On the q-Monotonicity Preservation of Durrmeyer-Type Operators

Abstract

We prove that various Durrmeyer-type operators preserve q-monotonicity in [0, 1] or $[0,\infty )$ as the case may be. Recall that a 1-monotone function is nondecreasing, a 2-monotone one is convex, and for $q>2$, a q-monotone function possesses a convex $(q-2)$nd derivative in the interior of the interval. The operators are the Durrmeyer versions of Bernstein (including genuine Bernstein–Durrmeyer), Szász and Baskakov operators. As a byproduct we have a new type of characterization of continuous q-monotone functions by the behavior of the integrals of the function with respect to measures that are related to the fundamental polynomials of the operators.

1 Introduction

Let I be an interval on the real line, finite or infinite, open or closed or half open. For a function f, defined on I, denote

$$\begin{aligned} \Delta ^1_h f(x):=\Delta _h f(x):={\left\{ \begin{array}{ll} f(x+h)-f(x), &{} \qquad x,x+h\in I, \\ 0, &{} \qquad \text {otherwise}, \end{array}\right. } \end{aligned}$$

and for $q\ge 1$,

$$\begin{aligned} \Delta _h^{q+1} f(x):=\Delta _h^q(\Delta _hf(x)). \end{aligned}$$

That is,

$$\begin{aligned} \Delta _h^q f(x)= {\left\{ \begin{array}{ll} \sum _{i=0}^q (-1)^{q-i}\left( {\begin{array}{c}q\\ i\end{array}}\right) f(x+ih), &{} \qquad x,x+qh\in I, \\ 0, &{} \qquad \text {otherwise}. \end{array}\right. } \end{aligned}$$

A function f defined on I, is called q-monotone there, if $\Delta ^q_h f(x)\ge 0$, $x\in I$, for all $h>0$. In particular a 1-monotone function is nondecreasing and a 2-monotone one is convex. It is well-known (it goes back to T. Popoviciu, Mathematica 8 (1934), and R. P. Boas Jr. and D. V. Widder, Duke Math. J. 7 (1940)) that for $q>2$, f is q-monotone in I, implies that f possesses a convex $(q-2)$nd derivative in the interior of I.

The classical Bernstein polynomials associated with a function f which is defined on [0, 1], are defined by,

$$\begin{aligned} (B_nf)(x):=\sum _{j=0}^{n}p_{n,j}(x){f}\left( \frac{j}{n}\right) ,\qquad x\in [0,1],\quad n\ge 1, \end{aligned}$$

where

$$\begin{aligned} p_{n,j}(x):=\left( {\begin{array}{c}n\\ j\end{array}}\right) x^{j}(1-x)^{n-j},\qquad 0\le j\le n. \end{aligned}$$

For the sake of simplicity we also denote $p_{0,0}(x)\equiv 1$ and $p_{n,j}(x)\equiv 0$, $j<0$ and $j>n$.

It is well-known (see, e.g., [11, 1.4(2)]) that the Bernstein polynomials preserve q-monotonicity of all orders $q\ge 1$.

Durrmeyer [6] and, independently, Lupaş [12] have modified the Bernstein polynomials and defined what we call the Bernstein–Durrmeyer polynomials. Namely (see [4]), for an integrable f on [0, 1], we set

$$\begin{aligned} (D_{n}f)(x):=(n+1)\sum _{j=0}^{n}p_{n,j}(x)\int _0^1p_{n,j}(t)f(t)dt,\qquad x\in [0,1],\quad n\ge 0. \end{aligned}$$

It is well-known that if $f\in C[0,1]$, then the Bernstein–Durrmeyer polynomials uniformly approximate f in [0, 1], as $n\rightarrow \infty$. As operators they are contractions in $L_p[0,1]$ $1\le p\le \infty$ (where by $L_\infty [0,1]$ we mean C[0, 1]), and are self-adjoint in $L_2[0,1]$. The interested reader may see details in [4] (see also [5]). We are interested in something else, namely, preservation of q-monotonicity.

Adell and de la Cal [2] proved that the Bernstein–Durrmeyer polynomials preserve monotonicity and convexity, applying techniques from probability and stochastic processes. In 2007 Attalienti and Raşa [3] proved that the Bernstein–Durrmeyer polynomials preserve q-monotonicity for all $q\ge 1$, using the heavy machinery of Karlin’s Total Positivity [10]. In Section 2, we will show by elementary means that the Bernstein–Durrmeyer polynomials preserve q-monotonicity for all $q\ge 1$. We will use the proof to prove that the genuine Bernstein–Durrmeyer polynomials [8] (see also [7]) have the same property.

In Section 3, we discuss the Szász–Durrmeyer operators [13] and in Section 4 the Baskakov–Durrmeyer operators [15]. We will prove that both preserve q-monotonicity. For functions differentiable sufficiently many times and all required integrals converge, the q-monotonicity of the Szász–Durrmeyer operators was proved by Păltănea [14] (see [9] for details and other operators). Finally, in Section 5 we present some Raşa–type inequalities for Durrmeyer-type operators.

2 q-monotonicity of the ordinary and genuine Bernstein–Durrmeyer polynomials

We first prove the following

Theorem 2.1

Let $q\ge 1$, and assume that $f\in C[0,1]$ is q-monotone. Then $D_nf$, $n\ge 1$, is q-monotone.

A crucial lemma is

Lemma 2.2

Given $q\ge 1$, let $f\in C[0,1]$ be q-monotone. Then, for every $n\ge q$ and $0\le k\le n-q$, we have

$$\begin{aligned} \int _0^1\left[ \sum _{i=0}^q(-1)^{q-i}\left( {\begin{array}{c}q\\ i\end{array}}\right) p_{n,k+i}(t)\right] f(t)dt\ge 0. \end{aligned}$$

(2.1)

An immediate consequence of Lemma 2.2 and the proof of Theorem 2.1 is a new type of characterization of continuous q-monotone functions in [0, 1], which may be interesting by itself.

Corollary 2.3

Let $q\ge 1$ and assume that $f\in C[0,1]$. Then f is q-monotone in [0, 1], if and only if, (2.1) holds for every $n\ge q$ and $0\le k\le n-q$.

Proof

If $f\in C[0,1]$ is q-monotone, then (2.1) follows by Lemma 2.2. Conversely, if (2.1) is valid for every $n\ge q$ and $0\le k\le n-q$, then by the proof of Theorem 2.1 it follows that $D_nf$, $n\ge 1$, is q-monotone. It is well-known (see, e.g., [4, Théorème II.2]) that $D_nf$ converges (uniformly) to f in [0, 1], so it follows that f is q-monotone. $\square$

Proof of Lemma 2.2

First, we observe that every q-monotone function $f\in C[0,1]$ may be uniformly approximated by q times continuously differentiable q-monotone functions on [0, 1]. For example, one may take the Bernstein polynomials associated with f. Hence, in order to prove (2.1) it suffices to assume that $f\in C^q[0,1]$ so that by assumption $f^{(q)}(x)\ge 0$, $x\in [0,1]$.

Denote

$$\begin{aligned} a_{n,k}:=\int _0^1p_{n,k}(t)f(t)dt, \end{aligned}$$

(2.2)

and

$$\begin{aligned} \Delta ^qa_{n,k}:=\sum _{i=0}^q(-1)^{q-i}\left( {\begin{array}{c}q\\ i\end{array}}\right) a_{n,k+i}. \end{aligned}$$

We will show that if $f\in C^q[0,1]$, then

$$\begin{aligned} \Delta ^qa_{n,k}=\frac{n!}{(n+q)!}\int _0^1p_{n+q,k+q}(t)f^{(q)}(t)dt,\quad 0\le k\le n-q. \end{aligned}$$

(2.3)

To this end, straightforward differentiation yields that for any $m,l\ge 0$,

$$\begin{aligned} p'_{m+1,l+1}(x)=(m+1)\left( p_{m,l}(x)-p_{m,l+1}(x)\right) . \end{aligned}$$

(2.4)

Hence, it readily follows that

$$\begin{aligned} \Delta ^1a_{n,k}=\frac{1}{n+1}\int _0^1p_{n+1,k+1}(t)f'(t)dt,\quad 0\le k\le n-1. \end{aligned}$$

(2.5)

Proceeding by induction, assuming (2.3) for $q-1$, and applying (2.4), we obtain (2.3) for q. Since $f^{(q)}(x)\ge 0$, our proof is complete. $\square$

Proof of Theorem 2.1

$D_nf$ is a polynomial of degree n. Thus, for $0\le n<q$ it is clearly q-monotone.

We write

$$\begin{aligned} (D_nf)(x)=(n+1)\sum _{k=0}^np_{n,k}(x)a_{n,k},\quad 0\le x\le 1, \end{aligned}$$

where $a_{n,k}$ is defined in (2.2).

Repeating, verbatim, the well-known computations for the Bernstein polynomials (see, e.g., [11, 1.4(2)]), we have, for $n\ge q$,

$$\begin{aligned} (D_nf)^{(q)}(x)=\frac{(n+1)!}{(n-q)!}\sum _{k=0}^{n-q}p_{n-q,k}(x)\Delta ^qa_{n,k},\quad 0\le x\le 1. \end{aligned}$$

(2.6)

Since by Lemma 2.2$\Delta ^qa_{n,k}\ge 0$, for all $0\le k\le n-q$, $n\ge q$, our proof is complete. $\square$

The genuine Bernstein–Durrmeyer polynomials associated with a function $f\in C[0,1]$, were defined by

$$\begin{aligned} (U_nf)(x)&:=f(0)p_{n,0}(x)+f(1)p_{n,n}(x)\\ {}&\quad +(n-1)\sum _{k=1}^{n-1}p_{n,k}(x)\int _0^1p_{n-2,k-1}(t)f(t)dt, \quad x\in [0,1],\quad n\ge 2. \end{aligned}$$

Recall that the advantage of the genuine Bernstein–Durrmeyer polynomials is that they preserve linear functions while the ordinary Bernstein–Durrmeyer polynomials only preserve constants.

We have the following.

Theorem 2.4

Let $q\ge 1$, and assume that $f\in C[0,1]$ is q-monotone. Then $U_nf$, $n\ge 2$, is q-monotone.

Proof

We have to separate the proof for $q=1$ and for $q\ge 2$, and we begin with the latter. Thus assume that $q\ge 2$. If $2\le n<q$, then there is nothing to prove. Thus, we assume that $n\ge q$.

Let L be the linear function interpolating f at $x=0,1$. Then, $g(x):=(f-L)(x)$ vanishes at both $x=0,1$ and, if f is q-monotone, so is g. Since $U_ng=U_nf-L$, if $U_ng$ is q-monotone, so is $U_nf$. Hence, without loss of generality, we may assume that $f(0)=f(1)=0$, so that

$$\begin{aligned} (U_nf)(x)&=(n-1)\sum _{k=1}^{n-1}p_{n,k}(x)\int _0^1p_{n-2,k-1}(t)f(t)dt\\&=(n-1)\sum _{k=1}^{n-1}p_{n,k}(x)a_{n-2,k-1},\quad x\in [0,1],\quad n\ge 2, \end{aligned}$$

where $a_{n,k}$ were defined in (2.2).

Denote $b_{n,k}:=a_{n-2,k-1}$, $1\le k\le n-1$, and $b_{n,0}=b_{n,n}=0$. Then we rewrite

$$\begin{aligned} (U_nf)(x)=(n-1)\sum _{k=0}^np_{n,k}(x)b_{n,k}. \end{aligned}$$

As in (2.6) we have

$$\begin{aligned} (U_nf)^{(q)}(x)=(n-1)\frac{n!}{(n-q)!}\sum _{k=0}^{n-q}p_{n-q,k}(x)\Delta ^qb_{n,k}. \end{aligned}$$

If $n\ge q+2$, then by virtue of Lemma 2.2, $\Delta ^qb_{n,k}=\Delta ^qa_{n-2,k-1}\ge 0$, $1\le k\le n-q-1$. If $n=q,q+1$, then this statement is empty. In both cases, we complete the proof if we show that

$$\begin{aligned} \Delta ^qb_{n,0}\ge 0\quad \text {and}\quad \Delta ^qb_{n,n-q}\ge 0. \end{aligned}$$

(2.7)

Recall that, as in the proof of Lemma 2.2, we may assume that $f\in C^q[0,1]$. First, assume that $n\ge q+1$. Since $f(0)=0$,

$$\begin{aligned} \Delta ^1b_{n,0}&=a_{n-2,0}=\int _0^1(1-t)^{n-2}f(t)dt\\ {}&=\frac{1}{n-1}\int _0^1(1-t)^{n-1}f'(t)dt=\frac{1}{n-1}\int _0^1p_{n-1,0}(t)f'(t)dt, \end{aligned}$$

and by (2.5),

$$\begin{aligned} \Delta ^1b_{n,1}=\Delta ^1a_{n-2,0}=\frac{1}{n-1}\int _0^1p_{n-1,1}(t)f'(t)dt. \end{aligned}$$

Hence, by (2.4),

$$\begin{aligned} \Delta ^2b_{n,0}=-\frac{1}{n(n-1)}\int _0^1p'_{n,1}(t)f'(t)dt=\frac{1}{n(n-1)}\int _0^1p_{n,1}(t)f''(t)dt. \end{aligned}$$

Assuming, by induction, that

$$\begin{aligned} \Delta ^{q-1}b_{n,0}=\frac{(n-2)!}{(n+q-3)!}\int _0^1p_{n+q-3,q-2}(t)f^{(q-1)}(t)dt, \end{aligned}$$

(2.8)

and noting that (2.3), for $q-1$, implies

$$\begin{aligned} \Delta ^{q-1}b_{n,1}=\Delta ^{q-1}a_{n-2,0}=\frac{(n-2)!}{(n+q-3)!}\int _0^1p_{n+q-3,q-1}(t)f^{(q-1)}(t)dt, \end{aligned}$$

we obtain, by (2.4) and integration by parts,

$$\begin{aligned} \Delta ^qb_{n,0}=\frac{(n-2)!}{(n+q-2)!}\int _0^1p_{n+q-2,q-1}(t)f^{(q)}(t)dt\ge 0, \end{aligned}$$

(2.9)

that proves the left hand inequality in (2.7).

Similarly, since $f(1)=0$,

$$\begin{aligned} \Delta ^1b_{n,n-1}&=-a_{n-2,n-2}=-\int _0^1t^{n-2}f(t)dt\\ {}&=\frac{1}{n-1}\int _0^1t^{n-1}f'(t)dt=\frac{1}{n-1}\int _0^1p_{n-1,n-1}(t)f'(t)dt, \end{aligned}$$

and by (2.5),

$$\begin{aligned} \Delta ^1b_{n,n-2}=\Delta ^1a_{n-2,n-3}=\frac{1}{n-1}\int _0^1p_{n-1,n-2}(t)f'(t)dt. \end{aligned}$$

Hence,

$$\begin{aligned} \Delta ^2b_{n,n-2}=-\frac{1}{n(n-1)}\int _0^1p'_{n,n-1}(t)f'(t)dt=\frac{1}{n(n-1)}\int _0^1p_{n,n-1}(t)f''(t)dt. \end{aligned}$$

Finally, again assuming, by induction, that

$$\begin{aligned} \Delta ^{q-1}b_{n,n-q+1}=\frac{(n-2)!}{(n+q-3)!}\int _0^1p_{n+q-3,n-1}(t)f^{(q-1)}(t)dt, \end{aligned}$$

(2.10)

we obtain as above,

$$\begin{aligned} \Delta ^qb_{n,n-q}=\frac{(n-2)!}{(n+q-2)!}\int _0^1p_{n+q-2,n-1}(t)f^{(q)}(t)dt\ge 0, \end{aligned}$$

(2.11)

that proves the right hand inequality in (2.7).

This completes the proof for $q\ge 2$ and $n\ge q+1$, so that we proceed to the case $n=q\ge 2$.

We begin with $n=q=2$, and observe that in this case,

$$\begin{aligned} (U_2f)(x)=2x(1-x)\int _0^1f(t)dt, \end{aligned}$$

which is convex, since $f(t)\le 0$, $0\le t\le 1$.

If $n=q>2$, then we have to investigate $\Delta ^qb_{n,0}=\Delta ^qb_{n,n-q}$, but both (2.9) and (2.11) are inapplicable. Thus we have to go back to (2.10) and (2.8).

We have

$$\begin{aligned} \Delta ^qb_{q,0}&=\Delta ^{q-1}b_{q,1}-\Delta ^{q-1}b_{q,0}\\&=\frac{(q-2)!}{(2q-3)!}\int _0^1\left( p_{2q-3,q-1}(t)-p_{2q-3,q-2}(t)\right) f^{(q-1)}(t)dt\\&=-\frac{(q-2)!}{(2q-2)!}\int _0^1p'_{2q-2,q-1}(t)f^{(q-1)}(t)dt\\&=\frac{(q-2)!}{(2q-2)!}\int _0^1p_{2q-2,q-1}(t)f^{(q)}(t)dt\ge 0. \end{aligned}$$

This completes the proof for all $n\ge q\ge 2$.

As for the case $q=1$, we observe that since $U_n$ preserves constants, we may assume that $f(0)=0$, so that, in turn,

$$\begin{aligned} 0\le \int _0^1f(t)dt\le f(1). \end{aligned}$$

(2.12)

First let $n>2$, and again denote $b_{n,0}=0$, $b_{n,n}=0$ and $b_{n,k}=a_{n-2,k-1}$, $1\le k\le n-1$. Then,

$$\begin{aligned} (U_nf)(x)=f(1)x^n+(n-1)\sum _{k=0}^np_{n,k}(x)b_{n,k}, \end{aligned}$$

and

$$\begin{aligned} \frac{d}{dx}(U_nf)(x)=nf(1)x^{n-1}+n(n-1)\sum _{k=0}^{n-1}p_{n-1,k}(x)\Delta ^1b_{n,k}. \end{aligned}$$

Now, $\Delta ^1b_{n,k}\ge 0$, $1\le k\le n-2$, so we only need to check $\Delta ^1b_{n,0}$ and $\Delta ^1b_{n,n-1}$.

To this end,

$$\begin{aligned} \Delta ^1b_{n,0}=b_{n,1}=a_{n-2,0}=\int _0^1(1-t)^{n-2}f(t)dt\ge 0, \end{aligned}$$

since $f(t)\ge 0$, $0\le t\le 1$.

On the other hand,

$$\begin{aligned} \Delta ^1b_{n,n-1}=-b_{n,n-1}=-a_{n-2,n-2}=-\int _0^1t^{n-2}f(t)dt\ge -\frac{f(1)}{n-1}, \end{aligned}$$

since f is nondecreasing in [0, 1].

Hence,

$$\begin{aligned} nf(1)x^{n-1}+n(n-1)p_{n-1,n-1}(x)\Delta ^1b_{n,n-1}\ge 0, \end{aligned}$$

and it follows that $(U_nf)'(x)\ge 0$, $x\in [0,1]$.

Finally, if $n=2$, then

$$\begin{aligned} \left( U_{2}f\right) (x)&=f(1)x^2+2x(1-x)\int _0^1f(t)dt\\&=\left( f(1)-\int _0^1f(t)dt\right) x^2+x(2-x)\int _0^1f(t)dt. \end{aligned}$$

Hence, by virtue of (2.12), $U_2f$ is nondecreasing in [0, 1]. This completes the proof. $\square$

3 The Szász–Durrmeyer operators

The Favard–Mirakyan–Szász operators associated with a function f defined in $[0,\infty )$, and such that $|f(x)|\le Ce^{Ax}$ for some constants $C,A>0$, are defined by

$$\begin{aligned} (S_nf)(x):=\sum _{k=0}^\infty s_k(nx)f\left( \frac{k}{n}\right) ,\qquad x\in [0,\infty ),\quad n\ge 1, \end{aligned}$$

where

$$\begin{aligned} s_k(x):=e^{-x}\frac{x^k}{k!},\quad 0\le x<\infty ,\quad k\ge 0. \end{aligned}$$

It is well-known that if, in addition, $f\in C[0,\infty )$, then the Favard-Mirakyan-Szász operators associated with f, approximate it uniformly on every compact subinterval of $[0,\infty )$. It is also known that they preserve q-monotonicity on $[0,\infty )$, which readily follows from the equation,

$$\begin{aligned} \frac{d^q}{dx^q}\left( S_nf\right) (x)=n^q\sum _{k=0}^\infty s_k(nx)\Delta ^qf\left( \frac{k}{n}\right) . \end{aligned}$$

(3.1)

In 1985, Mazhar and Totik [13] defined the Szász–Durrmeyer variant for continuous functions of the above type, by

$$\begin{aligned} (L_nf)(x):=n\sum _{k=0}^\infty s_k(nx)\int _0^\infty s_k(nt)f(t)dt,\qquad x\in [0,\infty ),\quad n>A. \end{aligned}$$

We will prove the following.

Theorem 3.1

Given $q\ge 1$, suppose that $f\in C[0,\infty )$ is q-monotone and there exist constants $C,A\ge 0$ such that $|f(x)|\le Ce^{Ax}$, $0\le x<\infty$. Then $L_nf$, $n>A$, is q-monotone in $[0,\infty )$.

A weaker result but easier to prove, and which goes along the lines of the proof of Lemma 2.2 is

Theorem 3.2

Given $q\ge 1$, suppose that $f\in C^q[0,\infty )$ is q-monotone and there exist constants $C,A\ge 0$ such that $|f^{(i)}(x)|\le Ce^{Ax}$, $0\le i\le q-1$, $0\le x<\infty$. Then $L_nf$, $n>A$, is q-monotone in $[0,\infty )$.

We may characterize q-monotone functions in $[0,\infty )$, with an analog of Corollary 2.3. For this we need the notation

$$\begin{aligned} a_{n,k}:=\int _0^\infty s_k(nt)f(t)dt. \end{aligned}$$

(3.2)

Corollary 3.3

Let $q\ge 1$ and assume that $f\in C[0,\infty )$ is such that $|f(x)|\le Ce^{Ax}$, $0\le x<\infty$, for some $C,A\ge 0$. Then, f is q-monotone in $[0,\infty )$, if and only if,

$$\begin{aligned} \Delta ^qa_{n,k}\ge 0, \quad n>A\,\,\text {and}\,\,k\ge 0. \end{aligned}$$

(3.3)

Proof

From the proof of Theorem 3.1 we conclude that if f is q-monotone, then (3.3) is valid. Conversely, we observe from the proof of Theorem 3.1 that (3.3) implies that $L_nf$, $n>A$ is q-monotone. Since $(L_nf)(x)\rightarrow f(x)$, $x\in [0,\infty )$, it follows that f is q-monotone in $[0,\infty )$. $\square$

We begin with the proof of Theorem 3.2.

Proof of Theorem 3.2

Similar to (3.1), we obtain,

$$\begin{aligned} \frac{d^q}{dx^q}(L_nf)(x)=n^{q+1}e^{-nx}\sum _{k=0}^\infty \frac{(nx)^k}{k!}\Delta ^q a_{n,k},\quad n>A, \end{aligned}$$

(3.4)

where $a_{n,k}$ are defined in (3.2).

Hence, our proof will be complete if we prove that $\Delta ^q a_{n,k}\ge 0$, $k=0,1,2\dots$, $n>A$.

To this end, since

$$\begin{aligned} s'_{k+1}(nt)=-n\left( s_{k+1}(nt)-s_k(nt)\right) , \end{aligned}$$

we get

$$\begin{aligned} \Delta ^1a_{n,k}=-\frac{1}{n}\int _0^\infty s'_{k+1}(nt)f(t)dt=\frac{1}{n}\int _0^\infty s_{k+1}(nt)f'(t)dt. \end{aligned}$$

Proceeding by induction we obtain

$$\begin{aligned} \Delta ^qa_{n,k}=\frac{1}{n^q}\int _0^\infty s_{k+q}(nt)f^{(q)}(t)dt\ge 0. \end{aligned}$$

Note that due to the integration by parts, we do not have to assume growth conditions on $f^{(q)}$, for the integral to converge. This completes the proof.

$\square$

Proof of Theorem 3.1

As above, we need to prove that $\Delta ^qa_{n,k}\ge 0$. We observe that

$$\begin{aligned} \Delta ^qa_{n,k}&=\int _0^\infty e^{-nt}\sum _{j=0}^{q}(-1)^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) \frac{\left( nt\right) ^{k+j}}{(k+j)!}f(t)dt\\&=\frac{(-1)^q}{n^q}\int _0^\infty \left( e^{-nt}\frac{(nt)^{k+q}}{(k+q)!}\right) ^{(q)}f(t)dt\\&=\frac{(-1)^q}{n^q}\int _0^\infty s^{(q)}_{k+q}(nt)f(t)dt. \end{aligned}$$

Denote $s_{k+q}(nt)=:g(t)$ and let $\Delta _h^qg(t):=\sum _{j=0}^q(-1)^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) g(t+jh)$. Rewrite

$$\begin{aligned} \Delta ^qa_{n,k}= & {} \frac{(-1)^q}{n^qh^q}\int _0^\infty \Delta _h^qg(t)f(t)dt\nonumber \\&+\left( \frac{(-1)^q}{n^q}\int _0^\infty g^{(q)}(t)f(t)dt-\frac{(-1)^q}{n^qh^q}\int _0^\infty \Delta _h^qg(t)f(t)dt\right) \nonumber \\=: & {} \frac{(-1)^q}{n^qh^q}\int _0^\infty \Delta _h^qg(t)f(t)dt+R_1. \end{aligned}$$

(3.5)

We will prove that $R_1\rightarrow 0$, as $h\rightarrow 0$. To this end, note that for some $\xi \in (t,t+qh)$,

$$\begin{aligned} |h^{-q}\Delta _h^qg(t)-g^{(q)}(t)|&=|g^{(q)}(\xi )-g^{(q)}(t)|\le |\xi -t|\max _{t\le u\le t+qh}|g^{(q+1)}(u)|\\&\le qh\max _{t\le u\le t+qh}|g^{(q+1)}(u)|. \end{aligned}$$

Hence,

$$\begin{aligned} |R_1|&\le qhn^{-q}\int _0^\infty \left[ \max _{t\le u\le t+qh}|g^{(q+1)}(u)|\right] |f(t)|dt\\&\le Cqhn\sum _{j=0}^{q+1}\left( {\begin{array}{c}q+1\\ j\end{array}}\right) \int _0^\infty \left[ \max _{t\le u\le t+qh}\left( e^{-nu} \frac{(nu)^{k+j}}{(k+j)!}\right) \right] e^{At}dt. \end{aligned}$$

Thus, it suffices to prove that for $m=0,1,2,\dots$,

$$\begin{aligned} \lim _{h\rightarrow 0}h\int _0^\infty \max _{t\le u\le t+qh}\left( e^{-nu}(nu)^m\right) e^{At}dt=0. \end{aligned}$$

Since $e^{-nu}(nu)^m\le m!$, and it is increasing in [0, m/n] and decreasing in $[m/n,\infty )$, we conclude that

$$\begin{aligned} \int _0^\infty \max _{t\le u\le t+qh}\left( e^{-nu}(nu)^m\right) e^{At}dt\le m!\int _0^{m/n}e^{At}dt+\int _{m/n}^\infty e^{-nt}(nt)^me^{At}dt<\infty . \end{aligned}$$

Hence, $R_1=O(h)$, as $h\rightarrow 0$.

Now,

$$\begin{aligned} \int _0^\infty \Delta _h^qg(t)f(t)dt= & {} \sum _{j=0}^q(-1)^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) \int _0^\infty g(t+jh)f(t)dt\nonumber \\= & {} \sum _{j=0}^q(-1)^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) \int _{jh}^\infty g(t)f(t-jh)dt\nonumber \\= & {} \sum _{j=0}^q(-1)^j\left( {\begin{array}{c}q\\ j\end{array}}\right) \int _{(q-j)h}^\infty g(t)f(t-qh+jh)dt\nonumber \\= & {} (-1)^q\int _{qh}^\infty g(t)\Delta _h^qf(t-qh)dt\nonumber \\&+\sum _{j=0}^q(-1)^j\left( {\begin{array}{c}q\\ j\end{array}}\right) \int _{(q-j)h}^{qh}g(t)f(t-qh+jh)dt\nonumber \\=: & {} (-1)^q\int _{qh}^\infty g(t)\Delta _h^qf(t-qh)dt+R_2. \end{aligned}$$

(3.6)

Now

$$\begin{aligned} |R_2|\le C(2^q-1)e^{qAh}\int _0^{qh}\frac{(nt)^{k+q}}{(k+q)!}dt=C(2^q-1)e^{qAh}\frac{n^{k+q}}{(k+q+1)!}{(qh)^{k+q+1}}. \end{aligned}$$

Finally, by (3.5), we have

$$\begin{aligned} \Delta ^qa_{n,k}=\frac{1}{n^qh^q}\int _{qh}^\infty e^{-nt}\frac{(nt)^{k+q}}{(k+q)!}\Delta _h^qf(t-qh)dt+R, \end{aligned}$$

where $R:=R_1+\frac{1}{(nh)^q}R_2=O(h)$, as $h\rightarrow 0$.

If f is q-monotone, then we get for all $h>0$ that

$$\begin{aligned} \Delta ^qa_{n,k}\ge R. \end{aligned}$$

Hence, as the left hand side is independent of h, we conclude that $\Delta ^qa_{n,k}\ge 0$, and the proof is complete. $\square$

4 The Baskakov–Durrmeyer operators

The Baskakov operators are defined for a function f of polynomial growth on $[0,\infty )$, by

$$\begin{aligned} (M_nf)(x):=\sum _{k=0}^\infty \left( {\begin{array}{c}n+k-1\\ k\end{array}}\right) x^k(1+x)^{-n-k}f\left( \frac{k}{n}\right) ,\quad x\in [0,\infty ),\quad n\ge 1. \end{aligned}$$

It is well-known that if, in addition, $f\in C[0,\infty )$, then the Baskakov operators associated with f, approximate it uniformly on every compact subinterval of $[0,\infty )$. It is also known that they preserve q-monotonicity on $[0,\infty )$, which readily follows from the equation,

$$\begin{aligned}&\frac{d^q}{dx^q}\left( M_nf\right) (x)\nonumber \\&\quad =\frac{(n+q-1)!}{(n-1)!}\sum _{k=0}^\infty \left( {\begin{array}{c}n+q+k-1\\ k\end{array}}\right) x^k(1+x)^{-n-q-k}\Delta ^qf\left( \frac{k}{n}\right) .\nonumber \\ \end{aligned}$$

(4.1)

Denoting

$$\begin{aligned} m_{n,k}(x):=\left( {\begin{array}{c}n+k-1\\ k\end{array}}\right) x^k(1+x)^{-n-k},\quad x\in [0,\infty ),\quad n\ge 1, k\ge 0, \end{aligned}$$

Sahai and Prasad [15] have defined the Baskakov–Durrmeyer variant of the above operators, for f such that $|f(x)|\le C(x+1)^A$, $C,A\ge 0$, by

$$\begin{aligned} (V_nf)(x):=(n-1)\sum _{k=0}^\infty m_{n,k}(x)\int _0^\infty m_{n,k}(t)f(t)dt,\quad x\in [0,\infty ),\quad n>A+1. \end{aligned}$$

Remark 4.1

In [15] the definition of the operators is only for integrable functions while one may define the operators for large enough n, for functions of polynomial growth. On the other hand, we prefer to assume that the functions are continuous in $[0,\infty )$, which, for $q\ge 2$, amounts to assuming continuity at $x=0$. This way we are guaranteed the approximation on compact subintervals of $[0,\infty )$.

For the Baskakov–Durrmeyer operators we have two results analogous to Theorems 3.1 and 3.2. Namely,

Theorem 4.2

Given $q\ge 1$, suppose that $f\in C[0,\infty )$ is q-monotone and there exist constants $C,A\ge 0$ such that $|f(x)|\le C(x+1)^A$, $0\le x<\infty$. Then $V_nf$, $n\ge 1$, $n>A+1$, is q-monotone in $[0,\infty )$.

Again, a weaker result, which is easier to prove, is

Theorem 4.3

Given $q\ge 1$, suppose that $f\in C^q[0,\infty )$ is q-monotone and there exist constants $C,A\ge 0$ such that $|f^{(i)}(x)|\le C(x+1)^A$, $0\le i\le q-1$, $0\le x<\infty$. Then $V_nf$, $n>A+q$, is q-monotone in $[0,\infty )$.

Yet another characterization of continuous q-monotone functions in $[0,\infty )$ is the next corollary that is proved exactly as Corollary 3.3, thus we omit the proof. Denote

$$\begin{aligned} a_{n,k}:=\int _0^\infty m_{n,k}(t)f(t)dt. \end{aligned}$$

(4.2)

Corollary 4.4

Let $q\ge 1$ and assume that $f\in C[0,\infty )$ is such that $|f(x)|\le C(x+1)^A$, $0\le x<\infty$, for some $C,A\ge 0$. Then, f is q-monotone in $[0,\infty )$, if and only if, (3.3) is valid, but this time, for $n>A+1$.

We begin with the proof of Theorem 4.3.

Proof of Theorem 4.3

The proof follows the lines of the proof of Theorem 3.1. Similar to (4.1), we obtain,

$$\begin{aligned} \frac{d^q}{dx^q}\left( V_nf\right) (x)=\frac{(n+q-1)!}{(n-2)!}\sum _{k=0}^\infty m_{n+q,k}(x)\Delta ^qa_{n,k}. \end{aligned}$$

Thus, we will conclude if we show that $\Delta ^qa_{n,k}\ge 0$. To this end, we observe that

$$\begin{aligned} m'_{n,k+1}(x)=-n\left( m_{n+1,k+1}(x)-m_{n+1,k}(x)\right) . \end{aligned}$$

Hence, for $n\ge 2$,

$$\begin{aligned} \Delta ^1a_{n,k}=-\frac{1}{n-1}\int _0^\infty m'_{n-1,k+1}(t)f(t)dt=\frac{1}{n-1}\int _0^\infty m_{n-1,k+1}(t)f'(t)dt. \end{aligned}$$

and by induction, for $n\ge q+1$,

$$\begin{aligned} \Delta ^qa_{n,k}=\frac{(n-q-1)!}{(n-1)!}\int _0^\infty m_{n-q,k+q}(t)f^{(q)}(t)dt\ge 0. \end{aligned}$$

Again, note that we do not need to assume growth conditions on $f^{(q)}$, for the integral to converge. This completes the proof. $\square$

Proof of Theorem 4.2

The proof is along the lines of the proof of Theorem 3.1 with the obvious changes. Similar to (3.5), we may write

$$\begin{aligned} \Delta ^qa_{n,k}=(-1)^q\frac{(n-q-1)!}{(n-1)!h^q}\int _0^\infty \Delta _h^qg(t)f(t)dt+R_1, \end{aligned}$$

where now $g(t):=m_{n-q,k+q}(t)$, and

$$\begin{aligned} |R_1|\le qh\frac{(n-q-1)!}{(n-1)!}\int _0^\infty \left[ \max _{t\le u\le t+qh}|g^{(q+1)}(u)|\right] |f(t)|dt. \end{aligned}$$

Since

$$\begin{aligned} m^{(q+1)}_{n-q,k+q}(u)=(-1)^{q+1}\frac{n!}{(n-q-1)!}\sum _{i=0}^{q+1}(-1)^i\left( {\begin{array}{c}q+1\\ i\end{array}}\right) m_{n+1,k+i-1}(u), \end{aligned}$$

(where $m_{n+1,-1}(u)\equiv 0$), all we have to do is to prove that for any pair (n, k),

$$\begin{aligned} \lim _{h\rightarrow 0}h\int _0^\infty \left[ \max _{t\le u\le t+qh}m_{n+1,k}(u)\right] (t+1)^Adt=0. \end{aligned}$$

Now, $m_{n+1,k}(u)\le 1$, and it is increasing in $[0,k/(n+1)]$ and decreasing in

$[k/(n+1),\infty )$. Thus,

$$\begin{aligned}&\int _0^\infty \left[ \max _{t\le u\le t+qh}m_{n+1,k}(u)\right] (t+1)^Adt\\&\quad \le \int _0^{k/(n+1)}(t+1)^Adt +\left( {\begin{array}{c}n+k\\ k\end{array}}\right) \int _{k/(n+1)}^\infty t^k(t+1)^A(1+t)^{-n-k-1}dt<\infty . \end{aligned}$$

Hence, $R_1=O(h)$, as $h\rightarrow 0$.

By virtue of (3.6), we may write

$$\begin{aligned} \int _0^\infty \Delta ^q_hg(t)f(t)dt=(-1)^q\int _{qh}^\infty g(t)\Delta ^q_hf(t-qh)dt+R_2. \end{aligned}$$

Again,

$$\begin{aligned} |R_2|\le C\left( {\begin{array}{c}n+k-1\\ k+q\end{array}}\right) (2^q-1)(qh+1)^A\frac{(qh)^{k+q+1}}{k+q+1}. \end{aligned}$$

Therefore, as before,

$$\begin{aligned} \Delta ^qa_{n,k}=\frac{(n-q-1)!}{(n-1)!h^q}\int _{qh}^\infty m_{n-q,k+q}(t)\Delta ^q_hf(t-qh)dt+R, \end{aligned}$$

where $R:=R_1+\frac{(n-q-1)!}{(n-1)!h^q}R_2=O(h)$, as $h\rightarrow 0$.

If f is q-monotone, then

$$\begin{aligned} \Delta ^qa_{n,k}\ge R. \end{aligned}$$

Hence, as the left hand side is independent of h, we conclude that $\Delta ^qa_{n,k}\ge 0$, and the proof is complete. $\square$

5 Concluding comments

In [1] the authors prove a qth version of Raşa inequality for the ordinary Bernstein, Favard–Mirakyan–Szász and Baskakov operators. We state the analogous results for the Durrmeyer variants of those operators. One readily observes that the proofs depend only on the appropriate $\Delta ^q$ differences being nonnegative. Hence, one may follow those in [1] verbatim, so we skip them here. We have,

Theorem 5.1

Let $q,n\in {\mathbb {N}}$ and assume that $f\in C[0,1]$ is q-monotone. Then for all $x,y\in [0,1]$,

$$\begin{aligned}&\mathrm {sgn}(x-y)^q\sum \limits _{\nu _1,\dots ,\nu _q=0}^{n}\sum _{j=0}^q(-1)^{q-j} \left( {\begin{array}{c}q\\ j\end{array}}\right) \left( \prod \limits _{i=1}^{j}p_{n,\nu _{i}}(x)\right) \left( \prod \limits _{i=j+1}^{q}p_{n,\nu _{i}}(y)\right) \\&\quad \times \int _0^\infty p_{qn,\nu _{1}+\cdots +\nu _{q}}(t)f(t)dt\ge 0. \end{aligned}$$

Theorem 5.2

Let $q\ge 1$ and assume that $f\in C[0,\infty )$ is q-monotone and is such that $|f(x)|\le Ce^{Ax}$, $x\in [0,\infty )$. Then for all $x,y\in [0,\infty )$ and each $n>A/q$,

$$\begin{aligned}&\mathrm {sgn}(x-y)^q\sum \limits _{\nu _1,\dots ,\nu _q=0}^{n}\sum _{j=0}^q(-1)^{q-j} \left( {\begin{array}{c}q\\ j\end{array}}\right) e^{-n(jx+(q-j)y)}\frac{(nx)^{\nu _1+\cdots +\nu _j}(ny)^{\nu _{j+1}+\cdots +\nu _q}}{\nu _1!\cdots \nu _q!}\\&\quad \times \int _0^\infty e^{-qnt}\frac{(nt)^{\nu _1+\cdots +\nu _q}}{(\nu _1+\cdots +\nu _q)!}f(t)dt\ge 0. \end{aligned}$$

And finally,

Theorem 5.3

Let $q\ge 1$ and assume that $f\in C[0,\infty )$ is q-monotone and is such that $|f(x)|\le C(x+1)^A$, $x\in [0,\infty )$. Then for all $x,y\in [0,\infty )$ and each $n>(A+1)/q$,

$$\begin{aligned}&\mathrm {sgn}(x-y)^q\sum \limits _{\nu _1,\dots ,\nu _q=0}^{n}\sum _{j=0}^q(-1)^{q-j} \left( {\begin{array}{c}q\\ j\end{array}}\right) \left( \prod \limits _{i=1}^{j}m_{n,\nu _{i}}(x)\right) \left( \prod \limits _{i=j+1}^{q}m_{n,\nu _{i}}(y)\right) \\&\quad \times \int _0^\infty m_{qn,\nu _{1}+\cdots +\nu _{q}}(t)f(t)dt\ge 0. \end{aligned}$$