Simultaneous approximation by operators of exponential type

Ulrich Abel
Technische Hochschule Mittelhessen
Fachbereich MND
Wilhelm-Leuschner-Straße 13, 61169 Friedberg,
Germany
<EMAIL_ADDRESS>
Abstract.

There are many results on the simultaneous approximation by sequences of
special positive linear operators. In the year 1978, Ismail and May as well as
Volkov independently studied operators of exponential type covering the most
classical approximation operators. In this paper we study asymptotic
properties of these class of operators. We prove that under certain
conditions, asymptotic expansions for sequences of operators belonging to a
slightly larger class of operators, can be differentiated term-by-term. This
general theorem contains several results which were previously obtained by
several authors for concrete operators. One corollary states, that the
complete asymptotic expansion for the Bernstein polynomials can be
differentiated term-by-term. This implies a well-known result on the
Voronovskaja formula obtained by Floater.

_Mathematics Subject Classification (2020):_ 41A36, 41A25, 41A60.

_Keywords:_ Approximation by positive operators, operators of exponential
type, rate of convergence, degree of approximation, asymptotic expansions.

##  1 Introduction

Let $e_{r}$ $\left(r=0,1,2,\ldots\right)$ denote the monomials defined by
$e_{r}\left(x\right)=x^{r}$. For each real number $x$, define the function
$\psi_{x}$ by $\psi_{x}=e_{1}-xe_{0}$.

Nearly all approximation operators $L_{n}$ possess asymptotic properties. The
most prominent instance are the Bernstein polynomials $B_{n}$. In 1932,
Voronovoskaja [20] showed the celebrated formula

$\lim_{n\rightarrow\infty}n\left(\left(B_{n}f\right)\left(x\right)-f\left(x\right)\right)=\frac{1}{2}x\left(1-x\right)f^{\prime\prime}\left(x\right),$
(1)

for all real functions $f$ bounded on $\left[0,1\right]$ which have a second
order derivative. In that same year, Bernstein [4] proved, for each
$x\in\left(0,1\right)$ and $q\in\mathbb{N}$, for which
$f^{\left(2q\right)}\left(x\right)$ exists, that

$\left(B_{n}f\right)\left(x\right)=\sum_{s=0}^{2q}\left(B_{n}\psi_{x}^{s}\right)\left(x\right)\frac{f^{\left(s\right)}\left(x\right)}{s!}+o\left(n^{-q}\right)\text{
}\qquad\left(n\rightarrow\infty\right).$ (2)

Perhaps this is the first asymptotic expansion for any polynomial operator.
Noting that $\left(B_{n}\psi_{x}^{0}\right)\left(x\right)=1$,
$\left(B_{n}\psi_{x}^{1}\right)\left(x\right)=0$,
$\left(B_{n}\psi_{x}^{2}\right)\left(x\right)=x\left(1-x\right)/n$, reveals
that Eq. $\left(\ref{voronovskaja}\right)$ is the special case $q=1$ of Eq.
$\left(\ref{Bernstein-asymptotic}\right)$. In other words, it is an asymptotic
expansion of order $1$ (as
$\left(B_{n}f\right)\left(x\right)=f\left(x\right)+x\left(1-x\right)f^{\prime\prime}\left(x\right)/\left(2n\right)+o\left(n^{-1}\right)$)
and the constructive Weierstrass theorem is one of order $0$ (as
$\left(B_{n}f\right)\left(x\right)=f\left(x\right)+o\left(1\right)$) as
$n\rightarrow\infty$. A more detailled inspection of the central moments
$B_{n}\psi_{x}^{s}$ shows that Eq. $\left(\ref{Bernstein-asymptotic}\right)$
can be written as a complete asymptotic expansion in the more explicit form

$\left(B_{n}f\right)\left(x\right)\sim\sum_{k=0}^{\infty}\frac{a_{k}\left(f,x\right)}{n^{k}}\text{
}\qquad\left(n\rightarrow\infty\right),$ (3)

where the coefficients $a_{k}\left(f,x\right)$ are independent of $n$. They
are linear combinations of $f$ and finitely many derivatives of $f$ multiplied
by certain polynomials. Explicit expressions for $a_{k}\left(f,x\right)$
involving Stirling numbers of the first and second kind can be derived [2].
Eq. $\left(\ref{Bernstein-asymptotic}\right)$ means that, for each
$q=0,1,\ldots,$

$\left(B_{n}f\right)\left(x\right)=\sum_{k=0}^{q}\frac{a\left(f,x\right)}{n^{k}}+o\left(n^{-q}\right)\text{
}\qquad\left(n\rightarrow\infty\right).$

Several authors studied asymptotic relations for the derivatives
$\left(B_{n}f\right)^{\left(r\right)}$.

If a sequence of operators $L_{n}$ has a complete asymptotic expansion of the
form

$\left(L_{n}f\right)\left(x\right)\sim\sum_{k=0}^{\infty}\frac{a_{k}\left(f,x\right)}{n^{k}}\text{
}\qquad\left(n\rightarrow\infty\right),$ (4)

it is quite natural to expect that the derivatives
$\left(L_{n}f\right)^{\left(r\right)}$ have a complete asymptotic expansion
which can be derived by term-by-term differentiation of the coefficients
$a_{k}\left(f,x\right)$ with respect to $x$, i.e.,

$\left(L_{n}f\right)^{\left(r\right)}\left(x\right)\sim\sum_{k=0}^{\infty}\frac{a_{k}^{\left[r\right]}\left(f,x\right)}{n^{k}}\text{
}\qquad\left(n\rightarrow\infty\right)$ (5)

with
$a_{k}^{\left[r\right]}\left(f,x\right)=\left(d/dx\right)^{r}a_{k}\left(f,x\right)$.
One of the first such formulas were shown for the first derivative of the
Bernstein–Kantorovich operators [1, Theorem 3]. The further step came in 2002
when A.-J. López-Moreno, J. Martínez-Moreno, and F.-J. Muñoz-Delgado [16]
established the asymptotic expansion for the derivatives of the $B_{n}f$,
namely

$\left(B_{n}f\right)^{\left(r\right)}\left(x\right)=f^{\left(r\right)}\left(x\right)+\sum_{s=1}^{2q}\frac{1}{s!}\left(\left(B_{n}\psi_{x}^{s}\right)\left(x\right)\cdot
f^{\left(s\right)}\left(x\right)\right)^{\left(r\right)}+o\left(n^{-q}\right)\text{
}\qquad\left(n\rightarrow\infty\right),$

provided that $f$ is sufficiently smooth on the interval $\left[0,1\right]$.
The essential property was the so-called $\varphi$-convexity of the operators.
In [17] A.-J. López-Moreno and F.-J. Muñoz-Delgado extended the concept of
$\varphi$-convexity in order to prove general results that allow to calculate
the asymptotic formulae of the partial derivatives of sequences of
multivariate linear operators. In 2005, Floater [6, Theorem 2] showed that
Voronovskaja’s formula $\left(\ref{voronovskaja}\right)$ for the Bernstein
polynomials can be differentiated, i.e.,

$\lim_{n\rightarrow\infty}n\left(\left(B_{n}f\right)^{\left(r\right)}\left(x\right)-f^{\left(r\right)}\left(x\right)\right)=\left(\frac{x\left(1-x\right)f^{\prime\prime}\left(x\right)}{2}\right)^{\left(r\right)},$
(6)

uniformly on $\left[0,1\right]$, for functions $f\in C^{r+2}\left[0,1\right]$.
This result is a special case of [17, Theorem 2] which was found after
completion of [6]. However, Floater used a completely different approach,
viz., remainder formulas for the Bernstein error in terms of divided
differences. This technique was refined by Abel [3] for simultaneous
approximation by Bernstein–Chlodovsky polynomials. Further results on
asymptotic properties of sequences of operators can be found in [10], [17],
[21]. Results on the differentiation of Voronovskaja’s formula, for general
classes of operators including that of exponential type, can be found in the
recent work by [11], [12], [13]. In the recent 2022 article Xiang studies
term-by-term differentiation of asymptotic expansions for certain Bernstein
type operators by probabilistic methods.

The purpose of this paper is to prove that operators of exponential type
possess, for sufficiently smooth functions $f$, a complete asymptotic
expansion $\left(\ref{Lnf-asymptotic}\right)$, which can be differentiated
term-by-term, i.e., $\left(\ref{Lnf-asymptotic-simultan}\right)$ holds with
$a_{k}^{\left[r\right]}\left(f,x\right)=\left(d/dx\right)^{r}a_{k}^{\left[0\right]}\left(f,x\right)$,
for each positive integer $r$. The most prominent approximation operators are
of exponential type. Operators of exponential type preserve linear functions.
In fact, we consider a more general class of operators, which not necessarily
have this property, but preserve constant functions.

Using $a_{0}\left(f,x\right)=f\left(x\right)$, a direct consequence of our
results is the fact that the Voronovskaja-type formula, i.e.,

$\lim_{n\rightarrow\infty}n\left(\left(L_{n}f\right)\left(x\right)-f\left(x\right)\right)=a_{1}\left(f,x\right),$

can by differentiated term-by-term, i.e.,

$\lim_{n\rightarrow\infty}n\left(\left(L_{n}f\right)^{\left(r\right)}\left(x\right)-f^{\left(r\right)}\left(x\right)\right)=\left(\frac{d}{dx}\right)^{r}a_{1}\left(f,x\right).$

##  2 Operators of exponential type

Let $I$ be a (finite or infinite) real interval. Recall that throughout the
paper we use the denotation $e_{r}\left(t\right)=t^{r}$ and
$\psi_{x}\left(t\right)=t-x$ $\left(t,x\in\mathbb{R}\right)$. In 1978, Ismail
and May [15] as well as Volkov [19] studied linear operators
$S_{n}:C\left(I\right)\rightarrow C\left(I\right)$ which preserve constant
functions and satisfy the differential equation

$\left(S_{n}f\right)^{\prime}\left(x\right)=\frac{n}{\varphi\left(x\right)}\left(S_{n}\left(\psi_{x}f\right)\right)\left(x\right),$
(7)

where $\varphi$ is an infinitely often differentiable function with
$\varphi\left(x\right)\neq 0$ on $I$. If $I$ is an infinite interval, the
class $C\left(I\right)$ may be restricted to functions satisfying a certain
growth condition. Such operators $S_{n}$ are called operators of exponential
type. There is a huge literature dealing with them. Operators of exponential
type preserve linear functions, i.e., $S_{n}e_{r}=e_{r}$, for
$r\in\left\\{0,1\right\\}$. The second moment $S_{n}e_{2}$, which is crucial
for the approximation properties of $S_{n}f$ as $n\rightarrow\infty$, is given
by $S_{n}e_{2}=e_{2}+\varphi/n$ (see [15, Proposition 2.1]). Denote the
central moments of the operators $S_{n}$ by

$\mu_{n,s}\left(x\right)=\left(S_{n}\psi_{x}^{s}\right)\left(x\right).$

In particular, we have $\mu_{n,0}=1$, $\mu_{n,1}=0$, and
$\mu_{n,2}=\varphi/n$. The recursive formula

$\mu_{n,s+1}=\frac{\varphi}{n}\left(s\mu_{n,s-1}+\mu_{n,s}^{\prime}\right)$
(8)

is an easy consequence of the differential equation $\left(\ref{ODE-
exponential-type}\right)$ (see [19, Eq. (3)], [15, Eq. (2.2)]). It easily
delivers, for small values of $s$ the following instances:

$\displaystyle\mu_{n,3}$ $\displaystyle=$
$\displaystyle\varphi\varphi^{\prime}n^{-2}$ $\displaystyle\mu_{n,4}$
$\displaystyle=$ $\displaystyle
3\varphi^{2}n^{-2}+\left(\varphi\left(\varphi^{\prime}\right)^{2}+\varphi^{2}\varphi^{\prime\prime}\right)n^{-3}$
$\displaystyle\mu_{n,5}$ $\displaystyle=$ $\displaystyle
10\varphi^{2}\varphi^{\prime}n^{-3}+\left(\varphi\left(\varphi^{\prime}\right)^{3}+4\varphi^{2}\varphi^{\prime}\varphi^{\prime\prime}+\varphi^{3}\varphi^{\prime\prime\prime}\right)n^{-4}$
$\displaystyle\mu_{n,6}$ $\displaystyle=$ $\displaystyle
15\varphi^{3}n^{-3}+\left(\varphi^{2}\left(\varphi^{\prime}\right)^{2}+15\varphi^{3}\varphi^{\prime\prime}\right)n^{-4}$
$\displaystyle+\left(\varphi\left(\varphi^{\prime}\right)^{4}+11\varphi^{2}\left(\varphi^{\prime}\right)^{2}\varphi^{\prime\prime}+4\varphi^{3}\left(\varphi^{\prime\prime}\right)^{2}+7\varphi^{3}\varphi^{\prime}\varphi^{\prime\prime\prime}+\varphi^{4}\varphi^{\prime\prime\prime\prime}\right)n^{-5}$

If $\varphi$ is a quadratic polynomial then $\mu_{n,s}\left(x\right)$ is a
polynomial in $x$ of degree at most $s$. For each $x\in I$, the recursive
formula implies that

$\mu_{n,s}\left(x\right)=O\left(n^{-\left\lfloor\left(s+1\right)/2\right\rfloor}\right)\text{
}\qquad\left(n\rightarrow\infty\right)\qquad\text{for }s=0,1,\ldots.$

A further consequence of the recursive formula is the representation

$\mu_{n,s}\left(x\right)=\sum_{j=\left\lfloor\left(s+1\right)/2\right\rfloor}^{s}g_{s,j}\left(x\right)n^{-j},$
(9)

where $g_{s,j}$ are polynomials of degree at most $s$. In particular,
$g_{0,0}\left(x\right)=1$, $g_{1,1}\left(x\right)=0$. Ismail and May [15,
Proposition 2 (iii)] showed that $g_{2s,2s}\left(x\right)$ is a constant
multiple of $\varphi^{s}\left(x\right)$ and $g_{2s+1,2s+1}\left(x\right)$ is a
constant multiple of
$\varphi^{s}\left(x\right)\varphi^{\prime}\left(x\right)$. In fact, it can be
shown that

$\displaystyle g_{2s,2s}\left(x\right)$ $\displaystyle=$
$\displaystyle\frac{\left(2s\right)!}{2^{s}s!}\varphi^{s}\left(x\right),$
$\displaystyle g_{2s+1,2s+1}\left(x\right)$ $\displaystyle=$
$\displaystyle\frac{s\cdot\left(2s+1\right)!}{3\cdot
2^{s}s!}\varphi^{s}\left(x\right)\varphi^{\prime}\left(x\right).$

By Eq. $\left(\ref{representation-mu-by-g}\right)$, the relation

$\left(S_{n}f\right)\left(x\right)=\sum_{s=0}^{2q}\left(\mu_{n,s}\left(x\right)\frac{f^{\left(s\right)}\left(x\right)}{s!}\right)+o\left(n^{-q}\right)\text{
}\qquad\left(n\rightarrow\infty\right),$

implies the complete asymptotic expansion

$\left(S_{n}f\right)\left(x\right)=\sum_{k=0}^{q}\frac{1}{n^{k}}\sum_{s=k}^{2k}\frac{f^{\left(s\right)}\left(x\right)}{s!}g_{s,k}\left(x\right)+o\left(n^{-q}\right)\text{
}\qquad\left(n\rightarrow\infty\right),$

for appropriate functions sufficiently smooth at $x$.

##  3 Main result

Recently, Holhoş [11, Eq. (2)] (see also [12, Eq. (2.1)]) extended the class
of exponential type operators, by considering the more general differential
equation

$\left(S_{n}f\right)^{\prime}\left(x\right)=\frac{\lambda_{n}}{\varphi\left(x\right)}\left(\left(S_{n}\left(\psi_{x}f\right)\right)\left(x\right)-\left(S_{n}\psi_{x}\right)\left(x\right)\cdot\left(S_{n}f\right)\left(x\right)\right),$
(10)

where $\left(\lambda_{n}\right)$ is a sequence of positive real numbers
tending to infinity. If $\varphi\left(x\right)$ is a quadratic polynomial then
$\mu_{n,s}\left(x\right)$ is a polynomial in $x$ of degree at most $s$. Holhoş
[11, Lemma 2] (cf. [12, Lemma 2.3]) proved that, for each $x\in I$ and
$s=0,1,\ldots$, the central moments $\mu_{n,s}$ of operators satisfying the
differential equation $\left(\ref{ODE-Holhos}\right)$ have finite limits
$\lim_{n\rightarrow\infty}\left(\lambda_{n}^{s}\cdot\mu_{n,2s}\left(x\right)\right)$
and
$\lim_{n\rightarrow\infty}\left(\lambda_{n}^{s+1}\cdot\mu_{n,2s+1}\left(x\right)\right)$
and determined their values. As a consequence, one has

$\mu_{n,s}\left(x\right)=O\left(\lambda_{n}^{-\left\lfloor\left(s+1\right)/2\right\rfloor}\right)\text{
}\qquad\left(n\rightarrow\infty\right).$

If the sequence $\left(\lambda_{n}\right)$ satisfies the inequalities
$0<\alpha<\lambda_{n}/n<\beta$, for certain constants $\alpha,\beta$, it
follows that

$\mu_{n,s}\left(x\right)=O\left(n^{-\left\lfloor\left(s+1\right)/2\right\rfloor}\right)\text{
}\qquad\left(n\rightarrow\infty\right),\qquad\text{for }s=0,1,\ldots.$ (11)

Let $s$ be a positive integer. If $I$ is a finite interval, let
$C_{s}\left(I\right)$ be the space of all bounded functions on $I$. If $I$ is
an unbounded interval, let $C_{s}\left(I\right)$ be the space of all locally
bounded functions $f$ on $I$ which satisfy the growth condition
$f\left(t\right)=O\left(\left|t\right|^{s}\right)$ as
$\left|t\right|\rightarrow+\infty$. For $x\in I$, let $C_{s,x}\left(I\right)$
denote the space of all functions in $C_{s}\left(I\right)$ having a derivative
of order $s$ at $x$. By Sikkema’s theorem (Lemma 4), it follows that

$\left(S_{n}f\right)\left(x\right)=\sum_{s=0}^{2q}\left(\mu_{n,s}\left(x\right)\frac{f^{\left(s\right)}\left(x\right)}{s!}\right)+o\left(n^{-q}\right)\text{
}\qquad\left(n\rightarrow\infty\right),$ (12)

for each function $f\in C_{2q,x}\left(I\right)$ and all positive integers $q$.
Already Volkov [19, THEOREM 2] showed for operators of exponential type the
asymptotic relation

$\left|\left(S_{n}f\right)\left(x\right)-\sum_{s=0}^{2q}\mu_{n,s}\left(x\right)\frac{f^{\left(s\right)}\left(x\right)}{s!}\right|\leq
K_{m}\left(x\right)n^{-q}\omega\left(f^{\left(2q\right)},n^{-1/2}\right)\text{
}\qquad\left(n\in\mathbb{N}\right),$

where $K_{m}\left(x\right)$ is a constant independent of $n$, and $\omega$
denotes the ordinary modulus of continuity. We remark that in [5] these
expansions were used to improve the rate of convergence for operators of
exponential type by applying extrapolation techniques.

As our main result we show that under appropriate conditions the asymptotic
expansion $\left(\ref{asymptotic-expansion}\right)$ can be differentiated
term-by-term.

###### Theorem 1

Let $r\in\mathbb{N}_{0}$, $q\in\mathbb{N}$ and $x\in I$. Suppose that the
operators $S_{n}$ satisfy the differential equation $\left(\ref{ODE-
Holhos}\right)$ such that, for certain constants $\alpha,\beta$, holds
$0<\alpha<\lambda_{n}/n<\beta$. Then, for each function $f\in
C_{2\left(q+r\right),x}\left(I\right)$, the operators $S_{n}$ possess the
(pointwise) asymptotic expansion

$\left(S_{n}f\right)^{\left(r\right)}\left(x\right)=\sum_{s=0}^{2q}\left(\mu_{n,s}\left(x\right)\frac{f^{\left(s\right)}\left(x\right)}{s!}\right)^{\left(r\right)}+o\left(n^{-q}\right)\text{
}\qquad\left(n\rightarrow\infty\right),$ (13)

In the special case $q=1$, we obtain the following Voronovskaja-type formula.

###### Corollary 2

Let $r\in\mathbb{N}_{0}$ and $x\in I$. For each function $f\in
C_{2r+2,x}\left(I\right)$, the operators $S_{n}$ satisfying the differential
equation $\left(\ref{ODE-Holhos}\right)$ with $0<\alpha<\lambda_{n}/n<\beta$,
have the (pointwise) asymptotic relation

$\left(S_{n}f\right)^{\left(r\right)}\left(x\right)=\left(f\left(x\right)+\mu_{n,1}\left(x\right)f^{\prime}\left(x\right)+\mu_{n,2}\left(x\right)f^{\prime\prime}\left(x\right)\right)^{\left(r\right)}+o\left(n^{-q}\right)\text{
}\qquad\left(n\rightarrow\infty\right).$

Because operators of exponential type preserve linear functions, they
trivially satisfy the differential equation $\left(\ref{ODE-Holhos}\right)$.
Therefore, the conclusion of Theorem 1 is valid also for operators of
exponential type. Noting that they satisfy $\mu_{n,1}\left(x\right)=0$ and
$\mu_{n,2}=\varphi/n$, we obtain the following corollary.

###### Corollary 3

Let $S_{n}$ be operators of exponential type, $r\in\mathbb{N}_{0}$ and $x\in
I$. For each function $f\in C_{2r+2,x}\left(I\right)$, it holds

$\lim_{n\rightarrow\infty}n\left(\left(S_{n}f\right)^{\left(r\right)}\left(x\right)-f^{\left(r\right)}\left(x\right)\right)=\frac{1}{2}\left(\varphi\left(x\right)f\left(x\right)\right)^{\left(r\right)}.$

In the Bernstein case $\varphi\left(x\right)=x\left(1-x\right)$, Corollary 3
recovers Floater’s celebrated formula $\left(\ref{Floater-Bernstein}\right)$.

##  4 Auxiliary results and proofs

Firstly, we recall the following general approximation theorem of Sikkema [18,
Theorem 1]. It provides an asymptotic expansion for a sequence of positive
linear operators $L_{n}$ under a certain condition on the central moments
$\left(L_{n}\psi_{x}^{s}\right)\left(x\right)$.

###### Lemma 4

Let $I$ be a real interval, fix $x\in I$ and let $q\in\mathbb{N}$ be even.
Furthermore, let $L_{n}:C\left(I\right)\rightarrow C\left(I\right)$ be a
sequence of positive linear operators which are applicable to polynomials.
Suppose that

$\left(L_{n}\psi_{x}^{s}\right)\left(x\right)=O\left(n^{-\left\lfloor\left(s+1\right)/2\right\rfloor}\right)\text{
}\qquad\left(n\rightarrow\infty\right)\qquad\text{for }s=0,1,\ldots,q+2.$ (14)

Then, for each function $f\in C_{q,x}\left(I\right)$, possessing a derivative
of order $q$ at $x$, it holds

$\left(L_{n}f\right)\left(x\right)=\sum_{s=0}^{q}\frac{f^{\left(s\right)}\left(x\right)}{s!}\left(L_{n}\psi_{x}^{s}\right)\left(x\right)+o\left(n^{-q}\right)\text{
}\qquad\left(n\rightarrow\infty\right).$ (15)

Furthermore, if $f^{\left(q+2\right)}\left(x\right)$ exists, the term
$o\left(n^{-q}\right)$ in Eq. $\left(\ref{sikkema2}\right)$ can be replaced by
$O\left(n^{-\left(q+1\right)}\right)$.

Holhoş [12, Page 3] proved the following extension of the recursive formula
$\left(\ref{recursive-exponential-type}\right)$.

###### Lemma 5

For positive integers $s$, the central moments $\mu_{n,s}$ satisfy the
recursive formula

$\frac{\lambda_{n}}{\varphi}\left(\mu_{n,s+1}-\mu_{n,1}\mu_{n,s}\right)=s\mu_{n,s-1}+\mu_{n,s}^{\prime}.$

###### Lemma 6

For nonnegative integers $m$, it holds

$\left(S_{n}\left(\psi_{x}^{m}f\right)\right)^{\prime}\left(x\right)=\frac{\lambda_{n}}{\varphi\left(x\right)}\left(S_{n}\left(\psi_{x}^{m+1}f\right)\left(x\right)-\left(S_{n}\psi_{x}\right)\left(x\right)\left(S_{n}f\right)\left(x\right)\right)-m\left(S_{n}\left(\psi_{x}^{m-1}f\right)\right)\left(x\right).$

Proof of Lemma 6. The formula follows from the differential equation
$\left(\ref{ODE-Holhos}\right)$.

The next lemma directly follows by an application of the Leibniz rule for
derivatives.

###### Lemma 7

For nonnegative integers $m,s$, it holds
$\left(\psi_{x}^{m}f\right)^{\left(s\right)}\left(x\right)=\binom{s}{m}m!f^{\left(s-m\right)}\left(x\right)$,
in particular

$\left(\psi_{x}f\right)^{\left(s\right)}\left(x\right)=sf^{\left(s-1\right)}\left(x\right),$
(16)

Now we are in position to prove the main theorem.

Proof of Theorem 1. We show that, for each integer
$j\in\left\\{0,1,\ldots,r\right\\}$,

$\left(S_{n}f\right)^{\left(j\right)}\left(x\right)=\sum_{s=0}^{2\left(q+r-j\right)}\left(\mu_{n,s}\left(x\right)\frac{f^{\left(s\right)}\left(x\right)}{s!}\right)^{\left(j\right)}+o\left(n^{-\left(q+r-j\right)}\right)\text{
}\qquad\left(n\rightarrow\infty\right).$ (17)

By Sikkema’s theorem (Lemma 4) or Volkov [19, THEOREM 2], the formula is
valid, for $j=0$. The proof uses mathematical induction. Suppose that
$\left(\ref{assertion-j}\right)$ is valid, for $0\leq j\leq m$, with a certain
$m\in\left\\{0,1,\ldots,r-1\right\\}$. Then, by Lemma 5 and the Leibniz rule
for differentiation,

$\displaystyle\left(S_{n}f\right)^{\left(m+1\right)}\left(x\right)$
$\displaystyle=\left[\frac{\lambda_{n}}{\varphi}\left(S_{n}\left(\psi_{x}f\right)-\mu_{n,1}\left(S_{n}f\right)\right)\right]^{\left(m\right)}\left(x\right)$
$\displaystyle=\sum_{j=0}^{m}\binom{m}{j}\left(\frac{\lambda_{n}}{\varphi\left(x\right)}\right)^{\left(m-j\right)}\left(S_{n}\left(\psi_{x}f\right)-\mu_{n,1}\left(S_{n}f\right)\right)^{\left(j\right)}\left(x\right).$
(18)

For $0\leq j\leq m$, we conclude, by induction hypothesis and Lemma 7,

$\displaystyle\left(S_{n}\left(\psi_{x}f\right)\right)^{\left(j\right)}\left(x\right)$
$\displaystyle=$
$\displaystyle\sum_{s=0}^{2\left(q+r-j\right)}\left(\mu_{n,s}\frac{\left(\psi_{x}f\right)^{\left(s\right)}}{s!}\right)^{\left(j\right)}\left(x\right)+o\left(n^{-\left(q+r-j\right)}\right)$
$\displaystyle=$
$\displaystyle\sum_{s=0}^{2\left(q+r-j\right)-1}\left(\mu_{n,s+1}\frac{f^{\left(s\right)}}{s!}\right)^{\left(j\right)}\left(x\right)+o\left(n^{-\left(q+r-j\right)}\right)$
$\displaystyle=$
$\displaystyle\sum_{s=0}^{2\left(q+r-m\right)-1}\left(\mu_{n,s+1}\frac{f^{\left(s\right)}}{s!}\right)^{\left(j\right)}\left(x\right)+o\left(n^{-\left(q+r-m\right)}\right)\text{
}\qquad\left(n\rightarrow\infty\right),$

since
$\left(\mu_{n,s+1}\frac{f^{\left(s\right)}}{s!}\right)^{\left(j\right)}\left(x\right)=O\left(\mu_{n,s+1}\left(x\right)\right)=O\left(n^{-\left\lfloor\left(s+2\right)/2\right\rfloor}\right)=O\left(n^{-\left(q+r-m+1\right)}\right)$
as $n\rightarrow\infty$, if $s\geq 2\left(q+r-m\right)$. Furthermore, for
$0\leq j\leq m$, we have

$\displaystyle\left(\mu_{n,1}\left(S_{n}f\right)\right)^{\left(j\right)}\left(x\right)$
$\displaystyle=$
$\displaystyle\sum_{i=0}^{j}\binom{j}{i}\mu_{n,1}^{\left(j-i\right)}\left(x\right)\left(S_{n}f\right)^{\left(i\right)}\left(x\right)$
$\displaystyle=$
$\displaystyle\sum_{i=0}^{j}\binom{j}{i}\mu_{n,1}^{\left(j-i\right)}\left(x\right)\left(\sum_{s=0}^{2\left(q+r-i\right)}\left(\mu_{n,s}\frac{f^{\left(s\right)}}{s!}\right)^{\left(i\right)}\left(x\right)+o\left(n^{-\left(q+r-i\right)}\right)\right)$
$\displaystyle=$
$\displaystyle\sum_{s=0}^{2\left(q+r-j\right)}\sum_{i=0}^{j}\binom{j}{i}\mu_{n,1}^{\left(j-i\right)}\left(x\right)\left(\mu_{n,s}\frac{f^{\left(s\right)}}{s!}\right)^{\left(i\right)}\left(x\right)+o\left(n^{-\left(q+r-j\right)-1}\right)$
$\displaystyle=$
$\displaystyle\sum_{s=0}^{2\left(q+r-j\right)}\left(\mu_{n,1}\mu_{n,s}\frac{f^{\left(s\right)}}{s!}\right)^{\left(j\right)}\left(x\right)+o\left(n^{-\left(q+r-j+1\right)}\right)$
$\displaystyle=$
$\displaystyle\sum_{s=0}^{2\left(q+r-m\right)-1}\left(\mu_{n,1}\mu_{n,s}\frac{f^{\left(s\right)}}{s!}\right)^{\left(j\right)}\left(x\right)+o\left(n^{-\left(q+r-m\right)}\right)\text{
}\qquad\left(n\rightarrow\infty\right),$

since
$\left(\mu_{n,1}\mu_{n,s}\frac{f^{\left(s\right)}}{s!}\right)^{\left(i\right)}\left(x\right)=O\left(\mu_{n,1}\left(x\right)\mu_{n,s}\left(x\right)\right)=O\left(n^{-1}n^{-\left\lfloor\left(s+1\right)/2\right\rfloor}\right)=O\left(n^{-1}n^{-\left(q+r-m\right)}\right)=O\left(n^{-\left(q+r-m+1\right)}\right)$
as $n\rightarrow\infty$, if $s\geq 2\left(q+r-m\right)$. Inserting into Eq.
$\left(\ref{formula-Snf-derivative-m+1}\right)$ we obtain

$\displaystyle\left(S_{n}f\right)^{\left(m+1\right)}$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{m}\binom{m}{j}\left(\frac{\lambda_{n}}{\varphi\left(x\right)}\right)^{\left(m-j\right)}\left(\sum_{s=0}^{2\left(q+r-m\right)-1}\left(\mu_{n,s+1}-\mu_{n,1}\mu_{n,s}\right)\frac{f^{\left(s\right)}}{s!}\right)^{\left(j\right)}\left(x\right)$
$\displaystyle+\lambda_{n}\cdot o\left(n^{-\left(q+r-m\right)}\right)$

as $n\rightarrow\infty$. By the Leibniz rule for differentiation,

$\left(S_{n}f\right)^{\left(m+1\right)}\left(x\right)=\sum_{s=0}^{2\left(q+r-m\right)-1}\left(\frac{\lambda_{n}}{\varphi}\left(\mu_{n,s+1}-\mu_{n,1}\mu_{n,s}\right)\frac{f^{\left(s\right)}}{s!}\right)^{\left(m\right)}\left(x\right)+\lambda_{n}\cdot
o\left(n^{-\left(q+r-m\right)}\right)$

as $n\rightarrow\infty$. By the recursive formula for the central moments
(Lemma 5),

$\displaystyle\left(S_{n}f\right)^{\left(m+1\right)}\left(x\right)$
$\displaystyle=$
$\displaystyle\sum_{s=0}^{2\left(q+r-m\right)-1}\left(\left(s\mu_{n,s-1}+\mu_{n,s}^{\prime}\right)\frac{f^{\left(s\right)}}{s!}\right)^{\left(m\right)}\left(x\right)+\lambda_{n}\cdot
o\left(n^{-\left(q+r-m\right)}\right)$ $\displaystyle=$
$\displaystyle\sum_{s=0}^{2\left(q+r-m\right)-2}\left(\mu_{n,s}\frac{f^{\left(s+1\right)}}{s!}+\mu_{n,s}^{\prime}\frac{f^{\left(s\right)}}{s!}\right)^{\left(m\right)}\left(x\right)+\lambda_{n}\cdot
o\left(n^{-\left(q+r-m\right)}\right)$

as $n\rightarrow\infty$, since
$\left(\mu_{n,s}^{\prime}\frac{f^{\left(s\right)}}{s!}\right)^{\left(m\right)}\left(x\right)=O\left(\mu_{n,s}\left(x\right)\right)=O\left(n^{-\left\lfloor\left(s+1\right)/2\right\rfloor}\right)=O\left(n^{-\left(q+r-m\right)}\right)$,
for $s=2\left(q+r-m\right)-1$. Since
$\mu_{n,s}f^{\left(s+1\right)}+\mu_{n,s}^{\prime}f^{\left(s\right)}=\left(\mu_{n,s}f^{\left(s\right)}\right)^{\prime}$,
we obtain

$\left(S_{n}f\right)^{\left(m+1\right)}\left(x\right)=\sum_{s=0}^{2\left(q+r-\left(m+1\right)\right)}\left(\mu_{n,s}\frac{f^{\left(s\right)}}{s!}\right)^{\left(m+1\right)}\left(x\right)+o\left(n^{-\left(q+r-\left(m+1\right)\right)}\right)$

as $n\rightarrow\infty$. Here we used the condition
$0<\alpha<\lambda_{n}/n<\beta$. This completes the proof.

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