Some general Kantorovich type operators

Petru I. Braica\(^\ast \) Ovidiu T. Pop\(^\S \)

August 8, 2012.

\(^\ast \) Secondary School “Grigore Moisil", 1 Mileniului 1 St. 440037 Satu Mare, Romania, e-mail: petrubr@yahoo.com

\(^\S \) National College “Mihai Eminescu", 5 Mihai Eminescu St., 440014 Satu Mare, Romania, e-mail: ovidiutiberiu@yahoo.com

A general class of linear and positive operators of Kantorovich-type is constructed. The operators of this type which preserve exactly two test functions from the set \(\{ e_0, e_1, e_2\} \) are determined and their approximation properties and convergence theorems are studied.

MSC. 41A10, 41A35, 41A36.

Keywords. Kantorovich operators, approximation and convergence theorem, Voronovskaja-type theorem.

1 Introduction

Let \(\mathbb N\) be the set of positive integers and \(\mathbb N_0=\mathbb N\cup \{ 0\} \). In 1930, L. V. Kantorovich in [3] introduced the linear and positive operators \(K_m:L_1([0,1])\to C([0,1])\), \(m\in \mathbb N_0\), defined for any \(f\in L_1([0,1])\), \(x\in [0,1]\) and \(m\in \mathbb N_0\) by

\begin{equation} \label{1.1} (K_m f)(x)=(m+1)\sum ^m_{k=0}\tbinom {m}{k} x^k(1-x)^{m-k} \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}f(t){\rm d}t. \end{equation}
1

The operators from (1) are called Kantorovich operators. We remark that these operators preserve only the test function \(e_0\).

The aim of this paper is to construct and study a general class of linear positive operators which preserve exactly two test functions from the set \(\{ e_0, e_1, e_2\} \).

In [4], J. P. King introduced and studied a Bernstein type operator, which preserves only the test functions \(e_0\) and \(e_2\). Therefore, we say that the operators constructed in this paper are operators of King’s type.

In Section 2, we recall some results from [6], which we use for obtaining the main results of this paper.

In Section 3, respectively 4, we determine the unique operators from Section 2, which preserve exactly the test functions \(e_0\) and \(e_1\), respectively \(e_0\) and \(e_2\). In the Sections 4 and 5, we give approximation, convergence and Voronovskaja’s type theorems for the operators obtained.

2 Preliminaries

Let \(\mathbb N\) be the set of positive integers and \(\mathbb N_0=\mathbb N\cup \{ 0\} \). In this section we recall some results from [8], which we shall use in the present paper. Let \(I, J\) be real intervals with the property \(I\cap J\neq \emptyset \). For any \(m,k\in \mathbb N_0\), \(m\neq 0\), we consider the functions \(\varphi _{m,k}:J\to \mathbb R\), with the property that \(\varphi _{m,k}(x)\geq 0\), for any \(x\in J\) and the linear positive functionals \(A_{m,k}:E(I)\to \mathbb R\).

For any \(m\in \mathbb N\) we define the operator \(L_m:E(I)\to F(J)\), by

\begin{equation} \label{2.1} (L_m f)(x)=\sum ^\infty _{k=0}\varphi _{m,k}(x) A_{m,k}(f), \end{equation}
2

for any \(x\in J\), where \(E(I)\) and \(F(J)\) are linear subspaces of real valued functions defined on \(I\), resp. \(J\), for which the sequence \((L_n)_{n\geq 0}\) defined above is convergent (in the topology of \(F(J)\)).

Remark 1

The operators \((L_m)_{m\in \mathbb N}\) are linear and positive on \(E(I\cap J)\).â–¡

For \(m\in \mathbb N\) and \(i\in \mathbb N_0\), we define \((T_{m,i})\) by

\begin{equation} \label{2.2} (T_{m,i}L_m)(x)=m^i(L_m\psi ^i_x)(x)=m^i\sum ^\infty _{k=0} \varphi _{m,k}(x)A_{m,k}(\psi ^i_x) \end{equation}
3

for any \(x\in I\cap J\), \(\psi ^i_x(t)=(t-x)^i\), \(t\in I\).

In that follows \(s\in \mathbb N_0\) is even and we assume that the next two conditions:

\(\bullet \) there exist the smallest \(\alpha _s, \alpha _{s+2}\in [0,+\infty )\), so that

\begin{equation} \label{2.3} \lim _{m\to \infty }\tfrac {(T_{m,j}L_m)(x)}{m^{\alpha _j}}= B_j(x)\in \mathbb R\end{equation}
4

for any \(x\in I\cap J\) and \(j\in \{ s, s+2\} \)

\begin{equation} \label{2.4} \alpha _{s+2}<\alpha _s+2. \end{equation}
5

\(\bullet \) \(I\cap J\) is an interval.

Theorem 2

(see [8]) Let \(f\in E(I)\) be a function. If \(x\in I\cap J\) and \(f\) is \(s\) times differentiable in a neighborhood of \(x\), \(f^{(s)}\) is continuous in \(x\), then

\begin{equation} \label{2.5} \lim _{m\to \infty }m^{s-\alpha _s}\left((L_m f)(x)-\sum ^s_{i=0} \tfrac {f^{(i)}(x)}{m^i i!}\, (T_{m,j} L_m)(x)\right)=0. \end{equation}
6

Assume that \(f\) is \(s\) times differentiable on \(I\), \(f^{(s)}\) is continuous on \(I\) and there exists a compact interval \(K\subset I\cap J\), such that there exists \(m(s)\in \mathbb N\) and constant \(k_j\in \mathbb R\) depending on \(K\), so for \(m\geq m(s)\) and \(x\in K\) the following inequalities

\begin{equation} \label{2.6} \tfrac {(T_{m,j}L_m)(x)}{m^{\alpha _j}}\leq k_j, \end{equation}
7

hold for \(j\in \{ s,s+2\} \).
Then the convergence expressed by \((\ref{2.5})\) is uniform on \(K\) and

\begin{gather} \label{2.7} m^{s-\alpha _s}\left|(L_m f)(x)-\sum ^s_{i=0} \tfrac {f^{(i)}(x)}{m^i i!}\, (T_{m,i} L_m)(x)\right|\\ \leq \tfrac {1}{s!}\, (k_s+k_{s+2})\omega \left(f^{(s)}; \tfrac {1}{\sqrt{m^{2+\alpha _s-\alpha _{s+2}}}}\right),\nonumber \end{gather}

for any \(x\in K\), \(m\geq m(s)\), where \(\omega (f;\cdot )\) denotes the modulus of continuity of the function \(f\).

Corollary 3

Let \(f:I\to \mathbb R\) be a \(s\) times differentiable function on \(I\cap J\) with \(f^{(s)}\) continuous on \(I\cap J\). Then

\begin{equation} \label{2.8} \lim _{m\to \infty } (L_m f)(x)=B_0(x) f(x) \end{equation}
9

if \(s=0\) and \(\alpha _0=0\), where \(B_0\) is defined by \((\ref{2.3})\). If \(s\geq 2\), then

\begin{equation} \label{2.9} \lim _{m\to \infty }m^{s-\alpha _s} \left[\! (L_m f)(x)-\! \sum ^{s-1}_{i=0}\tfrac {1}{m^i i!} \, (T_{m,i}L_m)(x)f^{(i)}(x)\right]\! \! =\! \tfrac {1}{s!}\, B_s(x)f^{(s)}(x), \end{equation}
10

where \(B_s\) are defined by \((\ref{2.3})\).

If \(f\) is a \(s\) times differentiable function on \(I\cap J\), with \(f^{(s)}\) continuous and bounded on \(I\cap J\) and (7) takes place for an interval \(K\subset I\cap J\), then the convergence in (9) and (10) are uniform on \(K\).

3 The construction of a general linear and positive operators

Let \(J\subset R\) be an interval, \(m_0\in \mathbb N_0\), \(m_0\geq 2\) given, \(\mathbb N_1=\{ m\in \mathbb N|m\geq m_0\} \), the function \(\alpha _m, \beta _m:J\to \mathbb R\), \(\alpha _m(x)\geq 0\), \(\beta _m(x)\geq 0\) for any \(x\in J\) and \(m\in \mathbb N_1\).

Definition 4

For \(m\in \mathbb N_1\), we define the operator of the following form

\begin{equation} \label{3.1} (K^*_m f)(x)=(m+1)\sum ^m_{k=0}\tbinom {m}{k} \alpha ^k_m(x)\beta ^{m-k}_m(x)\int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}f(t){\rm d}t \end{equation}
11

for any \(f\in L_1([0,1])\) and \(x\in J\).

Lemma 5

The following identities

\begin{equation} \label{3.2} (K^*_m e_0)(x)=(\alpha _m (x)+\beta _m(x))^m, \end{equation}
12

\begin{equation} \label{3.3} (K^*_m e_1)(x)=\tfrac {(\alpha _m (x)+\beta _m(x))^{m-1}}{2(m+1)}\, ((2m+1)\alpha _m(x)+\beta _m(x)), \end{equation}
13

and

\begin{align} \label{3.4} (K^*_m e_2)(x) & =\tfrac {(\alpha _m(x)+\beta _m(x))^{m-2}}{3(m+1)^2}\, \big(3m(m-1)\alpha ^2_m(x)\\ & \quad +6m\alpha _m(x)(\alpha _m(x)+\beta _m(x))+ (\alpha _m(x)+\beta _m(x))^2\big),\nonumber \end{align}

hold, for any \(x\in J\) and any \(m\in \mathbb N_1\).

Proof â–¼
For \(m\in \mathbb N_1\) and \(k\in \{ 0,1,\dots , m\} \) we have

\begin{align*} & \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}e_0(t){\rm d}t=\tfrac {1}{m+1}\, ,\quad \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}e_1(t){\rm d}t=\tfrac {2k+1}{(m+1)^2}\\ & \text{and}\; \; \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}e_2(t){\rm d}t=\tfrac {3k^2 +3k+1}{(m+1)^3}\, . \end{align*}

Then

\begin{align*} (K^*_m e_0)(x) & =(m+1)\sum ^m_{k=0}\tbinom {m}{k}\alpha ^k_m(x)\beta ^{m-k}_m(x) \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{k+1}}e_0(t){\rm d}t\\ & =\sum ^m_{k=0}\tbinom {m}{k}\alpha _m^k(x)\beta ^{m-k}_m(x), \end{align*}

so (12) holds;

\begin{align*} (K^*_m e_1)(x)& =(m+1)\sum ^m_{k=0}\tbinom {m}{k}\alpha ^k_m(x)\beta ^{m-k}_m(x) \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}e_1(t){\rm d}t\\ & =\tfrac {1}{2(m+1)}\bigg(2m\alpha _m(x)\sum ^m_{k=1}\tbinom {m-1}{k-1}\alpha ^{k-1}_m(x)\beta ^{m-k}_m(x)\\ & \qquad +\sum ^m_{k=0}\tbinom {m}{k}\alpha _m^k(x)\beta ^{m-1}_m(x)\bigg), \end{align*}

so (13) holds and

\begin{gather*} (K^*_m e_2)(x)=(m+1)\sum ^m_{k=0}\tbinom {m}{k}\alpha ^k_m(x)\beta ^{m-k}_m(x)\int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}} e_2(t){\rm d}t\\ =\tfrac {1}{3(m+1)^2}\bigg(3m(m+1)\alpha ^2_m(x)\sum ^m_{k=2}\tbinom {m-2}{k-2}\alpha ^{k-2}_m(x) \beta ^{m-k}_m(x)\\ +6m\alpha _m(x)\sum ^m_{k=1}\tbinom {m-1}{k-1}\alpha ^{k-1}_m(x)\beta ^{m-k}_m(x) +\sum ^m_{k=0} \tbinom {m}{k}\alpha ^k_m(x)\beta ^{m-k}_m(x)\bigg), \end{gather*}

from where (14) follows.

Proof â–¼
Remark 6

In the following, we will use Theorem 2, where \(I=[0,1]\),

\begin{equation} \label{3.5} \varphi _{m,k}(x)=(m+1)\tbinom {m}{k}\alpha ^k_m(x)\beta ^{m-k}_m(x) \end{equation}
15

and

\begin{equation} \label{3.6} A_{m,k}(x)=\int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}f(t){\rm d}t \end{equation}
16

for any \(x\in J\), \(f\in L_1([0,1])\), \(m\in \mathbb N_1\) and \(k\in \{ 0,1,\dots ,m\} \).â–¡

4 Kantorovich-type operators preserving the test
functions \(e_0\) and \(e_1\)

In this case, we impose the conditions \((K^*_m e_0)(x)=e_0(x)\) and \((K^*_m e_1)(x)=e_1(x)\), for any \(x\in J\) and \(m\in \mathbb N_1\). From the conditions above, taking (12) and (13) into account, we have

\begin{align*} & (\alpha _m (x)+\beta (x))^m=1\; \; \text{and}\\ & \tfrac {(\alpha _m(x)+\beta _m(x))^{m-1}}{2(m+1)}\, ((2m+1)\alpha _m(x)+\beta _m(x))=x, \end{align*}

from where

\begin{equation} \label{4.1} \alpha _m(x)=\tfrac {2(m+1)x-1}{2m}\, , \end{equation}
17

\begin{equation} \label{4.2} \beta _m(x)=\tfrac {2m+1-2(m+1)x}{2m}\, , \end{equation}
18

for any \(x\in [0,1]\) and \(m\in \mathbb N_1\).
From \(\alpha _m(x)\geq 0\) and \(\beta _m(x)\geq 0\), for any \(m\in \mathbb N_1\), we have

\begin{equation} \label{4.3} \tfrac {1}{2(m+1)}\leq x\leq \tfrac {2m+1}{2(m+1)}\, . \end{equation}
19

Lemma 7

The following

\begin{equation} \label{4.4} \left[\tfrac {1}{2(m_0+1)}\, ; \tfrac {2m_0+1}{2(m_0+1)}\right]\subset \left[\tfrac {1}{2(m+1)}\, ;\tfrac {2m+1}{2(m+1)}\right]\subset [0,1] \end{equation}
20

hold for any \(m\in \mathbb N_1\).

Proof â–¼
Because the function \(\tfrac {1}{2(m+1)}\) is decreasing and the function
\(\tfrac {2m+1}{2(m+1)}\)  is increasing, relation (20) follows.
Proof â–¼
Taking the remarks above, we construct the sequence of operators \((K^*_{1,m})_{m\geq m_0}\).
Definition 8

If \(m\in \mathbb N_1\), we define the operator

\begin{gather} \label{4.5} (K^*_{1,m}f)(x)\\ =\tfrac {m+1}{(2m)^m}\sum ^m_{k=0}\tbinom {m}{k} (2(m+1)x-1)^k(2m+1-2(m+1)x)^{m-k}\int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}f(t){\rm d}t\nonumber \end{gather}
for any \(f\in L_1([0,1])\) and any \(x\in \bigg[\tfrac {1}{2(m_0+1)}\, ;\tfrac {2m_0+1}{2(m_0+1)}\bigg]\).

Remark 9

In this case, we note \(J=\bigg[\tfrac {1}{2(m_0+1)}\, ;\tfrac {2m_0+1}{2(m_0+1)}\bigg]{=}I^{(1)}_{(m_0)}\).â–¡

Lemma 10

We have

\begin{equation} \label{4.6} (K^*_{1,m}e_0)(x)=1, \end{equation}
22

\begin{equation} \label{4.7} (K^*_{1,m}e_1)(x)=x \end{equation}
23

and

\begin{equation} \label{4.8} (K^*_{1,m}e_2)(x)=\tfrac {m-1}{m}\, x^2+\tfrac {1}{m}\, x-\tfrac {5m+3}{12m(m+1)^2} \end{equation}
24

for any \(x\in I^{(1)}_{(m_0)}\) and \(m\in \mathbb N_1\).

Proof â–¼
Results immediately from the condition above and (14).
Proof â–¼
Lemma 11

The following identities

\begin{equation} \label{4.9} (T_{m,0} K^*_{1,m})(x)=1, \end{equation}
25

\begin{equation} \label{4.10} (T_{m,1}K^*_{1,m})(x)=0 \end{equation}
26

and

\begin{equation} \label{4.11} (T_{m,2}K^*_{1,m})(x)=mx(1-x)-\tfrac {m(5m+3)}{12(m+1)^2} \end{equation}
27

hold, for any \(x\in I^{(1)}_{(m_0)}\) and \(m\in \mathbb N_1\)

Proof â–¼
By using Lemma 10 and relation (3), we have

\begin{align*} (T_{m,0}K^*_{1,m})(x) & =(K^*_{1,m}e_0)(x)=1,\\ (T_{m,1}K^*_{1,m})(x) & =m(K^*_{1,m}\psi _x)(x) =m((K^*_{1,m}e_1)(x)-x(K^*_{1,m}e_0)(x))=0 \end{align*}

and

\[ (T_{m,2}K^*_{1,m}(x)=m^2(K^*_{1,m}\psi ^2_x)(x)=m^2\big((K^*_{1,m}e_1)(x)+x^2 (K^*_{1,m}e_0)(x)\big), \]

from where (27) follows.

Proof â–¼
Lemma 12

We have that

\begin{equation} \label{4.12} \lim _{m\to \infty }(T_{m,0}K^*_{1,m})(x)=1, \end{equation}
28

\begin{equation} \label{4.13} \lim _{m\to \infty }\tfrac {(T_{m,2}K^*_{1,m})(x)}{m}=x(1-x) \end{equation}
29

for any \(x\in I^{(1)}_{(m_0)}\) and \(m(0)\in \mathbb N\) exists such that

\begin{equation} \label{4.14} \tfrac {(T_{m,2}K^*_{1,m})(x)}{m}\leq \tfrac {5}{4} \end{equation}
30

for any \(x\in I^{(1)}_{(m_0)}\) and \(m\in \mathbb N_1\), \(m\geq m(0)\).

Proof â–¼
The relation (28) and (29) results taking (25) and (28) into account. By using the definition of limit a function and because \(x(1-x)\leq \tfrac {1}{4}\)  for any \(x\in [0,1]\), from (29) the inequality (30) is obtained.
Proof â–¼
Theorem 13

Let \(f:[0,1]\to \mathbb R\) be a continuous function on \([0,1]\). Then

\begin{equation} \label{4.15} \lim _{m\to \infty }K^*_{1,m} f=f \end{equation}
31

uniformly on \(I^{(1)}_{(m_0)}\) and there exists \(m(0)\in \mathbb N_1\) such that

\begin{equation} \label{4.16} \left|(K^*_{1,m}f)(x)-f(x)\right|\leq \tfrac {9}{4}\, \omega \left(f; \tfrac {1}{\sqrt{m}}\right) \end{equation}
32

for any \(x\in I^{(1)}_{(m_0)}\) and \(m\in \mathbb N_1\), \(m\geq m_0\).

Proof â–¼
We apply Theorem 2 and Corollary 3 for \(s=0\), \(\alpha _0=0\), \(\alpha _2=1\), \(k_0=1\) and \(k_2=\tfrac {9}{4}\) .
Proof â–¼
Theorem 14

If \(f\in C([0,1])\), \(x\in I^{(1)}_{(m_0)}\), \(f\) is two times differentiable in neighborhood of \(x\) and \(f^{(2)}\) is continuous on \(x\), then

\begin{equation} \label{4.17} \lim _{m\to \infty }m\left((K^*_{1,m}f)(x)-f(x)\right)= \tfrac {1}{2}\, x(1-x)f^{(2)}(x). \end{equation}
33

Proof â–¼
We use the results from Corollary 3 for \(s=2\).
Proof â–¼

5 Kantorovich-type operators preserving the test
functions \(e_0\) and \(e_2\)

In this section, we impose the conditions \((K^*_m e_0)(x)=e_0(x)\) and
\((K^*_m e_2)(x)=e_2(x)\), for any \(x\in J\) and \(m\in \mathbb N_1\). Then, taking (12) and (14) into account, we have \((\alpha _m(x)+\beta _m(x))^m=1\) and

\begin{align*} & \tfrac {(\alpha _m(x)+\beta _m(x))^{m-2}}{3(m+1)^2}\, \big(3m(m-1)\alpha ^2_m(x)+6m\alpha _m(x)(\alpha _m(x)+\beta _m(x))\\ & \quad + (\alpha _m(x)+\beta _m(x))^2\big)=x^2, \end{align*}

from where

\begin{equation} \label{5.1} \alpha _m(x)+\beta _m(x)=1 \end{equation}
34

and

\begin{equation} \label{5.2} 3m(m-1)\alpha ^2_m(x)+6m\alpha _m(x)+1-3(m+1)^2 x^2=0. \end{equation}
35

The discriminant of the equation (35) is

\[ \Delta _m=12 m(2m+1+3(m-1)(m+1)^2 x^2)\geq 0, \]

for any \(x\in J\) and any \(m\in \mathbb N_1\), \(m\geq 2\) and we note

\begin{equation} \label{5.3} \delta _m(x)=3m(2m+1+3(m-1)(m+1)^2 x^2), \end{equation}
36

\(x\in J\), \(m\in \mathbb N_1\), \(m\geq 2\).

If \(m\in \mathbb N_1\), \(m\geq 2\), then for

\begin{equation} \label{5.4} x\geq \tfrac {1}{(m+1)\sqrt{3}} \end{equation}
37

the inequality \(\tfrac {1-3(m+1)^2 x^2}{3m(m-1)}\leq 0\) is true, so the equation from (35) has exactly one positive solution. This is

\begin{equation} \label{5.5} \alpha _m(x)=\tfrac {-3m+\sqrt{\delta _m(x)}}{3m(m-1)} \end{equation}
38

and then

\begin{equation} \label{5.6} \beta _m(x)=\tfrac {3m^2-\sqrt{\delta _m(x)}}{3m(m-1)} \end{equation}
39

where \(x\in J\), and to satisfy (37), \(m\in \mathbb N_1\), \(m\geq 2\).

Lemma 15

Let \(m\in \mathbb N_1\), \(m\geq 2\). Then \(\beta _m(x)\geq 0\), \(x\geq 0\) if and only if

\begin{equation} \label{5.7} 0\leq x\leq \tfrac {\sqrt{3m^2+3m+1}}{(m+1)\sqrt{3}}\, . \end{equation}
40

Proof â–¼
From \(\beta _m(x)\geq 0\) we have \(3m^2\geq \sqrt{\delta _m(x)}\), equivalent after calculus to \(x^2\leq \tfrac {3m^2+3m+1}{3(m+1)^2}\) , from where (40) follows.
Proof â–¼
Lemma 16

Let \(m\in \mathbb N_1\), \(m\geq 2\). If \(x\in \bigg[\tfrac {1}{(m+1)\sqrt{3}}\, ;\tfrac {\sqrt{3m^2+3m+1}}{(m+1)\sqrt{3}}\bigg]\), then \(\alpha _m(x)\geq 0\) and \(\beta _m(x)\geq 0\).

Proof â–¼
Results immediately from (37) and (40).
Proof â–¼

Lemma 17

The following

\begin{equation} \label{5.8} \left[\tfrac {1}{(m_0\! +\! 1)\sqrt{3}}\, ; \tfrac {\sqrt{3m^2_0\! +\! 3m_0\! +\! 1}}{(m_0+1)\sqrt{3}}\right] \! \subset \! \left[\tfrac {1}{(m\! +\! 1)\sqrt{3}}\, ;\tfrac {\sqrt{3m^2\! +\! 3m\! +\! 1}}{(m+1)\sqrt{3}}\right] \subset [0,1] \end{equation}
41

hold, for any \(m\in \mathbb N_1\).

Proof â–¼
By using that the functions \(\tfrac {1}{(m+1)\sqrt{3}}\) and \(\tfrac {\sqrt{3m^2+3m+1}}{(m+1)\sqrt{3}}\) are decreasing, relations from (41) follows.
Proof â–¼
Definition 18

If \(m\in \mathbb N_1\), we define the operator \(K^*_{2,m}\) by

\begin{align} \label{5.9} (K^*_{2,m}f)(x) & =\tfrac {m+1}{3m(m-1))^m}\sum ^m_{k=0}\tbinom {m}{k} \left(-3m+\sqrt{\delta _m(x)}\right)^k\\ & \quad \times \left(3m^2-\sqrt{\delta _m(x)}\right)^{m-k} \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}f(t)dt\nonumber \end{align}
for any \(f\in L_1([0,1])\) and any \(x\in \bigg[\tfrac {1}{(m_0+1)\sqrt{3}}\, ;\tfrac {\sqrt{3m_0^2+3m_0+1}}{(m_0+1)\sqrt{3}}\bigg]\).

Remark 19

In this section, we note

\[ J=\bigg[\tfrac {1}{(m_0+1)\sqrt{3}}\, ;\tfrac {\sqrt{3m_0^2+3m_0+1}}{(m_0+1)\sqrt{3}}\bigg] {=}I^{(2)}_{(m_0)}.\hfil \qed \]

Lemma 20

We have

\begin{equation} \label{5.10} (K^*_{2,m}e_0)(x)=1, \end{equation}
43

\begin{equation} \label{5.11} (K^*_{2,m}e_1)(x)=\tfrac {2\sqrt{\delta _m(x)}-3m-3}{6(m-1)(m+1)} \end{equation}
44

and

\begin{equation} \label{5.12} (K^*_{1,m} e_2)(x)=x^2 \end{equation}
45

for any \(x\in I^{(2)}_{(m_0)}\) and \(m\in \mathbb N_1\).

Proof â–¼
It is inferred from the conditions above and (13).
Proof â–¼
Lemma 21

The following identities

\begin{equation} \label{5.13} (T_{m,0}K^*_{1,m})(x)=1, \end{equation}
46

\begin{equation} \label{5.14} (T_{m,1}K^*_{2,m})(x)=m\left(\tfrac {2\sqrt{\delta _m(x)}-3m-3}{6(m-1)(m+1)}-x\right) \end{equation}
47

and

\begin{equation} \label{5.15} (T_{m,2}K^*_{2,m})(x)=2m^2 x\left(x-\tfrac {2\sqrt{\delta _m(x)}-3m-3}{6(m-1)(m+1)}\right). \end{equation}
48

hold, for any \(x\in I^{(2)}_{(m_0)}\) and \(m\in \mathbb N_1\)

Proof â–¼
By using Lemma 12 and (3), we have that

\begin{align*} (T_{m,0} K^*_{2,m})(x) & =(K^*_{2,m} e_0)(x)=1,\\ (T_{m,1}K^*_{2,m})(x) & =m(K^*_{2,m}\psi _x)(x)=m\left((K^*_{2,m}e_1)(x)-x(K^*_{2,m}e_0)(x)\right), \end{align*}

so (47) holds and

\begin{align*} (T_{m,2}K^*_{2,m})(x) & =m^2(K^*_{2,m}\psi ^2_x)(x)\\ & =m^2\left((K^*_{2,m}e_2)(x)-2x (K^*_{2,m}e_1)(x)+x^2(K_{2,m}e_0)(x)\right), \end{align*}

from where (48) is obtained.

Proof â–¼
Lemma 22

The following identity

\begin{equation} \label{5.16} \lim _{m\to \infty }m\left(\tfrac {2\sqrt{\delta _m(x)}-3m-3}{6(m-1)(m+1)}-x\right)= \tfrac {x-1}{2} \end{equation}
49

holds for any \(x\in I^{(2)}_{(m_0)}\).

Proof â–¼
We have

\begin{gather*} \lim _{m\to \infty }\left(\tfrac {m^2}{(m-1)(m+1)}\cdot \tfrac {\sqrt{\delta _m(x)}-3(m-1)(m+1)x}{3m} -\tfrac {m}{2(m-1)}\right)\\ =-\tfrac {1}{2}+\lim _{m\to \infty }\tfrac {\sqrt{\delta _m(x)}-3(m-1)(m+1)x}{3m}\\ =-\tfrac {1}{2}+\lim _{m\to \infty }\tfrac {\delta _m(x)-9(m-1)^2(m+1)^2 x^2} {3m\left(\sqrt{\delta _m(x)}+3(m-1)(m+1)x\right)} \end{gather*}

and after a few calculations, identity (49) follows.

Proof â–¼
Lemma 23

We have that

\begin{equation} \label{5.17} \lim _{m\to \infty }(T_{m,0}K^*_{2,m})(x)=1, \end{equation}
50

\begin{equation} \label{5.18} \lim _{m\to \infty }\tfrac {(T_{m,2}K^*_{2,m})(x)}{m}=x(1-x) \end{equation}
51

for any \(x\in I^{(2)}_{(m_0)}\) and \(m(0)\in \mathbb N\) exists such that

\begin{equation} \label{5.19} \tfrac {(T_{m,2}K^*_{2,m})(x)}{m}\leq \tfrac {5}{4} \end{equation}
52

for any \(x\in I^{(2)}_{(m_0)}\) and \(m\in \mathbb N_1\), \(m\geq m(0)\).

Proof â–¼
The relations (50) and (51) imply (46), (48) and (49). By using the definition of the limit of a function and because \(x(1-x)\leq \tfrac {1}{4}\)  for any \(x\in [0,1]\), from (51) the relation (52) is obtained.
Proof â–¼
Theorem 24

Let \(f:[0,1]\to \mathbb R\) be a continuous function on \([0,1]\). Then

\begin{equation} \label{5.20} \lim _{m\to \infty }K^*_{2,m} f=f \end{equation}
53

uniformly on \(x\in I^{(2)}_{(m_0)}\) and there exists \(m(0)\in \mathbb N\) such that

\begin{equation} \label{5.21} \left|(K_{2,m}^* f)(x)-f(x)\right|\leq \tfrac {9}{4}\, \omega \left(f; \tfrac {1}{\sqrt{m}}\right) \end{equation}
54

for any \(x\in I^{(2)}_{(m_0)}\) and any \(m\in \mathbb N_1\), \(m\geq m(0)\).

Proof â–¼
Theorem 24 is a results from Theorem 2 and Corollary 3 for \(s=0\), \(\alpha _0=0\), \(\alpha _2=1\), \(k_0=1\) and \(k_2=\tfrac {5}{4}\) .
Proof â–¼

Theorem 25

If \(f\in C([0,1])\), \(x\in I^{(2)}_{(m_0)}\), \(f\) is two times differentiable in a neighborhood of \(x\), \(f^{(2)}\) is continuous in \(x\), then

\begin{equation} \label{5.22} \lim _{m\to \infty }m\left((K^*_{2,m}f)(x)-f(x)\right)= \tfrac {x-1}{2}\, f^{(1)}(x)+\tfrac {x(1-x)}{2}\, f^{(2)}(x). \end{equation}
55

Proof â–¼
Taking Lemma 22 into account and applying Theorem 2 for \(s=2\), we obtain the relation (55).
Proof â–¼

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