A Class of Integral Operators that Fix Exponential Functions

Abstract

In this paper we introduce a general class of integral operators that fix exponential functions, containing several recent modified operators of Gauss–Weierstrass, or Picard or moment type operators. Pointwise convergence theorems are studied, using a Korovkin-type theorem and a Voronovskaja-type formula is obtained.

1 Introduction

The classical Bohman–Korovkin theorem (see [10, 22, 23]) is one of the pivotal results of approximation theory and several convergence theorems known in literature employ this basic tool. It states that a sequence of positive linear operators $T_nf$ acting on the set of the continuous functions over a compact interval of the real line converges to the identity operator only if it converges on a finite number of test functions which form a so-called Chebyshev system. A complete treatment of the Korovkin theorem can be found in the monographies [2, 3].

In this respect, if a sequence of operators $T_nf$ is such that $T_n\varphi = \varphi$ for some continuous function $\varphi ,$ then to obtain the convergence appears very simple, if the functions $\varphi$ belong to a Chebyshev system. Thus, it is of interest to define sequences of operators that have this property. In literature the so-called King type operators, have this property, especially in case of discrete operators (see e.g. [21]). In this paper we define an entire class of positive linear integral operators which fix exponential functions. This kind of results were obtained by Agratini, Aral and Deniz [1], by Aral [6] and recently by Yilmaz, Uysal and Aral [26], by considering modifications of specific operators (Picard, Gauss–Weiertsrass and moment-type operators). All the operators considered in these papers are special case of our present theory. Our approach includes also all the integral operators having a compactly supported kernel. Thus it applies for example to spline-type operators [11, 24]. Related results can be found in [16, 18, 19] and [17].

For our operators, we apply the Gadjiev version of Bohman–Korovkin theorem (see [14, 15]) in case of unbounded domains, for functions belonging to certain weighted spaces of continuous functions. Setting for every $n \in {\mathbb {N}}$ and $a>0,$

$$\begin{aligned} A_{a,n}:= \int _{-\infty }^\infty \exp (at) K_n(t) \mathrm{d}t,\quad \lambda _n(x):= x - \frac{1}{a}\log A_{a,n}, \end{aligned}$$

we define

$$\begin{aligned} (T_nf)(x) = A_{a,n}\int _{-\infty }^\infty \exp (-at) f\left( \lambda _n(x) +t\right) K_n(t)\mathrm{d}t, \end{aligned}$$

where $\{K_n\}$ is a family of non-negative functions (kernel) belonging to a suitable function space. Setting for $a>0,$ $\exp _a (x) := e^{ax}$ for $x\in {\mathbb {R}},$ we show that $(T_n \exp _a)(x) = \exp (ax)$ and $(T_n \exp _{2a})(x) = \exp (2ax)$. Moreover we show that $T_ne_j \rightarrow e_j$ where $e_j(x) = x^j,$ $j=0,1,2,$ so obtaining two uniform convergence theorems in weighted spaces of continuous functions. Then, using certain moduli of continuity we obtain certain quantitative estimates of the convergence and finally a Voronovskaja-type asymptotic formula. These kinds of asymptotic formulae are very useful also for applications, especially for discrete operators, like e.g. sampling-type operators (see, e.g., [11,12,13]. For integral operators of Mellin type see [7, 8]). The last section is devoted to several examples.

We recall here, that general approaches to convergence of integral operators was recently given in [9], and more recently in [20] and [25] in the frame of nonlinear operators.

2 Basic Notations

Let us denote by ${\mathbb {R}}$ the set of real numbers, and by ${\mathbb {N}}$ the set of the positive integers. By $L^1({\mathbb {R}})$ we denote the space comprising all the Lebesgue integrable functions on ${\mathbb {R}}$ with respect to the Lebesgue measure, and by $L^\infty ({\mathbb {R}})$ the space comprising all the essentially bounded functions on ${\mathbb {R}}.$ By $C({\mathbb {R}})$ we denote the space of all the continuous functions defined on ${\mathbb {R}}.$ Finally, for $r \in {\mathbb {N}},$ we will say that a function $f: {\mathbb {R}} \rightarrow {\mathbb {R}}$ is locally of class $C^r$ at a point $x\in {\mathbb {R}}$ if there is a neighbourhood U of x such that f is $(r-1)$-fold continuously differentiable in U and $f^{(r)}(x)$ exists.

3 The Class of Integral Operators

For any $a>0,$ we define the function $\exp _a(t):= e^{at},$ for $t \in {\mathbb {R}}.$

We define now the function space

$$\begin{aligned} L_{\exp _a}({\mathbb {R}}):= \{g: {\mathbb {R}} \rightarrow {\mathbb {R}} : g(\cdot ) e^{a|\cdot |} \in L^1({\mathbb {R}}) \}. \end{aligned}$$

Let now $\{K_n\}$ be a non negative kernel, that is, for every $n \in {\mathbb {N}},$ $K_n(t) \ge 0,$ for every $t \in {\mathbb {R}},$ $K_n \in L^1({\mathbb {R}}),$ and

$$\begin{aligned} \Vert K_n\Vert _1 = \int _{-\infty }^\infty K_n(t)\mathrm{d}t = 1. \end{aligned}$$

In what follows we assume that $K_n \in L_{\exp _a}({\mathbb {R}}),$ for every sufficiently large $n \in {\mathbb {N}},$ namely for $n\ge n_0,$ with $n_0\in {\mathbb {N}}.$ We set, for $n\ge n_0,$

$$\begin{aligned} A_{a,n}:= \int _{-\infty }^\infty \exp _a(t) K_n(t)\mathrm{d}t,\qquad \lambda _n(x):= x - \frac{1}{a}\log A_{a,n}. \end{aligned}$$

Note that from the assumption $K_n \in L_{\exp _a}({\mathbb {R}}),$ we have $A_{a,n}< + \infty ,\, A_{-a,n} < +\infty$ and if moreover for every $n \in {\mathbb {N}},$ $K_n$ is an even function, that is $K_n(t) = K_n(-t),$ for every $t \in {\mathbb {R}},$ one has easily $A_{a,n} = A_{-a,n}.$ We introduce now the sequence of integral operators defined by

$$\begin{aligned} (T_nf)(x):= & {} A_{a,n} \int _{-\infty }^\infty \exp _{-a}(t) f(\lambda _n(x) + t) K_n(t)\mathrm{d}t\nonumber \\= & {} A_{a,n} \int _{-\infty }^\infty e^{-at} f(\lambda _n(x) +t) K_n(t)\mathrm{d}t, \end{aligned}$$

(3.1)

for every function f belonging to the domain $D:= \bigcap _{n \ge n_0} \mathrm{dom}\,T_n,$ where $\mathrm{dom}\,T_n$ denotes the set of all the Lebesgue measurable functions f such that $(T_n |f|)(x)$ is convergent for almost all $x \in {\mathbb {R}}.$

Note that if $f \in L^\infty ({\mathbb {R}}),$ then $f \in D.$ Moreover, the functions $\exp _a$ and $\exp _{2a}$ both belong to D. We have the following

Proposition 3.1

Under the established assumptions on the kernel $\{K_n\},$ we have

$$\begin{aligned} (T_n \exp _a)(x) = \exp _a(x),\qquad (T_n\exp _{2a})(x) = \exp _{2a}(x). \end{aligned}$$

(3.2)

Proof

We have

$$\begin{aligned} (T_n \exp _a)(x)&= A_{a,n} \int _{-\infty }^\infty e^{-at} e^{a(\lambda _n(x) +t)} K_n(t) \mathrm{d}t = A_{a,n} e^{a\lambda _n(x)}\int _{-\infty }^\infty K_n(t) \mathrm{d}t \\&=A_{a,n} e^{a\lambda _n(x)} = \exp _a(x). \end{aligned}$$

Analogously, we have

$$\begin{aligned} (T_n\exp _{2a})(x)&= A_{a,n} \int _{-\infty }^\infty e^{-at} e^{2a(\lambda _n(x) + t)} K_n(t) \mathrm{d}t \\&= A_{a,n} e^{2a \lambda _n(x)}\int _{-\infty }^\infty e^{at} K_n(t)\mathrm{d}t\\&= A_{a,n} \exp _{2a}(x) \frac{A_{a,n}}{(A_{a,n})^2} = \exp _{2a}(x), \end{aligned}$$

that is the assertion $\square$

Remark 3.2

In order to establish a connection with classical integral operators of convolution type of the form

$$\begin{aligned} (T^0_nf)(x) := \int _{-\infty }^\infty f(x+t) K_n(t) \mathrm{d}t, \end{aligned}$$

for continuous and bounded functions $f \in D$ we have

$$\begin{aligned} \lim _{a \rightarrow 0^+} (T_nf)(x) = (T^0_nf)(x). \end{aligned}$$

whenever

$$\begin{aligned} \lim _{a \rightarrow 0^+} \frac{1}{a} \log A_{a,n} = 0. \end{aligned}$$

Indeed, under the above assumptions on the kernel $\{K_n\},$ it is easy to show that $\lim _{a \rightarrow 0^+} A_{a,n} = 1,$ and $\lim _{a \rightarrow 0^+} \lambda _n(x) = x.$ Thus under the assumptions on the function f the assertion follows by the Lebesgue theorem of dominated convergence.

4 Pointwise and Uniform Convergence

In this section we will study the pointwise convergence of $T_nf$ to f,  where f belongs to a suitable weighted space of continuous functions, as introduced in [14], using a Korovkin-type theorem, established in [14] (see also [15]). In order to do that, we introduce the following constants

$$\begin{aligned} B_{-a,n}:= & {} \int _{-\infty }^\infty t e^{-at} K_n(t) \mathrm{d}t\\ C_{-a,n}:= & {} \int _{-\infty }^\infty t^2 e^{-at} K_n(t) \mathrm{d}t. \end{aligned}$$

We call these constants the exponential moments of orders 1 and 2 respectively. At the same way, we refer to $A_{-a,n}$ as the exponential moment of order 0. We introduce the following subspace of $L_{\exp _a}({\mathbb {R}}),$ in which the above moments are well-defined

$$\begin{aligned} L^*_{\exp _a}({\mathbb {R}}):= \{g \in L_{\exp _a}({\mathbb {R}}): (c + d|\cdot |)^2 e^{a|\cdot |}g(\cdot ) \in L^1({\mathbb {R}})\}, \end{aligned}$$

with c, d positive constants. If $K_n \in L^*_{\exp _a}({\mathbb {R}}),$ then $B_{-a,n}, \,C_{-a,n}$ exist finite.

Let us introduce now the test functions $e_0(t) = 1,\,e_1(t) = t,\,e_2(t) = t^2,\,\,t \in {\mathbb {R}}.$ Obviously, $e_j \in D,$ for $j=0,1,2.$

We have the following

Proposition 4.1

Let $K_n \in L^*_{\exp _a}({\mathbb {R}}),$ for sufficiently large $n \in {\mathbb {N}}.$ Then we have

$$\begin{aligned} (T_n e_0)(x)= & {} A_{a,n}\,A_{-a,n}\\ (T_ne_1)(x)= & {} A_{a,n}\,A_{-a,n} x - \frac{1}{a}A_{a,n}\,A_{-a,n} \log A_{a,n} + A_{a,n}\,B_{-a,n} \end{aligned}$$

and

$$\begin{aligned} (T_ne_2)(x)&= A_{a,n}\,A_{-a,n} x^2 + \left( 2A_{a,n}\,B_{-a,n} -2A_{a,n}\,A_{-a,n} \frac{1}{a}\log A_{a,n}\right) x\\&\quad + A_{a,n}\,A_{-a,n} \frac{1}{a^2}\log ^2 A_{a,n} + A_{a,n}\,C_{-a,n} - 2A_{a,n}\,B_{-a,n} \frac{1}{a} \log A_{a,n}. \end{aligned}$$

Proof

We have

$$\begin{aligned} (T_ne_0)(x) = A_{a,n} \int _{-\infty }^\infty e^{-at} K_n(t) \mathrm{d}t = A_{a,n}\,A_{-a,n}. \end{aligned}$$

Next,

$$\begin{aligned} (T_ne_1)(x)&=A_{a,n}\int _{-\infty }^\infty e^{-at} (\lambda _n(x) + t) K_n(t) \mathrm{d}t\\&= A_{a,n}\,\lambda _n(x) \int _{-\infty }^\infty e^{-at} K_n(t) \mathrm{d}t + A_{a,n}\int _{-\infty }^\infty t e^{-at} K_n(t) \mathrm{d}t \\&= A_{a,n}\,A_{-a,n} x - A_{a,n}\,A_{-a,n} \frac{1}{a} \log A_{a,n} + A_{a,n}\,B_{-a,n}. \end{aligned}$$

Finally,

$$\begin{aligned} (T_ne_2)(x)&= A_{a,n} \int _{-\infty }^\infty e^{-at} (\lambda _n(x) + t)^2 K_n(t) \mathrm{d}t \\&= A_{a,n}\,(\lambda _n(x))^2 A_{-a,n} + A_{a,n}\,C_{-a,n} + 2 A_{a,n}\,B_{-a,n} \lambda _n(x) \\&= A_{a,n}\,A_{-a,n} x^2 + 2 A_{a,n} \left( B_{-a,n} - \frac{1}{a} A_{-a,n} \log A_{a,n}\right) x \\&\quad + A_{a,n}\left( A_{-a,n} \frac{1}{a^2} \log ^2 A_{a,n} + C_{-a,n} - \frac{2}{a} B_{-a,n} \log A_{a,n}\right) . \end{aligned}$$

The proof is completed $\square$

Corollary 4.2

Under the assumptions of Proposition 4.1, if moreover

$$\begin{aligned} \lim _{n \rightarrow +\infty } A_{a,n} = \lim _{n \rightarrow +\infty } A_{-a,n} = 1, \end{aligned}$$

(4.1)

and

$$\begin{aligned} \lim _{n \rightarrow +\infty } B_{-a,n} = \lim _{n \rightarrow +\infty } C_{-a,n} = 0, \end{aligned}$$

then

$$\begin{aligned} \lim _{n \rightarrow +\infty } (T_n e_j)(x) = e_j(x) \qquad (j=0,1,2). \end{aligned}$$

Proof

It is an immediate consequence of Proposition 4.1$\square$

Using Proposition 4.1 and Corollary 4.2, we now give a uniform convergence result for the operators $(T_nf)$ when f belongs to suitable weighted spaces of continuous functions. The key tool is a Korovkin-type theorem proved in [14] (see also [15]).

Given a continuous, strictly increasing function $\varphi : {\mathbb {R}} \rightarrow {\mathbb {R}},$ we define the function $\rho (x) := 1 + \varphi ^2(x),$ and assume that $\lim _{x \rightarrow \pm \infty } \rho (x) = +\infty .$

We will consider two particular cases: $\varphi _1(x) = x,$ and $\varphi _2(x) = \exp _a(x),$ and we set $\rho _1(x) := 1 + x^2$ and $\rho _2(x) := 1 + \exp _{2a}(x).$

Let us consider the spaces, for $j=1,2$

$$\begin{aligned} C^0_{\rho _j}({\mathbb {R}}):= \left\{ f\in C({\mathbb {R}}): \lim _{x \rightarrow \pm \infty }\frac{f(x)}{\rho _j(x)} =\ell _f \in {\mathbb {R}} \right\} . \end{aligned}$$

Note that if $f \in C^0_{\rho _j}({\mathbb {R}}),$ then $|f(x)| \le M_f \rho _j(x),$ for a suitable constant $M_f >0$ and $x \in {\mathbb {R}}.$ We define a norm on the space $C^0_{\rho _j}({\mathbb {R}})$ on setting

$$\begin{aligned} \Vert f\Vert _{\rho _j}:= \left\| \frac{f}{\rho _j}\right\| _\infty . \end{aligned}$$

We are ready to prove the main theorem of this section.

Theorem 4.3

Let $j=1,2$ and let $f \in C^0_{\rho _j}({\mathbb {R}}).$ Then, under the assumptions of Proposition 4.1 and Corollary 4.2, we have

$$\begin{aligned} \lim _{n \rightarrow +\infty } \Vert T_nf - f\Vert _{\rho _j} = 0. \end{aligned}$$

Proof

First, consider the case $j=1.$ The test functions $e_j$ obviously belong to $C^0_{\rho _1}({\mathbb {R}})$ and moreover

$$\begin{aligned} \left\| \frac{T_ne_0 - e_0}{\rho _1}\right\| _\infty = \left\| \frac{A_{a,n}\,A_{-a,n} - 1}{\rho _1}\right\| _\infty \le |A_{a,n}\,A_{-a,n} - 1|, \end{aligned}$$

(4.2)

and the last term tends to 0 as $n \rightarrow +\infty .$ Next,

$$\begin{aligned} \left\| \frac{T_ne_1 - e_1}{\rho _1}\right\| _\infty&\le |A_{a,n}\,A_{-a,n} - 1| \sup _{x \in {\mathbb {R}}}\frac{|x|}{x^2+1} \\&\quad +\frac{1}{a} A_{a,n}\,A_{-a,n} |\log A_{a,n}| + A_{a,n}\,|B_{-a,n}|, \end{aligned}$$

and so

$$\begin{aligned} \lim _{n \rightarrow +\infty } \left\| \frac{T_ne_1 - e_1}{\rho _1}\right\| _\infty = 0. \end{aligned}$$

Analogously, one can see that

$$\begin{aligned} \lim _{n \rightarrow +\infty }\left\| \frac{T_ne_2 - e_2}{\rho _1}\right\| _\infty = 0. \end{aligned}$$

For $j=2,$ taking into account of (4.2) and Proposition 3.1, we obtain again the convergence on the test functions $\exp ^k(ax),$ for $k=0,1,2.$ Applying Theorem 2 in [14] we obtain the assertion $\square$

5 Quantitative Estimates

For $K_n \in L^*_{\exp _a}({\mathbb {R}}),$ we define the absolute exponential moment of order 1 of $K_n$ as

$$\begin{aligned} {\widetilde{B}}_{-a,n}:= \int _{-\infty }^\infty e^{-at} |t| K_n(t)\mathrm{d}t. \end{aligned}$$

In the spaces $C^0_{\rho _j}({\mathbb {R}}),$ $j=1,2$ we can define various modulus of continuity. We begin with an estimate of the convergence expressed by Theorem 4.3, in terms of the classical modulus of continuity $\omega$ defined by

$$\begin{aligned} \omega (f, \delta ) := \sup _{|h| \le \delta }|f(x+h) - f(x)|\qquad (\delta >0). \end{aligned}$$

As it is well-known, for any uniformly continuous function f one has $\lim _{\delta \rightarrow 0}\omega (f,\delta ) = 0.$ We have the following estimate

Theorem 5.1

Let $K_{n}\in L^*_{\exp _a}({\mathbb {R}}),$ for sufficiently large values of $n\in {\mathbb {N}},$ and let $f\in C^0_{\rho _{j}} ( {\mathbb {R}}).$ Then for every $\delta >0,$ we have

$$\begin{aligned} \left\| T_{n}f-f\right\| _{\rho _{j}}&\le \omega \left( f,\delta \right) A_{a,n}\left( A_{-a,n}+\frac{1}{\delta }\overset{\sim }{B}_{-a,n}+\frac{1}{a\delta }A_{-a,n}\left| \log A_{a,n}\right| \right) \\&\quad +\left\| f\right\| _{\rho _{j}}\left| A_{a,n}A_{-a,n}-1\right| . \end{aligned}$$

Proof

For every $x\in {\mathbb {R}}$ and $\delta >0$ we have

$$\begin{aligned} \left| \left( T_{n}f\right) \left( x\right) -f\left( x\right) \right|&\le A_{a,n} \int ^{\infty }_{-\infty } e^{-at}\omega \left( f,\left| t-\frac{1}{a}\log A_{a,n}\right| \right) K_{n}\left( t\right) \mathrm{d}t \\&\quad + \left| f\left( x\right) \right| \left| A_{a,n}A_{-a,n}-1\right| =:I_{1}+I_{2}. \end{aligned}$$

Thus we have to estimate only $I_{1}.$ In order to do that, we use the following well-known property of $\omega$ (see e.g. [4])

$$\begin{aligned} \omega \left( f,\lambda \delta \right) \le \left( 1+\lambda \right) \omega \left( f,\delta \right) \quad \left( \lambda ,\delta >0\right) . \end{aligned}$$

Setting $\lambda :=\frac{\left| t-\frac{1}{a}\log A_{a,n}\right| }{ \delta },$ we have

$$\begin{aligned} I_{1}&\le A_{a,n}\omega \left( f,\delta \right) \int ^{\infty }_{-\infty } e^{-at}\left( 1+\frac{\left| t-\frac{1}{a} \log A_{a,n}\right| }{\delta }\right) K_{n}\left( t\right) \mathrm{d}t \\&\le \omega \left( f,\delta \right) A_{a,n}\left( A_{-a,n}+\frac{1}{\delta } \int ^{\infty }_{-\infty } e^{-at}\left( \left| t\right| + a^{-1} \left| \log A_{a,n}\right| \right) K_{n}\left( t\right) \mathrm{d}t\right) \\&\le \omega \left( f,\delta \right) A_{a,n}\left( A_{-a,n}+\frac{1}{\delta }\overset{\sim }{B}_{-a,n}+\frac{1}{a\delta }A_{-a,n}\left| \log A_{a,n}\right| \right) . \end{aligned}$$

Passing to norm, we get the desired result

$$\begin{aligned} \left\| T_{n}f-f\right\| _{\rho _{j}}&\le \omega \left( f,\delta \right) A_{a,n}\left( A_{-a,n}+\frac{1}{\delta }\overset{\sim }{B}_{-a,n}+\frac{1}{a\delta }A_{-a,n}\left| \log A_{a,n}\right| \right) \\&\quad +\left\| f\right\| _{\rho _{j}}\left| A_{a,n}A_{-a,n}-1\right| \,\,\,\, \end{aligned}$$

$\square$

Corollary 5.2

Let the assumptions of Theorem 5.1 be satisfied, and (4.1) holds. If moreover there is $\alpha >0$ such that $(n^\alpha {\widetilde{B}}_{-a,n})_n,$ and $(n^\alpha |\log A_{a,n}|)_n$ are bounded sequences, then there exists an absolute constant $M>0$ such that

$$\begin{aligned} \Vert T_nf - f\Vert _{\rho _j} \le M\omega \left( f, \frac{1}{n^\alpha }\right) + |A_{a,n}\,A_{-a,n} -1|\,\Vert f\Vert _{\rho _j}, \end{aligned}$$

for sufficiently large values of $n \in {\mathbb {N}}.$

Proof

It is a consequence of Theorem 5.1 on setting $\delta = n^{-\alpha } \,\,$ $\square$

Remark 5.3

Note that, under the assumptions of Corollary 5.2, if the function f satisfies a Lipschitz condition of order 1, and $|A_{a,n}\,A_{-a,n} -1|= {{\mathcal {O}}}(n^{-\alpha })$ as $n \rightarrow +\infty ,$ then

$$\begin{aligned} \Vert T_nf - f\Vert _{\rho _j} = {{\mathcal {O}}}(n^{-\alpha }) \qquad (n \rightarrow +\infty ). \end{aligned}$$

Now we state another estimate using a suitable weighted modulus of continuity. In order to do that, we introduce the exponential moment of order 4 of $K_{n} \in L^*_{\exp _{a}}({\mathbb {R}})$ setting

$$\begin{aligned} E_{-a,n}:=\int _{-\infty }^{\infty } e^{-at}t^{4}K_{n}(t)\mathrm{d}t. \end{aligned}$$

In order to establish rate of convergence, we will use special kind of modulus of continuity ${\widetilde{\omega }}$ which is compatible with the space $C_{\rho _{1}}^{0} ({\mathbb {R}})$. This weighted modulus of continuity was first introduced in [27] for $f\in C_{\rho _{1}}^{0} ({\mathbb {R}} _{0}^{+} )$, then considered in [5] for $f\in C_{\rho _{1}}^{0} ({\mathbb {R}})$ as follows:

$$\begin{aligned} {\widetilde{\omega }} ( f,\delta ) = \sup _{x \in {\mathbb {R}} ,|h|\le \delta }\frac{| f(x+h)-f(x)| }{1+(h+x)^{2}}. \end{aligned}$$

(5.1)

For any function $f\in C_{\rho _{1}}^{0} ( {\mathbb {R}}),$ $m\in {\mathbb {N}}$ and $\lambda ,\delta \in {\mathbb {R}} ^{+},$ ${\widetilde{\omega }}$ has the following properties (see [5, 27]):

• 1. ${\widetilde{\omega }}\left( f,\delta \right)$ is a monotonically increasing function of $\delta ,$

• 2. $\lim _{\delta \rightarrow 0^{+}}{\widetilde{\omega }}\left( f,\delta \right) =0,$

• 3. ${\widetilde{\omega }}\left( f,m\delta \right) \le m \widetilde{\omega }\left( f,\delta \right) ,$ for every $m\in {\mathbb {N}},$

• 4. ${\widetilde{\omega }}\left( f,\lambda \delta \right) \le \left( 1+\lambda \right) {\widetilde{\omega }}\left( f,\delta \right)$ for every $\lambda >0.$

Theorem 5.4

Let $K_n \in L^*_{\exp _a}$ be such that $E_{-a,n}$ exists finite. For $f\in C_{\rho _{1}}^{0} ( {\mathbb {R}})$, there holds

$$\begin{aligned}&\left\| T_{n}f-f\right\| _{\rho _{1}} \\&\quad \le 16 A_{a,n} {\widetilde{\omega }}\left( f,\sqrt{C_{-a,n}}\right) \\&\qquad \times \left\{ C_{-a,n}+\left( 1+C_{-a,n}\right) A_{-a,n}+\sqrt{ E_{-a,n}+A_{-a,n}\left( 1+ C^2_{-a,n}\right) } \right\} \\&\qquad +\left\| f\right\| _{\rho _{1}}\left| A_{a,n}A_{-a,n}- 1 \right| \end{aligned}$$

provided that $\left| a^{-1}\log A_{a,n}\right| \le \sqrt{C_{-a,n}}$ for sufficiently large $n\in {\mathbb {N}}.$

Proof

In view of Proposition 4.1, we have

$$\begin{aligned} \left| \left( T_{n}f\right) \left( x\right) -f(x)\right|&\le A_{a,n}\left| \int ^{\infty }_{-\infty } e^{-at}f\left( \lambda _{n}\left( x\right) +t\right) K_{n}\left( t\right) \mathrm {d}t-f(x) \int ^{\infty }_{-\infty } e^{-at}K_{n}\left( t\right) \mathrm {d}t\right| \\ {}&\quad +\left| f(x)\right| \left| (T_{n} e_{0})\left( x\right) -1\right| \\ {}&=A_{a,n}\left| \int ^{\infty }_{-\infty } \left[ f\left( \lambda _{n}\left( x\right) +t\right) -f(x)\right] e^{-at}K_{n}\left( t\right) \mathrm {d}t\right| \\ {}&\quad +\left| f(x)\right| \left| A_{a,n}A_{-a,n} -1\right| . \end{aligned}$$

For any $\delta >0,$ properties of ${\widetilde{\omega }}$ enable us to write

$$\begin{aligned}&\left| f\left( \lambda _{n}\left( x\right) +t\right) -f(x)\right| =\left| f\left( x+t- {a}^{-1} \log A_{a,n}\right) -f\left( x\right) \right| \\&\quad \le \left( 1+\left( x+t- {a}^{-1} \log A_{a,n}\right) ^{2}\right) {\widetilde{\omega }}\left( f,\left| t-a^{-1}\log A_{a,n}\right| \right) \\&\quad \le \left( 1+\left( x+ t- {a}^{-1} \log A_{a,n}\right) ^{2}\right) \left( 1+\frac{\left| t-a^{-1}\log A_{a,n}\right| }{\delta }\right) {\widetilde{\omega }}\left( f,\delta \right) . \end{aligned}$$

In view of this observation, there holds

$$\begin{aligned}&\left| ( T_{n}f ) (x) - f(x) \right| \le {\widetilde{\omega }} ( f, \delta ) A_{a,n}\\&\quad \times \int ^\infty _{-\infty } \left( 1+\left( x+t- {a}^{-1} \log A_{a,n}\right) ^{2}\right) \left( 1+\frac{\left| t-a^{-1}\log A_{a,n}\right| }{\delta }\right) e^{-at} K_{n} ( t ) \mathrm{d}t \\&\quad +\left| f(x)\right| \left| A_{a,n}A_{-a,n} -1\right| \le {\widetilde{\omega }} \left( f,\delta \right) A_{a,n}\\&\quad \times \int ^\infty _{-\infty } \left( 1+\left( x+t-\frac{1}{a}\log A_{a,n}\right) ^{2}\right) \left( 1+\frac{\left| a^{-1}\log A_{a,n}\right| }{\delta }\right) e^{-at}K_{n}\left( t\right) \mathrm{d}t \\&\quad +{\widetilde{\omega }}\left( f,\delta \right) A_{a,n} \int ^\infty _{-\infty } \left( 1+\left( x+t-\frac{1}{a}\log A_{a,n}\right) ^{2}\right) \frac{\left| t\right| }{\delta } e^{-at}K_{n}\left( t\right) \mathrm{d}t \\&\quad +\left| f(x)\right| \left| A_{a,n}A_{-a,n} -1\right| := J_{1}+J_{2}+\left| f(x)\right| \left| A_{a,n}A_{-a,n} -1\right| . \end{aligned}$$

For $J_{1},$ using the well-known inequality $\left| a-b\right| ^{p}\le 2^{p-1}\left( \left| a\right| ^{p}+\left| b\right| ^{p}\right) ,$ $a,b \in {\mathbb {R}},$ $p\ge 1$, we have

$$\begin{aligned} J_{1}&\le {\widetilde{\omega }}\left( f,\delta \right) A_{a,n}\\&\quad \times \left( 1+ \frac{\left| a^{-1}\log A_{a,n}\right| }{\delta }\right) \left\{ A_{-a,n}+2C_{-a,n}+2\left( x-a^{-1}\log A_{a,n}\right) ^{2}A_{-a,n}\right\} \\&\le 4A_{a,n}{\widetilde{\omega }}\left( f,\delta \right) \left( 1+ \frac{\left| a^{-1}\log A_{a,n}\right| }{\delta }\right) \left\{ C_{-a,n}+\left( 1+x^{2}+a^{-2}\log ^{2}A_{a,n}\right) A_{-a,n}\right\} . \end{aligned}$$

By the Cauchy–Schwarz inequality for $J_2,$ we have

$$\begin{aligned} J_2&= {\widetilde{\omega }} ( f,\delta ) A_{a,n} \int ^\infty _{-\infty } \left( 1+ ( x + t- a^{-1} \log A_{a,n} )^2 \right) \frac{ |t | }{\delta } e^{-at}K_n (t ) \mathrm{d}t \\&\le {\widetilde{\omega }} (f,\delta ) A_{a, n} \int _{-\infty }^\infty ( 1+ (x+t - a^{-1} \log A_{a,n} )^2 ) ^2 e^{-at} K_n (t) \mathrm{d}t )^{1/2} \frac{(C_{-a, n})^{1/2}}{\delta }. \end{aligned}$$

Repeated application of the inequality $\left| a-b\right| ^{p}\le 2^{p-1}\left( \left| a\right| ^{p}+\left| b\right| ^{p}\right) ,$ $a,b \in {\mathbb {R}},$ $p\ge 1$ yields

$$\begin{aligned}&\left( 1+\left( x+t- a^{-1}\log A_{a,n}\right) ^{2}\right) ^{2} \le \left( 1+2\left( t^{2}+\lambda _{n}^2 ( x) \right) \right) ^{2}\\&\quad \le 2\left( 1+4\left( t^{2}+\lambda _{n}^{2}\left( x\right) \right) ^{2}\right) \le 2+16\left( t^{4}+\lambda _{n}^{4}\left( x\right) \right) \\&\quad =2+16t^{4}+16\left( x-a^{-1}\log A_{a,n}\right) ^{4} \\&\quad \le 2+ 16 t^{4} + 128 \left( x^{4}+a^{-4}\log ^{4}A_{a,n}\right) . \end{aligned}$$

Using above inequality, we have

$$\begin{aligned} J_{2}\le 16 {\widetilde{\omega }}\left( f,\delta \right) A_{a,n}\left( E_{-a,n}+A_{-a,n}\left( 1+x^{4}+a^{-4}\log ^{4}A_{a,n}\right) \right) ^{1/2} \frac{(C_{-a, n})^{1/2}}{\delta }. \end{aligned}$$

Collecting all inequalities we get

$$\begin{aligned}&\left| \left( T_{n}f\right) (x) -f(x) \right| \\ {}&\quad \le 4 A_{a,n}{\widetilde{\omega }}\left( f,\delta \right) \left( 1+ \frac{\left| a^{-1}\log A_{a,n}\right| }{\delta }\right) \left\{ C_{-a,n}+\left( 1+x^{2}+a^{-2}\log ^{2}A_{a,n}\right) A_{-a,n}\right\} \\&\qquad + 16 {\widetilde{\omega }}\left( f,\delta \right) A_{a,n}\left( E_{-a,n}+A_{-a,n}\left( 1+x^{4}+a^{-4}\log ^{4}A_{a,n}\right) \right) ^{1/2} \frac{(C_{-a, n})^{1/2}}{\delta } \\ {}&\qquad +\left| f(x)\right| \left| A_{a,n}A_{-a,n} -1\right| . \end{aligned}$$

Choosing $\delta =\left( C_{-a,n}\right) ^{1/2}$ with $\left| a^{-1}\log A_{a,n}\right| \le \delta$ for sufficiently large n, dividing both sides by $1+x^{2}$ and taking supremum over all x,  we obtain

$$\begin{aligned}&\left\| T_{n}f-f\right\| _{\rho _{1}} \\ {}&\quad \le 16A_{a,n}\overset{\sim }{\omega }\left( f,\sqrt{C_{-a,n}}\right) \left\{ C_{-a,n}+\left( 1+a^{-2}\log ^{2}A_{a,n}\right) A_{-a,n}\right. \\&\qquad \left. + \sqrt{ E_{-a,n}+A_{-a,n}\left( 1+a^{-4}\log ^{4}A_{a,n}\right) } \right\} +\left| f(x)\right| \left| A_{a,n}A_{-a,n}-1\right| \\ {}&\quad \le 16A_{a,n}\overset{\sim }{\omega }\left( f,\sqrt{C_{-a,n}}\right) \left\{ C_{-a,n}+\left( 1+C_{-a,n}\right) A_{-a,n}\right. \\ {}&\qquad \left. + \sqrt{ E_{-a,n}+A_{-a,n}\left( 1+\left( C_{-a,n}\right) ^{2}\right) } \right\} \\ {}&\qquad +\left\| f\right\| _{\rho _{1}}\left| A_{a,n}A_{-a,n}-1\right| \,\,\, \end{aligned}$$

that is the assertion $\square$

Remark 5.5

The estimate of Theorem 5.4 gives an evaluation of the convergence in terms of the modulus of continuity with parameter $\sqrt{C_{-a,n}},$ under the further assumptions that $C_{-a,n} \rightarrow 0$ and $|A_{a,n}\,A_{-a,n} - 1| \rightarrow 0$ as $n \rightarrow +\infty .$ Thus one can obtain a corresponding corollary, as for Theorem 5.1 (see Corollary 5.2 and Remark 5.3).

6 An Asymptotic Formula

Here we establish two asymptotic formulae of Voronovskaya type for functions $f \in C^0_{\rho _j}({\mathbb {R}}),$ $j=1,2,$ which are locally of class $C^2$ at a point x. These kinds of formulae give an exact evaluation of the order of pointwise convergence. We examine the case $j=1.$

Theorem 6.1

Let $f \in C^0_{\rho _1}({\mathbb {R}})$ be locally of class $C^2$ at a point $x \in {\mathbb {R}}.$ Let $K_n \in L^*_{\exp _a}({\mathbb {R}}),$ for sufficiently large values of $n \in {\mathbb {N}},$ be a kernel satisfying the assumptions of Corollary 4.2. Assume that there exist $\alpha >0$ such that the kernel $\{K_n\}$ satisfies further the following conditions:

• (i) $\lim _{n \rightarrow +\infty } n^\alpha (A_{a,n}\,A_{-a,n} - 1) = \ell _0,$  $\lim _{n \rightarrow +\infty } n^\alpha B_{-a,n} = \ell _1,$  $\lim _{n \rightarrow +\infty } n^\alpha C_{-a,n} = \ell _2$ and $\lim _{n \rightarrow +\infty } n^\alpha \frac{1}{a}\log A_{a,n} = \ell _3,$ with $\ell _j \in {\mathbb {R}},$ for $j=0,1,2,3.$

• (ii) For every $\eta >0,$

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n^\alpha \int _{|t| \ge \eta } e^{-at}\left( t - \frac{1}{a}\log A_{a,n}\right) ^2 K_n(t)\mathrm{d}t = 0. \end{aligned}$$

Then,

$$\begin{aligned} \lim _{n \rightarrow +\infty } n^\alpha [(T_nf)(x) - f(x)] = \ell _0 f(x) + (\ell _1 - \ell _3) f'(x) + \frac{\ell _2}{2} f''(x). \end{aligned}$$

(6.1)

Proof

Since f has a polynomial growth of order 2,  and locally of class $C^2$ at the point x,  using a local Taylor formula of the second order, we can write

$$\begin{aligned} f(\lambda _n(x) + t)= & {} f\left( x - \frac{1}{a}\log A_{a,n} + t\right) = f(x) + f'(x) \left( t - \frac{1}{a}\log A_{a,n}\right) \nonumber \\&+ \frac{f''(x)}{2} \left( t - \frac{1}{a}\log A_{a,n}\right) ^2 + r_x \left( t - \frac{1}{a}\log A_{a,n}\right) \nonumber \\&\times \left( t - \frac{1}{a}\log A_{a,n}\right) ^2, \end{aligned}$$

(6.2)

where $r_x(y)$ is a bounded function such that $\lim _{y \rightarrow 0} r_x(y) = 0.$

We write

$$\begin{aligned} (T_nf)(x) - f(x)&= A_{a,n}\int _{-\infty }^\infty e^{-at} [f(\lambda _n(x) + t) - f(x)] K_n(t) \mathrm{d}t \\&\quad + (A_{a,n}\,A_{-a,n} - 1) f(x) =: I_1 + I_2. \end{aligned}$$

As to $I_2$ we have by assumptions that $n^\alpha I_2 \rightarrow \ell _0 f(x)$ as $n \rightarrow +\infty .$ Therefore we evaluate now the term $I_1.$ Inserting (6.2) in $I_1$, we can write

$$\begin{aligned} I_1= & {} A_{a,n} \left( B_{-a,n} - A_{-a,n} \frac{1}{a} \log A_{a,n}\right) f'(x) \nonumber \\&+ A_{a,n} \left( C_{-a,n} + \frac{A_{-a,n}}{a^2} \log ^2 A_{a,n} - \frac{2}{a} B_{-a,n} \log A_{a,n}\right) \frac{f''(x)}{2} + R_x \nonumber \\=: & {} I_1^1 + I_1^2 + R_x, \end{aligned}$$

(6.3)

where

$$\begin{aligned} R_x:= A_{a,n} \int _{-\infty }^\infty e^{-at} r_x\left( t- \frac{1}{a} \log A_{a,n}\right) \left( t - \frac{1}{a} \log A_{a,n}\right) ^2 K_n(t) \mathrm{d}t. \end{aligned}$$

Now,

$$\begin{aligned} \lim _{n \rightarrow +\infty } n^\alpha I_1^1 = (\ell _1 - \ell _3) f'(x),\quad \lim _{n \rightarrow +\infty } n^\alpha I_1^2 = \frac{\ell _2}{2}f''(x). \end{aligned}$$

Thus we have to estimate the remainder term $R_x.$

Since $\lim _{y \rightarrow 0}r_x(y) = 0,$ given an arbitrary $\varepsilon >0,$ there is $\delta \in ]0,1[$ such that $|r_x(y)| < \varepsilon$ whenever $|y| \le \delta .$ We take now an index ${\overline{n}}$ such that for $n \ge {\overline{n}},$ one has $a^{-1}|\log A_{a,n}| < \delta /2.$ Thus for $|t| < \delta /2,$ we have also $|t- a^{-1} \log A_{a,n}| < \delta .$ Thus

$$\begin{aligned} \left| r_x\left( t - \frac{1}{a}\log A_{a,n}\right) \right|< \varepsilon \qquad (|t| < \delta /2). \end{aligned}$$

(6.4)

Writing

$$\begin{aligned} R_x&= A_{a,n}\left\{ \int _{-\delta /2}^{\delta /2} + \int _{|t| \ge \delta /2} \right\} e^{-at} r_x\left( t - \frac{1}{a}\log A_{a,n}\right) \,\left( (t - \frac{1}{a}\log A_{a,n}\right) ^2 K_n(t)\mathrm{d}t \\&=: R_1 + R_2, \end{aligned}$$

we have, by (6.4),

$$\begin{aligned} |R_1| \le A_{a,n} \varepsilon \left( C_{-a,n} + A_{-a,n}\frac{1}{a^2} \log ^2 A_{a,n} - \frac{2 B_{-a,n}}{a}\log A_{a,n}\right) . \end{aligned}$$

Therefore using the assumptions (i) we have, for a suitable absolute positive constant M, 

$$\begin{aligned} \limsup _{n \rightarrow +\infty } n^\alpha |R_1| \le M \varepsilon . \end{aligned}$$

As to the term $R_2$, using the boundedness of $r_x$ we can write

$$\begin{aligned} |R_2| \le A_{a,n} \Vert r_x\Vert _\infty \int _{|t| \ge \delta /2} e^{-at} \left( t - \frac{1}{a}\log A_{a,n}\right) ^2 K_n(t)\mathrm{d}t, \end{aligned}$$

and by (ii) we obtain

$$\begin{aligned} \lim _{n \rightarrow +\infty } n^\alpha |R_2| = 0. \end{aligned}$$

This implies

$$\begin{aligned} \limsup _{n \rightarrow +\infty } n^\alpha |R_x| \le M\varepsilon , \end{aligned}$$

that is $|R_x | \rightarrow 0$ as $n \rightarrow +\infty .$ Thus the theorem is completely proved $\square$

Theorem 5.4 works perfectly in several particular examples, as we will see in the next section. But in certain situations, as for example, the moment-type operators, formula (6.1) becomes

$$\begin{aligned} \lim _{n \rightarrow +\infty } n^\alpha [(T_nf)(x) - f(x)] = 0, \end{aligned}$$

for a suitable constant $\alpha$ for which all the assumptions are satisfied. In case of the moment-type operator, we have $\alpha = 1.$ If we try to take $\alpha =2$ some of the assumptions (i) of Theorem 5.4 are not satisfied. This result is however interesting, but it gives no exact information about the pointwise order of approximation at a point x. Therefore, we will formulate a slight generalization of the above theorem, in order to include also the case of moment-type operators, by changing assumptions (i). We have the following

Theorem 6.2

Let $f \in C^0_{\rho _1}({\mathbb {R}})$ be locally of class $C^2$ at a point $x \in {\mathbb {R}}.$ Let $K_n \in L^*_{\exp _a}({\mathbb {R}}),$ for sufficiently large values of $n \in {\mathbb {N}},$ be a kernel satisfying the assumptions of Corollary 4.2. Assume that there exist $\alpha >0$ such that the kernel $\{K_n\}$ satisfies assumption (ii) of Theorem 5.4 and the following conditions:

• (j) $\lim _{n \rightarrow +\infty } n^\alpha (A_{a,n}\,A_{-a,n} - 1) = \ell _0,$   $\lim _{n \rightarrow +\infty } n^\alpha \left( B_{-a,n}-A_{-a,n}\frac{1}{a}\log A_{a,n}\right) = \lambda _1,$

• (jj) $\lim _{n \rightarrow +\infty } n^\alpha \left( C_{-a,n} + \frac{A_{-a,n}}{a^2} \log ^2 A_{a,n} - \frac{2}{a} B_{-a,n} \log A_{a,n}\right) = \lambda _2$

Then,

$$\begin{aligned} \lim _{n \rightarrow +\infty } n^\alpha [(T_nf)(x) - f(x)] = \ell _0 f(x) + \lambda _1 f'(x) + \frac{\lambda _2}{2} f''(x). \end{aligned}$$

(6.5)

Proof

The proof is clearly exactly the same. $\square$

7 Examples

In this section we discuss some examples of kernels $\{K_n\}$ for which the theory developed can be applied.

• (1) The Gauss–Weierstrass kernel For $n \in {\mathbb {N}},$ let us consider the kernel $\{K_n\}$ with

  $$\begin{aligned} K_n(t) = \sqrt{\frac{n}{\pi }} e^{-n t^2} \qquad (t \in {\mathbb {R}}). \end{aligned}$$

  This kernel is defined by even functions. First we evaluate the exponential moments of orders 0, 1, 2. In order to do that, we first calculate the coefficients $A_{a,n}$ using a differentiation under the integral. By solving a simple first order linear differential equation, we obtain

  $$\begin{aligned} A_{a,n} = A_{-a,n} = \sqrt{\frac{n}{\pi }} \int _{-\infty }^\infty e^{at} e^{-nt^2}\mathrm{d}t = e^{a^2/(4n)}. \end{aligned}$$

  Hence $A_{a,n}\,A_{-a,n} = e^{a^2/(2n)}.$ Taking $\alpha = 1,$ we obtain

  $$\begin{aligned} \lim _{n \rightarrow + \infty } n (A_{a,n}\,A_{-a,n} - 1) = \lim _{n \rightarrow +\infty } n(e^{a^2/(2n)} - 1) = \frac{a^2}{2}. \end{aligned}$$

  Moreover, using partial integrations, one has

  $$\begin{aligned} B_{-a,n} = -\frac{a}{2n} e^{a^2/4n},\,\,C_{-a,n} = \frac{1}{2n} A_{a,n} + \frac{a^2}{4n^2}A_{a,n}, \end{aligned}$$

  and so

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n B_{-a,n} = \frac{1}{2},\,\,\lim _{n \rightarrow +\infty }nC_{-a,n} = \frac{1}{2}. \end{aligned}$$

  Next, we have

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n \log A_{a,n} = \frac{a^2}{4}, \end{aligned}$$

  therefore we have also

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n \log ^2 A_{a,n} = 0. \end{aligned}$$

  Therefore, $\ell _1 -\ell _3 = -3a/4,\,\, \ell _2 = 1/2.$ Now we prove that the Gauss–Weierstrass kernel satisfies also assumption (ii) of Theorem 5.4. Let $\eta >0$ be fixed. Let us set

  $$\begin{aligned} J:= n \int _{|t| \ge \eta } e^{-at}\left( t - \frac{1}{a}\log A_{a,n}\right) ^2 K_n(t) \mathrm{d}t. \end{aligned}$$

  Then

  $$\begin{aligned} J&\le n \int _{|t| \ge \eta } e^{-at} t^2 K_n(t) \mathrm{d}t + n \left( \frac{1}{a^2}\log ^2 A_{a,n}\right) \int _{|t| \ge \eta } e^{-at} K_n(t) \mathrm{d}t \\&\quad + 2n \frac{1}{a}|\log A_{a,n}|\int _{|t| \ge \eta } e^{-at} |t| K_n(t) \mathrm{d}t =: J_1+J_2+J_3. \end{aligned}$$

  As to $J_1$ we can write

  $$\begin{aligned} J_1&= \frac{n \sqrt{n}}{\sqrt{\pi }}\int _{|t| \ge \eta } e^{-at} t^2 e^{-nt^2}\mathrm{d}t\\&= \frac{n \sqrt{n}}{\sqrt{\pi }}\left\{ \int _\eta ^\infty e^{-at} t^2 e^{-nt^2}\mathrm{d}t + \int _\eta ^\infty e^{at} t^2 e^{-nt^2}\mathrm{d}t \right\} \\&=: J_1^1 + J_1^2. \end{aligned}$$

  For $J_1^1$ we have by a suitable substitution,

  $$\begin{aligned} J^1_1 \le \frac{e^{-a\eta }}{\sqrt{\pi }}\int _{\sqrt{n} \eta }^\infty v^2 e^{-v^2}dv, \end{aligned}$$

  and so by the absolute continuity of the Lebesgue integral, we obtain

  $$\begin{aligned} \lim _{n \rightarrow +\infty } J^1_1 = 0. \end{aligned}$$

  Analogously,

  $$\begin{aligned} J_1^2 \le \frac{1}{\sqrt{\pi }}\int _{\sqrt{n}\eta }^\infty e^{av} v^2 e^{-v^2}dv, \end{aligned}$$

  and again, since the integrand in the right-hand side is Lebesgue integrable, we obtain $J_1^2 \rightarrow 0$ as $n \rightarrow +\infty .$ Next, we evaluate $J_2.$ We have immediately

  $$\begin{aligned} J_2 \le \frac{n}{a^2}\log ^2 A_{a,n} A_{-a,n} \rightarrow 0, \quad (n \rightarrow +\infty ). \end{aligned}$$

  Finally, we evaluate $J_3.$ Using the same reasonings as for the estimate of $J_1,$ we have

  $$\begin{aligned} J_3&\le \frac{a \sqrt{n}}{2 \sqrt{\pi }}\left\{ e^{-a\eta }\int _\eta ^\infty t e^{-nt^2}\mathrm{d}t +\int _\eta ^\infty e^{at}t e^{-nt^2}\mathrm{d}t \right\} \\&=: J_3^1 + J_3^2. \end{aligned}$$

  As for $J_1$ we obtain easily that $J_3^1 \rightarrow 0,\,J_3^2 \rightarrow 0$ as $n\rightarrow +\infty .$ Concluding, we obtain assumption (ii) of Theorem 5.4. Therefore all the assumptions introduced are satisfied with $\alpha =1.$ The asymptotic formula of Theorem 5.4 reads

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n ((T_nf)(x) - f(x) ) = \frac{a^2}{2}f(x) - \frac{3a}{4} f'(x) + \frac{1}{4} f''(x), \end{aligned}$$

  which is a result of [26].

• (2) The Picard Kernel For $n \in {\mathbb {N}},$ let us consider the kernel $\{K_n\}$ with

  $$\begin{aligned} K_n(t) = \frac{\sqrt{n}}{2}e^{-\sqrt{n}|t|} \quad (t \in {\mathbb {R}}). \end{aligned}$$

  Also this kernel is defined by even functions. Let us evaluate the exponential moments of order 0, 1, 2. First we have

  $$\begin{aligned} A_{a,n} = A_{-a,n} = \frac{n}{n-a^2},\,\,\text{ and }\,\,A^2_{a,n} = \left( \frac{n}{n-a^2}\right) ^2 \end{aligned}$$

  and from sufficiently large values of n,  $A_{a,n}$ is positive. Moreover

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n(A_{a,n} A_{-a,n} - 1) = 2a^2,\,\,\lim _{n \rightarrow +\infty } n \log A_{a,n} = a^2. \end{aligned}$$

  Now, we evaluate the moment $B_{-a,n}.$ Using elementary calculation, based on partial integration, we can see that

  $$\begin{aligned} B_{-a,n} = -\frac{2an}{(n-a^2)^2},\,\, C_{-a,n} = \sqrt{n} \left( \frac{1}{(a + \sqrt{n})^3} + \frac{1}{(\sqrt{n}-a)^3}\right) . \end{aligned}$$

  Therefore,

  $$\begin{aligned} \lim _{n \rightarrow +\infty } nB_{-a,n} = -2a,\,\,\lim _{n \rightarrow +\infty } n C_{-a,n} = 2. \end{aligned}$$

  Finally we check assumption (ii) of Theorem 5.4. We proceed as in the previous example. Let $\eta >0$ be fixed and set

  $$\begin{aligned} J:= n\int _{|t| \ge \eta }e^{-at} \left( t-\frac{1}{a}\log A_{a,n}\right) ^2 K_n(t) \mathrm{d}t. \end{aligned}$$

  We have

  $$\begin{aligned} J&\le \frac{n\sqrt{n}}{2}\left\{ \int _{|t| \ge \eta } e^{-at} t^2 e^{-\sqrt{n}|t|}\mathrm{d}t + \frac{1}{a^2}\log ^2 \frac{n}{n-a^2}\int _{|t| \ge \eta } e^{-at} e^{-\sqrt{n}|t|}\mathrm{d}t\right. \\&\quad \left. + \frac{2}{a}\log \frac{n}{n-a^2}\int _{|t| \ge \eta } e^{-at} |t| e^{-\sqrt{n}|t|} \mathrm{d}t\right\} =: J_1 + J_2 + J_3. \end{aligned}$$

  Using now elementary calculations, based on suitable substitutions, it is easy to see that

  $$\begin{aligned} J_1 \le \frac{e^{-a\eta }}{2}\int _{\sqrt{n} \eta }^\infty u^2 e^{-u/2} du + \frac{1}{2}\int _{\sqrt{n}\eta }^\infty u^2 e^{-u/2}du, \end{aligned}$$

  for every n such that $1- a/\sqrt{n} > 1/2.$ Thus, $J_1 \rightarrow 0$ as $n \rightarrow +\infty .$ Next, let us consider $J_2.$ We have easily

  $$\begin{aligned} J_2 \le \frac{n}{a^2} \log ^2 \frac{n}{n-a^2} \,A_{-a,n}, \end{aligned}$$

  and so $J_2 \rightarrow 0$ as $n \rightarrow +\infty .$ Finally, as to $J_3$ we have

  $$\begin{aligned} J_3 \le \frac{e^{-a\eta }\sqrt{n}}{a}\log \frac{n}{n-a^2}\int _{\sqrt{n}\eta }^\infty v e^{-v}dv + \frac{\sqrt{n}}{a} \log \frac{n}{n-a^2}\int _{\sqrt{n}\eta }^\infty u e^{-u/2} du, \end{aligned}$$

  for every n such that $1- a/\sqrt{n} > 1/2.$ Therefore, $J_3 \rightarrow 0$ as $n \rightarrow +\infty ,$ and assumption (ii) is satisfied. Therefore all the assumptions introduced are satisfied with $\alpha =1.$ The asymptotic formula of Theorem 5.4 reads

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n ((T_nf)(x) - f(x) ) = 2a^2 f(x) - 3a f'(x) + f''(x) \end{aligned}$$

  which is a result of [6].

• (3) The moment kernel For $n \in {\mathbb {N}}$ let us consider the kernel $\{K_n\}$ with

  $$\begin{aligned} K_n(t) = n \chi _{[0, 1/n]}(t) \qquad (t \in {\mathbb {R}}). \end{aligned}$$

  This kernel is not even, and the functions $K_n$ have compact support. We calculate now the coefficients $A_{a,n},\,A_{-a,n},\, B_{-a,n}$ and $C_{-a,n}$ and some their properties. We have

  $$\begin{aligned} A_{a,n} = n \int _0^{1/n} e^{at} \mathrm{d}t = \frac{n}{a}(e^{a/n} -1), \, A_{-a,n} = \frac{n}{a}(1 - e^{-a/n}), \end{aligned}$$

  and so

  $$\begin{aligned} \lim _{n \rightarrow +\infty } A_{a,n} = \lim _{n \rightarrow +\infty } A_{-a,n} = 1, \quad \lim _{n \rightarrow +\infty }\frac{1}{a} n \log A_{a,n} = \frac{1}{2}. \end{aligned}$$

  Moreover,

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n (A_{a,n} A_{-a,n} - 1) = 0. \end{aligned}$$

  Next,

  $$\begin{aligned} B_{-a,n} = -\frac{e^{-a/n}}{a} + \frac{n}{a^2} (1 - e^{-a/n}),\,\, C_{-a,n} = -\frac{e^{-a/n}}{an} + \frac{2}{a}B_{-a,n}, \end{aligned}$$

  and both tend to zero as $n \rightarrow +\infty .$ Now,

  $$\begin{aligned} \lim _{n \rightarrow +\infty } nB_{-a,n} = 1/2,\,\,\lim _{n \rightarrow +\infty } n C_{a,n} = 0. \end{aligned}$$

  Concerning assumption (ii) of Theorem 5.4, since the functions $K_n$ have supports [0, 1/n] for any given $\eta >0,$ one can find an integer $n_0$ such that for any $n \ge n_0$ the set $\{t: |t| \ge \eta \}$ does not intersect the interval [0, 1/n],  and so denoting $H_{n,\eta } := \{t : |t| \ge \eta \} \cap [0, 1/n],$ we have

  $$\begin{aligned} \int _{H_n} e^{-at} \left( t - \frac{1}{a} \log A_{a,n}\right) ^2 \mathrm{d}t = 0 \end{aligned}$$

  This implies that assumption (ii) is trivially satisfied. Therefore all the assumptions introduced are satisfied with $\alpha =1.$ The asymptotic formula of Theorem 5.4 reads

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n ((T_nf)(x) - f(x) ) = 0. \end{aligned}$$

  This result is not fully satisfactory, due to the fact that we have no a precise order of pointwise approximation. Therefore we now employ Theorem 6.2 with $\alpha = 2.$ In order to do that, we have to check the assumptions (j) and (jj). As to (j), we first calculate the limit

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n^2\left( B_{-a,n} - \frac{1}{a} \log A_{a,n}\right) , \end{aligned}$$

  which can be written as

  $$\begin{aligned} \lim _{n \rightarrow +\infty }\frac{n^2}{a^2}\left( -a e^{-a/n} + a \frac{n}{a} (1 - e^{-a/n}) - a \log \left( \frac{n}{a}(e^{a/n} -1)\right) \right) . \end{aligned}$$

  This can be interpreted as a restriction of the limit

  $$\begin{aligned} \lim _{x \rightarrow 0} \frac{a}{x^2}\left( -e^{-x} + \frac{1-e^{-x}}{x} - \log \left( \frac{e^x -1}{x}\right) \right) . \end{aligned}$$

  The above limit can be solved using elementary techniques, based on the L’Hospital rule and its value is given by $-3a/8.$ Now, in order to calculate $\lambda _1$ we write

  $$\begin{aligned}&n^2\left( B_{-a,n} - A_{-a,n}\frac{1}{a}\log A_{a,n}\right) \\&\quad = n^2\left( B_{-a,n} - \frac{1}{a}\log A_{a,n}\right) + n^2 \frac{1}{a}\log A_{a,n}\,(1 - A_{-a,n}), \end{aligned}$$

  thus we consider only the limit

  $$\begin{aligned} \lim _{n \rightarrow +\infty }n^2 \frac{1}{a}\log A_{a,n}\,(1 - A_{-a,n}). \end{aligned}$$

  Since as we have seen before $\lim _{n \rightarrow +\infty } (n/a) \log A_{a,n} = 1/2,$ we have to calculate only $\lim _{n \rightarrow +\infty } n( 1- A_{-a,n})$ and it is not difficult to see that its value is given by a/2. Thus, finally we obtain $\lambda _1 = -a/8.$ Now we proceed to the calculation of $\lambda _2.$ At first we calculate the limit

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n^2 C_{-a,n}. \end{aligned}$$

  Again this limit can be considered as a restriction of the limit

  $$\begin{aligned} \lim _{x \rightarrow 0} \frac{1}{x^2}\left( -xe^{-x} - 2e^{-x} + 2 \frac{1-e^{-x}}{x}\right) , \end{aligned}$$

  and using an analogous elementary approach, its value is given by 1/3. Next, using the previous results, we have immediately

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n^2 \frac{A_{-a,n}}{a^2} \log ^2A_{a,n} = \frac{1}{4}, \end{aligned}$$

  and

  $$\begin{aligned} \lim _{n \rightarrow +\infty } \frac{2}{a} n^2 B_{-a,n}\log A_{a,n} = 1/2. \end{aligned}$$

  Thus, we have $\lambda _2 = 1/12.$ The last limit $\ell _0 = \lim _{n \rightarrow +\infty } n^2 (A_{a,n}\,A_{-a,n} -1)$ can be obtained with the same reasonings as a restriction of the limit

  $$\begin{aligned} \lim _{x \rightarrow 0} a^2 \frac{e^x + e^{-x} -2 -x^2}{x^4} \end{aligned}$$

  which is given by $a^2/12.$ Concluding, the Voronovskaya formula (6.5) with $\alpha =2$ for the moment type operator is given by (see also [26])

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n^2 [(T_nf)(x) - f(x)] = \frac{a^2}{12} f(x) -\frac{a}{8} f'(x) + \frac{1}{24} f''(x). \end{aligned}$$

• (4) A spline kernel For $n \in {\mathbb {N}}$ let us consider the kernel $\{K_n\}$ with

  $$\begin{aligned} K_n(t) = n (1-|nt|)\chi _{[-1/n, 1/n]}(t)\qquad (t \in {\mathbb {R}}). \end{aligned}$$

  For this kernel we have, by elementary calculations

  $$\begin{aligned} A_{a,n} = A_{-a,n} = n\int _{-1/n}^{1/n} e^{at} (1 - n|t|)\mathrm{d}t = \frac{2n^2}{a^2}(\cosh (a/n) -1), \end{aligned}$$

  and so

  $$\begin{aligned} \lim _{n \rightarrow +\infty } A_{\pm a,n} = 1,\quad \lim _{n \rightarrow +\infty } n^2( A_{a,n} A_{-a,n} -1) = \frac{a^2}{6}, \end{aligned}$$

  and $\lim _{n \rightarrow +\infty } n^2\log A_{a,n} = a^2/12.$ Next,

  $$\begin{aligned} B_{-a,n} = n\int _{-1/n}^{1/n} t e^{-at}(1-n|t|) \mathrm{d}t = -\frac{2n}{a^2}\sinh (a/n) + \frac{4n^2}{a^3}(\cosh (a/n) -1), \end{aligned}$$

  and

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n^2 B_{-a,n} = -\frac{a}{6}. \end{aligned}$$

  As to $C_{-a,n}$ we have

  $$\begin{aligned} C_{-a,n}&= n\int _{-1/n}^{1/n} t^2 (1-n|t|) e^{-at}\mathrm{d}t \\&= \frac{2}{a^2}\cosh (a/n) - \frac{8n}{a^3} \sinh (a/n) + \frac{12 n^2}{a^4}(\cosh (a/n) -1), \end{aligned}$$

  and so

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n^2 C_{-a,n} = \frac{1}{6}. \end{aligned}$$

  Finally, since the functions $K_n$ have compact supports $[-1/n, 1/n],$ we easily see, as in the previous example, that (ii) of Theorem 5.4 is satisfied. Thus, all the assumptions used in the previous theory are satisfied with $\alpha = 2.$ The corresponding asymptotic formula is given by

  $$\begin{aligned} \lim _{n \rightarrow +\infty } n^2 ((T_nf)(x) - f(x) ) = \frac{a^2}{6} f(x) - \frac{a}{4} f'(x) + \frac{1}{12} f''(x). \end{aligned}$$

Remark 7.1

The kernel of Example (4) is generated by the spline function of second order, defined by

$$\begin{aligned} \beta _2 (x) = (1-|x|)\chi _{[-1,1]}(x) \qquad (x \in {\mathbb {R}}). \end{aligned}$$

The functions $K_n$ are given by $K_n(x) = n \beta _2 (nx),$ for $x \in {\mathbb {R}}.$ The spline functions of order k are given by the formula (see e.g. [11, 24])

$$\begin{aligned} \beta _k(x) = \frac{1}{(k-1)!} \sum _{j=0}^k (-1)^j {k \atopwithdelims ()j} \left( \frac{k}{2} + x-j\right) ^{k-1}_+, \end{aligned}$$

where for any real number r,  $r_+$ denotes its positive part. The theory may be applied also to any kernel generated by the spline $\beta _k.$

Remark 7.2

Note that the kernels of any of the previous examples satisfy the assumptions of Theorem 5.4, in particular the inequality

$$\begin{aligned} \frac{1}{a}|\log A_{a,n}| \le \sqrt{C_{-a,n}}, \end{aligned}$$

for every $a>0$ and sufficiently large $n \in {\mathbb {N}}.$ First, note that since in any of the above examples $A_{a,n} \ge 1,$ we have $\log A_{a,n} \ge 0.$ For the modified Gauss–Weierstrass kernel, one has easily

$$\begin{aligned} \sqrt{C_{-a,n}}> \frac{a}{2n} > \frac{1}{a}\log A_{a,n}. \end{aligned}$$

For the modified Picard kernel, one can employ the calculations of the limits in Example (2): since $n C_{-a,n} \rightarrow 2$ as $n \rightarrow +\infty ,$ and

$$\begin{aligned} \lim _{n \rightarrow +\infty }\frac{\sqrt{n}}{a}\log A_{a,n} = 0, \end{aligned}$$

for sufficiently large values of n we obtain the assertion. Similar arguments can be used for the remaining examples.

Remark 7.3

Further examples can be obtained using Phillips type operators and Post-Widder type operators which act on functions defined over the positive real axis, (see [16, 18, 19] in which some modified version are introduced).

Remark 7.4

Our approach may be applied also in case of general exponential functions of the form $a^{-x}$ for $a>1$ as studied in [17] for Baskakov-type operators that act on functions defined over the positive real axis.