# Degree of Approximation by Hybrid Operators

Authors: Naokant Deo; Hee Sun Jung; Ryozi Sakai
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We consider hybrid (Szász-beta) operators, which are a general sequence of integral type operators including beta function, and we give the degree of approximation by these Szász-beta-Durrmeyer operators.


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## Body

## 1. Introduction

The Lupaş-Durrmeyer operators were introduced by Sahai and Prasad [1] who studied the asymptotic formula for simultaneous approximation, and many mathematicians have given different results for the Durrmeyer operators (see [2–6]). Now we consider here a sequence of linear positive operators, which was introduced by Gupta et al. [7] as follows. Let n and β be positive integers. For f∈C[0,∞) satisfying ∫0∞‍f(t)/(1+t)n+β+1dt<∞,
(1)Bn,β[f](x)=∑k=0∞‍sn,k(x)∫0∞‍bn+β,k(t)f(t)dt,00000000000000x∈[0,∞),n+β+1>0,
where β is a positive integer,
(2)sn,k(x)=e-nx(nx)kk!,bn,k(t)=1B(k+1,n)tk(1+t)n+k+1,B(k+1,n)=Γ(k+1)Γ(n)Γ(k+n+1).
Let 0<p≤∞. For a function f on [0,∞), we define the norm by
(3)∥f∥Lp([0,∞))={(∫[0,∞)|f(t)|pdt|f(t)|p)1/p,0<p<∞,sup[0,∞)|f(t)|,p=∞.
Recently Jung and Sakai [8] investigated the Lupaş-Durrmeyer operators and studied the circumstances of convergence. Motivated with the idea of Jung and Sakai [8], we give the degree of approximation by Szász-Beta-Durrmeyer operators in this paper.


## 2. Basic Results

Lemma 1 (cf. [7]).
Letα, m, n, and r be integers with m≥0, r≥1, and n+α>m:
(4)Rn,m,r(α;x)∶=∑k=0∞‍sn,k(x)∫0∞‍bn+α,k+r(y)(y-x)mdy.
Then one has(i)
Rn,0,r(α;x)=1 and Rn,1,r(α;x)=((-α+1)x+r+1)/(n+α-1),(ii)
form⩾1(5)(n+α-m-1)Rn,m+1,r(α;x)=xRn,m,r′(α;x)+((2m-α+1)x+m+r+1)Rn,m,r(α;x)+mx(x+2)Rn,m-1,r(x),(iii)
(6)Rn,m,r(α;x)=O(1n[(m+1)/2])gn,m,r(α;x),where gn,m,r(α;x) is a polynomial of degree ⩽m such that the coefficients of gn,m,r(α;x) are bounded independently of n.Proof.
LetRn,m,r(x):=Rn,m,r(α;x). Then (i)
(7)Rn,0,r(x)=∑k=0∞‍sn,k(x)∫0∞‍bn+α,k+r(y)dy=∑k=0∞‍sn,k(x)=1.
Using
(8)∫0∞‍xbn,k(x)dx=k+1n-1,∑k=0∞‍ksn,k(x)=nx,
we see that
(9)Rn,1,r(x)=∑k=0∞‍sn,k(x)∫0∞‍bn+α,k+r(y)(y-x)dy=∑k=0∞‍sn,k(x)k+r+1n+α-1-x=(-α+1)x+r+1n+α-1.
(ii) Using xsn,k′(x)=(k-nx)sn,k(x), we obtain
(10)x(Rn,m,r′(x)+mRn,m-1,r(x))=∑k=0∞‍xsn,k′(x)∫0∞‍bn+α,k+r(y)(y-x)mdy=∑k=0∞‍sn,k(x)∫0∞‍(k-nx)bn+α,k+r(y)(y-x)mdy.
Since we know that
(11)y(1+y)bn+α,k+r′(y)=(k+r-(n+α+1)y)bn+α,k+r(y),k-nx=(k+r-(n+α+1)y)-(r-(α+1)x)+(n+α+1)(y-x),
we have
(12)(k-nx)bn+α,k+r(y)=y(1+y)bn+α,k+r′(y)-(r-(α+1)x)bn+α,k+r(y)+(n+α+1)bn+α,k+r(y)(y-x).
Then substituting (12) into (10), we consider the following:(13)x(Rn,m,r′(x)+mRn,m-1,r(x))=∑k=0∞‍sn,k(x)∫0∞‍(y(1+y)bn+α,k+r′(y)0000000000000-(r-(α+1)x)bn+α,k+r(y)0000000000000+(n+α+1)bn+α,k+r(y)(y-x)bn+α,k+r′)00000000000000×(y-x)mdy∶=A1+A2+A3.
Then since we see
(14)∫0∞‍y(1+y)bn+α,k+r′(y)(y-x)mdy=∫0∞‍((y-x)2+(1+2x)(y-x)+x(1+x))000×bn+α,k+r′(y)(y-x)mdy=∫0∞‍bn+α,k+r′(y)(y-x)m+2dy+(1+2x)∫0∞‍bn+α,k+r′(y)(y-x)m+1dy+x(1+x)∫0∞‍bn+α,k+r′(y)(y-x)mdy=-(m+2)∫0∞‍bn+α,k+r(y)(y-x)m+1dy-(1+2x)(m+1)∫0∞‍bn+α,k+r(y)(y-x)mdy-x(1+x)m∫0∞‍bn+α,k+r(y)(y-x)m-1dy,
we have
(15)A1=∑k=0∞‍sn,k(x)∫0∞‍y(1+y)bn+α,k+r′(y)(y-x)mdy=-(m+2)Rn,m+1,r(x)-(1+2x)(m+1)Rn,m,r(x)-x(1+x)mRn,m-1,r(x).
Here the last equation follows from integration by parts. Furthermore, we easily see
(16)A2+A3=-(r-(α+1)x)Rn,m,r(x)+(n+α+1)Rn,m+1,r(x).
Therefore, we conclude
(17)x(Rn,m,r′(x)+mRn,m-1,r(x))=-(m+2)Rn,m+1,r(x)-(1+2x)(m+1)Rn,m,r(x)-x(1+x)mRn,m-1,r(x)-(r-(α+1)x)Rn,m,r(x)+(n+α+1)Rn,m+1,r(x)=(n+α-m-1)Rn,m+1,r(x)-((2m-α+1)x+m+r+1)Rn,m,r(x)-x(1+x)mRn,m-1,r(x).
Consequently, (ii) is proved.
(iii) Form=1, (6) holds. Let us assume (6) for m⩾1. We note
(18)Rn,m,r′(x)=O(1n[(m+1)/2])gn,m,r′(x),gn,m,r′(x)∈𝒫m-1.
So, we have, by the assumption of induction,
(19)(n+α-m-1)Rn,m+1,r(α;x)=O(1n[(m+1)/2])xgn,m,r′(x)+((2m-α+1)x+m+r+1)O(1n[(m+1)/2])×gn,m,r(x)+mx(x+2)O(1n[m/2])gn,m-1,r(x).
Here, if m is even, then
(20)[m+12]+1=m2+1=m+22=[m+22],[m2]+1=m2+1=m+22=[m+22],
and if m is odd, then
(21)[m+12]+1=m+12+1=[m+22],[m2]+1=m-12+1=m+12=[m+22].
Hence we have
(22)Rn,m+1,r(x)=O(1n[(m+2)/2])gn,m+1,r(x),
and here we see that gn,m+1,r(x) is a polynomial of degree ⩽m+1 such that the coefficients of gn,m+1,r(x) are bounded independently of n.Lemma 2 (cf. [7]).
Letn, β, and r be integers with r≥0. Let f∈C(r)[0,∞) satisfy for a positive integer δ(23)|f(r)(x)|≤O(1)(x+1)δ.
Then one has, for n+β-r>δ,
(24)λn,β,r(Bn,β[f])(r)(x)=∑k=0∞‍sn,k(x)∫0∞‍bn+β-r,k+r(y)f(r)(y)dy,
where
(25)λn,β,r∶=(n+β-1)!nr(n+β-r-1)!.Proof.
Using(26)bn+β-r,k+r(r)(y)=(n+β-1)!(n+β-r-1)!∑i=0r‍(ri)(-1)ibn+β,k+i(y),
we have
(27)(Bn,β[f])(r)(x)=∑k=0∞‍sn,k(r)(x)∫0∞‍bn+β,k(y)f(y)dy=∑i=0r‍∑k=i∞‍(ri)(-1)r(-1)inrsn,k-i(x)×∫0∞‍bn+β,k(y)f(y)dy=nr∑k=0∞‍sn,k(x)∫0∞‍∑i=0r‍(ri)×(-1)r(-1)ibn+β,k+i(y)f(y)dy=nr(n+β-r-1)!(n+β-1)!∑k=0∞‍sn,k(x)×∫0∞‍(-1)rbn+β-r,k+r(r)(y)f(y)dy=1λn,β,r∑k=0∞‍sn,k(x)×∫0∞‍bn+β-r,k+r(y)f(r)(y)dy.


## 3. Main Results

Theorem 3.
Let0<p≤∞, and let δ and r be nonnegative integers. Let n and β be integers with n+β-r>δ. Let f∈C(r+1)[0,∞) satisfy
(28)|f(r)(x)|≤O(1)(x+1)δ,|f(r+1)(x)|≤O(1)(x+1)δ+2.
Then one has uniformly, for f and n,
(29)|λn,β,r(Bn,β[f])(r)(x)-f(r)(x)|≤O(1n1/3)(x+1)δ+2.Proof.
Let|t-x|<ε and x<ξ<t. By the second inequality of (28),
(30)|f(r)(t)-f(r)(x)|=|t-x||f(r+1)(ξ)|≤O(1)|t-x|(x+1)δ+2.
Let ε=n-ν,0<ν<1. Then using Lemma 2 and
(31)∑k=0∞‍sn,k(x)∫0∞‍bn+β-r,k+r(y)f(r)(x)dy=f(r)(x),
we have
(32)|λn,β,r(Bn,β[f])(r)(x)-f(r)(x)|=|∑k=0∞‍sn,k(x)(∫|t-x|<ε‍bn+β-r,k+r(t)|f(r)(t)-f(r)(x)|dt+∫|t-x|≥ε‍bn+β-r,k+r(t)|f(r)(t)-f(r)(x)|dt)|=E1+E2.
From (30) and Lemma 1, we have
(33)E1=O(1)|∑k=0∞‍sn,k(x)∫|t-x|<ε‍bn+β-r,k+r(t)0000000000000000×|t-x|(x+1)δ+2dt∑k=0∞‍|≤O(1)ε|Rn,0,r(β-r;x)|(x+1)δ+2=O(1)ε(x+1)δ+2.
Next, we estimate E2. By the use of the first inequality in (28), we have
(34)E2=C|∑k=0∞‍sn,k(x)∫|t-x|≥ε‍bn+β-r,k+r(t)00000000000×(|f(r)(t)|+|f(r)(x)|)dt∑k=0∞‍|≤C|∑k=0∞‍sn,k(x)∫|t-x|≥ε‍bn+β-r,k+r(t)00000000000000×((t+1)δ+(x+1)δ)dt∑k=0∞‍|.
Now using (t+1)δ=((t-x)+x+1)δ=∑i=0δ(δi)(t-x)i(x+1)δ-iand the notation
(35)〈i〉={1,i:odd,0,i:even,
we have
(36)E2≤C∑k=0∞‍sn,k(x)∫|t-x|≥ε‍bn+β-r,k+r(t)00000000000×(∑i=0δ‍(δi)|t-x|i|t-xε|〈i〉(x+1)δ-i)dt+C∑k=0∞‍sn,k(x)∫|t-x|≥ε‍bn+β-r,k+r(t)(t-xε)2(x+1)δdt=E21+E22.
Then, with ε=n-ν,
(37)E21≤C(∑i=1δ‍(δi)|Rn,i+〈i〉,r(β-r;x)|(1ε)〈i〉(x+1)δ-i)≤C∑i=1δ‍(δi)O(nν〈i〉n[(i+〈i〉+1)/2])0000×|gn,i+〈i〉,r(β-r;x)|(x+1)δ-i≤O(1n[(i+〈i〉+1)/2]-ν〈i〉)(x+1)δ+1≤O(1n1-ν)(x+1)δ+1.
Here for i≥1, we get
(38)[i+〈i〉+12]-ν〈i〉≥1-ν,
because
(39)[i+〈i〉+12]-ν〈i〉={i+12-ν,i:odd,i2-ν,i:even,E22≤C|Rn,2,r(β-r;x)|(1ε)2(x+1)δ≤O(1n[3/2])|gn,2,r(β-r;x)|(1ε)2(x+1)δ≤O(n2νn[3/2])|gn,2,r(β-r;x)|(x+1)δ≤O(1n1-2ν)(x+1)δ+2.
Finally we get
(40)E2≤O(1n1-2ν)(x+1)δ+2.
From (32),
(41)|λn,β,r(Bn,β[f])(r)(x)-f(r)(x)|≤O(1nν)(x+1)δ+2+O(1n1-2ν)(x+1)δ+2.
If we put ν=1/3, then we get
(42)|λn,β,r(Bn,β[f])(r)(x)-f(r)(x)|≤O(1n1/3)(x+1)δ+2.In the following, we letϕ(t)=1/(1+t),t∈[0,∞).Theorem 4.
Letr and γ be nonnegative integers. Let n and β be integers with n+β-r>2γ+1. Let f∈C(r+2)[0,∞) satisfy
(43)∥f(r+1)(x)ϕ2γ+1(x)∥L∞[0,∞)<∞,∥f(r+2)(x)ϕ2γ(x)∥L∞[0,∞)<∞.
Then one has uniformly, for f and n,
(44)|λn,β,r(Bn,β[f])(r)(x)-f(r)(x)|ϕ2(γ+1)(x)≤O(1n)(∥f(r+1)(x)ϕ2γ+1(x)∥L∞[0,∞)00000000000000+∥f(r+2)(x)ϕ2γ(x)∥L∞[0,∞)).Proof.
Forf∈C(r+2)[0,∞), we have
(45)f(r)(y)=f(r)(x)+f(r+1)(x)(y-x)+∫xy‍(y-u)f(r+2)(u)du,(46)|∫xy‍(y-u)f(r+2)(u)du|≤C∥f(r+2)(x)ϕ2γ(x)∥L∞[0,∞)×((1+x)2γ+(1+y)2γ)(y-x)2.
From (45), (46), and Lemma 2, we get
(47)λn,β,r(Bn,β[f])(r)(x)=f(r)(x)+f(r+1)(x)Rn,1,r(β-r;x)+∑k=0∞‍sn,k(x)∫0∞‍bn+β-r,k+r(t)0000000000×∫xt‍(t-u)f(r+2)(u)dudt,(48)|∑k=0∞‍sn,k(x)∫0∞‍bn+β-r,k+r(t)∫xt‍(t-u)f(r+2)(u)dudt|≤∥f(r+2)(x)ϕ2γ(x)∥L∞[0,∞)×(∫0∞‍(1+x)2γ|Rn,2,r(β-r;x)|00000000+∑k=0∞‍sn,k(x)∫0∞‍bn+β-r,k+r(t)(1+t)2γ(t-x)2dt).
Using (1+t)2γ≤C((t-x)2γ+(1+x)2γ), we obtain
(49)|∑k=0∞‍sn,k(x)∫0∞‍bn+β-r,k+r(t)(1+t)2γ(t-x)2dt|≤C(|Rn,2γ+2,r(β-r;x)|+(1+x)2γ|Rn,2,r(β-r;x)|).
Therefore, we have
(50)|λn,β,r(Bn,β[f])(r)(x)-f(r)(x)|ϕ2(γ+1)(x)≤|f(r+1)(x)ϕ2γ+1(x)||Rn,1,r(β-r;x)|ϕ(x)+C∥f(r+2)(x)ϕ2γ(x)∥L∞[0,∞)(1+x)2γ×|Rn,2,r(β-r;x)|ϕ2γ+2(x)+C∥f(r+2)(x)ϕ2γ(x)∥L∞[0,∞)×|Rn,2(γ+1),r(β-r;x)|ϕ2(γ+1)(x)≤O(1n)|f(r+1)(x)ϕ2γ+1(x)||gn,1,r(β-r;x)|ϕ(x)+O(1n)∥f(r+2)(x)ϕ2γ(x)∥L∞[0,∞)(1+x)2γ×|gn,2,r(β-r;x)|ϕ2(γ+1)(x)+O(1nγ+1)∥f(r+2)(x)ϕ2γ(x)∥L∞[0,∞)×|gn,2(γ+1),r(β-r;x)|ϕ2(γ+1)(x).
For x∈[0,∞), we have |gn,1,r(β-r;x)|ϕ(x)≤C, (1+x)2γ|gn,2,r(β-r;x)|ϕ2(γ+1)(x)≤C, and |gn,2(γ+1),r(β-r;x)|ϕ2(γ+1)(x)≤C. Hence
(51)|λn,β,r(Bn,β[f])(r)(x)-f(r)(x)|ϕ2(γ+1)(x)≤O(1n)(∥f(r+1)(x)ϕ2γ+1(x)∥L∞[0,∞)000000000+∥f(r+2)(x)ϕ2γ(x)∥L∞[0,∞)).Let us define the weighted modulus of smoothness by(52)ωk(f;η;t)∶=sup0⩽h⩽t∥Δhkf(·)η(·)∥L∞([0,∞)),t⩾0,k=1,2,
where
(53)Δh1f(x)=f(x+h)-f(x),(54)Δh2f(x)=f(x)-2f(x+h)+f(x+2h).Theorem 5.
Letr and γ be nonnegative integers. Let n and β be integers with n+β-r>2γ+2. Then one has, for f∈Cr([0,∞)),
(55)∥(λn,β,r(Bn,β[f])(r)(x)-f(r)(x))ϕ2γ+2(x)∥L∞([0,∞))⩽C(1nω1(f(r);ϕ2γ+1;1n)+ω2(f(r);ϕ2γ;1n)).To prove Theorem5, we need the following theorem.Theorem 6.
Letr and γ be nonnegative integers. Let n and β be integers with n+β-r>2γ. Let f∈C(r)([0,∞)) satisfy
(56)∥f(r)ϕ2γ∥L∞([0,∞))<∞.
Then one has uniformly, for n, f, and x∈[0,∞),
(57)|λn,β,r(Bn,β[f])(r)(x)ϕ2γ(x)|⩽C∥f(r)(x)ϕ2γ(x)∥L∞([0,∞)).Proof.
Using(1+y)2γ⩽C((1+x)2γ+(y-x)2γ), we have
(58)|∑k=0∞‍sn,k(x)∫0∞‍bn+β-r,k+r(y)(1+y)2γdy|⩽C|∑k=0∞‍sn,k(x)∫0∞‍bn+β-r,k+r(y)000000000000000×((1+x)2γ+(y-x)2γ)dy∑k=0∞‍|⩽C(Rn,0,r(β-r;x)ϕ-2γ(x)+Rn,2γ,r(β-r;x)).
Therefore, by Lemma 1 (6), we have
(59)|∑k=0∞‍sn,k(x)∫0∞‍bn+β-r,k+r(y)(1+y)2γdy|⩽C(ϕ-2γ(x)+O(1nγ)gn,2γ,r(β-r;x)).
Since |gn,2γ,r(β-r;x)ϕ2γ(x)| is uniformly bounded on [0,∞), we have with Lemma 2 and (59)
(60)|λn,β,r(Bn,β[f])(r)(x)ϕ2γ(x)|⩽|∑k=0∞‍sn,k(x)∫0∞‍bn+β-r,k+r(y)(1+y)2γdyϕ2γ(x)|×∥f(r)(x)ϕ2γ(x)∥L∞([0,∞))⩽C∥f(r)(x)ϕ2γ(x)∥L∞([0,∞)).
Therefore, we have the result.The Steklov function[f]h(x) for f∈C([0,∞)) is defined as follows:
(61)[f]h(x)∶=4h2∫0h/2‍∫0h/2‍[2f(x+s+t)-f(x+2(s+t))]dsdt,00000000000000000000000000000000x⩾0,h>0.
Then for the Steklov function [f]h(x) with respect to f∈C([0,∞)), we have the following properties.Lemma 7 (see [8, Lemma 2.4]).
Letf(x)∈C([0,∞)), and let η(x) be a positive and nonincreasing function on [0,∞). Then(i)
[f]h(x)∈C2([0,∞)),(ii)
(62)∥([f]h(x)-f(x))η(x)∥L∞([0,∞))⩽ω2(f;η;h2),(iii)
(63)∥[f]h′(x)η(x)∥L∞([0,∞))⩽4hω1(f;η;h2)η(x)η(x+h/2)+1hω1(f;η;h)η(x)η(x+h),(iv)
(64)∥[f]h′′(x)η(x)∥L∞([0,∞))⩽4h2[2ω2(f;η;h2)+14ω2(f;η;h)].Now, we prove Theorem5.Proof of Theorem5.
We know that, forf(x)∈Cr([0,∞)),
(65)[f]h(r)(x)=[f(r)]h(x),[f]h(r+1)(x)=[f(r)]h′(x),[f]h(r+2)(x)=[f(r)]h′′(x).
Then first, we split it as follows:
(66)∥(λn,β,r(Bn,β[f])(r)(x)-f(r)(x))ϕ2γ+2(x)∥L∞([0,∞))⩽∥λn,β,r(Bn,β[f-[f]h])(r)(x)ϕ2γ+2(x)∥L∞([0,∞))+∥(λn,β,r(Bn,β[[f]h])(r)(x)-[f]h(r)(x))0000×ϕ2γ+2(x)(λn,β,r(Bn,β[[f]h])(r)(x)-[f]h(r)(x))∥L∞([0,∞))+∥([f]h(r)(x)-f(r)(x))ϕ2γ+2(x)∥L∞([0,∞)).
Then for the first term, we have, using Theorem 6, (62), and (65),
(67)∥λn,β,r(Bn,β[f-[f]h])(r)(x)ϕ2γ+2(x)∥L∞([0,∞))⩽C∥[f(r)(x)-[f(r)]h(x)]ϕ2γ+2(x)∥L∞([0,∞))⩽Cω2(f(r);ϕ2γ+2(x);h).
Here, we suppose 0<h⩽1, and then we know that
(68)ϕ(x)ϕ(x+h)⩽2,ϕ(x)ϕ(x+h/2)⩽2.
For the second term, from Theorem 4, (65), (63), and (64) of Lemma 7,
(69)∥(λn,β,r(Bn,β[[f]h])(r)(x)-[f]h(r)(x))ϕ2γ+2(x)∥L∞([0,∞))⩽O(1n)(∥[f]h(r+1)(x)ϕ2γ+1(x)∥L∞([0,∞))0000000+∥[f]h(r+2)(x)ϕ2γ(x)∥L∞([0,∞)))⩽O(1n)(1hω1(f(r);ϕ2γ+1;h)+1h2ω2(f(r);ϕ2γ;h)).
Therefore, we have
(70)∥(λn,β,r(Bn,β[[f]h])(r)(x)-f(r)(x))ϕ2γ+2(x)∥L∞([0,∞))⩽O(1n)(1hω1(f(r);ϕ2γ+1;h)+1h2ω2(f(r);ϕ2γ;h))+ω2(f(r);ϕ2γ+2;h).
If we let h=1/n, then
(71)∥(λn,β,r(Bn,β[[f]h])(r)(x)-f(r)(x))ϕ2γ+2(x)∥L∞([0,∞))⩽C(1nω1(f(r);ϕ2γ+1;1n)+ω2(f(r);ϕ2γ;1n)),
because ω2(f(r);ϕ2γ+2;1/n)⩽ω2(f(r);ϕ2γ;1/n).



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*Source: 732069-2013-08-20.xml*