Introduction

In the year 1995, Lupaṣ [9] proposed the Lupaṣ operators:

$$\displaystyle \begin{aligned} \begin{array}{rcl} L_n(f,x)=\sum_{k=0}^{\infty}\frac{2^{-nx} (nx)_k}{k! 2^k}f\left(\frac{k}{n}\right), \end{array} \end{aligned} $$

where (nx)k is the rising factorial given by

$$\displaystyle \begin{aligned}(nx)_k=nx(nx+1)(nx+2)\cdots(nx+k-1), \, \, \, (nx)_0=1.\end{aligned}$$

Four years later, Agratini [2] introduced the Kantorovich-type generalization of the operators L n. After a decade Erençin and Taşdelen [4] considered a generalization of the operators discussed in [2] based on some parameters and established some approximation properties. We start here with the Kantorovich variant of Lupaṣ operators defined by

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} K_n(f,x)=n\sum_{k=0}^{\infty}\frac{2^{-na_n(x)} (na_n(x))_k}{k! 2^k}\int_{k/n}^{(k+1)/n}f(t)dt \end{array} \end{aligned} $$
(1)

with the hypothesis that these operators preserve the function e −x. Then using

$$\displaystyle \begin{aligned}\sum_{k=0}^{\infty} \frac{(a)_k}{k!}z^k=(1-z)^{-a},|z|<1,\end{aligned}$$

we write

$$\displaystyle \begin{aligned} \begin{array}{rcl} e^{-x} &\displaystyle =&\displaystyle n\sum_{k=0}^{\infty}\frac{2^{-na_n(x)} (na_n(x))_k}{k! 2^k}\int_{k/n}^{(k+1)/n}e^{-t}dt\\ &\displaystyle =&\displaystyle n\sum_{k=0}^{\infty}\frac{2^{-na_n(x)} (na_n(x))_k}{k! 2^k}e^{-k/n}(1-e^{-1/n})\\ &\displaystyle =&\displaystyle n(1-e^{-1/n})(2-e^{-1/n})^{-na_n(x)}, \end{array} \end{aligned} $$

which concludes

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} a_n(x)=\frac{x+\ln\left(n(1-e^{-1/n})\right)}{n \ln\left(2-e^{-1/n}\right)}. \end{array} \end{aligned} $$
(2)

Therefore the operators defined by (1) take the following alternate form

$$\displaystyle \begin{aligned} \begin{array}{rcl} K_n(f,x)&\displaystyle =&\displaystyle n\sum_{k=0}^{\infty}\frac{1} {k! 2^k}2^{-\frac{x+\ln\left(n(1-e^{-1/n})\right)}{ \ln\left(2-e^{-1/n}\right)}}\left(\frac{x+\ln\left(n(1-e^{-1/n})\right)}{ \ln\left(2-e^{-1/n}\right)}\right)_k\\ &\displaystyle &\displaystyle \int_{k/n}^{(k+1)/n}f(t)dt. \end{array} \end{aligned} $$

These operators preserve constant and the function e −x. The quantitative direct estimate for a sequence of linear positive operators was discussed and proved in [8] as the following result:

Theorem A ([8])

If a sequence of linear positive operators L n : C ∗[0, ∞) → C ∗[0, ∞), (where C ∗[0, ∞) be the subspace of all real-valued continuous functions, which has finite limit at infinity) satisfy the equalities

$$\displaystyle \begin{aligned} \begin{array}{rcl} ||L_n(e_0)-1||{}_{[0,\infty)}&\displaystyle =&\displaystyle \alpha _n\\ ||L_n(e^{-t})-e^{-x}||{}_{[0,\infty)}&\displaystyle =&\displaystyle \beta _n\\ ||L_n(e^{-2t})-e^{-2x}||{}_{[0,\infty)}&\displaystyle =&\displaystyle \gamma _n \end{array} \end{aligned} $$

then

$$\displaystyle \begin{aligned} \begin{array}{rcl} ||L_nf-f||{}_{[0,\infty)}\leq 2\omega ^*\left(f,\sqrt{ \alpha _n +2\beta _n+\gamma _n}\right), f\in C^*[0,\infty), \end{array} \end{aligned} $$

where the norm is the uniform norm and the modulus of continuity is defined by

$$\displaystyle \begin{aligned} \begin{array}{rcl} \omega ^*(f,\delta)= \sup_{|e^{-x}-e^{-t}|\leq \delta , x,t>0}|f(t)-f(x)|. \end{array} \end{aligned} $$

Very recently Acar et al. [1] used the above theorem and established quantitative estimates for the modification of well-known Szász–Mirakyan operators, which preserve the function e 2ax, a > 0. Actually such a modification may be important to discuss approximation properties, but if the operators preserve e −x or e −2x, then such results may provide better approximation in the sense of reducing the error. In the present paper, we study Kantorovich variant of Lupaṣ operators defined by (1) with a n(x) as given by (2) preserving e −x. We calculate a uniform estimate and establish a quantitative asymptotic result for the modified operators.

Auxiliary Results

In order to prove the main results, the following lemmas are required.

Lemma 1

The following representation holds

$$\displaystyle \begin{aligned} \begin{array}{rcl} K_n(e^{At},x)=\frac{n(e^{A/n}-1)}{A}(2-e^{A/n})^{-na_n(x)}. \end{array} \end{aligned} $$

Proof

We have

$$\displaystyle \begin{aligned} \begin{array}{rcl} K_n(e^{At},x)&\displaystyle =&\displaystyle n \sum_{k=0}^{\infty}\frac{2^{-na_n(x)} (na_n(x))_k}{k! 2^k}\int_{k/n}^{(k+1)/n}e^{At}dt\\ &\displaystyle =&\displaystyle n \sum_{k=0}^{\infty}\frac{2^{-na_n(x)} (na_n(x))_k}{k! 2^k}\left[e^{At}(e^{A/n}-1)\right]\\ &\displaystyle =&\displaystyle \frac{n(e^{A/n}-1)}{A}\left(2-e^{A/n}\right)^{-na_n(x)}. \end{array} \end{aligned} $$

Lemma 2

If e r(t) = t r, r ∈ N 0, then the moments of the operators (1) are given as follows:

$$\displaystyle \begin{aligned} \begin{array}{rcl} K_n(e_0,x)&\displaystyle =&\displaystyle 1,\\K_n(e_1,x)&\displaystyle =&\displaystyle a_n(x)+\frac{1}{2n},\\ K_n(e_2,x)&\displaystyle =&\displaystyle (a_n(x))^2+\frac{3a_n(x)}{n}+\frac{1}{3n^2},\\ K_n(e_3,x) &\displaystyle =&\displaystyle (a_n(x))^3+\frac{15(a_n(x))^2}{2n}+\frac{10a_n(x)}{n^2}+\frac{1}{4n^3},\\ K_n(e_4,x) &\displaystyle =&\displaystyle (a_n(x))^4+\frac{14(a_n(x))^3}{n}+\frac{50(a_n(x))^2}{n^2}+\frac{53a_n(x)}{n^3}+\frac{1}{5n^4}. \end{array} \end{aligned} $$

Lemma 3

If \(\mu _{n,m}(x)=K_n\left ((t-x)^m,x\right ),\) then by using Lemma 2 , we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mu_{n,0}(x)&\displaystyle =&\displaystyle 1,\\ \mu_{n,1}(x)&\displaystyle =&\displaystyle a_n(x)+\frac{1}{2n}-x,\\ \mu_{n,2}(x)&\displaystyle =&\displaystyle \left(a_n(x)-x\right)^2+\frac{3a_n(x)}{n}-\frac{x}{n}+\frac{1}{3n^2},\\ \mu_{n,4}(x)&\displaystyle =&\displaystyle \left(a_n(x)-x\right)^4+\frac{14(a_n(x))^3-30x(a_n(x))^2+18x^2a_n(x)-2x^3}{n}\\ &\displaystyle &\displaystyle +\frac{50(a_n(x))^2-40xa_n(x)+2x^2}{n^2}+\frac{53a_n(x)-x}{n^3}+\frac{1}{5n^4}. \end{array} \end{aligned} $$

Furthermore,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n \rightarrow \infty} n\left[\frac{x+\ln\left(n(1-e^{-1/n})\right)}{n\ln\left(2-e^{-1/n}\right)}+\frac{1}{2n}-x\right]=x \end{array} \end{aligned} $$

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \lim_{n \rightarrow \infty} n\left[\left(\frac{x+\ln\left(n(1-e^{-1/n})\right)}{n\ln\left(2-e^{-1/n}\right)}-x\right)^2 +\frac{3\left[x+\ln\left(n(1-e^{-1/n})\right)\right]}{n^2\ln\left(2-e^{-1/n}\right)}\right.\\ &\displaystyle &\displaystyle \quad \left. -\frac{x}{n}+\frac{1}{3n^2}\right]=2x. \end{array} \end{aligned} $$

Main Results

In this section, we present the quantitative estimates.

Theorem 1

For f ∈ C ∗[0, ∞), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} || K_nf-f||{}_{[0,\infty)}\leq 2\omega ^*\left(f,\sqrt{\gamma _n}\right), \end{array} \end{aligned} $$

where

$$\displaystyle \begin{aligned} \begin{array}{rcl} \gamma _n &\displaystyle =&\displaystyle || K_n(e^{-2t})-e^{-2x}||{}_{[0,\infty)}\\ &\displaystyle =&\displaystyle \left|\left|\frac{2xe^{-2x}}{n}+\frac{(24x^2-48x-11)e^{-2x}}{12n^2} +O\left(\frac{1}{n^3}\right)\right|\right|{}_{[0,\infty)}. \end{array} \end{aligned} $$

Proof

The operators K n preserve the constant and e −x. Thus α n = β n = 0. We only have to evaluate γ n. In view of Lemma 1, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} K_n(e^{-2t},x)&\displaystyle =&\displaystyle n \sum_{k=0}^{\infty}\frac{2^{-na_n(x)} (na_n(x))_k}{k! 2^k}\int_{k/n}^{(k+1)/n}e^{-2t}dt\\ &\displaystyle =&\displaystyle \frac{n(1-e^{-2/n})}{2}\left(2-e^{-2/n}\right)^{-na_n(x)}, \end{array} \end{aligned} $$

where a n(x) is given as

$$\displaystyle \begin{aligned} \begin{array}{rcl} a_n(x)=\frac{x+\ln\left(n(1-e^{-1/n})\right)}{n \ln\left(2-e^{-1/n}\right)}. \end{array} \end{aligned} $$

Thus using the software Mathematica, we get at once

$$\displaystyle \begin{aligned} \begin{array}{rcl} K_n(e^{-2t},x) &\displaystyle =&\displaystyle \frac{n(1-e^{-2/n})}{2}\left(2-e^{-2/n}\right)^{\left[-\frac{x+\ln\left(n(1-e^{-1/n})\right)}{ \ln\left(2-e^{-1/n}\right)}\right]}\\ &\displaystyle =&\displaystyle e^{-2x}+\frac{2xe^{-2x}}{n}+\frac{(24x^2-48x-11)e^{-2x}}{12n^2}+O\left(\frac{1}{n^3}\right). \end{array} \end{aligned} $$

This completes the proof of the theorem.

Theorem 2

Let f, f ′′∈ C ∗[0, ∞). Then the inequality

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \left|n\left[K_n(f,x)-f(x)\right]-x[f^{\prime}(x)+f^{\prime \prime}(x)]\right|\\ &\displaystyle &\displaystyle \quad \leq |p_n(x)||f^{\prime}|+|q_n(x)||f^{\prime \prime}|+2\left(2q_n(x)+2x+r_n(x)\right)\omega ^*\left(f^{\prime \prime},n^{-1/2}\right) \end{array} \end{aligned} $$

holds for any x ∈ [0, ∞), where

$$\displaystyle \begin{aligned} \begin{array}{rcl} p_n(x)&\displaystyle =&\displaystyle n \mu_{n,1}(x)-x,\\ q_n(x) &\displaystyle =&\displaystyle \frac{1}{2}\left(n\mu_{n,2}(x)-2x\right),\\ r_n(x) &\displaystyle =&\displaystyle n^2 \sqrt{K_n\left((e^{-x}-e^{-t})^4,x\right)}\sqrt{\mu_{n,4}(x)}, \end{array} \end{aligned} $$

and μ n,1(x), μ n,2(x), and μ n,4(x) are given in Lemma 3.

Proof

By Taylor’s expansion, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} f(t)=f(x)+(t-x)f^\prime(x)+\frac{1}{2}(t-x)^2 f^{\prime \prime}(x)+\varepsilon(t,x)(t-x)^2, \end{array} \end{aligned} $$
(3)

where

$$\displaystyle \begin{aligned}\varepsilon(t,x)=\frac{f^{\prime\prime}(\eta)-f^{\prime\prime}(x)}{2}\end{aligned}$$

and η is a number lying between x and t. If we apply the operator K n to both sides of (3), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl}&\displaystyle &\displaystyle \left|K_n(f,x)-f(x)-\mu_{n,1}(x)f^\prime(x)-\frac{1}{2}\mu_{n,2}(x) f^{\prime\prime}(x)\right|\\ &\displaystyle &\displaystyle \quad \le |K_n(\varepsilon(t,x)(t-x)^2,x)|, \end{array} \end{aligned} $$

Applying Lemma 2, we get

$$\displaystyle \begin{aligned} \begin{array}{rcl}&\displaystyle &\displaystyle \left|n[K_n(f,x)-f(x)]-x[f^\prime(x)+ f^{\prime\prime}(x)]\right|\\ &\displaystyle &\displaystyle \quad \le\left|n\mu_{n,1}(x)-x\right||f^\prime(x)|+\frac{1}{2}\left|n\mu_{n,2}(x)-2x\right||f^{\prime\prime}(x)|\\ &\displaystyle &\displaystyle \qquad +|nK_n(\varepsilon(t,x)(t-x)^2,x)|. \end{array} \end{aligned} $$

Put p n(x) := nμ n,1(x) − x and \(q_n(x):=\frac {1}{2}[n\mu _{n,2}(x)-2x].\) Thus

$$\displaystyle \begin{aligned} \begin{array}{rcl}&\displaystyle &\displaystyle \left|n[K_n(f,x)-f(x)]-x[f^\prime(x)+ f^{\prime\prime}(x)]\right|\\ &\displaystyle &\displaystyle \quad \le |p_n(x)|.|f^\prime (x)|+|q_n(x)|.|f^{\prime\prime}(x)|+|nK_n(\varepsilon(t,x)(t-x)^2,x)|. \end{array} \end{aligned} $$

In order to complete the proof of the theorem, we must estimate the term |nK n(ε(t, x)(t − x)2, x)|. Using the property

$$\displaystyle \begin{aligned}|f(t)-f(x)|\le \left(1+\frac{(e^{-t}-e^{-x})^2}{\delta^2}\right)\omega^*(f,\delta),\delta>0,\end{aligned}$$

we get

$$\displaystyle \begin{aligned}|\varepsilon(t,x)|\le \left(1+\frac{(e^{-t}-e^{-x})^2}{\delta^2}\right)\omega^*(f^{\prime\prime},\delta).\end{aligned}$$

For |e −x − e −t|≤ δ, one has |ε(t, x)|≤ 2ω ∗(f ′′, δ). In case |e −x − e −t| > δ, then \(|\varepsilon (t,x)|< 2\frac {(e^{-x}-e^{-t})^2}{\delta ^2}\omega ^*(f^{\prime \prime },\delta ).\) Thus

$$\displaystyle \begin{aligned}|\varepsilon(t,x)|\le 2\left(1+\frac{(e^{-x}-e^{-t})^2}{\delta^2}\omega^*(f^{\prime\prime},\delta)\right).\end{aligned}$$

Obviously using this and Cauchy–Schwarz inequality after choosing δ = n −1∕2, we get

$$\displaystyle \begin{aligned} \begin{array}{rcl}nK_n(|\varepsilon(t,x)|(t-x)^2,x)&\displaystyle \le&\displaystyle 2\omega^*(f^{\prime\prime}(x),n^{-1/2})\left[n\mu_{n,2}(x)+r_n(x)\right]\\ &\displaystyle =&\displaystyle 2\omega^*(f^{\prime\prime}(x),n^{-1/2})\left[2q_n(x)+2x+r_n(x)\right], \end{array} \end{aligned} $$

where r n(x) = n 2[K n((e −x − e −t)4, x).μ n,4(x)]1∕2 and

$$\displaystyle \begin{aligned} \begin{array}{rcl}K_n((e^{-x}-e^{-t})^4,x)&\displaystyle =&\displaystyle -\frac{n}{4}(e^{-4/n}-1)(2-e^{-4/n})^{-na_n(x)}\\ &\displaystyle &\displaystyle +\frac{4n}{3}e^{-x}(e^{-3/n}-1)(2-e^{-3/n})^{-na_n(x)}\\ &\displaystyle &\displaystyle -3ne^{-2x}(e^{-2/n}-1)(2-e^{-2/n})^{-na_n(x)}\\ &\displaystyle &\displaystyle +4n e^{-3x}(e^{-1/n}-1)(2-e^{-1/n})^{-na_n(x)}+e^{-4x}. \end{array} \end{aligned} $$

This completes the proof of the result.

Remark 1

From the Lemma 3, p n(x) → 0, q n(x) → 0 as n →∞ and using Mathematica, we get

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n \rightarrow \infty} n^2 \mu_{n,4}(x)=12x^2. \end{array} \end{aligned} $$

Furthermore

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n \rightarrow \infty} n^2 K_n\left((e^{-t}-e^{-x})^4,x\right)= 12 e^{-4x}x^2. \end{array} \end{aligned} $$

Thus in the above Theorem 2, convergence occurs for sufficiently large n.

Corollary 1

Let f, f ′′∈ C ∗[0, ∞). Then, the inequality

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n \rightarrow \infty} n\left[K_n(f,x)-f(x)\right]=x[f^{\prime}(x)+f^{\prime \prime}(x)] \end{array} \end{aligned} $$

holds for any x ∈ [0, ∞).

Remark 2

In case the operators (1) preserve the function e −2x, then in that case using Lemma 1, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} e^{-2x} &\displaystyle =&\displaystyle \frac{n(1-e^{-2/n})}{2}\left(2-e^{-2/n}\right)^{-na_n(x)}, \end{array} \end{aligned} $$

which implies

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} a_n(x)&\displaystyle =&\displaystyle \frac{2x+\ln\left(\frac{n(1-e^{-2/n})}{2}\right)}{n \ln(2-e^{-2/n})} \end{array} \end{aligned} $$
(4)

Also, for this preservation corresponding limits of Lemma 3 takes the following forms:

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n \rightarrow \infty} n\left[\frac{2x+\ln\left(\frac{n(1-e^{-2/n})}{2}\right)}{n \ln(2-e^{-2/n})}+\frac{1}{2n}-x\right]=2x \end{array} \end{aligned} $$

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n \rightarrow \infty} n\left[\!\!\left(\!\frac{2x+\ln\left(\frac{n(1-e^{-2/n})}{2}\right)}{n \ln(2-e^{-2/n})}-x\!\right)^2\!{+}\frac{3(2x+\ln\left(\frac{n(1-e^{-2/n}))}{2}\right)}{n^2 \ln(2-e^{-2/n})}-\frac{x}{n}+\frac{1}{3n^2}\!\!\right]{=}\,2x \end{array} \end{aligned} $$

and we have the following Theorems 1and 2 and Corollary 1 taking the following forms:

Theorem 3

For f ∈ C ∗[0, ∞), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} || K_nf-f||{}_{[0,\infty)}\leq 2\omega ^*\left(f,\sqrt{2\beta_n}\right), \end{array} \end{aligned} $$

where

$$\displaystyle \begin{aligned} \begin{array}{rcl} \beta _n &\displaystyle =&\displaystyle || K_n(e^{-t})-e^{-x}||{}_{[0,\infty)}\\ &\displaystyle =&\displaystyle \left|\left|\frac{-xe^{-x}}{n}+\frac{(12x^2+24x+11)e^{-x}}{24n^2}+O\left(\frac{1}{n^3}\right)\right|\right|{}_{[0,\infty)}. \end{array} \end{aligned} $$

Theorem 4

Let f, f ′′∈ C ∗[0, ∞). Then the inequality

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \left|n\left[K_n(f,x)-f(x)\right]-x[2f^{\prime}(x)+f^{\prime \prime}(x)]\right|\\ &\displaystyle &\displaystyle \quad \leq |\hat{p}_n(x)||f^{\prime}|+|\hat{q}_n(x)||f^{\prime \prime}|+2\left(2\hat{q}_n(x)+2x+\hat{r}_n(x)\right)\omega ^*\left(f^{\prime \prime},n^{-1/2}\right) \end{array} \end{aligned} $$

holds for any x ∈ [0, ∞), where

$$\displaystyle \begin{aligned} \begin{array}{rcl} \hat{p}_n(x)&\displaystyle =&\displaystyle n \mu_{n,1}(x)-x,\\ \hat{q}_n(x) &\displaystyle =&\displaystyle \frac{1}{2}\left(n\mu_{n,2}(x)-4x\right),\\ \hat{r}_n(x) &\displaystyle =&\displaystyle n^2 \sqrt{K_n\left((e^{-x}-e^{-t})^4,x\right)}\sqrt{\mu_{n,4}(x)}. \end{array} \end{aligned} $$

and μ n,1(x), μ n,2(x) and μ n,4(x) are given in Lemma 3 , with values of a n(x), given by (4).

Corollary 2

Let f, f ′′∈ C ∗[0, ∞). Then, the inequality

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n \rightarrow \infty} n\left[K_n(f,x)-f(x)\right]=x[2f^{\prime}(x)+f^{\prime \prime}(x)] \end{array} \end{aligned} $$

holds for any x ∈ [0, ∞).

Remark 3

Several other operators, which are linear and positive, can be applied to establish analogous results. Also, some other approximation properties for the operators studied in [3, 5,6,7, 10] and references therein may be considered for these operators.