Local saturation of a positive linear convolution operator

Abstract

Let $\{H_n(t)\}$ be a sequence of non-negative, even, and continuous functions on ℝ. In this paper, we consider a convolution operator $J_n(f;x)={\int }_0^{\mathrm{\infty }}f(t)H_n(t-x)dt$, $f\in L_p({\mathbb{R}}^{+})$, and then investigate the local saturation of $J_n(f;x)$.

MSC:44A35.

1 Introduction and theorems

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

$$\underset{|t|⩾T}{sup}{H}_n(t)⩽M,n⩾N,$$

(1.1)

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{H}_n(t)dt=1,n=1,2,\dots ,$$

(1.2)

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{t}^2{H}_n(t)dt={\mu }_n\to 0\text{as }n\to \mathrm{\infty }$$

(1.3)

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

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|t|}^{\alpha }{H}_n(t)dt\le C{\mu }_n^{\beta }.$$

(1.4)

As an example for $H_n(t)$ we give the following:

Let

$$H_n(t)=\frac{n}{\sqrt{\pi }}{e}^{-{(nt)}^2}.$$

Then it satisfies (1.2),

$${\mu }_n=\frac{1}{n^2}\to 0\text{as }n\to \mathrm{\infty }\text{and}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{t}^4{H}_n(t)dt=\frac{3}{4}\frac{1}{n^4}.$$

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

$$J_n(f;x):={\int }_0^{\mathrm{\infty }}f(t)H_n(t-x)dt,n=1,2,\dots .$$

(1.5)

Swetits and Wood [1] studied the operators on a finite interval $[0,r]$;

$$K_n(f;x):={\int }_0^{r}f(t)H_n^{\ast }(t-x)dt,f\in L_p([0,r]),n=1,2,\dots ,$$

where $H_n^{\ast }$ is defined on $[-r,r]$ 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 $[0,\mathrm{\infty })$ (e.g. see [2–4] and the references cited therein).

In this paper we extend [1] to the infinite interval ${\mathbb{R}}^{+}$. Then we use a similar methods as [1]. For $1<p<\mathrm{\infty }$ and $c⩾0$, we define $L_p^2([c,\mathrm{\infty }))$ as the space of those functions f such that $f\in L_p({\mathbb{R}}^{+})$ and $f^{\prime }$ is a locally absolutely continuous function on $[c,\mathrm{\infty })$, with $f^{\prime }\in L_p([c,\mathrm{\infty }))$ and $f^{″}\in L_p([c,\mathrm{\infty }))$. Let $C_c^2({\mathbb{R}}^{+})$ 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 $[a,b]\subset \mathbb{R}$ is the quantity

$$V_a^b(f)=\underset{P\in \mathcal{P}}{sup}\sum _{i=0}^{n_P-1}|f(x_{i+1})-f(x_i)|.$$

Here the supremum is taken over the set $\mathcal{P}$ of all partitions $P=\{x_0,x_1,\dots ,x_{n_P}\}$ of the interval considered. If $F:[b,\mathrm{\infty })\to \mathbb{R}$ and $x\in [b,\mathrm{\infty })$, we define

$$T_F(x)=sup\left\{\sum _1^n\right|F(x_j)-F(x_{j-1})|:b=x_0<x_1<\cdots <x_n=x,n\in \mathbb{N}\}.$$

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

Then we first give the pointwise convergence theorem.

Theorem 1.1 (cf. [5])

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

$$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\mu }_n}(J_n(f;x)-f(x))=\frac{1}{2}{f}^{″}(x).$$

(1.6)

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

Then the following is a direct convergence theorem.

Theorem 1.2 Let $0<b<a<\mathrm{\infty }$.

• (i) If $1<p<\mathrm{\infty }$, then we have for $f\in L_p^2([b,\mathrm{\infty }))$

  $${\parallel J_n(f;x)-f(x)\parallel }_{L_p([a,\mathrm{\infty }))}=O({\mu }_n)\mathit{\text{as }}n\to \mathrm{\infty }.$$

  (1.7)

• (ii) If $p=1$, then we have for $f^{\prime }\in BV([b,\mathrm{\infty }))$ with $f\in L_1([0,\mathrm{\infty }))$

  $${\parallel J_n(f;x)-f(x)\parallel }_{L_1([a,\mathrm{\infty }))}=O({\mu }_n)\mathit{\text{as }}n\to \mathrm{\infty }.$$

  (1.8)

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

  $${\parallel J_n(f;x)-f(x)\parallel }_{L_p([a,\mathrm{\infty }))}=o({\mu }_n)\mathit{\text{as }}n\to \mathrm{\infty }.$$

  (1.9)

Finally, we give an inverse theorem as follows.

Theorem 1.3 Let $0<b<a<\mathrm{\infty }$. Let $f\in L_p({\mathbb{R}}^{+})$ and $f^{\prime }\in L_p({\mathbb{R}}^{+})$.

• (i) For $1<p<\mathrm{\infty }$, the condition (1.7) implies $f\in L_p^2([a,\mathrm{\infty }))$.

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

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

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 $f(x)$. 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)

$${\int }_{|u|⩾\delta }{|u|}^{\ell }{H}_n(u)du\le C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha -\ell }}.$$

(2.1)

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

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

 □

Lemma 2.2 Let $e_0(x):=1$, $e_1(x):=x$ and let δ be a positive constant. Then for $n=1,2,\dots$ , we have

$$\left|J_n((\cdot -x);x)\right|⩽C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha -1}},0<\delta ⩽x$$

(2.2)

and

$$\left|J_n({(\cdot -x)}^2;x)\right|⩽{\mu }_n,x⩾0.$$

(2.3)

Moreover, we have for $0<\delta ⩽x$

$$|J_n(e_0;x)-e_0|⩽C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha }}\mathit{\text{and}}|J_n(e_1;x)-e_1|⩽C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha -1}}.$$

(2.4)

Here, α and β are defined in (1.4).

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

$$\begin{array}{rcl}\left|J_n((\cdot -x);x)\right|& =& |{\int }_0^{\mathrm{\infty }}(t-x)H_n(t-x)dt|=|{\int }_{-x}^{\mathrm{\infty }}y{H}_n(y)dy|\\ =& |{\int }_x^{\mathrm{\infty }}y{H}_n(y)dy|\because H_n(t)\text{ is even}\\ ⩽& {\int }_{y⩾\delta }y{H}_n(y)dy⩽C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha -1}}.\end{array}$$

For (2.3), we have from (1.3)

$$J_n({(\cdot -x)}^2;x)={\int }_0^{\mathrm{\infty }}{(t-x)}^2{H}_n(t-x)dt={\int }_{-x}^{\mathrm{\infty }}{y}^2{H}_n(y)dy⩽{\mu }_n.$$

Since we know for $0<\delta ⩽x$

$$J_n(e_0;x)={\int }_0^{\mathrm{\infty }}{H}_n(t-x)dt={\int }_{-x}^{\mathrm{\infty }}{H}_n(y)dy=1-{\int }_{-\mathrm{\infty }}^{-x}{H}_n(y)dy=1-{\int }_x^{\mathrm{\infty }}{H}_n(y)dy,$$

and from (2.1)

$$|{\int }_x^{\mathrm{\infty }}{H}_n(y)dy|⩽C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha }},$$

(2.5)

we have for $0<\delta ⩽x$,

$$|J_n(e_0;x)-e_0|⩽C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha }}.$$

Next, we give an estimate for $e_1$. Since

$$\begin{array}{rcl}{J}_n(e_1;x)& =& {\int }_0^{\mathrm{\infty }}t{H}_n(t-x)dt={\int }_{-x}^{\mathrm{\infty }}(x+y)H_n(y)dy\\ =& x(1-{\int }_{-\mathrm{\infty }}^{-x}{H}_n(y)dy)+{\int }_x^{\mathrm{\infty }}y{H}_n(y)dy\\ =& x-x{\int }_x^{\mathrm{\infty }}{H}_n(y)dy+{\int }_x^{\mathrm{\infty }}y{H}_n(y)dy\end{array}$$

and by (2.1), for $0<\delta ⩽x$,

$$|x{\int }_x^{\mathrm{\infty }}{H}_n(y)dy|+|{\int }_x^{\mathrm{\infty }}y{H}_n(y)dy|⩽2{\int }_x^{\mathrm{\infty }}y{H}_n(y)dy⩽C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha -1}},$$

we have for $0<\delta ⩽x$,

$$|J_n(e_1;x)-e_1|⩽C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha -1}}.$$

 □

Let $b>0$ and then we define

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

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

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

Proof Let $p=1$. Then we have

$$\begin{array}{rcl}{\parallel J_n\left(\chi ([0,b])f\right)\parallel }_{L_1([a,\mathrm{\infty }))}& =& {\int }_a^{\mathrm{\infty }}|{\int }_0^{\mathrm{\infty }}\chi ([0,b];t)f(t)H_n(t-x)dt|dx\\ \le & {\int }_a^{\mathrm{\infty }}{\int }_0^b|f(t)|H_n(t-x)dtdx\\ \le & {\int }_0^b|f(t)|{\int }_a^{\mathrm{\infty }}{H}_n(t-x)dxdt\\ \le & (\underset{t\in [0,b]}{sup}{\int }_a^{\mathrm{\infty }}{H}_n(t-x)dx){\parallel f\parallel }_{L_1({\mathbb{R}}^{+})}.\end{array}$$

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

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

(2.6)

Hence we have

$${\parallel J_n\left(\chi ([0,b])f\right)\parallel }_{L_1[a,\mathrm{\infty })}\le C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha }}{\parallel f\parallel }_{L_1({\mathbb{R}}^{+})}.$$

Let $1<p<\mathrm{\infty }$, $0<b<a\le x$ and $p+q=pq$. By Hölder’s inequality,

$$\begin{array}{rcl}\left|J_n(\chi ([0,b])f,x)\right|& =& |{\int }_0^{\mathrm{\infty }}\chi ([0,b];t)f(t)H_n(t-x)dt|\\ \le & C{({\int }_0^{\mathrm{\infty }}\chi ([0,b];t)H_n(t-x)dt)}^{1/q}\\ \times {\left({\int }_0^{\mathrm{\infty }}\chi ([0,b];t)\right|f(t){|}^p{H}_n(t-x)dt)}^{1/p}.\end{array}$$

Here, from (2.1) we have

$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\chi ([0,b];t)H_n(t-x)dt& =& {\int }_0^b{H}_n(t-x)dt={\int }_{x-b}^x{H}_n(y)dy\\ \le & {\int }_{\delta }^{\mathrm{\infty }}{H}_n(y)dy\le C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha }}.\end{array}$$

(2.7)

Hence,

$$\begin{array}{c}{\parallel J_n\left(\chi ([0,b])f\right)\parallel }_{L_p[a,\mathrm{\infty })}\hfill \\ \le C{\left(\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha }}\right)}^{1/q}{\left({\int }_a^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\chi ([0,b];t)\right|f(t){|}^p{H}_n(t-x)dtdx)}^{1/p}.\hfill \end{array}$$

(2.8)

Also, we see by (2.6)

$$\begin{array}{c}{\int }_a^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\chi ([0,b];t)|f(t){|}^p{H}_n(t-x)dtdx\hfill \\ \le {\int }_0^{\mathrm{\infty }}{\int }_a^{\mathrm{\infty }}\chi ([0,b];t)|f(t){|}^p{H}_n(t-x)dxdt\hfill \\ \le \underset{t\in [0,b]}{sup}{\int }_a^{\mathrm{\infty }}{H}_n(t-x)dx{\int }_0^{\mathrm{\infty }}|f(t){|}^{p}dt\le C{\parallel f\parallel }_{L_p({\mathbb{R}}^{+})}^p\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha }}.\hfill \end{array}$$

Thus, by (2.8) we conclude. □

3 Proof of theorems

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

$$f(t)=f(x)+f^{\prime }(x)(t-x)+\frac{f^{″}(\eta )}{2}{(t-x)}^2,x\lessgtr \eta \lessgtr t.$$

Then we see

$$\begin{array}{c}{J}_n(f;x)-f(x)-\frac{1}{2}{f}^{″}(x){\mu }_n\hfill \\ =f(x)(J_n(e_0;x)-e_0)+f^{\prime }(x){\int }_0^{\mathrm{\infty }}(t-x)H_n(t-x)dt\hfill \\ +\frac{1}{2}[{\int }_0^{\mathrm{\infty }}{f}^{″}(\eta ){(t-x)}^2{H}_n(t-x)dt-f^{″}(x){\mu }_n],x\lessgtr \eta \lessgtr t\hfill \\ :=J_1+J_2+J_3.\hfill \end{array}$$

For $J_1$, we have from (2.4)

$$|J_1|=|f(x)|O\left(\frac{{\mu }_n^{\beta }}{a^{\alpha }}\right),$$

and for $J_2$, we have from (2.2)

$$|J_2|=|f^{\prime }(x)J_n((\cdot -x);x)|=|f^{\prime }(x)|O\left(\frac{{\mu }_n^{\beta }}{a^{\alpha -1}}\right).$$

Now, we will estimate $J_3$. We have

$$\begin{array}{c}{\int }_0^{\mathrm{\infty }}{f}^{″}(\eta ){(t-x)}^2{H}_n(t-x)dt-f^{″}(x){\mu }_n\hfill \\ ={\int }_0^{\mathrm{\infty }}(f^{″}(\eta )-f^{″}(x)){(t-x)}^2{H}_n(t-x)dt-f^{″}(x){\int }_{-\mathrm{\infty }}^0{(t-x)}^2{H}_n(t-x)dt.\hfill \end{array}$$

For the second term, we have by (2.1)

$$|f^{″}(x){\int }_{-\mathrm{\infty }}^0{(t-x)}^2{H}_n(t-x)dt|\le |f^{″}(x)|{\int }_a^{\mathrm{\infty }}{u}^2{H}_n(u)du\le C|f^{″}(x)|\frac{{\mu }_n^{\beta }}{a^{\alpha -2}}.$$

For a given $\epsilon >0$, there exists a positive constant ${\delta }_1>0$ (depending only on ε) such that for $|x-\eta |<{\delta }_1$

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

(3.1)

For the first term, we have using (1.3)

$$\begin{array}{c}|{\int }_{|x-t|⩽{\delta }_1}(f^{″}(\eta )-f^{″}(x)){(t-x)}^2{H}_n(t-x)dt|\hfill \\ ⩽\epsilon {\int }_{|u|⩽{\delta }_1}{u}^2{H}_n(u)du\le \epsilon {\mu }_n,\hfill \end{array}$$

and by (2.1)

$$\begin{array}{c}|{\int }_{|x-t|⩾{\delta }_1}(f^{″}(\eta )-f^{″}(x)){(t-x)}^2{H}_n(t-x)dt|\hfill \\ ⩽2{\parallel f^{″}\parallel }_{L_{\mathrm{\infty }}(\mathbb{R})}{\int }_{|u|⩾{\delta }_1}{u}^2{H}_n(u)du⩽{\parallel f^{″}\parallel }_{L_{\mathrm{\infty }}(\mathbb{R})}O\left(\frac{{\mu }_n^{\beta }}{{\delta }_1^{\alpha -2}}\right).\hfill \end{array}$$

Then we have for some positive constants $C_1$, $C_2$, and $C_3$

$$\begin{array}{rcl}|J_n(f;x)-f(x)-\frac{1}{2}{f}^{″}(x){\mu }_n|& \le & C_1|f(x)|\frac{{\mu }_n^{\beta }}{a^{\alpha }}+C_2|f^{\prime }(x)|\frac{{\mu }_n^{\beta }}{a^{\alpha -1}}\\ +\epsilon {\mu }_n+C_3{\parallel f^{″}\parallel }_{L_{\mathrm{\infty }}(\mathbb{R})}\frac{{\mu }_n^{\beta }}{{\delta }_1^{\alpha -2}}.\end{array}$$

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

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

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

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 $1<p<\mathrm{\infty }$. Let $1<p<\mathrm{\infty }$ and let $\chi :=\chi ([0,b])$ and let ${\chi }_1(t)=1-\chi (t)$. Since

$$f(t)=(\chi f)(t)+({\chi }_{1}f)(t),f(x)=({\chi }_{1}f)(x),x\in [a,\mathrm{\infty }),$$

we have

$${\parallel J_n(f)-f\parallel }_{L_p([a,\mathrm{\infty }))}\le {\parallel J_n(\chi f)\parallel }_{L_p([a,\mathrm{\infty }))}+{\parallel J_n({\chi }_{1}f)-{\chi }_{1}f\parallel }_{L_p([a,\mathrm{\infty }))}=:K_1+K_2.$$

(3.2)

By Lemma 2.3,

$$K_1={\parallel J_n(\chi f)\parallel }_{L_p([a,\mathrm{\infty }))}\le C{\parallel f\parallel }_{L_p({\mathbb{R}}^{+})}\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha }}.$$

(3.3)

We estimate $K_2$. Let $f\in L_p^2([a,\mathrm{\infty }))$. Then

$$\begin{array}{rcl}J& :=& {\parallel J_n(({\chi }_{1}f)(t)-({\chi }_{1}f)(x),x)\parallel }_{L_p([a,\mathrm{\infty }))}\\ =& {\left({\int }_a^{\mathrm{\infty }}\right|{\int }_0^{\mathrm{\infty }}({\chi }_1(t)f(t)-f(x))H_n(t-x)dt{|}^{p}dx)}^{1/p}\\ \le & {\left({\int }_a^{\mathrm{\infty }}\right|({\int }_0^b+{\int }_b^{\mathrm{\infty }})({\chi }_1(t)f(t)-f(x))H_n(t-x)dt{|}^{p}dx)}^{1/p}\\ \le & {\left({\int }_a^{\mathrm{\infty }}\right|f(x){|}^p|{\int }_0^b{H}_n(t-x)dt{|}^{p}dx)}^{1/p}\\ +{\left({\int }_a^{\mathrm{\infty }}\right|{\int }_b^{\mathrm{\infty }}(f(t)-f(x))H_n(t-x)dt{|}^{p}dx)}^{1/p}\\ =:& I_1+I_2.\end{array}$$

(3.4)

Here, for the first term, using

$${\int }_0^b{H}_n(t-x)dt\le C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha }},$$

which is shown in (2.7), we have

$$I_1={\left({\int }_a^{\mathrm{\infty }}\right|f(x){|}^p|{\int }_0^b{H}_n(t-x)dt{|}^{p}dx)}^{1/p}\le C{\parallel f\parallel }_{L_p([a,\mathrm{\infty }))}\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha }}.$$

(3.5)

From this we suppose $1<p<\mathrm{\infty }$. By

$$f(t)-f(x)=f^{\prime }(x)(t-x)+{\int }_x^t(t-u)f^{″}(u)du,$$

we have

$$\begin{array}{rcl}{I}_2& =& {\left({\int }_a^{\mathrm{\infty }}\right|{\int }_b^{\mathrm{\infty }}(f(t)-f(x))H_n(t-x)dt{|}^{p}dx)}^{1/p}\\ \le & {\left({\int }_a^{\mathrm{\infty }}\right|{\int }_b^{\mathrm{\infty }}{f}^{\prime }(x)(t-x)H_n(t-x)dt{|}^{p}dx)}^{1/p}\\ +{\left({\int }_a^{\mathrm{\infty }}\right|{\int }_b^{\mathrm{\infty }}{\int }_x^t(t-u)f^{″}(u)du{H}_n(t-x)dt{|}^{p}dx)}^{1/p}\\ =:& I_{2,1}+I_{2,2}.\end{array}$$

Here we note that $y{H}_n(y)$ is an odd function. Then we have by (2.1)

$$\begin{array}{rcl}|{\int }_b^{\mathrm{\infty }}(t-x)H_n(t-x)dt|& =& |{\int }_{b-x}^{\mathrm{\infty }}y{H}_n(y)dy|={\int }_{x-b}^{\mathrm{\infty }}y{H}_n(y)dy\\ \le & {\int }_{y\ge \delta }y{H}_n(y)dy\le C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha -1}}\end{array}$$

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

$$\begin{array}{rcl}{I}_{2,1}& =& {\left({\int }_a^{\mathrm{\infty }}{\left(\right|f^{\prime }(x)\left|\right|{\int }_b^{\mathrm{\infty }}(t-x)H_n(t-x)dt\left|\right)}^{p}dx\right)}^{1/p}\\ \le & C\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha -1}}{\left({\int }_a^{\mathrm{\infty }}{|f^{\prime }(x)|}^{p}dx\right)}^{1/p}\le C{\parallel f^{\prime }\parallel }_{L_p([a,\mathrm{\infty }))}\frac{{\mu }_n^{\beta }}{{\delta }^{\alpha -1}}.\end{array}$$

(3.6)

Now we set

$$\theta (f^{″},x):=\underset{b\le t,t\ne x}{sup}\frac{1}{t-x}{\int }_x^t|f^{″}(u)|du,a\le x,$$

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

$${\parallel \theta (f^{″},x)\parallel }_{L_p([a,\mathrm{\infty }))}\le A_p{\parallel f^{″}\parallel }_{L_p([b,\mathrm{\infty }))},$$

(3.7)

where $A_p>0$ depend only on p [[6], Theorem 1, p.201]. Since

$$|{\int }_x^t(t-u)f^{″}(u)du|\le \pm {\int }_x^t|t-u||f^{″}(u)|du\le {(t-x)}^2\frac{1}{t-x}{\int }_x^t|f^{″}(u)|du,$$

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

$$\begin{array}{rcl}{I}_{2,2}& \le & {\left({\int }_a^{\mathrm{\infty }}\right|{\int }_b^{\mathrm{\infty }}\left({(t-x)}^2\frac{1}{t-x}{\int }_x^t\right|f^{″}(u)\left|du\right)H_n(t-x)dt{|}^{p}dx)}^{1/p}\\ \le & {\left({\int }_a^{\mathrm{\infty }}\right|\theta (f^{″};x){|}^p|{\int }_b^{\mathrm{\infty }}{(t-x)}^2{H}_n(t-x)dt{|}^{p}dx)}^{1/p}\\ \le & {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{u}^2{H}_n(u)du{\left({\int }_a^{\mathrm{\infty }}\right|\theta (f^{″};x){|}^{p}dx)}^{1/p}\\ \le & A_p{\mu }_n{\parallel f^{″}\parallel }_{L_p([b,\mathrm{\infty }))}.\end{array}$$

(3.8)

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

$$I_2\le I_{2,1}+I_{2,2}\le C({\parallel f^{\prime }\parallel }_{L_p([a,\mathrm{\infty }))}+{\parallel f^{″}\parallel }_{L_p([b,\mathrm{\infty }))}){\mu }_n.$$

(3.9)

So, from (3.5) and (3.9),

$$J\le I_1+I_2\le C({\parallel f\parallel }_{L_p({\mathbb{R}}^{+})}+{\parallel f^{\prime }\parallel }_{L_p([a,\mathrm{\infty }))}+{\parallel f^{″}\parallel }_{L_p([b,\mathrm{\infty }))}){\mu }_n.$$

(3.10)

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

$$\begin{array}{rcl}{K}_2& =& {\parallel J_n({\chi }_{1}f)-({\chi }_{1}f)\parallel }_{L_p([a,\mathrm{\infty }))}\\ \le & {\parallel J_n(({\chi }_{1}f)(t)-({\chi }_{1}f)(x),x)\parallel }_{L_p([a,\mathrm{\infty }))}+{\parallel ({\chi }_{1}f)(x)(J_n(e_0,x)-e_0)\parallel }_{L_p([a,\mathrm{\infty }))}\\ =& J+{\parallel ({\chi }_{1}f)(x)(J_n(e_0,x)-e_0)\parallel }_{L_p([a,\mathrm{\infty }))}\\ \le & C({\parallel f\parallel }_{L_p({\mathbb{R}}^{+})}+{\parallel f^{\prime }\parallel }_{L_p([a,\mathrm{\infty }))}+{\parallel f^{″}\parallel }_{L_p([b,\mathrm{\infty }))}){\mu }_n\\ +{\parallel J_n(e_0,x)-e_0\parallel }_{L_{\mathrm{\infty }}([a,\mathrm{\infty }))}{\parallel {\chi }_{1}f\parallel }_{L_p([a,\mathrm{\infty }))}\\ \le & C_1({\parallel f\parallel }_{L_p({\mathbb{R}}^{+})}+{\parallel f^{\prime }\parallel }_{L_p([a,\mathrm{\infty }))}+{\parallel f^{″}\parallel }_{L_p([b,\mathrm{\infty }))}){\mu }_n=O({\mu }_n).\end{array}$$

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

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

$$f(t)-f(x)=f^{\prime }(x)(t-x)+{\int }_x^t(t-u)d{f}^{\prime }(u).$$

(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

$$I_2^{\prime }:={\int }_a^{\mathrm{\infty }}|{\int }_b^{\mathrm{\infty }}(f(t)-f(x))H_n(t-x)dtdx|$$

(3.12)

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

$$\begin{array}{rcl}{I}_2^{\prime }& \le & |{\int }_a^{\mathrm{\infty }}{\int }_b^{\mathrm{\infty }}{f}^{\prime }(x)(t-x)H_n(t-x)dtdx|\\ +|{\int }_a^{\mathrm{\infty }}{\int }_b^{\mathrm{\infty }}{\int }_x^t(t-u)d{f}^{\prime }(u)H_n(t-x)dtdx|\\ =:& I_{2,1}^{\prime }+I_{2,2}^{\prime }.\end{array}$$

(3.13)

Now, we see that

$$\begin{array}{rcl}{I}_{2,1}^{\prime }& =& |{\int }_a^{\mathrm{\infty }}{\int }_b^{\mathrm{\infty }}{f}^{\prime }(x)(t-x)H_n(t-x)dtdx|\\ =& |{\int }_a^{\mathrm{\infty }}{f}^{\prime }(x){\int }_{x-b}^{\mathrm{\infty }}u{H}_n(u)dudx|\\ =& |{\int }_{a-b}^{\mathrm{\infty }}u{H}_n(u){\int }_a^{u+b}{f}^{\prime }(x)dxdu|\\ \le & \underset{a\le x<\mathrm{\infty }}{sup}|f^{\prime }(x)||{\int }_{\delta }^{\mathrm{\infty }}(u-\delta )u{H}_n(u)du|\\ \le & \underset{[b,\mathrm{\infty })}{sup}|f^{\prime }(x)|O({\mu }_n)\end{array}$$

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

$$\begin{array}{rcl}{I}_{2,2}^{\prime }& =& |{\int }_a^{\mathrm{\infty }}{\int }_b^{\mathrm{\infty }}{\int }_x^t(t-u)d{f}^{\prime }(u)H_n(t-x)dtdx|\\ =& |{\int }_a^{\mathrm{\infty }}{\int }_b^{\mathrm{\infty }}{\int }_0^{t-x}(t-x-y)d{f}^{\prime }(x+y)H_n(t-x)dtdx|\\ \le & 2{\int }_a^{\mathrm{\infty }}{\int }_b^{\mathrm{\infty }}{\int }_0^{|t-x|}|t-x||d{f}^{\prime }(x+y)|H_n(t-x)dtdx\because |t-x-y|\le 2|t-x|\\ \le & 2{\int }_a^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}{\int }_{j\eta \le |t-x|\le (j+1)\eta }{\int }_0^{(j+1)\eta }|d{f}^{\prime }(x+y)||t-x|H_n(t-x)dtdx\\ \le & 2\sum _{j=0}^{\mathrm{\infty }}{\int }_{j\eta \le |u|\le (j+1)\eta }|u|H_n(u)du{\int }_a^{\mathrm{\infty }}{\int }_0^{(j+1)\eta }|d{f}^{\prime }(x+y)|dx.\end{array}$$

Then

$$\begin{array}{c}{\int }_a^{\mathrm{\infty }}{\int }_0^{(j+1)\eta }|d{f}^{\prime }(x+y)|dx\hfill \\ ={\int }_a^{\mathrm{\infty }}{\int }_x^{x+(j+1)\eta }|d{f}^{\prime }(v)|dx\hfill \\ ={\int }_{a\le v\le a+(j+1)\eta }{\int }_{a\le x\le v}dx|d{f}^{\prime }(v)|+{\int }_{a+(j+1)\eta \le v<\mathrm{\infty }}{\int }_{v-(j+1)\eta \le x\le v}dx|d{f}^{\prime }(v)|\hfill \\ :=A_1+A_2.\hfill \end{array}$$

Since we can see

$$A_1={\int }_{a\le v\le a+(j+1)\eta }{\int }_{a\le x\le v}dx|d{f}^{\prime }(v)|={\int }_{a\le v\le a+(j+1)\eta }(v-a)|d{f}^{\prime }(v)|\le (j+1)\eta BV(f^{\prime };{\mathbb{R}}^{+})$$

and

$$A_2={\int }_{a+(j+1)\eta \le v<\mathrm{\infty }}{\int }_{v-(j+1)\eta \le x\le v}dx|d{f}^{\prime }(v)|\le (j+1)\eta BV(f^{\prime };{\mathbb{R}}^{+}),$$

we have

$${\int }_a^{\mathrm{\infty }}{\int }_0^{(j+1)\eta }|d{f}^{\prime }(x+y)|dx\le 2(j+1)\eta BV(f^{\prime };{\mathbb{R}}^{+}).$$

Now, we estimate ${\int }_{j\eta \le |u|\le (j+1)\eta }|u|H_n(u)du$ for a non-negative integer j. Let $j=0$. Then from (1.2) and (1.3), we have

$${\int }_{0\le |u|\le \eta }|u|H_n(u)du\le {({\int }_{0\le |u|\le \eta }{H}_n(u)du)}^{1/2}{({\int }_{0\le |u|\le \eta }{u}^2{H}_n(u)du)}^{1/2}=O\left({\mu }_n^{1/2}\right).$$

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

$${\int }_{j\eta \le |u|\le (j+1)\eta }|u|H_n(u)du\le C\frac{{\mu }_n^{\beta }}{{(j\eta )}^{\alpha -1}}.$$

Therefore, we have

$$\begin{array}{rcl}{I}_{2,2}^{\prime }& \le & {\int }_{0\le |u|\le \eta }|u|H_n(u)du{\int }_a^{\mathrm{\infty }}{\int }_0^{\eta }|d{f}^{\prime }(x+y)|dx\\ +\sum _{j=1}^{\mathrm{\infty }}{\int }_{j\eta \le |u|\le (j+1)\eta }|u|H_n(u)du{\int }_a^{\mathrm{\infty }}{\int }_0^{(j+1)\eta }|d{f}^{\prime }(x+y)|dx\\ \le & O\left({\mu }_n^{1/2}\right)\eta BV(f^{\prime };{\mathbb{R}}^{+})+\sum _{j=1}^{\mathrm{\infty }}\frac{1}{{(j\eta )}^{\alpha -1}}O\left({\mu }_n^{\beta }\right)(j+1)\eta BV(f^{\prime };{\mathbb{R}}^{+})\\ =& O\left({\mu }_n^{1/2}\right)\eta BV(f^{\prime };{\mathbb{R}}^{+})+O\left({\mu }_n^{\beta }\right)\frac{1}{{\eta }^{\alpha -2}}BV(f^{\prime };{\mathbb{R}}^{+})(\because \alpha >3).\end{array}$$

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

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

$${\parallel J_n(f;x)-f(x)\parallel }_{L_p([a,\mathrm{\infty }))}=O\left({\mu }_n^{\beta }\right)=o({\mu }_n).$$

 □

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

$$\psi (a)={\psi }^{\prime }(a)={\psi }^{″}(a)=0,\underset{r\to \mathrm{\infty }}{lim}{\psi }^{(i)}(r)=0,i=0,1,2$$

and

$$\psi (x)=0\text{for }x\notin [a,\mathrm{\infty }).$$

We use the bilinear functional;

$$A_n(f,\psi )=\frac{1}{{\mu }_n}{\int }_0^{\mathrm{\infty }}(J_n(f,x)-f(x))\psi (x)dx.$$

We will show that for fixed ψ, $A_n(\cdot ,\psi )$ is uniformly bounded on $L_p({\mathbb{R}}^{+})$. We see

$${\int }_0^{\mathrm{\infty }}{J}_n(f,x)\psi (x)dx={\int }_0^{\mathrm{\infty }}\psi (x){\int }_0^{\mathrm{\infty }}f(t)H_n(t-x)dtdx.$$

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

$$\psi (x)=\psi (t)+{\psi }^{\prime }(t)(x-t)+{\int }_x^t(x-u){\psi }^{″}(u)du.$$

Hence,

$$\begin{array}{c}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}f(t)\psi (x)H_n(t-x)dxdt\hfill \\ ={\int }_0^{\mathrm{\infty }}f(t)\psi (t){\int }_0^{\mathrm{\infty }}{H}_n(t-x)dxdt\hfill \\ +{\int }_0^{\mathrm{\infty }}f(t){\psi }^{\prime }(t){\int }_0^{\mathrm{\infty }}(x-t)H_n(t-x)dxdt\hfill \\ +{\int }_0^{\mathrm{\infty }}f(t){\int }_0^{\mathrm{\infty }}{\int }_x^t(x-u){\psi }^{″}(u)du{H}_n(t-x)dxdt\hfill \\ =I_1^{″}+I_2^{″}+I_3^{″}.\hfill \end{array}$$

From (1.2)

$$I_1^{″}={\int }_0^{\mathrm{\infty }}f(t)\psi (t){\int }_{-\mathrm{\infty }}^t{H}_n(u)dudt={\int }_0^{\mathrm{\infty }}f(t)\psi (t)(1-{\int }_t^{\mathrm{\infty }}{H}_n(u)du)dt.$$

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

$$\begin{array}{c}|{\int }_0^{\mathrm{\infty }}f(t)\psi (t){\int }_t^{\mathrm{\infty }}{H}_n(u)dudt|\hfill \\ \le |{\int }_0^{\mathrm{\infty }}f(t)\psi (t){\int }_t^{\mathrm{\infty }}{\left(\frac{u}{t}\right)}^{\alpha }{H}_n(u)dudt|\because {(u/t)}^{\alpha }\ge 1\hfill \\ \le O\left({\mu }_n^{\beta }\right){\int }_a^{\mathrm{\infty }}|f(t)\psi (t)|\frac{1}{t^{\alpha }}dt\text{by (1.4)}\hfill \\ \le O\left({\mu }_n^{\beta }\right)\underset{a\le t<\mathrm{\infty }}{sup}|\psi (t)|{\int }_a^{\mathrm{\infty }}|f(t)|\frac{1}{t^{\alpha }}dt\hfill \\ \le O\left({\mu }_n^{\beta }\right)\underset{a\le t<\mathrm{\infty }}{sup}|\psi (t)|{\parallel f\parallel }_{L_p([a,\mathrm{\infty }))}{\parallel t^{-\alpha }\parallel }_{L_q([a,\mathrm{\infty }))},\hfill \end{array}$$

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

$$I_1^{″}={\int }_0^{\mathrm{\infty }}f(t)\psi (t)dt+O\left({\mu }_n^{\beta }\right){\parallel f\parallel }_{L_p([a,\mathrm{\infty }))}.$$

We estimate $I_2^{″}$. We may $a\le t$, because $\psi (t)=0$ for $t\notin [a,\mathrm{\infty })$. Noting (1.4) and $\alpha <3$,

$$\begin{array}{rcl}\left|I_2^{″}\right|& =& {\int }_a^{\mathrm{\infty }}|f(t){\psi }^{\prime }(t)|{\int }_t^{\mathrm{\infty }}u{H}_n(u)dudt\\ \le & {\int }_a^{\mathrm{\infty }}|f(t){\psi }^{\prime }(t)|{\int }_t^{\mathrm{\infty }}\frac{u^{\alpha }}{t^{\alpha -1}}{H}_n(u)dudt\\ =& O\left({\mu }_n^{\beta }\right){\int }_a^{\mathrm{\infty }}|f(t){\psi }^{\prime }(t)|\frac{1}{t^{\alpha -1}}dt\\ =& O\left({\mu }_n^{\beta }\right){\parallel {\psi }^{\prime }\parallel }_{L_{\mathrm{\infty }}([a,\mathrm{\infty }))}{\parallel f\parallel }_{L_p([a,\mathrm{\infty }))}{\parallel t^{1-\alpha }\parallel }_{L_q([a,\mathrm{\infty }))}\\ =& O\left({\mu }_n^{\beta }\right){\parallel f\parallel }_{L_p([a,\mathrm{\infty }))}.\end{array}$$

Finally, we estimate $I_3^{″}$. Let $1<p<\mathrm{\infty }$ and $1/p+1/q=1$. As the estimation for $I_{2,2}$, we have by (1.3)

$$\begin{array}{rcl}\left|I_3^{″}\right|& =& |{\int }_0^{\mathrm{\infty }}f(t){\int }_0^{\mathrm{\infty }}{\int }_x^t(x-u){\psi }^{″}(u)du{H}_n(t-x)dxdt|\\ =& |{\int }_0^{\mathrm{\infty }}f(t){\int }_0^{\mathrm{\infty }}\frac{1}{t-x}{\int }_x^t{\psi }^{″}(u)du{(t-x)}^2{H}_n(t-x)dxdt|\\ \le & {\int }_a^{\mathrm{\infty }}|f(t)|\theta ({\psi }^{″},t){\int }_0^{\mathrm{\infty }}{(t-x)}^2{H}_n(t-x)dxdt\\ =& O({\mu }_n){\int }_a^{\mathrm{\infty }}|f(t)|\theta ({\psi }^{″},t)dt\le O({\mu }_n){\parallel f\parallel }_{L_p([a,\mathrm{\infty }))}{\parallel {\psi }^{″}\parallel }_{L_q([a,\mathrm{\infty }))}\\ =& O({\mu }_n){\parallel f\parallel }_{L_p([a,\mathrm{\infty }))}.\end{array}$$

We estimate $I_3^{″}$ for $p=1$. There exists η between x and t such that

$$\begin{array}{rcl}\left|I_3^{″}\right|& \le & |{\int }_0^{\mathrm{\infty }}f(t){\int }_0^{\mathrm{\infty }}{\psi }^{″}(\eta ){(t-x)}^2{H}_n(t-x)dxdt|\\ \le & \underset{a\le t<\mathrm{\infty }}{sup}|{\psi }^{″}(t)|{\int }_a^{\mathrm{\infty }}|f(t)|{\int }_0^{\mathrm{\infty }}{(t-x)}^2{H}_n(t-x)dxdt\\ \le & O({\mu }_n)\underset{a\le t<\mathrm{\infty }}{sup}|{\psi }^{″}(t)|{\parallel f\parallel }_{L_1([a,\mathrm{\infty }))}=O({\mu }_n){\parallel f\parallel }_{L_1([a,\mathrm{\infty }))}.\end{array}$$

Here we used (1.3). Consequently, it follows that $|A_n(f,\psi )|$ is uniformly bounded on $L_p({\mathbb{R}}^{+})$. Next, from Theorem 1.1 we see that for $f\in C_c^2({\mathbb{R}}^{+})$,

$$\underset{n\to \mathrm{\infty }}{lim}{A}_n(f,\psi )=\frac{1}{2}{\int }_a^{\mathrm{\infty }}{f}^{″}(x)\psi (x)dx=\frac{1}{2}{\int }_a^{\mathrm{\infty }}f(x){\psi }^{″}(x)dx.$$

(3.14)

Since $\{A_n(\cdot ,\psi )\}$ is uniformly bounded on $L_p({\mathbb{R}}^{+})$, and $C_c^2({\mathbb{R}}^{+})$ is dense in $L_p({\mathbb{R}}^{+})$, (3.14) yields

$$\underset{n\to \mathrm{\infty }}{lim}{A}_n(f,\psi )=\frac{1}{2}{\int }_a^{\mathrm{\infty }}f(x){\psi }^{″}(x)dx$$

(3.15)

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

$$\underset{j\to \mathrm{\infty }}{lim}{A}_{n_j}(f,\psi )=\{\begin{array}{cc}{\int }_0^{\mathrm{\infty }}h(x)\psi (x)dx,\hfill & p>1,\hfill \\ {\int }_0^{\mathrm{\infty }}\psi (x)dh(x),\hfill & p=1.\hfill \end{array}$$

(3.16)

From (3.15) and (3.16) we obtain

$$\frac{1}{2}{\int }_0^{\mathrm{\infty }}f(x){\psi }^{″}(x)dx=\{\begin{array}{cc}{\int }_0^{\mathrm{\infty }}h(x)\psi (x)dx,\hfill & p>1,\hfill \\ {\int }_0^{\mathrm{\infty }}\psi (x)dh(x),\hfill & p=1.\hfill \end{array}$$

(3.17)

A particular solution to (3.17) is

$$\frac{1}{2}f(x)=\{\begin{array}{cc}{\int }_{\eta }^x{\int }_{\eta }^{\xi }h(\mu )d\mu d\xi ,\hfill & p>1,\hfill \\ {\int }_{\eta }^x{\int }_{\eta }^{\xi }dh(\mu )d\xi ,\hfill & p=1.\hfill \end{array}$$

The homogeneous problem

$${\int }_a^{\mathrm{\infty }}f(x){\psi }^{″}(x)dx=0$$

has the general solution $f(x)=C_{1}x+C_2$ for $a\le x<\mathrm{\infty }$, since we can take $\psi \in C^2([a,\mathrm{\infty }))$ arbitrarily as ${\psi }^{″}\in L_q([a,\mathrm{\infty }))$, $\psi (a)={\psi }^{\prime }(a)=0$, ${lim}_{r\to \mathrm{\infty }}{\psi }^{(i)}(r)=0$, $i=0,1$. Hence, if $1<p<\mathrm{\infty }$, $f\in L_p^2([a,\mathrm{\infty }))$, and if $p=1$ then $f^{\prime }\in BV([a,\mathrm{\infty }))$. Hence, (i) and (ii) hold. We will show (iii). Now, if

$${\parallel J_n(f)-f\parallel }_{L_p([a,\mathrm{\infty }))}=o({\mu }_n),n\to \mathrm{\infty },$$

then

$$\begin{array}{rcl}|A_n(f,\psi )|& \le & \frac{1}{{\mu }_n}{\int }_a^{\mathrm{\infty }}|J_n(f,x)-f(x)||\psi (x)|dx\\ \le & \left(\underset{a\le x<\mathrm{\infty }}{sup}\right|\psi (x)\left|\right)\frac{C_p}{{\mu }_n}{\parallel J_n(f)-f\parallel }_{L_p([a,\mathrm{\infty }))},\end{array}$$

where $C_p>0$ is independent of n. Hence

$$\underset{n\to \mathrm{\infty }}{lim}{A}_n(f,\psi )=0.$$

(3.18)

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

$${\int }_a^{\mathrm{\infty }}f(x){\psi }^{″}(x)dx=0,$$

and so

$${\int }_a^{\mathrm{\infty }}{f}^{″}(x)\psi (x)dx=0,$$

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