Bernstein-Type Operators
Kawa Sardar Mohammad Ali
Submitted to the
Institute of Graduate Studies and Research
in partial fulfillment of the requirements for the degree of
Master of Science
in
Mathematics
Eastern Mediterranean University
June 2016
Approval of the Institute of Graduate Studies and Research
Prof. Dr. Cem Tanova Acting Director
I certify that this thesis satisfies the requirements as a thesis for the degree of Master of Science in Mathematics.
Prof. Dr. Nazim Mahmudov Chair, Department of Mathematics
We certify that we have read this thesis and that in our opinion it is fully adequate in
scope and quality as a thesis for the degree of Master of Science in Mathematics.
Prof. Dr. Sonuç Zorlu Oğurlu Supervisor
Examining Committee
1. Prof. Dr. Nazim Mahmudov
2. Prof. Dr. Sonuç Zorlu Oğurlu
iii
ABSTRACT
In this work we are interested in the approximation of some type of operators called
Bernstein-type. For this purpose, the operator Ln
f x,
f C
0, called the
Bernstein-type approximation operator is considered. The aim is to use someprobabilistic properties to improve and sharp to operator defined above. Also, the rates
of convergence as well as the continuity of the operator are studied. Various methods
of approaching the problem are evaluated in this study.
Keywords: Bernstein type operator, probabilistic approach, binomial distribution, rates of convergence.
iv
ÖZ
Bu çalışmada Bernstein - tipi operatörlerin yaklaşımlarıyla ilgilenilimiştir.Bunun için
,
0,
n
L f x f C olarak belirtilen Bernstein tipi yaklaşım operatörü ele
alınmıştır. Yukarıda verilen operatör için bazı olasılıksal metodlar kullanılarak yaklaşım özellikleri çalışılmıştır. Ayrıca, yakınsama hızı yanı sıra operatörün
sürekliliği incelenmiş olup farklı yöntemlerle yaklaşım problemi de bu çalışmada değerlendirilmiştir.
Anahtar kelimeler: Bernstein tipi operatörler, olasılıksal yaklaşım, binom dağılımı, yaklaşım hızı.
v
ACKNOWLEDGMENT
I give all thanks to God Almighty for making this thesis a success. May his name be
praised. I also want to appreciate my loving parents who supported me in all areas to
make this thesis a success. I wish to acknowledge, with gratitude the contribution of
my most humble and understanding supervisor Prof. Dr. Sonuc Zorlu Oğurlu for her
constant help, encouragement and suggestions. I cannot appreciate you enough but
God bless and keep your family save. I will also like to appreciate my friends who by
vi
TABLE OF CONTENTS
ABSTRACT………... iii ÖZ... iv ACKNOWLEDGMENT………. v 1 INTRODUCTION………...………. 1 2 BERNSTEIN OPERATOR……….. 3 2.1 INTRODUCTION………. 32.2 Evaluation of Convergence Rates………. 4
2.3 Limitation Property of 𝐿𝐾………... 15
3 APPROXIMATION PROPERTIES OF BERNSTEIN POLYNOMIALS VIA PROBABILISTIC TOOLS………. 25
3.1 Introduction……..……….…….. 25
3.2 Investigation of no Gibbs Phenomenon of Bernstein Polynomials... 27
3.3 The Convergence Speed of
Bk
t Towards
t ……….. 314 LIMITING PROPERTIES OF SOME BERNSTEIN TYPE OPERATORS……..39
4.1 Introduction……… 39
4.2 Limiting Properties………. 41
4.3 Convergence Rates………. 43
5 CONCLUSION………... 48
1
Chapter 1
INTRODUCTION
This work is divided in three main chapters as well as the introductory part and the
conclusion. An overview of what is developed in each of the chapter is given here.
The idea in chapter 2 is the following. Considering C
0,
to be the space of continuous functions on the half open interval
0, , , and given a function
0,
C
, Butzer, Hahn and Bleimann
1 introduced an approximation operator called the Bernstein-type operator for approximation and defined it as follows:
0 1 , , 1 1 k k n k k n n n L t C t k n t
k .
1.1They later proved the convergence of Lk
,t
t (ask ) for each t
0,
. The convergence of the function mentioned above is investigated by the estimation ofthe quantity Lk
,t t . This investigation is actually possible provided that the continuity of the function CB
0, The notation
. CB
0, is set for the space of
all the functions which converge uniformly and are bounded on the half open interval
0, Some probabilistic arguments are used in what will follows to sharp, and add
. accuracy to the results mentioned above. The first two module of the continuity are2
In the third chapter, the idea developed is the studies of two related approximation
problems of the Bernstein polynomials type Bk
t if given continuous function on the closed interval
0,1 : The following two results are investigated. The first is that there is no any Gibbs phenomenon at any jump type discontinuity points of and second result is the convergence of the first derivative
Bk
t of the initial polynomial Bk
t towards the first derivative
t . The well-known and classical probabilistic arguments are used to prove the above results. The second result forinstance is obtained based on the computation of the expectation of a function of
random variables.
The fourth chapter which is the third main chapter of this work is centered on the
following idea. From the poineenring work of Bernstein, various research results have
proven that probabilistic arguments are suitable for any approximation problem which
is based on positive linear operators.
The problem is usually defined as follows, given I a real valued interval and given
1 2
, R , R ,..., t t t
R I-valued random variables with their density depending on a parameter
.
tI Let us consider two linear operators Y and Y positively defined associated to k the random variable Rt and t
k
R respectively. By the mean of
,
, tY t E R Yk
,t E
Rkt , CB
I , t I,with E being the mathematical expectation of a random variable. CB
I is the space of all real valued continuous functions which are bounded on the interval I.3
Chapter 2
BERNSTEIN OPERATOR
2.1 Introduction
Consider the space C
0, of real valued continuous functions defined on the half
open interval
0, Let us consider also the function
. C
0, The
. approximation operator of the Bernstein type initially defined by Hahn, Butzer, andBleimann, is given by
0 1 , , 1 1 k k n k k n n n L t C t k n t
k .
2.1They proved that Lk
,t
t (ask ) for any t
0, they also
, established that the convergence rate can be investigated by estimating
,k
L t t of a continuity function CB
0, Some probabilistic arguments
. are used in what will follows to sharp, and add accuracy to the results mentioned above.The first two module of the continuity are used to establish the rates of convergence.
We prove the relation Lk
,t t 3
, t
1t
2 k
, where
,
stands for the first modulus continuity of . What follows is an improvement of the above theorem (inequality)
2
2
2 1 1 , 2 , , k t t t t L t t C k k 4
where 2
,
is the second modulus of continuity of CB[0, ), and 0,
supt ( ) .t
An approximation of the limit of L is the so called Szasz k operator.
2.2 Evaluation of Convergence Rates
Let W W W1, 2, 3,... be independent random variables with some probability distribution
such that P(Wi 1) p, P(Wi 0) where q, p t / (1 and t) q 1/ (1t), [0, ).
t In order to avoid the case where t 0,assume that t 0. The summation
1 ...
k k
S W W follows a binomial distribution b(n, , )k p whose parameters are k
and p, and P( ) ( ) , n 0,1, 2,..., . k n k n k n S n p q k
2.2Set Tk Sk / (k Sk 1), k1, 2,..., it follows from
2.1 that Lk( , ) t E(Tk), with the character E being the expectation operator. The convergence ofk
T p q in probability as t k implies that , Lk
,t ( ) as kt by the law of large numbers for C
0, To get an accurate result, we first calculate
.k
ET and 2 k
ET andsecondly, we estimate the quantity 2
( ) ( ) .
k k
e t E T t It is an easy task to prove that
k k
ET t tp t , as k .
2.3It follows from T and k
2.2 that2 2 2 0 ( ) ( 1) k k k n n ET p S n k n
5
2 1 ! 0 1 1 ! ! k n k n n n k p q k n k n k n n
2 1 ! 1 1 ! 1 ! k n k n n n k p q k n k n k n n n
1
! 1 1 ! 1 ! k n k n n n k p q k n n k n k n
1
! 1 1 ! 1 ! k n k n n n k p q k n n k n
Letting nm1
1 1 1 0 1 ! 1 1 1 1 ! 1 1 ! m k m k m m k p k m m k m q
1 1 1 0 ( 1) ! ( ) ! ( )! m k m k m m k p k m m k m q
1 1 1 1 1 1 0 0 ! 1 ! ! ! ! ! m m k k m k m k m m m k p k p k m m k m q k m m k m q
1 1 1 1 1 1 1 0 ! ! 0 ! ! ! ! m m k k m k m k m m m k p k p k m m k m q k m m k m q
1 1 1 1 1 1 1 0 ! ! 1 ! ! ! ! m m k k m k m k m m m k p k p k m m m k m q k m m k m q
1 1 1 1 1 1 1 0 ! ! 1 ! ! ! ! m m k k m k m k m m k p k p k m m k m q k m m k m q
1 1 1 0 1 0 1 1 ! ! 1 ! ! 0 0! 0 ! k m k m k m k k p q p q k m m k m k k
1 1 1 1 ! ! ! k m k m m k p q k m m k m
1 1 1 1 1 ! 1 ! ! k k m k m m pq k p q k k m m k m
6
1 1 1 1 ! . ! ! k m k m m k p q k m m k m
1 1 1 1 1 1 1 ! 1 1 1 1 ! ! k k m k m m t k t t t p q k k m m k m
1 1 1 1 ! ! ! k m k m m k p q k m m k m
1
1 1 1 1 1 ! 1 1 1 1 ! ! k m k m k m t k t p q t t k k m m k m
1 1 1 1 ! ! ! k m k m m k p q k m m k m
1 1 1 1 1 ! 1 ! ! k m k m k m t t p k k q k m m k m
1 1 1 1 ! . ! ! m k m k m p k q k m m k m
Letting nm1 and exploiting the binomial distribution condition p 1 q / (1 ) t it follows that t
1 1 2 2 1 1 0 ! 1 ! 1 1 ! 1 1 n k k k n k n t k k n p ET k n n k n k n q k t
1 1 2 1 1 0 ! 1 ! 1 ! 1 n k n k n k k n p k n n k n k n q
2 2 2 0 ! 1 ! ! 1 n k k n k n t k k n p k n n k n q k t
2 2 2 0 ! 1 1 ! ! n k n k n k k n p k n n k n n q
7
2 2 2 2 2 2 0 0 1 1 1 1 k k n k n k k n n k n n k n n t t k n p k n p k k n q k n n q
2
2 2 0 (1 ) 1 ( 1)( 1) k k n k n n k n t t p k n k n k q k n n k n
2
2 2 0 1 (1 ) 1 1 k k n k n n k n k n n k n t t p k q k n n
2
2 2 0 1 1 (1 ) 1 1 k k n k n n k n k n n t t p k q k n n
2
2 2 0 2 (1 ) . 1 1 k k n k n n k n n k n t t p k q n k n
Since
2 1 1 ( 2)( ) ( 1)( 1) 1 1 n k n n k n n k n k n n
1 1 1 1 1 1 1 1 1 n k n n k n k n n k n n k n n k n
1 1 1 1 1 1 1 n k n k n n k n n k n
1 1 1 1 1 1 1 1 1 n k n n k n n k n n k n n k n
1 1 1 1 1 1 1 k n k n n k n
1 1 1 1 1 1 1 k n k n n k n
1 1 1 1 1 1 1 1 k n n n k n 8
1 1 1 1 1 1 1 n n k n n
2 1 1 1 1 1 1 1 n n n k n
1 1 1 1 1 1 1 1 1 n n n k n n k n
1 1 1 1 1 1 1 1 1 1 1 1 k n n k n n n k n n k n
2 1 1 , 1 1 1 1 k n n k n k n n k n then we have,
2 2 2 2 2 2 2 0 0 ( 1) ! (1 ) ( )! ! 1 ( 1) k k k n k n n k n k k n n n t k k ET p q p q k t k n n n k n
2 2 2 2 2 2 0 0 ( 1)! (1 ) ( )!(n 1)! 1 k k k n k n n k n n k n n t k p q p q k t k n k n
2 2 1 2 1 2 2 2 1 0 (1 ) k k n k n k k k k k k k k n k k k n t p q p q p q k t
2 2 2 0 ( 1)! ( )! 1 !( 1) k n k n n k p q k n n k n
2
1 2 0 ! (1 ) 1 ! 1 ! n k k k n k n k n t p k p k t q k k k q
2 2 2 2 2 0 ! ( 1)! ! ! ( )! 1 !( 1) n k n n k n k p k p k k k q k n n k n q
2
2 1 1 2 2 0 1 ! ! (1 ) 1! 1 ! 0! ! n n k k k n k n k n k k t p k p p q k t q k k q
2 2 2 0 ( 1)! ( )! 1 !( 1) n k n k n k p k n n k n q
9
2 2 2 2 1 1 0 (1 ) k k n k n k k n k n t p q p q kp q k t
2 2 2 0 ( 1)! ( )! 1 !( 1) k n k n n k p q k n n k n
2 2 2 2 0 1 1 (1 ) 1 1 1 1 k k k n k n n k n t t t t p q k t t t t
1 1 2 2 2 0 1 ( 1)! 1 1 ( )! 1 !( 1) k k n k n n t k k p q t t k n n k n
2 2 2 2 2 1 2 1 (1 ) 1 1 1 k k k t t t t t t k t t t t
2
2 2 0 ( 1)! 1 ( )! 1 !( 1) 1 1 k k n k n k n tt k k t p q k n n k n t t
2 2 , (1 ) 1 1 k k k t t t t t kt R k t t t
2.4 where
2 2 2 0 ( 1)! . ( )! 1 !( 1) k n k n n k R p q k n n k n
Let us now analyze the term R Letting . n 1 m we obtain
1 1 1 1 1 1 1 m 1 1 1 ! . 1 ! ! m k m m k m k k k m m k p q p q R k m m k m k m
Since
1 2 1 1 2 2 2 k m k m k m k m k m k m k m
1
2
1 2 2 k m 2 1 , 1 -1 k m k m m k k m k m k m 10
1 2
2 1 3 2 , 1 k m k m k k
1 3 2 , k m k m
k 2 ,
we obtain
1 1
1 1 1 1
1
1 1 1 1 1 1 2 1 1 0 3 3 3 2 2 2 m k m m k m m k m k k k k k k m m m m m m p q p q p p q R k m k m q k m
1 1 1 1 2 0 3 1 2 , 1 1 k k m k m m m t t p q k m t
with
1 0 ; 1, 1 k m b m k p
, considering the binomial distribution b k p
,
2
1 3 1 2 1 t R t E k m t .
1
1 3 1 1 1 3 1 1 ,R t t E k m t t E where W follows binomial distribution b k
1,p
, and k 1 W with b k
1,q
, q Hence a result 1 p. of Chao and Strawderman [2, p. 430] gives
2 3 1 1 2 k t t p R k q
2 3 1 1 1 2 1 1 k t t t t k t
2 2 3 1 1 1 2 1 k k t t t t k t
2 2 2 1 1 3 1 2 1 k k k t t t t t k t
2 2 2 2 2 2 3 1 1 3 1 1 1 2 1 2 k k k k k k t t t t t t t t t k t k 11
2 3 2 3 3 1 3 3 1 3 1 2 1 2 1 2 k k k k k k k t t t t t t t k t k t k
2 3 2 3 1 3 3 1 2 1 2 2 k k t t t t t k t k k
2 3 1 . 2 t t R k
2.5Letting ek
t E T
k t
2 it follows from
2.3 , 2.4 , and
2.5 that
2 2 2 2 k k k k e t E T t E T tT t
2 2 2 k k k e t ET tET t
1
2 2 1 1 2
2 k k k k k t t t e t t t kt R t t tp t t t k t
1
2 2 1 1 2 2 2 2 1 2 k k k k t t t t e t t t kt R t t t t t t k t
1
2 1 1 k k k k t t t e t t kt R t t k t
2 2 3 1 1 2 1 1 k k k k k t k t t t e t kt t t k k t t
2 2 1 2 3 1 . 1 2 1 k k k t t t k t t k t t k k k t Since
t
1t
k
1 t
k, we have
2 2 1 3 1 k t t t t t e t k k k 12
2 3 1 1 1 t t t t t k k
2 2 3 1 1 t t t t t k k
2 2 3 1 1 2 t t t t t k k
2
2
2 1 3 1 4 1 t t t t t t k k k
2 4 1 . k t t e t k
2.6Consider the space CB
0, of continuous function which is bounded on the half open
interval
0, It is clear that the space,
. CB
0,
defined previously. To produce our first result, let us consider CB
0,
and set
,
sup
t w :t w , ,t w
0,
, 0. The convergence rate is obtained in terms of the first modulus of continuity
,
, as defined in the following theorem.Theorem 1. Consider Lk
,t be defined by
2.1 and CB
0, Then
.
2 , 3 , 1 , k 1. k L t t t t k
2.7Proof. Let and 0 Tk t ,
a stands for the greatest integer a.Obviously,
Tk t
,
1
, and13
The inequality E E2 ek
t 2 add to
2.6 lead to
2 , , 1 , k k L t t e t
2
2 4 1 , , 1 k k t t L t t
,
,
1 2 1
k t t L t t k and
2.7 is obtained by setting
1 t
t k.
2 1 , , 1 1 k t t L t t t t k k Lk
f x,
t
,
1 2
Lk
,t t 3
,
2 , 3 , 1 . k L t t t t kThe second modulus of continuity can also be exploited to obtain Theorem 1. (cf. [1]).
Let supt 0,
t , where CB
0, with
,
2 2 2 B 0, . t y t y t C Let us define the second modulus of continuity by
2 2 : , sup , y y 0.14 Eg ETk t , with k k ET t tp t as k . k Eg t tp t
1
k Eg t t t t t
1
k Eg t t t
1
k Eg t t t
1
. 1 k t t Eg t t t k
2.8An improved version of a result established by Bleimann, Butzer, and Hahn is given
below by dropping the condition k N t
24 1
in Theorem t
2 of
1 . Let us consider the following trivial result. Consider hCB
0,
h and hCB
0,
.With g Tk t we note that
0
g k h T h t
h t y dy u h t
y
du h t
y dy
dv 1 v y
0
0
g g k h T h t
h t y dy gh t g
yh t y dy
2
0 2 g k y h T h t gh t g h t y
2 1 2 k h T h t gh t g g h t g .15
Taking expectation and using
2.6 and
2.8 it is easy to see that
2 1 , 2 k L h t h t Eg h Eg h
1
1
2 , 2 k k t t L h t h t h E t t h k where ek
t E t
k t
2
1
1 , 2 k k t t L h t h t h e t h k where
2 4 1 k t t e t k
2 1 14 1 , . 2 k t t t t L h t h t h h k k It follows that
2
2 1 , . k t t L h t h t h h k
2.9Using
2.9 the following stronger version of the theorem is obtained. Theorem 2. Consider CB
0,
, t
0,
. Then for k 1, 2, ...,
2
2
2 1 1 , 2 , , k t t t t L t t C k k where C is a constant, with the saturation condition given by
10
16
The saturation properties depend on and , it is not an improvement of Theorem 2.
2.3 Limitation Property of
LkConsider the function CB
0, and define the Szasz operator by
0 , , ! n jt j n jt n S t e j n
t 0,
2.10
with j being a positive and fixed integer. Sj
,t is a suitable limit function of Lkis an interesting consequence. The limiting property established is proved via the
following lemma.
Lemma. Let n
jk t,
njk
t k n 1t k
jk , n0, 1, ..., jk, and
exp
!, n n jt jt jt n n 0, 1, 2, ... Then
i n
jt exp
n n
1
jk n 1
n
jk t,
2 exp , n jt jt n t
ii 0
,
0 jk n n n jk t jt
as k ,
iii 0max n jk n
jk t,
n
jt as 0 k .Proof. Since ey 1 y
0 y 1 ,
it follows that
,
1 n jk jk n n t t jk t k k
,
1 n n jk jk n n t k t t jk t k k t k
,
n n jk jk n n t k t k t jk t k t k k 17
,
n n jk jk n n t k k jk t k t k t k t
,
n jk n jk n n t k t t jk t k t k t
,
n jk n jk n n t k t t jk t k t k t k t
,
1 1 n n jk jk n n t t t jk t k t k t k t
,
1 n n jk jk n n t k t t t jk t k t k t k t
, 1 n jk n jk n n n n k t t t jk t k k t k t
,
! 1 ! ! jk n n n jk t t jk t jk n n k k t Since
jk ! jk n
! jk n, we have
,
1
! ! n n jk jk n n n jk t t k t t jt jk t n k k t k t n
,
1
, ! n jk n jt t jk t k t n where
, ! n jt n jt jt e n n 0, 1, 2, ...
! n jt n jt jt e n
,
1 . jk jt k n t jk t jt e k t 18 n
jk t,
n
jt exp
jt exp
t jk
k t
n
jt exp
jt jkt
k t
n
jt exp
jt k
t
jkt
k t
2 exp n jt jkt jt jkt k t
2 , exp . n jk t n jt jt k t Since
1
exp
1
0
1 ,
it follows that
,
1 n jk jk n n t t jk t n k
! , ! ! jk n n n jk t k t jk t jk n n k k
,
! ! ! jk n n n n n jk j t k jk t jk n n j k k t
! , ! ! jk n n n n n jk j t k t t jk t n jk n j k k t
! , ! ! n jk n n jt jk k t t jk t n jk n jk k t k t
1 ... 1 ! , 1 ! ! n jk n n jt jk jk jk n jk n t jk t n jk n jk k t
1 ... 1 , 1 ! n jk n n jt jk jk jk n t jk t n jk k t
1
1
, 1 ! n jk n jt jk jk jk n t jk t n jk jk jk k t 19
1 1 , 1 1 ! n n jk n i jt i t jk t n jk k t
1 , 1 1 . ! n n jk n jt n t jk t n jk k t Since n
jt exp
jt jt n n!, n being a positive integer, Thus,
exp ! n n jt jt jt n .
1 , exp 1 1 n jk n n n t jk t jt jt jk k t .Since
1
exp
1
,for 0 1,
1
1 , exp exp 1 n n n n n jk t jt jt jk jk
exp t jk 1 t k t k t
1
1 , exp exp n n n n jk n jk t jt jt jk jk
exp t jk k k t k t
,
exp
exp
1
1 n n n n jk jk t jt jt jk jk n
exp jkt k t k t k 20
,
exp
exp
1
exp
1 n n n n jk t jt jt jt jk n n
jk t,
n
jt exp
n n
1
jk n 1 .
To prove
ii let un n
jk t,
n
jt ,
n
n
jt
exp
jt2
k t
1 ,
and
1 exp
1
1
.n n jt n n jk n
Since n un nfrom
i , and un n n, we have0 0 0 . jk jk jk n n n n n n u
2.11
It is obvious that
2 exp 1 n n jt jt k t
2
0 0 exp 1 jk jk n n n n jt jt k t
2
0 0 exp 1 jk jk n n n n jt k t jt
2
0 0 exp 1 exp ! jk jk n n n n jt k t jk jt n
2
0 0 exp 1 exp ! jk jk n n n n jt k t jk jt n
2
0 0 exp 1 exp ! jk n n n n jt k t jk jt n
, where
0 ! exp n n jt n jt
2
0exp 1 exp exp
jk n n jt k t jk jt
21
2
0 exp 1 0 jk n n jt k t
as k .
2.12
1 exp
1
1 n n n n jt jk n
0 0 1 1 exp 1 jk jk n n n n n n jt jk n
,since exp
1 min 1,
0 ,
it follows that
0 0 1 1 1 min 1, 1 jk jk n n n n n n jt jk n
0 1 1 1 min 1, 1 jk n n n n jt jk n
0 0 1 min 1, 1 jk jk n n n n n n jt jk n
0 1 1 exp , 1 jk n n n n jt jk n
where
1
1 1 n n jk n . Using, n n
1
jk n 1
0 n jk 1 we have
1 0 2 1 0 1 jk jk n n n n n n jt jk n
1 0 2 1 1 1 1 0 1 1 jk jk n n n n n jk n n n n jt jt jk n jk n
22
1 0 2 1 1 1 1 0 exp , 1 ! 1 n jk jk n n n n jk n n jt n n jt jk jk n n jk n
where
exp
! n jt P U n jt n
1 0 2 1 1 , 1 jk jk k n n n n n jt P U jk jk n
with U being the Poisson random variable with mean .jt One can easily check the following relation
2 0 1 0 1 1 jk n n jt P U jk jk jk
, as k .
2.13
ii results from
2.11 ,
2.12 , and
2.13 , and
ii implies
iii .Theorem 3. Consider L and k Sj defined by
2.1 and
2.10
respectively for
0,
.B C
It follows that for each t
0, and for any fixed integer ,
j
, , 1 jk j kt t L S t t k , as k .
2.14
Proof. From
2.1 and
2.10 we have
