High order Approximation theory
for Banach space valued functions

George A. Anastassiou$^\ast $

February 17, 2017.

$^\ast $Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA, e-mail: ganastss@memphis.edu.

Here we study quantitatively the high degree of approximation of sequences of linear operators acting on Banach space valued differentiable functions to the unit operator. These operators are bounded by real positive linear companion operators. The Banach spaces considered here are general and no positivity assumption is made on the initial linear operators whose we study their approximation properties. We derive pointwise and uniform estimates which imply the approximation of these operators to the unit assuming differentiability of functions. At the end we study the special case where the high order derivative of the on hand function fulfills a convexity condition resulting into sharper estimates.

MSC. 41A17, 41A25, 41A36.

Keywords. Banach space valued differentiable functions, positive linear operator, convexity, modulus of continuity, rate of convergence.

1 Motivation

Let $\left( X,\left\Vert \cdot \right\Vert \right) $ be a Banach space, $N\in \mathbb {N}$. Consider $g\in C\left( \left[ 0,1\right] \right) $ and the classic Bernstein polynomials

\begin{equation} \big( \widetilde{B}_{N}g\big) \left( t\right) =\sum _{k=0}^{N}g\big( \tfrac {k}{N}\big) \tbinom {N}{k}t^{k}\left( 1-t\right) ^{N-k},\text{ \ }\forall \text{ }t\in \left[ 0,1\right] . \label{1}\end{equation}

1.1

Let also $f\in C\left( \left[ 0,1\right] ,X\right) $ and define the vector valued in $X$ Bernsein linear operators

\begin{equation} \big( B_{N}f\big) \left( t\right) =\sum _{k=0}^{N}f\big(\tfrac {k}{N}\big) \tbinom {N}{k}t^{k}\left( 1-t\right) ^{N-k},\text{ \ }\forall \text{ }t\in \left[ 0,1\right] . \label{2}\end{equation}

1.2

That is $\left( B_{N}f\right) \left( t\right) \in X.$

Clearly here $\left\| f\right\| \in C\left( \left[ 0,1\right] \right) $.

We notice that

\begin{equation} \left\Vert \left( B_{N}f\right) \left( t\right) \right\Vert \leq \sum _{k=0}^{N}\big\Vert f\big( \tfrac {k}{N}\big) \big\Vert \tbinom {N}{k}t^{k}\left( 1-t\right) ^{N-k}=\big( \widetilde{B}_{N}\left( \left\Vert f\right\Vert \right) \big) \left( t\right) ,\forall t\in \left[ 0,1\right] \label{3}\end{equation}

1.3

The property

\begin{equation} \left\| \left( B_{N}f\right) \left( t\right) \right\| \leq \big( \widetilde{B}_{N}\left( \left\| f\right\| \right) \big) \left( t\right) ,\text{ \ }\forall t\in \left[ 0,1\right] , \label{4}\end{equation}

1.4

is shared by almost all summation/integration similar operators and motivates our work here.

If $f\left( x\right) =c\in X$ the constant function, then

\begin{equation} \left( B_{N}c\right) =c. \label{5}\end{equation}

1.5

If $g\in C\left( \left[ 0,1\right] \right) $ and $c\in X$, then $cg\in C\left( \left[ 0,1\right] ,X\right) $ and

\begin{equation} \left( B_{N}\left( cg\right) \right) =c\widetilde{B}_{N}\left( g\right) . \label{6}\end{equation}

1.6

Again (1.5), (1.6) are fulfilled by many summation/integration operators.

In fact here (1.6) implies (1.5), when $g\equiv 1$.

The above can be generalized from $\left[ 0,1\right] $ to any interval $\left[ a,b\right] \subset \mathbb {R}$. All this discussion motivates us to consider the following situation.

Let $L_{N}:C\left( \left[ a,b\right] ,X\right) \hookrightarrow C\left( \left[ a,b\right] ,X\right) $, $\left( X,\left\| \cdot \right\| \right) $ a Banach space, $L_{N}$ is a linear operator, $\forall $ $N\in \mathbb {N}$, $x_{0}\in \left[ a,b\right] $. Let also $\widetilde{L}_{N}:C\left( \left[ a,b\right] \right) \hookrightarrow C\left( \left[ a,b\right] \right) $, a sequence of positive linear operators, $\forall $ $N\in \mathbb {N}$.

We assume that

\begin{equation} \left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) \right\| \leq \big( \widetilde{L}_{N}\left( \left\| f\right\| \big) \right) \left( x_{0}\right) ,\text{ \ } \label{7}\end{equation}

1.7

$\forall $ $N\in \mathbb {N}$, $\forall $ $x_{0}\in X$, $\forall $ $f\in C\left( \left[ a,b\right] ,X\right) .$

When $g\in C\left( \left[ a,b\right] \right) $, $c\in X$, we assume that

\begin{equation} \left( L_{N}\left( cg\right) \right) =c\widetilde{L}_{N}\left( g\right) . \label{8}\end{equation}

1.8

The special case of

\begin{equation} \widetilde{L}_{N}\left( 1\right) =1, \label{9.}\end{equation}

1.9

implies

\begin{equation} L_{N}\left( c\right) =c,\text{ \ }\forall \text{ }c\in X. \label{10}\end{equation}

1.10

We call $\widetilde{L}_{N}$ the companion operator of $L_{N}.$

Based on the above fundamental properties we study the high order approximation properties of the sequence of linear operators $\left\{ L_{N}\right\} _{N\in \mathbb {N}}$, i.e. their high speed convergence to the unit operator. No kind of positivity property of $\left\{ L_{N}\right\} _{N\in \mathbb {N}}$ is assumed. Other important motivation comes from [ 1 ] , [ 2 ] , [ 3 ] - [ 6 ] .

Our vector valued differentiation here resembles completely the numerical one, see [ 7 , pp. 83-84 ] .

2 Main Results

We need vector Taylor’s formula

Theorem 2.1

[ 7 , pp. 93 ] . Let $n\in \mathbb {N}$ and $f\in C^{n}\left( \left[ a,b\right] ,X\right) $, where $\left[ a,b\right] \subset \mathbb {R}$ and $X$ is a Banach space. Then

\begin{equation} f\left( b\right) =\sum _{i=0}^{n-1}\tfrac {\left( b-a\right) ^{i}}{i!}f^{\left( i\right) }\left( a\right) +\tfrac {1}{\left( n-1\right) !}\int _{a}^{b}\left( b-t\right) ^{n-1}f^{\left( n\right) }\left( t\right) dt. \label{11.}\end{equation}

2.11

Above the integral is the usual vector valued Riemann integral, see [ 7 , p. 86 ] .

We also need

Theorem 2.2

Let $n\in \mathbb {N}$ and $f\in C^{n}\left( \left[ a,b\right] ,X\right) $, where $\left[ a,b\right] \subset \mathbb {R}$ and $X$ is a Banach space. Then

\begin{equation} f\left( a\right) =\sum _{i=0}^{n-1}\tfrac {\left( a-b\right) ^{i}}{i!}f^{\left( i\right) }\left( b\right) +\tfrac {1}{\left( n-1\right) !}\int _{b}^{a}\left( a-t\right) ^{n-1}f^{\left( n\right) }\left( t\right) dt. \label{12.}\end{equation}

2.12

Proof â–¼

Let

\[ F\left( x\right) :=\sum _{i=0}^{n-1}\tfrac {\left( a-x\right) ^{i}}{i!}f^{\left( i\right) }\left( x\right) ,\text{ \ }x\in \left[ a,b\right] . \]

Here $F\in C\left( \left[ a,b\right] ,X\right) .$ Notice that $F\left( a\right) =f\left( a\right) $, and

\begin{equation} F\left( b\right) =\sum _{i=0}^{n-1}\tfrac {\left( a-b\right) ^{i}}{i!}f^{\left( i\right) }\left( b\right) . \label{13.}\end{equation}

2.13

We have

\[ F^{\prime }\left( x\right) =\tfrac {\left( a-x\right) ^{n-1}}{\left( n-1\right) !}f^{\left( n\right) }\left( x\right) ,\text{ \ }\forall \text{ }x\in \left[ a,b\right] . \]

Clearly $F^{\prime }\in C\left( \left[ a,b\right] ,X\right) .$

By [ 7 , pp. 92 ] we get

\begin{equation} F\left( b\right) -F\left( a\right) =\int _{a}^{b}F^{\prime }\left( t\right) dt. \label{14.}\end{equation}

2.14

That is we have

\begin{equation} \sum _{i=0}^{n-1}\tfrac {\left( a-b\right) ^{i}}{i!}f^{\left( i\right) }\left( b\right) -f\left( a\right) =\int _{a}^{b}\tfrac {\left( a-t\right) ^{n-1}}{\left( n-1\right) !}f^{\left( n\right) }\left( t\right) dt= -\int _{b}^{a}\tfrac {\left( a-t\right) ^{n-1}}{\left( n-1\right) !}f^{\left( n\right) }\left( t\right) dt, \label{15.}\end{equation}

2.15

proving (2.12).

Proof â–¼

Based on the above Theorems 2.1, 2.2, we have

Corollary 2.3

Let $\left( X,\left\Vert \cdot \right\Vert \right) $ be a Banach space and $f\in C^{n}\left( \left[ a,b\right] ,X\right) $, then we have the vector valued Taylor’s formula

\begin{equation} f\left( y\right) -\sum _{i=0}^{n-1}f^{\left( i\right) }\left( x\right) \tfrac {\left( y-x\right) ^{i}}{i!}=\tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}f^{\left( n\right) }\left( t\right) dt, \label{16.}\end{equation}

2.16

$\forall $ $x,y\in \left[ a,b\right] $ or

\begin{equation} f\left( y\right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x\right) \tfrac {\left( y-x\right) ^{i}}{i!}=\tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \big) dt, \label{17.}\end{equation}

2.17

$\forall $ $x,y\in \left[ a,b\right] .$

We need

Definition 2.4

Let $f\in C\left( \left[ a,b\right] ,X\right) ,$ where $\left( X,\left\| \cdot \right\| \right) $ is a Banach space. We define

\begin{equation} \omega _{1}\left( f,\delta \right) :=\underset {\left| x-y\right| \leq \delta }{\underset {x,y:}{\sup }}\left\| f\left( x\right) -f\left( y\right) \right\| \text{, \ }0<\delta \leq b-a, \label{18.}\end{equation}

2.18

the first modulus of continuity of $f$.

Remark 2.5

We study the remainder of (2.17):

\begin{equation} R_{n}\left( x,y\right) :=\tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \big) dt, \label{19.}\end{equation}

2.19

$\forall $ $x,y\in \left[ a,b\right] .$

We estimate $R_{n}\left( x,y\right) $.

Case of $y\geq x$. We have

\begin{align} \left\Vert R_{n}\left( x,y\right) \right\Vert & =\tfrac {1}{\left( n-1\right) !}\Big\Vert \int _{x}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left(n\right)}\left( x\right) \big) dt\Big\Vert \overset {\text{\cite[p. 88]{7}}}{\leq }\nonumber \\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\left\Vert f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \right\Vert dt\label{20.}\\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\omega _{1}\big( f^{\left( n\right) },\left\vert t-x\right\vert \big) dt \overset {\text{let}\ h{\gt}0}{=}\nonumber \\ & =\tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\omega _{1}\big( f^{\left( n\right) },\tfrac {\left\vert t-x\right\vert }{h}h\big) dt\nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\big( 1+\tfrac {\left\vert t-x\right\vert }{h}\big) dt & \label{21.}\\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\big( 1+\tfrac {\left( t-x\right) }{h}\big) dt\nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\Big[ \int _{x}^{y}\left( y-t\right) ^{n-1}dt+\tfrac {1}{h}\int _{x}^{y}\left( y-t\right) ^{n-1}\left( t-x\right) ^{2-1}dt\Big] \nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\Big[ \tfrac {\left( y-x\right) ^{n}}{n}+\tfrac {1}{h}\tfrac {\Gamma \left( n\right) \Gamma \left( 2\right) }{\Gamma \left( n+2\right) }\left( y-x\right) ^{n+1}\Big] \label{22}\\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\left[ \tfrac {\left( y-x\right) ^{n}}{n}+\tfrac {\left( y-x\right) ^{n+1}}{n\left( n+1\right) h}\right] =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left( y-x\right) ^{n}\left[ 1+\tfrac {\left( y-x\right) }{\left( n+1\right) h}\right] .\nonumber \end{align}

We have found that

\begin{equation} \left\Vert R_{n}\left( x,y\right) \right\Vert \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left( y-x\right) ^{n}\Big[ 1+\tfrac {\left( y-x\right) }{\left( n+1\right) h}\Big] , \label{23.}\end{equation}

2.23

for $y\geq x$, and $h{\gt}0.$

Case of $y\leq x$.

Then

\begin{align} \left\| R_{n}\left( x,y\right) \right\| & =\tfrac {1}{\left( n-1\right) !}\left\| \int _{x}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \big) dt\right\| \nonumber \\ & =\tfrac {1}{\left( n-1\right) !}\left\| \int _{y}^{x}\left( t-y\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \big) dt\right\| \label{24.}\\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{y}^{x}\left( t-y\right) ^{n-1}\left\| f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \right\| dt\nonumber \\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{y}^{x}\left( t-y\right) ^{n-1}\omega _{1}\big( f^{\left( n\right) },\tfrac {\left| t-x\right| }{h}h\big) dt\nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\int _{y}^{x}\left( t-y\right) ^{n-1}\big( 1+\tfrac {\left( x-t\right) }{h}\big) dt\nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\left[ \int _{y}^{x}\left( t-y\right) ^{n-1}dt+\tfrac {1}{h}\int _{y}^{x}\left( x-t\right) ^{2-1}\left( t-y\right) ^{n-1}dt\right] \label{25.}\\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\left[ \tfrac {\left( x-y\right) ^{n}}{n}+\tfrac {\left( x-y\right) ^{n+1}}{n\left( n+1\right) h}\right] \nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left( x-y\right) ^{n}\Big[ 1+\tfrac {\left( x-y\right) }{\left( n+1\right) h}\Big] .\nonumber \end{align}

Hence

\begin{equation} \left\| R_{n}\left( x,y\right) \right\| \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left( x-y\right) ^{n}\Big[ 1+\tfrac {\left( x-y\right) }{\left( n+1\right) h}\Big] , \label{26}\end{equation}

2.26

when $y\leq x$, $h{\gt}0$.

We have proved that

\begin{align} \left\| R_{n}\left( x,y\right) \right\| & =\left\| \tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \big) dt\right\| \leq \nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left| x-y\right| ^{n}\left[ 1+\tfrac {\left| x-y\right| }{\left( n+1\right) h}\right] , \label{27.}\end{align}

$\forall $ $x,y\in \left[ a,b\right] $, $h{\gt}0$.

We have established

Theorem 2.6

Let $\left( X,\left\| \cdot \right\| \right) $ be a Banach space and $f\in C^{n}\left( \left[ a,b\right] ,X\right) $, $n\in \mathbb {N}$. Then

\begin{equation} \bigg\| f\left( y\right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x\right) \tfrac {\left( y-x\right) ^{i}}{i!}\bigg\| \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left| x-y\right| ^{n}\left[ 1+\tfrac {\left| x-y\right| }{\left( n+1\right) h}\right] , \label{28.}\end{equation}

2.28

$\forall $ $x,y\in \left[ a,b\right] $, $h{\gt}0$.

It follows our first main result

Theorem 2.7

Let $N\in \mathbb {N}$ and $L_{N}:C\left( \left[ a,b\right] ,X\right) \rightarrow C\left( \left[ a,b\right] ,X\right) $, where $\left( X,\left\| \cdot \right\| \right) $ is a Banach space and $L_{N}$ is a linear operator. Let the positive linear operators $\widetilde{L}_{N}:C\left( \left[ a,b\right] \right) \hookrightarrow C\left( \left[ a,b\right] \right) $, such that

\begin{equation} \left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) \right\| \leq \big( \widetilde{L}_{N}\left( \left\| f\right\| \right) \big) \left( x_{0}\right) ,\text{ \ } \label{29.}\end{equation}

2.29

$\forall $ $N\in \mathbb {N},$ $\forall $ $f\in C\left( \left[ a,b\right] ,X\right) $, $\forall $ $x_{0}\in \left[ a,b\right] .$

Furthermore assume that

\begin{equation} L_{N}\left( cg\right) =c\widetilde{L}_{N}\left( g\right) ,\text{ \ }\forall \text{ }g\in C\left( \left[ a,b\right] \right) \text{, }\forall \text{ }c\in X. \label{30.}\end{equation}

2.30

Let $n\in \mathbb {N}$, here we deal with $f\in C^{n}\left( \left[ a,b\right] ,X\right) .$

Then

\begin{align} & \bigg\Vert \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{n\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{i}\big) \big) \left( x_{0}\right) \bigg\Vert \leq \nonumber \\ & \leq \tfrac {\omega _{1}\big( f^{(n)},\big( \big( \widetilde{L}_{N}\left( \left\vert \cdot -x_{0} \right\vert ^{n+1}\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{n+1}}\big) }{n!} \left( \big( \widetilde{L}_{N}\big( \left\vert \cdot -x_{0}\right\vert ^{n+1}\big) \big) \left( x_{0}\right) \right) ^{\left( \frac{n}{n+1}\right) }\times \nonumber \\ & \quad \times \Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{n+1}}+\tfrac {1}{n+1}\Big] , \label{31.}\end{align}

\begin{align} & \left\Vert \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\Vert \leq \nonumber \\ & \leq \left\Vert f\left( x_{0}\right) \right\Vert \big\vert \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -1\big\vert +\nonumber \\ & \quad +\sum _{k=0}^{n} \tfrac {\left\Vert f^{(k)}(x_{0}) \right\Vert }{k!} \big( \big( \widetilde{L}_{N}( \vert \cdot -x_{0}\vert ^{k}) \big) (x_{0}) \big) +\nonumber \\ & \quad +\frac{\omega _{1}\big( f^{\left( n\right) },\big( \big( \widetilde{L}_{N}\big( \left\vert \cdot -x_{0}\right\vert ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\frac{1}{n+1}}\big) }{n!}\big( \big( \widetilde{L}_{N}\big( \big\vert \cdot -x_{0}\big\vert ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\times \nonumber \\ & \quad \times \Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{n+1}}+\tfrac {1}{n+1}\Big] , \label{32.}\end{align}

from (2.32), and as $\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \rightarrow 1$, $\big( \widetilde{L}_{N}\left( \big\vert \cdot -x_{0}\right\vert ^{n+1}\big) \big) \left( x_{0}\right) \rightarrow 0$, we obtain $\left( L_{N}\left( f\right) \right) \left( x_{0}\right) \rightarrow f\left( x_{0}\right) $, as $N\rightarrow \infty ,$

3) if $f^{\left( k\right) }\left( x_{0}\right) =0$, $k=0,1,...,n,$ we get that

\begin{align} & \left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\| \leq \nonumber \\ & \leq \tfrac {\omega _{1}\big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{n+1}}\big) }{n!}\times \nonumber \\ & \quad \times \big( \big(\widetilde{L}_{N}(| \cdot -x_{0}|^{n+1}) \big) (x_{0}) \big) ^{(\frac{n}{n+1})} \Big[ \big( \big( \widetilde{L}_{N}(1) \big) (x_{0}) \big) ^{\frac{1}{n+1}}+t\tfrac {1}{n+1} \Big] , \label{33.}\end{align}

an extreme high speed of convergence,

4) one also derives

\begin{align} & \left\| \left\| \left( L_{N}\left( f\right) \right) -f\right\| \right\| _{\infty ,\left[ a,b\right] }\leq \nonumber \\ & \leq \left\| \left\| f\right\| \right\| _{\infty ,\left[ a,b\right] }\left\| \big( \widetilde{L}_{N}\left( 1\right) \big) -1\right\| _{\infty ,\left[ a,b\right]}+\nonumber \\ & \quad +\sum _{k=0}^{n}\tfrac {\left\| \left\| f^{\left( k\right) }\right\| \right\| _{\infty ,\left[ a,b\right] }}{k!}\left\| \big( \widetilde{L}_{N}\big( | \cdot -x_{0}| ^{k}\big) \big) ( x_{0}) \right\| _{\infty ,x_{0}\in \left[ a,b\right] }+\nonumber \\ & \quad +\tfrac {\omega _{1}\big( f^{\left( n\right) },\left\| \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right\| _{\infty ,x_{0}\in \left[ a,b\right] }^{\frac{1}{n+1}}\big) }{n!} \times \nonumber \\ & \quad \times \left\| \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \right\| _{\infty ,x_{0}\in \left[ a,b\right] }^{\left( \frac{n}{n+1}\right) }\big[ \big\| \widetilde{L}_{N}\left( 1\right) \big\| ^{\frac{1}{n+1}}+\tfrac {1}{n+1}\big] , \label{34}\end{align}

if $\widetilde{L}_{N}\left( 1\right) \overset {u}{\rightarrow }1$, uniformly, and $\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \overset {u}{\rightarrow }0$, uniformly in $x_{0}\in \left[ a,b\right] $, by (2.34), we obtain $L_{N}\left( f\right) \overset {u}{\rightarrow }f$, uniformly, as $N\rightarrow \infty .$

Proof â–¼

1) One can rewrite (2.28) as follows

\begin{equation} \Big\| f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big\| \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left[ \left| \cdot -x_{0}\right| ^{n}+\tfrac {\left| \cdot -x_{0}\right| ^{n+1}}{\left( n+1\right) h}\right] , \label{35.}\end{equation}

2.35

for a fixed $x_{0}\in \left[ a,b\right] $, $h{\gt}0$.

We observe that ($N\in \mathbb {N}$)

\begin{align} & \bigg\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{i}\big) \big) \left( x_{0}\right) \bigg\| =\nonumber \\ & =\bigg\| \big( L_{N}\Big[ f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big] \big) \left( x_{0}\right) \bigg\| \label{36}\\ & \leq \Big( \widetilde{L}_{N}\Big( \Big\| f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big\| \Big) \Big) \left( x_{0}\right)\overset {\text{(by (\ref{35.}))}}{\leq }\nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\Big[ \big( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n}\right) \big) \left( x_{0}\right) +\tfrac {\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) }{\left( n+1\right) h}\Big] =:\left( \xi _{1}\right) . \label{37.}\end{align}

Above notice that $\big( f\left( \cdot \right) -\sum _{i=0}^{n}\frac{f^{\left( i\right) }\left( x_{0}\right) }{i!}\left( \cdot -x_{0}\right) ^{i}\big) \in C\left( \left[ a,b\right] ,X\right) $.

By Hölder’s inequality and Riesz representation theorem we obtain

\begin{equation} \big( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n}\right) \big) \left( x_{0}\right) \leq \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1} \big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}. \label{38}\end{equation}

2.38

Therefore

\begin{align} \left( \xi _{1}\right) & \leq \tfrac {\omega _{1}\big( f^{\left( n\right) },h\big) }{n!}\left[ \big( \big( \widetilde{L}_{N}\left( \big| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}\right.\nonumber \\ & \quad \left. +\tfrac {1}{h}\tfrac {\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) }{\left( n+1\right) }\right] =:\left( \xi _{2}\right) . \label{39.}\end{align}

We choose

\begin{equation} h:=\big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}, \label{40.}\end{equation}

2.40

in case of $\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) {\gt}0.$

Then it holds

\begin{align} \left( \xi _{2}\right) & =\tfrac {\omega _{1}\Big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\Big) }{n!} \label{41}\\ & \quad \times \bigg[ \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}\nonumber \\ & \quad +\tfrac {\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\left( \frac{n}{n+1}\right) }}{\left( n+1\right) }\bigg] =\nonumber \\ & =\tfrac {\omega _{1}\Big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\Big) }{n!}\nonumber \\ & \quad \times \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\Big] . \label{42}\end{align}

We have proved that

\begin{align} & \bigg\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{i}\big) \big) \left( x_{0}\right) \bigg\| \leq \nonumber \\ & \leq \tfrac {\omega _{1}\Big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\Big) }{n!}\nonumber \\ & \quad \times \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right)\big) ^{\left( \frac{n}{n+1}\right) }\Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\Big] . \label{43}\end{align}

By Riesz representation theorem we have

\begin{equation} \big( \widetilde{L}_{N}\left( g\right) \big) \left( x_{0}\right) =\int _{\left[ a,b\right] }g\left( t\right) d\mu _{x_{0}}\left( t\right) ,\text{ \ }\forall \text{ }g\in C\left( \left[ a,b\right] \right) , \label{44}\end{equation}

2.44

where $\mu _{x_{0}}$ is a positive finite measure on $\left[ a,b\right] .$

That is

\begin{equation} \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) =\mu _{x_{0}}\left( \left[ a,b\right] \right) =:M. \label{45}\end{equation}

2.45

We have that $\mu _{x_{0}}\left( \left[ a,b\right] \right) {\gt}0$, because otherwise, if $\mu _{x_{0}}\left( \left[ a,b\right] \right) =0$, then $\big( \widetilde{L}_{n}\left( g\right) \big) \left( x_{0}\right) =0$, $\forall $ $g\in C\left( \left[ a,b\right] \right) $, and the whole theory here becomes trivial.

Therefore it holds $\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) {\gt}0.$

In case of

\begin{equation} \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) =0, \label{46}\end{equation}

2.46

we have

\begin{equation} \int _{\left[ a,b\right] }\left| t-x_{0}\right| ^{n+1}d\mu _{x_{0}}\left( t\right) =0. \label{47}\end{equation}

2.47

The last implies $\left| t-x_{0}\right| ^{n+1}=0$, a.e, hence $\left| t-x_{0}\right| =0$, a.e, then $t-x_{0}=0$ a.e., and $t=x_{0}$, a.e. on $\left[ a,b\right] $. Consequently $\mu _{x_{0}}\left( \left\{ t\in \left[ a,b\right] :t\neq x_{0}\right\} \right) =0$. That is $\mu _{x_{0}}=\delta _{x_{0}}M$, where $\delta _{x_{0}}$ is the Dirac measure at $\left\{ x_{0}\right\} $. In that case holds

\begin{equation} \big( \widetilde{L}_{N}\left( g\right) \big) \left( x_{0}\right) =g\left( x_{0}\right) M, \qquad \forall g\in C\left( \left[ a,b\right] \right) . \label{48}\end{equation}

2.48

Under (2.46), the right hand side of (2.43) equals zero. Furthermore it holds

\begin{equation} \Big( \widetilde{L}_{N}\Big(\Big\| f\left(\cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big\| \Big) \Big) \left( x_{0}\right) \overset {\text{(\ref{48})}}{=}\left\| f\left( x_{0}\right) -f\left( x_{0}\right) \right\| M=0. \label{49}\end{equation}

2.49

So that by (2.36) to have

also implying

\begin{equation} \left( L_{N}\left( f\right) \right) \left( x_{0}\right) =Mf\left( x_{0}\right) . \label{51}\end{equation}

2.51

So we have proved that inequality (2.43) will be always true.

2) Next we see that

\begin{align} & \big\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \big\| =\nonumber \\ & =\left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\left( \big( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) +\right.\nonumber \\ & \quad +\left. \sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) -f\left( x_{0}\right) \right\| \leq \label{52}\\ & \leq \bigg\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) \bigg\| +\nonumber \\ & \quad +\bigg\| \sum _{k=1}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) +f\left( x_{0}\right) \big( \widetilde{L}_{N}\left( 1\big) \right) \left( x_{0}\right) -f\big( x_{0}\big) \bigg\| \overset {\text{(\ref{43})}}{\leq }\nonumber \\ & \leq \left\| f\left( x_{0}\right) \right\| \big| \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -1\big| +\sum _{k=1}^{n}\tfrac {\left\| f^{\left( k\right) }\left( x_{0}\right) \right\| }{k!}\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) +\nonumber \\ & \quad +\tfrac {\omega _{1}\Big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\Big) }{n!}\nonumber \\ & \quad \times \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\Big] . \label{53}\end{align}

We have proved that

\begin{align} & \left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\| \nonumber \\ & \leq \left\| f\left( x_{0}\right) \right\| \left| \left( \widetilde{L}_{N}\left( 1\right) \right) \left( x_{0}\right) -1\right| +\nonumber \\ & \quad +\sum _{k=1}^{n}\tfrac {\left\| f^{\left( k\right) }\left( x_{0}\right) \right\| }{k!}\big( \widetilde{L}_{N}\big( \big| \cdot -x_{0}\big| ^{k}\big) \big) \left( x_{0}\right) +\nonumber \\ & \quad +\tfrac {\omega _{1}\big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\big) }{n!} \label{54}\\ & \quad \times \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\Big].\nonumber \end{align}

By Hölder’s inequality for $k=1,...,n,$ we obtain

\begin{equation} \big( \widetilde{L}_{N}\big( \big| \cdot -x_{0}\big| ^{k}\big) \big) \left( x_{0}\right) \leq \big( \big( \widetilde{L}_{N}\big( \big| \cdot -x_{0}\big| ^{n+1}\big) \big) \big( x_{0}\big) \big) ^{\left( \frac{k}{n+1}\right) }\big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\left( \frac{n+1-k}{n+1}\right) }. \label{55}\end{equation}

2.55

Clealry by (2.54) and (2.55), when $\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \rightarrow 1$ and $\big( \widetilde{L}_{N}\big( \big| \cdot -x_{0}\big| ^{n+1}\big) \big) \left( x_{0}\right) \rightarrow 0$, we obtain $\left( L_{N}\left( f\right) \right) \left( x_{0}\right) \rightarrow f\left( x_{0}\right) $, as $N\rightarrow \infty $. Notice that $\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) $ will be bounded.

3) If $f^{\left( k\right) }\left( x_{0}\right) =0$, $k=0,1,...,n,$ we get that

\begin{align} & \left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\| \overset {\text{(54)}}{\leq }\nonumber \\ & \leq \tfrac {\omega _{1}\Big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\Big) }{n!}\nonumber \\ & \quad \times \Big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \Big) ^{\left( \frac{n}{n+1}\right) }\Big[ \big( \big( \widetilde{L}_{N}\left( 1\big) \right) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\Big] , \label{56}\end{align}

an extreme high speed of convergence.

4) One also derives from (2.54) that

\begin{align} & \left\| \left\| \left( L_{N}\left( f\right) \right) -f\right\| \right\| _{\infty ,\left[ a,b\right] }\leq \nonumber \\ & \leq \left\| \left\| f\right\| \right\| _{\infty ,\left[ a,b\right] }\left\| \widetilde{L}_{N}\left( 1\right) -1\right\| _{\infty ,\left[ a,b\right] }+\nonumber \\ & \quad +\sum _{k=1}^{n}\tfrac {\left\| \left\| f^{\left( k\right) }\right\| \right\| _{\infty ,\left[ a,b\right] }}{k!}\left\| \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) \right\| _{\infty ,x_{0}\in \left[ a,b\right] }+\nonumber \\ & \quad +\tfrac {\omega _{1}\Big( f^{\left( n\right) },\left\| \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right\| _{\infty ,x_{0}\in \left[ a,b\right] }^{\frac{1}{\left( n+1\right) }}\Big) }{n!}\nonumber \\ & \quad \times \left\| \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \right\| _{\infty ,x_{0}\in \left[ a,b\right] }^{\left( \frac{n}{n+1}\right) }\big[ \| \widetilde{L}_{N}\left( 1\right) \| ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\big] . \label{57}\end{align}

Inequality (2.55), for $k=1,...,n$, implies

\begin{align} & \left\| \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) \right\| _{\infty ,x_{0}\in \left[ a,b\right] }\nonumber \\ & \leq \left\| \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \right\| _{\infty ,x_{0}\in \left[ a,b\right] }^{\left( \frac{k}{n+1}\right) }\left\| \widetilde{L}_{N}\left( 1\right) \right\| _{\infty ,x_{0}\in \left[ a,b\right] }^{\left( \frac{n+1-k}{n+1}\right) }. \label{58}\end{align}

Consequently, if $\widetilde{L}_{N}\left( 1\right) \overset {u}{\rightarrow }1 $, uniformly, and $\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \overset {u}{\rightarrow }0$, uniformly in $x_{0}\in \left[ a,b\right] $, by (2.57) and (2.58), we obtain $L_{N}\left( f\right) \overset {u}{\rightarrow }f$, uniformly, as $N\rightarrow \infty .$

Here the assumption $\widetilde{L}_{N}\left( 1\right) \overset {u}{\rightarrow }1$, uniformly, as $N\rightarrow \infty $, implies that $\big\| \widetilde{L}_{N}\left( 1\right) \big\| $ is bounded.

The proof of the theorem now is complete.

Proof â–¼

We make

Remark 2.8

Let $\left( X,\left\| \cdot \right\| \right) $ be a Banach space and $f\in C^{n}\left( \left[ a,b\right] ,X\right) $, $n\in \mathbb {N}$, and $x_{0}\in \left( a,b\right) $ be fixed. Then

\begin{align} f\left( y\right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( y-x_{0}\right) ^{i}}{i!}& =\tfrac {1}{\left( n-1\right) !}\int _{x_{0}}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right)} \left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big) dt\nonumber \\ & =:R_{n}\left( x_{0},y\right) \text{, \ }\forall \text{ }y\in \left[ a,b\right] \label{59}\end{align}

We assume that $g\left( t\right) :=\big\| f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big\| $ is convex in $t\in \left[ a,b\right] .$

We consider $0{\lt}h\leq \min \left( x_{0}-a,b-x_{0}\right) .$ Obviously $g\left( x_{0}\right) =0$. Then by Lemma 8.1.1, p. 243 of [ 1 ] , we obtain

\begin{equation} g\left( t\right) \leq \tfrac {\omega _{1}\left( g,h\right) }{h}\left| t-x_{0}\right| \text{, \ }\forall \text{ }t\in \left[ a,b\right] . \label{60}\end{equation}

2.60

For any $t_{1},t_{2}\in \left[ a,b\right] :\left| t_{1}-t_{2}\right| \leq h$ we get

\begin{align} & \Big| \big\| f^{\left( n\right) }\left( t_{1}\right) -f^{\left( n\right) }\left( x_{0}\right) \big\| -\big\| f^{\left( n\right) }\left( t_{2}\right) -f^{\left( n\right) }\left( x_{0}\right) \big\| \Big|\nonumber \\ & \leq \big\| f^{\left( n\right) }\left( t_{1}\right) -f^{\left( n\right) }\left( t_{2}\right) \big\| \leq \omega _{1}\left( f^{n},h\right) . \label{61}\end{align}

That is

\begin{equation} \omega _{1}\left( g,h\right) \leq \omega _{1}\big( f^{\left( n\right) },h\big) . \label{62}\end{equation}

2.62

The last implies

\begin{equation} \big\| f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big\| \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h}\left| t-x_{0}\right| \text{, \ \ }\forall \text{ }t\in \left[ a,b\right] . \label{63}\end{equation}

2.63

We estimate $R_{n}\left( x_{0},y\right) .$

Case of $y\geq x_{0}.$ We have

\begin{align} \left\| R_{n}\left( x_{0},y\right) \right\| & =\tfrac {1}{\left( n-1\right) !}\left\| \int _{x_{0}}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big) dt\right\| \overset {\text{(\cite[pp. 88]{7})}}{\leq }\nonumber \\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{x_{0}}^{y}\left( y-t\right) ^{n-1}\left\| f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \right\| dt\overset {\text{(\ref{63})}}{\leq }\nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !h}\int _{x_{0}}^{y}\left( y-t\right) ^{n-1}\left( t-x_{0}\right) ^{2-1}dt=\nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !h}\tfrac {\Gamma \left( n\right) \Gamma \left( 2\right) }{\Gamma \left( n+2\right) }\left( y-x_{0}\right) ^{n+1}=\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n+1\right) !h}\left( y-x_{0}\right) ^{n+1}. \label{64}\end{align}

We proved that

\begin{equation} \left\| R_{n}\left( x_{0},y\right) \right\| \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n+1\right) !h}\left( y-x_{0}\right) ^{n+1},\text{ \ }\forall \text{ }y\geq x_{0}. \label{65}\end{equation}

2.65

Case of $y\leq x_{0}$. Then

\begin{align} \left\| R_{n}\left( x_{0},y\right) \right\| & =\tfrac {1}{\left( n-1\right) !}\left\| \int _{x_{0}}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big) dt\right\| \nonumber \\ & =\tfrac {1}{\left( n-1\right) !}\left\| \int _{y}^{x_{0}}\left( t-y\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big) dt\right\| \nonumber \\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{y}^{x_{0}}\left( t-y\right) ^{n-1}\left\| f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \right\| dt\overset {\text{(\ref{63})}}{\leq }\nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n-1\right) !}\int _{y}^{x_{0}}\left( x_{0}-t\right) ^{2-1}\left( t-y\right) ^{n-1}dt \label{66} \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\left( x_{0}-y\right) ^{n+1}.\nonumber \end{align}

That is proving

\begin{equation} \left\| R_{n}\left( x_{0},y\right) \right\| \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\left( x_{0}-y\right) ^{n+1},\text{ \ }\forall \text{ }y\in \left[ a,b\right] :y\leq x_{0}. \label{67}\end{equation}

2.67

We have established that

\begin{equation} \left\| R_{n}\left( x_{0},y\right) \right\| \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\left| y-x_{0}\right| ^{n+1}\text{, \ }\forall \text{ }y\in \left[ a,b\right] , \label{68}\end{equation}

2.68

where $0{\lt}h\leq \min \left( x_{0}-a,b-x_{0}\right) $, $x_{0}\in \left( a,b\right) $, and $\left\| f\left( \cdot \right) -f\left( x_{0}\right) \right\| $ is convex over $\left[ a,b\right] .$

We have proved

Theorem 2.9

Let $\left( X,\left\| \cdot \right\| \right) $ be a Banach space and $f\in C^{n}\left( \left[ a,b\right] ,X\right) $, $n\in \mathbb {N}$, and $x_{0}\in \left( a,b\right) $ be fixed. Let $0{\lt}h\leq \min \left( x_{0}-a,b-x_{0}\right) $, and assume that $\left\| f^{\left( n\right) }\left( \cdot \right) -f^{\left( n\right) }\left( x_{0}\right) \right\| $ is convex over $\left[ a,b\right] $. Then

\begin{equation} \bigg\| f\left( y\right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( y-x_{0}\right) ^{i}}{i!}\bigg\| \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\left| y-x_{0}\right| ^{n+1}\text{, \ }\forall \text{ }y\in \left[ a,b\right] . \label{69}\end{equation}

2.69

We give our second main result under convexity.

Theorem 2.10

2.70

$\forall $ $N\in \mathbb {N},$ $\forall $ $f\in C\left( \left[ a,b\right] ,X\right) $, where $x_{0}\in \left( a,b\right) $ is fixed$.$

Furthermore assume that

\begin{equation} L_{N}\left( cg\right) =c\widetilde{L}_{N}\left( g\right) ,\text{ \ }\forall \text{ }g\in C\left( \left[ a,b\right] \right) \text{, }\forall \text{ }c\in X. \label{71}\end{equation}

2.71

Let $n\in \mathbb {N}$, here we deal with $f\in C^{n}\left( \left[ a,b\right] ,X\right) .$

We further assume that $\big\| f^{\left( n\right) }\left( \cdot \right) -f^{\left( n\right) }\left( x_{0}\right) \big\| $ is convex over $\left[ a,b\right] $, and

\begin{equation} 0\leq \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \leq \min \left( x_{0}-a,b-x_{0}\right) . \label{72}\end{equation}

2.72

Then

\begin{align} \left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\| & \leq \left\| f\left( x_{0}\right) \right\| \left| \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -1\right| +\nonumber \\ & \quad +\sum _{k=1}^{n}\tfrac {\left\| f^{\left( k\right) }\left( x_{0}\right) \right\| }{k!}\big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) \big) +\nonumber \\ & \quad +\tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right)!}, \label{74}\end{align}

as $\big( \widetilde{L}_{N}\left(1\right) \big) \left( x_{0}\right) \rightarrow 1$, and $\big( \widetilde{L}_{N} \big( \left| \cdot -x_{0}\right|^{n+1}\big) \big) \left( x_{0}\right) \rightarrow 0$, we obtain $\left( L_{N}\left( f\right) \right) \left( x_{0}\right) \rightarrow f\left( x_{0}\right) $, as $N\rightarrow \infty ,$

3) if $f^{\left( k\right) }\left( x_{0}\right) =0$, $k=0,1,...,n,$ we get that

\begin{equation} \left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\| \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}, \label{75}\end{equation}

2.75

a high speed of convergence.

Proof â–¼

1) One can rewrite (2.69) as follows

\begin{equation} \bigg\| f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\bigg\| \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\left| \cdot -x_{0}\right| ^{n+1}\text{, \ }\forall \text{ }y\in \left[ a,b\right] . \label{76}\end{equation}

2.76

We observe that

\begin{align} & \bigg\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{i}\big) \big) \left( x_{0}\right) \bigg\| = \label{77}\\ & =\bigg\| \Big( L_{N}\Big[ f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big] \Big) \left( x_{0}\right) \bigg\| \leq \nonumber \\ & \leq \Big(\widetilde{L}_{N}\Big( \Big\| f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big\| \Big) \Big) \left( x_{0}\right) \overset {\text{(by (\ref{76}))}}{\leq } \label{78}\\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right)=\nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}, \label{79}\end{align}

by choosing

\begin{equation} h:=\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) , \label{80}\end{equation}

2.80

if $\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) {\gt}0.$

We have proved that

By Riesz representation theorem we have

2.82

where $\mu _{x_{0}}$ is a positive finite measure on $\left[ a,b\right] .$

That is

\begin{equation} \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) =\mu _{x_{0}}\left( \left[ a,b\right] \right) =:M. \label{83}\end{equation}

2.83

Without loss of generality we assume that $M{\gt}0$, if $M=0$, then our theory is trivial.

In case of

\begin{equation} \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) =0, \label{84}\end{equation}

2.84

we have

\begin{equation} \int _{\left[ a,b\right] }\left| t-x_{0}\right| ^{n+1}d\mu _{x_{0}}\left( t\right) =0. \label{85}\end{equation}

2.85

\begin{equation} \big( \widetilde{L}_{N}\left( g\right) \big) \left( x_{0}\right) =g\left( x_{0}\right) M, \label{86}\end{equation}

2.86

$\forall $ $g\in C\left( \left[ a,b\right] \right) .$

Under (2.84), the right hand side of (2.81) equals zero. Furthermore it holds

\begin{align} \Big( \widetilde{L}_{N}\Big( \Big\| f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big\| \Big) \Big) \left( x_{0}\right) \overset {\text{(\ref{86})}}{=}\left\| f\left( x_{0}\right) -f\left( x_{0}\right) \right\| M=0. \label{87}\end{align}

So that by (2.77) to have

\begin{equation} \bigg\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\left( \widetilde{L}_{N}\left( \left( \cdot -x_{0}\right) ^{i}\right) \right) \left( x_{0}\right) \bigg\| =0. \label{88}\end{equation}

2.88

Therefore inequality (2.81) is true again and always.

2) Next again we see that

\begin{align} & \left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\| =\nonumber \\ & =\left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) +\right.\nonumber \\ & \quad +\left. \sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) -f\left( x_{0}\right) \right\| \leq \label{89}\\ & \leq \bigg\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) \bigg\| +\nonumber \\ & \quad +\bigg\| \sum _{k=1}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) +f\left( x_{0}\right) \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -f\left( x_{0}\right) \bigg\| \overset {\text{(\ref{81})}}{\leq }\nonumber \\ & \leq \big\| f\left( x_{0}\right) \big\| \left| \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -1\right| +\sum _{k=1}^{n}\tfrac {\left\| f^{\left( k\right) }\left( x_{0}\right) \right\| }{k!}\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) +\nonumber \\ & \quad +\tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}. \label{90}\end{align}

We have proved that

\begin{align} \left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\| \leq & \left\| f\left( x_{0}\right) \right\| \left| \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -1\right| +\nonumber \\ & \quad +\sum _{k=1}^{n}\tfrac {\left\| f^{\left( k\right) }\left( x_{0}\right) \right\| }{k!}\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) +\nonumber \\ & \quad +\tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}. \label{91}\end{align}

Clearly by (2.55) and (2.91), when $\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \rightarrow 1$ and $\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \rightarrow 0$, we obtain $\left( L_{N}\left( f\right) \right) \left( x_{0}\right) \rightarrow f\left( x_{0}\right) $, as $N\rightarrow \infty $. Notice that $\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) $ will be bounded.

3) If $f^{\left( k\right) }\left( x_{0}\right) =0$, $k=0,1,...,n,$ we get that

\begin{align} \left\| \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\| \overset {\text{(\ref{91})}}{\leq } \tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}, \label{92}\end{align}

a high speed of convergence.

The theorem is proved.

Proof â–¼

Theorem 2.11

All as in Theorem 2.10. Consider $n\in \mathbb {N}$ an odd number. Then inequalities (2.73)–(2.75) are sharp, in fact they are attained by $f_{\ast }\left( t\right) =\vec{i}\left| t-x_{0}\right| ^{n+1}$, where $\vec{i}\in X$, $\big\| \vec{i}\big\| =1$, $\forall $ $t\in \left[ a,b\right] .$

Proof â–¼

Let $n$ an odd natural number, then $n+1$ is even. We consider $f_{\ast }\left( t\right) =\vec{i}\left| t-x_{0}\right| ^{n+1}=\vec{i}\left( t-x_{0}\right) ^{n+1}\in X$, where $\vec{i}\in X$, $\big\| \vec{i}\big\| =1$. We have that $f_{\ast }\in C^{n}\left( \left[ a,b\right] ,X\right) $ and

\begin{equation} f_{\ast }^{\left( n\right) }\left( t\right) =\vec{i}\left( n+1\right) !\left( t-x_{0}\right) \text{, \ }\forall t\in \left[ a,b\right] , \label{93}\end{equation}

2.93

along with $f_{\ast }^{\left( k\right) }\left( x_{0}\right) =0$, $k=0,1,...,n.$

Furthermore it holds

\begin{equation} \left\| f_{\ast }^{\left( n\right) }\left( t\right) -f_{\ast }^{\left( n\right) }\left( x_{0}\right) \right\| =\left( n+1\right) !\left| t-x_{0}\right| , \label{94}\end{equation}

2.94

which is a convex function in $t\in \left[ a,b\right] $. So we apply $f_{\ast }$ to inequalities (2.73)-(2.75).

1) The left hand side of (2.73) equals $\left( \widetilde{L}_{N}( \left| t-x_{0}\right| ^{n+1}) \right) \left( x_{0}\right) $. The right hand side of (2.73) is

\begin{align} & \tfrac {\omega _{1}\big( f_{\ast }^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \big) }{\left( n+1\right) !}= \label{95}\\[2mm]& =\tfrac {\omega _{1}\left( \vec{i}\left( n+1\right) !\left( t-x_{0}\right) ,\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !} \nonumber \\[2mm]& =\end{align}

=
=c t_1,t_2 a,b :
t_1-t_2 ≤( L_N( ·-x_0 ^n+1) ) x_0 t_1-t_2 = L_N ·-x_0 ^n+1 x_0 .

\end{align}

Hence the right hand side of (2.73) equals also $\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) $. That is (2.73) is an attained inequality by $f_{\ast }.$

2) The left hand side of (2.74) equals $\big( \widetilde{L}_{N}\big( \left| t-x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) $, which equals its right hand side. That is (2.74) is an attained inequality by $f_{\ast }.$

3) Same as above, inequality (2.75) is attained by $f_{\ast }.$

We have proved that (2.73)-(2.75) are sharp inequalities.

Proof â–¼

Corollary 2.12

(to Theorem 2.7) Let $\left( X,\left\| \cdot \right\| \right) $ be a Banach space and any $f\in C^{1}\left( \left[ 0,1\right] ,X\right) $. Then

\begin{align} \left\| \left( B_{N}\left( f\right) \right) \left( x\right) -f\left( x\right) \right\| \leq \tfrac {3}{2}\omega _{1}\left( f^{\prime },\sqrt{\tfrac {x\left( 1-x\right) }{N}}\right) \sqrt{\tfrac {x\left( 1-x\right) }{N}} & \leq \tfrac {0.75}{\sqrt{N}}\omega _{1}\big( f^{\prime },\tfrac {1}{2\sqrt{N}}\big),\label{97} \end{align}

$\forall x\in \left[ 0,1\right] , \forall N\in \mathbb {N}$, and

\begin{equation} \left\| \left\| B_{N}\left( f\right) -f\right\| \right\| _{\infty ,\left[ 0,1\right] }\leq \tfrac {0.75}{\sqrt{N}}\omega _{1}\big( f^{\prime },\tfrac {1}{2\sqrt{N}}\big) . \label{98}\end{equation}

2.98

Hence $B_{n}\left( f\right) \overset {u}{\rightarrow }f$, uniformly, as $N\rightarrow \infty .$

Proof â–¼

The operators $B_{N}$, $\widetilde{B}_{N}$ fulfill (2.29), (2.30). We have that $\widetilde{B}_{N}\left( 1\right) =1$, $\big( \widetilde{B}_{N}\left( id\right) \big) \left( x\right) =x$,

\begin{equation} \big( \widetilde{B}_{N}\left( \left( \cdot -x\right) \right) \big) \left( x\right) =0\text{,} \label{99}\end{equation}

2.99

and

\[ \big( \widetilde{B}_{N}\big( \left( \cdot -x\right) ^{2}\big) \big) \left( x\right) =\tfrac {x\left( 1-x\right) }{N}\leq \tfrac {1}{4N},\text{ \ }\forall \text{ }x\in \left[ 0,1\right] . \]

We use (2.31) for $n=1$. We have (by use of (2.99)) that

\begin{align} & \left\| \left( B_{N}\left( f\right) \right) \left( x\right) -f\left( x\right) \right\| \leq \nonumber \\ & \leq \tfrac {3}{2}\omega _{1}\big( f^{\prime },\big( \big( \widetilde{B}_{N}\big( \left( \cdot -x\right) ^{2}\big) \big) \left( x\right) \big) ^{\frac{1}{2}}\big) \big( \big( \widetilde{B}_{N}\left( \big( \cdot -x\right) ^{2}\big) \big) \left( x\right) \big) ^{\frac{1}{2}}= \label{100}\\ & =\tfrac {3}{2}\omega _{1}\left( f^{\prime },\sqrt{\tfrac {x\left( 1-x\right) }{N}}\right) \sqrt{\tfrac {x\left( 1-x\right) }{N}}\leq \tfrac {3}{2}\omega _{1}\left( f^{\prime },\tfrac {1}{2\sqrt{N}}\right) \tfrac {1}{2\sqrt{N}}=\nonumber \\ & =\tfrac {3}{4\sqrt{N}}\omega _{1}\left( f^{\prime },\tfrac {1}{2\sqrt{N}}\right) =\tfrac {0.75}{\sqrt{N}}\omega _{1}\left( f^{\prime },\tfrac {1}{2\sqrt{N}}\right) . \label{101}\end{align}

Proof â–¼

We finish with

Corollary 2.13

(to Theorem 2.10) Let $\left( X,\left\| \cdot \right\| \right) $ be a Banach space and any $f\in C^{1}\left( \left[ 0,1\right] ,X\right) $ such that $\left\| f^{\prime }\left( t\right) -f^{\prime }\left( x_{0}\right) \right\| $ is convex function in $t\in \left[ 0,1\right] $, where $x_{0}\in \left( 0,1\right) $ is a fixed number. Then

\begin{align} \left\| \left( B_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\| & \leq \tfrac {1}{2}\omega _{1}\big( f^{\prime },\tfrac {x_{0}\left( 1-x_{0}\right) }{N}\big) \leq \tfrac {1}{2}\omega _{1}\left( f^{\prime },\tfrac {1}{4N}\right) ,\text{ \ }\forall \text{ }N\in \mathbb {N}. \label{102}\end{align}

Above notice the high speed of convergence $\frac{1}{N}$ under the convexity assumption.

Inequalities (2.102) are sharp. The first part of (2.102) is attained by $\vec{i}\left( t-x_{0}\right) ^{2}$, $\vec{i}\in X$, $\big\| \vec{i}\big\| =1$, $\forall $ $t\in \left[ 0,1\right] $. The second part of (2.102) is equality at $x_{0}=\frac{1}{2}.$

As $N\rightarrow \infty $, we have that $\left( B_{N}\left( f\right) \right) \left( x_{0}\right) \rightarrow f\left( x_{0}\right) $.

Proof â–¼

Let $x_{0}\in \left( 0,1\right) $, then $x_{0}\left( 1-x_{0}\right) \leq x_{0}$ and $x_{0}\left( 1-x_{0}\right) \leq 1-x_{0}$, hence $x_{0}\left( 1-x_{0}\right) \leq \min \left( x_{0},1-x_{0}\right) $ and

\begin{equation} \tfrac {x_{0}\left( 1-x_{0}\right) }{N}\leq \min \left( x_{0},1-x_{0}\right) \text{, \ }\forall \text{ }N\in \mathbb {N}. \label{103}\end{equation}

2.103

The last shows that (see (2.99))

\begin{equation} 0<\big( \widetilde{B}_{N}\big( \left( \cdot -x_{0}\right) ^{2}\big) \big) \left( x_{0}\right) \leq \min \left( x_{0},1-x_{0}\right) \text{, \ }\forall \text{ }N\in \mathbb {N}. \label{104}\end{equation}

2.104

Let here $f\in C^{1}\left( \left[ 0,1\right] ,X\right) $ such that $\left\Vert f^{\prime }\left( \cdot \right) -f^{\prime }\left( x_{0}\right) \right\Vert $ is convex over $\left[ 0,1\right] .$ Then, by (2.73), we get

\begin{align} \left\Vert \left( B_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\Vert & \leq \tfrac {\omega _{1}\left( f^{\prime },\left( \widetilde{B}_{N}\left( \left( \cdot -x_{0}\right) ^{2}\right) \right) \left( x_{0}\right) \right) }{2} \label{105}\\ & =\tfrac {\omega _{1}\big( f^{\prime },\frac{x_{0}\left( 1-x_{0}\right) }{N}\big) }{2}\leq \tfrac {1}{2}\omega _{1}\big( f^{\prime },\tfrac {1} {4N}\nonumber \big) . \end{align}

Proof â–¼

Bibliography

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2: G.A. Anastassiou, Lattice homomorphism - Korovkin type inequalities for vector valued functions, Hokkaido Mathematical Journal, Vol. 26 (1997), pp. 337–364. $\includegraphics[scale=0.1]{ext-link.png}$
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High order Approximation theory for Banach space valued functions

1 Motivation

2 Main Results

Bibliography

High order Approximation theory
for Banach space valued functions