1 Introduction

A generalization of Bernstein polynomials based on q-integers was proposed by Lupaş in 1987 in [1]. However, the Lupaş q-Bernstein operators are rational functions rather than polynomials. In 1997, Phillips [2] proposed the Phillips q-Bernstein polynomials, and for decades thereafter the application of q integers in positive linear operators became a hot topic in approximation theory, such as generalized q-Bernstein polynomials [36], Durrmeyer-type q-Bernstein operators [79], Kantorovich-type q-Bernstein operators [1013], etc. As we know, q integers play important roles not only in approximation theory, but also in CAGD. Based on the Phillips q-Bernstein polynomials [2], which are generalizations of Bernstein polynomials, generalized Bézier curves and surfaces were introduced in [1416]. In [14], Oruç and Phillips constructed q-Bézier curves using the basis functions of Phillips q-Bernstein polynomials. Dişibüyük and Oruç [15, 16] defined the q generalization of rational Bernstein–Bézier curves and tensor product q-Bernstein–Bézier surfaces. Moreover, Simeonov et al. [17] introduced a new variant of the blossom, the q blossom, which is specifically adapted to developing identities and algorithms for q-Bernstein bases and q-Bézier curves. In 2014, Han et al. [18] proposed a generalization of q-analog Bézier curves with one shape parameter, and established degree evaluation and de Casteljau algorithms and some other properties. In 2016, Han et al. [19] introduced a new generalization of weighted rational Bernstein–Bézier curves based on q integers, and investigated the generalized rational Bézier curve from a geometric point of view, obtaining degree evaluation and de Casteljau algorithms, etc.

Recently, Chen et al. [20] introduced a new family of α-Bernstein operators, and investigated some approximation properties, such as the rate of convergence, Voronovskaja-type asymptotic formulas, etc. They also obtained the monotonic and convex properties. For \(f(x)\in [0,1]\), \(n\in \mathbb{N}\), and any fixed real α, the α-Bernstein operators they introduced are defined as

$$\begin{aligned} T_{n,\alpha }=\sum_{i=0}^{n}f_{i}p_{n,i}^{(\alpha )}(x), \end{aligned}$$
(1)

where \(f_{i}=f ( \frac{i}{n} ) \). For \(i=0,1,\ldots,n\), the α-Bernstein polynomial \(p_{n,i}^{\alpha }(x)\) of degree n is defined by \(p_{1,0}^{(\alpha )}(x)=1-x\), \(p_{1,1}^{(\alpha )}(x)=x\) and

p n , i ( α ) ( x ) = [ ( n 2 i ) ( 1 α ) x + ( n 2 i 2 ) ( 1 α ) ( 1 x ) + ( n i ) α x ( 1 x ) ] × x i 1 ( 1 x ) n 1 i ,
(2)

where \(n\geq 2\).

Motivated by above research, in this paper we propose the q analogue of α-Bernstein operators, called \((\alpha , q)\)-Bernstein operators, which are defined as

$$ T_{n,q,\alpha }(f;x)=\sum_{i=0}^{n}f_{i}p_{n,q,i}^{(\alpha )}(x), $$
(3)

where \(q\in (0,1]\), \(f_{i}=f ( \frac{[i]_{q}}{[n]_{q}} ) \), \(i=0,1,2,\ldots,n\), \(p_{1,q,0}^{(\alpha )}(x)=1-x\), \(p_{1,q,1}^{(\alpha )}(x)=x\), and

p n , q , i ( α ) ( x ) = ( [ n 2 i ] q ( 1 α ) x + [ n 2 i 2 ] q ( 1 α ) q n i 2 ( 1 q n i 1 x ) + [ n i ] q α x ( 1 q n i 1 x ) ) x i 1 ( 1 x ) q n i 1 ( n 2 ) .
(4)

By simple computations, we can also express the \((\alpha , q)\) operators (3) as

T n , q , α ( f ; x ) = ( 1 α ) i = 0 n 1 g i [ n 1 i ] q x i ( 1 x ) q n 1 i + α i = 0 n f i [ n i ] q x i ( 1 x ) q n i ,
(5)

where

$$\begin{aligned} g_{i}= \biggl( 1-\frac{q^{n-1-i}[i]_{q}}{[n-1]_{q}} \biggr) f_{i}+ \frac{q ^{n-1-i}[i]_{q}}{[n-1]_{q}}f_{i+1}. \end{aligned}$$
(6)

Here, we mention some definitions based on q integers, the details of which can be found in [21, 22]. For any fixed real number \(0< q\leq 1\) and each non-negative integer k, we denote q-integers by \([k]_{q}\), where

$$ [k]_{q}:= \textstyle\begin{cases} \frac{1-q^{k}}{1-q}, &q\neq 1, \\ k, &q=1. \end{cases} $$

Also, q-factorial and q-binomial coefficients are defined as follows:

[ k ] q ! : = { [ k ] q [ k 1 ] q [ 1 ] q , k = 1 , 2 , , 1 , k = 0 , [ n k ] q : = [ n ] q ! [ k ] q ! [ n k ] q ! ( n k 0 ) .

The q-analog of \((1+x)^{n}\) is defined by \((1+x)_{q}^{n}:=\prod_{s=0}^{n-1} ( 1+q^{s}x ) \). The q derivative and q derivative of the product are defined as \(D_{q}f(x):=\frac{d_{q}f(x)}{d _{q}x}=\frac{f(qx)-f(x)}{(q-1)x} \) and \(D_{q}(f(x)g(x)):=f(qx)D_{q}g(x)+g(x)D _{q}f(x)\), respectively. We also have \(D_{q}x^{n}=[n]_{q}x^{n-1}\) and \(D_{q}(1-x)_{q}^{n}=-[n]_{q}(1-qx)_{q}^{n-1}\).

The rest of this paper is organized as follows. In the next section, we give some basic properties of the operators \(T_{n,q,\alpha }(f)\), such as the moments and central moments for proving the convergence theorems, the forward difference form of \(T_{n,q,\alpha }(f)\) for proving shape-preserving properties, etc. In Sect. 3, we obtain the convergence property and the rate of convergence theorem. In Sect. 4, we investigate some shape-preserving properties, such as monotonicity- and convexity-preserving properties with respect to \(f(x)\), and also we study the monotonicity with n and q of \(T_{n,q,\alpha }(f)\).

2 Auxiliary results

For proving the main results, we require the following lemmas.

Lemma 2.1

We have the following equalities:

$$\begin{aligned} T_{n,q,\alpha }(1;x)=1, \quad\quad T_{n,q,\alpha }(t;x)=x. \end{aligned}$$
(7)

Proof

By (5), we have

T n , q , α ( 1 ; x ) = ( 1 α ) i = 0 n 1 [ n 1 i ] q x i ( 1 x ) q n 1 i + α i = 0 n [ n i ] q x i ( 1 x ) q n i = 1 .

However,

T n , q , α ( t ; x ) = ( 1 α ) i = 0 n 1 [ ( 1 q n 1 i [ i ] q [ n 1 ] q ) [ i ] q [ n ] q + q n 1 i [ i ] q [ n 1 ] q [ i + 1 ] q [ n ] q ] [ n 1 i ] q x i ( 1 x ) q n 1 i + α i = 0 n [ i ] q [ n ] q [ n i ] q x i ( 1 x ) q n i = ( 1 α ) i = 0 n 1 [ i ] q [ n 1 ] q [ n 1 i ] q x i ( 1 x ) q n 1 i + α i = 0 n [ i ] q [ n ] q [ n i ] q x i ( 1 x ) q n i = ( 1 α ) x + α x = x .

Lemma 2.1 is proved. □

Remark 2.2

From Lemma 2.1, we know that the \((\alpha , q)\)-Bernstein operators \(T_{n,q,\alpha }(f;x)\) reproduce linear functions; that is,

$$\begin{aligned} T_{n,q,\alpha }(at+b;x)=ax+b, \end{aligned}$$

for all real numbers a and b.

We immediately obtain Lemma 2.3 from (5) and Lemma 2.1.

Lemma 2.3

For all functions f and g defined in \([0,1]\), \(x\in [0,1]\), real numbers λ, μ defined in \([0,1]\), and \(q\in (0,1]\), the following statements hold true.

  1. (i)

    Endpoint interpolation: \(T_{n,q,\alpha }(f;0)=f(0)\) and \(T_{n,q,\alpha }(f;1)=f(1)\).

  2. (ii)

    Linearity: \(T_{n,q,\alpha }(\lambda f+\mu g;x)= \lambda T_{n,q,\alpha }(f;x)+\mu T_{n,q,\alpha }(g;x)\).

  3. (iii)

    Non-negative: For \(0\leq \alpha \leq 1\) and \(0< q<1\), if f is non-negative on \([0,1]\), so is \((\alpha , q)\)-Bernstein operators \(T_{n,q,\alpha }(f;x)\).

  4. (iv)

    Monotone: For fixed \(0\leq \alpha \leq 1\) and \(0< q<1\), if \(f\geq g\), then \(T_{n,q,\alpha }(f;x)\geq T_{n,q,\alpha }(g;x)\).

Lemma 2.4

  1. (i)

    The \((\alpha , q)\)-Bernstein operators may be expressed in the form

    T n , q , α (f;x)= r = 0 n ( ( 1 α ) [ n 1 r ] q q r g 0 + α [ n r ] q q r f 0 ) x r ,
    (8)

    where [ n 1 n ] q =0, \(\triangle_{q}^{r}f_{j}=\triangle_{q}^{r-1}f_{j+1}-q ^{r-1}\triangle_{q}^{r-1}f_{j}\), \(r\geq 1\), with \(\triangle_{q}^{0}f _{j}=f_{j}=f ( \frac{[j]_{q}}{[n]_{q}} ) \).

  2. (ii)

    The higher-order forward difference of \(g_{i}\) may be expressed in the form

    $$\begin{aligned} \triangle_{q}^{r}g_{i}= \biggl( 1- \frac{q^{n-i-1}[i]_{q}}{[n-1]_{q}} \biggr) \triangle_{q}^{r}f_{i}+ \frac{q^{n-i-1-r}[i+r]_{q}}{[n-1]_{q}}\triangle _{q}^{r}f_{i+1}, \end{aligned}$$
    (9)

    where \(\triangle_{q}^{0}g_{i}=g_{i}\), which is defined in (6).

Proof

We can obtain (8) easily by [2]. Next, in order to prove (9), we use induction on r. It is clear that (9) holds for \(r=0\). Let us assume that (9) holds for some \(r=k\geq 0\). For \(r=k+1\), we have

$$\begin{aligned}& \triangle_{q}^{k+1}g_{i} \\& \quad = \triangle_{q}^{k}g_{i+1}-q^{k} \triangle_{q}^{k}g_{i} \\& \quad = \biggl( 1-\frac{q^{n-i-2}[i+1]_{q}}{[n-1]_{q}} \biggr) \triangle_{q} ^{k}f_{i+1}+\frac{q^{n-i-2-k}[i+k+1]_{q}}{[n-1]_{q}}\triangle_{q}^{k}f _{i+2} \\& \quad \quad {} -q^{k} \biggl[ \biggl( 1-\frac{q^{n-i-1}[i]_{q}}{[n-1]_{q}} \biggr) \triangle_{q}^{k}f_{i}+\frac{q^{n-i-k-1}[i+k]_{q}}{[n-1]_{q}} \triangle _{q}^{k}f_{i+1} \biggr] \\& \quad = \biggl[ 1-\frac{q^{n-i-2} ( 1+q[i]_{q} ) }{[n-1]_{q}} \biggr] \triangle_{q}^{k}f_{i+1}- \biggl( 1-\frac{q^{n-i-1}[i]_{q}}{[n-1]_{q}} \biggr) q ^{k}\triangle_{q}^{k}f_{i} \\& \quad \quad {} -\frac{q^{n-i-1}[i+k]_{q}}{[n-1]_{q}}\triangle_{q}^{k}f_{i+1}+ \frac{q ^{n-i-2-k}[i+k]_{q}}{[n-1]_{q}}\triangle_{q}^{k}f_{i+2} \\& \quad = \biggl( 1-\frac{q^{n-i-1}[i]_{q}}{[n-1]_{q}} \biggr) \triangle_{q}^{k+1}f _{i}-\frac{q^{n-i-2}}{[n-1]_{q}}\triangle_{q}^{k}f_{i+1}- \frac{q^{n-i-1}[i+k]_{q}}{[n-1]_{q}}\triangle_{q}^{k}f_{i+1} \\& \quad \quad {} +\frac{q^{n-i-2-k}[i+k+1]_{q}}{[n-1]_{q}}\triangle_{q}^{k}f_{i+2} \\& \quad = \biggl( 1-\frac{q^{n-i-1}[i]_{q}}{[n-1]_{q}} \biggr) \triangle_{q}^{k+1}f _{i}-\frac{q^{n-i-2}[i+k+1]_{q}}{[n-1]_{q}}\triangle_{q}^{k}f_{i+1}+ \frac{q ^{n-i-1-k}[i+k+1]_{q}}{[n-1]_{q}}\triangle_{q}^{k}f_{i+2} \\& \quad = \biggl( 1-\frac{q^{n-i-1}[i]_{q}}{[n-1]_{q}} \biggr) \triangle_{q}^{k+1}f _{i}+\frac{q^{n-i-k-2}[i+k+1]_{q}}{[n-1]_{q}} \bigl( \triangle_{q}^{k}f _{i+2}-q^{k}\triangle_{q}^{k}f_{i+1} \bigr) \\& \quad = \biggl( 1-\frac{q^{n-i-1}[i]_{q}}{[n-1]_{q}} \biggr) \triangle_{q}^{k+1}f _{i}+\frac{q^{n-i-k-2}[i+k+1]_{q}}{[n-1]_{q}}\triangle_{q}^{k+1}f_{i+1}. \end{aligned}$$

This shows that (9) holds when k is replaced by \(k+1\), and this completes the proof of Lemma 2.4. □

Since \(f [ \frac{[j]_{q}}{[n]_{q}},\frac{[j+1]_{q}}{[n]_{q}},\ldots, \frac{[j+k]_{q}}{[n]_{q}} ] =\frac{[n]_{q}^{k}\triangle_{q}^{k}f _{j}}{q^{\frac{k(2j+k-1)}{2}}[k]_{q}!}=\frac{f^{(k)}(\xi )}{k!}\), where \(\xi \in ( \frac{[j]_{q}}{[n]_{q}},\frac{[j+k]_{q}}{[n]_{q}} ) \), the q differences of the monomial \(x^{k}\) of order greater than k are zero. We see from Lemma 2.4 that, for all \(n\geq k\), \(T_{n,q,\alpha } ( t^{k};x ) \) is a polynomial of degree k. Actually, the \((\alpha , q)\)-Bernstein operators are degree-reducing on polynomials; that is, if f is a polynomial of degree m, and then \(T_{n,q,\alpha }(f)\) is a polynomial of degree \(\leq \min\{m,n\}\). In particular, we have the following results.

Lemma 2.5

Letting \(f(t)=t^{k}\), \(n-1\geq k\geq 2\), we have

$$\begin{aligned} T_{n,q,\alpha }\bigl(t^{k};x\bigr)=a_{k}x^{k}+a_{k-1}x^{k-1}+ \cdots+a_{1}x+a_{0}, \end{aligned}$$

where \(a_{k}=\frac{q^{\frac{k(k-1)}{2}}[n-2]_{q}!}{[n-k]_{q}![n]_{q} ^{k}} \{ (1-\alpha )[n-k]_{q}[n-1+k]_{q}+\alpha [n]_{q}[n-1]_{q} \} \).

Proof

Indeed, from (9) and \(\triangle_{q}^{k}f_{j}=\frac{q^{ \frac{k(2j+k-1)}{2}}[k]_{q}!f^{(k)}(\xi )}{k![n]_{q}^{k}}\), we have

$$\begin{aligned} \triangle_{q}^{k}g_{0}=\triangle_{q}^{k}f_{0}+ \frac{q^{n-1-k}[k]_{q}}{[n-1]_{q}}\triangle_{q}^{k}f_{1},\quad\quad \triangle_{q}^{k}f_{0}=\frac{q^{\frac{k(k-1)}{2}}[k]_{q}!}{[n]_{q} ^{k}},\quad\quad \triangle_{q}^{k}f_{1}=\frac{q^{\frac{k(k+1)}{2}}[k]_{q}!}{[n]_{q} ^{k}}. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} \triangle_{q}^{k}g_{0}= \biggl( 1+ \frac{q^{n-1}[k]_{q}}{[n-1]_{q}} \biggr) \frac{q ^{\frac{k(k-1)}{2}}[k]_{q}!}{[n]_{q}^{k}}= \frac{[n-1+k]_{q}}{[n-1]_{q}} \frac{q^{\frac{k(k-1)}{2}}[k]_{q}!}{[n]_{q} ^{k}}. \end{aligned}$$

Hence, using (8), we have

a k = [ ( 1 α ) [ n 1 k ] q [ n 1 + k ] q [ n 1 ] q + α [ n k ] q ] q k ( k 1 ) 2 [ k ] q ! [ n ] q k .

We then obtain the proof of Lemma 2.5 by simple computations. □

Lemma 2.6

The following equalities hold true:

$$\begin{aligned}& T_{n,q,\alpha } \bigl( t^{2};x \bigr) =x^{2}+ \frac{x(1-x)}{[n]_{q}}+\frac{(1- \alpha )q^{n-1}[2]_{q}x(1-x)}{[n]_{q}^{2}}, \end{aligned}$$
(10)
$$\begin{aligned}& T_{n,q,\alpha } \bigl( (t-x)^{2};x \bigr) =\frac{x(1-x)}{[n]_{q}}+ \frac{(1- \alpha )q^{n-1}[2]_{q}x(1-x)}{[n]_{q}^{2}}. \end{aligned}$$
(11)

Proof

For \(f(t)=t^{2}\), we have \(\triangle_{q}^{0}f_{0}=f_{0}=0\), \(\triangle_{q}^{1}f_{0}=f_{1}-f_{0}=\frac{1}{[n]_{q}^{2}}\), \(\triangle_{q}^{1}f_{1}=f_{2}-f_{1}=\frac{2q+q^{2}}{[n]_{q}^{2}}\), \(\triangle_{q}^{2}f_{0}=\triangle_{q}^{1}f_{1}-q\triangle_{q}^{1}f _{0}=f_{2}-[2]_{q}f_{1}+qf_{0}=\frac{q[2]_{q}}{[n]_{q}^{2}}\), and \(\triangle_{q}^{2}f_{1}=f_{3}-[2]_{q}f_{2}+qf_{1}=\frac{q^{3}+q^{4}}{[n]_{q} ^{2}}\). By (9), we have \(\triangle_{q}^{0}g_{0}=0\), and

$$\begin{aligned}& \triangle_{q}^{1}g_{0}=\triangle_{q}^{1}f_{0}+ \frac{q^{n-2}}{[n-1]_{q}}\triangle_{q}^{1}f_{1}= \frac{1}{[n]_{q}^{2}}+\frac{2q ^{n-1}+q^{n}}{[n-1]_{q}[n]_{q}^{2}}, \\& \triangle_{q}^{2}g_{0}=\triangle_{q}^{2}f_{0}+ \frac{q^{n-3}[2]_{q}}{[n-1]_{q}}\triangle_{q}^{2}f_{1}= \frac{q[2]_{q}}{[n]_{q} ^{2}}+\frac{[2]_{q} ( q^{n}+q^{n+1} ) }{[n-1]_{q}[n]_{q}^{2}}. \end{aligned}$$

From (8), we have

$$\begin{aligned}& T_{n,q,\alpha } \bigl( t^{2};x \bigr) \\& \quad = (1-\alpha )\triangle_{q}^{0}g_{0}+\alpha \triangle_{q}^{0}f_{0}+ \bigl[ (1-\alpha )[n-1]_{q}\triangle_{q}^{1}g_{0}+ \alpha [n]_{q} \triangle_{q}^{1}f_{0} \bigr] x \\& \quad \quad {} + \biggl[ (1-\alpha )\frac{[n-1]_{q}[n-2]_{q}}{[2]_{q}}\triangle_{q} ^{2}g_{0}+\alpha \frac{[n]_{q}[n-1]_{q}}{[2]_{q}}\triangle_{q}^{2}f _{0} \biggr] x^{2} \\& \quad = \biggl[ \frac{(1-\alpha )[n-1]_{q}}{[n]_{q}^{2}}+\frac{(1-\alpha ) ( 2q^{n-1}+q^{n} ) }{[n]_{q}^{2}}+\frac{\alpha }{[n]_{q}} \biggr] x \\& \quad \quad {} + \biggl[ \frac{(1-\alpha )q[n-1]_{q}[n-2]_{q}}{[n]_{q}^{2}}+\frac{(1- \alpha )[n-2]_{q} ( q^{n}+q^{n+1} ) }{[n]_{q}^{2}}+\frac{ \alpha q[n-1]_{q}}{[n]_{q}} \biggr] x^{2} \\& \quad = \frac{[n]_{q}+(1-\alpha )q^{n-1}[2]_{q}}{[n]_{q}^{2}}x+ \biggl( 1- \frac{1}{[n]_{q}}-\frac{(1-\alpha )q^{n-1}[2]_{q}}{[n]_{q}^{2}} \biggr) x ^{2} \\& \quad = x^{2}+\frac{x(1-x)}{[n]_{q}}+\frac{(1-\alpha )q^{n-1}[2]_{q}x(1-x)}{[n]_{q} ^{2}}. \end{aligned}$$

Hence, (10) is proved. Finally, using Lemma 2.1, we obtain

$$ T_{n,q,\alpha } \bigl( (t-x)^{2};x \bigr) =T_{n,q,\alpha } \bigl( t^{2};x \bigr) -2xT _{n,q,\alpha }(t;x)+x^{2}T_{n,q,\alpha }(1;x)=T_{n,q,\alpha } \bigl( t ^{2};x \bigr) -x^{2}. $$

Then (11) is proved by (10). This completes the proof of Lemma 2.6. □

3 Convergence properties

We now state the well-known Bohman–Korovkin theorem, followed by a proof based on that given by Cheney [23].

Theorem 3.1

Let \(\{L_{n}\}\) denote a sequence of monotone linear operators that map a function \(f\in C[a,b]\) to a function \(L_{n}f\in C[a,b]\), and let \(L_{n}f\rightarrow f\) uniformly on \([a,b]\) for \(f=1, t\) and \(t^{2}\). Then \(L_{n}f\rightarrow f\) uniformly on \([a,b]\) for all \(f\in C[a,b]\).

Theorem 3.1 leads to the following theorem on the convergence of \((\alpha , q)\)-Bernstein operators.

Theorem 3.2

Let \(q:=\{q_{n}\}\) denote a sequence such that \(q_{n}\in (0,1)\) and \(\lim_{n\rightarrow \infty }q_{n}=1\). Then, for any \(f\in C[0,1]\) and \(\alpha \in [0,1]\), \(T_{n,q,\alpha }(f;x)\) converges uniformly to \(f(x)\) on \([0,1]\).

Proof

From Lemma 2.1, we see that \(T_{n,q,\alpha }(f;x)=f(x)\) for \(f(t)=1\) and \(f(t)=t\). Since \(\lim_{n\rightarrow \infty }q_{n}=1\), we see from (10) that \(T_{n,q,\alpha }(f;x)\) converges uniformly to \(f(x)\) for \(f(t)=t^{2}\) as \(n\rightarrow \infty \). It also follows that \(T_{n,q,\alpha }\) is a monotone operator by Lemma 2.3; the proof is then completed by applying the Bohman–Korovkin theorem 3.1. □

As we know, the space \(C{[0,1]}\) of all continuous functions on \([0,1]\) is a Banach space with sup-norm \(\Vert f \Vert :=\sup_{x\in [0,1]} \vert f(x) \vert \). Letting \(f\in C{[0,1]}\), the Peetre K functional is defined by \(K_{2}(f;\delta ):=\inf_{g\in C^{2}{[0,1]}} \{ \Vert f-g \Vert +\delta \Vert g'' \Vert \}\), where \(\delta >0\) and \(C^{2}{[0,1]}:=\{g \in C{[0,1]}: g', g''\in C{[0,1]}\}\). By [24], there exists an absolute constant \(C>0\), such that

$$ K_{2}(f;\delta )\leq C\omega_{2} ( f;\sqrt{\delta } ) , $$
(12)

where \(\omega_{2}(f;\delta ):=\sup_{0< h\leq \delta } \sup_{x,x+h,x+2h\in [0,1]} \vert f(x+2h)-2f(x+h)+f(x) \vert \) is the second-order modulus of smoothness of \(f\in C{[0,1]}\).

Theorem 3.3

For \(f\in C{[0,1]}\), \(\alpha \in [0,1]\), \(q\in (0,1)\), we have

$$\begin{aligned} \bigl\vert T_{n,q,\alpha }(f;x)-f(x) \bigr\vert \leq C\omega_{2} \biggl( f;\frac{\sqrt{2[n]_{q}+(1- \alpha )2[2]_{q}q^{n-1}}}{4[n]_{q}} \biggr) , \end{aligned}$$

where C is a positive constant.

Proof

Letting \(g\in C^{2}{[0,1]}\), \(x,t\in [0,1]\), by Taylor’s expansion we have

$$\begin{aligned} g(t)=g(x)+g'(x) (t-x)+ \int_{x}^{t}(t-u)g''(u) \,du. \end{aligned}$$

Using Lemma 2.1, we obtain

$$\begin{aligned} T_{n,q,\alpha }(g;x)=g(x)+T_{n,q,\alpha } \biggl( \int_{x}^{t}(t-u)g''(u) \,du;x \biggr) . \end{aligned}$$

Thus, we have

$$\begin{aligned} \bigl\vert T_{n,q,\alpha }(g;x)-g(x) \bigr\vert =& \biggl\vert T_{n,q,\alpha } \biggl( \int _{x}^{t}(t-u)g''(u) \,du;x \biggr) \biggr\vert \\ \leq &T_{n,q,\alpha } \biggl( \biggl\vert \int_{x}^{t}(t-u) \bigl\vert g''(u) \bigr\vert \,du \biggr\vert ;x \biggr) \\ \leq &T_{n,q,\alpha } \bigl( (t-x)^{2};x \bigr) \bigl\Vert g'' \bigr\Vert \\ \leq &\frac{[n]_{q}+(1-\alpha )q^{n-1}[2]_{q}}{4[n]_{q}^{2}} \bigl\Vert g'' \bigr\Vert . \end{aligned}$$
(13)

However, using Lemma 2.1, we have

$$\begin{aligned} \bigl\vert T_{n,q,\alpha }(f;x) \bigr\vert \leq \Vert f \Vert . \end{aligned}$$
(14)

Now, (13) and (14) imply

$$\begin{aligned} \bigl\vert T_{n,q,\alpha }(f;x)-f(x) \bigr\vert \leq & \bigl\vert T_{n,q,\alpha }(f-g;x)-(f-g) (x) \bigr\vert + \bigl\vert T_{n,q,\alpha }(g;x)-g(x) \bigr\vert \\ \leq &2 \Vert f-g \Vert + \frac{[n]_{q}+(1-\alpha )q^{n-1}[2]_{q}}{4[n]_{q}^{2}} \bigl\Vert g'' \bigr\Vert . \end{aligned}$$

Hence, taking the infimum on the right-hand side over all \(g\in C^{2} {[0,1]}\), we obtain

$$\begin{aligned} \bigl\vert T_{n,q,\alpha }(f;x)-f(x) \bigr\vert \leq 2K_{2} \biggl( f;\frac{[n]_{q}+(1- \alpha )q^{n-1}[2]_{q}}{8[n]_{q}^{2}} \biggr) . \end{aligned}$$

By (12), we obtain

$$\begin{aligned} \bigl\vert T_{n,q,\alpha }(f;x)-f(x) \bigr\vert \leq C\omega_{2} \biggl( f;\frac{\sqrt{2[n]_{q}+(1- \alpha )2[2]_{q}q^{n-1}}}{4[n]_{q}} \biggr) , \end{aligned}$$

where C is a positive constant. Theorem 3.3 is proved. □

Remark 3.4

Letting \(q:=\{q_{n}\}\) denote a sequence such that \(q_{n}\in (0,1)\) and \(\lim_{n\rightarrow \infty }q_{n}=1\), we know that, under the conditions of theorem 3.3, the convergence rate of the operators \(T_{n,q, \alpha }(f)\) to f is \(1/\sqrt{[n]_{q}}\) as \(n\rightarrow \infty \). This convergence rate can be improved depending on the choice of q, at least as fast as \(1/\sqrt{n}\).

Example 3.5

Letting \(f(x) = 1 - \cos(4e^{x})\), the graphs of \(f(x)\) and \(T_{n,q,0.9}(f;x)\) with different values of n and q are shown in Fig. 1. Figure 2 shows the graphs of \(f(x)\) and \(T_{10,0.9,\alpha }(f;x)\) with \(\alpha =0.6\) and \(\alpha =0.9\).

Figure 1
figure 1

Convergence of \(T_{n,q,\alpha }(f;x)\) to \(f(x)\) for fixed \(\alpha =0.9\)

Figure 2
figure 2

Convergence of \(T_{n,q,\alpha }(f;x)\) to \(f(x)\) for fixed \(q=0.9\)

4 Shape-preserving properties

The \((\alpha , q)\)-Bernstein operators \(T_{n,q,\alpha }(f;x)\) have a monotonicity-preserving property.

Theorem 4.1

Let \(f\in C{[0,1]}\). If f is a monotonically increasing or monotonically decreasing function on \([0,1]\), so are all its \((\alpha , q)\)-Bernstein operators for fixed \(q\in (0,1)\) and \(\alpha \in [0,1]\).

Proof

From (5), we have

T n + 1 , q , α (f;x)=(1α) i = 0 n g i [ n i ] q x i ( 1 x ) q n i +α i = 0 n + 1 f i [ n + 1 i ] q x i ( 1 x ) q n + 1 i ,

where \(f_{i}=\frac{[i]_{q}}{[n+1]_{q}}\), \(g_{i}= ( 1- \frac{q^{n-i}[i]_{q}}{[n]_{q}} ) f_{i}+ \frac{q^{n-i}[i]_{q}}{[n]_{q}}f_{i+1}\). Then the q derivative of \(T_{n+1,q,\alpha }(f;x)\) is

D q [ T n + 1 , q , α ( f ; x ) ] = ( 1 α ) i = 0 n g i [ n i ] q D q [ x i ( 1 x ) q n i ] + α i = 0 n + 1 f i [ n + 1 i ] q D q [ x i ( 1 x ) q n + 1 i ] ,

and we denote the first and second parts of the right-hand side of the last equation by \(\Lambda_{1}\) and \(\Lambda_{2}\), respectively. We then have

Λ 1 = ( 1 α ) i = 0 n g i [ n i ] q [ [ i ] q x i 1 ( 1 q x ) q n i [ n i ] q x i ( 1 q x ) q n i 1 ] = ( 1 α ) [ n ] q [ i = 1 n g i [ n 1 i 1 ] q x i 1 ( 1 q x ) q n i i = 0 n 1 g i [ n 1 i ] q x i ( 1 q x ) q n i 1 ] = ( 1 α ) [ n ] q i = 0 n 1 [ n 1 i ] q x i ( 1 q x ) q n i 1 q 1 g i .

Using (9), we obtain

$$\begin{aligned} \triangle_{q}^{1}g_{i}= \biggl( 1- \frac{q^{n-i}[i]_{q}}{[n]_{q}} \biggr) \triangle_{q}^{1}f_{i}+ \frac{q^{n-i-1}[i+1]}{[n]_{q}}\triangle_{q} ^{1}f_{i+1}. \end{aligned}$$

Thus, we have

Λ 1 = ( 1 α ) i = 0 n 1 [ ( [ n ] q q n i [ i ] q ) q 1 f i + q n i 1 [ i + 1 ] q q 1 f i + 1 ] [ n 1 i ] q × x i ( 1 q x ) q n i 1 .
(15)

Similarly, we can obtain

Λ 2 =α [ n + 1 ] q i = 0 n [ n i ] q x i ( 1 q x ) q n i q 1 f i .
(16)

Therefore, by using (15) and (16), the derivative of \(( \alpha , q)\)-Bernstein operators \(T_{n,q,\alpha }(f;x)\) may be expressed in the form

D q [ T n , q , α ( f ; x ) ] = ( 1 α ) i = 0 n 1 [ ( [ n ] q q n i [ i ] q ) q 1 f i + q n i 1 [ i + 1 ] q q 1 f i + 1 ] [ n 1 i ] q × x i ( 1 q x ) q n i 1 + α [ n + 1 ] q i = 0 n [ n i ] q x i ( 1 q x ) q n i q 1 f i .

Since if f is monotonically increasing on \([0,1]\), the forward differences \(\triangle_{q}^{1}f_{i}\) and \(\triangle_{q}^{1}f_{i+1}\) are non-negative, and so is \(D_{q} [ T_{n,q,\alpha }(f;x) ] \). Hence, \((\alpha , q)\)-Bernstein operators \(T_{n,q,\alpha }(f;x)\) are monotonically increasing on \([0,1]\) for fixed \(q\in (0,1)\) and \(\alpha \in [0,1]\). On the contrary, if f is monotonically decreasing on \([0,1]\), then operators \(T_{n,q,\alpha }(f;x)\) are monotonically decreasing on \([0,1]\) for fixed \(q\in (0,1)\) and \(\alpha \in [0,1]\). Theorem 4.1 is proved. □

The \((\alpha , q)\)-Bernstein operators \(T_{n,q,\alpha }(f;x)\) have a convexity-preserving property

Theorem 4.2

Let \(f\in C{[0,1]}\). If f is convex on \([0,1]\), so are all of its \(( \alpha , q)\)-Bernstein operators \(T_{n,q,\alpha }(f;x)\) for fixed \(q\in (0,1)\) and \(\alpha \in [0,1]\).

Proof

From (5), we obtain

T n + 2 , q , α ( f ; x ) = ( 1 α ) i = 0 n + 1 g i [ n + 1 i ] q x i ( 1 x ) q n i + 1 + α i = 0 n + 2 f i [ n + 2 i ] q x i ( 1 x ) q n + 2 i ,

where \(f_{i}=\frac{[i]_{q}}{[n+2]_{q}}\), \(g_{i}= ( 1- \frac{q^{n-i+1}[i]_{q}}{[n+1]_{q}} ) f_{i}+ \frac{q^{n-i+1}[i]_{q}}{[n+1]_{q}}f_{i+1}\). The q-derivative of \(T_{n+2,q,\alpha }(f;x)\) can easily obtained by the proof theorem 4.1, which may be expressed as

D q [ T n + 2 , q , α ( f ; x ) ] = ( 1 α ) [ n + 1 ] q i = 0 n [ n i ] q x i ( 1 q x ) q n i ( g i + 1 g i ) + α [ n + 2 ] q i = 0 n + 1 [ n + 1 i ] q x i ( 1 q x ) q n i + 1 ( f i + 1 f i ) .

Then we have

D q 2 [ T n + 2 , q , α ( f ; x ) ] = ( 1 α ) [ n + 1 ] q i = 0 n [ n i ] q ( g i + 1 g i ) D q [ x i ( 1 q x ) q n i ] + α [ n + 2 ] q i = 0 n + 1 [ n + 1 i ] q ( f i + 1 f i ) D q [ x i ( 1 q x ) q n i 1 ] .

By some easy computations, we obtain

D q 2 [ T n + 2 , q , α ( f ; x ) ] = ( 1 α ) [ n + 1 ] q [ n ] q i = 0 n 1 [ n 1 i ] q x i ( 1 q 2 x ) q n i 1 q 2 g i + α [ n + 2 ] q [ n + 1 ] q i = 0 n [ n i ] q x i ( 1 q 2 x ) q n i q 2 f i ,

where \(\triangle_{q}^{2}g_{i}= ( 1- \frac{q^{n-i+1}[i]_{q}}{[n+1]_{q}} ) \triangle_{q}^{2}f_{i}+\frac{q ^{n-i-1}[i+2]_{q}}{[n+1]_{q}}\triangle_{q}^{2}f_{i+1}\). By the connection between the second-order q differences and convexity, we know that \(\triangle_{q}^{2}f_{i}\) and \(\triangle_{q}^{2}f_{i+1}\) are all non-negative since f is convex on \([0,1]\). Hence, we obtain \(D_{q}^{2} [ T_{n+2,q,\alpha }(f;x) ] \geq 0\), and then the convexity-preserving property of \(T_{n,q,\alpha }(f;x)\). Theorem 4.2 is proved. □

Next, if \(f(x)\) is convex, the \((\alpha , q)\)-Bernstein operators \(T_{n,q,\alpha }(f;x)\), for n fixed, are monotonic in q.

Theorem 4.3

For \(0< q_{1}\leq q_{2}\leq 1\), \(\alpha \in [0,1]\) and for \(f(x)\) convex on \([0,1]\), then \(T_{n,q_{2},\alpha }(f;x)\leq T_{n,q_{1},\alpha }(f;x)\).

Proof

In the following main proof of our results, we must introduce a linear polynomial function:

$$\begin{aligned} g(x)=\frac{f_{i+1}-f_{i}}{\frac{[i+1]_{q}}{[n]_{q}}- \frac{[i]_{q}}{[n]_{q}}} \biggl( x-\frac{[i]_{q}}{[n]_{q}} \biggr) +f_{i}, \end{aligned}$$
(17)

where \(\frac{[i]_{q}}{[n]_{q}}\leq x<\frac{[i+1]_{q}}{[n]_{q}}\), \(f _{i}=f ( \frac{[i]_{q}}{[n]_{q}} ) \), \(i=0,\ldots,n-1\). Then it is straightforward to check that \(g_{i}=g ( \frac{[i]_{q}}{[n-1]_{q}} ) \). Since f is convex on \([0,1]\), the intrinsic linear polynomial function \(g(x)\) must be convex on \([0,1]\) as well. Therefore, by the classical results of q-Bernstein operators (see [3]), we note that

$$\begin{aligned} T_{n,q,\alpha }(f;x)=(1-\alpha )B_{n-1}^{q}(g;x)+\alpha B_{n}^{q}(f;x). \end{aligned}$$
(18)

We have \(B_{n-1}^{q_{2}}(g;x)\leq B_{n-1}^{q_{1}}(g;x)\) and \(B_{n}^{q_{2}}(f;x)\leq B_{n}^{q_{1}}(f;x)\), and the desired result is obvious. Theorem 4.3 is proved. □

Finally, if \(f(x)\) is convex, we give the monotonicity of \((\alpha , q)\)-Bernstein operators \(T_{n,q,\alpha }(f; x)\) with n.

Theorem 4.4

If \(f(x)\) is convex on \([0,1]\), for fixed \(q\in (0,1)\) and \(\alpha \in [0,1]\), we have

$$\begin{aligned} T_{n-1,q,\alpha }(f;x)-T_{n,q,\alpha }(f;x)\geq 0\quad (n\geq 2). \end{aligned}$$

Proof

Combining (17) and (18), and the fact that if f and g are convex on \([0,1]\), then

$$\begin{aligned} B_{n-2}^{q}(g;x)\geq B_{n-1}^{q}(g;x), \quad\quad B_{n-1}^{q}(f;x)\geq B_{n}^{q}(f;x) \end{aligned}$$

(see [25]). The desired result is obvious. □

Example 4.5

Letting the convex function \(f(x) = 1 - \sin(\pi x)\), \(x\in [0,1]\), the graphs of \(f(x)\) and \(T_{n,0.9,0.9}(f;x)\) with different values of \(n=10, 15, 20, 30\) are shown in Fig. 3. Figure 4 shows the graphs of \(f(x) = 1 - \sin(\pi x)\) and \(T_{10,q,0.9}(f;x)\) with \(q=0.6, 0.7, 0.8, 0.9\).

Figure 3
figure 3

Monotonicity of \(T_{n,q,\alpha }(f;x)\) in the parameter n

Figure 4
figure 4

Monotonicity of \(T_{n,q,\alpha }(f;x)\) in the parameter q

5 Conclusion

In this paper, we proposed a new family of generalized Bernstein operators, named \((\alpha , q)\)-Bernstein operators, and denoted by \(T_{n,q,\alpha }(f)\). We study the rate of convergence of these operators, investigate their monotonicity-, convexity-preserving properties with respect to \(f(x)\), and also obtain their monotonicity with n and q of \(T_{n,q,\alpha }(f)\).