1 Introduction

Ismail and May [9] studied exponential type operators \(L_{n}\). These operators have the form

$$\begin{aligned} \left( L_{n}f\right) \left( x\right) =\int _{0}^{\infty }k_{n}\left( x,t\right) f\left( t\right) dt \end{aligned}$$
(1)

with kernels \(k_{n}\left( x,t\right) \) satisfying the partial differential equation

$$\begin{aligned} \frac{\partial }{\partial x}k_{n}\left( x,t\right) =\frac{n}{p\left( x\right) }k_{n}\left( x,t\right) \left( t-x\right) . \end{aligned}$$

The special case \(p\left( x\right) =2x^{3/2}\) leads to the kernel

$$\begin{aligned} k_{n}\left( x,t\right) =e^{-n\sqrt{x}}\left( ne^{-nt/\sqrt{x} }t^{-1/2}I_{1}\left( 2n\sqrt{t}\right) +\delta \left( t\right) \right) , \end{aligned}$$
(2)

where \(\delta \left( \cdot \right) \) denotes the Dirac delta function and \( I_{1}\) the modified Bessel function of first kind given by

$$\begin{aligned} I_{1}(z)=\sum _{k=0}^{\infty }\frac{\left( \frac{z}{2}\right) ^{1+2k}}{ k!\varGamma \left( k+2\right) }. \end{aligned}$$

The corresponding operator [9, (3.16)] has the explicit representation

$$\begin{aligned} \left( T_{n}f\right) \left( x\right) =e^{-n\sqrt{x}}\left\{ n\int _{0}^{\infty }e^{-nt/\sqrt{x}}t^{-1/2}I_{1}\left( 2n\sqrt{t}\right) f\left( t\right) dt+f\left( 0\right) \right\} , \end{aligned}$$
(3)

where \(x\in \left( 0,\infty \right) \).

Note that the operators (3) can alternatively be written in the form

$$\begin{aligned} \left( T_{n}f\right) \left( x\right) =\frac{n}{\sqrt{x}}\sum _{k=1}^{\infty }s_{k}\left( n\sqrt{x}\right) \int _{0}^{\infty }s_{k-1}\left( \frac{nt}{ \sqrt{x}}\right) f\left( t\right) dt+e^{-n\sqrt{x}}f\left( 0\right) , \end{aligned}$$
(4)

where \(x\in \left( 0,\infty \right) \) and

$$\begin{aligned} s_{k}\left( x\right) =e^{-x}\frac{x^{k}}{k!}. \end{aligned}$$

We observe that the operators \(T_{n}\) are closely related to the well-known Phillips operators [10] (see [11, Eq. (1.1), case c=0]) given by

$$\begin{aligned} \left( P_{n}f\right) \left( x\right) =n\sum _{k=0}^{\infty }s_{k}\left( nx\right) \int _{0}^{\infty }s_{k-1}\left( nt\right) f\left( t\right) dt+e^{-nx}f\left( 0\right) . \end{aligned}$$

If we substitute n in the Phillips operators by the real parameter \(n/ \sqrt{x}\), we immediately get the operators (3) by Ismail and May in the form (4). More precisely, we have \(\left( P_{n/ \sqrt{x}}f\right) \left( x\right) =\left( T_{n}f\right) \left( x\right) \). However, unlike the operators \(T_{n}\), the Phillips operators \(P_{n}\) are not exponential type operators. Many other integral type operators (see, e.g., [1, 2, 6, 7]) including the Phillips operators are not of exponential type.

Recently Gupta [5] gave some direct results including an estimate in terms of the second order modulus of continuity and a Voronovskaja-type result for these operators. The recent paper [8] contains a quantitative asymptotic formula in terms of the modulus of continuity with exponential growth, a Korovkin-type result for exponential functions and also a Voronovskaja-type asymptotic formula in the simultaneous approximation.

The present paper deals with the rate of convergence of the operators \(T_{n}\) for functions of bounded variation.

2 Some Lemmas

Let \(e_{r}\) \(\left( r=0,1,2,\ldots \right) \) denote the monomials \( e_{r}\left( x\right) =x^{r}\). For real A, put \(h_{A}\left( t\right) =e^{At} \). The moments \(T_{n}e_{r}\) of the operators (3) are encoded in the moment generating function (m.g.f.) defined by \(\left( T_{n}h_{A}\right) \left( x\right) \). Gupta [5, Remark 1] derived the following formula.

Lemma 1

The moment generating function of the operators (3) is given by

$$\begin{aligned} \left( T_{n}h_{A}\right) \left( x\right) =\exp \left( \frac{nAx}{n-A\sqrt{x}} \right) \qquad \left( n>\left| A\right| \sqrt{x}\right) . \end{aligned}$$

Expanding in powers of A, we obtain, for \(\left| A\right| <n/\sqrt{ x}\),

$$\begin{aligned} \left( T_{n}h_{A}\right) \left( x\right)= & {} 1+xA+\left( \frac{2x^{3/2}}{n} +x^{2}\right) \frac{A^{2}}{2!}+\frac{\left( 6x^{2}+6nx^{5/2}+n^{2}x^{3}\right) }{n^{2}}\frac{A^{3}}{3!} \\&+\frac{\left( 24x^{5/2}+36nx^{3}+12n^{2}x^{7/2}+n^{3}x^{4}\right) }{n^{3}} \frac{A^{4}}{4!}+\cdots \end{aligned}$$

Obviously, the coefficients of \(A^{r}/r!\) provide the r-th order moments \( \left( T_{n}e_{r}\right) \left( x\right) \). Moreover, we have

$$\begin{aligned} \left( T_{n}e_{r}\right) \left( x\right) =x^{r}+\frac{r\left( r-1\right) }{n} x^{(2r-1)/2}+\frac{r\left( r-1\right) ^{2}\left( r-2\right) }{2n^{2}} x^{r-1}+O\left( n^{-3}\right) \end{aligned}$$

as \(n\rightarrow \infty \).

For \(x\in \mathbb {R}\), define \(\psi _{x}\left( t\right) =t-x\). The central moments \(\mu _{n,r}\left( x\right) :=\left( T_{n}\psi _{x}^{r}\right) \left( x\right) \) are crucial for the approximation properties of the operators (3).

Remark 1

The central moments \(\mu _{n,r}\left( x\right) =\left( T_{n}\psi _{x}^{r}\right) \left( x\right) \) of the operators (3) are the coefficients of \(A^{r}/r!\) in the expansion

$$\begin{aligned}&\exp \left( \frac{nAx}{n-A\sqrt{x}}-Ax\right) \\&\quad =1+\frac{2x^{3/2}}{n}\frac{A^{2}}{2!}+\frac{6x^{2}}{n^{2}}\frac{A^{3}}{3!} +\left( \frac{24x^{5/2}}{n^{3}}+\frac{12x^{3}}{n^{2}}\right) \frac{A^{4}}{4!} +\cdots \end{aligned}$$

In particular, we have \(\mu _{n,0}\left( x\right) =1\), \(\mu _{n,1}\left( x\right) =0\), \(\mu _{n,2}\left( x\right) =2x^{3/2}/n\).

Lemma 2

Let \(x\in (0,\infty )\) and let the kernel \(k_{n}(x,t)\) be as defined in (2). Then we have

$$\begin{aligned} \eta _{n}\left( x,y\right) =\int _{0}^{y}k_{n}\left( x,t\right) dt \le \frac{2x^{3/2}}{n\left( x-y\right) ^{2}}, \qquad 0<y<x \end{aligned}$$

and

$$\begin{aligned} 1-\eta _{n}\left( x,z\right) =\int _{z}^{\infty }k_{n}\left( x,t\right) dt \le \frac{2x^{3/2}}{n\left( z-x\right) ^{2}}, \qquad x<z<\infty . \end{aligned}$$

Proof

For \(0<y<x\) in view of Remark 1, we have

$$\begin{aligned} \int _{0}^{y}k_{n}\left( x,t\right) dt\le & {} \int _{0}^{y}k_{n}\left( x,t\right) \frac{(x-t)^2}{(x-y)^2} dt\\\le & {} (x-y)^{-2}\mu _{n,2}\left( x\right) \\= & {} \frac{ 2x^{3/2}}{n\left( x-y\right) ^{2}},\qquad 0<y<x. \end{aligned}$$

The other estimate follows analogously. \(\square \)

For the proof of the main result we need an estimation of \(\left( T_{n}\text { {sgn }}\psi _{x}\right) \left( x\right) \). It is easy to see that

$$\begin{aligned} \left( T_{n}\text {{sgn }}\psi _{x}\right) \left( x\right) =1-2\left( T_{n}\chi _{\left( 0,x\right) }\right) \left( x\right) , \end{aligned}$$

where \(\chi _{\left( 0,x\right) }\) denotes the characteristic function of the interval \(\left( 0,x\right) \). Therefore, we consider

$$\begin{aligned} \left( T_{n}\chi _{\left( 0,x\right) }\right) \left( x\right) =\int _{0}^{x}k_{n}(x,t)dt. \end{aligned}$$

The next result describes the asymptotic behaviour of the latter expression.

Lemma 3

For each \(x\in (0,\infty )\), we have

$$\begin{aligned} \int _{0}^{x}k_{n}(x,t)dt=\frac{1}{2}+\frac{1}{4\root 4 \of {x}\sqrt{\pi n}}+ \frac{1}{64\root 4 \of {x^{3}}\sqrt{\pi }n^{3/2}}+O\left( n^{-2}\right) \qquad \left( n\rightarrow \infty \right) . \end{aligned}$$

Proof

From the definition (3) we obtain

$$\begin{aligned} \left( T_{n}\chi _{\left( 0,x\right) }\right) \left( x\right) =ne^{-n\sqrt{x} }\int _{0}^{x}e^{-nt/\sqrt{x}}t^{-1/2}I_{1}\left( 2n\sqrt{t}\right) dt. \end{aligned}$$

The change of variable replacing t with \(\left( \sqrt{x}-t\right) ^{2}\) leads to the representation

$$\begin{aligned} \left( T_{n}\chi _{\left( 0,x\right) }\right) \left( x\right) =2ne^{-n\sqrt{x }}\int _{0}^{\sqrt{x}}e^{-n\left( \sqrt{x}-t\right) ^{2}/\sqrt{x}}I_{1}\left( 2n\left( \sqrt{x}-t\right) \right) dt. \end{aligned}$$

Taking advantage of the integral representation [3, (9.6.19)]

$$\begin{aligned} I_{1}\left( t\right) =\frac{1}{\pi }\int _{0}^{\pi }e^{z\cos \vartheta }\cos \vartheta d\vartheta \end{aligned}$$

we obtain

$$\begin{aligned} \left( T_{n}\chi _{\left( 0,x\right) }\right) \left( x\right) =\frac{2n}{\pi }\int _{0}^{\sqrt{x}}\int _{0}^{\pi }e^{-nh\left( t,\vartheta \right) }\cos \vartheta d\vartheta dt, \end{aligned}$$

where

$$\begin{aligned} h\left( t,\vartheta \right)= & {} \sqrt{x}+\frac{\left( \sqrt{x}-t\right) ^{2}}{ \sqrt{x}}-2\left( \sqrt{x}-t\right) \cos \vartheta \\= & {} \frac{t^{2}}{\sqrt{x}}+2\left( \sqrt{x}-t\right) \left( 1-\cos \vartheta \right) . \end{aligned}$$

Note that \(h\left( 0,0\right) =0\). For a sufficiently small positive real \( \delta \), there exists \(\varepsilon \in \left( 0,\sqrt{x}\right) \), such that \(h\left( t,\vartheta \right) >\delta \) if \(t\in \left[ \varepsilon , \sqrt{x}\right] \) and \(\vartheta \in \left[ 0,\pi \right] \). This implies that, for arbitrary small \(a>0\), there is a constant \(c>0\) such that

$$\begin{aligned} \left( T_{n}\chi _{\left( 0,x\right) }\right) \left( x\right) =\frac{2n}{\pi }\int _{0}^{a}\int _{0}^{a}e^{-nh\left( t,\vartheta \right) }\cos \vartheta d\vartheta dt+O\left( e^{-cn}\right) \qquad \left( n\rightarrow \infty \right) . \end{aligned}$$

We study

$$\begin{aligned} J_{n}\left( x\right) :=\frac{2n}{\pi }\int _{0}^{a}\int _{0}^{a}e^{-nh\left( t,\vartheta \right) }\cos \vartheta d\vartheta dt. \end{aligned}$$

This equation can be rewritten in the form

$$\begin{aligned} J_{n}\left( x\right) =\frac{2n}{\pi }\int _{0}^{a}\int _{0}^{a}\exp \left( -n\left( \frac{t^{2}}{\sqrt{x}}+\sqrt{x}\vartheta ^{2}\right) \right) g_{n}\left( t,\vartheta \right) d\vartheta dt, \end{aligned}$$

where

$$\begin{aligned} g_{n}\left( t,\vartheta \right) =e^{-n\left( 2\left( \sqrt{x}-t\right) \left( 1-\cos \vartheta \right) -\sqrt{x}\vartheta ^{2}\right) }\cos \vartheta . \end{aligned}$$

A further change of variable replacing t with \(\sqrt{x/n}t\) and replacing \( \vartheta \) with \(\vartheta /\sqrt{n}\) yields

$$\begin{aligned} J_{n}\left( x\right) =\frac{2\sqrt{x}}{\pi }\int _{0}^{a\sqrt{n}}\int _{0}^{a \sqrt{n}}\exp \left( -\sqrt{x}\left( t^{2}+\vartheta ^{2}\right) \right) g_{n}\left( \sqrt{\frac{x}{n}}t,\frac{\vartheta }{\sqrt{n}}\right) d\vartheta dt. \end{aligned}$$

Noting that \(-2\left( 1-\cos \vartheta \right) +\vartheta ^{2}=:\vartheta ^{4}F\left( \vartheta \right) \) and \(2\left( 1-\cos \vartheta \right) =:\vartheta ^{2}G\left( \vartheta \right) \), where F and G are entire functions, we see that

$$\begin{aligned} g_{n}\left( \sqrt{x/n}t,\vartheta /\sqrt{n}\right)= & {} \exp \left( \frac{ \sqrt{x}}{n}\vartheta ^{4}F\left( \vartheta \right) +\frac{t}{\sqrt{n}} \vartheta ^{2}G\left( \frac{\vartheta }{\sqrt{n}}\right) \right) \cos \frac{ \vartheta }{\sqrt{n}} \\= & {} \sum _{i_{1},i_{2},i_{3}\ge 0}a\left( i_{1},i_{2},i_{3}\right) \left[ \frac{\sqrt{x}}{n}\vartheta ^{4}F\left( \frac{\vartheta }{\sqrt{n}}\right) \right] ^{i_{1}}\left[ \frac{t}{\sqrt{n}}\vartheta ^{2}G\left( \frac{ \vartheta }{\sqrt{n}}\right) \right] ^{i_{2}}\left[ \frac{\vartheta }{\sqrt{n }}\right] ^{i_{3}}. \end{aligned}$$

Therefore, we have

$$\begin{aligned}&g_{n}\left( \sqrt{x/n}t,\vartheta /\sqrt{n}\right) \\&\quad =\sum _{\begin{array}{c} i_{1},i_{2},i_{3}\ge 0, \\ 2i_{1}+i_{2}+i_{3}\le 3 \end{array}} n^{-\left( 2i_{1}+i_{2}+i_{3}\right) /2}a\left( i_{1},i_{2},i_{3}\right) \left[ \sqrt{x}F\left( \frac{\vartheta }{\sqrt{n}}\right) \right] ^{i_{1}} \\&\quad \left[ G\left( \frac{\vartheta }{\sqrt{n}}\right) \right] ^{i_{2}}t^{i_{2}} \vartheta ^{4i_{1}+2i_{2}+i_{3}}+O\left( n^{-2}\right) \end{aligned}$$

as \(n\rightarrow \infty \). After some calculations we arrive at

$$\begin{aligned} J_{n}\left( x\right)= & {} \frac{1}{2}+\frac{1}{4\root 4 \of {x}\sqrt{\pi n}}+\frac{1}{64\root 4 \of {x^{3}} \sqrt{\pi }n^{3/2}}+O\left( n^{-2}\right) \end{aligned}$$

as \(n\rightarrow \infty \). \(\square \)

3 Rate of convergence

Theorem 1

Let f be a function of bounded variation on each finite subinterval of \(\left( 0,\infty \right) \). Suppose that f satisfies the growth condition \(\left| f\left( t\right) \right| \le C e^{\gamma t}\) , for \(t>0\). Then, it exists a sequence \(\left( \varepsilon _{n}\left( x\right) \right) \) with \(\varepsilon _{n}\left( x\right) \rightarrow 0\) as \( n\rightarrow \infty \), such that, for \(n>2\gamma \sqrt{x}\),

$$\begin{aligned}&\left| \left( T_{n}f\right) \left( x\right) -\frac{1}{2}\left( f\left( x+\right) + f\left( x-\right) \right) \right| \\&\quad \le \left( \frac{1}{2\root 4 \of {x}\sqrt{\pi n}}+\frac{1}{32\root 4 \of {x^{3}} \sqrt{\pi }n^{3/2}}\right) \left| f\left( x+\right) -f\left( x-\right) \right| +\frac{2+\sqrt{x}}{n\sqrt{x}}\sum _{k=1}^{n}V_{x-x/\sqrt{k}}^{x+x/ \sqrt{k}}(f_{x}) \\&\qquad +\frac{2Ce^{\gamma x}}{n\sqrt{x}} +\frac{\sqrt{2}Cx^{3/4}}{\sqrt{n}}\exp \left( \frac{n\gamma x}{n-2\gamma \sqrt{x}}\right) +\frac{\varepsilon _{n}\left( x\right) }{n^{2}}, \end{aligned}$$

where

$$\begin{aligned} f_{x}\left( t\right) =\left\{ \begin{array}{lrl} f\left( t\right) -f\left( x-\right) &{} &{} \left( 0<t<x\right) , \\ 0 &{} &{} \left( t=x\right) , \\ f\left( t\right) -f\left( x+\right) &{} &{} \left( x<t<\infty \right) . \end{array} \right. \end{aligned}$$

Proof

Let \(x\in (0,\infty )\). Our starting point is the inequality

$$\begin{aligned}&\left| \left( T_{n}f\right) \left( x\right) -\frac{1}{2}\left( f\left( x+\right) + f\left( x-\right) \right) \right| \\&\quad \le \frac{1}{2}\left| f\left( x+\right) -f\left( x-\right) \right| \cdot \left| \left( T_{n}\text {{sgn }}\psi _{x}\right) \left( x\right) \right| +\left| \left( T_{n}f_{x}\right) \left( x\right) \right| . \end{aligned}$$

Firstly, by Lemma 3, we obtain

$$\begin{aligned} \left( T_{n}\text {{sgn }}\psi _{x}\right) \left( x\right)= & {} 1-2\left( T_{n}\chi _{\left( 0,x\right) }\right) \left( x\right) =1-2\left( \frac{1}{2} +\frac{1}{4\root 4 \of {x}\sqrt{\pi n}}+\frac{1}{64\root 4 \of {x^{3}}\sqrt{\pi } n^{3/2}}+O\left( n^{-2}\right) \right) \\= & {} \frac{-1}{2\root 4 \of {x}\sqrt{\pi n}}+\frac{-1}{32\root 4 \of {x^{3}}\sqrt{\pi } n^{3/2}}+O\left( n^{-2}\right) \end{aligned}$$

as \(n\rightarrow \infty \). Next we estimate \(\left( T_{n}f_{x}\right) \left( x\right) \) as follows:

$$\begin{aligned} \left( T_{n}f_{x}\right) \left( x\right)= & {} \int _{0}^{\infty }k_{n}\left( x,t\right) f_{x}\left( t\right) dt \\= & {} \left( \int _{0}^{x-x/\sqrt{n}}+\int _{x-x/\sqrt{n}}^{x+x/\sqrt{n} }+\int _{x+x/\sqrt{n}}^{\infty }\right) k_{n}\left( x,t\right) f_{x}\left( t\right) dt \\= & {} :E_{1}+E_{2}+E_{3}. \end{aligned}$$

Integrating the first term by parts we obtain

$$\begin{aligned} E_{1}= & {} \int _{0}^{x-x/\sqrt{n}}f_{x}\left( t\right) d_{t}(\eta _{n}\left( x,t\right) ) \\= & {} f_{x}\left( x-\frac{x}{\sqrt{n}}\right) \eta _{n}\left( x,x-\frac{x}{ \sqrt{n}}\right) -\int _{0}^{x-x/\sqrt{n}}\eta _{n}\left( x,t\right) d_{t}\left( f_{x}\left( t\right) \right) . \end{aligned}$$

Since \(\left| f_{x}\left( y\right) \right| \le V_{y}^{x}\left( f_{x}\right) \), we have

$$\begin{aligned} \left| E_{1}\right| \le V_{x-x/\sqrt{n}}^{x}\left( f_{x}\right) \cdot \eta _{n}\left( x,x-\frac{x}{\sqrt{n}}\right) +\int _{0}^{x-x/\sqrt{n} }\eta _{n}\left( x,t\right) d_{t}\left( -V_{t}^{x}\left( f_{x}\right) \right) . \end{aligned}$$

Applying Lemma 2, and in the next step integrating by parts, we get

$$\begin{aligned} \left| E_{1}\right|\le & {} \frac{2x^{3/2}}{n[x-\left( x-x/\sqrt{n} \right) ]^{2}}V_{x-x/\sqrt{n}}^{x}(f_{x})+\frac{2x^{3/2}}{n}\int _{0}^{x-x/ \sqrt{n}}\frac{1}{\left( x-t\right) ^{2}}d_{t}(-V_{t}^{x}(f_{x})) \\= & {} \frac{2x^{3/2}}{n}\left[ \frac{1}{x^{2}}V_{0}^{x}\left( f_{x}\right) +2\int _{0}^{x-x/\sqrt{n}}\frac{1}{\left( x-t\right) ^{3}}V_{t}^{x}\left( f_{x}\right) dt\right] \\\le & {} \frac{2}{n\sqrt{x}}\sum _{k=1}^{n}V_{x-x/\sqrt{k}}^{x}\left( f_{x}\right) . \end{aligned}$$

Next for \(t\in \left[ x-x/\sqrt{n},x+x/\sqrt{n}\right] \) and by the fact \( \int _{x-x/\sqrt{n}}^{x+x/\sqrt{n}}d_{t}(\eta _{n}\left( x,t\right) )\le 1\), we conclude that

$$\begin{aligned} \left| E_{2}\right| \le \frac{1}{n}\sum _{k=1}^{n}V_{x-x/\sqrt{k} }^{x+x/\sqrt{k}}\left( f_{x}\right) . \end{aligned}$$

Finally,

$$\begin{aligned} E_{3}= & {} \int _{x+x/\sqrt{n}}^{\infty }k_{n}\left( x,t\right) f_{x}\left( t\right) dt \\= & {} \left( \int _{x+x/\sqrt{n}}^{2x}+\int _{2x}^{\infty }\right) k_{n}\left( x,t\right) f_{x}\left( t\right) dt \\=: & {} E_{31}+E_{32}. \end{aligned}$$

Arguing analogously as in estimate of \(E_{1}\), we have

$$\begin{aligned} \left| E_{31}\right| \le \frac{2}{n\sqrt{x}} \sum _{k=1}^{n}V_{x}^{x+x/\sqrt{k}}\left( f_{x}\right) . \end{aligned}$$

By assumption, we have \(\left| f_{x}\left( t\right) \right| \le \left| f\left( x+\right) \right| +\left| f\left( t\right) \right| \le C\left( e^{\gamma x}+e^{\gamma t}\right) \), such that

$$\begin{aligned} \left| E_{32}\right| \le \int _{2x}^{\infty }k_{n}\left( x,t\right) \left| f_{x}\left( t\right) \right| dt\le C\int _{2x}^{\infty }k_{n}\left( x,t\right) \left( e^{\gamma x}+e^{\gamma t}\right) dt. \end{aligned}$$

Since \(\left| t-x\right| /x\ge 1\), for \(t\ge 2x\), we conclude that

$$\begin{aligned} \left| E_{32}\right|\le & {} \frac{Ce^{\gamma x}}{x^{2}} \int _{0}^{\infty }k_{n}\left( x,t\right) \left( t-x\right) ^{2}dt+\frac{C}{x} \int _{0}^{\infty }k_{n}\left( x,t\right) e^{\gamma t}\left| t-x\right| dt \\= & {} \frac{Ce^{\gamma x}}{x^{2}}\mu _{n,2}\left( x\right) +\frac{C}{x}\left( \mu _{n,2}(x)\cdot \left( T_{n}e^{2\gamma \cdot }\right) \left( x\right) \right) ^{1/2} \\= & {} \frac{2Ce^{\gamma x}}{n\sqrt{x}}+C\left( \frac{2x^{3/2}}{n}e^{\frac{ 2n\gamma x}{n-2\gamma \sqrt{x}}}\right) ^{1/2}. \end{aligned}$$

Collecting the estimates of \(E_{1},E_{2},E_{3}\), we get the desired result. \(\square \)

Remark 2

One may extend the present result to generalized bounded variations of integral type (see [4]) and for \(\bigwedge BV^{(p)}\) and \(\varphi \bigwedge BV\) discussed in [12]. This may be treated as further research in this direction.