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# Different types of quantum integral inequalities via $$(\alpha ,m)$$-convexity

Journal of Inequalities and Applications20182018:264

https://doi.org/10.1186/s13660-018-1860-2

• Received: 19 May 2018
• Accepted: 20 September 2018
• Published:

## Abstract

In this paper, based on $$(\alpha,m)$$-convexity, we establish different type inequalities via quantum integrals. These inequalities generalize some results given in the literature.

## Keywords

• Quantum integral inequalities
• $$(\alpha, m)$$-convex functions
• Hermite–Hadamard’s inequality
• Simpson’s inequality

• 34A08
• 26A51
• 26D10
• 26D15

## 1 Introduction and preliminaries

Throughout the paper, let $$I:= [a, b]\subseteq\mathbb{R}$$ with $$0\leq a< b$$ be an interval, $$I^{\circ}$$ be the interior of I and let $$0< q<1$$ be a constant.

Let $$f:I\rightarrow\mathbb{R}$$ be convex on I, then the Hermite–Hadamard inequality holds:
$$f \biggl(\frac{a+b}{2} \biggr)\leq\frac{1}{b-a} \int^{b}_{a} f(x)\, \mathrm{d}x\leq \frac{f(a)+f(b)}{2}.$$
(1.1)
If $$f:I\rightarrow\mathbb{R}$$ is four times continuously differentiable on $$I^{\circ}$$ and $$\|f^{(4)} \|_{\infty}= \sup_{x\in (a,b)} |f^{(4)}(x) |<\infty$$, then the Simpson inequality holds:
$$\biggl\vert \frac{1}{3}\biggl[\frac{f(a)+f(b)}{2}+2f \biggl(\frac{a+b}{2} \biggr)\biggr]-\frac{1}{b-a} \int^{b}_{a} f(x)\, \mathrm{d}x \biggr\vert \leq \frac {1}{2880} \bigl\Vert f^{(4)} \bigr\Vert _{\infty}(b-a)^{4}.$$
(1.2)

Many researchers generalized the inequalities (1.1) and (1.2). For more details on these inequalities, see [58, 1014, 16, 17, 22, 24, 25].

In 2014, Tariboon and Ntouyas defined the q-derivative and q-integral as follows.

### Definition 1.1

()

Let $$f: I\rightarrow\mathbb{R}$$ be a continuous function and let $$x\in I$$. Then the q-derivative on I of f at x is defined as
$${}_{a}D_{q}f(x)=\frac{f(x)-f(qx+(1-q)a)}{(1-q)(x-a)}, \quad x\neq a,\qquad {}_{a}D_{q}f(a)={\lim_{x\to a}} {}_{a}D_{q}f(x).$$

### Definition 1.2

()

Let $$f: I\rightarrow\mathbb{R}$$ be a continuous function. Then the q-integral on I is defined as
$$\int^{x}_{a}f(t)\, {}_{a} \mathrm{d}_{q}t=(1-q) (x-a)\sum^{\infty}_{n=0}q^{n}f \bigl(q^{n}x+\bigl(1-q^{n}\bigr)a \bigr)$$
for $$x\in I$$. Moreover, if $$c\in(a,x)$$, then the q-integral on I is defined as
$$\int^{x}_{c}f(t)\, {}_{a} \mathrm{d}_{q}t= \int^{x}_{a}f(t)\, {}_{a} \mathrm{d}_{q}t- \int ^{c}_{a}f(t)\, {}_{a} \mathrm{d}_{q}t.$$

In the same paper, they also proved the following q-Hölder inequality.

### Theorem 1.1

()

Let $$f,g: I\rightarrow\mathbb{R}$$ be two continuous functions. Then the inequality
$$\int^{x}_{a} \bigl\vert f(t) \bigr\vert \bigl\vert g(t) \bigr\vert \, {}_{a}\mathrm{d}_{q}t\leq \biggl( \int ^{x}_{a} \bigl\vert f(t) \bigr\vert ^{r_{1}}\, {}_{a}\mathrm{d}_{q}t \biggr)^{\frac{1}{r_{1}}} \biggl( \int^{x}_{a} \bigl\vert g(t) \bigr\vert ^{r_{2}}\, {}_{a}\mathrm{d}_{q}t \biggr)^{\frac{1}{r_{2}}}$$
holds for all $$x\in I$$ and $$r_{1},r_{2}>1$$ with $$r_{1}^{-1}+r_{2}^{-1}=1$$.

In 2018, Alp et al. generalized the Hermite–Hadamard inequality to the form of q-integrals as follows.

### Theorem 1.2

()

Let $$f:I\rightarrow\mathbb{R}$$ be convex and differentiable on I with $$0< q<1$$. Then we have
$$f \biggl(\frac{qa+b}{1+q} \biggr)\leq\frac{1}{b-a} \int_{a}^{b}f(x)\, {}_{a}\mathrm {d}_{q}x\leq\frac{qf(a)+f(b)}{1+q}.$$
(1.3)

For more details on the inequality (1.3), see [15, 18, 20, 21, 23]. For other type quantum integral inequalities, the interested reader can refer to [3, 4, 27, 29, 31].

In 1993, Miheşan gave the definition of $$(\alpha,m)$$-convex functions as follows.

### Definition 1.3

()

For $$b^{*}>0$$, the function $$f: [0,b^{*}]\rightarrow\mathbb{R}$$ is named $$(\alpha,m)$$-convex with $$\alpha,m\in(0,1]$$ if the inequality
$$f \bigl(tx+m(1-t)y \bigr)\leq t^{\alpha}f(x)+m\bigl(1-t^{\alpha} \bigr)f(y)$$
holds for all $$x,y\in[0,b^{*}]$$ and $$t\in[0,1]$$.

This paper aims to establish different types of quantum integral inequalities via $$(\alpha,m)$$-convexity. Some relevant connections of the results obtained in this paper with previous ones are also pointed out.

## 2 Auxiliary results

For proving main results, we need the following lemma.

### Lemma 2.1

Let $$f: I\rightarrow\mathbb{R}$$ be a continuous and q-differentiable function on $$I^{\circ}$$ with $$0< q <1$$. Then the identity
\begin{aligned} &\lambda \bigl[\mu f(b)+(1-\mu)f(a) \bigr]+(1-\lambda)f \bigl(\mu b+(1-\mu )a \bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \\ &\quad =(b-a) \biggl\{ \int_{0}^{\mu}(qt+\lambda\mu-\lambda){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} + \int_{\mu}^{1} (qt+\lambda\mu-1){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm {d}_{q}t \biggr\} \end{aligned}
holds for all $$\lambda,\mu\in[0,1]$$ if $${}_{a}D_{q}f$$ is integrable on I.

### Proof

By an identical transformation, we get
\begin{aligned} &(b-a) \biggl\{ \int_{0}^{\mu}(qt+\lambda\mu-\lambda){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} + \int_{\mu}^{1} (qt+\lambda\mu-1){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm {d}_{q}t \biggr\} \\ &\quad =(b-a) \biggl\{ \int_{0}^{1} (qt+\lambda\mu-1){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} + \int_{0}^{\mu}(1-\lambda) {}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm {d}_{q}t \biggr\} . \end{aligned}
(2.1)
From Definition 1.1, we get
\begin{aligned} {}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)&=\frac {f(tb+(1-t)a)-f(q[tb+(1-t)a]+(1-q)a)}{(1-q)(tb+(1-t)a-a)} \\ &=\frac{f(tb+(1-t)a)-f(qtb+(1-qt)a)}{t(1-q)(b-a)}. \end{aligned}
Utilizing the above calculation and Definition 1.2, we have
\begin{aligned} & \int_{0}^{1}t {}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\quad = \int_{0}^{1} \frac{f(tb+(1-t)a)-f(qtb+(1-qt)a)}{(1-q)(b-a)}\,{}_{0} \mathrm {d}_{q}t \\ &\quad =\frac{1}{b-a} \Biggl\{ \sum_{n=0}^{\infty}q^{n} f \bigl(q^{n} b+\bigl(1-q^{n}\bigr)a \bigr) \\ &\qquad {} -\sum_{n=0}^{\infty}q^{n} f \bigl(q^{n+1} b+\bigl(1-q^{n+1}\bigr)a \bigr) \Biggr\} \\ &\quad =\frac{1}{b-a} \Biggl\{ \sum_{n=0}^{\infty}q^{n} f \bigl(q^{n} b+\bigl(1-q^{n}\bigr)a \bigr) \\ &\qquad {} -\frac{1}{q}\sum_{n=0}^{\infty}q^{n+1} f \bigl(q^{n+1} b+\bigl(1-q^{n+1}\bigr)a \bigr) \Biggr\} \\ &\quad =\frac{1}{b-a} \Biggl\{ f(b)+ \biggl(1-\frac{1}{q} \biggr)\sum _{n=1}^{\infty }q^{n} f \bigl(q^{n} b+\bigl(1-q^{n}\bigr)a \bigr) \Biggr\} \\ &\quad =\frac{1}{b-a} \Biggl\{ \frac{1}{q}f(b)-\frac{1-q}{q}\sum _{n=0}^{\infty }q^{n} f \bigl(q^{n} b+\bigl(1-q^{n}\bigr)a \bigr) \Biggr\} \\ &\quad =\frac{f(b)}{q(b-a)}-\frac{1}{q(b-a)^{2}} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x, \end{aligned}
(2.2)
\begin{aligned} & \int_{0}^{1}{}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\quad = \int_{0}^{1} \frac{f(tb+(1-t)a)-f(qtb+(1-qt)a)}{t(1-q)(b-a)}\,{}_{0} \mathrm {d}_{q}t \\ &\quad =\frac{1}{b-a} \Biggl\{ \sum_{n=0}^{\infty} f \bigl(q^{n} b+\bigl(1-q^{n}\bigr)a \bigr)-\sum _{n=0}^{\infty} f \bigl(q^{n+1} b+ \bigl(1-q^{n+1}\bigr)a \bigr) \Biggr\} \\ &\quad =\frac{f(b)-f(a)}{b-a} \end{aligned}
(2.3)
and
\begin{aligned} & \int_{0}^{\mu}{}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\quad = \int_{0}^{\mu}\frac {f(tb+(1-t)a)-f(qtb+(1-qt)a)}{t(1-q)(b-a)}\,{}_{0} \mathrm{d}_{q}t \\ &\quad =\frac{1}{b-a} \Biggl\{ \sum_{n=0}^{\infty} f \bigl(q^{n}\mu b+\bigl(1-q^{n}\mu \bigr)a \bigr) -\sum _{n=0}^{\infty} f \bigl(q^{n+1}\mu b+ \bigl(1-q^{n+1}\mu\bigr)a \bigr) \Biggr\} \\ &\quad =\frac{f(\mu b+(1-\mu)a)-f(a)}{b-a}. \end{aligned}
(2.4)
Substituting (2.2), (2.3) and (2.4) into (2.1), we can obtain the desired result. This ends the proof. □

### Remark 2.1

In Lemma 2.1, if one takes $$q\rightarrow1^{-}$$, one has [9, Lemma 2].

### Remark 2.2

Consider Lemma 2.1.
1. (i)
Putting $$\mu=0$$, we have
\begin{aligned}& f(a)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \\& \quad =(b-a) \int_{0}^{1} (qt-1){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t. \end{aligned}
(2.5)

2. (ii)
Putting $$\mu=1$$, we have
$$f(b)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x =(b-a) \int_{0}^{1} qt {}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t.$$
(2.6)

3. (iii)
Putting $$\mu=\frac{1}{1+q}$$, we have
\begin{aligned} &\lambda\frac{qf(a)+f(b)}{1+q}+(1-\lambda)f \biggl(\frac{qa+b}{1+q} \biggr)- \frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \\ &\quad =(b-a) \biggl\{ \int_{0}^{\frac{1}{1+q}} \biggl(qt-\frac{\lambda q}{1+q} \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr) \,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} + \int_{\frac{1}{1+q}}^{1} \biggl(qt+\frac{\lambda}{1+q}-1 \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr) \,{}_{0}\mathrm{d}_{q}t \biggr\} . \end{aligned}
(2.7)

### Remark 2.3

Consider Lemma 2.1.

(i) Putting $$\lambda=0$$, we get
\begin{aligned} &f \bigl(\mu b+(1-\mu)a \bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \\ &\quad =(b-a) \biggl\{ \int_{0}^{\mu}qt {}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm {d}_{q}t \\ &\qquad {} + \int_{\mu}^{1} (qt-1){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \biggr\} . \end{aligned}
(2.8)
Specially, taking $$\mu=\frac{1}{1+q}$$, we obtain the midpoint-like integral identity
\begin{aligned} &f \biggl({\frac{qa+b}{1+q}} \biggr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a}\mathrm {d}_{q}x \\ &\quad =(b-a) \biggl\{ \int_{0}^{\frac{1}{1+q}} qt {}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} + \int_{\frac{1}{1+q}}^{1} (qt-1){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm {d}_{q}t \biggr\} , \end{aligned}
which is presented by Alp et al. in [2, Lemma 11].
(ii) Putting $$\lambda=\frac{1}{3}$$, we get
\begin{aligned} &\frac{1}{3} \bigl[\mu f(b)+(1-\mu)f(a)+2f \bigl(\mu b+(1-\mu)a \bigr) \bigr]-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \\ &\quad =(b-a) \biggl\{ \int_{0}^{\mu}\biggl(qt+\frac{1}{3}\mu- \frac{1}{3} \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} + \int_{\mu}^{1} \biggl(qt+\frac{1}{3}\mu-1 \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr) \,{}_{0}\mathrm{d}_{q}t \biggr\} . \end{aligned}
(2.9)
Specially, taking $$\mu=\frac{1}{1+q}$$, we obtain the Simpson-like integral identity
\begin{aligned} &\frac{1}{3} \biggl[\frac{qf(a)+f(b)}{1+q}+2f \biggl(\frac{qa+b}{1+q} \biggr) \biggr]-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \\ &\quad =(b-a) \biggl\{ \int_{0}^{\frac{1}{1+q}} \biggl(qt-\frac{q}{3+3q} \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr) \,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} + \int_{\frac{1}{1+q}}^{1} \biggl(qt+\frac{1}{3+3q}-1 \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr) \,{}_{0}\mathrm{d}_{q}t \biggr\} . \end{aligned}
(2.10)
(iii) Putting $$\lambda=\frac{1}{2}$$, we get
\begin{aligned} &\frac{1}{2} \bigl[\mu f(b)+(1-\mu)f(a)+f \bigl(\mu b+(1-\mu)a \bigr) \bigr]-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \\ &\quad =(b-a) \biggl\{ \int_{0}^{\mu}\biggl(qt+\frac{1}{2}\mu- \frac{1}{2} \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} + \int_{\mu}^{1} \biggl(qt+\frac{1}{2}\mu-1 \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr) \,{}_{0}\mathrm{d}_{q}t \biggr\} . \end{aligned}
(2.11)
Specially, taking $$\mu=\frac{1}{1+q}$$, we obtain the averaged midpoint-trapezoid-like integral identity
\begin{aligned} &\frac{1}{2} \biggl[\frac{qf(a)+f(b)}{1+q}+f \biggl(\frac{qa+b}{1+q} \biggr) \biggr]-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \\ &\quad =(b-a) \biggl\{ \int_{0}^{\frac{1}{1+q}} \biggl(qt-\frac{q}{2+2q} \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr) \,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} + \int_{\frac{1}{1+q}}^{1} \biggl(qt+\frac{1}{2+2q}-1 \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr) \,{}_{0}\mathrm{d}_{q}t \biggr\} . \end{aligned}
(2.12)
(iv) Putting $$\lambda=1$$, we get
\begin{aligned} &\mu f(b)+(1-\mu)f(a)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \\ &\quad =(b-a) \int_{0}^{1} (qt+\mu-1){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t. \end{aligned}
(2.13)
Specially, taking $$\mu=\frac{1}{1+q}$$, we obtain the trapezoid-like integral identity
\begin{aligned} &\frac{qf(a)+f(b)}{1+q}-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \\ &\quad =(b-a) \int_{0}^{1} \biggl(qt+\frac{1}{1+q}-1 \biggr){}_{a}D_{q}f \bigl(tb+(1-t)a \bigr) \,{}_{0}\mathrm{d}_{q}t, \end{aligned}
which is presented by Sudsutad et al. in [26, Lemma 3.1].

It is worth to mention here that to the best of our knowledge the obtained identities (2.5)–(2.13) are new in the literature.

Next we provide some calculations which will be used in this paper.

### Lemma 2.2

Let $$\mu\in[0,1]$$ and $$\tau\in[0,\infty)$$. From Definition 1.2, we have
$$\int_{0}^{\mu}t^{\tau} \,{}_{0} \mathrm{d}_{q}t=(1-q)\sum_{n=0}^{\infty} \mu^{\tau +1}q^{(\tau+1)n} =\frac{\mu^{\tau+1}(1-q)}{1-q^{\tau+1}}$$
and
$$\int_{0}^{\mu}(1-t)^{\tau} \,{}_{0}\mathrm{d}_{q}t=(1-q)\mu\sum _{n=0}^{\infty }q^{n} \bigl(1-q^{n}\mu \bigr)^{\tau}.$$

### Lemma 2.3

Let $$\lambda,\mu\in[0,1]$$ and $$\tau\in[0,\infty)$$. Then we have
\begin{aligned} & \int_{0}^{\mu}t^{\tau} \bigl\vert qt-( \lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \\ & \quad = \textstyle\begin{cases} \frac{\mu^{\tau+1}(1-q)(\lambda-\lambda\mu)}{1-q^{\tau+1}}-\frac{q\mu ^{\tau+2}(1-q)}{1-q^{\tau+2}}, &(\lambda+q)\mu\leq\lambda, \\ \frac{2(1-q)^{2}(\lambda-\lambda\mu)^{\tau+2}}{(1-q^{\tau+1})(1-q^{\tau +2})}+\frac{q\mu^{\tau+2}(1-q)}{1-q^{\tau+2}}-\frac{\mu^{\tau +1}(1-q)(\lambda-\lambda\mu)}{1-q^{\tau+1}}, &(\lambda+q)\mu>\lambda, \end{cases}\displaystyle \end{aligned}
and
\begin{aligned} & \int_{0}^{\mu}(1-t)^{\tau} \bigl\vert qt-( \lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \\ &\quad = \textstyle\begin{cases} (1-q)\mu\sum_{n=0}^{\infty}q^{n} (\lambda-\lambda\mu-q^{n+1}\mu ) (1-q^{n}\mu )^{\tau}, &(\lambda+q)\mu\leq\lambda, \\ \left [ \textstyle\begin{array}{l} 2(1-q)(\lambda-\lambda\mu)^{2}\sum_{n=0}^{\infty}q^{n-1} (1-q^{n} ) [1-q^{n-1}(\lambda-\lambda\mu) ]^{\tau} \\ \quad {}-(1-q)\mu\sum_{n=0}^{\infty}q^{n} (\lambda-\lambda\mu-q^{n+1}\mu ) (1-q^{n}\mu )^{\tau} \end{array}\displaystyle \right ], &(\lambda+q)\mu>\lambda. \end{cases}\displaystyle \end{aligned}

### Proof

When $$(\lambda+q)\mu\leq\lambda$$, making use of Lemma 2.2, we get
\begin{aligned} \int_{0}^{\mu}t^{\tau} \bigl\vert qt-( \lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t &= \int_{0}^{\mu}\bigl[(\lambda-\lambda \mu)t^{\tau}-qt^{\tau+1} \bigr]\,{}_{0} \mathrm{d}_{q}t \\ &=\frac{\mu^{\tau+1}(1-q)(\lambda-\lambda\mu)}{1-q^{\tau+1}}-\frac{q\mu ^{\tau+2}(1-q)}{1-q^{\tau+2}}. \end{aligned}
When $$(\lambda+q)\mu>\lambda$$, making use of Lemma 2.2 again, we get
\begin{aligned} & \int_{0}^{\mu}t^{\tau} \bigl\vert qt-( \lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \\ &\quad = \int_{0}^{\frac{\lambda-\lambda\mu}{q}} \bigl[(\lambda-\lambda\mu )t^{\tau}-qt^{\tau+1} \bigr]\,{}_{0} \mathrm{d}_{q}t + \int^{\mu}_{\frac{\lambda-\lambda\mu}{q}} \bigl[qt^{\tau+1}-(\lambda - \lambda\mu)t^{\tau} \bigr]\,{}_{0}\mathrm{d}_{q}t \\ &\quad =2 \int_{0}^{\frac{\lambda-\lambda\mu}{q}} \bigl[(\lambda-\lambda\mu )t^{\tau}-qt^{\tau+1} \bigr]\,{}_{0} \mathrm{d}_{q}t + \int^{\mu}_{0} \bigl[qt^{\tau+1}-(\lambda- \lambda\mu)t^{\tau} \bigr]\,{}_{0}\mathrm{d}_{q}t \\ &\quad =\frac{2(1-q)^{2}(\lambda-\lambda\mu)^{\tau+2}}{(1-q^{\tau +1})(1-q^{\tau+2})}+\frac{q\mu^{\tau+2}(1-q)}{1-q^{\tau+2}}-\frac{\mu ^{\tau+1}(1-q)(\lambda-\lambda\mu)}{1-q^{\tau+1}}. \end{aligned}
Similarly, we also get
\begin{aligned} & \int_{0}^{\mu}(1-t)^{\tau} \bigl\vert qt-( \lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \\ &\quad = \textstyle\begin{cases} (1-q)\mu\sum_{n=0}^{\infty}q^{n} (\lambda-\lambda\mu-q^{n+1}\mu ) (1-q^{n}\mu )^{\tau}, &(\lambda+q)\mu\leq\lambda, \\ \left [ \textstyle\begin{array}{l} 2(1-q)(\lambda-\lambda\mu)^{2}\sum_{n=0}^{\infty}q^{n-1} (1-q^{n} ) [1-q^{n-1}(\lambda-\lambda\mu) ]^{\tau}\\ \quad {}-(1-q)\mu\sum_{n=0}^{\infty}q^{n} (\lambda-\lambda\mu-q^{n+1}\mu ) (1-q^{n}\mu )^{\tau} \end{array}\displaystyle \right ], &(\lambda+q)\mu>\lambda. \end{cases}\displaystyle \end{aligned}
This completes the proof. □

The following results of Lemma 2.4, Lemma 2.5 and Lemma 2.6 are stated without proof.

### Lemma 2.4

Let $$\lambda,\mu\in[0,1]$$ and $$\tau\in[0,\infty)$$. Then we have
\begin{aligned} & \int_{0}^{1} t^{\tau} \bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &\quad = \textstyle\begin{cases} \frac{(1-q)(1-\lambda\mu)}{1-q^{\tau+1}}-\frac{q(1-q)}{1-q^{\tau+2}}, &\lambda\mu+q\leq1, \\ \frac{2(1-q)^{2}(1-\lambda\mu)^{\tau+2}}{(1-q^{\tau+1})(1-q^{\tau +2})}+\frac{q(1-q)}{1-q^{\tau+2}}-\frac{(1-q)(1-\lambda\mu)}{1-q^{\tau+1}}, &\lambda\mu+q>1, \end{cases}\displaystyle \end{aligned}
and
\begin{aligned} & \int_{0}^{1} (1-t)^{\tau} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &\quad = \textstyle\begin{cases} (1-q)\sum_{n=0}^{\infty}q^{n} (1-\lambda\mu-q^{n+1} ) (1-q^{n} )^{\tau}, &\lambda\mu+q\leq1, \\ \left [ \textstyle\begin{array}{l} 2(1-q)(1-\lambda\mu)^{2}\sum_{n=0}^{\infty}q^{n-1} (1-q^{n} ) [1-q^{n-1}(1-\lambda\mu) ]^{\tau}\\ \quad {}-(1-q)\sum_{n=0}^{\infty}q^{n} (1-\lambda\mu-q^{n+1} ) (1-q^{n} )^{\tau} \end{array}\displaystyle \right ], &\lambda\mu+q>1. \end{cases}\displaystyle \end{aligned}

### Lemma 2.5

Let $$\lambda,\mu\in[0,1]$$ and $$\tau\in[0,\infty)$$. Then we have
\begin{aligned} & \int_{0}^{\mu}t^{\tau} \bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &\quad = \textstyle\begin{cases} \frac{\mu^{\tau+1}(1-\lambda\mu)(1-q)}{1-q^{\tau+1}}-\frac{q\mu^{\tau +2}(1-q)}{1-q^{\tau+2}}, &(\lambda+q)\mu\leq1, \\ \frac{2(1-q)^{2}(1-\lambda\mu)^{\tau+2}}{(1-q^{\tau+1})(1-q^{\tau +2})}+\frac{q\mu^{\tau+2}(1-q)}{1-q^{\tau+2}}-\frac{\mu^{\tau +1}(1-\lambda\mu)(1-q)}{1-q^{\tau+1}}, &(\lambda+q)\mu> 1, \end{cases}\displaystyle \end{aligned}
and
\begin{aligned} & \int_{0}^{\mu}(1-t)^{\tau} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &\quad = \textstyle\begin{cases} (1-q)\mu\sum_{n=0}^{\infty}q^{n} (1-\lambda\mu-q^{n+1}\mu ) (1-q^{n}\mu )^{\tau}, &(\lambda+q)\mu\leq1, \\ \left [ \textstyle\begin{array}{l} 2(1-q)(1-\lambda\mu)^{2}\sum_{n=0}^{\infty}q^{n-1} (1-q^{n} ) [1-q^{n-1}(1-\lambda\mu) ]^{\tau}\\ \quad {}-(1-q)\mu\sum_{n=0}^{\infty}q^{n} (1-\lambda\mu-q^{n+1}\mu ) (1-q^{n}\mu )^{\tau} \end{array}\displaystyle \right ], &(\lambda+q)\mu>1. \end{cases}\displaystyle \end{aligned}

### Lemma 2.6

Let $$\lambda,\mu\in[0,1]$$ and $$\theta\in[1,\infty)$$. Then we have
\begin{aligned} & \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert ^{\theta}\,{}_{0}\mathrm{d}_{q}t \\ &\quad = \textstyle\begin{cases} (1-q)\sum_{n=0}^{\infty}q^{n} (1-\lambda\mu-q^{n+1} )^{\theta}, &0\leq\lambda\mu\leq1-q, \\ \left [ \textstyle\begin{array}{l} (1-q)(1-\lambda\mu)^{\theta+1}\sum_{n=0}^{\infty}q^{n-1} (1-q^{n} )^{\theta}\\ \quad {}+(1-q)\sum_{n=0}^{\infty}q^{n} (q^{n+1}-1+\lambda\mu )^{\theta}\\ \quad {}-(1-q)(1-\lambda\mu)^{\theta+1}\sum_{n=0}^{\infty}q^{n-1} (q^{n}-1 )^{\theta}\end{array}\displaystyle \right ], &1-q< \lambda\mu\leq1. \end{cases}\displaystyle \end{aligned}

## 3 Main results

In 2018, Alp et al. established the q-Hermite–Hadamard inequality in . Here we give a new proof, which is more concise.

### Theorem 3.1

Let $$f:I\rightarrow\mathbb{R}$$ be a convex function on $$[a,b]$$ with $$0< q<1$$. Then we have
$$f \biggl(\frac{qa+b}{1+q} \biggr)\leq\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a}\mathrm {d}_{q}x\leq\frac{qf(a)+f(b)}{1+q}.$$

### Proof

It is obvious that $$\sum_{n=0}^{\infty}(1-q)q^{n}=1$$, $$0< q<1$$. Since Jensen’s inequality defined on convex sets for infinite sums still remains true, utilizing this fact and Definition 1.2, we have
\begin{aligned} f \biggl(\frac{qa+b}{1+q} \biggr) &=f \Biggl(\sum _{n=0}^{\infty}(1-q)q^{n} \bigl(q^{n}b+ \bigl(1-q^{n}\bigr)a \bigr) \Biggr) \\ &\leq\sum_{n=0}^{\infty}(1-q)q^{n}f \bigl(q^{n}b+\bigl(1-q^{n}\bigr)a \bigr) \\ &=\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x. \end{aligned}
Using Definition 1.2 and the convexity of f, we get
\begin{aligned} \frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x &= \sum_{n=0}^{\infty}(1-q)q^{n}f \bigl(q^{n}b+\bigl(1-q^{n}\bigr)a \bigr) \\ &\leq\sum_{n=0}^{\infty}(1-q)q^{n} \bigl[q^{n}f(b)+\bigl(1-q^{n}\bigr)f(a) \bigr] \\ &=\frac{qf(a)+f(b)}{1+q}. \end{aligned}
The proof is completed. □

Using Lemma 2.1, we can obtain the following theorem.

### Theorem 3.2

For $$0\leq a< b$$ and some fixed $$m\in(0,1]$$, let $$f: [a,\frac {b}{m} ]\rightarrow\mathbb{R}$$ be a continuous and q-differentiable function on $$(a,\frac{b}{m} )$$, and let $${}_{a}D_{q}f$$ be integrable on $$[a,\frac{b}{m} ]$$ with $$0< q <1$$. Then the inequality
\begin{aligned} & \biggl\vert \lambda \bigl[\mu f(b)+(1-\mu)f(a) \bigr]+(1-\lambda)f \bigl(\mu b+(1-\mu)a \bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq\min \bigl\{ \mathcal{H}_{1}(\lambda,\mu,\alpha,m),\mathcal {H}_{2}(\lambda,\mu,\alpha,m) \bigr\} \end{aligned}
holds for all $$\lambda,\mu\in[0,1]$$ if $$|{}_{a}D_{q}f|$$ is $$(\alpha ,m)$$-convex on $$[a,\frac{b}{m} ]$$ with $$\alpha,m\in(0,1]^{2}$$, where
\begin{aligned}& \begin{aligned} \mathcal{H}_{1}(\lambda,\mu,\alpha,m) &=(b-a) \biggl\{ \bigl[\Phi_{1}(\lambda,\mu,\alpha)+\Phi_{2}( \lambda,\mu,\alpha )-\Phi_{3}(\lambda,\mu,\alpha) \bigr] \bigl\vert {}_{a}D_{q}f(b) \bigr\vert \\ &\quad {} +m \bigl[\Phi_{4}(\lambda,\mu)+\Phi_{5}(\lambda, \mu)-\Phi_{6}(\lambda,\mu )-\Phi_{1}(\lambda,\mu,\alpha) \\ &\quad {} -\Phi_{2}(\lambda,\mu,\alpha)+\Phi_{3}(\lambda, \mu,\alpha) \bigr] \biggl\vert {}_{a}D_{q}f \biggl( \frac{a}{m} \biggr) \biggr\vert \biggr\} , \end{aligned} \\& \begin{aligned} \mathcal{H}_{2}(\lambda,\mu,\alpha,m) &=(b-a) \biggl\{ \bigl[\Phi_{7}(\lambda,\mu,\alpha)+\Phi_{8}(\lambda,\mu, \alpha )-\Phi_{9}(\lambda,\mu,\alpha)\bigr] \bigl\vert {}_{a}D_{q}f(a) \bigr\vert \\ &\quad {} +m\bigl[\Phi_{4}(\lambda,\mu)+\Phi_{5}(\lambda, \mu)-\Phi_{6}(\lambda,\mu )-\Phi_{7}(\lambda,\mu,\alpha) \\ &\quad {} -\Phi_{8}(\lambda,\mu,\alpha)+\Phi_{9}(\lambda, \mu,\alpha)\bigr] \biggl\vert {}_{a}D_{q}f\biggl( \frac{b}{m}\biggr) \biggr\vert \biggr\} , \end{aligned} \\ & \begin{aligned}\Phi_{1}(\lambda,\mu,\alpha)&= \int_{0}^{\mu}t^{\alpha} \bigl\vert qt-( \lambda -\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &= \textstyle\begin{cases} \frac{\mu^{\alpha+1}(1-q)(\lambda-\lambda\mu)}{1-q^{\alpha+1}}-\frac {q\mu^{\alpha+2}(1-q)}{1-q^{\alpha+2}}, &(\lambda+q)\mu\leq\lambda, \\ \frac{2(1-q)^{2}(\lambda-\lambda\mu)^{\alpha+2}}{(1-q^{\alpha +1})(1-q^{\alpha+2})}+\frac{q\mu^{\alpha+2}(1-q)}{1-q^{\alpha+2}}-\frac {\mu^{\alpha+1}(1-q)(\lambda-\lambda\mu)}{1-q^{\alpha+1}}, &(\lambda+q)\mu>\lambda, \end{cases}\displaystyle \end{aligned} \\ & \begin{aligned}[b] \Phi_{2}(\lambda,\mu,\alpha)&= \int_{0}^{1} t^{\alpha} \bigl\vert qt-(1- \lambda\mu ) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &= \textstyle\begin{cases} \frac{(1-q)(1-\lambda\mu)}{1-q^{\alpha+1}}-\frac{q(1-q)}{1-q^{\alpha+2}}, &\lambda\mu+q\leq1, \\ \frac{2(1-q)^{2}(1-\lambda\mu)^{\alpha+2}}{(1-q^{\alpha+1})(1-q^{\alpha +2})}+\frac{q(1-q)}{1-q^{\alpha+2}}-\frac{(1-q)(1-\lambda\mu )}{1-q^{\alpha+1}}, &\lambda\mu+q>1, \end{cases}\displaystyle \end{aligned} \end{aligned}
(3.1)
\begin{aligned}& \begin{aligned} \Phi_{3}(\lambda,\mu,\alpha)&= \int_{0}^{\mu}t^{\alpha} \bigl\vert qt-(1- \lambda\mu ) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &= \textstyle\begin{cases} \frac{\mu^{\alpha+1}(1-\lambda\mu)(1-q)}{1-q^{\alpha+1}}-\frac{q\mu ^{\alpha+2}(1-q)}{1-q^{\alpha+2}}, &(\lambda+q)\mu\leq1, \\ \frac{2(1-q)^{2}(1-\lambda\mu)^{\alpha+2}}{(1-q^{\alpha+1})(1-q^{\alpha +2})}+\frac{q\mu^{\alpha+2}(1-q)}{1-q^{\alpha+2}}-\frac{\mu^{\alpha +1}(1-\lambda\mu)(1-q)}{1-q^{\alpha+1}}, &(\lambda+q)\mu> 1, \end{cases}\displaystyle \end{aligned} \\ & \begin{aligned} \Phi_{4}(\lambda,\mu)&= \int_{0}^{\mu}\bigl\vert qt-(\lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &= \textstyle\begin{cases} \lambda\mu(1-\mu)-\frac{q\mu^{2}}{1+q}, &(\lambda+q)\mu\leq\lambda, \\ \frac{2(\lambda-\lambda\mu)^{2}}{1+q}+\frac{q\mu^{2}}{1+q}- \lambda\mu (1-\mu), &(\lambda+q)\mu>\lambda, \end{cases}\displaystyle \end{aligned} \\ & \begin{aligned}[b] \Phi_{5}(\lambda,\mu)&= \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \\ &= \textstyle\begin{cases} \frac{1}{1+q}-\lambda\mu, &\lambda\mu+q\leq1, \\ \frac{2(1-\lambda\mu)^{2}}{1+q}+\lambda\mu-\frac{1}{1+q}, &\lambda\mu+q>1, \end{cases}\displaystyle \end{aligned} \end{aligned}
(3.2)
\begin{aligned}& \begin{aligned} \Phi_{6}(\lambda,\mu) & = \int_{0}^{\mu}\bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ & = \textstyle\begin{cases} \mu(1-\lambda\mu)-\frac{q\mu^{2}}{1+q}, &(\lambda+q)\mu\leq1, \\ \frac{2(1-\lambda\mu)^{2}}{1+q}+\frac{q\mu^{2}}{1+q}-\mu(1-\lambda\mu), &(\lambda+q)\mu> 1, \end{cases}\displaystyle \end{aligned} \\ & \begin{aligned} &\Phi_{7}(\lambda,\mu,\alpha) \\ &\quad = \int_{0}^{\mu}(1-t)^{\alpha} \bigl\vert qt-( \lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \\ &\quad = \textstyle\begin{cases} (1-q)\mu\sum_{n=0}^{\infty}q^{n} (\lambda-\lambda\mu-q^{n+1}\mu ) (1-q^{n}\mu )^{\alpha}, &(\lambda+q)\mu\leq\lambda, \\ \left [ \textstyle\begin{array}{l} 2(1-q)(\lambda-\lambda\mu)^{2}\sum_{n=0}^{\infty}q^{n-1} (1-q^{n} ) [1-q^{n-1}(\lambda-\lambda\mu) ]^{\alpha}\\ \quad {}-(1-q)\mu\sum_{n=0}^{\infty}q^{n} (\lambda-\lambda\mu-q^{n+1}\mu ) (1-q^{n}\mu )^{\alpha} \end{array}\displaystyle \right ], &(\lambda+q)\mu>\lambda, \end{cases}\displaystyle \end{aligned} \\ & \begin{aligned}[b] &\Phi_{8}(\lambda,\mu,\alpha) \\ &\quad = \int_{0}^{1} (1-t)^{\alpha} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \\ &\quad = \textstyle\begin{cases} (1-q)\sum_{n=0}^{\infty}q^{n} (1-\lambda\mu-q^{n+1} ) (1-q^{n} )^{\alpha}, &\lambda\mu+q\leq1, \\ \left [ \textstyle\begin{array}{l} 2(1-q)(1-\lambda\mu)^{2}\sum_{n=0}^{\infty}q^{n-1} (1-q^{n} ) [1-q^{n-1}(1-\lambda\mu) ]^{\alpha}\\ \quad {}-(1-q)\sum_{n=0}^{\infty}q^{n} (1-\lambda\mu-q^{n+1} ) (1-q^{n} )^{\alpha} \end{array}\displaystyle \right ], &\lambda\mu+q>1, \end{cases}\displaystyle \end{aligned} \end{aligned}
(3.3)
and
\begin{aligned} &\Phi_{9}(\lambda,\mu,\alpha) \\ &\quad = \int_{0}^{\mu}(1-t)^{\alpha} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \\ &\quad = \textstyle\begin{cases} (1-q)\mu\sum_{n=0}^{\infty}q^{n} (1-\lambda\mu-q^{n+1}\mu ) (1-q^{n}\mu )^{\alpha}, &(\lambda+q)\mu\leq1, \\ \left [ \textstyle\begin{array}{l} 2(1-q)(1-\lambda\mu)^{2}\sum_{n=0}^{\infty}q^{n-1} (1-q^{n} ) [1-q^{n-1}(1-\lambda\mu) ]^{\alpha}\\ \quad {}-(1-q)\mu\sum_{n=0}^{\infty}q^{n} (1-\lambda\mu-q^{n+1}\mu ) (1-q^{n}\mu )^{\alpha} \end{array}\displaystyle \right ], &(\lambda+q)\mu>1. \end{cases}\displaystyle \end{aligned}

### Proof

From Lemma 2.1, utilizing the property of the modulus and the $$(\alpha,m)$$-convexity of $$|{}_{a}D_{q}f|$$, we have
\begin{aligned} & \biggl\vert \lambda\bigl[\mu f(b)+(1-\mu)f(a)\bigr]+(1-\lambda)f\bigl(\mu b+(1-\mu)a\bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq(b-a) \biggl\{ \int_{0}^{\mu} \vert qt+\lambda\mu-\lambda \vert \bigl\vert {}_{a}D_{q}f\bigl(tb+(1-t)a\bigr) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} + \int_{\mu}^{1} \vert qt+\lambda\mu-1 \vert \bigl\vert {}_{a}D_{q}f\bigl(tb+(1-t)a \bigr) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \biggr\} \\ &\quad \leq(b-a) \biggl\{ \int_{0}^{\mu}\bigl\vert qt-(\lambda-\lambda\mu) \bigr\vert \biggl[t^{\alpha}\bigl\vert {}_{a}D_{q}f(b) \bigr\vert +m\bigl(1-t^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac {a}{m}\biggr) \biggr\vert \biggr]\,{}_{0} \mathrm{d}_{q}t \\ &\qquad {} + \int_{\mu}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \biggl[t^{\alpha}\bigl\vert {}_{a}D_{q}f(b) \bigr\vert +m\bigl(1-t^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{a}{m}\biggr) \biggr\vert \biggr]\,{}_{0} \mathrm{d}_{q}t \biggr\} \\ &\quad =(b-a) \biggl\{ \int_{0}^{\mu}\bigl\vert qt-(\lambda-\lambda\mu) \bigr\vert \biggl[t^{\alpha}\bigl\vert {}_{a}D_{q}f(b) \bigr\vert +m\bigl(1-t^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac {a}{m}\biggr) \biggr\vert \biggr]\,{}_{0} \mathrm{d}_{q}t \\ &\qquad {} + \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \biggl[t^{\alpha}\bigl\vert {}_{a}D_{q}f(b) \bigr\vert +m\bigl(1-t^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{a}{m}\biggr) \biggr\vert \biggr]\,{}_{0} \mathrm{d}_{q}t \\ &\qquad {} - \int_{0}^{\mu}\bigl\vert qt-(1-\lambda\mu) \bigr\vert \biggl[t^{\alpha}\bigl\vert {}_{a}D_{q}f(b) \bigr\vert +m\bigl(1-t^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{a}{m}\biggr) \biggr\vert \biggr]\,{}_{0} \mathrm{d}_{q}t \biggr\} \\ &\quad =(b-a) \biggl\{ \biggl[ \int_{0}^{\mu}t^{\alpha}\bigl\vert qt-( \lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t + \int_{0}^{1} t^{\alpha}\bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} - \int_{0}^{\mu}t^{\alpha}\bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \biggr] \bigl\vert {}_{a}D_{q}f(b) \bigr\vert \\ &\qquad {} +m \biggl[ \int_{0}^{\mu}\bigl\vert qt-(\lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t + \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} - \int_{0}^{\mu}\bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t - \int_{0}^{\mu}t^{\alpha}\bigl\vert qt-( \lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \\ &\qquad {} - \int_{0}^{1} t^{\alpha}\bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t + \int_{0}^{\mu}t^{\alpha}\bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \biggr] \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert \biggr\} . \end{aligned}
Similarly, we get
\begin{aligned} & \biggl\vert \lambda\bigl[\mu f(b)+(1-\mu)f(a)\bigr]+(1-\lambda)f\bigl(\mu b+(1-\mu)a\bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq(b-a) \biggl\{ \biggl[ \int_{0}^{\mu}(1-t)^{\alpha}\bigl\vert qt-( \lambda-\lambda \mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t + \int_{0}^{1} (1-t)^{\alpha}\bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} - \int_{0}^{\mu}(1-t)^{\alpha}\bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \biggr] \bigl\vert {}_{a}D_{q}f(a) \bigr\vert \\ &\qquad {} +m\biggl[ \int_{0}^{\mu}\bigl\vert qt-(\lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t + \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \\ &\qquad {} - \int_{0}^{\mu}\bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t - \int_{0}^{\mu}(1-t)^{\alpha}\bigl\vert qt-( \lambda-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \\ &\qquad {} - \int_{0}^{1} (1-t)^{\alpha}\bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t + \int_{0}^{\mu}(1-t)^{\alpha}\bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \biggr] \biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m} \biggr) \biggr\vert \biggr\} . \end{aligned}
Using Lemma 2.3, Lemma 2.4 and Lemma 2.5, we get the desired result. This completes the proof. □

### Corollary 3.1

In Theorem 3.2, putting $$\mu=\frac{1}{1+q}$$, we have
\begin{aligned} & \biggl\vert \lambda\frac{qf(a)+f(b)}{1+q}+(1-\lambda)f \biggl(\frac {qa+b}{1+q} \biggr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq\min \biggl\{ \mathcal{H}_{1} \biggl(\lambda, \frac{1}{1+q},\alpha,m \biggr),\mathcal{H}_{2} \biggl(\lambda, \frac{1}{1+q},\alpha,m \biggr) \biggr\} . \end{aligned}

### Remark 3.1

Consider Corollary 3.1.

(i) Putting $$\lambda=0$$, we get the midpoint-like integral inequality
\begin{aligned} & \biggl\vert f \biggl(\frac{qa+b}{1+q} \biggr)-\frac{1}{b-a} \int _{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq\min \biggl\{ \mathcal{H}_{1} \biggl(0,\frac{1}{1+q}, \alpha,m \biggr),\mathcal{H}_{2} \biggl(0,\frac{1}{1+q},\alpha,m \biggr) \biggr\} , \end{aligned}
where
\begin{aligned} &\mathcal{H}_{1} \biggl(0,\frac{1}{1+q},\alpha,m \biggr) \\ &\quad =(b-a) \biggl\{ \frac{[(1+q)^{\alpha+2}-(1+q^{\alpha +2})](1-q)^{2}}{(1+q)^{\alpha+2}(1-q^{\alpha+1})(1-q^{\alpha+2})} \bigl\vert {}_{a}D_{q}f(b) \bigr\vert \\ &\qquad {} +m \biggl[\frac{2q}{(1+q)^{3}}-\frac{[(1+q)^{\alpha+2}-(1+q^{\alpha +2})](1-q)^{2}}{(1+q)^{\alpha+2}(1-q^{\alpha+1})(1-q^{\alpha+2})} \biggr] \biggl\vert {}_{a}D_{q}f \biggl(\frac{a}{m} \biggr) \biggr\vert \biggr\} \end{aligned}
and
\begin{aligned} &\mathcal{H}_{2}\biggl(0,\frac{1}{1+q},\alpha,m\biggr) \\ &\quad = (b-a)\biggl\{ \biggl[\Phi_{7}\biggl(0,\frac{1}{1+q}, \alpha\biggr)+\Phi _{8}\biggl(0,\frac{1}{1+q},\alpha\biggr)- \Phi_{9}\biggl(0,\frac{1}{1+q},\alpha \biggr)\biggr] \bigl\vert {}_{a}D_{q}f(a) \bigr\vert \\ &\qquad {} +m\biggl[\frac{2q}{(1+q)^{3}}-\Phi_{7}\biggl(0, \frac{1}{1+q},\alpha \biggr)-\Phi_{8}\biggl(0,\frac{1}{1+q}, \alpha\biggr) \\ &\qquad {} +\Phi_{9}\biggl(0,\frac{1}{1+q},\alpha\biggr)\biggr] \biggl\vert {}_{a}D_{q}f \biggl(\frac{b}{m}\biggr) \biggr\vert \biggr\} . \end{aligned}
Specially, taking $$\alpha=1=m$$, we obtain
\begin{aligned} & \biggl\vert f \biggl(\frac{qa+b}{1+q} \biggr)-\frac{1}{b-a} \int _{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq(b-a) \biggl\{ \frac{3q}{(1+q)^{3}(1+q+q^{2})} \bigl\vert {}_{a}D_{q}f(b) \bigr\vert +\frac{-q+2q^{2}+2q^{3}}{(1+q)^{3}(1+q+q^{2})} \bigl\vert {}_{a}D_{q}f(a) \bigr\vert \biggr\} , \end{aligned}
which is established by Alp et al. in [2, Theorem 13].
(ii) Putting $$\lambda=\frac{1}{3}$$ and $$\alpha=1=m$$, we get the Simpson-like integral inequality
\begin{aligned} & \biggl\vert \frac{1}{3} \biggl[\frac{qf(a)+f(b)}{1+q}+2f \biggl( \frac {qa+b}{1+q} \biggr) \biggr]-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq\min \biggl\{ \mathcal{H}_{1} \biggl(\frac{1}{3}, \frac{1}{1+q},1,1 \biggr),\mathcal{H}_{2} \biggl( \frac{1}{3},\frac{1}{1+q},1,1 \biggr) \biggr\} . \end{aligned}
Specially, if $$q\rightarrow1^{-}$$, then we obtain
$$\biggl\vert \frac{1}{3} \biggl[\frac{f(a)+f(b)}{2}+2f \biggl( \frac{a+b}{2} \biggr) \biggr]-\frac{1}{b-a} \int_{a}^{b}f(x)\, \mathrm{d}x \biggr\vert \leq \frac{5(b-a)}{72} \bigl[ \bigl\vert f'(b) \bigr\vert + \bigl\vert f'(a) \bigr\vert \bigr],$$
which is established by Alomari et al. in [1, Corollary 1].
(iii) Putting $$\lambda=\frac{1}{2}$$ and $$\alpha=1=m$$, we get the averaged midpoint-trapezoid-like integral inequality
\begin{aligned} & \biggl\vert \frac{1}{2} \biggl[\frac{qf(a)+f(b)}{1+q}+f \biggl( \frac {qa+b}{1+q} \biggr) \biggr]-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq\min \biggl\{ \mathcal{H}_{1} \biggl(\frac{1}{2}, \frac{1}{1+q},1,1 \biggr),\mathcal{H}_{2} \biggl( \frac{1}{2},\frac{1}{1+q},1,1 \biggr) \biggr\} . \end{aligned}
Specially, if $$q\rightarrow1^{-}$$, then we obtain
$$\biggl\vert \frac{1}{2}\biggl[\frac{f(a)+f(b)}{2}+f\biggl( \frac{a+b}{2} \biggr)\biggr]-\frac{1}{b-a} \int_{a}^{b}f(x)\mathrm{d}x \biggr\vert \leq \frac{b-a}{16} \bigl[ \bigl\vert f'(b) \bigr\vert + \bigl\vert f'(a) \bigr\vert \bigr],$$
which is established by Xi and Qi in [30, Corollary 3.4].
(iv) Putting $$\lambda=1$$, we get the trapezoid-like integral inequality
\begin{aligned} & \biggl\vert \frac{qf(a)+f(b)}{1+q}-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a}\mathrm {d}_{q}x \biggr\vert \\ &\quad \leq\min \biggl\{ \mathcal{H}_{1}\biggl(1,\frac{1}{1+q}, \alpha,m \biggr),\mathcal{H}_{2}\biggl(1,\frac{1}{1+q},\alpha,m \biggr) \biggr\} , \end{aligned}
where
\begin{aligned} &\mathcal{H}_{1}\biggl(1,\frac{1}{1+q},\alpha,m\biggr) \\ &\quad = (b-a) \biggl\{ \frac{2q^{\alpha+2}(1-q)^{2}+q^{2}(1+q)^{\alpha +1}(1-q)(1-q^{\alpha})}{(1+q)^{\alpha+2}(1-q^{\alpha+1})(1-q^{\alpha +2})} \bigl\vert {}_{a}D_{q}f(b) \bigr\vert \\ &\qquad {} +m\biggl[\frac{2q^{2}}{(1+q)^{3}}-\frac{2q^{\alpha +2}(1-q)^{2}+q^{2}(1+q)^{\alpha+1}(1-q)(1-q^{\alpha})}{(1+q)^{\alpha +2}(1-q^{\alpha+1})(1-q^{\alpha+2})}\biggr] \biggl\vert {}_{a}D_{q}f \biggl(\frac {a}{m} \biggr) \biggr\vert \biggr\} \end{aligned}
and
\begin{aligned} &\mathcal{H}_{2} \biggl(1,\frac{1}{1+q},\alpha,m \biggr) \\ &\quad = (b-a)\biggl\{ \biggl[\Phi_{7} \biggl(1,\frac{1}{1+q}, \alpha \biggr)+\Phi _{8} \biggl(1,\frac{1}{1+q},\alpha \biggr)- \Phi_{9} \biggl(1,\frac{1}{1+q},\alpha \biggr) \biggr] \bigl\vert {}_{a}D_{q}f(a) \bigr\vert \\ &\qquad {} +m \biggl[\frac{2q^{2}}{(1+q)^{3}}-\Phi_{7} \biggl(1, \frac{1}{1+q},\alpha \biggr)-\Phi_{8} \biggl(1,\frac{1}{1+q}, \alpha \biggr) \\ &\qquad {} +\Phi_{9} \biggl(1,\frac{1}{1+q},\alpha \biggr) \biggr] \biggl\vert {}_{a}D_{q}f \biggl(\frac{b}{m} \biggr) \biggr\vert \biggr\} . \end{aligned}
Specially, taking $$\alpha=1=m$$, we obtain
\begin{aligned} & \biggl\vert \frac{qf(a)+f(b)}{1+q}-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a}\mathrm {d}_{q}x \biggr\vert \\ &\quad \leq(b-a) \biggl\{ \frac{q^{2}(1+4q+q^{2})}{(1+q)^{4}(1+q+q^{2})} \bigl\vert {}_{a}D_{q}f(b) \bigr\vert +\frac{q^{2}(1+3q^{2}+2q^{3})}{(1+q)^{4}(1+q+q^{2})} \bigl\vert {}_{a}D_{q}f(a) \bigr\vert \biggr\} , \end{aligned}
which is established by Sudsutad et al. in [26, Theorem 4.1].

If $$|{}_{a}D_{q}f|^{r}$$ for $$r\geq1$$ is $$(\alpha,m)$$-convex, then the following theorem can be obtained.

### Theorem 3.3

For $$0\leq a< b$$ and some fixed $$m\in(0,1]$$, let $$f: [a,\frac {b}{m} ]\rightarrow\mathbb{R}$$ be a continuous and q-differentiable function on $$(a,\frac{b}{m} )$$, and let $${}_{a}D_{q}f$$ be integrable on $$[a,\frac{b}{m} ]$$ with $$0< q <1$$. Then the inequality
\begin{aligned} & \biggl\vert \lambda \bigl[\mu f(b)+(1-\mu)f(a) \bigr]+(1- \lambda)f \bigl(\mu b+(1-\mu)a \bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq(b-a)\min \bigl\{ \mathcal{J}_{1}(\lambda,\mu,\alpha,m,r), \mathcal {J}_{2}(\lambda,\mu,\alpha,m,r) \bigr\} \end{aligned}
holds for all $$\lambda,\mu\in[0,1]$$ if $$|{}_{a}D_{q}f|^{r}$$ for $$r\geq1$$ is $$(\alpha,m)$$-convex on $$[a,\frac{b}{m} ]$$ with $$\alpha,m\in (0,1]^{2}$$, where
\begin{aligned}& \begin{aligned} &\mathcal{J}_{1}( \lambda,\mu,\alpha,m,r) \\ &\quad =\Phi_{5}^{1-\frac{1}{r}}(\lambda,\mu) \biggl[ \Phi_{2}(\lambda,\mu,\alpha ) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r} +m\bigl(\Phi_{5}(\lambda,\mu)- \Phi_{2}(\lambda,\mu,\alpha)\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{a}{m}\biggr) \biggr\vert ^{r} \biggr]^{\frac{1}{r}} \\ &\qquad {} +(1-\lambda)\mu^{1-\frac{1}{r}} \biggl[\Upsilon_{1}(\mu, \alpha) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r}+m\Upsilon_{2}(\mu,\alpha) \biggl\vert {}_{a}D_{q}f\biggl(\frac {a}{m}\biggr) \biggr\vert ^{r} \biggr]^{\frac{1}{r}}, \end{aligned} \\& \begin{aligned} &\mathcal{J}_{2}(\lambda,\mu, \alpha,m,r) \\ &\quad =\Phi_{5}^{1-\frac{1}{r}}(\lambda,\mu) \biggl[ \Phi_{8}(\lambda,\mu,\alpha ) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r} +m\bigl(\Phi_{5}(\lambda,\mu)- \Phi_{8}(\lambda,\mu,\alpha)\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{b}{m}\biggr) \biggr\vert ^{r} \biggr]^{\frac{1}{r}} \\ &\qquad {} +(1-\lambda)\mu^{1-\frac{1}{r}} \biggl[\Upsilon_{3}(\mu, \alpha) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r}+m\Upsilon_{4}(\mu,\alpha) \biggl\vert {}_{a}D_{q}f\biggl(\frac {b}{m}\biggr) \biggr\vert ^{r} \biggr]^{\frac{1}{r}}, \end{aligned} \\& \Upsilon_{1}(\mu,\alpha)= \int_{0}^{\mu}t^{\alpha} \,{}_{0} \mathrm{d}_{q}t=\frac {\mu^{\alpha+1}(1-q)}{1-q^{\alpha+1}}, \end{aligned}
(3.4)
\begin{aligned}& \Upsilon_{2}(\mu,\alpha)= \int_{0}^{\mu} \bigl(1-t^{\alpha} \bigr) \,{}_{0}\mathrm {d}_{q}t=\mu-\frac{\mu^{\alpha+1}(1-q)}{1-q^{\alpha+1}}, \end{aligned}
(3.5)
\begin{aligned}& \Upsilon_{3}(\mu,\alpha)= \int_{0}^{\mu}(1-t)^{\alpha} \,{}_{0}\mathrm {d}_{q}t=(1-q)\mu\sum _{n=0}^{\infty}q^{n} \bigl(1-q^{n}\mu \bigr)^{\alpha}, \end{aligned}
(3.6)
\begin{aligned}& \Upsilon_{4}(\mu,\alpha)= \int_{0}^{\mu} \bigl(1-(1-t)^{\alpha} \bigr) \,{}_{0}\mathrm{d}_{q}t=\mu-(1-q)\mu\sum _{n=0}^{\infty}q^{n} \bigl(1-q^{n}\mu \bigr)^{\alpha}, \end{aligned}
(3.7)
and $$\Phi_{2}(\lambda,\mu,\alpha)$$, $$\Phi_{5}(\lambda,\mu)$$, $$\Phi_{8}(\lambda ,\mu,\alpha)$$ are defined by (3.1), (3.2) and (3.3), respectively.

### Proof

Using Lemma 2.1 and the power mean inequality, we have
\begin{aligned} & \biggl\vert \lambda\bigl[\mu f(b)+(1-\mu)f(a)\bigr]+(1-\lambda)f\bigl(\mu b+(1-\mu)a\bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq(b-a) \biggl\{ \biggl( \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t\biggr)^{1-\frac{1}{r}} \\ &\qquad {} \times\biggl( \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \bigl\vert {}_{a}D_{q}f \bigl(tb+(1-t)a\bigr) \bigr\vert ^{r}\,{}_{0}\mathrm{d}_{q}t \biggr)^{\frac{1}{r}} \\ &\qquad {} +(1-\lambda) \biggl( \int_{0}^{\mu}1 \,{}_{0} \mathrm{d}_{q}t\biggr)^{1-\frac{1}{r}} \biggl( \int_{0}^{\mu}\bigl\vert {}_{a}D_{q}f \bigl(tb+(1-t)a\bigr) \bigr\vert ^{r}\,{}_{0}\mathrm {d}_{q}t\biggr)^{\frac{1}{r}} \biggr\} . \end{aligned}
(3.8)
Utilizing the $$(\alpha,m)$$-convexity of $$|{}_{a}D_{q}f|^{r}$$, we get
\begin{aligned} & \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \bigl\vert {}_{a}D_{q}f\bigl(tb+(1-t)a \bigr) \bigr\vert ^{r}\,{}_{0}\mathrm{d}_{q}t \\ &\quad \leq \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \biggl[t^{\alpha} \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r}+m\bigl(1-t^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m} \biggr) \biggr\vert ^{r} \biggr]\,{}_{0}\mathrm{d}_{q}t \\ & \quad = \biggl( \int_{0}^{1}t^{\alpha}\bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \biggr) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r} \\ &\qquad {} +m \biggl( \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t- \int _{0}^{1}t^{\alpha} \bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert ^{r} \end{aligned}
(3.9)
and
\begin{aligned} & \int_{0}^{\mu}\bigl\vert {}_{a}D_{q}f \bigl(tb+(1-t)a\bigr) \bigr\vert ^{r}\,{}_{0} \mathrm{d}_{q}t \\ &\quad \leq \int_{0}^{\mu}\biggl[t^{\alpha} \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r}+m \bigl(1-t^{\alpha }\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{a}{m}\biggr) \biggr\vert ^{r} \biggr] \,{}_{0}\mathrm{d}_{q}t \\ &\quad = \biggl( \int_{0}^{\mu}t^{\alpha}\,{}_{0} \mathrm{d}_{q}t \biggr) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r} +m \biggl( \int_{0}^{\mu}\bigl(1-t^{\alpha}\bigr) \,{}_{0}\mathrm{d}_{q}t \biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert ^{r}. \end{aligned}
(3.10)
Using (3.9) and (3.10) in (3.8), we get
\begin{aligned} & \biggl\vert \lambda\bigl[\mu f(b)+(1-\mu)f(a)\bigr]+(1-\lambda)f\bigl(\mu b+(1-\mu)a\bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq(b-a)\biggl\{ \biggl[ \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t\biggr]^{1-\frac{1}{r}} \\ &\qquad {} \times\biggl[\biggl( \int_{0}^{1}t^{\alpha}\bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t\biggr) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r} \\ &\qquad {} +m\biggl( \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t- \int _{0}^{1}t^{\alpha} \bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t\biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}} \\ &\qquad {} +(1-\lambda)\mu^{1-\frac{1}{r}}\biggl[\biggl( \int_{0}^{\mu}t^{\alpha }\,{}_{0} \mathrm{d}_{q}t\biggr) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r} \\ &\qquad {} +m\biggl( \int_{0}^{\mu}\bigl(1-t^{\alpha}\bigr) \,{}_{0}\mathrm{d}_{q}t\biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}}\biggr\} . \end{aligned}
(3.11)
Similarly, we get
\begin{aligned} & \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \bigl\vert {}_{a}D_{q}f\bigl(tb+(1-t)a \bigr) \bigr\vert ^{r}\,{}_{0}\mathrm{d}_{q}t \\ &\quad \leq \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \biggl[(1-t)^{\alpha} \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r}+m\bigl(1-(1-t)^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f\biggl(\frac {b}{m}\biggr) \biggr\vert ^{r} \biggr]\,{}_{0}\mathrm{d}_{q}t \\ &\quad = \biggl( \int_{0}^{1}(1-t)^{\alpha}\bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t \biggr) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r} \\ &\qquad {} +m \biggl( \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t- \int _{0}^{1}(1-t)^{\alpha} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \biggr) \\ &\qquad {}\times\biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m} \biggr) \biggr\vert ^{r} \end{aligned}
(3.12)
and
\begin{aligned} & \int_{0}^{\mu}\bigl\vert {}_{a}D_{q}f \bigl(tb+(1-t)a\bigr) \bigr\vert ^{r}\,{}_{0} \mathrm{d}_{q}t \\ &\quad \leq \int_{0}^{\mu}\biggl[(1-t)^{\alpha} \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r}+m \bigl(1-(1-t)^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{b}{m}\biggr) \biggr\vert ^{r} \biggr] \,{}_{0}\mathrm{d}_{q}t \\ &\quad = \biggl( \int_{0}^{\mu}(1-t)^{\alpha}\,{}_{0} \mathrm{d}_{q}t \biggr) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r} \\ &\qquad {} +m \biggl( \int_{0}^{\mu}\bigl(1-(1-t)^{\alpha}\bigr) \,{}_{0}\mathrm{d}_{q}t \biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m}\biggr) \biggr\vert ^{r}. \end{aligned}
(3.13)
Using (3.12) and (3.13) in (3.8), we get
\begin{aligned} & \biggl\vert \lambda\bigl[\mu f(b)+(1-\mu)f(a)\bigr]+(1-\lambda)f\bigl(\mu b+(1-\mu)a\bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ & \quad \leq(b-a) \biggl\{ \biggl[ \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm {d}_{q}t\biggr]^{1-\frac{1}{r}} \\ &\qquad {} \times\biggl[\biggl( \int_{0}^{1}(1-t)^{\alpha}\bigl\vert qt-(1- \lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t\biggr) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r} \\ &\qquad {} +m\biggl( \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t- \int _{0}^{1}(1-t)^{\alpha} \bigl\vert qt-(1-\lambda\mu) \bigr\vert \,{}_{0}\mathrm{d}_{q}t \biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m} \biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}} \\ &\qquad {} +(1-\lambda)\mu^{1-\frac{1}{r}}\biggl[\biggl( \int_{0}^{\mu}(1-t)^{\alpha }\,{}_{0} \mathrm{d}_{q}t\biggr) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r} \\ &\qquad {} +m\biggl( \int_{0}^{\mu}\bigl(1-(1-t)^{\alpha}\bigr) \,{}_{0}\mathrm{d}_{q}t \biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}} \biggr\} . \end{aligned}
(3.14)
From (3.11) and (3.14), utilizing (3.1), (3.2), (3.3) and Lemma 2.2, we can deduce the desired result. The proof is complete. □

If $$|{}_{a}D_{q}f|^{r}$$ for $$r>1$$ is $$(\alpha,m)$$-convex, then the following theorem can be obtained.

### Theorem 3.4

For $$0\leq a< b$$ and some fixed $$m\in(0,1]$$, let $$f: [a,\frac {b}{m} ]\rightarrow\mathbb{R}$$ be a continuous and q-differentiable function on $$(a,\frac{b}{m} )$$, and let $${}_{a}D_{q}f$$ be integrable on $$[a,\frac{b}{m} ]$$ with $$0< q <1$$. Then the inequality
\begin{aligned} & \biggl\vert \lambda \bigl[\mu f(b)+(1-\mu)f(a) \bigr]+(1-\lambda)f \bigl(\mu b+(1-\mu)a \bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq(b-a)\min \bigl\{ \mathcal{K}_{1}(\lambda,\mu,\alpha,m), \mathcal {K}_{2}(\lambda,\mu,\alpha,m) \bigr\} \end{aligned}
holds for all $$\lambda,\mu\in[0,1]$$ if $$|{}_{a}D_{q}f|^{r}$$ for $$r> 1$$ with $$r^{-1}+s^{-1}=1$$ is $$(\alpha,m)$$-convex on $$[a,\frac{b}{m} ]$$ with $$\alpha,m\in(0,1]^{2}$$, where
\begin{aligned}& \begin{aligned} &\mathcal{K}_{1}(\lambda,\mu,\alpha,m) \\ &\quad =\Psi_{1}^{\frac{1}{s}}(\lambda,\mu) \biggl[ \Psi_{2}(\alpha) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r}+m\bigl(1-\Psi_{2}(\alpha )\bigr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}} \\ &\qquad {} +(1-\lambda)\mu^{\frac{1}{s}} \biggl[\Upsilon_{1}(\mu, \alpha) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r}+m\Upsilon_{2}(\mu ,\alpha) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}}, \end{aligned} \\& \begin{aligned} &\mathcal{K}_{2}(\lambda,\mu,\alpha,m) \\ &\quad =\Psi_{1}^{\frac{1}{s}}(\lambda,\mu) \biggl[ \Psi_{3}(\alpha) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r}+m\bigl(1-\Psi_{3}(\alpha )\bigr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}} \\ &\qquad {} +(1-\lambda)\mu^{\frac{1}{s}} \biggl[\Upsilon_{3}(\mu, \alpha) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r}+m\Upsilon_{4}(\mu ,\alpha) \biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}}, \end{aligned} \\& \begin{aligned} \Psi_{1}(\lambda,\mu)&= \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert ^{s}\,{}_{0}\mathrm{d}_{q}t \\ &= \textstyle\begin{cases} (1-q)\sum_{n=0}^{\infty}q^{n}(1-\lambda\mu-q^{n+1})^{s}, &0\leq\lambda\mu\leq1-q, \\ \left [ \textstyle\begin{array}{l} (1-q)(1-\lambda\mu)^{s+1}\sum_{n=0}^{\infty}q^{n-1}(1-q^{n})^{s}\\ \quad {}+(1-q)\sum_{n=0}^{\infty}q^{n}(q^{n+1}-1+\lambda\mu)^{s}\\ \quad {}-(1-q)(1-\lambda\mu)^{s+1}\sum_{n=0}^{\infty}q^{n-1}(q^{n}-1)^{s} \end{array}\displaystyle \right ], &1-q< \lambda\mu\leq1, \end{cases}\displaystyle \end{aligned} \\& \Psi_{2}(\alpha)= \int_{0}^{1} t^{\alpha} \,{}_{0} \mathrm{d}_{q}t=\frac {1-q}{1-q^{\alpha+1}}, \\& \Psi_{3}(\alpha)= \int_{0}^{1} (1-t)^{\alpha} \,{}_{0}\mathrm{d}_{q}t=(1-q)\sum _{n=0}^{\infty}q^{n}\bigl(1-q^{n} \bigr)^{\alpha}, \end{aligned}
and $$\Upsilon_{1}(\mu,\alpha)$$, $$\Upsilon_{2}(\mu,\alpha)$$, $$\Upsilon_{3}(\mu ,\alpha)$$, $$\Upsilon_{4}(\mu,\alpha)$$ are defined by (3.4), (3.5), (3.6) and (3.7), respectively.

### Proof

Using Lemma 2.1 and the Hölder inequality, we have
\begin{aligned} & \biggl\vert \lambda\bigl[\mu f(b)+(1-\mu)f(a)\bigr]+(1-\lambda)f\bigl(\mu b+(1-\mu)a\bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq(b-a) \biggl\{ \biggl( \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert ^{s}\,{}_{0}\mathrm{d}_{q}t \biggr)^{\frac{1}{s}} \\ &\qquad {} \times\biggl( \int_{0}^{1} \bigl\vert {}_{a}D_{q}f \bigl(tb+(1-t)a\bigr) \bigr\vert ^{r}\,{}_{0} \mathrm{d}_{q}t\biggr)^{\frac{1}{r}} \\ &\qquad {} +(1-\lambda) \biggl( \int_{0}^{\mu} 1^{s} \,{}_{0} \mathrm{d}_{q}t\biggr)^{\frac{1}{s}} \biggl( \int_{0}^{\mu} \bigl\vert {}_{a}D_{q}f \bigl(tb+(1-t)a\bigr) \bigr\vert ^{r}\,{}_{0}\mathrm {d}_{q}t\biggr)^{\frac{1}{r}} \biggr\} . \end{aligned}
(3.15)
Utilizing the $$(\alpha,m)$$-convexity of $$|{}_{a}D_{q}f|^{r}$$, we get
\begin{aligned} & \int_{0}^{1} \bigl\vert {}_{a}D_{q}f \bigl(tb+(1-t)a\bigr) \bigr\vert ^{r}\,{}_{0} \mathrm{d}_{q}t \\ &\quad \leq \int_{0}^{1} \biggl[t^{\alpha} \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r}+m \bigl(1-t^{\alpha }\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{a}{m}\biggr) \biggr\vert ^{r} \biggr] \,{}_{0}\mathrm{d}_{q}t \\ &\quad = \biggl( \int_{0}^{1} t^{\alpha} \,{}_{0} \mathrm{d}_{q}t \biggr) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r}+m \biggl( \int_{0}^{1} \bigl(1-t^{\alpha}\bigr) \,{}_{0}\mathrm {d}_{q}t \biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert ^{r} \end{aligned}
(3.16)
and
\begin{aligned} & \int_{0}^{\mu} \bigl\vert {}_{a}D_{q}f \bigl(tb+(1-t)a\bigr) \bigr\vert ^{r}\,{}_{0}\mathrm {d}_{q}t \\ &\quad \leq \int_{0}^{\mu} \biggl[t^{\alpha} \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r}+m \bigl(1-t^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{a}{m}\biggr) \biggr\vert ^{r} \biggr] \,{}_{0}\mathrm{d}_{q}t \\ &\quad = \biggl( \int_{0}^{\mu}t^{\alpha} \,{}_{0} \mathrm{d}_{q}t \biggr) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r}+m \biggl( \int_{0}^{\mu}\bigl(1-t^{\alpha} \bigr) \,{}_{0}\mathrm{d}_{q}t \biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert ^{r}. \end{aligned}
(3.17)
Using (3.16) and (3.17) in (3.15), we get
\begin{aligned} & \biggl\vert \lambda\bigl[\mu f(b)+(1-\mu)f(a)\bigr]+(1-\lambda)f\bigl(\mu b+(1-\mu)a\bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq(b-a) \biggl\{ \biggl( \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert ^{s}\,{}_{0}\mathrm{d}_{q}t \biggr)^{\frac{1}{s}} \\ &\qquad {} \times\biggl[\biggl( \int_{0}^{1} t^{\alpha} \,{}_{0} \mathrm{d}_{q}t\biggr) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r}+m\biggl( \int_{0}^{1} \bigl(1-t^{\alpha}\bigr) \,{}_{0}\mathrm {d}_{q}t\biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac {1}{r}} \\ &\qquad {} +(1-\lambda)\mu^{\frac{1}{s}} \biggl[\biggl( \int_{0}^{\mu}t^{\alpha} \,{}_{0} \mathrm{d}_{q}t\biggr) \bigl\vert {}_{a}D_{q}f(b) \bigr\vert ^{r} \\ &\qquad {} +m\biggl( \int_{0}^{\mu}\bigl(1-t^{\alpha}\bigr) \,{}_{0}\mathrm{d}_{q}t\biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{a}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}} \biggr\} . \end{aligned}
(3.18)
Similarly, we get
\begin{aligned} & \int_{0}^{1} \bigl\vert {}_{a}D_{q}f \bigl(tb+(1-t)a\bigr) \bigr\vert ^{r}\,{}_{0} \mathrm{d}_{q}t \\ &\quad \leq \int_{0}^{1} \biggl[(1-t)^{\alpha} \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r}+m \bigl(1-(1-t)^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{b}{m}\biggr) \biggr\vert ^{r} \biggr] \,{}_{0}\mathrm{d}_{q}t \\ &\quad = \biggl( \int_{0}^{1} (1-t)^{\alpha} \,{}_{0}\mathrm{d}_{q}t \biggr) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r} \\ &\qquad {} +m \biggl( \int_{0}^{1} \bigl(1-(1-t)^{\alpha}\bigr) \,{}_{0}\mathrm{d}_{q}t \biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m}\biggr) \biggr\vert ^{r} \end{aligned}
(3.19)
and
\begin{aligned} & \int_{0}^{\mu} \bigl\vert {}_{a}D_{q}f \bigl(tb+(1-t)a\bigr) \bigr\vert ^{r}\,{}_{0}\mathrm {d}_{q}t \\ &\quad \leq \int_{0}^{\mu} \biggl[(1-t)^{\alpha} \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r}+m \bigl(1-(1-t)^{\alpha}\bigr) \biggl\vert {}_{a}D_{q}f \biggl(\frac{b}{m}\biggr) \biggr\vert ^{r} \biggr] \,{}_{0}\mathrm{d}_{q}t \\ &\quad = \biggl( \int_{0}^{\mu}(1-t)^{\alpha} \,{}_{0}\mathrm{d}_{q}t \biggr) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r} \\ &\qquad {} +m \biggl( \int_{0}^{\mu}\bigl(1-(1-t)^{\alpha}\bigr) \,{}_{0}\mathrm{d}_{q}t \biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m}\biggr) \biggr\vert ^{r}. \end{aligned}
(3.20)
Using (3.19) and (3.20) in (3.15), we get
\begin{aligned} & \biggl\vert \lambda\bigl[\mu f(b)+(1-\mu)f(a)\bigr]+(1-\lambda)f\bigl(\mu b+(1-\mu)a\bigr)-\frac{1}{b-a} \int_{a}^{b}f(x)\,{}_{a} \mathrm{d}_{q}x \biggr\vert \\ &\quad \leq(b-a) \biggl\{ \biggl( \int_{0}^{1} \bigl\vert qt-(1-\lambda\mu) \bigr\vert ^{s}\,{}_{0}\mathrm{d}_{q}t \biggr)^{\frac{1}{s}} \\ &\qquad {} \times\biggl[\biggl( \int_{0}^{1} (1-t)^{\alpha} \,{}_{0}\mathrm{d}_{q}t\biggr) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r} \\ &\qquad {} +m\biggl( \int_{0}^{1} \bigl(1-(1-t)^{\alpha}\bigr) \,{}_{0}\mathrm{d}_{q}t\biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}} \\ &\qquad {} +(1-\lambda)\mu^{\frac{1}{s}} \biggl[\biggl( \int_{0}^{\mu}(1-t)^{\alpha} \,{}_{0}\mathrm{d}_{q}t\biggr) \bigl\vert {}_{a}D_{q}f(a) \bigr\vert ^{r} \\ &\qquad {} +m\biggl( \int_{0}^{\mu}\bigl(1-(1-t)^{\alpha}\bigr) \,{}_{0}\mathrm{d}_{q}t \biggr) \biggl\vert {}_{a}D_{q}f\biggl(\frac{b}{m}\biggr) \biggr\vert ^{r}\biggr]^{\frac{1}{r}} \biggr\} . \end{aligned}
(3.21)
From (3.18) and (3.21), utilizing (3.4), (3.5), (3.6), (3.7), Lemma 2.2 and Lemma 2.6, we can deduce the desired result. The proof is completed. □

### Remark 3.2

For $$\mu=\frac{1}{1+q}$$, if we put $$\lambda=0$$, $$\lambda=\frac{1}{3}$$, $$\lambda=\frac{1}{2}$$ and $$\lambda=1$$ in Theorem 3.3 and Theorem 3.4, then we can get the midpoint-like integral inequality, the Simpson-like integral inequality, the averaged midpoint-trapezoid-like integral inequality and the trapezoid-like integral inequality, respectively.

Next we establish the q-integral inequalities involving the product of two $$(\alpha,m)$$-convex functions.

### Theorem 3.5

For $$0\leq a< b$$ and some fixed $$m\in(0,1]$$, let $$f,g: [a,\frac {b}{m} ]\rightarrow\mathbb{R}$$ be continuous and nonnegative functions. Then the inequality
$$\frac{1}{b-a} \int_{a}^{b} f(x)g(x)\,{}_{a} \mathrm{d}_{q}x\leq\min \bigl\{ \mathcal {L}_{1}( \alpha_{1},\alpha_{2},m),\mathcal{L}_{2}( \alpha_{1},\alpha_{2},m) \bigr\}$$
holds if f and g are $$(\alpha_{1},m)$$-convex and $$(\alpha _{2},m)$$-convex on $$[a,\frac{b}{m} ]$$ with $$\alpha_{1},\alpha_{2}\in (0,1]^{2}$$, respectively, where
\begin{aligned} &\mathcal{L}_{1}(\alpha_{1},\alpha_{2},m) \\ &\quad = \biggl[\frac{1-q}{1-q^{\alpha_{1}+\alpha_{2}+1}}-\frac{1-q}{1-q^{\alpha _{1}+1}}-\frac{1-q}{1-q^{\alpha_{2}+1}}+1 \biggr]m^{2}f \biggl(\frac{a}{m} \biggr)g \biggl( \frac{a}{m} \biggr) \\ &\qquad {} +\frac{1-q}{1-q^{\alpha_{1}+\alpha_{2}+1}}f(b)g(b)+ \biggl[\frac {1-q}{1-q^{\alpha_{2}+1}}- \frac{1-q}{1-q^{\alpha_{1}+\alpha_{2}+1}} \biggr]mf \biggl(\frac{a}{m} \biggr)g(b) \\ &\qquad {} + \biggl[\frac{1-q}{1-q^{\alpha_{1}+1}}-\frac{1-q}{1-q^{\alpha_{1}+\alpha _{2}+1}} \biggr]mf(b)g \biggl( \frac{a}{m} \biggr), \\ &\mathcal{L}_{2}(\alpha_{1},\alpha_{2},m) \\ &\quad = \bigl[\Theta(\alpha_{1},\alpha_{2})-\Theta( \alpha_{1})-\Theta(\alpha _{2})+1 \bigr]m^{2}f \biggl(\frac{b}{m} \biggr)g \biggl(\frac{b}{m} \biggr) \\ &\qquad {} +\Theta(\alpha_{1},\alpha_{2})f(a)g(a)+ \bigl[ \Theta(\alpha_{1})-\Theta(\alpha _{1},\alpha_{2}) \bigr]mf(a)g \biggl(\frac{b}{m} \biggr) \\ &\qquad {} + \bigl[\Theta(\alpha_{2})-\Theta(\alpha_{1}, \alpha_{2}) \bigr]mf \biggl(\frac {b}{m} \biggr)g(a), \\ &\Theta(\alpha_{1},\alpha_{2})= \int_{0}^{1} (1-t)^{\alpha_{1}+\alpha_{2}} \,{}_{0}\mathrm{d}_{q}t=(1-q)\sum _{n=0}^{\infty}q^{n} \bigl(1-q^{n} \bigr)^{\alpha _{1}+\alpha_{2}}, \end{aligned}
and
$$\Theta(\alpha_{i})= \int_{0}^{1} (1-t)^{\alpha_{i}} \,{}_{0}\mathrm{d}_{q}t=(1-q)\sum _{n=0}^{\infty}q^{n} \bigl(1-q^{n} \bigr)^{\alpha_{i}},\quad i=1,2.$$

### Proof

Using the $$(\alpha_{1},m)$$-convexity of f and the $$(\alpha _{2},m)$$-convexity of g, respectively, for all $$t\in[0,1]$$, we have
$$f \bigl(tb+(1-t)a \bigr)\leq t^{\alpha_{1}}f(b)+m \bigl(1-t^{\alpha_{1}}\bigr)f \biggl(\frac {a}{m} \biggr)$$
(3.22)
and
$$g \bigl(tb+(1-t)a \bigr)\leq t^{\alpha_{2}}g(b)+m \bigl(1-t^{\alpha_{2}}\bigr)g \biggl(\frac {a}{m} \biggr).$$
(3.23)
Multiplying (3.22) with (3.23), we get
\begin{aligned} &f \bigl(tb+(1-t)a \bigr)g \bigl(tb+(1-t)a \bigr) \\ &\quad \leq t^{\alpha_{1}+\alpha_{2}}f(b)g(b)+\bigl(1-t^{\alpha_{1}}\bigr) \bigl(1-t^{\alpha _{2}}\bigr)m^{2}f \biggl(\frac{a}{m} \biggr)g \biggl(\frac{a}{m} \biggr) \\ &\qquad {} +t^{\alpha_{2}}\bigl(1-t^{\alpha_{1}}\bigr)mf \biggl( \frac{a}{m} \biggr)g(b)+t^{\alpha _{1}}\bigl(1-t^{\alpha_{2}} \bigr)mf(b)g \biggl(\frac{a}{m} \biggr). \end{aligned}
(3.24)
Taking the q-integral for (3.24) with respect to t on $$(0,1)$$ and using Lemma 2.2, we obtain
\begin{aligned} & \int_{0}^{1} f \bigl(tb+(1-t)a \bigr)g \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\quad \leq \biggl[\frac{1-q}{1-q^{\alpha_{1}+\alpha_{2}+1}}-\frac{1-q}{1-q^{\alpha _{1}+1}}-\frac{1-q}{1-q^{\alpha_{2}+1}}+1 \biggr]m^{2}f \biggl(\frac{a}{m} \biggr)g \biggl( \frac{a}{m} \biggr) \\ &\qquad {} +\frac{1-q}{1-q^{\alpha_{1}+\alpha_{2}+1}}f(b)g(b)+ \biggl[\frac {1-q}{1-q^{\alpha_{2}+1}}- \frac{1-q}{1-q^{\alpha_{1}+\alpha_{2}+1}} \biggr]mf \biggl(\frac{a}{m} \biggr)g(b) \\ &\qquad {} + \biggl[\frac{1-q}{1-q^{\alpha_{1}+1}}-\frac{1-q}{1-q^{\alpha_{1}+\alpha _{2}+1}} \biggr]mf(b)g \biggl( \frac{a}{m} \biggr). \end{aligned}
(3.25)
Similarly, we get
\begin{aligned} & \int_{0}^{1} f \bigl(tb+(1-t)a \bigr)g \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t \\ &\quad \leq \biggl( \int_{0}^{1}(1-t)^{\alpha_{1}+\alpha_{2}}\,{}_{0} \mathrm{d}_{q}t- \int _{0}^{1}(1-t)^{\alpha_{1}}\,{}_{0} \mathrm{d}_{q}t \\ &\qquad {} - \int_{0}^{1}(1-t)^{\alpha_{2}}\,{}_{0} \mathrm{d}_{q}t+1 \biggr)m^{2}f \biggl(\frac {b}{m} \biggr)g \biggl(\frac{b}{m} \biggr) \\ &\qquad {} + \biggl( \int_{0}^{1}(1-t)^{\alpha_{1}+\alpha_{2}}\,{}_{0} \mathrm{d}_{q}t \biggr)f(a)g(a) \\ &\qquad {} + \biggl( \int_{0}^{1}(1-t)^{\alpha_{1}}\,{}_{0} \mathrm{d}_{q}t- \int_{0}^{1}(1-t)^{\alpha _{1}+\alpha_{2}}\,{}_{0} \mathrm{d}_{q}t \biggr)mf(a)g \biggl(\frac{b}{m} \biggr) \\ &\qquad {} + \biggl( \int_{0}^{1}(1-t)^{\alpha_{2}}\,{}_{0} \mathrm{d}_{q}t- \int _{0}^{1}(1-t)^{\alpha_{1}+\alpha_{2}}\,{}_{0} \mathrm{d}_{q}t \biggr)mf \biggl(\frac {b}{m} \biggr)g(a). \end{aligned}
(3.26)
A simple calculation shows that
$$\int_{0}^{1} f \bigl(tb+(1-t)a \bigr)g \bigl(tb+(1-t)a \bigr)\,{}_{0}\mathrm{d}_{q}t = \frac{1}{b-a} \int_{a}^{b} f(x)g(x)\,{}_{a} \mathrm{d}_{q}x.$$
(3.27)
From (3.25), (3.26) and (3.27), we obtain the desired result. This ends the proof. □

### Corollary 3.2

In Theorem 3.5, choosing $$\alpha_{1}=\alpha_{2}=\alpha$$, we obtain
$$\frac{1}{b-a} \int_{a}^{b} f(x)g(x)\,{}_{a} \mathrm{d}_{q}x\leq\min \bigl\{ \mathcal {T}_{1}(\alpha,m), \mathcal{T}_{2}(\alpha,m) \bigr\} ,$$
where
\begin{aligned} \mathcal{T}_{1}(\alpha,m) &= \frac{1-q}{1-q^{2\alpha+1}}f(b)g(b)+ \biggl[ \frac{1-q}{1-q^{2\alpha +1}}-\frac{2(1-q)}{1-q^{\alpha+1}}+1 \biggr]m^{2}f \biggl( \frac{a}{m} \biggr)g \biggl(\frac{a}{m} \biggr) \\ &\quad {} +\frac{q^{\alpha+1}(1-q)(1-q^{\alpha})}{(1-q^{\alpha+1})(1-q^{2\alpha +1})}m \biggl[f \biggl(\frac{a}{m} \biggr)g(b)+f(b)g \biggl(\frac{a}{m} \biggr) \biggr] \end{aligned}
and
\begin{aligned} \mathcal{T}_{2}(\alpha,m) &= \Biggl[(1-q)\sum _{n=0}^{\infty}q^{n} \bigl(1-q^{n} \bigr)^{2\alpha}-2(1-q)\sum_{n=0}^{\infty}q^{n} \bigl(1-q^{n} \bigr)^{\alpha}+1 \Biggr]m^{2}f \biggl( \frac {b}{m} \biggr)g \biggl(\frac{b}{m} \biggr) \\ &\quad {} +(1-q)\sum_{n=0}^{\infty}q^{n} \bigl(1-q^{n} \bigr)^{2\alpha}f(a)g(a) \\ &\quad {} + \Biggl[(1-q)\sum_{n=0}^{\infty}q^{n} \bigl(1-q^{n} \bigr)^{\alpha}-(1-q)\sum _{n=0}^{\infty}q^{n} \bigl(1-q^{n} \bigr)^{2\alpha} \Biggr] \\ &\quad {} \times \biggl[mf(a)g \biggl(\frac{b}{m} \biggr)+mf \biggl( \frac{b}{m} \biggr)g(a) \biggr]. \end{aligned}
Further, taking $$\alpha=1=m$$, we get
\begin{aligned} &\frac{1}{b-a} \int_{a}^{b} f(x)g(x)\,{}_{a} \mathrm{d}_{q}x \\ &\quad \leq \frac{1}{1+q+q^{2}}f(b)g(b)+\frac{q(1+q^{2})}{(1+q)(1+q+q^{2})}f(a)g(a) \\ &\qquad {} +\frac{q^{2}}{(1+q)(1+q+q^{2})} \bigl[f(a)g(b)+f(b)g(a) \bigr], \end{aligned}
which is established by Sudsutad et al. in [26, Theorem 4.3].

## 4 Conclusions

In the present research, based on a new quantum integral identity with multiple parameters, we have developed some quantum error estimations of different type inequalities through $$(\alpha,m)$$-convexity, such as the midpoint-like inequalities, the Simpson-like inequalities, the averaged midpoint-trapezoid-like inequalities and the trapezoid-like inequalities. The inequalities derived in this work are very helpful in error estimations involved in various approximation processes. We expect that the ideas of this article will facilitate further study concerning quantum integral inequalities.

## Declarations

### Funding

This work was partially supported by the National Natural Science Foundation of China (No. 61374028) and sponsored by Research Fund for Excellent Dissertation of China Three Gorges University (No. 2018SSPY132, No. 2018SSPY134).

### Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

### Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

## Authors’ Affiliations

(1)
Department of Mathematics, College of Science, China Three Gorges University, Yichang, China
(2)
Three Gorges Mathematical Research Center, China Three Gorges University, Yichang, China
(3)
Hubei Provincial Collaborative Innovation Center for New Energy Microgrid, China Three Gorges University, Yichang, China

## References

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