# Quantitative Global Estimates for Generalized Double Szasz-Mirakjan Operators

Authors: Mehmet Ali Özarslan; Hüseyin Aktuğlu
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We introduce the generalized double Szász-Mirakjan operators in this paper. We obtain several
quantitative estimates for these operators. These estimates help us to determine some function classes𝒮 (including some Lipschitz-type spaces) which provide uniform convergence on the whole domain
[0,∞)×[0,∞).


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## Body

## 1. Introduction

The well-known Szász-Mirakjan operators are defined on the space𝒜1 as follows:
(1)Sn(f;x)=e-nx∑k=0∞‍f(kn)(nx)kk!,
where 𝒜1 is the set of all real functions on [0,∞) such that the right-hand side in (1) make sense for all n>0 and  x∈[0,∞). By modifying the Szász-Mirakjan operators as
(2)Dn(f;x)=e-nun(x)∑k=0∞‍f(kn)(nun(x))kk!,
where {un(x)} is a sequence of real-valued, continuous functions defined on [0,∞) with 0≤un(x)<∞, it has been shown in [1] that if one let
(3)un*(x):=-1+4n2x2+12n,n∈ℕ,
then the operators defined by
(4)Dn*(f;x):=Sn(f;un*(x))
preserve the test function e2(x)=x2 and provide a better error estimation than the operators Sn(f;x) for all f∈CB([0,∞)) and for each x∈[0,∞). Note that CB([0,∞)) denotes the space of all bounded and continuous functions on [0,∞). On the other hand, by letting
(5)vn(x):=x-12n;n∈ℕ,
it has been shown in [2] that the operators defined by
(6)Vn*(f;x):=Sn(f;vn(x))
do not preserve the test functions e1(x)=x and e2(x)=x2 but provide the best error estimation among all the Szász-Mirakjan operators for all f∈CB([0,∞)) and for each x∈[1/2,∞). For the other linear positive operator families which preserve e2(x)=x2, we refer [3–9]. On the other hand, in [10, 11] the authors considered some operators preserving e1(x)=x.Favard was the first to introduce the double Szász-Mirakjan operators [12]:
(7)Sn(f;x,y)=e-n(x+y)∑k,l=0∞‍f(kn,ln)(nx)kk!(ny)ll!,f∈𝒜2,
where 𝒜2 is the set of all real functions on [0,∞)×[0,∞) such that the right-hand side in (7) has a meaning for all n>0 and  x,y∈[0,∞). Recently, Dirik and Demirci have introduced and investigated different variants of the general double Szász-Mirakjan operators:
(8)Dn(f;x,y):Sn(f;un(x),vn(y))=e-n(un(x)+vn(y))×∑k,l=0∞‍f(kn,ln)×(nun(x))kk!(nvn(y))ll!,cccccccccccccccccccccccccccccf∈𝒜2.
In [13], they considered the case of operators
(9)un(1)(x):=-1+4n2x2+12n,vn(1)(y):=-1+4n2y2+12n,ddddddddd2ddddddn∈ℕ,
which preserve the test function e2,0(x,y)+e0,2(x,y):=x2+y2 and provide a better error estimation than the operators Sn(f;x,y) for all f∈CB([0,∞)×[0,∞)) and for each x,y∈[0,∞). On the other hand, in [14], they considered the case
(10)un(2)(x,α):=-(nα+1)+4n2(x2+αx)+(nα+1)22n,vn(2)(y,β):=-(nβ+1)+4n2(y2+βy)+(nβ+1)22n,aaaaaaaaaaaaaaaaaaaaaaaaaaaaaan∈ℕ,α,β∈ℝ.
Note that for this case, the operators Dn(f;x,y) do not preserve any test function (i.e., e0,0(x,y)=1, e1,0(x,y)=x, e0,1(x,y)=y, and e2,0(x,y)+e0,2(x,y)=x2+y2)  but provide a better error estimation than the operators Sn(f;x,y) for all f∈CB([0,∞)×[0,∞)) and x,y∈[0,1].Finally, we should note that, following the similar arguments as used in [2], the best error estimation among all the general double Szász-Mirakjan operators can be obtained from the case:
(11)un(3)(x):=x-12n,vn(3)(y):=y-12n,n∈ℕ,
for all f∈CB([0,∞)×[0,∞)) and x,y∈[1/2,∞).For the operatorsDn(f;x,y) the following Lemma is straightforward.Lemma 1.
Letx=(x,y), t=(t,s), ei,j(x)=xiyj, i,j=0,1,2, andψx2(t)=∥t-x∥2. Then, for each x,y≥0 andn>1, one has (a)
Dn(e0,0;x,y)=1,(b)
Dn(e1,0;x,y)=un(x),Dn(e0,1;x,y)=vn(y),(c)
Dn(e2,0+e0,2;x,y)=un2(x)+vn2(y)+((un(x)+vn(y))/n),(d)
Dn(ψx2(t);x,y)=(un(x)-x)2+(vn(y)-y)2+((un(x)+vn(y))/n).


## 2. Global Results

In this section we first introduce the following Lipschitz-type space:(12)LipM*(α):={∥t-x∥α(∥t∥+x+y)α/2f∈C([0,∞)×[0,∞)):ggggg|f(t)-f(x)|≤M∥t-x∥α(∥t∥+x+y)α/2;ggg0000ggggvg00gggt,s;x,y∈(0,∞)∥t-x∥α(∥t∥+x+y)α/2f∈C([0,∞)×[0,∞))},
where t=(t,s), x=(x,y), M is any positive constant, and 0<α≤1.We should note that this space is the bivariate extension of Lipschitz-type space considered earlier by Szasz [15]. For the space LipM*(α) with 0<α≤1, we have the following approximation result.Theorem 2.
For anyf∈LipM*(α),α∈(0,1] and for each x,y∈(0,∞),n∈ℕ, one has
(13)|Dn(f;x,y)-f(x,y)|≤M(x+y)α/2[un(x)+vn(y)n(un(x)-x)2+(vn(y)-y)2vvvvvvvv000000vvvv+un(x)+vn(y)n]α/2.Proof.
Takeα=1. Then, for f∈LipM*(1) and for each x,y∈(0,∞), we get
(14)|Dn(f;x,y)-f(x,y)|≤Dn(|f(t,s)-f(x,y)|;x,y)≤MDn(∥t-x∥(∥t∥+x+y)1/2;x,y)≤M(x+y)1/2Dn(∥t-x∥;x,y).
Using the Cauchy-Schwarz inequality, we obtain
(15)|Dn(f;x,y)-f(x,y)|≤M(x+y)1/2Dn(ψx2(t);x,y)=M(x+y)1/2z00zzzz×(un(x)-x)2+(vn(y)-y)2+un(x)+vn(y)n.
Secondly let 0<α<1. Then, for f∈LipM*(α) and for each x,y∈(0,∞), we have
(16)|Dn(f;x,y)-f(x,y)|≤Dn(|f(t,s)-f(x,y)|;x,y)≤MDn(∥t-x∥α(∥t∥+x+y)α/2;x,y)≤M(x+y)α/2Dn(∥t-x∥α;x,y).
Applying the Hölder inequality with p=2/α and q=2/(2-α), we have, for any f∈LipM*(α),
(17)|Dn(f;x,y)-f(x,y)|≤M(x+y)α/2[Dn(ψx2(t);x,y)]α/2=M(x+y)α/20=×[un(x)+vn(y)n(un(x)-x)2+(vn(y)-y)2gg0000gg+un(x)+vn(y)n]α/2,
Hence, the result.The following lemma will be used in the rest of the paper.Lemma 3.
One has, for eachx,y>0,
(18)Dn((t-x)2+(s-y)2;x,y)≤1x(un(x)-x)2+un(x)n222022+1y(vn(y)-y)2+vn(y)n.Proof.
Using the fact thata+b≤a+b(a,b≥0), we get
(19)Dn((t-x)2+(s-y)2;x,y)=e-n(un(x)+vn(y))∑k,l=0∞‍(kn-x)2+(ln-y)2=×(nun(x))kk!(nvn(y))ll!≤e-nun(x)∑k=0∞‍|kn-x|(nun(x))kk!=+e-nvn(y)∑l=0∞‍|ln-y|(nvn(y))ll!=e-nun(x)∑k=0∞‍|k/n-x|k/n+x(nun(x))kk!=+e-nvn(y)∑l=0∞‍|l/n-y|l/n+y(nvn(y))ll!≤e-nun(x)x∑k=0∞‍|kn-x|(nun(x))kk!=+e-nvn(y)y∑l=0∞‍|ln-y|(nvn(y))ll!.
Finally, applying the Cauchy-Schwarz inequality, we write
(20)Dn((t-x)2+(s-y)2;x,y)≤1xe-nun(x)∑k=0∞‍(kn-x)2(nun(x))kk!≤+1ye-nvn(y)∑l=0∞‍(ln-y)2(nvn(y))ll!=1x(un(x)-x)2+un(x)n≤+1y(vn(y)-y)2+vn(y)n.
Using Lemma 1, we get the result.Recall that, for allf∈CB([0,∞)×[0,∞)), the modulus of f denoted by ω(f;δ) is defined as
(21)ω(f;δ):=sup{(t-x)2+(s-y)2|f(t,s)-f(x,y)|:00000000000000(t-x)2+(s-y)2<δ,00000000000000(t,s),(x,y)∈[0,∞)×[0,∞)(t-x)2+(s-y)2|f(t,s)-f(x,y)|}.Theorem 4.
Letf*(x,y)=f(x2,y2). Then one has, for each x,y>0,
(22)|Dn(f;x,y)-f(x,y)|≤2ω(f*;δn(x,y)),
where
(23)δn(x,y):=1x(un(x)-x)2+un(x)nδn(x,y):=+1y(vn(y)-y)2+vn(y)n.Proof.
We directly have(24)|Dn(f;x,y)-f(x,y)|≤Dn(|f(t,s)-f(x,y)|;x,y)=Dn(|f*(t,s)-f*(x,y)|;x,y)≤Dn(ω(f*;(t-x)2+(s-y)2);x,y)=e-n(un(x)+vn(y))ll×∑k,l=0∞‍ω(f*;(kn-x)2+(ln-y)2;x,y(kn-x)2+(ln-y)2)ll×(nun(x))kk!(nvn(y))ll!.
Therefore,
(25)|Dn(f;x,y)-f(x,y)|=e-n(un(x)+vn(y))∑k,l=0∞‍(nun(x))kk!(nvn(y))ll!=×ω(f*;(k/n-x)2+(l/n-y)2Dn((t-x)2+(s-y)2;x,y)00000000000×Dn(((t-x)2+(s-y)2)((s-y)(t-x)20000000000000000000+(s-y)2)1/2;x,y((t-x)2+(s-y)2));x,y(k/n-x)2+(l/n-y)2Dn((t-x)2+(s-y)2;x,y)).
Because of the fact that
(26)ω(f;λδ)≤(1+λ)ω(f;δ),
we have
(27)|Dn(f;x,y)-f(x,y)|≤ω(f*;Dn((t-x)2+(s-y)2;x,y))≤×e-n(un(x)+vn(y))≤×∑k,l=0∞‍[1+(k/n-x)2+(l/n-y)2Dn((t-x)2+(s-y)2;x,y)]≤×(nun(x))kk!(nvn(y))ll!,
and hence
(28)|Dn(f;x,y)-f(x,y)|≤2ω(f*;Dn((t-x)2+(s-y)2;x,y)).
Finally, using Lemma 3, the proof is completed.Theorem 5.
Letf*(x,y)=f(x2,y2). Let
(29)f*∈LipM(α):={f*∈CB([0,∞)×[0,∞)):00000000000000000|f*(t)-f*(x)|≤M∥t-x∥α;00000000000000000000000000t,s;x,y∈(0,∞)},
where t=(t,s),x=(x,y),M is any positive constant, and 0<α≤1. Then
(30)|Dn(f;x,y)-f(x,y)|≤Mδnα(x,y),
where δn(x,y) is the same as in Theorem 4.Proof.
We directly have(31)|Dn(f;x,y)-f(x,y)|≤Dn(|f(t,s)-f(x,y)|;x,y)=Dn(|f*(t,s)-f*(x,y)|;x,y)≤MDn(((t-x)2+(s-y)2)α/2;x,y)=Me-n(un(x)+vn(y))d×∑k,l=0∞‍((kn-x)2+(ln-y)2)α/2d×(nun(x))kk!(nvn(y))ll!.
Applying the Hölder inequality with p=1/α and q=1/(1-α),we have
(32)|Dn(f;x,y)-f(x,y)|≤M[Dn((t-x)2+(s-y)2;x,y)]α.
Using Lemma 3, we get the result.


## 3. Concluding Remarks

In this section we show that takingun(x)=x and vn(y)=y or un(x)=un(i)(x) and vn(y)=vn(i)(y),i=1,3, in Theorems 2, 4, and 5 gives global results. Also we present the results obtained by Theorems 2, 4, and 5 for un(x)=un(2)(x) and vn(y)=vn(2)(y).Corollary 6.
For anyf∈LipM*(α),α∈(0,1] and for all x,y∈(0,∞),n∈ℕ, one has
(33)|Dn(f;x,y)-f(x,y)|≤Mnα/2,
uniformly as n→∞, for the following pairs of un(x) and vn(x):(i)
un(x)=x and vn(y)=y,(ii)
un(x)=(-1+4n2x2+1)/2n and vn(y)=(-1+4n2y2+1)/2n,(iii)
un(x)=x-1/2n and vn(y)=y-1/2n.Proof.
(i) Takingun(x)=x and vn(y)=y in (13), we directly have
(34)|Dn(f;x,y)-f(x,y)|≤Mnα/2.
(ii) Taking un(x)=(-1+4n2x2+1)/2n and vn(y)=(-1+4n2y2+1)/2n in (13) gives
(35)|Dn(f;x,y)-f(x,y)|=M(x+y)α/2[1n(4n2y2+2nx2x+y-x4n2x2+1cccccccg00ccccc=-y4n2y2+1+2nx2+2ny2)]α/2≤M(x+y)α/2[1n(4n2y2+2nx2x+y-x4n2x2ccc00cccccccccc=-y4n2y2+2nx2+2ny2)]α/2=Mnα/2.
(iii) Taking un(x)=x-1/2n and vn(y)=y-1/2n in (13), we have
(36)|Dn(f;x,y)-f(x,y)|=M(x+y)α/2[(-12n)2+(-12n)2+x+y-1/nn]α/2≤M(x+y)α/2[12n2(2nx+2ny-1)]α/2=M(x+y)α/2[1n(x+y)]α/2=Mnα/2.Corollary 7.
Letf*(x,y)=f(x2,y2). Then one has
(37)|Dn(f;x,y)-f(x,y)|≤2ω(f*;2n),
uniformly as n→∞, for the following pairs of un(x) and vn(x):(i)
un(x)=x and vn(y)=y,(ii)
un(x)=(-1+4n2x2+1)/2n and vn(y)=(-1+4n2y2+1)/2n,(iii)
un(x)=x-1/2n and vn(y)=y-1/2n.Proof.
(i) Takingun(x)=x and vn(y)=y in (23), we directly have
(38)|Dn(f;x,y)-f(x,y)|≤2ω(f*;2n).
(ii) Taking un(x)=(-1+4n2x2+1)/2n and vn(y)=(-1+4n2y2+1)/2n in (23) gives
(39)|Dn(f;x,y)-f(x,y)|≤2ω(f*;δn(x,y)),
where
(40)δn(x,y):=1xy(y1n(x-x4n2x2+1+2nx2)hhhhhhhhhhhhhh+x1n(y-y4n2y2+1+2ny2))vvvv≤1xy(y1nx+x1ny)vvvv=2n.
(iii) Taking un(x)=x-1/2n and vn(y)=y-1/2n in (23), we have
(41)|Dn(f;x,y)-f(x,y)|≤2ω(f*;δn(x,y)),
where
(42)δn(x,y):=1x(12n)2+xn-12n2hhhhhhhh+1y(12n)2+yn-12n2δn(x,y):=12xy(x1n2(4ny-1)hhhhh0000hh0000hhh+y1n2(4nx-1))δn(x,y):≤1xy(x1ny+y1nx)δn(x,y):=2n.Corollary 8.
Letf*(x,y)=f(x2,y2), and let
(43)f*∈LipM(α):={f*∈CB([0,∞)×[0,∞)):zzzzzzzzzzzzzzzzz|f*(t)-f*(x)|≤M∥t-x∥α;zzzzzzzzzzzzzzzzzzzzzzzzzt,s;x,y∈(0,∞)([0,∞)×[0,∞))},
where t=(t,s),x=(x,y),M is any positive constant, and 0<α≤1. Then
(44)|Dn(f;x,y)-f(x,y)|≤M(4n)α/2,
uniformly as n→∞, for the following pairs of un(x) and vn(y):(i)
un(x)=x and vn(y)=y,(ii)
un(x)=(-1+4n2x2+1)/2n and vn(y)=(-1+4n2y2+1)/2n,(iii)
un(x)=x-1/2n and vn(y)=y-1/2n.Proof.
(i) Takingun(x)=x and vn(y)=y in (23), we directly have
(45)|Dn(f;x,y)-f(x,y)|≤M(4n)α/2.
(ii) Taking un(x)=(-1+4n2x2+1)/2n and vn(y)=(-1+4n2y2+1)/2n in (23) gives
(46)|Dn(f;x,y)-f(x,y)|≤Mδnα/2(x,y),
where
(47)δn(x,y):=δn(x,y)δn(x,y):=1xyδn(x,y):=×(y1n(x-x4n2x2+1+2nx2)δn000000(x,y):=+x1n(y-y4n2y2+1+2ny2))δn(x,y):≤1xy(y1nx+x1ny)δn(x,y):=2n.
(iii) Taking un(x)=x-1/2n and vn(y)=y-1/2n in (23), we have
(48)|Dn(f;x,y)-f(x,y)|≤Mδnα/2(x,y),
where
(49)δn(x,y):=1x(12n)2+xn-12n2δn(x,y):=+1y(12n)2+yn-12n2δn(x,y):=12xy(x1n2(4ny-1)δn(x,y):=cccccccccc+y1n2(4nx-1))δn(x,y):≤1xy(x1ny+y1nx)δn(x,y):=2n.Remark 9.
Corollaries7 and 8 conclude that f is a real continuous and bounded function on [0,∞)×[0,∞)and if f*(x,y)=f(x2,y2) is uniformly continuous on [0,∞)×[0,∞), then Dn(f) converges uniformly to f as n→∞. Note that the one variable version of Corollary 7 was given in [16].Corollary 10.
Take(50)un(x)=un(2)(x,γ)=-(nγ+1)+4n2(x2+γx)+(nγ+1)22n,vn(y)=vn(2)(y,β)=-(nβ+1)+4n2(y2+βy)+(nβ+1)22n,
where α,β∈ℝ.Then (i)
for anyf∈LipM*(α), α∈(0,1] and for each x,y∈(0,∞), n∈ℕ, one has
(51)|Dn(f;x,y)-f(x,y)|≤M[2n(x+y)]α/2[δ(x,γ)+δ(y,β)]α/2,where(52)δ(x,γ)=[n2(2x+γ)2+2nγ+1(2x+γ)h0hh×(nγ-n2(2x+γ)2+2nγ+1+2nx+1)],(ii)
letf*(x,y)=f(x2,y2). Then one has for each x,y>0(53)|Dn(f;x,y)-f(x,y)|≤2ω(f*;δ(x,γ)+δ(y,β)),where(54)δ(x,γ)=12x+hh000×((1n(2x+γ)(n2(2x+γ)2+2nγ+1nγ-n2(2x+γ)2+2nγ+1ccccccccccccccccccccccccc+2nx+1n2(2x+γ)2+2nγ+1))1/2,(iii)
letf*(x,y)=f(x2,y2), and let
(55)f*∈LipM(α):={f*∈CB([0,∞)×[0,∞)):fffffffddffffffff|f*(t)-f*(x)|≤M∥t-x∥α;jjjjjjjjjjjjjjjjjjjjddjjjjjjjjt,s;x,y∈(0,∞)([0,∞)×[0,∞))},wheret=(t,s), x=(x,y), M is any positive constant, and 0<α≤1. Then
(56)|Dn(f;x,y)-f(x,y)|≤M[δ(x,γ)+δ(y,β)]α/2,whereδ(x,γ) is the same given in Corollary 10(ii).It should be mentioned that, forα=0 and β=0, un(2)(x,0)=(-1+4n2x2+1)/2n and vn(2)(y,0)=(-1+4n2y2+1)/2n. Therefore, Corollary 10(i), Corollary 10(ii), and Corollary 10(iii) reduce to Corollary 6(ii), Corollary 7(ii), and Corollary 8(ii), respectively.



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