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CONVERGENCE OF CERTAIN NONLINEAR COUNTERPART OF THE BERNSTEIN OPERATORS
HARUN KARSLI AND H. ERHAN ALTIN
Abstract. The present paper concerns with the nonlinear Bernstein opera-tors N Bnf of the form
(N Bnf )(x) = n X k=0 Pn;k x; f k n ; 0 x 1 ; n 2 N; acting on bounded functions on an interval [0; 1] ; where Pn;k satisfy some
suitable assumptions. As a continuation of the very recent paper of the authors [13], we establish some pointwise convergence results for these type operators on the interval [0; 1] :
1. Introduction
We consider the problem of approximating a given real-valued function f , de-…ned on [0; 1], by means of a sequence of nonlinear Bernstein operators (N Bnf ).
Operators like positive linear, convolution, moment and sampling operators play an important role in several branches of Mathematics, for instance in reconstruction of signals and images, in Fourier analysis, operator theory, probability theory and approximation theory.
In this paper, we deal with a certain nonlinear counterpart of the Bernstein operators, considered in [13].
Let f be a function de…ned on the interval [0; 1] and let N := f1; 2; :::g : The classical Bernstein operators Bnf applied to f are de…ned as
(Bnf )(x) = n X k=0 f k n pn;k(x) ; 0 x 1 ; n 2 N; (1)
Received by the editors: September 12, 2014; Accepted: January 01, 2015 . 2010 Mathematics Subject Classi…cation. 41A35, 41A25, 47G10.
Key words and phrases. Nonlinear Bernstein operators, bounded variation, (L )Lipschitz condition, pointwise convergence.
c 2 0 1 5 A n ka ra U n ive rsity
where pn;k(x) =
n k x
k(1 x)n k is the Bernstein basis. These polynomials were
introduced by Bernstein [7] in 1912 to give the …rst constructive proof of the Weier-strass approximation theorem. Some properties of the polynomials (1) can be found in Lorentz [14].
We now state a brief and technical explanation of the relation between approx-imation by linear and nonlinear operators. Approxapprox-imation with nonlinear integral operators of convolution type was introduced by J. Musielak in [15] and widely developed in [5] ( and the references contained therein). In [15], the assumption of linearity of the singular integral operators was replaced by an assumption of a Lipschitz condition for the kernel function K (t; u) with respect to the second variable. Especially, nonlinear integral operators of type
(T f ) (x) =
b
Z
a
K (t x; f (t)) dt; x 2 (a; b) ;
and its special cases were studied by Bardaro-Karsli and Vinti [2], [3] and Karsli [9], [10] in some Lebesgue spaces.
For further reading, we also refer the reader to [1], [6], [11] and the very recent paper of the authors [13] as well as the monographs [5] and [8] where other kind of convergence results of linear and nonlinear operators in the Lebesgue spaces, Musielak-Orlicz spaces, BV -spaces and BV'-spaces have been considered.
Very recently, by using the techniques due to Musielak [15], Karsli-Tiryaki and Altin [13] introduced the following type nonlinear counterpart of the well-known Bernstein operators; (N Bnf )(x) = n X k=0 Pn;k x; f k n ; 0 x 1 ; n 2 N; (2)
acting on bounded functions f on an interval [0; 1] ; where Pn;k satisfy some
suit-able assumptions. They proved some existence and approximation theorems for the nonlinear Bernstein operators. In particular, they obtain some pointwise conver-gence for the nonlinear sequence of Bernstein operators (2) to some point x of f; as n ! 1:
As a continuation of the very recent paper of the authors [13], we estimate the rate of pointwise convergence for the nonlinear sequence of Bernstein operators (2) to the point x, at the Lebesgue points of f , as n ! 1.
An outline of the paper is as follows: The next section contains basic de…nitions and notations. In Section 3, the main approximation results of this study are given. In Section 4, we give some certain results which are necessary to prove the main result. The …nal section, that is Section 5, concerns with the proof of the main results presented in Section 3.
2. Preliminaries
In this section, we recall the following structural assumptions according to [13], which will be fundamental in proving our convergence theorems.
Let X be the set of all bounded Lebesgue measurable functions f : [0; 1] ! R. Let be the class of all functions : R+0 ! R+0 such that the function is continuous and concave with (0) = 0; (u) > 0 for u > 0.
We now introduce a sequence of functions: Let fPn;kgn2Nbe a sequence functions
Pn;k: [0; 1] x R! R de…ned by
Pn;k(t; u) = pn;k(t)Hn(u) (3)
for every t 2 [0; 1]; u 2 R, where Hn: R ! R is such that Hn(0) = 0 and pn;k(t) is
the Bernstein basis.
Throughout the paper we assume that : N ! R+is an increasing and continuous
function such that lim
n!1 (n) = 1:
First of all we assume that the following conditions hold: a ) Hn: R ! R is such that
jHn(u) Hn(v)j (ju vj) ; 2 ;
holds for every u; v 2 R, for every n 2 N: That is, Hn satis…es a (L ) Lipschitz
condition. b ) We now set Kn(x; u) := 8 > < > : P k nu pn;k(x) ; 0 < u 1 0 ; u = 0 (4) and Bn(x) := x+(1 x)=nZ = x x=n =
dt(Kn(x; t)) for any …xed x 2 (0; 1)
where > 0; 1 and n(x; t) := t Z 0 duKn(x; u) : (5)
Similar approach and some particular examples can be found in [6], [11], [12], [13] and [16].
c) Denoting by rn(u) := Hn(u) u, u 2 R and n 2 N. Assume that for n su¢ ciently large sup u jrn(u)j 1 (n); holds.
The symbol [a] will denote the greatest integer not greater than a. 3. Convergence Results
We will consider the following type nonlinear Bernstein operators, (N Bnf ) (x) = n X k=0 Pn;k x; f k n de…ned for every f 2 X for which NBnf is well-de…ned, where
Pn;k(x; u) = pn;k(x)Hn(u)
for every x 2 [0; 1]; u 2 R.
We are now ready to establish the main results of this study:
De…nition 1. A point x02 R is called a Lebesgue point of the function f, if
lim h!0+ 1 h h Z 0 jf (x0+ t) f (x0)j dt = 0; (6) holds.
Theorem 1. Let 2 and f 2 L1([0; 1]) be such that jfj 2 BV ([0; 1]).
Suppose that Pn;k(x; u) satis…es condition (a), (b) and (c). Then at each point
x 2 (0; 1) for which (6) holds we have for each > 0 and for su¢ ciently large n 2 N, j(NBnf ) (x) f (x)j Bn(x) n 1 +Bn(x) n 2 4 1 _ 0 (jfxj) + [n ] X k=1 x+(1 x)=k_ 1= x x=k1= (jfxj) 3 5 + 1 (n) where Bn(x) = Bn(x) max n x ; (1 x) o , ( > 0).
Theorem 2. Let 2 and f 2 L1([0; 1]) be such that jfj 2 BV ([0; 1]).
Suppose that Pn;k(x; u) satis…es condition (a), (b) and (c). Then at each point
x 2 (0; 1) for which (6) holds we have lim
Proof . From Theorem 1 and the de…nition of function we reach the result, by the arbitrariness of > 0.
Corollary 1 . Let 2 and f 2 L1([0; 1]) be such that jfj 2 BV ([0; 1]).
Suppose that Pn;k(x; u) satis…es condition (a), (b) and (c). Then
lim
n!1j(NBnf ) (x) f (x)j = 0
holds almost everywhere in (0; 1).
Since almost all x 2 (0; 1) are Lebesgue points of the function f, then the assertion follows by Theorem 2.
4. Auxiliary Result
In this section we give certain results, which are necessary to prove our theorems. Lemma 1. ([13], Lemma 2). For all x 2 (0; 1) and for each n 2 N; let
N Bn((t x) ; x) := 1 Z 0 ju xj du(Kn(x; u)) Bn(x) n = ; ( > 0) (7)
holds, where Bn(x) is as de…ned in Section 2. Then one has
n(x; t) =: t Z 0 du(Kn(x; u)) Bn(x) (x t) n = ; 0 t < x; (8) and 1 n(x; t) = 1 Z t du(Kn(x; u)) Bn(x) (t x) n = ; x < t < 1: (9)
The following lemma is the slight modi…cation of the Lemma 1 in [4].
Lemma 2 . Let 2 . Then, if x02 R is a Lebesgue point of the function f, we
have
h
Z
0
(jf (x0+ t) f (x0)j) dt = o (jhj) as h ! 0: (10)
Proof . In order to prove our lemma we will show the following two statements:
h Z 0 (jf (x0+ t) f (x0)j) dt = o (h) as h ! 0+; 0 Z h (jf (x0+ t) f (x0)j) dt = o ( h) as h ! 0 :
Since is concave, one has for h < 0 and h > 0, respectively, 1 h 0 Z h (jf (x0+ t) f (x0)j) dt 0 @ 1 h 0 Z h jf (x0+ t) f (x0)j dt 1 A and 1 h h Z 0 (jf (x0+ t) f (x0)j) dt 0 @1 h h Z 0 jf (x0+ t) f (x0)j dt 1 A :
Hence, by continuity of and (0) = 0, we reach the desired result.
5. Proof of the Theorems Proof of Theorem 1. Suppose that
x + < 1 ; x > 0 ; (11) for any 0 < . Let jIn(x)j = j(NBnf ) (x) f (x)j = n X k=0 Pn;k x; f k n f (x) :
From (2) and using triangle inequality, we can rewrite jIn(x)j as follows:
jIn(x)j n X k=0 Pn;k x; f k n n X k=0 Pn;k(x; f (x)) + n X k=0 Pn;k(x; f (x)) f (x) = In;1(x) + In;2(x)
From (c) it is easy to see that the second term of the right-hand-side of the above inequality is less than or equal to 1
(n). Indeed; In;2(x) = n X k=0 Pn;k(x; f (x)) f (x) = n X k=0 pn;k(x) Hn(f (x)) n X k=0 pn;k(x) f (x) = jHn(f (x)) f (x)j n X k=0 pk;n(x) 1 (n)
holds for n su¢ ciently large.
As to the …rst term, by (a) and using Lebesgue-Stieltjes integral representation of Bernstein polynomial, we have the following inequality,
In;1(x) n X k=0 f k n f (x) pn;k(x) = 1 Z 0 (jf (t) f (x)j) dt(Kn(x; t)) :
According to (b), we can split the last integral in three terms as follows:
In;1(x) 0 B B @ x x=nZ 0 + x+(1 x)=nZ x x=n + 1 Z x+(1 x)=n 1 C C A (jf (t) f (x)j) dt(Kn(x; t)) = I1(n; x) + I2(n; x) + I3(n; x) :
First, we estimate I2(n; x). We have for t 2 h x x=n ; x + (1 x) =n i jI2(n; x)j = x+(1 x)=nZ x x=n (jf (t) f (x)j) dt(Kn(x; t)) x Z x x=n (jf (t) f (x)j) dt(Kn(x; t)) + x+(1 x)=nZ x (jf (t) f (x)j) dt(Kn(x; t)) = I2;1(n; x) + I2;2(n; x) : Setting F (t) := x Z t (jf (y) f (x)j) dy;
then, according to Lemma 2, for each > 0 there exists a > 0 such that
F (t) (x t) (12)
for all 0 < x t .
We now …x this and estimate I2;1(n; x) and I2;2(n; x) respectively.
Now, we recall the Lebesgue-Stieltjes integral representation and using (5), we can write I2;1(n; x) as I2;1(n; x) = x Z x x=n (jf (t) f (x)j) dtKn(x; t) = x Z x x=n (jf (t) f (x)j) @ @t n(x; t) dt = x Z x x=n @ @t n(x; t) d ( F (t)) : (13)
Applying partial Lebesgue-Stieltjes integration (13) and using (12), we obtain, I2;1(n; x) = F x x=n @ @t n x; x x=n + x Z x x=n F (t) @ 2 @t2( n(x; t)) dt x=n @ @t n x; x x=n + x Z x x=n (x t) @ 2 @t2( n(x; t)) dt:
Integration by parts again gives
I2;1(n; x) = x=n @ @t n x; x x=n + 8 > > < > > : x=n @ @t n x; x x=n + x Z x x=n @ @t( n(x; t)) dt 9 > > = > > ; = x Z x x=n @ @t( n(x; t)) dt = x Z x x=n dt(Kn(x; t)) Bn(x) x n 1 :
We can use a similar method for I2;2(n; x). Then, we …nd the following inequality,
I2;2(n; x) x+(1 x)=nZ x dt(Kn(x; t)) Bn(x) (1 x) n 1 :
Next, we estimate I1(n; x). Using partial Lebesgue-Stieltjes integration, we obtain jI1(n; x)j = x x=nZ 0 (jf (t) f (x)j) dt(Kn(x; t)) = x x=nZ 0 (jfx(t)j) @ @t( n(x; t)) dt = fx x x n n x; x x n x x=nZ 0 n(x; t) dt( (jfx(t)j)) :
Let y = x x=n . By Lemma 1, it is clear that
n(x; y) Bn(x) (x y) n 1
: (14)
Here we note that fx x x n = fx x x n (jfx(x)j) x _ x x=n (jfxj) :
Using partial integration and applying (14), we obtain jI1(n; x)j x _ x x=n (jfxj) n x; x x n + x x=nZ 0 n(x; t) dt x _ t (jfxj) ! x _ x x=n (jfxj) Bn(x) x n 1 +Bn(x) n x x=nZ 0 (x t) dt x _ t (jfxj) !
= x _ x x=n (jfxj) Bn(x) x n 1 +Bn(x) n 2 6 4 x n x _ x x=n (jfxj) + x x _ 0 (jfxj) + x x=nZ 0 x _ t (jfxj) (x t) +1dt 3 7 7 5 = Bn(x) n 2 6 6 4x x _ 0 (jfxj) + x x=nZ 0 x _ t (jfxj) (x t) +1dt 3 7 7 5 : Changing the variable t by x x=u1= in the last integral, we have
x x=nZ 0 x _ t (jfxj) (x t) +1dt = 1 x n Z 1 x _ x x=u1= (jfxj) du 1 x [n ] X k=1 x _ x x=k1= (jfxj) : Consequently, we obtain jI1(n; x)j Bn(x) n x 2 4 x _ 0 (jfxj) + [n ] X k=1 x _ x x=k1= (jfxj) 3 5 : Using a similar method, we can …nd
jI3(n; x)j Bn(x) n (1 x) 2 4 1 _ x (jfxj) + [n ] X k=1 x+(1 x)=k1= _ x (jfxj) 3 5 : Collecting the above estimates we get the required result.
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Address : Abant Izzet Baysal University Faculty of Science and Arts Department of Mathe-matics 14280 Golkoy Bolu
E-mail : karsli_h@ibu.edu.tr E-mail : erhanaltin@ibu.edu.tr
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