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q-Hardy type inequalities for quantum integrals

Abstract

The aim of this work is to obtain quantum estimates for q-Hardy type integral inequalities on quantum calculus. For this, we establish new identities including quantum derivatives and quantum numbers. After that, we prove a generalized q-Minkowski integral inequality. Finally, with the help of the obtained equalities and the generalized q-Minkowski integral inequality, we obtain the results we want. The outcomes presented in this paper are q-extensions and q-generalizations of the comparable results in the literature on inequalities. Additionally, by taking the limit \(q\rightarrow 1^{-}\), our results give classical results on the Hardy inequality.

Introduction

Hardy’s integral inequality, proved by G.H. Hardy in 1920 [4] is

$$ \int _{0}^{\infty } \biggl( \frac{1}{x} \int _{0}^{x}f ( t ) \,dt \biggr) ^{p}\,dx\leq \biggl( \frac{p}{p-1} \biggr) ^{p} \int _{0}^{ \infty }f^{p} ( t ) \,dt, $$
(1.1)

where \(p>1\), \(x>0\), f is a nonnegative measurable function on \(( 0,\infty ) \) and \(\int _{0}^{\infty }f^{p} ( t ) \,dt\) is convergent. Also the constant \(( \frac{p}{p-1} ) ^{p}\) is the best possible.

Hardy’s type inequalities have been studied by a large number of authors during the 20th century and has motivated some important lines of study which are currently active. Over the last 20 years a large number of papers have appeared in the literature which deal with the simple proofs, various generalizations and discrete analogues of Hardy’s inequality and its generalizations; see [5, 8, 11, 12, 15, 1719].

The inequalities have become an important cornerstone in mathematical analysis and optimization and many uses of these inequalities have been discovered in a variety of settings. Recently, the Hermite–Hadamard type inequality has become the subject of intensive research. For recent results, refinements, counterparts, generalizations and new Hadamard’s-type inequalities, see [1, 7, 10, 14, 16, 20].

On the other hand, the study of calculus without limits is known as quantum calculus or q-calculus. The famous mathematician Euler initiated the study q-calculus in the 18th century by introducing the parameter q in Newton’s work of infinite series. In the early 20th century, Jackson [6] has started a symmetric study of q-calculus and introduced q-definite integrals. The subject of quantum calculus has numerous applications in various areas of mathematics and physics, such as number theory, combinatorics, orthogonal polynomials, basic hyper-geometric functions, quantum theory, mechanics and in theory of relativity. This subject has received outstanding attention by many researchers and hence it is considered as an in-corporative subject between mathematics and physics. The reader is referred to [2, 3, 9] for some current advances in the theory of quantum calculus and theory of inequalities in quantum calculus.

The purpose of this work is to establish quantum estimates for q-Hardy type integral inequalities on quantum calculus. For this, we establish new identities including quantum derivatives and quantum numbers. After that, we prove a generalized q-Minkowski integral inequality. Finally, with the help of the obtained equalities and the generalized q-Minkowski integral inequality, we obtain the results we want. The outcomes presented in this paper are q-extensions and q-generalizations of the comparable results in the literature on inequalities. In addition, by taking the limit \(q\rightarrow 1^{-}\), our results give classical results on the Hardy inequality.

Preliminaries and definitions of q-calculus

Throughout this paper, let \(a< b\) and \(0< q<1\) be a constant. The following definitions, notations and theorems for q-derivative and q-integral of a function f on \([ a,b ]\) are given in [2, 3, 9].

The notation \([ z ] _{q}\) is defined by

$$ [ z ] _{q}=\frac{1-q^{n}}{1-q}\quad \bigl( z\in \mathbb{C} ;q\in \mathbb{C} \backslash \{ 1 \} ;q^{z}\neq 1 \bigr). $$
(2.1)

A special case of (2.1) when \(z\in \mathbb{N} \) is

$$ [ n ] _{q}=\frac{1-q^{n}}{1-q}=1+q+q^{2}+\cdots +q^{n-1} \quad ( n\in \mathbb{N} ). $$

Also

$$ [ -n ] _{q}=-\frac{1}{q^{n}} [ n ] _{q} \quad ( n\in \mathbb{N} ). $$
(2.2)

Definition 1

Let \(f: [ a,b ] \rightarrow \mathbb{R} \) be a continuous function, then q-derivative of f at \(x\in [ a,b ] \) is characterized by the expression

$$ D_{q}f ( x ) = \frac{f ( x ) -f ( qx ) }{ ( 1-q ) x},\quad x\neq 0. $$
(2.3)

Since \(f: [ a,b ] \rightarrow \mathbb{R} \) is a continuous function, thus we have \(D_{q}f ( a ) =\underset{x\rightarrow a}{\lim }D_{q}f ( x ) \) The function f is said to be q- differentiable on \([ a,b ] \) if \(D_{q}f ( t ) \) exists for all \(x\in [ a,b ] \). Also \(\underset{q\rightarrow 1^{-}}{\lim }D_{q}f ( x ) =f^{\prime } ( x ) \) is classic derivative.

Theorem 1

Assume that \(f,g:I\subset \mathbb{R} \rightarrow \mathbb{R} \) are continuous functions, then we have the properties of the q-derivative:

$$\begin{aligned} (\mathrm{I}) &\quad D_{q} \bigl( af ( x ) \pm bg ( x ) \bigr) =aD_{q}f ( x ) \pm bD_{q}g ( x ) . \\ (\mathrm{II}) &\quad D_{q} \bigl( f ( x ) g ( x ) \bigr) =f ( qx ) D_{q}g ( x ) +g ( x ) D_{q}f ( x ) . \\ (\mathrm{III}) &\quad D_{q} \biggl( \frac{f ( x ) }{g ( x ) } \biggr) = \frac{f ( qx ) D_{q}g ( x ) +g ( x ) D_{q}f ( x ) }{g ( x ) g ( qx ) }. \end{aligned}$$

Definition 2

Suppose \(0< a< b\). The definite q-integral is defined as

$$ \int _{0}^{b}f ( t ) \,d_{q}t= ( 1-q ) b \sum_{n=0}^{\infty }q^{n}f \bigl( q^{n}b \bigr) $$
(2.4)

and

$$ \int _{a}^{b}f ( t ) \,d_{q}t= \int _{0}^{b}f ( t ) \,d_{q}t- \int _{0}^{a}f ( t ) \,d_{q}t, $$

where \(\sum_{n=0}^{\infty }q^{n}f ( q^{n}b ) \) and \(\sum_{n=0}^{\infty }q^{n}f ( q^{n}a ) \) are convergent.

Definition 3

([9])

The improper q-integral of \(f ( t ) \) on \([ 0,\infty ) \) is defined by

$$ \int _{0}^{\infty }f ( t ) \,d_{q}t=\sum _{n=- \infty }^{\infty } \int _{q^{n+1}}^{q^{n}}f ( t ) \,d_{q}t= ( 1-q ) \sum_{n=-\infty }^{\infty }q^{n}f \bigl( q^{n} \bigr) \quad ( 0< q< 1 ) $$

and

$$ \int _{0}^{\infty }f ( t ) \,d_{q}t=\sum _{n=- \infty }^{\infty } \int _{q^{n}}^{q^{n+1}}f ( t ) \,d_{q}t= \frac{q-1}{q}\sum_{n=-\infty }^{\infty }q^{n}f \bigl( q^{n} \bigr) \quad ( 1< q ) , $$

where \(\sum_{n=-\infty }^{\infty }q^{n}f ( q^{n} ) \) is convergent.

We have the following properties of the q-integral of (2.4):

$$\begin{aligned} (\mathrm{I}) &\quad D_{q} \int _{a}^{x}f ( t ) \,d_{q}t=f ( x ) . \\ (\mathrm{II}) &\quad \int _{a}^{x}D_{q}f ( t ) \,d_{q}t=f(x)-f(a). \\ (\mathrm{III}) & \quad \int _{a}^{x} \bigl[ f ( t ) \pm g ( t ) \bigr] \,d_{q}t= \int _{a}^{x}f ( t ) \,d_{q}t \pm \int _{a}^{x}g ( t ) \,d_{q}t. \\ (\mathrm{IV}) &\quad \int _{0}^{x}t^{\alpha }\,d_{q}t= \frac{x^{\alpha +1}}{ [ \alpha +1 ] _{q}}, \quad \text{for }\alpha \in \mathbb{R} \backslash \{ -1 \} . \\ (\mathrm{V}) & \quad \text{The integration by parts rule of the }q\text{-integral:} \\ &\quad \int _{c}^{x}f ( t ) D_{q}g ( t ) \,d_{q}t= f ( t ) g ( t ) \vert _{c}^{x}- \int _{c}^{x}g ( qt ) D_{q}f ( t ) \,d_{q}t. \end{aligned}$$
(2.5)

Theorem 2

(q-Hölder inequality)

Let f, g be q-integrable on \([ a,b ] \) and \(0< q<1\) and \(\frac{1}{s}+\frac{1}{r}=1\) with \(s>1\). Then we have

$$ \int _{a}^{b} \bigl\vert f ( t ) g ( t ) \bigr\vert \,d_{q}t\leq \biggl( \int _{a}^{b} \bigl\vert f ( t ) \bigr\vert ^{s}\,d_{q}t \biggr) ^{\frac{1}{s}} \biggl( \int _{a}^{b} \bigl\vert f ( t ) g ( t ) \bigr\vert ^{r}\,d_{q}t \biggr) ^{\frac{1}{r}}. $$

Auxiliary results

The following results which will be used. There is no general change of variables property for the q-integral. However, the variable can be changed as follows.

Lemma 1

(q-Change of variables property)

Let \(f:I\rightarrow \mathbb{R} \) be a function and \(0< q<1\). Then we have

$$ \int _{0}^{1}f ( sb ) \,d_{q}s= \frac{1}{b} \int _{0}^{b}f ( t ) \,d_{q}t, $$
(3.1)

where \(b\neq 0\) and \(\int _{0}^{b}f ( t ) \,d_{q}t\) is convergent.

Proof

From the definition of the q-integral, we have

$$\begin{aligned}& \int _{0}^{1}f ( sb ) \,d_{q}s \\& \quad = ( 1-q ) ( 1-0 ) \sum_{n=0}^{\infty }q^{n}f \bigl( \bigl[ q^{n}1+ \bigl( 1-q^{n} \bigr) 0 \bigr] b \bigr) \\& \quad = \frac{1}{b} \int _{0}^{b}f ( t ) \,d_{q}t \end{aligned}$$

as desired. □

A general chain rule for q-derivative does not exist. However, a chain rule of \(( h ( t ) ) ^{p}\) and \(( h ( t ) ) ^{\frac{1}{p}}\) can be calculated as follows.

Lemma 2

Let \(h:I\subset \mathbb{R} \rightarrow \mathbb{R} \) be a function \(p\in \mathbb{Z} \) and \(0< q<1\). Then we have

$$ D_{q} \bigl( h ( t ) \bigr) ^{p}= \Biggl( \sum _{i=0}^{p-1} \bigl[ h ( t ) \bigr] ^{p-1-i} \bigl[ h ( qt ) \bigr] ^{i} \Biggr) D_{q}h ( t ) . $$
(3.2)

In (3.2) if we choose \(q\rightarrow 1^{-}\) we have the classical derivative of \(( h ( t ) ) ^{p}\),

$$ \lim_{q\rightarrow 1^{-}}D_{q} \bigl( h ( t ) \bigr) ^{p}=p \bigl( h ( t ) \bigr) ^{p-1}h^{\prime } ( t ) = \bigl[ \bigl( h ( t ) \bigr) ^{p} \bigr] ^{\prime }. $$

Proof

By the definition of the q-derivative we have

$$\begin{aligned}& D_{q} \bigl( h ( t ) \bigr) ^{p} \\& \quad = \frac{ [ h ( t ) ] ^{p}- [ h ( qt ) ] ^{p}}{ ( 1-q ) t} \\& \quad = \frac{ ( h ( t ) -h ( qt ) ) }{ ( 1-q ) t}\sum_{i=0}^{p-1} \bigl[ h ( t ) \bigr] ^{p-1-i} \bigl[ h ( qt ) \bigr] ^{i} \\& \quad = \Biggl( \sum_{i=0}^{p-1} \bigl[ h ( t ) \bigr] ^{p-1-i} \bigl[ h ( qt ) \bigr] ^{i} \Biggr) D_{q}h ( t ) \end{aligned}$$

as desired. □

Lemma 3

Let \(h:I\subset \mathbb{R} \rightarrow \mathbb{R} \) be a function \(p\in \mathbb{Z} \) and \(0< q<1\). Then we have

$$ D_{q} \bigl( h ( t ) \bigr) ^{\frac{1}{p}}= \frac{D_{q}h ( t ) }{\sum_{i=0}^{p-1} ( h ( t ) ) ^{\frac{p-1-i}{p}} ( h ( qt ) ) ^{\frac{i}{p}}} . $$
(3.3)

In (3.3) if we choose \(q\rightarrow 1^{-}\) we have the classical derivative of \(( h ( t ) ) ^{\frac{1}{p}}\),

$$ \lim_{q\rightarrow 1^{-}}D_{q} \bigl( h ( t ) \bigr) ^{ \frac{1}{p}}=\frac{h^{\prime } ( t ) }{p ( h ( t ) ) ^{\frac{p-1}{p}}}= \bigl[ \bigl( h ( t ) \bigr) ^{\frac{1}{p}} \bigr] ^{\prime }. $$

Proof

We consider

$$\begin{aligned}& y ( t ) = \bigl( h ( t ) \bigr) ^{ \frac{1}{p}}, \\& \bigl( y ( t ) \bigr) ^{p} = \bigl( h ( t ) \bigr), \end{aligned}$$

such that

$$ D_{q} \bigl( y ( t ) \bigr) ^{p}=D_{q} \bigl( h ( t ) \bigr), $$
(3.4)

and from (3.2) we know

$$ D_{q} \bigl( y ( t ) \bigr) ^{p}= \Biggl( \sum _{i=0}^{p-1} \bigl[ y ( t ) \bigr] ^{p-1-i} \bigl[ y ( qt ) \bigr] ^{i} \Biggr) D_{q}y ( t ) =D_{q} \bigl( h ( t ) \bigr) . $$
(3.5)

Thus, we get

$$ D_{q}y ( t ) = \frac{D_{q}h ( t ) }{\sum_{i=0}^{p-1} ( h ( t ) ) ^{\frac{p-1-i}{p}} ( h ( qt ) ) ^{\frac{i}{p}}} $$

as desired. □

Similarly, we have more general result as follows.

Lemma 4

Let \(h:I\subset \mathbb{R} \rightarrow \mathbb{R} \) be a function \(\frac{n}{m}\in \mathbb{Q} \) and \(0< q<1\). Then we have

$$ D_{q} \bigl( h ( t ) \bigr) ^{\frac{n}{m}}= \frac{\sum_{i=0}^{n-1} ( h ( t ) ) ^{n-1-i} ( h ( qt ) ) ^{i}}{\sum_{i=0}^{m-1} ( h ( t ) ) ^{\frac{n ( m-1-i ) }{m}} ( h ( qt ) ) ^{\frac{ni}{m}}}D_{q}h ( t ). $$
(3.6)

In (3.6) if we choose \(q\rightarrow 1^{-}\) we have the classical derivative of \(( h ( t ) ) ^{\frac{n}{m}}\),

$$ \lim_{q\rightarrow 1^{-}}D_{q} \bigl( h ( t ) \bigr) ^{ \frac{n}{m}}=\frac{n}{m} \bigl( h ( t ) \bigr) ^{\frac{n}{m}-1}h^{ \prime } ( t ) = \bigl[ \bigl( h ( t ) \bigr) ^{\frac{n}{m}} \bigr] ^{\prime }. $$

Proof

We consider

$$\begin{aligned}& y ( t ) = \bigl( h ( t ) \bigr) ^{ \frac{n}{m}} , \\& \bigl( y ( t ) \bigr) ^{m} = \bigl( h ( t ) \bigr) ^{n}, \end{aligned}$$

such that

$$ D_{q} \bigl( y ( t ) \bigr) ^{m}=D_{q} \bigl( h ( t ) \bigr) ^{n}, $$

and from (3.2) we have

$$\begin{aligned}& \Biggl( \sum_{i=0}^{m-1} \bigl[ y ( t ) \bigr] ^{m-1-i} \bigl[ y ( qt ) \bigr] ^{i} \Biggr) D_{q}y ( t ) \\& \quad = \Biggl( \sum_{i=0}^{n-1} \bigl[ h ( t ) \bigr] ^{n-1-i} \bigl[ h ( qt ) \bigr] ^{i} \Biggr) D_{q} \bigl( h ( t ) \bigr) . \end{aligned}$$

Thus, we get

$$\begin{aligned} D_{q}y ( t ) =&\frac{\sum_{i=0}^{n-1} [ h ( t ) ] ^{n-1-i} [ h ( qt ) ] ^{i}}{\sum_{i=0}^{m-1} [ y ( t ) ] ^{m-1-i} [ y ( qt ) ] ^{i}}D_{q}h ( t ) \\ =&\frac{\sum_{i=0}^{n-1} ( h ( t ) ) ^{n-1-i} ( h ( qt ) ) ^{i}}{\sum_{i=0}^{m-1} ( h ( t ) ) ^{\frac{n ( m-1-i ) }{m}} ( h ( qt ) ) ^{\frac{ni}{m}}}D_{q}h ( t ) \end{aligned}$$

as desired. □

Main results

Firstly, we will prove the generalized q-Minkokski type integral inequality which will be used in the next theorem.

Theorem 3

(Generalized q-Minkowski integral inequality)

Let \(\alpha \in ( 0,1 ] \), \(1\leq p\leq \infty \), \(f: [ a,b ] \times [ c,d ] \rightarrow \mathbb{R} \) be a q-integrable function. Then the following inequality holds:

$$ \biggl( \int _{a}^{b} \biggl\vert \int _{c}^{d}f ( x,y ) \,d_{q}y \biggr\vert ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}\leq \int _{c}^{d} \biggl( \int _{a}^{b} \bigl\vert f ( x,y ) \bigr\vert ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}\,d_{q}y, $$
(4.1)

where \(q\in ( 0,1 ) \).

Proof

The case \(p=1\) corresponds to Fubini’s theorem. For the case \(p=\infty \) we just notice that

$$ \biggl( \int _{a}^{b} \biggl\vert \biggl( \int _{c}^{d}f ( x,y ) \,d_{q}y \biggr) _{q} \biggr\vert ^{p}\,d_{q}x \biggr) ^{ \frac{1}{p}}\leq \int _{c}^{d}\operatorname{ess}\sup_{x\in \mathbb{R} ^{n}} \bigl\vert f ( x,y ) \bigr\vert \,d_{q}y. $$

Now assume that \(1< p<\infty \) and we can write

$$\begin{aligned}& \int _{a}^{b} \biggl\vert \int _{c}^{d}f ( x,y ) \,d_{q}y \biggr\vert ^{p}\,d_{q}x \\& \quad = \int _{a}^{b} \biggl\vert \int _{c}^{d}f ( x,y ) \,d_{q}y \biggr\vert ^{p-1} \biggl\vert \int _{c}^{d}f ( x,y ) \,d_{q}y \biggr\vert \,d_{q}x \\& \quad \leq \int _{a}^{b} \biggl\vert \int _{c}^{d}f ( x,t ) \,d_{q}t \biggr\vert ^{p-1} \biggl( \int _{c}^{d} \bigl\vert f ( x,y ) \bigr\vert \,d_{q}y \biggr) \,d_{q}x \\& \quad = \int _{a}^{b} \biggl( \int _{c}^{d} \biggl\vert \int _{c}^{d}f ( x,t ) \,d_{q}t \biggr\vert ^{p-1} \bigl\vert f ( x,y ) \bigr\vert \,d_{q}y \biggr) \,d_{q}x \\& \quad = \int _{c}^{d} \biggl( \int _{a}^{b} \biggl\vert \int _{c}^{d}f ( x,t ) \,d_{q}t \biggr\vert ^{p-1} \bigl\vert f ( x,y ) \bigr\vert \,d_{q}x \biggr) \,d_{q}y \end{aligned}$$

the last step coming from Fubini’s theorem. By applying the q-Hölder inequality to the inner integral with respect to x, we have

$$\begin{aligned}& \int _{a}^{b} \biggl\vert \int _{c}^{d}f ( x,y ) \,d_{q}y \biggr\vert ^{p}\,d_{q}x \\& \quad \leq \int _{c}^{d} \biggl\{ \biggl( \int _{a}^{b} \biggl( \biggl\vert \int _{c}^{d}f ( x,t ) \,d_{q}t \biggr\vert ^{r ( p-1 ) }\,d_{q}x \biggr) ^{\frac{1}{r}} \biggl( \int _{a}^{b} \bigl\vert f ( x,y ) \bigr\vert ^{p}\,d_{q}x \biggr) ^{ \frac{1}{p}} \biggr) \biggr\} \,d_{q}y \\& \quad = \int _{c}^{d} \biggl\{ \biggl( \int _{a}^{b} \biggl( \biggl\vert \int _{c}^{d}f ( x,t ) \,d_{q}t \biggr\vert ^{p}\,d_{q}x \biggr) ^{\frac{1}{r}} \biggl( \int _{a}^{b} \bigl\vert f ( x,y ) \bigr\vert ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}} \biggr) \biggr\} \,d_{q}y \\& \quad = \biggl( \int _{a}^{b} \biggl\vert \int _{c}^{d}f ( x,t ) \,d_{q}t \biggr\vert ^{p}\,d_{q}x \biggr) ^{\frac{1}{r}} \int _{c}^{d} \biggl( \int _{a}^{b} \bigl\vert f ( x,y ) \bigr\vert ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}\,d_{q}y. \end{aligned}$$

Finally dividing both sides by \(\int _{a}^{b} ( \vert \int _{c}^{d}f ( x,t ) \,d_{q}t \vert ^{p}\,d_{q}x ) ^{\frac{1}{r}}\) we have

$$ \biggl( \int _{a}^{b} \biggl\vert \int _{c}^{d}f ( x,y ) \,d_{q}y \biggr\vert ^{p}\,d_{q}x \biggr) ^{1-\frac{1}{r}}\leq \int _{c}^{d} \biggl( \int _{a}^{b} \bigl\vert f ( x,y ) \bigr\vert ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}\,d_{q}y $$

i.e.

$$ \biggl( \int _{a}^{b} \biggl\vert \int _{c}^{d}f ( x,y ) \,d_{q}y \biggr\vert ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}\leq \int _{c}^{d} \biggl( \int _{a}^{b} \bigl\vert f ( x,y ) \bigr\vert ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}\,d_{q}y, $$

which gives the required inequality. □

Theorem 4

(q-Hardy inequality)

If f is a nonnegative function on \(( 0,\infty )\), \(p>1\) and \(\int _{0}^{\infty }f^{p} ( t ) \,d_{q}t\) is convergent, then the following inequality holds:

$$ \biggl( \int _{0}^{\infty } \biggl( \frac{1}{x} \int _{0}^{x}f ( t ) \,d_{q}t \biggr) ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}\leq \frac{1}{ [ \frac{p-1}{p} ] _{q}} \biggl( \int _{0}^{\infty }f^{p} ( t ) \,d_{q}t \biggr) ^{\frac{1}{p}}, $$
(4.2)

where \(q\in ( 0,1 ) \).

Proof

From (3.1) by the q-changing variables \(t=xs\) it follows that

$$ \frac{1}{x} \int _{0}^{x}f ( t ) \,d_{q}t= \int _{0}^{1}f ( sx ) \,d_{q}s. $$

Thus, we write

$$ \biggl( \int _{0}^{\infty } \biggl( \frac{1}{x} \int _{0}^{x}f ( t ) \,d_{q}t \biggr) ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}= \biggl( \int _{0}^{\infty } \biggl( \int _{0}^{1}f ( xs ) \,d_{q}s \biggr) ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}. $$
(4.3)

From the generalized q-Minkowski integral inequality and by using the q-changing variables \(xs=t\), we have

$$\begin{aligned}& \biggl( \int _{0}^{\infty } \biggl( \int _{0}^{1}f ( xs ) \,d_{q}s \biggr) ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}} \\& \quad \leq \int _{0}^{1} \biggl( \int _{0}^{\infty }f^{p} ( xs ) \,d_{q}x \biggr) ^{\frac{1}{p}}\,d_{q}s= \int _{0}^{1} \biggl( \int _{0}^{ \infty }\frac{1}{s}f^{p} ( t ) \,d_{q}t \biggr) ^{\frac{1}{p}}\,d_{q}s \\& \quad = \biggl( \int _{0}^{1}s^{-\frac{1}{p}}\,d_{q}s \biggr) \biggl( \int _{0}^{ \infty }f^{p} ( t ) \,d_{q}t \biggr) ^{\frac{1}{p}}= \frac{1}{ [ 1-\frac{1}{p} ] _{q}} \biggl( \int _{0}^{\infty }f^{p} ( t ) \,d_{q}t \biggr) ^{\frac{1}{p}} \end{aligned}$$
(4.4)

from (4.3) and (4.4)

$$ \biggl( \int _{0}^{\infty } \biggl( \frac{1}{x} \int _{0}^{x}f ( t ) \,d_{q}t \biggr) ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}\leq \frac{1}{ [ \frac{p-1}{p} ] _{q}} \biggl( \int _{0}^{\infty }f^{p} ( t ) \,d_{q}t \biggr) ^{\frac{1}{p}} $$

and the proof is completed. □

Remark 1

In (4.2) if we choose \(q\rightarrow 1^{-}\) we recapture the classical Hardy inequality.

The following theorem generalizes the q-Hardy type integral inequality by introducing power weights \(x^{r}\).

Theorem 5

If f is a nonnegative function on \(( 0,\infty )\), \(p\geq 1\), \(r< p-1\) and \(\int _{0}^{\infty }t^{r}f^{p} ( t ) \,d_{q}t\) is convergent, then the following inequality holds:

$$ \int _{0}^{\infty } \biggl( \frac{1}{x} \int _{0}^{x}f ( t ) \,d_{q}t \biggr) ^{p}x^{r}\,d_{q}x\leq \frac{1}{ [ \frac{p-r-1}{p} ] _{q}^{p}} \int _{0}^{\infty }t^{r}f^{p} ( t ) \,d_{q}t, $$

where \(q\in ( 0,1 ) \).

Proof

By the q-changing variables \(t=xs\) we get

$$ \biggl( \int _{0}^{\infty } \biggl( \frac{1}{x} \int _{0}^{x}f ( t ) \,d_{q}t \biggr) ^{p}x^{r}\,d_{q}x \biggr) ^{\frac{1}{p}}= \biggl( \int _{0}^{\infty } \biggl( \int _{0}^{1}f ( xs ) x^{ \frac{r}{P}}\,d_{q}s \biggr) ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}}. $$

So, from Minkowski q-integral inequality and by the changing variables \(xs=u\) the proof is completed as follows:

$$\begin{aligned}& \biggl( \int _{0}^{\infty } \biggl( \int _{0}^{1}f ( xs ) x^{ \frac{r}{P}}\,d_{q}s \biggr) ^{p}\,d_{q}x \biggr) ^{\frac{1}{p}} \\& \quad \leq \int _{0}^{1} \biggl( \int _{0}^{\infty }x^{r}f^{p} ( xs ) \,d_{q}x \biggr) ^{\frac{1}{p}}\,d_{q}s= \int _{0}^{1} \biggl( \int _{0}^{\infty }\frac{u^{r}}{s^{r+1}}f^{p} ( u ) \,d_{q}u \biggr) ^{ \frac{1}{p}}\,d_{q}s \\& \quad = \biggl( \int _{0}^{1}s^{\frac{-r-1}{p}}\,d_{q}s \biggr) \biggl( \int _{0}^{\infty }u^{r}f^{p} ( u ) \,d_{q}u \biggr) ^{ \frac{1}{p}} \\& \quad = \frac{1}{ [ \frac{p-r-1}{p} ] _{q}} \biggl( \int _{0}^{ \infty }u^{r}f^{p} ( u ) \,d_{q}u \biggr) ^{\frac{1}{p}}. \end{aligned}$$

 □

Remark 2

In Theorem 5 if we put \(r=0\) we obtain the inequality (4.2).

Definition 4

For a given weight r, we define the modified q-Hardy operator as

$$ H_{q,r}f ( x ) =\frac{1}{xr ( x ) } \int _{0}^{x}r ( t ) f ( t ) \,d_{q}t. $$

The following theorem will be proved using the q-Hardy operator.

Theorem 6

Assume f is a nonnegative function on \(( 0,\infty ) \), r being an absolutely continuous function on \(( 0,\infty )\), and \(p>1\). Also assume \(\int _{0}^{\infty }f^{p} ( x ) \,d_{q}x\) is convergent, and

$$ \frac{ [ p-1 ] _{q}}{p}+\frac{x}{p} \frac{D_{q}r ( x ) }{r ( qx ) }\sum _{i=0}^{p-1} \biggl[ \frac{h_{r,a}f ( qx ) }{h_{r,a}f ( x ) } \biggr] ^{i}\geq \frac{1}{\lambda }, $$
(4.5)

for almost every \(x>0\) and for some \(\lambda >0\). Then we have the following inequality:

$$ \int _{0}^{\infty } \bigl( H_{r}f ( x ) \bigr) ^{p}\,d_{q}x \leq \lambda ^{p}\beta ^{p} \int _{0}^{\infty }f^{p} ( x ) \,d_{q}x, $$

where

$$ H_{q,r}f ( x ) =\frac{1}{xr ( x ) } \int _{0}^{x}r ( t ) f ( t ) \,d_{q}t. $$

Proof

We assume \(0< a< b<\infty \) and

$$ h_{q,r,a}f ( x ) =\frac{1}{r ( x ) } \int _{a}^{x}r ( t ) f ( t ) \,d_{q}t. $$

Then, defining \(H_{r,a}f ( x ) =\frac{1}{x}h_{r,a}f ( x ) \), and integrating by parts from (2.5) with \(w= ( h_{r,a}f ( x ) ) ^{p}\) and \(D_{q}g ( x ) =x^{-p}\) noting that \(g(x)=\frac{x^{1-p}}{ [ 1-p ] _{q}}\), we get

$$\begin{aligned}& \int _{a}^{b} \bigl( H_{q,r,a}f ( x ) \bigr) ^{p}\,d_{q}x \\& \quad = \int _{a}^{b} \bigl( h_{q,r,a}f ( x ) \bigr) ^{p}x^{-p}\,d_{q}x \\& \quad = \int _{0}^{b} \bigl( h_{q,r,a}f ( x ) \bigr) ^{p}x^{-p}\,d_{q}x- \int _{0}^{a} \bigl( h_{q,r,a}f ( x ) \bigr) ^{p}x^{-p}\,d_{q}x \\& \quad = \int _{0}^{b} \bigl( h_{q,r,a}f ( x ) \bigr) ^{p}D_{q}\frac{x^{1-p}}{ [ 1-p ] _{q}}\,d_{q}x- \int _{0}^{a} \bigl( h_{q,r,a}f ( x ) \bigr) ^{p}D_{q} \frac{x^{1-p}}{ [ 1-p ] _{q}}\,d_{q}x \\& \quad = \bigl( h_{q,r,a}f ( x ) \bigr) ^{p} \frac{x^{1-p}}{ [ 1-p ] _{q}}\bigg\vert _{0}^{b}- \int _{0}^{b} \frac{ ( qx ) ^{1-p}}{ [ 1-p ] _{q}}D_{q} \bigl( h_{q,r,a}f ( x ) \bigr) ^{p}\,d_{q}x \\& \qquad {} - \bigl( h_{q,r,a}f ( x ) \bigr) ^{p} \frac{x^{1-p}}{ [ 1-p ] _{q}}\bigg\vert _{0}^{a}+ \int _{0}^{a} \frac{ ( qx ) ^{1-p}}{ [ 1-p ] _{q}}D_{q} \bigl( h_{q,r,a}f ( x ) \bigr) ^{p}\,d_{q}x \\& \quad = \bigl( h_{q,r,a}f ( b ) \bigr) ^{p} \frac{b^{1-p}}{ [ 1-p] _{q}} \\& \qquad {} -\frac{q^{1-p}}{ [ 1-p ] _{q}} \int _{0}^{b}x^{1-p}D_{q}h_{q,r,a}f ( x ) \Biggl( \sum_{i=0}^{p-1} \bigl[ h_{q,r,a}f ( x ) \bigr] ^{p-1-i} \bigl[ h_{q,r,a}f ( qx ) \bigr] ^{i} \Biggr) \,d_{q}x \\& \qquad {} +\frac{q^{1-p}}{ [ 1-p ] _{q}} \int _{0}^{a}x^{1-p}D_{q}h_{q,r,a}f ( x ) \Biggl( \sum_{i=0}^{p-1} \bigl[ h_{q,r,a}f ( x ) \bigr] ^{p-1-i} \bigl[ h_{q,r,a}f ( qx ) \bigr] ^{i} \Biggr) \,d_{q}x \\& \quad = \bigl( h_{q,r,a}f ( b ) \bigr) ^{p} \frac{b^{1-p}}{ [ 1-p ] _{q}} \\& \qquad {} -\frac{q^{1-p}}{ [ 1-p ] _{q}} \int _{a}^{b}x^{1-p}D_{q}h_{q,r,a}f ( x ) \Biggl( \sum_{i=0}^{p-1} \bigl[ h_{q,r,a}f ( x ) \bigr] ^{p-1-i} \bigl[ h_{q,r,a}f ( qx ) \bigr] ^{i} \Biggr) \,d_{q}x. \end{aligned}$$

We notice that from (2.2)

$$ \bigl( h_{q,r,a}f ( b ) \bigr) ^{p} \frac{b^{1-p}}{ [ 1-p ] _{q}}=-q^{p-1} \bigl( h_{q,r,a}f ( b ) \bigr) ^{p} \frac{b^{1-p}}{ [ p-1 ] _{q}} $$

is negative since \(p-1\in \mathbb{N} \), \(p-1>0\) and \(h_{q,r,a}f ( b ) >0\) with \(b>0\). Also, from the definition of \(h_{q,r,a}f ( x ) \) we have

$$\begin{aligned}& D_{q}h_{q,r,a}f ( x ) \\& \quad = D_{q} \biggl( \frac{1}{r ( x ) } \int _{a}^{x}r ( t ) f ( t ) \,d_{q}t \biggr) \\& \quad = D_{q} \biggl( \frac{1}{r ( x ) } \int _{0}^{x}r ( t ) f ( t ) \,d_{q}t \biggr) -D_{q} \biggl( \frac{1}{r ( x ) }\int _{0}^{a}r ( t ) f ( t ) \,d_{q}t \biggr) \\& \quad = \frac{1}{r ( qx ) }D_{q} \biggl( \int _{0}^{x}r ( t ) f ( t ) \,d_{q}t \biggr) + \biggl( \int _{0}^{x}r ( t ) f ( t ) \,d_{q}t \biggr) D_{q} \frac{1}{r ( x ) }- \biggl( \int _{0}^{a}r ( t ) f ( t ) \,d_{q}t \biggr) D_{q}\frac{1}{r ( x ) } \\& \quad = \frac{1}{r ( qx ) }D_{q} \biggl( \int _{0}^{x}r ( t ) f ( t ) \,d_{q}t \biggr) + \biggl( \int _{a}^{x}r ( t ) f ( t ) \,d_{q}t \biggr) D_{q} \frac{1}{r ( x ) } \\& \quad = \frac{r ( x ) }{r ( qx ) }f ( x ) + \biggl( \int _{a}^{x}r ( t ) f ( t ) \,d_{q}t \biggr) D_{q}\frac{1}{r ( x ) } \\& \quad = \frac{r ( x ) }{r ( qx ) }f ( x ) -h_{q,r,a}f ( x ) \frac{D_{q}r ( x ) }{r ( qx ) }. \end{aligned}$$

Hence, by \([ 1-p ] _{q}=-\frac{1}{q^{ ( p-1 ) }} [ ( p-1 ) ] _{q}\)

$$\begin{aligned}& [ p-1 ] _{q} \int _{a}^{b} \bigl( H_{q,r,a}f ( x ) \bigr) ^{p}\,d_{q}x \\& \quad \leq \int _{a}^{b}x^{1-p} \biggl( \frac{r ( x ) }{r ( qx ) }f ( x ) -h_{r,a}f ( x ) \frac{D_{q}r ( x ) }{r ( qx ) } \biggr) \Biggl( \sum_{i=0}^{p-1} \bigl[ h_{q,r,a}f ( x ) \bigr] ^{p-1-i} \bigl[ h_{q,r,a}f ( qx ) \bigr] ^{i} \Biggr) \,d_{q}x \\& \quad = \int _{a}^{b}x^{1-p} \frac{r ( x ) }{r ( qx ) }f ( x ) \bigl[ h_{q,r,a}f ( x ) \bigr] ^{p-1} \Biggl( \sum_{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}f ( qx ) }{h_{q,r,a}f ( x ) } \biggr] ^{i} \Biggr) \,d_{q}x \\& \qquad {} - \int _{a}^{b}x^{1-p} \bigl[ h_{q,r,a}f ( x ) \bigr] ^{p}\frac{D_{q}r ( x ) }{r ( qx ) } \Biggl( \sum_{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}f ( qx ) }{h_{q,r,a}f ( x ) } \biggr] ^{i} \Biggr) \,d_{q}x, \end{aligned}$$

or equivalently

$$\begin{aligned}& \int _{a}^{b} \Biggl[ [ p-1 ] _{q}+x \frac{D_{q}r ( x ) }{r ( qx ) } \Biggl( \sum _{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}f ( qx ) }{h_{q,r,a}f ( x ) } \biggr] ^{i} \Biggr) \Biggr] \bigl( H_{q,r,a}f ( x ) \bigr) ^{p}\,d_{q}x \\& \quad \leq \int _{a}^{b} \Biggl( \frac{r ( x ) }{r ( qx ) }\sum_{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}f ( qx ) }{h_{q,r,a}f ( x ) } \biggr] ^{i} \Biggr) f ( x ) \bigl( H_{q,r,a}f ( x ) \bigr) ^{p-1}\,d_{q}x. \end{aligned}$$

Now, using (4.5) and the q-Hölder inequality, we have

$$\begin{aligned}& \frac{p}{\lambda } \int _{a}^{b} \bigl( H_{q,r,a}f ( x ) \bigr) ^{p}\,d_{q}x \\& \quad \leq \Biggl( \int _{a}^{b} \Biggl( \frac{r ( x ) }{r ( qx ) }\sum _{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}f ( qx ) }{h_{q,r,a}f ( x ) } \biggr] ^{i} \Biggr) ^{p}f^{p} ( x ) \,d_{q}x \Biggr) ^{\frac{1}{p}} \biggl( \int _{a}^{b} \bigl[ H_{q,r,a}f ( x ) \bigr] ^{ ( p-1 ) p^{ \prime }}\,d_{q}x \biggr) ^{\frac{1}{p^{\prime }}}, \end{aligned}$$

where \(\frac{1}{p}+\frac{1}{p^{\prime }}=1\), that is,

$$ \int _{a}^{b} \bigl( H_{q,r,a}f ( x ) \bigr) ^{p}\,d_{q}x \leq \frac{\lambda ^{p}}{p^{p}} \int _{0}^{\infty } \Biggl( \frac{r ( x ) }{r ( qx ) }\sum _{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}f ( qx ) }{h_{q,r,a}f ( x ) } \biggr] ^{i} \Biggr) ^{p}f^{p} ( x ) \,d_{q}x. $$

If we take \(c>a\), then

$$\begin{aligned} \int _{c}^{b} \bigl( H_{q,r,a}f ( x ) \bigr) ^{p}\,d_{q}x \leq& \int _{a}^{b} \bigl( H_{q,r,a}f ( x ) \bigr) ^{p}\,d_{q}x \\ \leq& \frac{\lambda ^{p}}{p^{p}} \int _{0}^{\infty } \Biggl( \frac{r ( x ) }{r ( qx ) }\sum _{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}f ( qx ) }{h_{q,r,a}f ( x ) } \biggr] ^{i} \Biggr) ^{p}f^{p} ( x ) \,d_{q}x. \end{aligned}$$

Invoking the dominated convergence theorem, taking \(a\rightarrow \infty \), we get

$$ \int _{c}^{b} \bigl( H_{q,r}f ( x ) \bigr) ^{p}\,d_{q}x \leq \frac{\lambda ^{p}}{p^{p}} \int _{0}^{\infty } \Biggl( \frac{r ( x ) }{r ( qx ) }\sum _{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}f ( qx ) }{h_{q,r,a}f ( x ) } \biggr] ^{i} \Biggr) ^{p}f^{p} ( x ) \,d_{q}x $$

for all \(c,b>0\). Finally, letting \(b\rightarrow \infty \) and \(c\rightarrow 0\),

$$ \int _{0}^{\infty } \bigl( H_{q,r}f ( x ) \bigr) ^{p}\,d_{q}x \leq \frac{\lambda ^{p}}{p^{p}} \int _{0}^{\infty } \Biggl( \frac{r ( x ) }{r ( qx ) }\sum _{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}f ( qx ) }{h_{q,r,a}f ( x ) } \biggr] ^{i} \Biggr) ^{p}f^{p} ( x ) \,d_{q}x. $$

 □

In Theorem 6 if we take the limit \(q\rightarrow 1^{-}\) we obtain the following theorem, proved by N. Levinson in 1964 (cf. [13, Theorem 4]).

Remark 3

Let f be a nonnegative function on \(( 0,\infty ) \), r being absolutely continuous function on \(( 0,\infty ) \) and \(p>1\). Also assume \(\int _{0}^{\infty } ( f ( x ) ) ^{p}\,dx\) is convergent, and

$$ \frac{p-1}{p}+x\frac{r^{\prime }}{r}\geq \frac{1}{\lambda }, $$

for almost every \(x>0\) and for some \(\lambda >0\). Then we have the following inequality:

$$ \int _{0}^{\infty } \bigl( H_{r}f ( x ) \bigr) ^{p}\,dx \leq \lambda ^{p} \int _{0}^{\infty }f^{p} ( x ) \,dx, $$

where

$$ H_{r}f ( x ) =\frac{1}{xr ( x ) } \int _{0}^{x}r ( t ) f ( t ) \,dt. $$

Theorem 7

Assume f is a nonnegative function on \(( 0,\infty ) \), u is absolutely continuous function on \(( 0,\infty ) \) and \(p>1\). Also assume \(\int _{a}^{b} ( f ( x ) ) ^{p}\,d_{q}x\) is convergent, and

$$ \frac{ [ p-1 ] _{q}}{p}-\frac{x}{p} \frac{D_{q}u ( x ) }{u ( x ) }\sum _{i=0}^{p-1} \biggl( \frac{u ( qx ) }{u ( x ) } \biggr) ^{\frac{i}{p}}\sum_{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}g ( qx ) }{h_{q,r,a}g ( x ) } \biggr] ^{i} \geq \frac{1}{\lambda }, $$
(4.6)

for almost every \(x>0\) and for some \(\lambda >0\). Then we have the following inequality:

$$ \int _{0}^{\infty } \bigl( H_{q}f ( x ) \bigr) ^{p}u ( x ) \,d_{q}x\leq \frac{\lambda ^{p}}{p^{p}} \int _{0}^{ \infty } \Biggl( \sum _{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}g ( qx ) }{h_{q,r,a}g ( x ) } \biggr] ^{i} \Biggr) ^{p}f^{p} ( x ) u ( qx ) \,d_{q}x, $$
(4.7)

where

$$ H_{q}f ( x ) =\frac{1}{x} \int _{0}^{x}f ( t ) \,d_{q}t. $$

Proof

If we consider \(r ( x ) = ( \frac{1}{u ( x ) } ) ^{ \frac{1}{p}}\), then

$$ f ( x ) =r ( x ) g ( x ) = \biggl( \frac{1}{u ( x ) } \biggr) ^{\frac{1}{p}}g ( x ) $$

and we apply Theorem 6 to g, we assume \(0< a< b<\infty \) and

$$ h_{q,r,a}g ( x ) =\frac{1}{r ( x ) } \int _{a}^{x}r ( t ) g ( t ) \,d_{q}t= \bigl( u ( x ) \bigr) ^{\frac{1}{p}} \int _{a}^{x}f ( t ) \,d_{q}t. $$

Then, defining \(H_{q,r,a}g ( x ) =\frac{1}{x}h_{q,r,a}g ( x ) \), and integrating by parts from (2.5) with \(w= ( h_{q,r,a}g ( x ) ) ^{p}\) and \(D_{q}v ( x ) =x^{-p}\) noting that \(v(x)=\frac{x^{1-p}}{ [ 1-p ] _{q}}\) we get

$$\begin{aligned}& \int _{a}^{b} \bigl( H_{q,r,a}g ( x ) \bigr) ^{p}\,d_{q}x \\& \quad = \bigl( h_{q,r,a}g ( b ) \bigr) ^{p} \frac{b^{1-p}}{ [ 1-p ] _{q}} \\& \qquad {} -\frac{q^{1-p}}{ [ 1-p ] _{q}} \int _{0}^{b}x^{1-p}D_{q}h_{q,r,a}g ( x ) \Biggl( \sum_{i=0}^{p-1} \bigl[ h_{q,r,a}g ( x ) \bigr] ^{p-1-i} \bigl[ h_{q,r,a}g ( qx ) \bigr] ^{i} \Biggr) \,d_{q}x \\& \qquad {} +\frac{q^{1-p}}{ [ 1-p ] _{q}} \int _{0}^{a}x^{1-p}D_{q}h_{q,r,a}g ( x ) \Biggl( \sum_{i=0}^{p-1} \bigl[ h_{q,r,a}g ( x ) \bigr] ^{p-1-i} \bigl[ h_{q,r,a}g ( qx ) \bigr] ^{i} \Biggr) \,d_{q}x \\& \quad = \bigl( h_{q,r,a}g ( b ) \bigr) ^{p} \frac{b^{1-p}}{ [ 1-p ] _{q}} \\& \qquad {} -\frac{q^{1-p}}{ [ 1-p ] _{q}} \int _{a}^{b}x^{1-p}D_{q}h_{q,r,a}g ( x ) \Biggl( \sum_{i=0}^{p-1} \bigl[ h_{q,r,a}g ( x ) \bigr] ^{p-1-i} \bigl[ h_{q,r,a}g ( qx ) \bigr] ^{i} \Biggr) \,d_{q}x. \end{aligned}$$

We notice that from (2.2)

$$ \bigl( h_{q,r,a}g ( b ) \bigr) ^{p} \frac{b^{1-p}}{ [ 1-p ] _{q}}=-q^{p-1} \bigl( h_{q,r,a}g ( b ) \bigr) ^{p} \frac{b^{1-p}}{ [ p-1 ] _{q}} $$

is negative since \(p-1\in \mathbb{N} \), \(p-1>0\) and \(h_{q,r,a}g ( b ) >0\) with \(b>0\). Also, from the definition of \(h_{q,r,a}g ( x ) \) we have

$$\begin{aligned}& D_{q}h_{q,r,a}g ( x ) \\& \quad = D_{q} \biggl( \bigl( u ( x ) \bigr) ^{\frac{1}{p}} \int _{a}^{x}f ( t ) \,d_{q}t \biggr) \\& \quad = D_{q} \biggl( \bigl( u ( x ) \bigr) ^{\frac{1}{p}} \int _{0}^{x}f ( t ) \,d_{q}t \biggr) -D_{q} \biggl( \bigl( u ( x ) \bigr) ^{\frac{1}{p}} \int _{0}^{a}f ( t ) \,d_{q}t \biggr) \\& \quad = \bigl( u ( qx ) \bigr) ^{\frac{1}{p}}D_{q} \biggl( \int _{0}^{x}f ( t ) \,d_{q}t \biggr) + \biggl( \int _{0}^{x}f ( t ) \,d_{q}t \biggr) D_{q} \bigl( u ( x ) \bigr) ^{\frac{1}{p}}- \biggl( \int _{0}^{a}f ( t ) \,d_{q}t \biggr) D_{q} \bigl( u ( x ) \bigr) ^{\frac{1}{p}} \\& \quad = \bigl( u ( qx ) \bigr) ^{\frac{1}{p}}D_{q} \biggl( \int _{0}^{x}f ( t ) \,d_{q}t \biggr) + \biggl( \int _{a}^{x}f ( t ) \,d_{q}t \biggr) D_{q} \bigl( u ( x ) \bigr) ^{\frac{1}{p}} \\& \quad = \bigl( u ( qx ) \bigr) ^{\frac{1}{p}}f ( x ) + \frac{h_{q,r,a}g ( x ) }{ ( u ( x ) ) ^{\frac{1}{p}}}\frac{D_{q}u ( x ) }{\sum_{i=0}^{p-1} ( u ( x ) ) ^{\frac{p-1-i}{p}} ( u ( qx ) ) ^{\frac{i}{p}}} \\& \quad = \bigl( u ( qx ) \bigr) ^{\frac{1}{p}}f ( x ) +\frac{h_{q,r,a}g ( x ) }{u ( x ) }D_{q}u ( x ) \sum_{i=0}^{p-1} \biggl( \frac{u ( qx ) }{u ( x ) } \biggr) ^{\frac{i}{p}}. \end{aligned}$$

Hence, by \([ 1-p ] _{q}=-\frac{1}{q^{ ( p-1 ) }} [ ( p-1 ) ] _{q}\)

$$\begin{aligned}& [ p-1 ] _{q} \int _{a}^{b} \bigl( H_{q,r,a}g ( x ) \bigr) ^{p}\,d_{q}x \\& \quad \leq \int _{a}^{b}x^{1-p} \bigl( u ( qx ) \bigr) ^{\frac{1}{p}}f ( x ) \bigl[ h_{q,r,a}g ( x ) \bigr] ^{p-1} \sum_{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}g ( qx ) }{h_{q,r,a}g ( x ) } \biggr] ^{i}\,d_{q}x \\& \qquad {} + \int _{a}^{b}x^{1-p} \frac{ ( h_{q,r,a}g ( x ) ) ^{p}}{u ( x ) }D_{q}u ( x ) \sum_{i=0}^{p-1} \biggl( \frac{u ( qx ) }{u ( x ) } \biggr) ^{ \frac{i}{p}}\sum _{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}g ( qx ) }{h_{q,r,a}g ( x ) } \biggr] ^{i}\,d_{q}x \end{aligned}$$

or equivalently

$$\begin{aligned}& \int _{a}^{b} \Biggl[ [ p-1 ] _{q}-x \frac{D_{q}u ( x ) }{u ( x ) }\sum_{i=0}^{p-1} \biggl( \frac{u ( qx ) }{u ( x ) } \biggr) ^{\frac{i}{p}}\sum _{i=0}^{p-1} \biggl( \frac{h_{q,r,a}g ( qx ) }{h_{q,r,a}g ( x ) } \biggr) ^{i} \Biggr] \bigl( H_{q,r,a}g ( x ) \bigr) ^{p}\,d_{q}x \\& \quad \leq \int _{a}^{b} \bigl( u ( qx ) \bigr) ^{ \frac{1}{p}}\sum_{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}g ( qx ) }{h_{q,r,a}g ( x ) } \biggr] ^{i}f ( x ) \bigl[ H_{q,r,a}g ( x ) \bigr] ^{p-1}\,d_{q}x. \end{aligned}$$

Finally, by using (4.6) and the q-Hölder inequality, we have

$$ \int _{0}^{\infty } \bigl( H_{q,r}f ( x ) \bigr) ^{p}\,d_{q}x \leq \frac{\lambda ^{p}}{p^{p}} \int _{0}^{\infty } \Biggl( \sum _{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}g ( qx ) }{h_{q,r,a}g ( x ) } \biggr] ^{i} \Biggr) ^{p}f^{p} ( x ) u ( qx ) \,d_{q}x $$

and

$$ \int _{0}^{\infty } \bigl( H_{q}f ( x ) \bigr) ^{p}u ( x ) \,d_{q}x\leq \frac{\lambda ^{p}}{p^{p}} \int _{0}^{ \infty } \Biggl( \sum _{i=0}^{p-1} \biggl[ \frac{h_{q,r,a}g ( qx ) }{h_{q,r,a}g ( x ) } \biggr] ^{i} \Biggr) ^{p}f^{p} ( x ) u ( qx ) \,d_{q}x, $$

and this completes the proof. □

In Theorem 7 if we take the limit \(q\rightarrow 1^{-}\) we obtain the following result, proved by N. Levinson in 1964 [13] on continuous analysis.

Remark 4

Assume that f is a nonnegative function on \(( 0,\infty ) \), u is absolutely continuous function on \(( 0,\infty )\), and \(p>1\). Also assume \(\int _{a}^{b} ( f ( x ) ) ^{p}\,dx\) is convergent, and

$$ \frac{p-1}{p}-px\frac{u^{\prime }}{u}\geq \frac{1}{\lambda }, $$

for almost every \(x>0\) and for some \(\lambda >0\). Then we have the following inequality:

$$ \int _{0}^{\infty } \bigl( Hf ( x ) \bigr) ^{p}u ( x ) \,dx\leq \lambda ^{p} \int _{0}^{\infty }f^{p} ( x ) u ( x ) \,dx, $$

where

$$ Hf ( x ) =\frac{1}{x} \int _{0}^{x}f ( t ) \,dt. $$

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The authors would like to express their sincere thanks to the editor and the anonymous reviewers for their helpful comments and suggestions.

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Correspondence to Necmettin Alp.

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Alp, N., Sarikaya, M.Z. q-Hardy type inequalities for quantum integrals. Adv Differ Equ 2021, 355 (2021). https://doi.org/10.1186/s13662-021-03514-6

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  • DOI: https://doi.org/10.1186/s13662-021-03514-6

MSC

  • 34A08
  • 26A51
  • 26D15

Keywords

  • Hardy inequality
  • Opial inequality
  • Hölder’s inequality