Convergence and Voronovskaja-type theorems for derivatives of generalized Baskakov operators

Cent. Eur. J. Math. • 6(2) • 2008 • 325-334DOI: 10.2478/s11533-008-0025-9

Central European Journal of Mathematics

Convergence and Voronovskaja-type theorems forderivatives of generalized Baskakov operators

Research Article

Abdul Wafi∗, Salma Khatoon†

Department Of Mathematics, Faculty of Natural Sciences, Jamia Millia Islamia, New Delhi-110025, India

Received 27 November 2007; accepted 5 March 2008

Abstract: In the present paper our aim is to establish convergence and Voronovskaja-type theorems for first derivativesof generalized Baskakov operators for functions of one and two variables in exponential and polynomial weightspaces.

MSC: 41A36, 41A63

Keywords: exponential weight spaces • derivatives of linear positive operators • convergence and Voronovskaja-typetheorems© Versita Warsaw and Springer-Verlag Berlin Heidelberg.

1. IntroductionBecker [1, 2] examined the Szász-Mirakjan and Baskakov operators for functions of one variable in polynomial andexponential weight spaces. Corresponding generalized Baskakov operators for functions of one variable were alsostudied in polyomial and exponential weight spaces by us in [10, 11].Recently, many papers were published on approximation by modified Szász-Mirakjan and Baskakov operators for func-tions of one and two variables [3, 4, 6, 7, 12] which deals with convergence, degree of approximation, and Voronovskaja-type theorems as well as convergence of partial derivatives of these operators.In [9] we introduced generalized Baskakov operators for functions of two variables in polynomial and exponential weightspaces. We also investigated the degree of approximation for functions of one and two variables and obtained theconvergence and direct theorems.In the present paper we study the convergence of first derivatives of generalized Baskakov operators for functions ofone and two variables with polynomial and exponential growth and also obtain a Voronovskaja-type theorem.∗ E-mail: [email protected]† E-mail: [email protected]

325

Convergence and Voronovskaja-type theorems for derivatives of generalized Baskakov operators

2. NotationsLet us consider as in [1], for a fixed p ∈ W = N ∪ {0} and for all x ∈ R0 = [0,+∞), the weights, wp(x), as

w0(x) = 1, wp(x) = (1 + xp)−1, p > 0and polynomial weight spaces

Cp = {f : wpf is uniformly continuous and bounded on R0},where the norm in Cp is given by

‖f‖p = supx∈R0wp(x)|f(x)|.Moreover, letCmp = {f ∈ Cp : f (k) ∈ Cp, k = 1, 2, ..., m},

for fixed p and m ∈ N.Analogously, in [2], with the weights vq(x) = e−qx , q > 0, x ∈ R0, we can define the exponential weight spacesCq = {f : vqf is uniformly continuous and bounded on R0},

with the norm‖f‖q = supx∈R0vq(x)|f(x)|.and the class Cm

q for fixed q and m ∈ N.Mihesan [5] introduced the following generalized Baskakov operators with non-negative constant a ≥ 0 independent ofn

Ban(f ; x) = ∞∑k=0 pn,k (x, a)f(k/n), x ∈ R0, k = 0, 1, 2, ..., n = 1, 2, ..., (2.1)

wherepn,k (x, a) = e −ax1+x pk (n, a)

k! xk(1 + x)n+k (2.2)such that

∞∑k=0 pn,k (x, a) = 1 (2.3)

andpk (n, a) = ∑k

i=0(ki )(n)iak−i with (n)0 = 1, (n)i = n(n+ 1)...(n+ i− 1), for i ≥ 1,defined for f ∈ C (R0), the space of functions continuous on R0.It is clear from the definition of Ban(f ; x), wp(x) (vq(x)) and Cp (Cq) that Ban(f ; x) is a linear positive operator from Cp intoCp (Cq into Cq).Some properties of the operator (2.1) are stated in lemmas below.Lemma 2.1.For a, x ≥ 0, n = 1, 2, ..., we have

Ban(1; x) = 1; Ban(t; x) = x + axn(1 + x) , (2.4)

Ban(t2; x) = x2n + x

n + x2 + a2x2n2(1 + x)2 + 2ax2

n(1 + x) + axn2(1 + x) . (2.5)

We will give here the proof of (2.4) as provided by Mihesan in [5].326

Abdul Wafi, Salma Khatoon

Proof. Taking f(t) = 1 in (2.1) and using (2.3), we getBan(1; x) = ∞∑

k=0 pn,k (x, a) = 1.Now, taking f(t) = t in (2.1), we have

Ban(t; x) = ∞∑k=0 pn,k (x, a).t = 1

n

∞∑k=0 kpn,k (x, a),

making use of (2.2), we getBan(t; x) = e −ax1+x 1

n

∞∑k=0 k

pk (n, a)k! xk(1 + x)n+k

= e −ax1+x xn(1 + x) ∞∑

k=1pk (n, a)(k − 1)! xk−1(1 + x)n+k−1 ,

replacing k by k + 1in above, we obtainBan(t; x) = e −ax1+x x

n(1 + x) ∞∑k=0

pk+1(n, a)k! xk(1 + x)n+k . (2.6)

Now, we have from [5]∞∑k=0

pk+1(n, a)k! xk(1 + x)n+k = e ax1+x [a+ n(1 + x)]. (2.7)

Substituting (2.7) in (2.6), we getBan(t; x) = e −ax1+x x

n(1 + x)e ax1+x [a+ n(1 + x)] = x + axn(1 + x)

Formula (2.5) was proved by us in [8].Lemma 2.2.For a, x ≥ 0, n = 1, 2, ..., if a ≥ 0, a/n→ 0 as n→∞, we have,

Ban(t − x; x) = axn(1 + x) ≥ 0 (2.8)

andBan((t − x)2; x) = x(1 + x)

n + 1n2 ax1 + x

(a+ 1)x + 11 + x . (2.9)Equality (2.8) was proved in [8] and (2.9) was proved in [5]. One more result was mentioned by Mihesan [5] as follows

Ban(eqt ; x) = e ax1+x (eq/n−1)(1 + x − xeq/n)n . (2.10)We prove following lemma for later use.

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Lemma 2.3.For a, x ≥ 0, n = 1, 2, ..., we have

Ban((t − x)3; x) = x(1 + x)n

1 + 2xn + 3ax

n2 + 1n3 ( ax1 + x + 3a2x2(1 + x)2 + a3x3(1 + x)3 ), (2.11)

Ban((t − x)4; x) = 3x2(1 + x)2n2 + x(1 + x)(6x2 + 6x + 1)

n3 + 1n3 (10ax2 + 8ax3 + 6a2x31 + x )

+ 1n4 ( ax1 + x + 7a2x2(1 + x)2 + 6a3x3(1 + x)3 + a4x4(1 + x)4 ). (2.12)

Proof. Substituting f(t) = (t − x)3 and f(t) = (t − x)4 in (2.1), we can easily prove (2.11) and (2.12).With the use of Lemmas 2.1-2.3, we can easily show that:Lemma 2.4.For a, x ≥ 0, n = 1, 2, ..., if a ≥ 0, a/n→ 0 as n→∞, we have,

limn→∞

Ban(t − x; x) = 0, limn→∞

nBan(t − x; x) = ax1 + x , (2.13)limn→∞

nBan((t − x)2; x) = x(1 + x) (2.14)and lim

n→∞n2Ban((t − x)4; x) = 3x2(1 + x)2. (2.15)

Moreover, for every fixed ξ ∈ R0 there exists a positive constant M1(ξ) depending only on ξ such thatn4Ban((t − x)8; x) ≤ M1(ξ), for all n ∈ N. (2.16)

Mihesan [5] established uniform convergence of the operator (2.1) for functions with exponential growth as the following:Theorem 2.1.If f ∈ C (R0), |f(x)| ≤ eqx , x ≥ 0, q > 0, then for a ≥ 0, a/n→ 0 as n→∞, we have,lim

n→∞Ban(f ; x) = f(x)

holds uniformly on [0, x1] for each x1 > 0.

Similar results have also been mentioned for Cp and Cq in [9] (see corollaries of Theorems (2.3.2) and (4.2) in [9]).Also a Voronovskaja-type theorem for the operator (2.1) with functions of exponential growth was proved in [8].Theorem 2.2.If f ∈ C (R0), |f(x)| ≤ eqx , x ≥ 0, q > 0 and supposing that fq(x) exists at a certain point of R0, then

Ban(f ; x)− f(x) = 12 {x(1 + x)n + 1

n2 ax1 + x(a+ 1)x + 11 + x } fq(x)− 2x2 ε(a)

n (x)x

where, ε(a)n (x)→ 0 as n→∞.

In the next section we shall prove analogues of Theorems 2.1 and 2.2 for the first derivatives of generalized Baskakovoperators (2.1) of functions of one variable.328


3. Convergence and Voronovskaja-type Theorems for Functions of OneVariableBefore considering the main results of this section we will prove following lemma.Lemma 3.1.For f ∈ Cp (f ∈ Cq), a, x ≥ 0 and Ban(f ; x) as defined in (2.1), we have

(Ban(f))′x = nx(1 + x) [Ban(f(t)(t − x); x)− Ban(t − x; x)Ban(f(t); x)]. (3.1)

Proof. Differentiating (2.1) with respect to x, we obtain(Ban(f))′x = ∞∑

k=0ddx [pn,k (x, a)]f(k/n), (3.2)

using (2.2), we haveddx [pn,k (x, a)] = pk (n, a)

k! ddx [e −ax1+x xk(1 + x)n+k ],

= nx(1 + x) [( kn − x)− ax

n(1 + x) ]pn,k (x, a). (3.3)substituting (3.3) in (3.2), we obtain

(Ban(f))′x (x) = ∞∑k=0

nx(1 + x) [( kn − x)− ax

n(1 + x) ]pn,k (x, a)f(k/n),taking t = k/n and using (2.1) & (2.8), we arrive at (3.1).First we shall prove convergence of the first derivative of the operator (2.1) for functions of one variable having polynomialand exponential growthTheorem 3.1.For a ≥ 0, f ∈ C 1

p or C 1q with fixed p ∈ W or q > 0,

limn→∞

(Ban(f))′x (x) = f ′(x), for every x ∈ (0,+∞). (3.4)Proof. For x ∈ (0,+∞), f ∈ C 1

p , by Taylor’s Formula we havef(t) = f(x) + f ′(x)(t − x) + φ(t, x)(t − x), t ∈ R0, (3.5)

where the function φ(t) ≡ φ(t, x) ∈ Cp and limt→x φ(t) = 0.From (3.1) and (3.5), we have for n ∈ N(Ban(f))′x (x) = n

x(1 + x) [Ban((t − x){f(x) + f ′(x)(t − x) + φ(t, x)(t − x)}; x)−Ban(t − x; x)Ban({f(x) + f ′(x)(t − x) + φ(t, x)(t − x)}; x)]

= nx(1 + x) [f ′(x)Ban((t − x)2; x) + Ban(φ(t, x)(t − x)2; x)

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− f ′(x){Ban(t − x; x)}2 − Ban(t − x; x)Ban(φ(t, x)(t − x); x)] (3.6)By the properties of φ(t, x) and Theorem 2.2, we get

limn→∞

Ban(φ(t, x)(t − x); x) = 0, limn→∞

Ban(φ2(t, x); x) = 0. (3.7)Applying the Hölder inequality, we have

|Ban(φ(t, x)(t − x)2; x)| ≤ {Ban(φ2(t, x); x)}1/2{Ban((t − x)4; x)}1/2,using (3.7) in above, we get lim

n→∞Ban(φ(t, x)(t − x)2; x) = 0. (3.8)

Using (2.13), (2.14), (3.7) and (3.8) in (3.6), we arrive at (3.4). The proof of Theorem 3.1 is identical for f ∈ C 1q .

Now, we proceed to state and prove a Voronovskaja-type theorem for the first derivatives of generalized Baskakovoperators (2.1) for functions of one variable in polynomial and exponential weight spaces.Theorem 3.2.Let f ∈ C 3

p or C 3q with fixed p ∈ W or q > 0. Then for every x ∈ (0,+∞), a ≥ 0

limn→∞

n[(Ban(f))px (x)− f p(x)] = 1 + 2x2 fq(x) + x(1 + x)2 f(x) + a(1 + x)2 f p(x) + ax1 + x fq(x). (3.9)

Proof. Since the proofs for f ∈ C 3p and f ∈ C 3

q are identical, we will consider the case f ∈ C 3p only. Let x ∈ (0,+∞)be a fixed point, then Taylor’s formula implies that

f(t) = 3∑k=0

(t − x)kk! f (k)(x) + φ(t, x)(t − x)3, t ∈ R0, (3.10)

where, φ(t) ≡ φ(t, x) ∈ Cp and limt→x φ(t) = 0.From (3.1) and (3.10), we have

(Ban(f))′x (x) = nx(1 + x) [Ban((t − x){ 3∑

k=0(t − x)kk! f (k)(x) + φ(t, x)(t − x)3}; x)

− Ban(t − x; x)Ban({ 3∑k=0

(t − x)kk! f (k)(x) + φ(t, x)(t − x)3}; x)]

= nx(1 + x) [f ′(x)Ban((t − x)2; x) + 12! f ′′(x)Ban((t − x)3; x) + 13! f ′′′(x)Ban((t − x)4; x)− f ′(x){Ban(t − x; x)}2

− 12! f ′′(x)Ban(t − x; x)Ban((t − x)2; x)− 13! f ′′′(x)Ban(t − x; x)Ban((t − x)3; x)+ Ban(φ(t, x)(t − x)4; x)− Ban(t − x; x)Ban(φ(t, x)(t − x)3; x)]. (3.11)

By the properties of φ(t, x) and Theorem 2.1, we getlimn→∞

Ban(φ(t, x)(t − x)3; x) = 0, limn→∞

Ban(φ2(t, x); x) = 0 (3.12)

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and the Hölder inequality implies that|Ban(φ(t, x)(t − x)4; x)| ≤ {Ban(φ2(t, x); x)}1/2{Ban((t − x)8; x)}1/2,

using (2.16) and (3.12) in above, we havelimn→∞

n2Ban(φ(t, x)(t − x)4; x) = 0. (3.13)Using value from Lemma 2.4, (3.12) and (3.13) in (3.11), we obtain (3.9).4. Convergence and Voronovskaja-type theorems for Functions of Two Vari-ablesAs in section 2, we define for fixed p1, p2 ∈ W = N ∪ {0}, the weights wp1,p2 (x, y) as

wp1,p2 (x, y) = wp1 (x)wp2 (y), (x, y) ∈ R20 = [0,+∞)× [0,+∞),the polynomial weight spaces Cp1,p2 of all real-valued functions f continuous and bounded on R20 . The norm in Cp1,p2 isdefined by

‖f‖p1,p2 = sup(x,y)∈R20wp1,p2 (x, y)|f(x, y)|and let Cmp1,p2 , m ∈ N, be the class of all f ∈ Cm

p1,p2 having partial derivatives of the order m in Cp1,p2 .Analogously, we define the exponential weight spaces Cq1,q2 , q1, q2 > 0, of functions of two variables with the weightsvq1,q2 (x, y) = vq1 (x)vq2 (y) = e−q1(x)e−q2(x),

and with the norm‖f‖q1,q2 = sup(x,y)∈R20 vq1,q2 (x, y)|f(x, y)|and the class Cm

q1,q2 , m ∈ N.In the space Cp1,p2 , p1, p2 ∈ W (Cq1,q2 , q1, q2 > 0), we define the operators for functions of two variables [9] withnon-negative constants a, b ≥ 0 independent of n asBa,bm,n(f ; x, y) = ∞∑

j=0∞∑k=0 pm,j (x, a)pn,k (y, b)f(j/m, k/n), m, n ∈ N, (x, y) ∈ R20 , (4.1)

where, pm,j (x, a), pn,k (y, b) are defined by (2.2).We observed that Ba,bm,n(f ; x, y) are linear positive operators from Cp1,p2 (Cq1,q2 ) into Cp1,p2 (Cq1,q2 ), provided m, n are largeenough and∞∑j=0

∞∑k=0 pm,j (x, a)pn,k (y, b) = 1. (4.2)

Also, we haveBa,bm,n(1; x, y) = 1, for all m, n ∈ N, (x, y) ∈ R20 . (4.3)If f(x, y) = f1(x)f2(y) and f1 ∈ Cq1 , f2 ∈ Cq2 , then for all m,n ∈ N, (x, y) ∈ R20

Ba,bm,n(f ; x, y) = Bam(f1(t); x)Bbn(f2(z);y). (4.4)From [9], it is known that for f ∈ Cp1,p2 (Cq1,q2 ), a, b ≥ 0

limm,n→∞

Ba,bm,n(f ; x, y) = f(x, y). (4.5)From (4.1) we derive the formula for first partial derivatives of Ba,bm,n(f ; x, y) in the following lemma.

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Lemma 4.1.For f ∈ Cp1,p2 (Cq1,q2 ), a, b ≥ 0, (t, z) ∈ R20 and Ba,bm,n(f ; x, y) as defined in (4.1), we have

(Ba,bm,n(f))′x (x, y) = mx(1 + x) [Ba,bm,n(f(t, z)(t − x); x, y)− Bam(t − x; x)Ba,bm,n(f(t, z); x, y)] (4.6)

and (Ba,bm,n(f))′y(x, y) = ny(1 + y) [Ba,bm,n(f(t, z)(z − y); x, y)− Bbn(z − y;y)Ba,bm,n(f(t, z); x, y)] (4.7)

for all (x, y) ∈ R2+ = (0,+∞)× (0,+∞).Proof. Differentiating (4.1) partially with respect to ′x ′, we get

(Ba,bm,n(f))′x (x, y) = ∞∑j=0

∞∑k=0

ddx [pm,j (x, a)]pn,k (y, b)f(j/m, k/n).

Now, from (3.3), we obtainddx [pm,j (x, a)] = m

x(1 + x) [( jm − x)− axm(1 + x) ]pm,j (x, a).

This impliesBa,bm,n(f))′x (x, y) = ∞∑

j=0∞∑k=0

mx(1 + x) [( jm − x)− ax

m(1 + x) ] pm,j (x, a)pn,k (y, b)f(j/m, k/n)= mx(1 + x) ∞∑

j=0∞∑k=0 [( jm − x)pm,j (x, a)pn,k (y, b)f(j/m, k/n)− ax

m(1 + x)pm,j (x, a)pn,k (y, b)f(j/m, k/n)]= mx(1 + x) [ ∞∑

j=0∞∑k=0 (t − x)pm,j (x, a)pn,k (y, b)f(t, z)− ax

m(1 + x) ∞∑j=0

∞∑k=0 pm,j (x, a)pn,k (y, b)f(t, z)],

using (2.8) and (4.1), the above changes to (4.6).Differentiating (4.1) partially with respect to ′y′ and proceeding as above, one can easily prove (4.7).Now we shall state and prove the analogue of Theorem 3.1 asTheorem 4.1.For f ∈ C 1

p1,p2 or f ∈ C 1q1,q2 (p1, p2 ∈ W , q1, q2 > 0), a, b ≥ 0 and every (x, y) ∈ R2+

limn→∞

(Ba,bn,n(f))′x (x, y) = f px (x, y), (4.8)

limn→∞

(Ba,bn,n(f))py(x, y) = f ′y(x, y). (4.9)Proof. Since, the proof of (4.8) and (4.9) are identical, we shall only give the proof of (4.8) for f ∈ C 1

p1,p2 and a fixed(x, y) ∈ R2+. Taylor’s formula implies thatf(t, z) = f(x, y) + f ′x (x, y)(t − x) + f ′y(x, y)(z − y) + φ(t, z; x, y)√(t − x)2 + (z − y)2, (4.10)

for (t, z) ∈ R20 , where φ(t, z) ≡ φ(t, z; x, y) ∈ Cp1,p2 and limt→x,z→y φ(t, z) = 0.332


From (4.6) and (4.10), we get(Ba,bn,n(f))′x (x, y) = n

x(1 + x) [Ba,bn,n(f(t, z)(t − x); x, y)− Ban(t − x; x)Ba,bn,n(f(t, z); x, y)]= nx(1 + x) [f ′x (x, y)Ban((t − x)2; x, y) + Ba,bn,n(φ(t, z)(t − x)√(t − x)2 + (z − y)2; x, y)

− f ′x (x, y){Ban(t − x; x)}2 − Ban(t − x; x)Ba,bn,n(φ(t, z)√(t − x)2 + (z − y)2; x, y)]. (4.11)The properties of φ(t, z) and (4.5) imply that

limn→∞

Ba,bn,n(φ(t, z)√(t − x)2 + (z − y)2; x, y) = 0, (4.12)limn→∞

Ba,bn,n(φ2(t, z); x, y) = 0. (4.13)Also, by the Hölder inequality, (4.4) and (2.4), we have for n ∈ N|Ba,bn,n(φ(t, z)(t − x)√(t − x)2 + (z − y)2; x, y)|

≤ {Ba,bn,n(φ2(t, z); x, y)}1/2{Ban((t − x)4; x) + Ban((t − x)2; x)Bbn((z − y)2;y)}1/2,which by (2.14), (2.15) and (4.13) implies that

limn→∞

Ba,bn,n(φ(t, z)(t − x)√(t − x)2 + (z − y)2; x, y) = 0. (4.14)Hence, substituting the values from (2.13), (2.14), (4.12) and (4.14) into (4.11), we obtain (4.8).Similarly, we can prove convergence in exponential weight spaces.Next, we shall prove analogue of Theorem 3.2, i.e., a Voronovskaja-type theorem for functions of two variables.Theorem 4.2.Let f ∈ C 3

p1,p2 or f ∈ C 3q1,q2 ( p1, p2 ∈ W , q1, q2 > 0 ). Then for every (x, y) ∈ R2+, we have

limn→∞

n{(Ba,bn,n(f))′x (x, y)− f ′x (x, y)} (4.15)= 1 + 2x2 f ′′x2 (x, y) + y(1 + y)2 f ′′′xy2 (x, y) + x(1 + x)2 f ′′′x3 (x, y) + a(1 + x)2 f ′x (x, y) + ax1 + x f

′′x2 (x, y) + ay1 + yf

′′xy(x, y),

limn→∞

n{(Ba,bn,n(f))′y(x, y)− f ′y(x, y)} (4.16)= 1 + 2y2 f ′′y2 (x, y) + x(1 + x)2 f ′′′x2y(x, y) + y(1 + y)2 f ′′′y3 (x, y) + a(1 + y)2 f ′x (x, y) + ay1 + yf

′′y2 (x, y) + ax1 + x f

′′xy(x, y).

Proof. Applying Taylor’s formula to (4.6) and (4.7) and then making use of Lemma 2.4 and arguing similarly as inthe proof of the previous theorem, we can prove (4.15) and (4.16).AcknowledgementsWe are thankful to the referee for his valuable remarks and suggestions.

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Convergence and Voronovskaja-type theorems for derivatives of generalized Baskakov operators