ON THE RATE OF CONVERGENCE OF BASKAKOV-KANTOROVICH-B´EZIER OPERATORS FOR BOUNDED VARIATION FUNCTIONS

. In the present paper we introduce Baskakov-Kantorovich-B´ezier operators and study their rate of convergence for functions of bounded variation. MSC 2000. 41A36, 41A25, 26A45.


INTRODUCTION
Let W (0, ∞) be the class of functions f which are locally integrable on (0, ∞) and are of polynomial growth as t → ∞, i.e., for some positive r, there holds f (t) = O (t r ) , as t → ∞.We consider the Kantorovich variant V n of the Baskakov operators associating to each function f ∈ W (0, ∞) the series (1) V n (f ; x) = n ∞ k=0 v n+1,k (x) where I k = [k/n, (k + 1) /n] and Note that the operators (1) are well defined, for sufficiently large n, provided f ∈ W (0, ∞).
We mention a slightly different definition for the Kantorovich variant V * n of the Baskakov operators, given by (2) (see, e.g., [3, Eq. (9.2.3), p. 115]).The former definition (1) has the advantage to satisfy the relation where F = f is a primitive of f and V n denotes the ordinary Baskakov operators given by In the present paper we introduce the Bézier variant of the operators (1).For each function f ∈ W (0, ∞) and α ≥ 1, we introduce the Bézier type Baskakov-Kantorovich operators V n,α as where and is the Baskakov-Bézier basis function.It is obvious that V n,α are positive linear operators and V n,α (1; x) = 1.In the special case α = 1, the operators V n,α reduce to the operators V n ≡ V n,1 .Some basic properties of J n,k are as follows: Rates of convergence on functions of bounded variation, for different Bézier type operators, were studied in several papers, e.g., [6], [7], [8], [1].In the present paper we estimate the rate of convergence by the Baskakov-Kantorovich-Bézier operators (4).Furthermore, we find the limit of the sequence V n,α (f ; x) for bounded locally integrable functions f having a discontinuity of the first kind in x ∈ (0, ∞).

THE MAIN RESULTS
As main result we derive the following estimate on the rate of convergence.Theorem 1. Assume that f ∈ W (0, ∞) is a function of bounded variation on every finite subinterval of (0, ∞).Furthermore, let α ≥ 1, x ∈ (0, ∞) and λ > 1 be given.Then, for each r ∈ N, there exists a constant M (f, α, r, x), such that, for sufficiently large n, the Baskakov-Kantorovich-Bézier operators V n,α satisfy the estimate Remark 1.The exponent r in the O-term of Eq. ( 5) can be chosen arbitrary large.
As an immediate consequence of Theorem 1 we obtain in the special case α = 1 the following estimate for the Baskakov-Kantorovich operators V n .
Corollary 2. Under the assumptions of Theorem 1 there holds, for sufficiently large n, , where g x is as defined in Theorem 1.

AUXILIARY RESULTS
In order to prove our main result we shall need the following lemmas.Throughout the paper let e r denote the monomials e r (t) = t r , r = 0, 1, 2, . . ., and, for each real x, put ψ x (t) = t − x.Lemma 4. [9,Lm. 1].For all x > 0 and n, k ∈ N, there holds In particular, we have Remark 2. Note that, given any λ > 1 and any x > 0, for all n sufficiently large, we have the estimate Throughout the paper let where χ n,k denotes the characteristic function of the interval [k/n, (k + 1) /n] with respect to [0, ∞).Given a function f ∈ W (0, ∞), with this definition there holds, for all sufficiently large n, Note that, in particular, For each λ > 1, and for all sufficiently large n, we have: Proof.We first prove Eq. ( 9).There holds where we applied Lemma 4. Now Eq. ( 9) is a consequence of Remark 2. The proof of Eq. ( 10) is similar.
The following lemma is the well-known Berry-Esseen bound for the central limit theorem of probability theory.It can be used to estimate upper and lower bounds for the partial sums of Baskakov basis functions.
Lemma 7. (Berry-Esseen) [4, p. 300], [5, p. 342].Let (ξ k ) ∞ k=1 be a sequence of independent and identically distributed random variables with the expectation Then there exists an absolute constant C, 1/ √ 2π ≤ C < 0.82, such that for all t and n, there holds k=1 be a sequence of independent random variables with the same geometric distribution where x > 0 is a parameter.Then there holds Lemma 9.For all x ∈ (0, ∞), there holds Proof.We follow the proof of [9,Lm. 5].Let (ξ k ) ∞ k=1 be the sequence of independent random variables as defined in Lemma 8. Then the probability distribution of the random variable η n = n k=1 ξ k is Therefore, we have .
Application of Lemma 7 in combination with Lemma 8 implies Replacing n by n + 1 completes the proof.

PROOF OF THE MAIN RESULTS
Proof of Theorem 1.Our starting-point is the identity where First, we estimate V n,α (sign x (t) ; x) as follows.Choose k such that x ∈ [k /n, (k + 1) /n).Hence, Therefore, by Lemma 9 and Lemma 4, we obtain .
In order to complete the proof of the theorem we need an estimate of V n,α (g x ; x).We use the integral representation (7) and decompose [0, ∞) into three parts as follows and thus ( 14) Next we estimate I 1 .Put y = x − x/ √ n.Using integration by parts with Eq. ( 8) we have Since y = x − x/ √ n ≤ x, Eq. ( 9) of Lemma 6 implies, for each λ > 1 and n sufficiently large, Integrating the last term by parts, we obtain Replacing the variable y in the last integral by x − x/ √ n, we get Hence (15) Finally, we estimate I 3 .We put and divide I 3 = I 31 + I 32 , where With y = x + x/ √ n the first integral can be written in the form By Eq. ( 10) of Lemma 6, we conclude, for each λ > 1 and n sufficiently large, In a similar way as above we obtain 2x y which implies the estimate (16) Lastly, we estimate I 32 .By assumption, there exists an integer r, such that f (t) = O t 2r , as t → ∞.Thus, for a certain constant M > 0 depending only on f , x and r, we have Finally, combining (11), ( 12), (18), we obtain (5).This completes the proof of Theorem 1.
Proof of Theorem 3. Since the function ψ 2 x given by ψ 2 x (t) = (t − x) 2 is of bounded variation on every finite subinterval of [0, ∞), we deduce from Theorem 1 that, for all x ∈ (0, ∞), lim n→∞ V n,α (ψ 2 x ; x) = 0.If f ∈ L ∞ (0, ∞), then g x defined as in ( 6) is also bounded and is continuous at the point x.By the Korovkin theorem, we conclude lim n→∞ V n,α (g x ; x) = g x (x) = 0. Therefore, the right-hand side of Inequality (11) tends to zero as n → ∞.This completes the proof of Theorem 3.