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formerly Central European Journal of Physics

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Blending type approximation by Stancu-Kantorovich operators based on Pólya-Eggenberger distribution

Arun Kajla
/ Serkan Araci
• Corresponding author
• Department of Economics, Faculty of Economics, Administrative and Social Sciences, Hasan Kalyoncu University, TR-27410 Gaziantep, Turkey
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• Other articles by this author:
• De Gruyter OnlineGoogle Scholar
Published Online: 2017-06-14 | DOI: https://doi.org/10.1515/phys-2017-0037

Abstract

In the paper the authors introduce the Kantorovich variant of Stancu operators based on Pólya-Eggenberger distribution. By making use of this new operator, we obtain some indispensable auxiliary results. We also deal with a Voronovskaja type asymptotic formula and some estimates of the rate of approximation involving modulus of smoothness, such as Ditzian-Totik modulus of smoothness. The rate of convergence for differential functions whose derivatives are bounded is also obtained.

PACS: 02.30.Mv; 02.30.Tb

1 Introduction

In the year 1923, Eggenberger and Pólya introduced originally Pólya-Eggenberger urn model to study processes such as the spread of contagious diseases. In one of its simplest form, the Pólya-Eggenberger urn model contains w white balls and b black balls. A ball is drawn at random and then replaced together with s balls of the same color. This procedure is repeated n times and noting the distribution of the random variable X representing the number of times a white ball is drawn (see [13]).

The distribution of X is given by $Pr(X=k)=nkw(w+s)⋅…⋅(w+k−1¯s)b(b+s)⋅…⋅(b+n−k−1¯s)(w+b)(w+b+s)⋅…⋅(w+b+n−1¯s),$(1) for k = 0, 1, ..., n and $\overline{k-1}s=\left(k-1\right)s.$ The distribution (1) is known as Pólya-Eggenberger distribution with parameters (n, w, b, s) and contains binomial, respectively hypergeometric distribution as particular cases, cf. [8].

Using (1), Stancu [25] constructed a new class of linear positive operators associated to a real-valued function f : [0, 1] → ℝ as follows. $Pnαf;x=∑k=0npn,k[α](x)fkn=∑k=0nnk∏ν=0k−1(x+να)∏μ=0n−k−1(1−x+μα)(1+α)(1+2α)⋯(1+(n−1)α)fkn,$(2) where ${p}_{n,k}^{\left[\alpha \right]}$ are usual Stancu polynomials and α is a nonnegative parameter which may depend only on the natural number n. In the case when α = 0 operators (2) reduce to the known Bernstein operators [6] and for $\alpha =\frac{1}{n}$ we have $Pn1n(f;x)=∑k=0npn,k1n(x)fkn=2(n!)(2n)!∑k=0nnk∏ν=0k−1(nx+ν)∏μ=0n−k−1(n−nx+μ)fkn,$(3) given in [17]. Further information about the applications of (2) and (3), one can refer two recent papers [18], [19]. Taking into account the period in which the Stancu operators (2) were introduced, we remark that there exists a huge interest to study them. Some representative examples in this sense could be the papers of Razi [24], Finta [11], [12], Wang et al. [26], Abel et al. [1], Agrawal et al. [2], [3], [4], [5], Gupta et al. [15], [7], [16] and Deo et al [8].

For ρ > 0, Özarslan and Duman [21] introduced a sequence of modified Bernstein-Kantorovich operators as follows: $Kn,ρ(f;x)=∑k=0npn,k(x)∫01fk+tρn+1dtx∈I,$(4) known as modified Bernstein-Kantorovich operators, in which ${p}_{n,k}\left(x\right)=\left(\genfrac{}{}{0em}{}{n}{k}\right){x}^{k}\left(1-x{\right)}^{n-k}.$

Motivated by above articles, for fC[0, 1], let us introduce $Kn,ρ[α](f;x)=∑k=0npn,k[α](x)∫01fk+tρn+1dt.$ We now call it as the Stancu-Kantorovich type operators arising from Pólya-Eggenberger distribution, where ρ > 0, ${p}_{n,k}^{\left[\alpha \right]}\left(x\right)=\left(\genfrac{}{}{0em}{}{n}{k}\right)\frac{1}{{1}^{\left[n,-\alpha \right]}}{x}^{\left[k,-\alpha \right]}\left(1-x{\right)}^{\left[n-k,-\alpha \right]}$ are the known Stancu’s fundamental polynomials and t[n, h] = t(th) ·...· (t − (n − 1)h).

The aim of this paper is to introduce a new Kantorovich type modification of Stancu operators based on P′olya-Eggenberger distribution. For these new operators some indispensable auxiliary results are obtained in the second section. Our further study focuses on the qualitative part of these new operators involving the uniform convergence and asymptotic behavior. In order to get the degree of approximation, some quantitative theorems will be established. We are motivated to write this paper from Özarslan and Duman’s paper [21].

2 Auxiliary results

Throughout of the paper, we make use of the notations: ℕ denotes the set of positive integers and ℕ0 = ℕ ∪ {0}. The monomials ek(x) = xk, for k ∈ ℕ0 also called test functions play a key role in uniform approximation arising from linear positive operators. From (4) we present a useful form of these operators.

Lemma 1

For ρ > 0, α > 0 and x ∈ (0, 1), we get $Kn,ρ[α](f;x)=1βxα,1−xα∫01txα−1(1−t)1−xα−1Kn,ρ(f;t)dt,$ where Kn,ρf are defined by (4).

Proof

Using the relationship between Euler’s functions $β(x,y)=Γ(x)Γ(y)Γ(x+y),$ in which Γ(r) is usual Gamma function given by $Γ(r)=∫0∞ur−1e−udu,r>0,$ and, for n ∈ ℕ, it satisfies the following relation $Γ(r+n)=r(r+1)⋅…⋅(r+n−1)Γ(r)$ thus we get $βxα+k,1−xα+n−k=Γxα+kΓ1−xα+n−kΓ1α+n=pn,k[α](x)nk−1βxα,1−xα.$ Since $pn,k[α](x)=nkβxα,1−xα−1βxα+k,1−xα+n−k$ we readily see that $Kn,ρ[α](f;x)=∑k=0nnkβxα+k,1−xα+n−kβxα,1−xα∫01fk+sρn+1ds=1βxα,1−xα∑k=0nnk∫01txα+k−1(1−t)1−xα+n−k−1dt×∫01fk+sρn+1ds=1βxα,1−xα∫01txα−1(1−t)1−xα−1Kn,ρ(f;t)dt.$

Below, we present four results involving Stancu-Kantorovich type operators (4) without proof. The images of the test functions ek(x) = xk, for k ∈ ℕ0 by operators (4) are given in the following Lemma 2.

Lemma 2

For the Stancu-Kantorovich type operators hold $Kn,ρ[α](e0;x)=1;Kn,ρ[α](e1;x)=nxn+1+1(n+1)(1+ρ);Kn,ρ[α](e2;x)=x2n(n−1)(1+α)(1+n)2+nx(3+ρ+α(2+n+nρ))(1+ρ)(1+α)(1+n)2+1(1+n)2(1+2ρ);Kn,ρ[α](e3;x)=n(n−1)(n−2)x(x+α)(x+2α)(1+α)(1+2α)(1+n)3+3n(n−1)(2+ρ))x(x+α)(1+α)(1+ρ)(1+n)3+nx(7+2ρ(6+ρ))(1+ρ)(1+2ρ)(1+n)3+1(1+3ρ)(1+n)3;Kn,ρ[α]e4;x=n(n−1)(n−2)(n−3)x4(1+α)(1+2α)(1+3α)(1+n)4+2n(n−1)(n−2)x3(5+3ρ+3α(2+n+nρ))(1+α)(1+2α)(1+3α)(1+ρ)(1+n)4n(n−1)x2(25+ρ(51+14ρ)+α(65+(99−2ρ)ρ+6n(1+2ρ)(5+3ρ))(1+α)(1+2α)(1+3α)(1+ρ)(1+2ρ)(1+n)4+α2(36(1+ρ)+n(1+2ρ)(35−ρ+11n(1+ρ))))(1+α)(1+2α)(1+3α)(1+ρ)(1+2ρ)(1+n)4+nx(1+nα)(1+α(−1+6n(1+nα)))(1+α)(1+2α)(1+3α)(1+n)4+4(1+nα)(1+2nα)(1+α)(1+2α)(1+ρ)(1+n)4+4(1+3ρ)(1+n)4+6(1+nα)(1+α)(1+2ρ)(1+n)4+1(1+4ρ)(1+n)4.$ From the expression of (4), for brevity we will write in the sequel ${\mathcal{P}}_{n,\rho ,r}^{\left[\alpha \right]}\left(x\right):={K}_{n,\rho }^{\left[\alpha \right]}\left(\left({e}_{1}-x{\right)}^{r};x\right)$, where n ≥ 1, r ≥ 0 and x ∈ [0, 1].

Lemma 3

For the Stancu-Kantorovich type operators hold $Pn,ρ,1[α](x)=−xn+1+1(n+1)(1+ρ);Pn,ρ,2[α](x)=(1−n)(1+α+nα)x2(1+α)(1+n)2+x(n+1)2n(1+nα)(1+α)−2(1+ρ)+1(n+1)2(1+2ρ).$

Lemma 4

For any n ∈ ℕ, we have $Pn,ρ,2[α](x)=Kn,ρ[α]((e1−x)2;x)≤Dρ[α] x(1−x)(1+n),$ where ${\mathcal{D}}_{\rho }^{\left[\alpha \right]}$ is a positive fixed based on ρ and α.

Since α is a non-negative parameter which may depend only on the natural number n, we state the following Lemma.

Lemma 5

If α → 0 as n → ∞ and $\underset{n\to \mathrm{\infty }}{lim}n\alpha =c\in \mathbb{R},$ then $limn→∞n⋅Pn,ρ,1[α](x)=−x+11+ρ,limn→∞n⋅Pn,ρ,2[α](x)=(1+c)x(1−x),limn→∞n2⋅Pn,ρ,4[α](x)=3(1+c)2x2(1−x)2.$

3 Main results

Our studies focuses on the qualitative part of Stancu-Kantorovich type operators, involving the uniform convergence and asymptotic behavior.

Theorem 1

Let fC[0, 1] and α ∈ ℕ0 depending on n ∈ ℕ, with α → 0, as n → ∞, then $\underset{n\to \mathrm{\infty }}{lim}{K}_{n,\rho }^{\left[\alpha \right]}\left(f;x\right)=f\left(x\right)$ uniformly on [0, 1].

The next result provides a Voronovskaja type result for the Stancu-Kantorovich type operators.

Theorem 2

Let f : [0, 1] → ℝ, α → 0 as n → ∞ and $\underset{n\to \mathrm{\infty }}{lim}n\alpha =c\in \mathbb{R}.$ If fC2[0, 1], then $limn→∞nKn,ρ[α](f;x)−f(x)=−x+11+ρf′(x)+(1+c)x(1−x)2f′′(x).$

Proof

In order to prove this theorem, we first use Taylor’s expansion formula for a function f, as follows. $f(t)=f(x)+f′(x)(t−x)+12f′′(x)(t−x)2+ϖ(t,x)(t−x)2,$(5) where ϖ(t, x) := ϖ(tx) is a bounded function and $\underset{t\to x}{lim}\varpi \left(t,x\right)=0.$ By the linearity of Stancu-Kantorovich type operators, and then applying the operators ${K}_{n,\rho }^{\left[\alpha \right]}$ to the both side of above equation (5), we derive $Kn,ρ[α](f;x)−f(x)=Kn,ρ[α]((e1−x);x)f′(x)+12Kn,ρ[α](e1−x)2;xf′′(x)+Kn,ρ[α]ϖ(t,x)⋅(e1−x)2;x.$ Further, from Lemma 3, we obtain $limn→∞nKn,ρ[α](f;x)−f(x)=−x+11+ρf′(x)+(1+c)x(1−x)2f′′(x)+limn→∞nKn,ρ[α]ϖ(t,x)⋅(e1−x)2;x.$(6) Thus, we can estimate the last term on the right-hand side of the above equality, applying the Cauchy-Schwarz inequality, that is: $Kn,ρ[α]ϖ(t,x)⋅(e1−x)2;x≤Kn,ρ[α]ϖ2(t,x);xKn,ρ[α](e1−x)4;x.$(7) Since ϖ2(x, x) = 0 and ϖ2(•, x) ∈ C[0, 1], by Theorem 1, we get $limn→∞Kn,ρ[α]ϖ2(t,x);x=ϖ2(x,x)=0.$(8) Therefore, taking Lemma 5 into account and from (7), and (8) yields $limn→∞nKn,ρ[α]ϖ(t,x)⋅(e1−x)2;x=0$ and using (6), we arrive at the desired result (4). □

The main tools to measure the degree of approximation of linear positive operators towards the identity operators are moduli of smoothness. For fC[0, 1] and δ ≥ 0 we know the definition of the moduli of smoothness of first, and second order, given by $ω1(f,δ):=sup{|f(x+h)−f(x)|:x,x+h∈[0,1],0≤h≤δ}$ and $ω2(f,δ):=sup{|f(x+h)−2f(x)+f(x−h)|:x,x±h∈[0,1],0≤h≤δ}$ respectively.

Definition 1

Let fCB[0, 1] be the space of all real-valued functions continuous and bounded on [0, 1] endowed with the norm ||f|| = supx∈[0, 1]|f(x)| and let us consider Peetre’s K-functional $K2(f,δ)=inf{∥f−g∥+δ∥g′′∥:g∈C2[0,1]}, or δ>0.$(9) There exists an absolute fixed M > 0, such that $K2(f,δ)≤M⋅ω2f,δ,$(10) conformable ([9], p. 177, Theorem 2.4).

Proposition 1

Let f be a real-valued function continuous and bounded on [0, 1], with ||f|| = supx∈[0, 1]|f(x)|, then $Kn,ρ[α](f;x)≤∥f∥.$

Proof

It is proved by making use of the definition of Stancu-Kantorovich type operators and Lemma 2, as follows. $Kn,ρ[α](f;x)=∑k=0npn,k[α](x)∫01fk+tρn+1dt≤∥f∥⋅Kn,ρ[α](e0;x)=∥f∥.$

Theorem 3

For fC[0, 1] and x ∈ [0, 1]. Then, there exists a constant M > 0 such that $∣Kn,ρ[α](f;x)−f(x)∣≤Mω2f,(n+1)−1/2δn(x)+ωf,−x(n+1)+1(n+1)(1+ρ),$ where ${\delta }_{n,\rho }^{2}\left(x\right)=\frac{{\mathcal{D}}_{\rho }^{\left[\alpha \right]}\text{\hspace{0.17em}}x\left(1-x\right)}{\left(1+n\right)}+{\left(\frac{-x}{\left(n+1\right)}+\frac{1}{\left(n+1\right)\left(1+\rho \right)}\right)}^{2}.$

Proof

We first consider the auxiliary operator in this form: $Tn,ρ[α](f;x)=Kn,ρ[α](f;x)+f(x)−f(nx(n+1)+1(n+1)(1+ρ)).$ Then, by Corollary 3, it becomes $Tn,ρ[α](1;x)=Kn,ρ[α](1;x)=1$ and $Tn,ρ[α](t;x)=Kn,ρ[α](t;x)+x−nx(n+1)+1(n+1)(1+ρ)=x.$

Let gC2[0, 1] and t ∈ [0, 1]. Applying Taylor’s expansion we derive $g(t)=g(x)+(t−x)g′(x)+∫xt(t−u)g′′(u)du.$ Using the operator ${T}_{n,\rho }^{\left[\alpha \right]}$ on both sides of the above equation, we have $Tn,ρ[α](g;x)=g(x)+Tn,ρ[α](∫xt(t−u)g′′(u)du)=g(x)+Kn,ρ[α](∫xt(t−u)g′′(u)du,x)−∫xnx(n+1)+1(n+1)(1+ρ)(nx(n+1)+1(n+1)(1+ρ)−u)g′′(u)du.$ Hence $∣Tn,ρ[α](g;x)−g(x)∣≤Kn,ρ[α](|∫xt|t−u||g′′(u)|du|,x)+|∫xnx(n+1)+1(n+1)(1+ρ)|nx(n+1)+1(n+1)(1+ρ)−u||g′′(u)|du|≤{Kn,ρ[α]((t−x)2;x)+(nx(n+1)+1(n+1)(1+ρ)−x)2}||g′′||={Kn,ρ[α]((t−x)2;x)+(−x(n+1)+1(n+1)(1+ρ))2}||g′′||.$ From Lemma 4, we get $∣Tn,ρ[α](g;x)−g(x)∣≤Dρ[α] x(1−x)(1+n)+(−x(n+1)+1(n+1)(1+ρ))2$ Hence $∣Tn,ρ[α](g;x)−g(x)∣≤δn,ρ2(x)||g′′||,$ where${\delta }_{n,\rho }^{2}\left(x\right)=\frac{{\mathcal{D}}_{\rho }^{\left[\alpha \right]}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\left(1-x\right)}{\left(1+n\right)}+{\left(\frac{-x}{\left(n+1\right)}+\frac{1}{\left(n+1\right)\left(1+\rho \right)}\right)}^{2}.$ In view of Proposition 1, we have $∣Tn,ρ[α](f;x)∣≤3||f||,$ for all fC[0, 1].

Now, for fC[0, 1] and gC2[0, 1], we get $∣Kn,ρ[α](f;x)−f(x)∣≤∣Tn,ρ[α](f;x)−f(x)+f(nx(n+1)+1(n+1)(1+ρ))−f(x)∣≤|Tn,ρ[α](f−g;x)|+|Tn,ρ[α](g;x)−g(x)|+|g(x)−f(x)|+|f(nx(n+1)+1(n+1)(1+ρ))−f(x)|≤4||f−g||+Mn+1δn2(x)||g′′||+ω(f,|−x(n+1)+1(n+1)(1+ρ)|).$ Taking the infimum on the right hand side over all gC2[0, 1], we obtain $∣Kn,ρ[α](f;x)−f(x)∣≤4K2(f,1n+1δn2(x))+ω(f,|−x(n+1)+1(n+1)(1+ρ)|),$ and by the inequality (10), we get $∣Kn,ρ[α](f;x)−f(x)∣≤Mω2f,(n+1)−1/2δn(x)+ωf,−x(n+1)+1(n+1)(1+ρ),$ which completes the proof. □

4 Global approximation

Let fC[0, 1] and $\varphi \left(x\right)=\sqrt{x\left(1-x\right)},x\in \left[0,1\right].$ The second order Ditzian-Totik Modulus of smoothness and corresponding K- functional are given by, respectively, $ω2ϕ(f,δ)=sup00),$ where W2(φ) = {gC[0, 1] : g′ ∈ ACloc[0, 1], φ2g″ ∈ C[0, 1]} and g′ ∈ ACloc[0, 1] means that g is differentiable and g′ is absolutely continuous on every closed interval [a, b] ⊂ (0, 1). It is known ([10], Theorem 1.3.1) that there exists a positive constant C > 0, such that $K~2,ϕ(x)(f,δ)≤Cω2ϕ(f,δ).$(11) Also, the Ditzian -Totik moduli of first order is given by $ωψ→(f,δ)=sup0 where ψ is an admissible step -weight function on [0, 1].

Theorem 4

Let fC[0, 1]. Then, for x ∈ [0.1], $||Kn,ρ[α]f−f||≤Cω2ϕ(f,(n+1)−1/2)+ωψρ→f,(n+1)−1,$ where C > 0 is an absolute constant, $\varphi \left(x\right)=\sqrt{x\left(1-x\right)}$ and ψρ(x) = (ρ + 1)x + 1.

Proof

We introduce the auxiliary operators as follows: $Tn,ρ[α](f;x)=Kn,ρ[α](f;x)+f(x)−f(nx(n+1)+1(n+1)(1+ρ)).$ Let gW2(φ) then by using Taylor’s expansion of g, on proceeding as in the proof of Theorem 3, we get $∣ Tn,ρ[α](g;x)−g(x)∣≤Kn,ρ[α](|∫xt|t−u||g′′(u)|du|,x)+∫xnx(n+1)+1(n+1)(1+ρ)|nx(n+1)+1(n+1)(1+ρ)−u||g′′(u)|du.$(12) Now let, $ζn2(x):=x(1−x)+1(n+1).$ Because the function ${\zeta }_{n}^{2}$ is concave on x ∈ [0, 1], for u = λx + (1 - λ)t, λ ∈ [0, 1], we get $∣t−u∣ζn2(u)=λ∣t−x∣ζn2(λx+(1−λ)t)≤λ∣t−x∣ζn2(x)λ+ζn2(t)(1−λ)≤∣t−x∣ζn2(x).$ Thus, the inequality (12) leads us to $∣Tn,ρ[α](g;x)−g(x)∣≤Kn,ρ[α](|∫xt|t−u|ζn2(u)du|,x)||ζn2g′′||+(∫xnx(n+1)+1(n+1)(1+ρ)|nx(n+1)+1(n+1)(1+ρ)−u|ζn2(u)du)||ζn2g′′||.≤1ζn2(x)||ζn2g′′||Kn,ρ[α]((t−x)2;x)+−x(n+1)+1(n+1)(1+ρ)2.$(13) Since $∣ Tn,ρ[α](g;x)−g(x)∣≤Cn+1||ζn2g′′||≤Cn+1(||ϕ2g′′||+1n+1||g′′||).$ Using (13), we have for fC[0, 1], $∣Kn,ρ[α](f;x)−f(x)∣≤∣ Tn,ρ[α](f−g,x)∣+∣ Tn,ρ[α](g;x)−g(x)∣+∣g(x)−f(x)∣+|f(nx(n+1)+1(n+1)(1+ρ))−f(x)|≤4||f−g||+Cn+1||ϕ2g′′||+C(n+1)2||g′′||+|f(nx(n+1)+1(n+1)(1+ρ))−f(x)|$ Taking the infimum on the right hand side over all gW2(φ), we obtain $∣Kn,ρ[α](f;x)−f(x)∣≤CK~2,ϕ(f,1(n+1))+|f(nx(n+1)+1(n+1)(1+ρ))−f(x)|.$(14) On the other hand, $|f(nx(n+1)+1(n+1)(1+ρ))−f(x)|=|f(x+ψρ(x)nx(n+1)+1(n+1)(1+ρ)−xψρ(x))−f(x)|≤supt∈[0,1]|f(t+ψρ(x)1(1+ρ)−x(n+1)ψρ(x))−f(t)|≤ωψρ→f,1(1+ρ)−x(n+1)ψρ(x)≤ωψρ→f,1(n+1).$(15) Hence, combining (11), (14) and (15), the desired result is immediate. □

Let us give the Lipschitz-type space with two parameters defined in [20]: For a1 ≥ 0, a2 > 0 and η ∈ (0, 1], $LipM(a1,a2)(η):=f∈C[0,1]:|f(t)−f(x)|≤M|t−x|η(t+a1x2+a2x)η2;t∈[0,1],x∈(0,1],$ where M is a positive constant.

Theorem 5

If $f\in Li{p}_{M}^{\left({a}_{1},{a}_{2}\right)}\left(\eta \right)$ and x ∈ (0, 1], then we have $Kn,ρ[α](f;x)−f(x)≤MPn,ρ,2[α](x)a1x2+a2xη/2,$ where ${\mathcal{P}}_{n,\rho ,2}^{\left[\alpha \right]}\left(x\right)$ is given in Lemma 3.

Proof

Let we prove the theorem for the case 0 < η ≤ 1, applying Holder’s inequality with $p=\frac{2}{\eta },q=\frac{2}{2-\eta }$ $Kn,ρ[α](f;x)−f(x)≤∑k=0npn,k[α](x)∫01fk+tρn+1−f(x)dt≤∑k=0npn,k[α](x)∫01fk+tρn+1−f(x)2ηdtη2≤∑k=0npn,k[α](x)∫01fk+tρn+1−f(x)2ηdtη2×∑k=0npn,k[α](x)2−η2=∑k=0npn,k[α](x)∫01fk+tρn+1−f(x)2ηdtη2≤M∑k=0npn,k[α](x)∫01k+tρn+1−x2k+tρn+1+a1x2+a2xdtη2≤Ma1x2+a2xη2∑k=0npn,k[α](x)∫01k+tρn+1−x2dtη2=Ma1x2+a2xη2Kn,ρ[α]((t−x)2;x)η2=Ma1x2+a2xη2(Pn,ρ,2[α](x))η2.$ Therefore, the proof is completed. □

We establish the rate of convergence for differential functions whose derivatives are of bounded variation on [0, 1]. Let DBV[0, 1] be the class of differentiable functions f defined on [0, 1], whose derivatives f′ are of bounded variation on [0, 1]. The functions fDBV[0, 1] could be represented $f(x)=∫0xg(t)dt+f(0),$ where gBV[0, 1], which means that g is a function of bounded variation on [0, 1]. Also, the operators ${K}_{n,\rho }^{\left[\alpha \right]}f$ admit the integral representation $Kn,ρ[α](f;x)=∫01Sn,ρ[α](x,t)f(t)dt,$(16) where the kernel ${\mathcal{S}}_{n,\rho }^{\left[\alpha \right]}$ is given by $Sn,ρ[α](x,t)=∑k=0npn,k[α](x)χn,kρ(t),$ where ${\chi }_{n,k}^{\rho }\left(t\right)$ is the characteristic function of the interval [k/(n + 1), (k + 1)/(n + 1)] with respect to [0, 1].

Lemma 6

Let α be a non-negative parameter which may depend on n ∈ ℕ, with α → 0 as n → ∞ and $\underset{n\to \mathrm{\infty }}{lim}n\alpha =c\in \mathbb{R}$. For a fixed x ∈ (0, 1) and sufficiently large n, it follows

1. ${\lambda }_{n,\rho }^{\left[\alpha \right]}\left(x,y\right)=\underset{0}{\overset{y}{\int }}{\mathcal{S}}_{n,\rho }^{\left[\alpha \right]}\left(x,t\right)dt\le \frac{{\mathcal{D}}_{\rho }^{\left[\alpha \right]}}{\left(1+n\right)}\frac{x\left(1-x\right)}{\left(x-y{\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0\le y

2. $1-{\lambda }_{n,\rho }^{\left[\alpha \right]}\left(x,z\right)=\underset{z}{\overset{1}{\int }}{\mathcal{S}}_{n,\rho }^{\left[\alpha \right]}\left(x,t\right)dt\le \frac{{\mathcal{D}}_{\rho }^{\left[\alpha \right]}}{\left(1+n\right)}\frac{x\left(1-x\right)}{\left(z-x{\right)}^{2}},x

Proof

1. Using Lemma 4, we get $λn,ρ[α](x,y)=∫0ySn,ρ[α](x,t)dt≤∫0y(x−tx−y)2Sn,ρ[α](x,t)dt=1(x−y)2⋅Kn,ρ[α](e1−x)2;x≤Dρ[α](1+n)x(1−x)(x−y)2.$

2. The proof is immediately, hence the details are omitted. □

Theorem 6

Let fDBV[0, 1], α → 0 as n → ∞ and $\underset{n\to \mathrm{\infty }}{lim}n\alpha =c\in \mathbb{R}.$ Then for every x ∈ (0, 1) and sufficiently large n, we have $Kn,ρ[α](f;x)−f(x)≤−x+11+ρ|f′(x+)+f′(x−)|2+Dρ[α]x(1−x)(1+n)|f′(x+)−f′(x−)|2+Dρ[α](1−x)(1+n)∑k=1[n]⋁x−(x/k)x(fx′)+xn⋁x−(x/n)x(fx′)+Dρ[α]x(1+n)∑k=1[n]⋁xx+((1−x)/k)(fx′)+(1−x)n⋁xx+((1−x)/n)(fx′),$ where $\underset{a}{\overset{b}{\bigvee }}\left({f}_{x}^{\mathrm{\prime }}\right)$ denotes the total variation of f′x on [a, b] and f′x is defined by $fx′(t)=f′(t)−f′(x−),0≤t(17)

Proof

The Stancu-Durrmeyer type operators preserve constants and using (16), for every x ∈ (0, 1) we have $Kn,ρ[α](f;x)−f(x)=∫01Sn,ρ[α](x,t)(f(t)−f(x))dt=∫01Sn,ρ[α](x,t)∫xtf′(u)dudt.$(18) For any fDBV[0, 1], from (17) we may write $f′(u)=fx′(u)+f′(x+)+f′(x−)2+f′(x+)−f′(x−)2sgn(u−x)+δx(u)f′(u)−f′(x+)+f′(x−)2,$(19) where $δx(u)=1,u=x0,u≠x.$ Obviously, $∫01(∫xt(f′(u)−f′(x+)+f′(x−)2)δx(u)du)Sn,ρ[α](x,t)dt=0$ and $∫01(∫xtf′(x+)+f′(x−)2du)Sn,ρ[α](x,t)dt=f′(x+)+f′(x−)2∫01(t−x)Sn,ρ[α](x,t)dt=f′(x+)+f′(x−)2⋅Kn,ρ[α](e1−x;x).$ Applying Chauchy-Schwarz inequality for linear positive operators, it follows $∫01Sn,ρ[α](x,t)(∫xtf′(x+)−f′(x−)2sgn(u−x)du)dt≤f′(x+)−f′(x−)2∫01|t−x|Sn,ρ[α](x,t)dt≤f′(x+)−f′(x−)2⋅Kn,ρ[α](|t−x|;x)≤f′(x+)−f′(x−)2Kn,ρ[α]((t−x)2;x)1/2.$ Using Lemma 3, respectively Lemma 4 and the relations (18), (19) yields $Kn,ρ[α](f;x)−f(x)≤|f′(x+)−f′(x−)|2Dρ[α]x(1−x)(1+n)+|∫0x∫xtfx′(u)duSn,ρ[α](x,t)dt+∫x1∫xtfx′(u)duSn,ρ[α](x,t)dt|.$(20) Let be $Gn,ρ[α](fx′,x)=∫0x∫xtfx′(u)duSn,ρ[α](x,t)dt,Fn,ρ[α](fx′,x)=∫x1∫xtfx′(u)duSn,ρ[α](x,t)dt.$ To complete the proof, it is sufficient to estimate ${\mathcal{G}}_{n,\rho }^{\left[\alpha \right]}$ and ${\mathcal{F}}_{n,\rho }^{\left[\alpha \right]}.$ Since ${\int }_{e}^{f}{d}_{t}{\lambda }_{n,\rho }^{\left[\alpha \right]}\left(x,t\right)\le 1$ for all [a, b] ⊆ [0, 1], applying the integration formula by parts and using Lemma 6 with $y=x-\left(x/\sqrt{n}\right),$ we may write $Gn,ρ[α](fx′,x)=|∫0x∫xtfx′(u)dudtλn,ρ[α](x,t)|=|∫0xλn,ρ[α](x,t)fx′(t)dt|≤(∫0y+∫yx)|fx′(t)|⋅|λn,ρ[α](x,t)|dt≤Dρ[α]x(1−x)(1+n)∫0y⋁tx(fx′)(x−t)−2dt+∫yx⋁tx(fx′)dt≤Dρ[α]x(1−x)(1+n)∫0y⋁tx(fx′)(x−t)−2dt+xn⋁x−(x/n)x(fx′).$ By the substitution of u = x/(xt), we get $Dρ[α]x(1−x)(1+n)∫0x−(x/n)(x−t)−2⋁tx(fx′)dt=Dρ[α](1−x)(1+n)∫1n⋁x−(x/u)x(fx′)du≤Dρ[α](1−x)(1+n)∑k=1[n]∫kk+1⋁x−(x/k)x(fx′)du≤Dρ[α](1−x)(1+n)∑k=1[n]⋁x−(x/k)x(fx′).$ Thus $Gn,ρ[α](fx′,x)≤Dρ[α](1−x)(1+n)∑k=1[n]⋁x−(x/k)x(fx′)+xn⋁x−(x/n)x(fx′).$(21) Using the integration formula by parts and applying Lemma 6 with $z=x+\left(\left(1-x\right)/\sqrt{n}\right),$ we get $Fn,ρ[α](fx′,x)=|∫x1∫xtfx′(u)duSn,ρ[α](x,t)dt|=|∫xz∫xtfx′(u)dudt(1−λn,ρ[α](x,t))+∫z1∫xtfx′(u)dudt(1−λn,ρ[α](x,t))|=|[∫xtfx′(u)(1−λn,ρ[α](x,t))du]xz−∫xzfx′(t)(1−λn,ρ[α](x,t))dt+∫z1∫xtfx′(u)dudt(1−λn,ρ[α](x,t))|=|∫xzfx′(u)du(1−λn,ρ[α](x,z))−∫xzfx′(t)(1−λn,ρ[α](x,t))dt+[∫xtfx′(u)du(1−λn,ρ[α](x,t))]z1−∫z1fx′(t)(1−λn,ρ[α](x,t))dt|=|∫xzfx′(t)(1−λn,ρ[α](x,t))dt+∫z1fx′(t)(1−λn,ρ[α](x,t))dt|≤Dρ[α]x(1−x)(1+n)∫z1⋁xt(fx′)(t−x)−2dt+∫xz⋁xt(fx′)dt=Dρ[α]x(1−x)(1+n)∫x+((1−x)/n)1⋁xt(fx′)(t−x)−2dt+(1−x)n⋁xx+((1−x)/n)(fx′).$ By the substitution of v = (1 - x)/(tx), we get $Fn,ρ[α](fx′,x)≤Dρ[α]x(1−x)(1+n)∫1n⋁xx+((1−x)/v)(fx′)(1−x)−1dv+(1−x)n⋁xx+((1−x)/n)(fx′)≤Dρ[α]x(1+n)∑k=1[n]∫kk+1⋁xx+((1−x)/v)(fx′)dv+(1−x)n⋁xx+((1−x)/n)(fx′)=Dρ[α]x(1+n)∑k=1[n]⋁xx+((1−x)/k)(fx′)+(1−x)n⋁xx+((1−x))/n(fx′).$(22) Collecting the estimates (20)-(22), we get the required result. □

Acknowledgement

The second author of this paper is supported by the Research fund of Hasan Kalyoncu University in 2017.

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About the article

Received: 2017-01-21

Accepted: 2017-03-13

Published Online: 2017-06-14

Citation Information: Open Physics, Volume 15, Issue 1, Pages 335–343, ISSN (Online) 2391-5471,

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© 2017 A. Kajla and S. Araci. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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