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\title[Sharp estimates for the rates of convergence of the iterates]
 {Sharp inequalities for the rates of convergence of the iterates of some operators which preserve the constants}
\author{Marius Mihai Birou}
\address{Technical University of Cluj Napoca, \\ Department  of Mathematics \\
28, Memorandumului Street,\\
400114 Cluj-Napoca,\\
Romania} \email{Marius.Birou@math.utcluj.ro}
%

%
\subjclass{41A36, 41A25} \keywords{positive linear operators,
iterates, convergence, sharp inequalities}
\begin{abstract}
In this paper we give estimates for the rates of convergence for
the iterates of some positive linear operators which preserve only
the constants. We obtain sharp inequalities when we use both
continuous functions and differentiable functions. We present some
optimal results for the Cesaro, Stancu and Schurer operators.
\end{abstract}
\maketitle

\section{Introduction}

Starting with the articles ~\cite{kel} and ~\cite{karl} of R.P.
Kelisky, T.J. Rivlin and respectively S. Karlin, Z. Ziegler, the
iterates of the positive linear operators were intensively
studied.

The convergence of the sequence of the iterates of some positive
linear operators which preserve only the constants was proved in
~\cite{alto}, ~\cite{rus1}, ~\cite{gonska}, ~\cite{rasa2}, ~\cite{rus2}, ~\cite{galaz}, ~\cite%
{gavrea1}, ~\cite{gavrea2}, ~\cite{alto2}.

On the other hand, estimations of the rates of convergence for the
iterates of some positive operators preserving the constants were
given in ~\cite{mahm} using moduli of smoothness. In ~\cite{abel}
the authors got sharp inequalities for the iterates of the
Bernstein operators. In ~\cite{rasa1} the author obtained an estimate of the convergence rate for the iterations of linear and positive operators that reproduce linear functions in the case of differentiable functions.

In this note we obtain inequalities for the rates of convergence
of the iterates of some positive linear operators
$L:C[a,b]\rightarrow C[a,b]$
which preserve only the constants and have the interpolation point $x=a$ or $%
x=b$. In Section 2 we get these estimations both for continuous
functions (using  moduli of smoothness and divided difference) and
for differentiable functions. The inequalities ~(\ref{est11}),
~(\ref{est12}), ~(\ref{estc1}), ~(\ref{estc2}), ~(\ref{ec31}) and
~(\ref{ec32}) are sharp in sense that we get equality if we take
$f=e_{1}.$ In Section 3 we determine the best constants in some
inequalities involving the iterates of Cesaro, Stancu and Schurer
operators.

Throughout the paper we use the following notations and
definitions:

$\bullet$ the the monomial functions: $e_{i}:[a,b]\rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
,$ $e_{i}(x)=x^{i},$ $i=0,1,\ldots ;$

$\bullet$ the first and and the second moduli of smootness of the
functiom $f\in C[a,b]$:
\begin{equation*}
\omega _{1}(f,\delta )=\sup \left\{ f(x+h)-f(x):x,x+h\in \lbrack
a,b],\ 0\leq h\leq \delta \right\} ,
\end{equation*}%
and respectively%
\begin{equation*}
\omega _{2}(f,\delta )=\sup \left\{ f(x+h)-2f(x)+f(x-h):x,x\pm
h\in \lbrack a,b],\ 0\leq h\leq \delta \right\},
\end{equation*}
where $\delta \geq 0$,

$\bullet$ the divided difference of the function  $f\in C[a,b]$ on
the distinct points $x_{1},x_{2}\in \lbrack a,b]$:

\begin{equation*}
\lbrack x_{1},x_{2};f]=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}.
\end{equation*}

\section{Main results}

\begin{theorem}\label{thmmc1}
Let $L:C[a,b]\rightarrow C[a,b]$ be a positive linear operator which
preserves
only the constants and has interpolation point $x=a.$ If%
\begin{equation*}
L^{k}e_{1}(x)>a,\ x\in (a,b],
\end{equation*}%
then we have, for every $f\in C[a,b]$ and $x\in \lbrack a,b]$,
\begin{equation}
\left\vert L^{k}f(x)-f(a)\right\vert \leq \frac{4}{b-a}\lambda
_{k}(x)\omega _{1}\left( f,\lambda _{k}(x)\right) +3\omega
_{2}\left( f,\lambda _{k}(x)\right) ,  \label{est11}
\end{equation}%
where%
\begin{equation}
\lambda _{k}(x)=\frac{1}{2}\sqrt{(b-a)(L^{k}e_{1}(x)-a)}.
\label{lam}
\end{equation}%

\end{theorem}

\begin{proof} Let $f\in C[a,b]$ and $0<\delta \leq (b-a)/2.$ If $F$ is a positive linear
functional on $C[a,b],$ then from the optimal result of P\u{a}lt\u{a}nea ~%
\cite{palt} we have:
\begin{equation}
\left\vert f(x)-F(f)\right\vert \leq f(x)\left\vert F(e_{0})-1\right\vert +%
\frac{1}{\delta }\left\vert F(e_{1}-xe_{0})\right\vert \omega
_{1}\left( f,\delta \right)   \label{palt2}
\end{equation}%
\begin{equation*}
+\left( F(e_{0})+\frac{1}{2\delta ^{2}}F(e_{1}-xe_{0})^{2}\right)
\omega _{2}\left( f,\delta \right) ,\ x\in \lbrack a,b].
\end{equation*}%
Taking $F(f)=f(a)$ we get%
\begin{align*}
\left\vert f-f(a)\right\vert & \leq \frac{e_{1}-ae_{0}}{\delta
}\omega _{1}\left( f,\delta \right) +\left( e_{0}+\frac{\left(
e_{1}-ae_{0}\right)
^{2}}{2\delta ^{2}}\right) \omega _{2}\left( f,\delta \right)  \\
& \leq \frac{e_{1}-ae_{0}}{\delta }\omega _{1}\left( f,\delta
\right) +\left( e_{0}+\frac{(b-a)(e_{1}-ae_{0})}{2\delta
^{2}}\right) \omega _{2}\left( f,\delta \right) .
\end{align*}%
Since $L$ preserves the constant functions, it follows that
\begin{equation}
\left\vert L^{k}f-f(a)\right\vert \leq \frac{1}{\delta }\left(
L^{k}e_{1}-ae_{0}\right) \omega _{1}\left( f,\delta \right) +\left( e_{0}+%
\frac{(b-a)(L^{k}e_{1}-ae_{0})}{2\delta ^{2}}\right) \omega
_{2}\left( f,\delta \right) .  \label{estim2}
\end{equation}

If we take in ~(\ref{estim2})
\begin{equation*}
\delta =\lambda _{k}(x),\ x\in (a,b],
\end{equation*}%
where $\lambda _{k}$ is given by ~(\ref{lam}) we get that
~(\ref{est11}) holds for $x\in (a,b].$

For $x=a$, due the interpolation property of $L,$ we have
$L^{k}f(a)=f(a).$ Therefore ~(\ref{est11}) is also true for $x=a.$
This completes the proof.
\end{proof}

\begin{theorem}\label{thmmc2}
Let $L:C[a,b]\rightarrow C[a,b]$ be a positive linear operator which
preserves
only the constants and has interpolation point $x=b.$ If%
\begin{equation*}
L^{k}e_{1}(x)<b,\ x\in \lbrack a,b),
\end{equation*}%
then we have, for every $f\in C[a,b]$ and $x\in \lbrack a,b]$,
\begin{equation}
\left\vert L^{k}f(x)-f(b)\right\vert \leq \frac{4}{b-a}\mu
_{k}(x)\omega _{1}\left( f,\mu _{k}(x)\right) +3\omega _{2}\left(
f,\mu _{k}(x)\right) , \label{est12}
\end{equation}%
where%
\begin{equation*}
\mu _{k}(x)=\frac{1}{2}\sqrt{(b-a)(b-L^{k}e_{1}(x))}.
\end{equation*}
\end{theorem}

\begin{proof} Taking $F(f)=f(b)$ in ~(\ref{palt2}) we get%
\begin{align*}
\left\vert f-f(b)\right\vert & \leq \frac{be_{0}-e_{1}}{\delta
}\omega _{1}\left( f,\delta \right) +\left(
e_{0}+\frac{(be_{0}-e_{1})^{2}}{2\delta
^{2}}\right) \omega _{2}\left( f,\delta \right)  \\
& \leq \frac{be_{0}-e_{1}}{\delta }\omega _{1}\left( f,\delta
\right) +\left( e_{0}+\frac{(b-a)(be_{0}-e_{1})}{2\delta
^{2}}\right) \omega _{2}\left( f,\delta \right) .
\end{align*}


The conclusion follows analogous as in Theorem \ref{thmmc1}.
\end{proof}


\begin{theorem}
\label{thc1} Let $L:C[a,b]\rightarrow C[a,b]$ be a positive linear
operator which preserves constants and has the interpolation point
$x=a$. Then, for
every $f\in C[a,b]$ and $x\in \lbrack a,b]$ we have%
\begin{equation}\label{estc1}
m_{a}(L^{k}(e_{1})(x)-a)\leq L^{k}(f)(x)-f(a)\leq
M_{a}(L^{k}(e_{1})(x)-a),
\end{equation}%
where $m_{a},M_{a}\in
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ such that $m_{a}\leq \lbrack a,t;f]\leq M_{a}$ when $t\in (a,b]$
$.$
\end{theorem}

\begin{proof}
We have%
\begin{equation*}
f(x)-f(a)=\left\{
\begin{array}{cc}
\lbrack a,x;f](x-a), & x\in (a,b] \\
0, & x=a%
\end{array}%
\right.
\end{equation*}%
It follows
\begin{equation}
m_{a}(e_{1}-a)\leq f-f(a)\leq M_{a}(e_{1}-a).  \label{ec1}
\end{equation}%
Applying $k$ times the operator $L$ on ~(\ref{ec1}) we get the
conclusion.
\end{proof}

\begin{remark}
From Theorem \ref{thc1} we get the following criterion for the
convergence of the iterates (see also ~\cite[Corolar 2]{gavrea1}):
if $L:C[a,b]\rightarrow C[a,b]$ is a positive linear operator
which preserves the constants, has the interpolation point $x=a$
and satisfies the condition
\begin{equation*}
\lim_{k\rightarrow \infty }L^{k}e_{1}=a,\ \text{uniformly on
}[a,b],
\end{equation*}%
then for every $f\in C[a,b]$ we have
\begin{equation*}
\lim_{k\rightarrow \infty }L^{k}f=f(a),\text{ uniformly on }[a,b].
\end{equation*}
\end{remark}

\begin{theorem}
\label{thc2} Let $L:C[a,b]\rightarrow C[a,b]$ be a positive linear
operator which preserves constants and has the interpolation point
$x=b$. Then, for
every $f\in C[a,b]$ and $x\in \lbrack a,b]$ we have%
\begin{equation}\label{estc2}
m_{b}(b-L^{k}(e_{1})(x))\leq f(b)-L^{k}(f)(x)\leq
M_{b}(b-L^{k}(e_{1})(x)),
\end{equation}%
where $m_{b},M_{b}\in
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ such that $m_{b}\leq \lbrack t,b;f]\leq M_{b}$ for every $t\in
\lbrack a,b).$
\end{theorem}

The proof follows analogous with that of Theorem \ref{thc1} using
the
formula%
\begin{equation*}
f(b)-f(x)=\left\{
\begin{array}{cc}
\lbrack x,b;f](b-x), & x\in \lbrack a,b) \\
0, & x=b%
\end{array}%
\right.
\end{equation*}

\begin{remark}
From Theorem \ref{thc2} we get the following criterion for the
convergence of the iterates: if $L:C[a,b]\rightarrow C[a,b]$ is a
positive linear operator which preserves the constants, has the
interpolation point $x=b$ and satisfies the condition
\begin{equation*}
\lim_{k\rightarrow \infty }L^{k}e_{1}=b,\ \text{uniformly on
}[a,b],
\end{equation*}%
then for every $f\in C[a,b]$ we have
\begin{equation*}
\lim_{k\rightarrow \infty }L^{k}f=f(b),\ \text{ uniformly on
}[a,b].
\end{equation*}
\end{remark}

\begin{theorem}
\label{thdf1} Let $L:C[a,b]\rightarrow C[a,b]$ be a positive linear
operator which preserves constants and has the interpolation point
$x=a$. Then, for
every $f\in C^{1}[a,b]$ and $x\in \lbrack a,b]$ we have%
\begin{equation}\label{ec31}
m^{\prime}(L^{k}(e_{1})(x)-a)\leq L^{k}(f)(x)-f(a)\leq
M^{\prime}(L^{k}(e_{1})(x)-a),
\end{equation}%
where $m^{\prime},M^{\prime}\in
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ such that $m^{\prime}\leq f^{\prime }(t)\leq M^{\prime},$ $t\in \lbrack a,b]$ and%
\begin{equation*}
\left\vert L^{k}(f)(x)-f(a)\right\vert \leq \overline{M^{\prime}}%
(L^{k}(e_{1})(x)-a),
\end{equation*}%
where $\overline{M^{\prime}}=\max_{t\in \lbrack a,b]}\left\vert
f^{\prime }(t)\right\vert .$
\end{theorem}

\begin{proof}
If $x\in(a,b]$, then using the mean value theorem it follows that there
exists $\xi \in
( a,x)$ such that%
\begin{equation}\label{med}
f(x)-f(a)=(x-a)f^{\prime }(\xi ).
\end{equation}%
If $x=a$ the formula (\ref{med}) also holds for every $\xi \in
[a,b]$.

Therefore%
\begin{equation}
m^{\prime}(e_{1}-a)\leq f-f(a)\leq M^{\prime}(e_{1}-a).
\label{ec3}
\end{equation}%
Applying $k$ times the operator $L$ on ~(\ref{ec3}) we get
~(\ref{ec31}). The proof is ended.
\end{proof}

\begin{theorem}
\label{thdf2} Let $L:C[a,b]\rightarrow C[a,b]$ be a positive linear
operator which preserves constants and has the interpolation point
$x=b$. Then, for
every $f\in C^{1}[a,b]$ and $x\in \lbrack a,b]$ we have%
\begin{equation}\label{ec32}
m^{\prime}(b-L^{k}(e_{1})(x))\leq f(b)-L^{k}(f)(x)\leq
M^{\prime}(b-L^{k}(e_{1})(x)),
\end{equation}%
where $m^{\prime},M^{\prime}\in
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ such that $m^{\prime}\leq f^{\prime }(t)\leq M^{\prime},$ $t\in \lbrack a,b]$ and%
\begin{equation*}
\left\vert L^{k}(f)(x)-f(b)\right\vert \leq \overline{M^{\prime}}%
(b-L^{k}(e_{1})(x)),
\end{equation*}%
where $\overline{M^{\prime}}=\max_{t\in \lbrack a,b]}\left\vert
f^{\prime }(t)\right\vert .$
\end{theorem}

The proof follows analogous with that of Theorem \ref{thdf1} using
the mean value theorem:
\begin{equation*}
f(b)-f(x)=(b-x)f^{\prime }(\xi ),\ \xi \in ( a,b).
\end{equation*}

\section{Applications}

We consider the following positive linear operators which preserve
only the constants:

$\bullet$ Cesaro operator%
\begin{equation*}
C:C[0,1]\rightarrow C[0,1],\ C(f)(x)=\left\{
\begin{array}{cc}
f(0), & x=0 \\
\displaystyle\frac{1}{x}\int_{0}^{x}f(t)dt, & x>0%
\end{array}%
\right. ,\ x\in \lbrack 0,1]
\end{equation*}

$\bullet$ Bernstein-Stancu operators (see ~\cite{stancu})
\begin{equation*}
S_{n,\alpha }:C[0,1]\rightarrow C[0,1],\ S_{n,\alpha
}(f)(x)=\sum\limits_{i=0}^{n}\binom{n}{i}x^{i}(1-x)^{n-i}f\left( \frac{%
i+\alpha }{n+\alpha }\right) ,
\end{equation*}
\begin{equation*}
x\in \lbrack 0,1],\ n=0,1,\ldots ,\ \alpha >0,
\end{equation*}
and
\begin{equation*}
S_{n,\beta }:C[0,1]\rightarrow C[0,1],\ S_{n,\beta
}(f)(x)=\sum\limits_{i=0}^{n}\binom{n}{i}x^{i}(1-x)^{n-i}f\left( \frac{i}{%
n+\beta }\right) ,
\end{equation*}
\begin{equation*}
x\in \lbrack 0,1],\ n=0,1,\ldots ,\ \beta >0,
\end{equation*}

$\bullet$ Schurer operator%
\begin{equation*}
S_{n,p}:C[0,1]\rightarrow C[0,1],\ S_{n,p}(f)(x)=\sum\limits_{i=0}^{n-p}%
\binom{n-p}{i}x^{i}(1-x)^{n-p-i}f\left( \frac{i}{n}\right),
\end{equation*}
\begin{equation*}
x\in \lbrack 0,1],\ n=0,1,\ldots ,\ n,p\in
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
,\ n\geq p.
\end{equation*}

The operators $C$, $S_{n,\beta }$, $S_{n,p}$ have the interpolation point $%
x=0$ while the operator $S_{n,\alpha }$ interpolates the
continuous
functions at $x=1.$ For every $k\geq 0$ we have by induction (see also ~\cite%
{gavrea1} for the operators $C$, $S_{n,\beta }$, $S_{n,p}$):%
\begin{equation*}
C^{k}e_{1}=\frac{1}{2^{k}}e_{1},\ S_{n,\alpha }^{k}e_{1}=\left( \frac{n}{%
n+\beta }\right) ^{k}e_{1},S_{n,p}^{k}e_{1}=\left(
\frac{n-p}{n}\right) ^{k}e_{1},
\end{equation*}%
\begin{equation*}
S_{n,\alpha }^{k}e_{1}=e_{0}+\left( \frac{n}{n+\alpha }\right)
^{k}(e_{1}-e_{0}).
\end{equation*}

From Theorem \ref{thmmc1} and Theorem \ref{thmmc2} we have:

\begin{theorem}. For every $f\in C[0,1]$ and $x\in \lbrack 0,1]$ we
have:
\begin{enumerate}
\item \begin{equation*}\left\vert C^{k}f(x)-f(0)\right\vert \leq 2\sqrt{\frac{x}{2^{k}}}%
\cdot \omega _{1}\left( f,\frac{1}{2}\sqrt{\frac{x}{2^{k}}}\right)
+3\omega _{2}\left( f,\frac{1}{2}\sqrt{\frac{x}{2^{k}}}\right)
,\end{equation*}

\item \begin{equation*}\left\vert S_{n,\alpha }^{k}f(x)-f(0)\right\vert \leq \end{equation*}
\begin{equation*} 2\sqrt{\left(
\frac{n}{n+\beta }\right) ^{k}x}\cdot\omega _{1}\left(
f,\frac{1}{2}\sqrt{\left(
\frac{n}{n+\beta }\right) ^{k}x}\right) +3\omega _{2}\left( f,\frac{1}{2}%
\sqrt{\left( \frac{n}{n+\beta }\right) ^{k}x}\right)
\end{equation*},

\item \begin{equation*}\left\vert S_{n,p}^{k}f(x)-f(0)\right\vert \leq \end{equation*}
\begin{equation*}2\sqrt{\left( \frac{%
n-p}{n}\right) ^{k}x}\cdot\omega _{1}\left( f,\frac{1}{2}\sqrt{\left( \frac{n-p}{n%
}\right) ^{k}x}\right) +3\omega _{2}\left( f,\frac{1}{2}\sqrt{\left( \frac{%
n-p}{n}\right) ^{k}x}\right) \end{equation*},

\item \begin{equation*}\left\vert S_{n,\beta }^{k}f(x)-f(1)\right\vert \leq \end{equation*}
\begin{equation*}2\sqrt{\left(
\frac{n}{n+\alpha }\right) ^{k}x}\cdot\omega _{1}\left(
f,\frac{1}{2}\sqrt{\left(
\frac{n}{n+\alpha }\right) ^{k}x}\right) +3\omega _{2}\left( f,\frac{1}{2}%
\sqrt{\left( \frac{n}{n+\alpha }\right) ^{k}x}\right)
.\end{equation*}
\end{enumerate}
\end{theorem}

Using Theorem \ref{thc1}, Theorem \ref{thc2} and respectively Theorem \ref%
{thdf1}, Theorem \ref{thdf2} we get the following sharp estimates:

\begin{theorem}
\label{ths} Let $f\in C[0,1]$. If $m_{0},M_{0},m_{1},M_{1}\in
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ such that $m_{0}\leq \lbrack 0,t;f]\leq M_{0}$, $t\in(0,1]$ and
$m_{1}\leq \lbrack t,1;f]\leq M_{1}$, $t\in[0,1)$ then for every
$k\geq 0$ we have:

\begin{enumerate}
\item $m_{0}c_{1}(k)e_{1}\leq C^{k}(f)-f(0)e_{0}\leq M_{0}c_{1}(k)e_{1},$
where $c_{1}(k)=\frac{1}{2^{k}}$,

\item $m_{0}c_{2}(k,n,\beta )e_{1}\leq S_{n,\beta }^{k}(f)-f(0)e_{0}\leq
M_{0}c_{2}(k,n,\beta )e_{1},$ where $c_{2}(k,n,\beta )=\left( \frac{n}{%
n+\beta }\right) ^{k}$,

\item $m_{0}c_{3}(k,n,p)e_{1}\leq S_{n,p}^{k}(f)-f(0)e_{0}\leq
M_{0}c_{3}(k,n,p)e_{1},$ where $c_{3}(k,n,p)=\left(
\frac{n-p}{n}\right) ^{k} $,

\item $m_{1}c_{4}(k,n,\alpha )(e_{0}-e_{1})\leq f(1)e_{0}-S_{n,\alpha
}^{k}(f)\leq M_{1}c_{4}(k,n,\alpha )(e_{0}-e_{1}),$ where
$c_{4}(k,n,\alpha )=\left( \frac{n}{n+\alpha }\right) ^{k}$.
\end{enumerate}
\end{theorem}

\begin{theorem}
\label{ths2} Let $f\in C^{1}[0,1]$. If $m^{\prime},M^{\prime}\in
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ such that $m^{\prime}\leq f^{\prime }(t)\leq M^{\prime},$ $t\in
\lbrack 0,1]$, then for every $k\geq 0$ we have:

\begin{enumerate}
\item $m^{\prime}c_{1}(k)e_{1}\leq C^{k}(f)-f(0)e_{0}\leq M^{\prime}c_{1}(k)e_{1},$

\item $m^{\prime}c_{2}(k,n,\beta )e_{1}\leq S_{n,\beta }^{k}(f)-f(0)e_{0}\leq
M^{\prime}c_{2}(k,n,\beta )e_{1},$

\item $m^{\prime}c_{3}(k,n,p)e_{1}\leq S_{n,p}^{k}(f)-f(0)e_{0}\leq
M^{\prime}c_{3}(k,n,p)e_{1},$

\item $m^{\prime}c_{4}(k,n,\alpha )(e_{0}-e_{1})\leq f(1)e_{0}-S_{n,\alpha
}^{k}(f)\leq M^{\prime}c_{4}(k,n,\alpha )(e_{0}-e_{1}),$
\end{enumerate}
where the constants $c_{1}(k),$ $c_{2}(k,n,\beta ),$ $c_{3}(k,n,p),$ $%
c_{4}(k,n,\alpha )$ are given in Theorem \ref{ths}.

\end{theorem}

The constants $c_{1}(k),$ $c_{2}(k,n,\beta ),$ $c_{3}(k,n,p),$ $%
c_{4}(k,n,\alpha )$ in Theorem \ref{ths} and Theorem \ref{ths2}
are the best possible: for $f=e_{1}$ we get equality.

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