On approximation by some Bernstein–Kantorovich exponential-type polynomials

Abstract

Since the introduction of Bernstein operators, many authors defined and/or studied Bernstein type operators and their generalizations, among them are Morigi and Neamtu (Adv Comput Math 12:133–149, 2000). They proposed an analog of classical Bernstein operator and proved some convergence results for continuous functions.

Herein, we introduce their integral extensions in Kantorovich sense by replacing the usual differential and integral operators with their more general analogues. By means of these operators, we are able to reconstruct the functions which are not necessarily continuous. It is shown that the operators form an approximation process in both C[0,1] and Lp,μ[0,1], which is an exponentially weighted space.

Also, quantitative results are stated in terms of appropriate moduli of smoothness and K-functionals. Furthermore, a quantitative Voronovskaya type result is presented.

Authors

Ali Aral

Diana Otrocol
(Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy)

Diana Otrocol

Ioan Raşa

Keywords

Bernstein–Kantorovich operator; uniform convergence; modulus of continuity.

References

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A. Aral, D. Otrocol, I. Raşa, On approximation by some Bernstein–Kantorovich exponential-type polynomials, Periodica Mathematica Hungarica, 79 (2019) 2, pp. 236-254,
doi: 10.1007/s10998-019-00284-3

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Springer

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0031-5303

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∎

11institutetext: Ali Aral 22institutetext: Kırıkkale University, Faculty of Sciences and Arts, Department of Mathematics, 71450, Kırıkkale-Turkey

22email: aliaral73@yahoo.com
33institutetext: Diana Otrocol 44institutetext: Technical University of Cluj-Napoca, Faculty of Automation and Computer Science, Department of Mathematics, Memorandumului St. no. 28, Cluj-Napoca 400114, Romania; Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, P.O.Box. 68-1, 400110, Cluj-Napoca, Romania

44email: Diana.Otrocol@math.utcluj.ro
55institutetext: Ioan Raşa 66institutetext: Technical University of Cluj-Napoca, Faculty of Automation and Computer Science, Department of Mathematics, Memorandumului St. no. 28, Cluj-Napoca 400114, Romania

66email: ioan.rasa@math.utcluj.ro

On approximation by some Bernstein-Kantorovich exponential-type polynomials

Ali Aral    Diana Otrocol and Ioan Raşa
(Received: date / Accepted: date)
Abstract

Since the introduction of Bernstein operators, many authors defined and/or studied Bernstein type operators and their generalizations, among them are Morigi and Neamtu in 2000. They proposed an analog of classical Bernstein operator and proved some convergence results for continuous functions. Herein, we introduce their integral extensions in Kantorovich sense by replacing the usual differential and integral operators with their more general analogs. By means of these operators, we are able to reconstruct the functions which are not necessarily continuous. It is shown that the operators form an approximation process in both C​[0,1]C\left[0,1\right] and Lp,μ​[0,1]L_{p,\mu}\left[0,1\right], which is an exponentially weighted space. Also, quantitative results are stated in terms of appropriate moduli of smoothness and KK-functionals. Furthermore, a quantitative Voronovskaya type result is presented.

1 Introduction

The classical Bernstein operators associate the polynomial Bn​fB_{n}f to any function f∈C​[0,1]f\in C[0,1] and the association is defined by

Bn​(f)​(x)=∑k=0nf​(kn)​pn,k​(x),n∈ℕ,B_{n}\left(f\right)\left(x\right)=\sum\limits_{k=0}^{n}f\left(\frac{k}{n}\right)p_{n,k}\left(x\right),\ n\in\mathbb{N}, (1)

where

pn,k​(x)=(nk)​xk​(1−x)n−k.p_{n,k}(x)={n\choose k}x^{k}(1-x)^{n-k}.

In fact, it defines a linear approximation process in the space of all continuous functions on [0,1].\left[0,1\right]. To obtain an approximation process for Lebesgue integrable functions, the Kantorovich version of (1) was defined in Kantorovich by replacing the sample values f​(k/n)f(k/n) with the mean values of ff in the interval [kn,k+1n]\left[\frac{k}{n},\frac{k+1}{n}\right], that is

Kn​(f)​(x)=(n+1)​∑k=0npn,k​(x)​∫kn+1k+1n+1f​(t)​𝑑t,x∈[0,1],n∈ℕ,K_{n}\left(f\right)\left(x\right)=\left(n+1\right)\sum\limits_{k=0}^{n}p_{n,k}\left(x\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}f\left(t\right)dt,x\in\left[0,1\right],\ n\in\mathbb{N},

where f:[0,1]→ℝf:\left[0,1\right]\rightarrow\mathbb{R} is a locally integrable function. We note that BnB_{n} and KnK_{n} are connected by the relation

Kn=D∘Bn+1∘I,K_{n}=D\circ B_{n+1}\circ I, (2)

where DD is the differentiation operator (i.e. D​(f)=f′D\left(f\right)=f^{{}^{\prime}}, f∈C1​[0,1]f\in C^{1}\left[0,1\right]) and II is the antiderivative operator (i.e. I​(f;x)=∫0xf​(t)​𝑑t,I\left(f;x\right)=\int_{0}^{x}f\left(t\right)dt, f∈C​[0,1]f\in C\left[0,1\right]) and x∈[0,1]x\in\left[0,1\right]. These operators allow us to reconstruct a Lebesgue integrable function by means of its mean values on the sets [kn,k+1n]\left[\frac{k}{n},\frac{k+1}{n}\right]. It is well known that Bernstein operators reproduce ei​(x)=xi,e_{i}\left(x\right)=x^{i}, i=0,1i=0,1 whereas Bernstein-Kantorovich operators reproduce only e0​(x)=1.e_{0}\left(x\right)=1.

Recently, Aral et al. Ar-Card-Gar studied a sequence of linear positive operators which generalize the classical Bernstein operators and perform better than BnB_{n} under sufficient conditions. These operators reproduce the exponential functions exp⁡(μ​t)\exp(\mu t) and exp⁡(2​μ​t)\exp(2\mu t), μ>0,\mu>0, and are defined by

Gn​f​(x)=Gn​(f;x)=∑k=0nf​(kn)​e−μ​k/n​eμ​x​pn,k​(an​(x)),x∈[0,1],n∈ℕ,G_{n}f\left(x\right)=G_{n}\left(f;x\right)=\sum\limits_{k=0}^{n}f\left(\frac{k}{n}\right)e^{-\mu k/n}e^{\mu x}p_{n,k}\left(a_{n}\left(x\right)\right),x\in\left[0,1\right],n\in\mathbb{N},

where

an​(x)=eμ​x/n−1eμ/n−1.a_{n}\left(x\right)=\frac{e^{\mu x/n}-1}{e^{\mu/n}-1}. (3)

They have close connection with the Bernstein operators that is given by

Gn​f​(x)=expμ⁡(x)​Bn​(fexpμ;an​(x)),G_{n}f\left(x\right)=\exp_{\mu}\left(x\right)B_{n}\left(\frac{f}{\exp_{\mu}};a_{n}\left(x\right)\right), (4)

where for a fixed real parameter μ>0\mu>0, the exponential function is defined as expμ⁡(x)=eμ​x\exp_{\mu}\left(x\right)=e^{\mu x}. We note that the generalization is a special case of the modification introduced by Morigi and Neamtu in Morigi-Nematu .

In order to give the generalization of the operators Kn​(f)K_{n}\left(f\right), in Radu2 Păltănea considered the operators Dμ:C1​[0,1]→C​[0,1]D_{\mu}:C^{1}\left[0,1\right]\rightarrow C\left[0,1\right] and Iμ:C​[0,1]→C1​[0,1]I_{\mu}:C\left[0,1\right]\rightarrow C^{1}\left[0,1\right], and defined the operators KnμK_{n}^{\mu} as

Knμ=Dμ∘Bn+1∘Iμ,K_{n}^{\mu}=D_{\mu}\circ B_{n+1}\circ I_{\mu}, (5)

where

Iμ​(f,x)=eμ​x​∫0xe−μ​t​f​(t)​𝑑t,f∈C​[0,1]​and​x∈[0,1],I_{\mu}\left(f,x\right)=e^{\mu x}\int_{0}^{x}e^{-\mu t}f\left(t\right)dt,f\in C\left[0,1\right]\mbox{and}\ x\in\left[0,1\right], (6)
Dμ​(f,x)=f′​(x)−μ​f​(x),f∈C1​[0,1]​and​x∈[0,1].D_{\mu}\left(f,x\right)=f^{{}^{\prime}}\left(x\right)-\mu f\left(x\right),f\in C^{1}\left[0,1\right]\mbox{and}\ x\in\left[0,1\right]. (7)

Using the technique given in Radu2 with the operators in (5), we aim to construct a generalization of the operator GnG_{n} and call it as generalized Bernstein Kantorovich operator. The construction is described in Section 2. Moreover, certain elementary properties including exponential moments are given in the same section. As we will see in Lemma 2 below, these operators reproduce only the function exp⁡(2​μ​t)\exp\left(2\mu t\right), μ>0.\mu>0. In the same section, we also show that the new operator can be defined with the help of classical Bernstein operator and Bernstein-Kantorovich operator. In Section 3, it is shown that these operators form an approximation process in C​[0,1]C\left[0,1\right] by obtaining uniform convergence of them and an estimate in terms of modulus of continuity. In Section 4, a quantitative Voronovskaya type result is presented. In Section 5, we show that the new operators converge to the function in Lp,μL_{p,\mu}-norm. Furthermore, the approximation error of the operators is expressed in terms of an appropriate integral modulus of continuity.

The obtained results show that our technique is more convenient in order to introduce a Kantorovich version of Gn.G_{n}.

We mention that Kantorovich type operators have been object of the investigation by several mathematicians. Altomare et al. in Altomare-Va introduced a unitary approach to the study of the approximation properties for a large class of Kantorovich type operators including the classical Bernstein-Kantorovich operators and their various generalizations. Also, more results related to the mentioned operators can be found in Altomare2 and Altomare 3 . Various results on Kantorovich-type operators were given in Zhou and Ivanov . It is worth mentioning that Kantorovich type operators in classical sense were considered in the aforementioned more general framework, as well as in the context of the qq-operators and (p,q)\left(p,q\right)-operators, a very active area of research (see Aral-Gupta , tuncer- aral and the references therein).

2 Preliminaries

Throughout this section, we introduce a new family of operators and present some elementary properties.

Definition 1

Let μ>0\mu>0. For any n∈ℕn\in\mathbb{N} and x∈[0,1]x\in\left[0,1\right] consider the operator K~n:C​[0,1]→C​[0,1]\widetilde{K}_{n}:C\left[0,1\right]\rightarrow C\left[0,1\right], defined by

K~n​f​(x)=an+1′​(x)​(n+1)​eμ​x​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1e−μ​t​f​(t)​𝑑t,\widetilde{K}_{n}f\left(x\right)=a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu t}f\left(t\right)dt, (8)

where an+1​(x)a_{n+1}\left(x\right) is given in (3).

To represent the operators in (8), we sometimes use the notation K~n​(f;x)\widetilde{K}_{n}\left(f;x\right).

For a given f∈L1​[0,1],f\in L_{1}\left[0,1\right], we define Fμ∈C​[0,1]F_{\mu}\in C\left[0,1\right] as

Fμ​(x)=∫0xe−μ​t​f​(t)​𝑑t.F_{\mu}\left(x\right)=\int_{0}^{x}e^{-\mu t}f\left(t\right)dt. (9)

Then, we have

K~n​f​(x)=an+1′​(x)​(n+1)​eμ​x​∑k=0npn,k​(an+1​(x))​[Fμ​(k+1n+1)−Fμ​(kn+1)].\widetilde{K}_{n}f\left(x\right)=a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\left[F_{\mu}\left(\frac{k+1}{n+1}\right)-F_{\mu}\left(\frac{k}{n+1}\right)\right].

Using the notation

δnμ​(F)​(x):=Fμ​(nn+1​x+1n+1)−Fμ​(nn+1​x)\delta_{n}^{\mu}\left(F\right)\left(x\right):=F_{\mu}\left(\frac{n}{n+1}x+\frac{1}{n+1}\right)-F_{\mu}\left(\frac{n}{n+1}x\right) (10)

the operators K~n\widetilde{K}_{n} can be represented as

K~n​f​(x)\displaystyle\widetilde{K}_{n}f\left(x\right) =an+1′​(x)​(n+1)​expμ⁡(x)​∑k=0npn,k​(an+1​(x))​δnμ​(F)​(kn)\displaystyle=a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)\exp_{\mu}\left(x\right)\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\delta_{n}^{\mu}\left(F\right)\left(\frac{k}{n}\right)
=an+1′​(x)​expμ⁡(x)​Kn​(expμ−1⁡(⋅)​f​(⋅);an+1​(x))\displaystyle=a_{n+1}^{{}^{\prime}}\left(x\right)\exp_{\mu}\left(x\right)K_{n}\left(\exp_{\mu}^{-1}\left(\cdot\right)f\left(\cdot\right);a_{n+1}\left(x\right)\right)
=an+1′​(x)​(n+1)​expμ⁡(x)​Bn​(δnμ​(F);an+1​(x)),\displaystyle=a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)\exp_{\mu}\left(x\right)B_{n}\left(\delta_{n}^{\mu}\left(F\right);a_{n+1}\left(x\right)\right), (11)

where BnB_{n}  and KnK_{n} denote Bernstein and Bernstein-Kantorovich operators, respectively. Note that K~n​f​(x)\widetilde{K}_{n}f(x) is an exponential polynomial, based on the representation of Bernstein polynomials.

When we compare with the definition of the operators in Radu2 , the above relationships show that the approach which will be used below is more convenient to apply to the generalized Bernstein operators Gn,G_{n}, which preserve exponential functions. As we can see in Radu2 , the operator KnμK_{n}^{\mu} preserves good properties of Kantorovich operator KnK_{n} . So we can expect that K~n\widetilde{K}_{n} also can preserve good properties of Kn.K_{n}.

Our process depends on the following lemma from Radu2 .

Lemma 1

Let n∈ℕn\in\mathbb{N} and x∈[0,1]x\in\left[0,1\right]. Then

  1. a)

    (Dμ∘Iμ)​(f)​(x)=f​(x)\left(D_{\mu}\circ I_{\mu}\right)\left(f\right)\left(x\right)=f\left(x\right), for all f∈C​[0,1]f\in C\left[0,1\right],

  2. b)

    (Iμ∘Dμ)​(f)​(x)=f​(x)\left(I_{\mu}\circ D_{\mu}\right)\left(f\right)\left(x\right)=f\left(x\right), for all f∈C1​[0,1]f\in C^{1}\left[0,1\right], with f​(0)=0f\left(0\right)=0.

Note that DμD_{\mu} and IμI_{\mu} in Lemma 1 are defined as in (6) and (7), respectively.

Theorem 2.1

Let n∈ℕn\in\mathbb{N} and x∈[0,1]x\in\left[0,1\right]. Then

K~n=Dμ∘Gn+1∘Iμ.\widetilde{K}_{n}=D_{\mu}\circ G_{n+1}\circ I_{\mu}.
Proof

Let f∈C​[0,1]f\in C\left[0,1\right] and x∈[0,1].x\in\left[0,1\right]. With the help of the relations in Lemma 1 and the representation given in (4), we obtain

(Dμ∘Gn+1∘Iμ)​(f)​(x)\displaystyle\left(D_{\mu}\circ G_{n+1}\circ I_{\mu}\right)\left(f\right)\left(x\right) =(Gn+1​(Iμ​(f);x))′−μ​Gn+1​(Iμ​(f);x)\displaystyle=\left(G_{n+1}\left(I_{\mu}\left(f\right);x\right)\right)^{{}^{\prime}}-\mu G_{n+1}\left(I_{\mu}\left(f\right);x\right)
=(expμ⁡(x)​Bn+1​(Fμ;an+1​(x)))′−μ​expμ⁡(x)​Bn+1​(Fμ;an+1​(x))\displaystyle\hskip-56.9055pt=\left(\exp_{\mu}\left(x\right)B_{n+1}\left(F_{\mu};a_{n+1}\left(x\right)\right)\right)^{{}^{\prime}}-\mu\exp_{\mu}\left(x\right)B_{n+1}\left(F_{\mu};a_{n+1}\left(x\right)\right)
=μ​expμ⁡(x)​Bn+1​(Fμ;an+1​(x))+expμ⁡(x)​Bn+1′​(Fμ;an+1​(x))\displaystyle\hskip-56.9055pt=\mu\exp_{\mu}\left(x\right)B_{n+1}\left(F_{\mu};a_{n+1}\left(x\right)\right)+\exp_{\mu}\left(x\right)B_{n+1}^{{}^{\prime}}\left(F_{\mu};a_{n+1}\left(x\right)\right)
−μ​expμ⁡(x)​Bn+1​(Fμ;an+1​(x)),\displaystyle-\mu\exp_{\mu}\left(x\right)B_{n+1}\left(F_{\mu};a_{n+1}\left(x\right)\right),

where FμF_{\mu} defined as in (9). Using the fact that

(pn+1,k​(an+1​(x)))′=(n+1)​an+1′​(x)​(pn,k−1​(an+1​(x))−pn,k​(an+1​(x)))\left(p_{n+1,k}\left(a_{n+1}\left(x\right)\right)\right)^{{}^{\prime}}=\left(n+1\right)a_{n+1}^{{}^{\prime}}\left(x\right)\left(p_{n,k-1}\left(a_{n+1}\left(x\right)\right)-p_{n,k}\left(a_{n+1}\left(x\right)\right)\right)

we have

(Dμ∘Gn+1∘Iμ)​(x)\displaystyle\left(D_{\mu}\circ G_{n+1}\circ I_{\mu}\right)\left(x\right)
=eμ​x​∑k=0n+1(pn+1,k​(an+1​(x)))′​Fμ​(kn+1)\displaystyle\hskip-71.13188pt=e^{\mu x}\sum\limits_{k=0}^{n+1}\left(p_{n+1,k}\left(a_{n+1}\left(x\right)\right)\right)^{{}^{\prime}}F_{\mu}\left(\frac{k}{n+1}\right)
=(n+1)​an+1′​(x)​eμ​x​∑k=0n+1(pn,k−1​(an+1​(x))−pn,k​(an+1​(x)))​Fμ​(kn+1)\displaystyle\hskip-71.13188pt=\left(n+1\right)a_{n+1}^{{}^{\prime}}\left(x\right)e^{\mu x}\sum\limits_{k=0}^{n+1}\left(p_{n,k-1}\left(a_{n+1}\left(x\right)\right)-p_{n,k}\left(a_{n+1}\left(x\right)\right)\right)F_{\mu}\left(\frac{k}{n+1}\right)
=(n+1)​an+1′​(x)​eμ​x​∑k=0npn,k​(an+1​(x))​[Fμ​(k+1n+1)−Fμ​(kn+1)]\displaystyle\hskip-71.13188pt=\left(n+1\right)a_{n+1}^{{}^{\prime}}\left(x\right)e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\left[F_{\mu}\left(\frac{k+1}{n+1}\right)-F_{\mu}\left(\frac{k}{n+1}\right)\right]
=an+1′​(x)​(n+1)​eμ​x​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1e−μ​t​f​(t)​𝑑t.\displaystyle\hskip-71.13188pt=a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu t}f\left(t\right)dt.

For this family of operators, we give here some of their properties and results.

Lemma 2

Let n∈ℕn\in\mathbb{N} and x∈[0,1].x\in\left[0,1\right]. The following equalities hold.

K~n​(e0;x)=eμ​(x−n−1)/(n+1)​eμ​x​(eμ/(n+1)+1−eμ​x/(n+1))n,\widetilde{K}_{n}\left(e_{0};x\right)=e^{\mu\left(x-n-1\right)/\left(n+1\right)}e^{\mu x}\left(e^{\mu/\left(n+1\right)}+1-e^{\mu x/\left(n+1\right)}\right)^{n}, (12)
K~n​(expμ;x)=μn+1​eμ​x/(n+1)eμ/(n+1)−1​eμ​x,\mathcal{\ }\widetilde{K}_{n}\left(\exp_{\mu};x\right)=\frac{\mu}{n+1}\frac{e^{\mu x/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1}e^{\mu x}, (13)
K~n​(expμ2;x)=e2​μ​x,\mathcal{\ }\widetilde{K}_{n}\left(\exp_{\mu}^{2};x\right)=e^{2\mu x}, (14)
K~n​(expμ3;x)\displaystyle\mathcal{\ }\widetilde{K}_{n}\left(\exp_{\mu}^{3};x\right) =12​eμ​x+μ​x/(n+1)​(1+eμ/(n+1))\displaystyle=\frac{1}{2}e^{\mu x+\mu x/\left(n+1\right)}\left(1+e^{\mu/\left(n+1\right)}\right) (15)
(−eμ/(n+1)+eμ​x/(n+1)+eμ/(n+1)+μ​x/(n+1))n,\displaystyle\hskip 28.45274pt\left(-e^{\mu/\left(n+1\right)}+e^{\mu x/\left(n+1\right)}+e^{\mu/\left(n+1\right)+\mu x/\left(n+1\right)}\right)^{n},
K~n​(expμ4;x)\displaystyle\mathcal{\ }\widetilde{K}_{n}\left(\exp_{\mu}^{4};x\right) =13​eμ​x+μ​x/(n+1)​(1+eμ/(n+1)+e2​μ/(n+1))\displaystyle=\frac{1}{3}e^{\mu x+\mu x/\left(n+1\right)}\left(1+e^{\mu/\left(n+1\right)}+e^{2\mu/\left(n+1\right)}\right) (16)
×(−eμ/(n+1)−e2​μ/(n+1)+eμ​x/(n+1)\displaystyle\times\left(-e^{\mu/\left(n+1\right)}-e^{2\mu/\left(n+1\right)}+e^{\mu x/\left(n+1\right)}\right.
+eμ/(n+1)+μ​x/(n+1)+e2​μ/(n+1)+μ​x/(n+1))n.\displaystyle\left.+e^{\mu/\left(n+1\right)+\mu x/\left(n+1\right)}+e^{2\mu/\left(n+1\right)+\mu x/\left(n+1\right)}\right)^{n}.
Proof

It follows from (8) that if f=e0,f=e_{0}, then

K~n​(e0;x)\displaystyle\widetilde{K}_{n}\left(e_{0};x\right) =μn+1​eμ​x/(n+1)eμ/(n+1)−1​eμ​x​(n+1)​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1e−μ​t​𝑑t\displaystyle=\frac{\mu}{n+1}\frac{e^{\mu x/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1}e^{\mu x}\left(n+1\right)\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu t}dt
=eμ​x/(n+1)eμ/(n+1)​eμ​x​∑k=0ne−μ​k/(n+1)​pn,k​(an+1​(x))\displaystyle=\frac{e^{\mu x/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}}e^{\mu x}\sum\limits_{k=0}^{n}e^{-\mu k/\left(n+1\right)}p_{n,k}\left(a_{n+1}\left(x\right)\right)
=eμ​x/(n+1)eμ/(n+1)​eμ​x​e−μ​n/(n+1)​(eμ/(n+1)+1−eμ​x/(n+1))n\displaystyle=\frac{e^{\mu x/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}}e^{\mu x}e^{-\mu n/\left(n+1\right)}\left(e^{\mu/\left(n+1\right)}+1-e^{\mu x/\left(n+1\right)}\right)^{n}
=eμ​x/(n+1)​eμ​(x−1)​(eμ/(n+1)+1−eμ​x/(n+1))n.\displaystyle=e^{\mu x/\left(n+1\right)}e^{\mu\left(x-1\right)}\left(e^{\mu/\left(n+1\right)}+1-e^{\mu x/\left(n+1\right)}\right)^{n}.

If f​(x)=eμ​x,f\left(x\right)=e^{\mu x}, then

K~n​(eμ​t;x)\displaystyle\widetilde{K}_{n}\left(e^{\mu t};x\right) =an+1′​(x)​eμ​x​(n+1)​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1𝑑t\displaystyle=a_{n+1}^{{}^{\prime}}\left(x\right)e^{\mu x}\left(n+1\right)\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}dt
=μn+1​eμ​x/(n+1)eμ/(n+1)−1​eμ​x​∑k=0npn,k​(an+1​(x))\displaystyle=\frac{\mu}{n+1}\frac{e^{\mu x/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1}e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)
=μn+1​eμ​x/(n+1)eμ/(n+1)−1​eμ​x.\displaystyle=\frac{\mu}{n+1}\frac{e^{\mu x/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1}e^{\mu x}.

Finally, for f​(x)=e2​μ​xf\left(x\right)=e^{2\mu x}, we get

K~n​(e2​μ​t;x)\displaystyle\widetilde{K}_{n}\left(e^{2\mu t};x\right) =an+1′​(x)​eμ​x​(n+1)​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1eμ​t​𝑑t\displaystyle=a_{n+1}^{{}^{\prime}}\left(x\right)e^{\mu x}\left(n+1\right)\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{\mu t}dt
=μn+1​eμ​x/(n+1)eμ/(n+1)−1​eμ​x​(n+1)​∑k=0npn,k​(an+1​(x))\displaystyle=\frac{\mu}{n+1}\frac{e^{\mu x/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1}e^{\mu x}\left(n+1\right)\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)
⋅(1μ​eμ​k/(n+1)​(eμ/(n+1)−1))\displaystyle\quad\cdot\left(\frac{1}{\mu}e^{\mu k/\left(n+1\right)}\left(e^{\mu/\left(n+1\right)}-1\right)\right)
=eμ​x/(n+1)​eμ​x​∑k=0npn,k​(an+1​(x))​eμ​k/(n+1)\displaystyle=e^{\mu x/\left(n+1\right)}e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)e^{\mu k/\left(n+1\right)}
=eμ​x/(n+1)​eμ​x​eμ​x​n/(n+1)=e2​μ​x.\displaystyle=e^{\mu x/\left(n+1\right)}e^{\mu x}e^{\mu xn/\left(n+1\right)}=e^{2\mu x}.

Other results are similar.

Lemma 3

Let expμ,x⁡(t)=eμ​t−eμ​x.\exp_{\mu,x}\left(t\right)=e^{\mu t}-e^{\mu x}.   For n∈ℕn\in\mathbb{N} and x∈[0,1],x\in\left[0,1\right], we have

limn→∞n​(K~n​(e0;x)−1)=μ​(μ​x−1)−μ​x​(μ​x−2),\lim_{n\rightarrow\infty}n\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right)=\mu\left(\mu x-1\right)-\mu x\left(\mu x-2\right), (17)
limn→∞n​(K~n​(expμ;x)−eμ​x)=μ2​(2​x−1)​eμ​x,\lim_{n\rightarrow\infty}n\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\right)=\frac{\mu}{2}\left(2x-1\right)e^{\mu x}, (18)

and

limn→∞n2​K~n​(expμ,x4⁡(x);x)=3​μ2​(1−x)2​x2​e4​μ​x.\lim_{n\rightarrow\infty}n^{2}\widetilde{K}_{n}\left(\exp_{\mu,x}^{4}\left(x\right);x\right)=3\mu^{2}\left(1-x\right)^{2}x^{2}e^{4\mu x}. (19)

Using the above limits, we get

limn→∞n​K~n​(expμ,x;x)\displaystyle\lim_{n\rightarrow\infty}n\widetilde{K}_{n}\left(\exp_{\mu,x};x\right) =limn→∞n​(K~n​(expμ;x)−eμ​x​K~n​(e0;x))\displaystyle=\lim_{n\rightarrow\infty}n\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\widetilde{K}_{n}\left(e_{0};x\right)\right)
=limn→∞n​(K~n​(expμ;x)−eμ​x)\displaystyle=\lim_{n\rightarrow\infty}n\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\right)
−limn→∞n​eμ​x​(K~n​(e0;x)−1)\displaystyle-\lim_{n\rightarrow\infty}ne^{\mu x}\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right)
=eμ​x​[μ2​(2​x−1)−μ​(μ​x−1)+μ​x​(μ​x−2)]\displaystyle=e^{\mu x}\left[\frac{\mu}{2}\left(2x-1\right)-\mu\left(\mu x-1\right)+\mu x\left(\mu x-2\right)\right] (20)

and

limn→∞n​K~n​(expμ,x2;x)\displaystyle\lim_{n\rightarrow\infty}n\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right) =−eμ​x​limn→∞n​(K~n​(expμ;x)−eμ​x)\displaystyle=-e^{\mu x}\lim_{n\rightarrow\infty}n\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\right)
−eμ​x​limn→∞n​(K~n​(expμ;x)−eμ​x​K~n​(e0;x))\displaystyle-e^{\mu x}\lim_{n\rightarrow\infty}n\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\widetilde{K}_{n}\left(e_{0};x\right)\right)
=−2​eμ​x​limn→∞n​(K~n​(expμ;x)−eμ​x)\displaystyle=-2e^{\mu x}\lim_{n\rightarrow\infty}n\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\right)
+e2​μ​x​limn→∞n​(K~n​(e0;x)−1)\displaystyle+e^{2\mu x}\lim_{n\rightarrow\infty}n\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right)
=e2​μ​x​(μ​(μ​x−1)−μ​x​(μ​x−2)−μ​(2​x−1)).\displaystyle=e^{2\mu x}\left(\mu\left(\mu x-1\right)-\mu x\left(\mu x-2\right)-\mu\left(2x-1\right)\right). (21)
Lemma 4

Let αn,μ:=‖K~n​e0−e0‖∞\alpha_{n,\mu}:={\left\|\widetilde{K}_{n}e_{0}-e_{0}\right\|}_{\infty}. The following relations hold as n→∞n\rightarrow\infty:

αn,μ→0,\alpha_{n,\mu}\rightarrow 0, (22)
‖K~n​(expμ)−expμ‖∞→0{\left\|\widetilde{K}_{n}(\exp_{\mu})-\exp_{\mu}\right\|}_{\infty}\rightarrow 0 (23)

and

βn,μ:=supx∈[0,1]K~n​(expμ,x2;x)→0.\beta_{n,\mu}:=\sup_{x\in\left[0,1\right]}\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right)\rightarrow 0. (24)
Proof

We have

(K~n​e0)′​(x)\displaystyle\left(\widetilde{K}_{n}e_{0}\right)^{{}^{\prime}}\left(x\right) =μeμ​(x−n−1)/(n+1)eμ​x(eμ/(n+1)+1−eμ​x/(n+1))n−1×\displaystyle=\mu e^{\mu\left(x-n-1\right)/\left(n+1\right)}e^{\mu x}\left(e^{\mu/\left(n+1\right)}+1-e^{\mu x/\left(n+1\right)}\right)^{n-1}\times
((n+2)n+1​(eμ/(n+1)+1)−2​eμ​x/(n+1)).\displaystyle\left(\frac{\left(n+2\right)}{n+1}\left(e^{\mu/\left(n+1\right)}+1\right)-2e^{\mu x/\left(n+1\right)}\right).

If

(K~n​e0)′​(xn)=0,\left(\widetilde{K}_{n}e_{0}\right)^{{}^{\prime}}\left(x_{n}\right)=0,

then

xn=(n+1)μ​ln⁡((n+2)(n+1)​(eμ/(n+1)+1)2).x_{n}=\frac{\left(n+1\right)}{\mu}\ln\left(\frac{\left(n+2\right)}{\left(n+1\right)}\frac{\left(e^{\mu/\left(n+1\right)}+1\right)}{2}\right).

Also, we can write

K~n​e0​(0)−1\displaystyle\widetilde{K}_{n}e_{0}\left(0\right)-1 =e−μ​eμ​n/(n+1)−1\displaystyle=e^{-\mu}e^{\mu n/\left(n+1\right)}-1 (25)
=e−μ/(n+1)−1,\displaystyle=e^{-\mu/\left(n+1\right)}-1,
K~n​e0​(1)−1=eμ/(n+1)−1\widetilde{K}_{n}e_{0}\left(1\right)-1=e^{\mu/\left(n+1\right)}-1 (26)

and

K~n​e0​(xn)−1\displaystyle\widetilde{K}_{n}e_{0}\left(x_{n}\right)-1 =(n+2)(n+1)​(eμ/(n+1)+1)2​((n+2)(n+1)​(eμ/(n+1)+1)2)n+1​e−μ\displaystyle=\frac{\left(n+2\right)}{\left(n+1\right)}\frac{\left(e^{\mu/\left(n+1\right)}+1\right)}{2}\left(\frac{\left(n+2\right)}{\left(n+1\right)}\frac{\left(e^{\mu/\left(n+1\right)}+1\right)}{2}\right)^{n+1}e^{-\mu}
×(eμ/(n+1)+1−(n+2)(n+1)​(eμ/(n+1)+1)2)n−1\displaystyle\quad\times\left(e^{\mu/\left(n+1\right)}+1-\frac{\left(n+2\right)}{\left(n+1\right)}\frac{\left(e^{\mu/\left(n+1\right)}+1\right)}{2}\right)^{n}-1
=(eμ/(n+1)+12)2​n+2​(n+2n+1)n+2​(nn+1)n​e−μ−1.\displaystyle=\left(\frac{e^{\mu/\left(n+1\right)}+1}{2}\right)^{2n+2}\left(\frac{n+2}{n+1}\right)^{n+2}\left(\frac{n}{n+1}\right)^{n}e^{-\mu}-1. (27)

Since K~n​e0​(x)−1\widetilde{K}_{n}e_{0}\left(x\right)-1   is a continuous function, it attains its extreme values at either endpoints 0 and 11 or critical point xnx_{n}.

Since the limits of (25), (26) and ( 27) are equal to zero, (22) is proved.

(23) and (24) follow similarly from

K~n​(expμ;x)−eμ​x=eμ​x​(μn+1​eμ​x/(n+1)eμ/(n+1)−1−1),\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}=e^{\mu x}\left(\frac{\mu}{n+1}\frac{e^{\mu x/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1}-1\right),

and

K~n​(expμ,x2;x)\displaystyle\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right) =2​eμ​x​(eμ​x−K~n​(expμ;x))+e2​μ​x​(K~n​(e0;x)−1)\displaystyle=2e^{\mu x}\left(e^{\mu x}-\widetilde{K}_{n}\left(\exp_{\mu};x\right)\right)+e^{2\mu x}\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right)
=2​e2​μ​x​(1−μn+1​eμ​x/(n+1)eμ/(n+1)−1)+e2​μ​x​(K~n​(e0;x)−1).\displaystyle=2e^{2\mu x}\left(1-\frac{\mu}{n+1}\frac{e^{\mu x/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1}\right)+e^{2\mu x}\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right).

3 Convergence in C​[0,1]C\left[0,1\right]

We shall denote by C​[0,1]C[0,1] the space of all real valued continuous functions on the interval [0,1][0,1] endowed with the sup-norm ∥⋅∥∞.\left\|\cdot\right\|_{\infty}.

Considering the calculations in the proof of Lemma 4, we get ‖K~n​e0‖∞=αn,μ+1\left\|\widetilde{K}_{n}e_{0}\right\|_{\infty}=\alpha_{n,\mu}+1, where αn,μ\alpha_{n,\mu} is defined as in that proof. So we can say that the linear operator K~n\widetilde{K}_{n} maps C​[0,1]C\left[0,1\right] into itself and it is continuous with respect to the sup-norm. Since the sequence αn,μ\alpha_{n,\mu} is convergent, we get ‖K~n‖≤d,\left\|\widetilde{K}_{n}\right\|\leq d, dd is a positive constant.

In the following theorem, we establish the convergence of the operators K~n\widetilde{K}_{n} towards the identity operator.

Theorem 3.1

If f∈C​[0,1],f\in C\left[0,1\right], then K~n​f\widetilde{K}_{n}f converges to ff uniformly on [0,1]\left[0,1\right].

Proof

Since {e0,expμ,expμ2}\left\{e_{0},\exp_{\mu},\exp_{\mu}^{2}\right\} is an extended complete Chebyshev system, from Korovkin’s theorem we have to prove that the thesis is fulfilled for the functions e0,expμ,expμ2.e_{0},\exp_{\mu},\exp_{\mu}^{2}. But this is a consequence of Lemma 2 and Lemma 4, and so the theorem is proved.

Theorem 3.2

For every n∈ℕn\in\mathbb{N} and f∈C​[0,1],f\in C\left[0,1\right], we have

‖K~n​f−f‖∞\displaystyle\left\|\widetilde{K}_{n}f-f\right\|_{\infty} ≤bn​eμ​‖f‖∞+2​eμ​ω​(expμ−1⁡f,nn+1​γn+1n+1)+\displaystyle\leq b_{n}e^{\mu}\left\|f\right\|_{\infty}+2e^{\mu}\omega\left(\exp_{\mu}^{-1}f,\frac{n}{n+1}\gamma_{n}+\frac{1}{n+1}\right)+
+3​eμ​ω​(expμ−1⁡f,n+1n+1)\displaystyle\quad+3e^{\mu}\omega\left(\exp_{\mu}^{-1}f,\frac{\sqrt{n}+1}{n+1}\right)

where

bn=μn+1​eμ/(n+1)eμ/(n+1)−1−1b_{n}=\frac{\mu}{n+1}\frac{e^{\mu/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1}-1

and

γn=(eμ/(n+1)−1)​(n+1)μ−1eμ/(n+1)−1−(n+1)μ​ln⁡[(eμ/(n+1)−1)​(n+1)μ].\gamma_{n}=\frac{\left(e^{\mu/\left(n+1\right)}-1\right)\frac{\left(n+1\right)}{\mu}-1}{e^{\mu/\left(n+1\right)}-1}-\frac{\left(n+1\right)}{\mu}\ln\left[\left(e^{\mu/\left(n+1\right)}-1\right)\frac{\left(n+1\right)}{\mu}\right]. (28)
Proof

Using the representation (11) and the fact that bn=supx∈[0,1](an+1′​(x)−1)=b_{n}=\sup_{x\in\left[0,1\right]}\!\!\left(a_{n+1}^{{}^{\prime}}(x)-1\right)==μn+1​eμ/(n+1)eμ/(n+1)−1−1,=\frac{\mu}{n+1}\frac{e^{\mu/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1}-1, we have

‖K~n​f−f‖∞\displaystyle\left\|\widetilde{K}_{n}f-f\right\|_{\infty} ≤supx∈[0,1](an+1′​(x)−1)​eμ​‖Bn​((n+1)​δnμ​(F);an+1​(⋅))‖∞\displaystyle\leq\sup_{x\in\left[0,1\right]}\left(a_{n+1}^{{}^{\prime}}\left(x\right)-1\right)e^{\mu}\left\|B_{n}\left(\left(n+1\right)\delta_{n}^{\mu}\left(F\right);a_{n+1}\left(\cdot\right)\right)\right\|_{\infty}
+(n+1)​eμ​‖Bn​(δnμ​(F);an+1​(⋅))−δnμ​(F)‖∞\displaystyle\quad+\left(n+1\right)e^{\mu}\left\|B_{n}\left(\delta_{n}^{\mu}\left(F\right);a_{n+1}\left(\cdot\right)\right)-\delta_{n}^{\mu}\left(F\right)\right\|_{\infty}
+eμ​‖(n+1)​δnμ​(F)−e−μ⁣⋅​f‖∞\displaystyle\quad+e^{\mu}\left\|\left(n+1\right)\delta_{n}^{\mu}\left(F\right)-e^{-\mu\cdot}f\right\|_{\infty}
≤bn​eμ​‖e−μ⁣⋅​f‖∞+(n+1)​eμ​‖Bn​(δnμ​(F);an+1​(⋅))−Bn​(δnμ​(F);⋅)‖∞\displaystyle\leq b_{n}e^{\mu}\left\|e^{-\mu\cdot}f\right\|_{\infty}+\left(n+1\right)e^{\mu}\left\|B_{n}\left(\delta_{n}^{\mu}\left(F\right);a_{n+1}\left(\cdot\right)\right)-B_{n}\left(\delta_{n}^{\mu}\left(F\right);\cdot\right)\right\|_{\infty}
+(n+1)​eμ​‖Bn​(δnμ​(F);⋅)−δnμ​(F)‖∞\displaystyle\quad+\left(n+1\right)e^{\mu}\left\|B_{n}\left(\delta_{n}^{\mu}\left(F\right);\cdot\right)-\delta_{n}^{\mu}\left(F\right)\right\|_{\infty}
+eμ​‖(n+1)​δnμ​(F)−e−μ⁣⋅​f‖∞\displaystyle\quad+e^{\mu}\left\|\left(n+1\right)\delta_{n}^{\mu}\left(F\right)-e^{-\mu\cdot}f\right\|_{\infty}
:=bn​eμ​‖f‖∞+I1+I2+I3.\displaystyle:=b_{n}e^{\mu}\left\|f\right\|_{\infty}+I_{1}+I_{2}+I_{3}.

To estimate I1,I_{1}, we use the following inequality given by (6.1.966.1.96) in Altomare Kitap for Bernstein operators in the case of an arbitrary continuous function on [0,1]\left[0,1\right]:

ω​(Bn​(f),δ)≤2​ω​(f,δ), ​δ>0.\omega\left(B_{n}\left(f\right),\delta\right)\leq 2\omega\left(f,\delta\right),\text{ \ \ \ }\delta>0.

Thus we have

|Bn​(δnμ​(F);an+1​(x))−Bn​(δnμ​(F);x)|\displaystyle\left|B_{n}\left(\delta_{n}^{\mu}\left(F\right);a_{n+1}\left(x\right)\right)-B_{n}\left(\delta_{n}^{\mu}\left(F\right);x\right)\right|
≤ω​(Bn​(δnμ​(F)),|an+1​(x)−x|)\displaystyle\leq\omega\left(B_{n}\left(\delta_{n}^{\mu}\left(F\right)\right),\left|a_{n+1}\left(x\right)-x\right|\right)
≤2​ω​(δnμ​(F),|an+1​(x)−x|)\displaystyle\leq 2\omega\left(\delta_{n}^{\mu}\left(F\right),\left|a_{n+1}\left(x\right)-x\right|\right)
≤2​ω​(δnμ​(F),γn),\displaystyle\leq 2\omega\left(\delta_{n}^{\mu}\left(F\right),\gamma_{n}\right), (29)

where

γn=maxx∈[0,1]​|an+1​(x)−x|.\gamma_{n}=\underset{x\in\left[0,1\right]}{\max}\left|a_{n+1}\left(x\right)-x\right|.

an+1​(x)−xa_{n+1}\left(x\right)-x attains its maximum value within [0,1]\left[0,1\right] at the point

x0=n+1μ​ln⁡[(eμ/(n+1)−1)​(n+1)μ]x_{0}=\frac{n+1}{\mu}\ln\left[\left(e^{\mu/\left(n+1\right)}-1\right)\frac{\left(n+1\right)}{\mu}\right]

and thus γn\gamma_{n} is defined as in (28). It is easily seen that

limn→∞γn=0.\lim_{n\rightarrow\infty}\gamma_{n}=\allowbreak 0.

Let us estimate ω​(δnμ​(F),δ).\omega\left(\delta_{n}^{\mu}\left(F\right),\delta\right). For fixed δ>0\delta>0 and x,y∈[0,1],x,y\in\left[0,1\right], |x−y|≤δ\left|x-y\right|\leq\delta then

|δnμ​(F)​(x)−δnμ​(F)​(y)|\displaystyle\left|\delta_{n}^{\mu}\left(F\right)\left(x\right)-\delta_{n}^{\mu}\left(F\right)\left(y\right)\right| =1n+1​|e−μ​ξn,x​f​(ξn,x)−e−μ​ηn,y​f​(ηn,y)|\displaystyle=\frac{1}{n+1}\left|e^{-\mu\xi_{n,x}}f\left(\xi_{n,x}\right)-e^{-\mu\eta_{n,y}}f\left(\eta_{n,y}\right)\right|
≤1n+1​ω​(expμ−1⁡f,|ξn,x−ηn,y|),\displaystyle\leq\frac{1}{n+1}\omega\left(\exp_{\mu}^{-1}f,\left|\xi_{n,x}-\eta_{n,y}\right|\right),

where ξn,x∈[nn+1​x,nn+1​x+1n+1]\xi_{n,x}\in\Big[\frac{n}{n+1}x,\frac{n}{n+1}x+\frac{1}{n+1}\Big] and ηn,y∈[nn+1​y,nn+1​y+1n+1].\eta_{n,y}\in\left[\frac{n}{n+1}y,\frac{n}{n+1}y+\frac{1}{n+1}\right]. Noting that

|ξn,x−ηn,y|≤nn+1​|x−y|+1n+1≤nn+1​δ+1n+1\left|\xi_{n,x}-\eta_{n,y}\right|\leq\frac{n}{n+1}\left|x-y\right|+\frac{1}{n+1}\leq\frac{n}{n+1}\delta+\frac{1}{n+1}

we get

|δnμ​(F)​(x)−δnμ​(F)​(y)|≤1n+1​ω​(expμ−1⁡f,nn+1​δ+1n+1).\left|\delta_{n}^{\mu}\left(F\right)\left(x\right)-\delta_{n}^{\mu}\left(F\right)\left(y\right)\right|\leq\frac{1}{n+1}\omega\left(\exp_{\mu}^{-1}f,\frac{n}{n+1}\delta+\frac{1}{n+1}\right).

So it follows that

ω​(δnμ​(F),δ)≤1n+1​ω​(expμ−1⁡f,nn+1​δ+1n+1).\omega\left(\delta_{n}^{\mu}\left(F\right),\delta\right)\leq\frac{1}{n+1}\omega\left(\exp_{\mu}^{-1}f,\frac{n}{n+1}\delta+\frac{1}{n+1}\right). (30)

Using (29) and (30), we get

I1\displaystyle I_{1} =(n+1)​eμ​‖Bn​(δnμ​(F);an+1​(⋅))−Bn​(δnμ​(F);⋅)‖∞\displaystyle=\left(n+1\right)e^{\mu}\left\|B_{n}\left(\delta_{n}^{\mu}\left(F\right);a_{n+1}\left(\cdot\right)\right)-B_{n}\left(\delta_{n}^{\mu}\left(F\right);\cdot\right)\right\|_{\infty}
≤2​eμ​(n+1)​ω​(δnμ​(F),γn)\displaystyle\leq 2e^{\mu}\left(n+1\right)\omega\left(\delta_{n}^{\mu}\left(F\right),\gamma_{n}\right)
≤2​eμ​ω​(expμ−1⁡f,nn+1​γn+1n+1).\displaystyle\leq 2e^{\mu}\omega\left(\exp_{\mu}^{-1}f,\frac{n}{n+1}\gamma_{n}+\frac{1}{n+1}\right).

Let us estimate I2I_{2}. Using (30), we have

I2\displaystyle I_{2} =eμ​(n+1)​‖Bn​(δnμ​(F);⋅)−δnμ​(F)‖∞\displaystyle=e^{\mu}\left(n+1\right)\left\|B_{n}\left(\delta_{n}^{\mu}\left(F\right);\cdot\right)-\delta_{n}^{\mu}\left(F\right)\right\|_{\infty}
≤2​eμ​(n+1)​ω​(δnμ​(F),1n)\displaystyle\leq 2e^{\mu}\left(n+1\right)\omega\left(\delta_{n}^{\mu}\left(F\right),\frac{1}{\sqrt{n}}\right)
≤2​eμ​ω​(expμ−1⁡f,n+1n+1).\displaystyle\leq 2e^{\mu}\omega\left(\exp_{\mu}^{-1}f,\frac{\sqrt{n}+1}{n+1}\right).

Now we proceed to estimate I3.I_{3}. For x∈[0,1],x\in\left[0,1\right], by using Lagrange’s theorem we have

(n+1)​δnμ​(F)​(x)\displaystyle\left(n+1\right)\delta_{n}^{\mu}\left(F\right)\left(x\right) =(n+1)​[Fμ​(nn+1​x+1n+1)−Fμ​(nn+1​x)]\displaystyle=\left(n+1\right)\left[F_{\mu}\left(\frac{n}{n+1}x+\frac{1}{n+1}\right)-F_{\mu}\left(\frac{n}{n+1}x\right)\right]
=Fμ′​(ξn,x)=e−μ​ξn,x​f​(ξn,x),\displaystyle=F_{\mu}^{{}^{\prime}}\left(\xi_{n,x}\right)=e^{-\mu\xi_{n,x}}f\left(\xi_{n,x}\right), (31)

where ξn,x∈[nn+1​x,nn+1​x+1n+1].\xi_{n,x}\in\left[\frac{n}{n+1}x,\frac{n}{n+1}x+\frac{1}{n+1}\right]. Since |ξn,x−x|<1n+1\left|\xi_{n,x}-x\right|<\frac{1}{n+1} we get

|(n+1)​δnμ​(F)​(x)−e−μ​x​f​(x)|\displaystyle\left|\left(n+1\right)\delta_{n}^{\mu}\left(F\right)\left(x\right)-e^{-\mu x}f\left(x\right)\right| ≤|e−μ​ξn,x​f​(ξn,x)−e−μ​x​f​(x)|\displaystyle\leq\left|e^{-\mu\xi_{n,x}}f\left(\xi_{n,x}\right)-e^{-\mu x}f\left(x\right)\right|
≤ω​(expμ−1⁡f,1n+1)\displaystyle\leq\omega\left(\exp_{\mu}^{-1}f,\frac{1}{n+1}\right)

and so

I3=eμ​‖(n+1)​δnμ​(F)−e−μ⁣⋅​f‖∞≤eμ​ω​(expμ−1⁡f,1n+1).I_{3}=e^{\mu}\left\|\left(n+1\right)\delta_{n}^{\mu}\left(F\right)-e^{-\mu\cdot}f\right\|_{\infty}\leq e^{\mu}\omega\left(\exp_{\mu}^{-1}f,\frac{1}{n+1}\right).

Collecting all the estimates of I1,I_{1}, I2​ I_{2\text{ }} and I3,I_{3}, we have the desired result.

Remark 1

It is easy to verify that the sequences bnb_{n} and γn\gamma_{n} in Theorem 3.2 satisfy bn=𝒪​(1n)b_{n}={\cal O}(\tfrac{1}{n}) and γn=𝒪​(1n)\gamma_{n}={\cal O}(\tfrac{1}{n}) as n→∞n\rightarrow\infty.

4 Quantitative Voronovskaya theorem

To describe the rate of pointwise convergence of the operators, we give a quantitative version of Voronovskaya theorem. Note that for a general linear positive operators similar results were given in Go-Pi-Ra .

For f∈C2​[0,1]f\in C^{2}\left[0,1\right] and x0,x_{0}, x∈[0,1],x\in\left[0,1\right], we use the following version of the Taylor formula:

f​(x)\displaystyle f\left(x\right) =(f∘logμ)​(eμ​x0)+(f∘logμ)′​(eμ​x0)​expμ,x0⁡(x)+\displaystyle=\left(f\circ\log_{\mu}\right)\left(e^{\mu x_{0}}\right)+\left(f\circ\log_{\mu}\right)^{{}^{\prime}}\left(e^{\mu x_{0}}\right)\exp_{\mu,x_{0}}\left(x\right)+
+12​(f∘logμ)′′​(eμ​x0)​expμ,x02⁡(x)+R​(f;x0;x),\displaystyle\quad+\frac{1}{2}\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\left(e^{\mu x_{0}}\right)\exp_{\mu,x_{0}}^{2}\left(x\right)+R\left(f;x_{0};x\right),

where logμ\log_{\mu} is the inverse function of eμe^{\mu} and the remainder R​(f;x0;x)R\left(f;x_{0};x\right) is

R​(f;x0;x)=12​expμ,x02⁡(x)​((f∘logμ)′′​(eμ​ξx)−(f∘logμ)′′​(eμ​x0)),R\left(f;x_{0};x\right)=\frac{1}{2}\exp_{\mu,x_{0}}^{2}\left(x\right)\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\left(e^{\mu\xi_{x}}\right)-\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\left(e^{\mu x_{0}}\right)\right),

with ξx\xi_{x} between xx and x0.x_{0}. Thus, we can write

|R​(f,x0;x)|\displaystyle\left|R\left(f,x_{0};x\right)\right| =12​expμ,x02⁡(x)​|(f∘logμ)′′​(eμ​ξx)−(f∘logμ)′′​(eμ​x)|\displaystyle=\frac{1}{2}\exp_{\mu,x_{0}}^{2}\left(x\right)\left|\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\left(e^{\mu\xi_{x}}\right)-\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\left(e^{\mu x}\right)\right|
≤12​expμ,x02⁡(x)​ω​((f∘logμ)′′;|expμ,x⁡(ξx)|)\displaystyle\leq\frac{1}{2}\exp_{\mu,x_{0}}^{2}\left(x\right)\omega\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}};\left|\exp_{\mu,x}\left(\xi_{x}\right)\right|\right)
≤12​expμ,x02⁡(x)​ω​((f∘logμ)′′;|expμ,x⁡(x0)|),\displaystyle\leq\frac{1}{2}\exp_{\mu,x_{0}}^{2}\left(x\right)\omega\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}};\left|\exp_{\mu,x}\left(x_{0}\right)\right|\right), (32)

where ω​(f;⋅)\omega\left(f;\cdot\right) is the classical modulus of continuity.

We will need the following KK-functional introduced by J. Peetre petre and defined by

K(f;ε,C,C1):=inf{∥f−g∥∞+ε∥g′∥∞:g∈C1}K\left(f;\varepsilon,C,C^{1}\right):=\inf\left\{\left\|f-g\right\|_{\infty}+\varepsilon\left\|g^{{}^{\prime}}\right\|_{\infty}:g\in C^{1}\right\} (33)

for f∈C​[0,1]f\in C\left[0,1\right] and ε≥0.\varepsilon\geq 0. The KK-functional and the modulus of continuity are related by the relation

K​(f;ε/2,C,C1)=12​ω~​(f;ε).K\left(f;\varepsilon/2,C,C^{1}\right)=\frac{1}{2}\widetilde{\omega}\left(f;\varepsilon\right). (34)

Here ω~​(f;⋅)\widetilde{\omega}\left(f;\cdot\right) denotes the least concave majorant of ω​(f;⋅),\omega\left(f;\cdot\right), see Lorentz .

The remainder R​(f,x0;x)R\left(f,x_{0};x\right) can be estimated in terms of ω~.\widetilde{\omega}.

Lemma 5

Let f∈C2​[0,1]f\in C^{2}\left[0,1\right] and x0,x_{0}, x∈[0,1].x\in\left[0,1\right]. Then we have

|R​(f,x0;x)|≤12​expμ,x02⁡(x)​ω~​((f∘logμ)′′;13​|expμ,x0⁡(x)|).\left|R\left(f,x_{0};x\right)\right|\leq\frac{1}{2}\exp_{\mu,x_{0}}^{2}\left(x\right)\widetilde{\omega}\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}};\frac{1}{3}\left|\exp_{\mu,x_{0}}\left(x\right)\right|\right).
Proof

From (32),we have

|R​(f,x0;x)|≤expμ,x02⁡(x)​‖(f∘logμ)′′‖∞.\left|R\left(f,x_{0};x\right)\right|\leq\exp_{\mu,x_{0}}^{2}\left(x\right)\left\|\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\right\|_{\infty}.

For g∈C3​[0,1],g\in C^{3}\left[0,1\right], using the Lagrange form of remainder we have

|R​(g,x0;x)|\displaystyle\left|R\left(g,x_{0};x\right)\right| =16​|expμ,x03⁡(x)|​|(g∘logμ)′′′​(θx)|\displaystyle=\frac{1}{6}\left|\exp_{\mu,x_{0}}^{3}\left(x\right)\right|\left|\left(g\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime\prime}}}\left(\theta_{x}\right)\right|
≤16​|expμ,x03⁡(x)|​‖(g∘logμ)′′′‖∞,\displaystyle\leq\frac{1}{6}\left|\exp_{\mu,x_{0}}^{3}\left(x\right)\right|\left\|\left(g\circ\log_{\mu}\right)^{{}^{{}^{{}^{\prime\prime\prime}}}}\right\|_{\infty},

where θx\theta_{x} is between xx and x0.x_{0}. Keeping ff fixed and letting gg arbitrary in C3​[0,1],C^{3}\left[0,1\right], we have

|R​(f,x0;x)|\displaystyle\left|R\left(f,x_{0};x\right)\right| ≤|R​(f−g,x0;x)|+|R​(g,x0;x)|\displaystyle\leq\left|R\left(f-g,x_{0};x\right)\right|+\left|R\left(g,x_{0};x\right)\right|
≤expμ,x02⁡(x)​‖(f∘logμ)′′−(g∘logμ)′′‖∞\displaystyle\leq\exp_{\mu,x_{0}}^{2}\left(x\right)\left\|\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}-\left(g\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\right\|_{\infty}
+16​|expμ,x03⁡(x)|​‖(g∘logμ)′′′‖∞.\displaystyle\quad+\frac{1}{6}\left|\exp_{\mu,x_{0}}^{3}\left(x\right)\right|\left\|\left(g\circ\log_{\mu}\right)^{{}^{{}^{{}^{\prime\prime\prime}}}}\right\|_{\infty}.

Considering (33), (34) and passing to infimum over g∈C3​[0,1]g\in C^{3}\left[0,1\right] yields

|R​(f,x0;x)|\displaystyle\left|R\left(f,x_{0};x\right)\right| ≤expμ,x02⁡(x)​K​((f∘logμ)′′;16​|expμ,x0⁡(x)|,C,C1)\displaystyle\leq\exp_{\mu,x_{0}}^{2}\left(x\right)K\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}};\frac{1}{6}\left|\exp_{\mu,x_{0}}(x)\right|,C,C^{1}\right)
=12​expμ,x02⁡(x)​ω~​((f∘logμ)′′;13​|expμ,x0⁡(x)|).\displaystyle=\frac{1}{2}\exp_{\mu,x_{0}}^{2}\left(x\right)\widetilde{\omega}\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}};\frac{1}{3}\left|\exp_{\mu,x_{0}}\left(x\right)\right|\right).
Theorem 4.1

If f∈C2​[0,1]f\in C^{2}\left[0,1\right] and x∈[0,1],x\in\left[0,1\right], then

|K~n(f;x)−f(x)−(K~n(e0;x)−1)12​μ2[f′′(x)−3μf′(x)+2μ2f(x)]\displaystyle\left|\widetilde{K}_{n}\left(f;x\right)-f\left(x\right)-\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right)\frac{1}{2\mu^{2}}\left[f^{{}^{\prime\prime}}\left(x\right)-3\mu f^{{}^{\prime}}\left(x\right)+2\mu^{2}f\left(x\right)\right]\right.
−e−μ​x(K~n(expμ;x)−eμ​x)1μ2[2μf′(x)−f′′(x)]|\displaystyle\left.-e^{-\mu x}\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\right)\frac{1}{\mu^{2}}\left[2\mu f^{{}^{\prime}}\left(x\right)-f^{{}^{\prime\prime}}\left(x\right)\right]\right|
≤12​K~n​(expμ,x4;x)​K~n​(e0;x)​ω~​((f∘logμ)′′;13​K~n​(expμ,x2;x)K~n​(e0;x)).\displaystyle\leq\frac{1}{2}\sqrt{\widetilde{K}_{n}\left(\exp_{\mu,x}^{4};x\right)}\sqrt{\widetilde{K}_{n}\left(e_{0};x\right)}\widetilde{\omega}\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}};\frac{1}{3}\frac{\sqrt{\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right)}}{\sqrt{\widetilde{K}_{n}\left(e_{0};x\right)}}\right).
Proof

For f∈C2​[0,1],f\in C^{2}\left[0,1\right], we can write

R​(f,x;t)=f​(t)−(f∘logμ)​(eμ​x)−(f∘logμ)′​(eμ​x)​expμ,x⁡(t)−12​(f∘logμ)′′​(eμ​x)​expμ,x2⁡(t).R\left(f,x;t\right)=f\left(t\right)-\left(f\circ\log_{\mu}\right)\left(e^{\mu x}\right)-\left(f\circ\log_{\mu}\right)^{{}^{\prime}}\left(e^{\mu x}\right)\exp_{\mu,x}\left(t\right)-\frac{1}{2}\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\left(e^{\mu x}\right)\exp_{\mu,x}^{2}\left(t\right).

This equality leads to

K~n​(R​(f,x;t);x)\displaystyle\widetilde{K}_{n}\left(R\left(f,x;t\right);x\right) =K~n​(f;x)−f​(x)​K~n​(e0;x)−(f∘logμ)′​(eμ​x)​K~n​(expμ,x;x)\displaystyle=\widetilde{K}_{n}\left(f;x\right)-f\left(x\right)\widetilde{K}_{n}\left(e_{0};x\right)-\left(f\circ\log_{\mu}\right)^{{}^{\prime}}\left(e^{\mu x}\right)\widetilde{K}_{n}\left(\exp_{\mu,x};x\right)
−12​(f∘logμ)′′​(eμ​x)​K~n​(expμ,x2;x)​.\displaystyle\text{\ }-\frac{1}{2}\left(f\circ\log_{\mu}\right)^{{}^{\prime\prime}}\left(e^{\mu x}\right)\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right)\text{.}

Since

(f∘logμ)′​(eμ​x)=e−μ​xμ​f′​(x)\left(f\circ\log_{\mu}\right)^{{}^{\prime}}\left(e^{\mu x}\right)=\frac{e^{-\mu x}}{\mu}f^{{}^{\prime}}\left(x\right)

and

(f∘logμ)′′​(eμ​x)=e−2​μ​x​(1μ2​f′′​(x)−1μ​f′​(x)),\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\left(e^{\mu x}\right)=e^{-2\mu x}\left(\frac{1}{\mu^{2}}f^{{}^{\prime\prime}}\left(x\right)-\frac{1}{\mu}f^{{}^{\prime}}\left(x\right)\right),

we can write

K~n​(R​(f,x;t);x)\displaystyle\widetilde{K}_{n}\left(R\left(f,x;t\right);x\right) =K~n​(f;x)−f​(x)​K~n​(e0;x)−e−μ​xμ​f′​(x)​K~n​(expμ,x;x)\displaystyle=\widetilde{K}_{n}\left(f;x\right)-f\left(x\right)\widetilde{K}_{n}\left(e_{0};x\right)-\frac{e^{-\mu x}}{\mu}f^{{}^{\prime}}\left(x\right)\widetilde{K}_{n}\left(\exp_{\mu,x};x\right)
−e−2​μ​x2​(1μ2​f′′​(x)−1μ​f′​(x))​K~n​(expμ,x2;x).\displaystyle-\frac{e^{-2\mu x}}{2}\left(\frac{1}{\mu^{2}}f^{{}^{\prime\prime}}\left(x\right)-\frac{1}{\mu}f^{{}^{\prime}}\left(x\right)\right)\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right).

It can be rearranged as

K~n​(R​(f,x;t);x)=\displaystyle\widetilde{K}_{n}\left(R\left(f,x;t\right);x\right)=
=K~n​(f;x)−f​(x)−f​(x)​(K~n​(e0;x)−1)\displaystyle=\widetilde{K}_{n}\left(f;x\right)-f\left(x\right)-f\left(x\right)\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right)
−1μ​f′​(x)​[e−μ​x​(K~n​(expμ;x)−eμ​x)−(K~n​(e0;x)−1)]\displaystyle\quad-\frac{1}{\mu}f^{{}^{\prime}}\left(x\right)\left[e^{-\mu x}\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\right)-\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right)\right]
−12​(1μ2​f′′​(x)−1μ​f′​(x))​[((K~n​(e0;x)−1)−2​e−μ​x​(K~n​(expμ;x)−eμ​x))]\displaystyle\quad-\frac{1}{2}\left(\frac{1}{\mu^{2}}f^{{}^{\prime\prime}}\left(x\right)-\frac{1}{\mu}f^{{}^{\prime}}\left(x\right)\right)\left[\left(\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right)-2e^{-\mu x}\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\right)\right)\right]
=K~n​(f;x)−f​(x)−(K~n​(e0;x)−1)​[f​(x)+12​(1μ2​f′′​(x)−1μ​f′​(x))−1μ​f′​(x)]\displaystyle=\widetilde{K}_{n}\left(f;x\right)-f\left(x\right)-\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right)\left[f\left(x\right)+\frac{1}{2}\left(\frac{1}{\mu^{2}}f^{{}^{\prime\prime}}\left(x\right)-\frac{1}{\mu}f^{{}^{\prime}}\left(x\right)\right)-\frac{1}{\mu}f^{{}^{\prime}}\left(x\right)\right]
−e−μ​x​(K~n​(expμ;x)−eμ​x)​[1μ​f′​(x)−(1μ2​f′′​(x)−1μ​f′​(x))]\displaystyle\quad-e^{-\mu x}\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\right)\left[\frac{1}{\mu}f^{{}^{\prime}}\left(x\right)-\left(\frac{1}{\mu^{2}}f^{{}^{\prime\prime}}\left(x\right)-\frac{1}{\mu}f^{{}^{\prime}}\left(x\right)\right)\right]
=K~n​(f;x)−f​(x)−(K~n​(e0;x)−1)​12​μ2​[f′′​(x)−3​μ​f′​(x)+2​μ2​f​(x)]\displaystyle=\widetilde{K}_{n}\left(f;x\right)-f\left(x\right)-\left(\widetilde{K}_{n}\left(e_{0};x\right)-1\right)\frac{1}{2\mu^{2}}\left[f^{{}^{\prime\prime}}\left(x\right)-3\mu f^{{}^{\prime}}\left(x\right)+2\mu^{2}f\left(x\right)\right]
−e−μ​x​(K~n​(expμ;x)−eμ​x)​1μ2​[2​μ​f′​(x)−f′′​(x)].\displaystyle\quad-e^{-\mu x}\left(\widetilde{K}_{n}\left(\exp_{\mu};x\right)-e^{\mu x}\right)\frac{1}{\mu^{2}}\left[2\mu f^{{}^{\prime}}\left(x\right)-f^{{}^{\prime\prime}}\left(x\right)\right].

On the other hand, from Lemma 5 we have

|K~n​(R​(f,x;t);x)|≤K~n​(12​expμ,x2⁡(t)​ω~​((f∘logμ)′′;13​|expμ,x⁡(t)|);x).\left|\widetilde{K}_{n}\left(R\left(f,x;t\right);x\right)\right|\leq\widetilde{K}_{n}\left(\frac{1}{2}\exp_{\mu,x}^{2}\left(t\right)\widetilde{\omega}\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}};\frac{1}{3}\left|\exp_{\mu,x}\left(t\right)\right|\right);x\right).

For arbitrary g∈C3​[0,1],g\in C^{3}\left[0,1\right], we get

|K~n​(R​(f,x;t);x)|=\displaystyle\left|\widetilde{K}_{n}\left(R\left(f,x;t\right);x\right)\right|=
=K~n​(expμ,x2⁡(t)​K​((f∘logμ)′′;16​|expμ,x⁡(t)|,C,C1);x)\displaystyle=\widetilde{K}_{n}\left(\exp_{\mu,x}^{2}\left(t\right)K\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}};\frac{1}{6}\left|\exp_{\mu,x}\left(t\right)\right|,C,C^{1}\right);x\right)
≤K~n​(expμ,x2⁡(t)​{‖(f∘logμ)′′−(g∘logμ)′′‖∞+|expμ,x⁡(t)|6​‖(g∘logμ)′′′‖∞};x)\displaystyle\leq\widetilde{K}_{n}\left(\exp_{\mu,x}^{2}\left(t\right)\left\{\left\|\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}-\left(g\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\right\|_{\infty}+\frac{\left|\exp_{\mu,x}\left(t\right)\right|}{6}\left\|\left(g\circ\log_{\mu}\right)^{{}^{{}^{{}^{\prime\prime\prime}}}}\right\|_{\infty}\right\};x\right)
=K~n​(expμ,x2;x)​‖(f∘logμ)′′−(g∘logμ)′′‖∞+16​K~n​(|expμ,x3|;x)​‖(g∘logμ)′′′‖∞\displaystyle=\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right)\left\|\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}-\left(g\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\right\|_{\infty}+\frac{1}{6}\widetilde{K}_{n}\left(\left|\exp_{\mu,x}^{3}\right|;x\right)\left\|\left(g\circ\log_{\mu}\right)^{{}^{{}^{{}^{\prime\prime\prime}}}}\right\|_{\infty}
≤K~n​(expμ,x4;x)K~n​(e0;x){∥(f∘logμ)′′−(g∘logμ)′′∥∞\displaystyle\leq\sqrt{\widetilde{K}_{n}\left(\exp_{\mu,x}^{4};x\right)}\sqrt{\widetilde{K}_{n}\left(e_{0};x\right)}\left\{\left\|\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}-\left(g\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}}\right\|_{\infty}\right.
+16K~n​(expμ,x2;x)K~n​(e0;x)∥(g∘logμ)′′′∥∞}.\displaystyle\quad+\frac{1}{6}\frac{\sqrt{\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right)}}{\sqrt{\widetilde{K}_{n}\left(e_{0};x\right)}}\left\|\left(g\circ\log_{\mu}\right)^{{}^{{}^{{}^{\prime\prime\prime}}}}\right\|_{\infty}\Bigg\}.

Passing to the infimum over g∈C3​[0,1]g\in C^{3}\left[0,1\right] again, we get

|K~n​(R​(f,x;t);x)|≤\displaystyle\left|\widetilde{K}_{n}\left(R\left(f,x;t\right);x\right)\right|\leq
≤K~n​(expμ,x4;x)​K~n​(e0;x)​K​((f∘logμ)′′;16​K~n​(expμ,x2;x)K~n​(e0;x),C,C1)\displaystyle\leq\sqrt{\widetilde{K}_{n}\left(\exp_{\mu,x}^{4};x\right)}\sqrt{\widetilde{K}_{n}\left(e_{0};x\right)}K\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}};\frac{1}{6}\frac{\sqrt{\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right)}}{\sqrt{\widetilde{K}_{n}\left(e_{0};x\right)}},C,C^{1}\right)
=12​K~n​(expμ,x4;x)​K~n​(e0;x)​ω~​((f∘logμ)′′;13​K~n​(expμ,x2;x)K~n​(e0;x)).\displaystyle=\frac{1}{2}\sqrt{\widetilde{K}_{n}\left(\exp_{\mu,x}^{4};x\right)}\sqrt{\widetilde{K}_{n}\left(e_{0};x\right)}\widetilde{\omega}\left(\left(f\circ\log_{\mu}\right)^{{}^{{}^{\prime\prime}}};\frac{1}{3}\frac{\sqrt{\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right)}}{\sqrt{\widetilde{K}_{n}\left(e_{0};x\right)}}\right).

Using (17), (18), (21) and (19) in Theorem 4.1, respectively, and considering

limn→∞K~n​(expμ,x2;x)K~n​(e0;x)=0\lim_{n\rightarrow\infty}\frac{\widetilde{K}_{n}\left(\exp_{\mu,x}^{2};x\right)}{\widetilde{K}_{n}\left(e_{0};x\right)}=0

we have:

Corollary 1

If f∈C2​[0,1]f\in C^{2}\left[0,1\right] and x∈[0,1],x\in\left[0,1\right], then

limn→∞2​n​μ2​(K~n​(f;x)−f​(x))=\displaystyle\lim_{n\rightarrow\infty}2n\mu^{2}\left(\widetilde{K}_{n}\left(f;x\right)-f\left(x\right)\right)=
=(μ​(μ​x−1)−μ​x​(μ​x−2))​[f′′​(x)−3​μ​f′​(x)+2​μ2​f​(x)]\displaystyle=\left(\mu\left(\mu x-1\right)-\mu x\left(\mu x-2\right)\right)\left[f^{{}^{\prime\prime}}\left(x\right)-3\mu f^{{}^{\prime}}\left(x\right)+2\mu^{2}f\left(x\right)\right]
+μ​(2​x−1)​[2​μ​f′​(x)−f′′​(x)].\displaystyle\quad+\mu\left(2x-1\right)\left[2\mu f^{{}^{\prime}}\left(x\right)-f^{{}^{\prime\prime}}\left(x\right)\right].

5 Convergence in Lp,μ​[0,1]L_{p,\mu}\left[0,1\right] and Lp​[0,1]L_{p}\left[0,1\right]

Let 1≤p<∞1\leq p<\infty be fixed and Lp,μ​[0,1]L_{p,\mu}\left[0,1\right] be the space of all functions for which expμ⁡f\exp_{\mu}f is Lebesgue integrable with the pp-power over [0,1].\left[0,1\right]. The norm in Lp,μ​[0,1]L_{p,\mu}\left[0,1\right] is defined as

‖f‖p,μ:=(∫01|e−μ​x​f​(x)|p​𝑑x)1/p.\left\|f\right\|_{p,\mu}:=\left(\int_{0}^{1}\left|e^{-\mu x}f\left(x\right)\right|^{p}dx\right)^{1/p}.

The norm of a linear operator LnL_{n} acting from the space Lp,μL_{p,\mu} to Lp,μL_{p,\mu} given by

‖Ln‖Lp,μ,Lp,μ:=sup‖f‖p,μ≠0​‖Ln​f‖p,μ‖f‖p,μ.\left\|L_{n}\right\|_{L_{p,\mu},L_{p,\mu}}:=\underset{\left\|f\right\|_{p,\mu}\neq 0}{\sup}\frac{\left\|L_{n}f\right\|_{p,\mu}}{\left\|f\right\|_{p,\mu}}.

Also, K~n\widetilde{K}_{n} maps the space L1,μ​[0,1]L_{1,\mu}\left[0,1\right] into the space C​[0,1].C\left[0,1\right].

Lemma 6

For f∈Lp,μ​[0,1],f\in L_{p,\mu}\left[0,1\right], we haveK~n​(f)∈Lp,μ​[0,1]\mathcal{\ }\widetilde{K}_{n}\left(f\right)\in L_{p,\mu}\left[0,1\right] and for all n∈ℕ,n\in\mathbb{N}, ‖K~n‖Lp,μ,Lp,μ≤eμ​(p−1)p.\left\|\widetilde{K}_{n}\right\|_{L_{p,\mu},L_{p,\mu}}\leq e^{\mu\frac{\left(p-1\right)}{p}}.

Proof

Applying the Jensen’s inequality to the measure (n+1)​d​t(n+1)dt, we get

|K~n​(f;x)|p\displaystyle\left|\widetilde{K}_{n}\left(f;x\right)\right|^{p} ≤(∑k=0npn,k​(an+1​(x))​an+1′​(x)​(n+1)​eμ​x​∫kn+1k+1n+1e−μ​t​|f​(t)|​𝑑t)p\displaystyle\leq\left(\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)e^{\mu x}\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu t}\left|f\left(t\right)\right|dt\right)^{p}
≤∑k=0npn,k​(an+1​(x))​(an+1′​(x)​(n+1)​eμ​x​∫kn+1k+1n+1e−μ​t​|f​(t)|​𝑑t)p\displaystyle\leq\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\left(a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)e^{\mu x}\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu t}\left|f\left(t\right)\right|dt\right)^{p}
≤maxx∈[0,1](an+1′(x))p−1∑k=0npn,k(an+1(x))an+1′(x)eμ​p​x(n+1)\displaystyle\leq\max_{x\in\left[0,1\right]}\left(a_{n+1}^{{}^{\prime}}\left(x\right)\right)^{p-1}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)a_{n+1}^{{}^{\prime}}\left(x\right)e^{\mu px}\left(n+1\right)
⋅∫kn+1k+1n+1e−μ​p​t|f(t)|pdt.\displaystyle\quad\cdot\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu pt}\left|f\left(t\right)\right|^{p}dt. (35)

If we take the integral on the interval [0,1]\left[0,1\right], we obtain

∫01|e−μ​x​K~n​(f;x)|p​𝑑x\displaystyle\int_{0}^{1}\left|e^{-\mu x}\widetilde{K}_{n}\left(f;x\right)\right|^{p}dx
≤maxx∈[0,1](an+1′(x))p−1∑k=0n∫01pn,k(an+1(x))an+1′(x)dx\displaystyle\leq\max_{x\in\left[0,1\right]}\left(a_{n+1}^{{}^{\prime}}\left(x\right)\right)^{p-1}\sum\limits_{k=0}^{n}\int_{0}^{1}p_{n,k}\left(a_{n+1}\left(x\right)\right)a_{n+1}^{{}^{\prime}}\left(x\right)dx
⋅[(n+1)​∫kn+1k+1n+1e−μ​p​t​|f​(t)|p​𝑑t]\displaystyle\quad\cdot\left[\left(n+1\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu pt}\left|f\left(t\right)\right|^{p}dt\right]
=maxx∈[0,1](an+1′(x))p−1∑k=0n∫kn+1k+1n+1e−μ​p​t|f(t)|pdt\displaystyle=\max_{x\in\left[0,1\right]}\left(a_{n+1}^{{}^{\prime}}\left(x\right)\right)^{p-1}\sum\limits_{k=0}^{n}\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu pt}\left|f\left(t\right)\right|^{p}dt
=maxx∈[0,1](an+1′(x))p−1∫01e−μ​p​t|f(t)|pdt.\displaystyle=\max_{x\in\left[0,1\right]}\left(a_{n+1}^{{}^{\prime}}\left(x\right)\right)^{p-1}\int_{0}^{1}e^{-\mu pt}\left|f\left(t\right)\right|^{p}dt.

Since

maxx∈[0,1]⁡(an+1′​(x))=μn+1​eμ/(n+1)eμ/(n+1)−1,\max_{x\in\left[0,1\right]}\left(a_{n+1}^{{}^{\prime}}\left(x\right)\right)=\frac{\mu}{n+1}\frac{e^{\mu/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1},

using the inequality u≤eu−1,u\leq e^{u}-1, u>0,u>0, we have

maxx∈[0,1]⁡(an+1′​(x))\displaystyle\max_{x\in\left[0,1\right]}\left(a_{n+1}^{{}^{\prime}}\left(x\right)\right) =μn+1​eμ/(n+1)eμ/(n+1)−1≤μn+1​(eμ/(n+1)μ/(n+1))\displaystyle=\frac{\mu}{n+1}\frac{e^{\mu/\left(n+1\right)}}{e^{\mu/\left(n+1\right)}-1}\leq\frac{\mu}{n+1}\left(\frac{e^{\mu/\left(n+1\right)}}{\mu/\left(n+1\right)}\right)
≤eμ/(n+1).\displaystyle\leq e^{\mu/\left(n+1\right)}. (36)

Consequently we get

‖K~n​(f)‖p,μ≤eμn+1​(p−1)p​‖f‖p,μ.\left\|\widetilde{K}_{n}\left(f\right)\right\|_{p,\mu}\leq e^{\frac{\mu}{n+1}\frac{\left(p-1\right)}{p}}\left\|f\right\|_{p,\mu}.

In view of this inequality, for all n∈ℕn\in\mathbb{N} we can write

‖K~n‖Lp,μ,Lp,μ≤eμ​(p−1)p.\left\|\widetilde{K}_{n}\right\|_{L_{p,\mu},L_{p,\mu}}\leq e^{\mu\frac{\left(p-1\right)}{p}}.
Theorem 5.1

Let f∈Lp,μ​[0,1]f\in L_{p,\mu}\left[0,1\right] for 1≤p<∞.1\leq p<\infty. Then we have

limn→∞‖K~n​(f)−f‖p,μ=0.\lim_{n\rightarrow\infty}\left\|\widetilde{K}_{n}\left(f\right)-f\right\|_{p,\mu}=0.
Proof

From the Luzin theorem for a given ε>0\varepsilon>0, there exists g∈C​[0,1]g\in C[0,1] such that

‖f−g‖p,μ<ε.\left\|f-g\right\|_{p,\mu}<\varepsilon.

On the other hand, since K~n​g\widetilde{K}_{n}g converges to gg uniformly on [0,1],\left[0,1\right], there exists an n0∈Nn_{0}\in N such that, for n≥n0n\geq n_{0}

‖K~n​(g)−g‖∞<ε.\left\|\widetilde{K}_{n}\left(g\right)-g\right\|_{\infty}<\varepsilon.

In view of the above inequalities

‖K~n​(f)−f‖p,μ\displaystyle\left\|\widetilde{K}_{n}\left(f\right)-f\right\|_{p,\mu} ≤‖K~n​(f)−K~n​(g)‖p,μ+‖K~n​(g)−g‖∞+‖f−g‖p,μ\displaystyle\leq\left\|\widetilde{K}_{n}\left(f\right)-\widetilde{K}_{n}\left(g\right)\right\|_{p,\mu}+\left\|\widetilde{K}_{n}\left(g\right)-g\right\|_{\infty}+\left\|f-g\right\|_{p,\mu}
≤(‖K~n‖Lp,μ,Lp,μ+1)​‖f−g‖p,μ+‖K~n​(g)−g‖∞\displaystyle\leq\left(\left\|\widetilde{K}_{n}\right\|_{L_{p,\mu},L_{p,\mu}}+1\right)\left\|f-g\right\|_{p,\mu}+\left\|\widetilde{K}_{n}\left(g\right)-g\right\|_{\infty}
≤(‖K~n‖Lp,μ,Lp,μ+2)​ε.\displaystyle\leq\left(\left\|\widetilde{K}_{n}\right\|_{L_{p,\mu},L_{p,\mu}}+2\right)\varepsilon.

for n≥n0.n\geq n_{0}. Thus, we have the desired result.

Now we want to give a quantitative approximation theorem for (8) in the space Lp​[0,1].L_{p}\left[0,1\right]. To describe our results we will use the following integral modulus of continuity defined by

ω(f;t)Lp=sup0≤h<t∥f(⋅+h)−f(⋅)∥Lp​[0,1].\omega\left(f;t\right)_{L_{p}}=\sup_{0\leq h<t}\left\|f\left(\cdot+h\right)-f\left(\cdot\right)\right\|_{L_{p}\left[0,1\right].}

∥⋅∥Lp\left\|\cdot\right\|_{L_{p}} denotes the usual LpL_{p} norm and the corresponding KK-functionals is defined by

Kp​(f;t):=infg∈A​C​[0,1],g′∈Lp{‖f−g‖Lp+t​‖g′‖Lp},K_{p}\left(f;t\right):=\inf_{g\in AC\left[0,1\right],g^{{}^{\prime}}\in L_{p}}\left\{\left\|f-g\right\|_{L_{p}}+t\left\|g^{{}^{\prime}}\right\|_{L_{p}}\right\},

where A​C​[0,1]AC\left[0,1\right] indicates the set of all absolute continuous functions on the interval [0,1]\left[0,1\right]. We know that Kp​(f;t)K_{p}\left(f;t\right) and ω​(f;t)Lp\omega\left(f;t\right)_{L_{p}} are equivalent (see (Devore, , Theorem 2.1)), i.e., there is a constant CC such that

C−1​ω​(f;t)Lp≤Kp​(f;t)≤C​ω​(f;t)Lp.C^{-1}\omega\left(f;t\right)_{L_{p}}\leq K_{p}\left(f;t\right)\leq C\omega\left(f;t\right)_{L_{p}}. (37)

For given f∈Lp​[0,1]f\in L_{p}\left[0,1\right], the Hardy-Littlewood maximum function is defined by

M​(f;x)=sup0≤t≤1t≠x​1t−x​∫xt|f​(u)|​𝑑u.M\left(f;x\right)=\underset{t\neq x}{\sup_{0\leq t\leq 1}}\frac{1}{t-x}\int_{x}^{t}\left|f\left(u\right)\right|du. (38)

It is well known that (see Stein )

‖M​(f)‖Lp≤Cp​‖f‖Lp.\left\|M\left(f\right)\right\|_{L_{p}}\leq C_{p}\left\|f\right\|_{L_{p}}. (39)

Also, as in (35), considering (36) we can write

|K~n​(f;x)|p\displaystyle\left|\widetilde{K}_{n}\left(f;x\right)\right|^{p} ≤maxx∈[0,1](an+1′(x))p−1∑k=0npn,k(an+1(x))an+1′(x)eμ​p​x(n+1)\displaystyle\leq\max_{x\in\left[0,1\right]}\left(a_{n+1}^{{}^{\prime}}\left(x\right)\right)^{p-1}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)a_{n+1}^{{}^{\prime}}\left(x\right)e^{\mu px}\left(n+1\right)
⋅∫kn+1k+1n+1e−μ​p​t|f(t)|pdt\displaystyle\quad\cdot\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu pt}\left|f\left(t\right)\right|^{p}dt
≤eμ​p​eμn+1​(p−1)​∑k=0npn,k​(an+1​(x))​an+1′​(x)​(n+1)​∫kn+1k+1n+1|f​(t)|p​𝑑t.\displaystyle\leq e^{\mu p}e^{\frac{\mu}{n+1}\left(p-1\right)}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}\left|f\left(t\right)\right|^{p}dt.

Integrating, we have

∫01|K~n​(f;x)|p​𝑑x\displaystyle\int_{0}^{1}\left|\widetilde{K}_{n}\left(f;x\right)\right|^{p}dx ≤eμ​p​eμn+1​(p−1)​∑k=0n((n+1)​∫01pn,k​(an+1​(x))​an+1′​(x)​𝑑x)\displaystyle\leq e^{\mu p}e^{\frac{\mu}{n+1}\left(p-1\right)}\sum\limits_{k=0}^{n}\left(\left(n+1\right)\int_{0}^{1}p_{n,k}\left(a_{n+1}\left(x\right)\right)a_{n+1}^{{}^{\prime}}\left(x\right)dx\right)
⋅∫kn+1k+1n+1|f(t)|pdt\displaystyle\quad\cdot\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}\left|f\left(t\right)\right|^{p}dt
≤eμ​p​eμn+1​(p−1)​∑k=0n∫kn+1k+1n+1|f​(t)|p​𝑑t\displaystyle\leq e^{\mu p}e^{\frac{\mu}{n+1}\left(p-1\right)}\sum\limits_{k=0}^{n}\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}\left|f\left(t\right)\right|^{p}dt

and

‖K~n​(f)‖p≤eμ​eμn+1​(p−1)p​‖f‖p.\left\|\widetilde{K}_{n}\left(f\right)\right\|_{p}\leq e^{\mu}e^{\frac{\mu}{n+1}\frac{\left(p-1\right)}{p}}\left\|f\right\|_{p}.

Consequently, for all n∈ℕn\in\mathbb{N} we can write

‖K~n‖Lp,Lp≤eμ​eμ​(p−1)p.\left\|\widetilde{K}_{n}\right\|_{L_{p},L_{p}}\leq e^{\mu}e^{\mu\frac{\left(p-1\right)}{p}}. (40)
Lemma 7

Let g∈A​C​[0,1]g\in AC[0,1] and g′∈Lp​[0,1],g^{{}^{\prime}}\in L_{p}[0,1], p>1.p>1. Then we have

‖K~n​(g)−g‖p≤αn,μ​‖g‖p+μ−1​(βn,μ)1/2​eμ2​(2+1n+1)​Cp​‖g′‖p.\left\|\widetilde{K}_{n}\left(g\right)-g\right\|_{p}\leq\alpha_{n,\mu}\left\|g\right\|_{p}+\mu^{-1}\left(\beta_{n,\mu}\right)^{1/2}e^{\frac{\mu}{2}\left(2+\frac{1}{n+1}\right)}C_{p}\left\|g^{{}^{\prime}}\right\|_{p}.

where αn,μ\alpha_{n,\mu}, βn,μ\beta_{n,\mu} are defined as in Lemma 4 and CpC_{p} is defined in (39).

Proof

Using the equality g​(t)=g​(x)+∫xtg′​(u)​𝑑u,g\left(t\right)=g\left(x\right)+\int_{x}^{t}g^{{}^{\prime}}\left(u\right)du, we can write

K~n​(g;x)−g​(x)+g​(x)−g​(x)​K~n​(e0;x)=\displaystyle\widetilde{K}_{n}\left(g;x\right)-g\left(x\right)+g\left(x\right)-g\left(x\right)\widetilde{K}_{n}\left(e_{0};x\right)=
=an+1′​(x)​(n+1)​eμ​x​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1e−μ​t​(g​(t)−g​(x))​𝑑t\displaystyle=a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu t}\left(g\left(t\right)-g\left(x\right)\right)dt
=an+1′​(x)​(n+1)​eμ​x​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1e−μ​t​∫xtg′​(u)​𝑑u​𝑑t.\displaystyle=a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}e^{-\mu t}\int_{x}^{t}g^{{}^{\prime}}\left(u\right)dudt.

Hence

|K~n​(g;x)−g​(x)|≤\displaystyle\left|\widetilde{K}_{n}\left(g;x\right)-g\left(x\right)\right|\leq
≤maxx∈[0,1]⁡|K~n​(e0;x)−1|​|g​(x)|\displaystyle\leq\max_{x\in\left[0,1\right]}\left|\widetilde{K}_{n}\left(e_{0};x\right)-1\right|\left|g\left(x\right)\right|
+M​(g′;x)​an+1′​(x)​(n+1)​eμ​x​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1|t−x|​𝑑t,\displaystyle\quad+M\left(g^{{}^{\prime}};x\right)a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}\left|t-x\right|dt,

where

M​(g′;x)=sup0≤t≤1x≠t​1t−x​∫xt|g′​(u)|​𝑑uM\left(g^{{}^{\prime}};x\right)=\underset{x\neq t}{\sup_{0\leq t\leq 1}}\frac{1}{t-x}\int_{x}^{t}\left|g^{{}^{\prime}}\left(u\right)\right|du

is defined as in (38). By Lagrange’s theorem for x∈(0,1)x\in\left(0,1\right) and t∈[0,1],t\in\left[0,1\right], we get

μ​|t−x|≤|expμ,x⁡(t)|.\mu\left|t-x\right|\leq\left|\exp_{\mu,x}\left(t\right)\right|.

Considering (22) and using the above inequality with Cauchy-Schwarz inequality, we get

|K~n​(g;x)−g​(x)|≤\displaystyle\left|\widetilde{K}_{n}\left(g;x\right)-g\left(x\right)\right|\leq
≤αn,μ​|g​(x)|+μ−1​M​(g′;x)​(an+1′​(x)​(n+1)​eμ​x​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1𝑑t)1/2\displaystyle\leq\alpha_{n,\mu}\left|g\left(x\right)\right|+\mu^{-1}M\left(g^{{}^{\prime}};x\right)\left(a_{n+1}^{{}^{\prime}}\left(x\right)\left(n+1\right)e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}dt\right)^{1/2}
×(an+1′​(x)​eμ​x​(n+1)​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1expμ,x2⁡(t)​𝑑t)1/2\displaystyle\quad\times\left(a_{n+1}^{{}^{\prime}}\left(x\right)e^{\mu x}\left(n+1\right)\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}\exp_{\mu,x}^{2}\left(t\right)dt\right)^{1/2}
≤αn,μ​|g​(x)|\displaystyle\leq\alpha_{n,\mu}\left|g\left(x\right)\right|
+μ−1​M​(g′;x)​(an+1′​(x)​eμ​x​∑k=0npn,k​(an+1​(x)))1/2\displaystyle\quad+\mu^{-1}M\left(g^{{}^{\prime}};x\right)\left(a_{n+1}^{{}^{\prime}}\left(x\right)e^{\mu x}\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\right)^{1/2}
×(an+1′​(x)​eμ​x​(n+1)​∑k=0npn,k​(an+1​(x))​∫kn+1k+1n+1expμ,x2⁡(t)​𝑑t)1/2.\displaystyle\times\left(a_{n+1}^{{}^{\prime}}\left(x\right)e^{\mu x}\left(n+1\right)\sum\limits_{k=0}^{n}p_{n,k}\left(a_{n+1}\left(x\right)\right)\int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}\exp_{\mu,x}^{2}\left(t\right)dt\right)^{1/2}.

Using (36) we have

|K~n​(g;x)−g​(x)|≤αn,μ​|g​(x)|+μ−1​M​(g′;x)​eμ2​(2+1n+1)​(K~n​(expμ,x2⁡(t);x))1/2.\left|\widetilde{K}_{n}\left(g;x\right)-g\left(x\right)\right|\leq\alpha_{n,\mu}\left|g\left(x\right)\right|+\mu^{-1}M\left(g^{{}^{\prime}};x\right)e^{\frac{\mu}{2}\left(2+\frac{1}{n+1}\right)}\left(\widetilde{K}_{n}\left(\exp_{\mu,x}^{2}\left(t\right);x\right)\right)^{1/2}.

From (24) we can write

|K~n​(g;x)−g​(x)|≤αn,μ​|g​(x)|+μ−1​M​(g′;x)​eμ2​(2+1n+1)​(βn,μ)1/2.\left|\widetilde{K}_{n}\left(g;x\right)-g\left(x\right)\right|\leq\alpha_{n,\mu}\left|g\left(x\right)\right|+\mu^{-1}M\left(g^{{}^{\prime}};x\right)e^{\frac{\mu}{2}\left(2+\frac{1}{n+1}\right)}\left(\beta_{n,\mu}\right)^{1/2}.

Considering (39), we get

‖K~n​(g)−g‖p≤αn,μ​‖g‖p+μ−1​(βn,μ)1/2​eμ2​(2+1n+1)​Cp​‖g′‖p.\left\|\widetilde{K}_{n}\left(g\right)-g\right\|_{p}\leq\alpha_{n,\mu}\left\|g\right\|_{p}+\mu^{-1}\left(\beta_{n,\mu}\right)^{1/2}e^{\frac{\mu}{2}\left(2+\frac{1}{n+1}\right)}C_{p}\left\|g^{{}^{\prime}}\right\|_{p}.
Theorem 5.2

Let f∈Lp​[0,1],f\in L_{p}\left[0,1\right], p>1.p>1. Then we have

‖K~n​(f)−f‖p≤αn,μ​‖f‖p+C​K​ω​(f;(βn,μ)1/2)Lp,\left\|\widetilde{K}_{n}\left(f\right)-f\right\|_{p}\leq\alpha_{n,\mu}\left\|f\right\|_{p}+CK\omega\left(f;\left(\beta_{n,\mu}\right)^{1/2}\right)_{L_{p}},

where K:=max⁡{eμ​eμ​(p−1)p+αn,μ+1,μ−1​eμ2​(2+1n+1)​Cp}.K:=\max\left\{e^{\mu}e^{\mu\frac{\left(p-1\right)}{p}}+\alpha_{n,\mu}+1,\mu^{-1}e^{\frac{\mu}{2}\left(2+\frac{1}{n+1}\right)}C_{p}\right\}.

Proof

For g∈A​C​[0,1],g′∈Lpg\in AC\left[0,1\right],\ g^{\prime}\in L_{p}, we can write

‖K~n​(f)−f‖p\displaystyle\left\|\widetilde{K}_{n}\left(f\right)-f\right\|_{p} =‖K~n​(f−g+g)−f‖p\displaystyle=\left\|\widetilde{K}_{n}\left(f-g+g\right)-f\right\|_{p}
≤‖K~n​(f−g)−(f−g)+(K~n​(g)−g)‖p\displaystyle\leq\left\|\widetilde{K}_{n}\left(f-g\right)-\left(f-g\right)+\left(\widetilde{K}_{n}\left(g\right)-g\right)\right\|_{p}
≤‖K~n​(f−g)‖p+‖f−g‖p+‖K~n​(g)−g‖p\displaystyle\leq\left\|\widetilde{K}_{n}\left(f-g\right)\right\|_{p}+\left\|f-g\right\|_{p}+\left\|\widetilde{K}_{n}\left(g\right)-g\right\|_{p}
≤(‖K~n‖Lp,Lp+1)​‖f−g‖p+‖K~n​(g)−g‖p\displaystyle\leq\left(\left\|\widetilde{K}_{n}\right\|_{L_{p},L_{p}}+1\right)\left\|f-g\right\|_{p}+\left\|\widetilde{K}_{n}\left(g\right)-g\right\|_{p}

Using the fact that ‖K~n‖Lp,Lp≤eμ​eμ​(p−1)p\left\|\widetilde{K}_{n}\right\|_{L_{p},L_{p}}\leq e^{\mu}e^{\mu\frac{\left(p-1\right)}{p}} (see (40)) and Lemma 7, we get

‖K~n​(f)−f‖p≤\displaystyle\left\|\widetilde{K}_{n}\left(f\right)-f\right\|_{p}\leq
≤(‖K~n‖Lp,Lp+1)​‖f−g‖p+αn,μ​‖g‖p+μ−1​(βn,μ)1/2​eμ2​(2+1n+1)​Cp​‖g′‖p\displaystyle\leq\left(\left\|\widetilde{K}_{n}\right\|_{L_{p},L_{p}}+1\right)\left\|f-g\right\|_{p}+\alpha_{n,\mu}\left\|g\right\|_{p}+\mu^{-1}\left(\beta_{n,\mu}\right)^{1/2}e^{\frac{\mu}{2}\left(2+\frac{1}{n+1}\right)}C_{p}\left\|g^{{}^{\prime}}\right\|_{p}
≤(eμ​eμ​(p−1)p+1)​‖f−g‖p+αn,μ​‖f−g‖p+αn,μ​‖f‖p+\displaystyle\leq\left(e^{\mu}e^{\mu\frac{\left(p-1\right)}{p}}+1\right)\left\|f-g\right\|_{p}+\alpha_{n,\mu}\left\|f-g\right\|_{p}+\alpha_{n,\mu}\left\|f\right\|_{p}+
+μ−1​(βn,μ)1/2​eμ2​(2+1n+1)​Cp​‖g′‖p.\displaystyle\quad+\mu^{-1}\left(\beta_{n,\mu}\right)^{1/2}e^{\frac{\mu}{2}\left(2+\frac{1}{n+1}\right)}C_{p}\left\|g^{{}^{\prime}}\right\|_{p}.

Define max⁡{eμ​eμ​(p−1)p+αn,μ+1,μ−1​eμ2​(2+1n+1)​Cp}:=K\max\left\{e^{\mu}e^{\mu\frac{\left(p-1\right)}{p}}+\alpha_{n,\mu}+1,\mu^{-1}e^{\frac{\mu}{2}\left(2+\frac{1}{n+1}\right)}C_{p}\right\}:=K, we have (see (5.3))

‖K~n​(f)−f‖p\displaystyle\left\|\widetilde{K}_{n}\left(f\right)-f\right\|_{p} ≤αn,μ​‖f‖p+K​(‖f−g‖p+(βn,μ)1/2​‖g′‖p)\displaystyle\leq\alpha_{n,\mu}\left\|f\right\|_{p}+K\left(\left\|f-g\right\|_{p}+\left(\beta_{n,\mu}\right)^{1/2}\left\|g^{{}^{\prime}}\right\|_{p}\right)
≤αn,μ​‖f‖p+C​K​ω​(f;(βn,μ)1/2)Lp.\displaystyle\leq\alpha_{n,\mu}\left\|f\right\|_{p}+CK\omega\left(f;\left(\beta_{n,\mu}\right)^{1/2}\right)_{L_{p}}.
Remark 2

The sequences βn,μ\beta_{n,\mu} and αn,μ\alpha_{n,\mu} in Theorem 5.2 satisfy βn,μ=𝒪​(1n)\beta_{n,\mu}={\cal{O}}(\tfrac{1}{n}) and αn,μ=𝒪​(1n)\alpha_{n,\mu}={\cal{O}}(\tfrac{1}{n}) as n→∞n\rightarrow\infty.

Acknowledgements.
We are grateful to the referee for very helpful comments and suggestions.

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