{array}{rcl}|{A}_{n}\left(f,\psi \right)|& \le & \frac{1}{{\mu }_{n}}{\int }_{a}^{\mathrm{\infty }}|{J}_{n}\left(f,x\right)-f\left(x\right)||\psi \left(x\right)|\phantom{\rule{0.2em}{0ex}}dx\\ \le & \left(\underset{a\le x<\mathrm{\infty }}{sup}|\psi \left(x\right)|\right)\frac{{C}_{p}}{{\mu }_{n}}{\parallel {J}_{n}\left(f\right)-f\parallel }_{{L}_{p}\left(\left[a,\mathrm{\infty }\right)\right)},\end{array}$

where ${C}_{p}>0$ is independent of n. Hence

$\underset{n\to \mathrm{\infty }}{lim}{A}_{n}\left(f,\psi \right)=0.$
(3.18)

Considering (3.15), (3.16), (3.17), and (3.18), we obtain

${\int }_{a}^{\mathrm{\infty }}f\left(x\right){\psi }^{″}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx=0,$

and so

${\int }_{a}^{\mathrm{\infty }}{f}^{″}\left(x\right)\psi \left(x\right)\phantom{\rule{0.2em}{0ex}}dx=0,$

consequently, we have ${f}^{″}\left(x\right)=0$, that is, f is linear on $\left[a,\mathrm{\infty }\right)$. □

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## Acknowledgements

The authors thank the referees for many valuable comments and corrections.

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Correspondence to Hee Sun Jung.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors conceived of the study, participated in its design and coordination, drafted the manuscript and participated in the sequence alignment. All authors read and approved the final manuscript.

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