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Some new fractional q-integral Grüss-type inequalities and other inequalities

Abstract

In this paper, we employ a fractional q-integral on the specific time scale, T t 0 ={t:t= t 0 q n ,n a nonnegative integer}{0}, where t 0 R and 0<q<1, to establish some new fractional q-integral Grüss-type inequalities by using one or two fractional parameters. Furthermore, other fractional q-integral inequalities are also obtained.

MSC:26D10, 26A33.

1 Introduction

In the past several years, by using the Riemann-Liouville fractional integrals, the fractional integral inequalities and applications have been addressed extensively by several researchers. For example, we refer the reader to [16] and the references cited therein. Dahmani et al. [7] gave the following fractional integral inequalities by using the Riemann-Liouville fractional integrals. Let f and g be two integrable functions on [0,) satisfying the following conditions:

φ 1 f(x) φ 2 , ψ 1 g(x) ψ 2 , φ 1 , φ 2 , ψ 1 , ψ 2 R,x[0,).

For all t>0, α>0 and β>0, then

| t α Γ ( α + 1 ) J α (fg)(t) J α f(t) J α g(t)| ( t α Γ ( α + 1 ) ) 2 ( φ 2 φ 1 )( ψ 2 ψ 1 )

and

To the best of authors’ knowledge, only some fractional q-integral inequalities have been established in recent years. That is, only Öğünmez and Özkan [8], Bohner and Ferreira [9] and Yang [10] obtained some fractional q-integral inequalities. With motivation from the papers [7, 11, 12], the main purpose of this article is to establish some new fractional q-integral inequalities. First of all, by using one or two fractional parameters, we establish some new fractional q-integral Grüss-type inequalities on the specific time scale T t 0 ={t:t= t 0 q n ,n a nonnegative integer}{0}, where t 0 R and 0<q<1. In general, a time scale is an arbitrary nonempty closed subset of real numbers [13]. Furthermore, other fractional q-integral inequalities are also obtained.

2 Description of fractional q-calculus

In this section, we introduce the basic definitions on fractional q-calculus. More results concerning fractional q-calculus can be found in [1417].

Let t 0 R and define T t 0 ={t:t= t 0 q n ,n a nonnegative integer}{0}, 0<q<1. For a function f: T t 0 R, the nabla q-derivative of f

q f(t)= f ( q t ) f ( t ) ( q 1 ) t

for all t T t 0 {0}. The q-integral of f is

0 t f(s)s=(1q)t i = 0 q i f ( t q i ) .

The q-factorial function is defined in the following way: if n is a positive integer, then

( t s ) ( n ) ̲ =(ts)(tqs) ( t q 2 s ) ( t q n 1 s ) .

If n is not a positive integer, then

( t s ) ( n ) ̲ = t n k = 0 1 ( s / t ) q k 1 ( s / t ) q n + k .

The q-derivative of the q-factorial function with respect to t is

q ( t s ) ( n ) ̲ = 1 q n 1 q ( t s ) ( n 1 ) ̲ ,

and the q-derivative of the q-factorial function with respect to s is

q ( t s ) ( n ) ̲ = 1 q n 1 q ( t q s ) ( n 1 ) ̲ .

The q-exponential function is defined as

e q (t)= k = 0 ( 1 q k t ) , e q (0)=1.

Define the q-gamma function by

Γ q (ν)= 1 1 q 0 1 ( t 1 q ) ν 1 e q (qt)t,ν R + .

Note that

Γ q (ν+1)= [ ν ] q Γ q (ν),ν R + ,

where [ ν ] q :=(1 q ν )/(1q). The fractional q-integral is defined as

q ν f(t)= 1 Γ q ( ν ) 0 t ( t q s ) ( ν 1 ) ̲ f(s)s.

Note that

q ν (1)= 1 Γ q ( ν ) q 1 q ν 1 t ( ν ) ̲ = 1 Γ q ( ν + 1 ) t ( ν ) ̲ .

3 Fractional q-integral Grüss-type inequalities

To state the main results in this paper, we employ the following lemmas. For the sake of convenience, we use the following assumption (A) in this section:

φ 1 f(x) φ 2 , ψ 1 g(x) ψ 2 , φ 1 , φ 2 , ψ 1 , ψ 2 R,x T t 0 .

Lemma 1 Let φ 1 , φ 2 R and f be a function defined on T t 0 . Then, for all t>0 and ν>0, we have

t ( ν ) ̲ Γ q ( ν + 1 ) q ν f 2 ( t ) ( q ν f ( t ) ) 2 = ( φ 2 t ( ν ) ̲ Γ q ( ν + 1 ) q ν f ( t ) ) ( q ν f ( t ) φ 1 t ( ν ) ̲ Γ q ( ν + 1 ) ) t ( ν ) ̲ Γ q ( ν + 1 ) q ν ( φ 2 f ( t ) ) ( f ( t ) φ 1 ) .
(1)

Proof Let φ 1 , φ 2 R and f be a function defined on T t 0 . For any τ>0 and ρ>0, we have

(2)

Multiplying both sides of (2) by ( t q τ ) ( ν 1 ) ̲ / Γ q (ν) and integrating the resulting identity with respect to τ from 0 to t, we get

(3)

Multiplying both sides of (3) by ( t q ρ ) ( ν 1 ) ̲ / Γ q (ν) and integrating the resulting identity with respect to ρ from 0 to t, we obtain

which implies (1). □

Lemma 2 Let f and g be two functions defined on T t 0 . Then, for all t>0, μ>0 and ν>0, we have

(4)

Proof In order to prove Lemma 2, we firstly prove that the following inequality (i.e., Cauchy-Schwarz inequality for double q-integrals) holds. Let f(x,y), g(x,y) and h(x,y) be three functions defined on T t 0 2 with h(x,y)0. Then we have

According to the definition of q-integral, it is easy to obtain that double q-integral is

Due to discrete Cauchy-Schwarz inequality with weight coefficient, we have

Next, we prove that Lemma 2 holds. Let H(τ,ρ) be defined by

H(τ,ρ)= ( f ( τ ) f ( ρ ) ) ( g ( τ ) g ( ρ ) ) ,t>0,τ>0,ρ>0.
(5)

Multiplying both sides of (5) by ( t q τ ) ( ν 1 ) ̲ ( t q ρ ) ( μ 1 ) ̲ /( Γ q (ν) Γ q (μ)) and integrating the resulting identity with respect to τ and ρ from 0 to t, then applying the Cauchy-Schwarz inequality for double q-integrals, we obtain (4). □

Lemma 3 Let φ 1 , φ 2 R and f be a function defined on T t 0 . Then, for all t>0 and ν>0, we have

(6)

Proof Multiplying both sides of (3) by ( t q ρ ) ( μ 1 ) ̲ / Γ q (μ) and integrating the resulting identity with respect to ρ from 0 to t, we obtain

which implies (6). □

Theorem 1 Let f and g be two functions defined on T t 0 satisfying (A). Then, for all t>0 and ν>0, we have

| t ( ν ) ̲ Γ q ( ν + 1 ) q ν (fg)(t) q ν f(t) q ν g(t)| ( t ( ν ) ̲ 2 Γ q ( ν + 1 ) ) 2 ( φ 2 φ 1 )( ψ 2 ψ 1 ).
(7)

Proof Let f and g be two functions defined on T t 0 satisfying (A). Multiplying both sides of (6) by ( t q τ ) ( ν 1 ) ̲ ( t q ρ ) ( ν 1 ) ̲ / Γ q 2 (ν) and integrating the resulting identity with respect to τ and ρ from 0 to t, we can state that

(8)

Applying the Cauchy-Schwarz inequality for double q-integrals, we have

(9)

Since ( φ 2 f(x))(f(x) φ 1 )0 and ( ψ 2 g(x))(g(x) ψ 1 )0, we have

Thus,

t ( ν ) ̲ Γ q ( ν + 1 ) q ν f 2 ( t ) ( q ν f ( t ) ) 2 ( φ 2 t ( ν ) ̲ Γ q ( ν + 1 ) q ν f ( t ) ) ( q ν f ( t ) φ 1 t ( ν ) ̲ Γ q ( ν + 1 ) ) , t ( ν ) ̲ Γ q ( ν + 1 ) q ν g 2 ( t ) ( q ν g ( t ) ) 2 ( ψ 2 t ( ν ) ̲ Γ q ( ν + 1 ) q ν g ( t ) ) ( q ν g ( t ) ψ 1 t ( ν ) ̲ Γ q ( ν + 1 ) ) .
(10)

Combining (9) and (10), from Lemma 1, we deduce that

(11)

Now by using the elementary inequality 4xy ( x + y ) 2 , x,yR, we can state that

4 ( φ 2 t ( ν ) ̲ Γ q ( ν + 1 ) q ν f ( t ) ) ( q ν f ( t ) φ 1 t ( ν ) ̲ Γ q ( ν + 1 ) ) ( t ( ν ) ̲ Γ q ( ν + 1 ) ( φ 2 φ 1 ) ) 2 , 4 ( ψ 2 t ( ν ) ̲ Γ q ( ν + 1 ) q ν g ( t ) ) ( q ν g ( t ) ψ 1 t ( ν ) ̲ Γ q ( ν + 1 ) ) ( t ( ν ) ̲ Γ q ( ν + 1 ) ( ψ 2 ψ 1 ) ) 2 .
(12)

From (11) and (12), we obtain (7). □

Theorem 2 Let f and g be two functions defined on T t 0 satisfying (A). Then, for all t>0, μ>0 and ν>0, we have

Proof Since ( φ 2 f(x))(f(x) φ 1 )0 and ( ψ 2 g(x))(g(x) ψ 1 )0, then we can write

t ( ν ) ̲ Γ q ( ν + 1 ) q μ ( φ 2 f ( x ) ) ( f ( x ) φ 1 ) t ( μ ) ̲ Γ q ( μ + 1 ) q ν ( φ 2 f ( x ) ) ( f ( x ) φ 1 ) 0 , t ( ν ) ̲ Γ q ( ν + 1 ) q μ ( ψ 2 g ( x ) ) ( g ( x ) ψ 1 ) t ( μ ) ̲ Γ q ( μ + 1 ) q ν ( ψ 2 g ( x ) ) ( g ( x ) ψ 1 ) 0 .
(13)

Applying Lemma 3 to f and g, then by using Lemma 2 and the formula (13), we obtain Theorem 2. □

4 The other fractional q-integral inequalities

For the sake of simplicity, we always assume that q ν ϕ denotes q ν ϕ(t) and all of fractional q-integrals are finite in this section.

Theorem 3 Let f and g be two functions defined on T t 0 and α,β>1 satisfying 1/α+1/β=1. Then the following inequalities hold:

  1. (a)

    1 α q ν ( | f | α )+ 1 β q ν ( | g | β ) Γ q ( ν + 1 ) t ( ν ) ̲ q ν (|f|) q ν (|g|).

  2. (b)

    1 α q ν ( | f | α ) q ν ( | g | α )+ 1 β q ν ( | f | β ) q ν ( | g | β ) ( q ν ( | f g | ) ) 2 .

  3. (c)

    1 α q ν ( | f | α ) q ν ( | g | β )+ 1 β q ν ( | f | β ) q ν ( | g | α ) q ν (|f| | g | α 1 ) q ν (|f| | g | β 1 ).

  4. (d)

    q ν ( | f | α ) q ν ( | g | β ) q ν (|fg|) q ν ( | f | α 1 | g | β 1 ).

Proof According to the well-known Young inequality,

1 α x α + 1 β y β xy,x,y0,α,β>1, 1 α + 1 β =1.

Putting x=f(τ) and y=g(ρ), τ,ρ>0, we have

1 α | f ( τ ) | α + 1 β | g ( ρ ) | β | f ( τ ) | | g ( ρ ) | ,τ,ρ>0.
(14)

Multiplying both sides of (6) by ( t q τ ) ( ν 1 ) ̲ ( t q ρ ) ( ν 1 ) ̲ / Γ q 2 (ν), we obtain

Integrating the preceding identity with respect to τ and ρ from 0 to t, we can state that

1 α t ( ν ) ̲ Γ q ( ν + 1 ) q ν ( | f ( t ) | α ) + 1 β t ( ν ) ̲ Γ q ( ν + 1 ) q ν ( | g ( t ) | β ) q ν ( | f ( t ) | ) q ν ( | g ( t ) | ) ,

which implies (a). The rest of inequalities can be proved in the same manner by the next choice of the parameters in the Young inequality:

  1. (b)

    x=|f(τ)||g(ρ)|, y=|f(ρ)||g(τ)|.

  2. (c)

    x=|f(τ)|/|g(τ)|, y=|f(ρ)|/|g(ρ)|, (g(τ)g(ρ)0).

  3. (d)

    x=|f(ρ)|/|f(τ)|, y=|g(ρ)|/|g(τ)|, (f(τ)g(ρ)0).

Repeating the foregoing arguments, we obtain (b)-(d). □

Theorem 4 Let f and g be two functions defined on T t 0 and α,β>1 satisfying 1/α+1/β=1. Then the following inequalities hold:

  1. (a)

    1 α q ν ( | f | α ) q ν ( | g | 2 )+ 1 β q ν ( | f | 2 ) q ν ( | g | β ) q ν (|fg|) q ν ( | f | 2 / β | g | 2 / α ).

  2. (b)

    1 α q ν ( | f | 2 ) q ν ( | g | β )+ 1 β q ν ( | f | β ) q ν ( | g | 2 ) q ν ( | f | 2 / α | g | 2 / β ) q ν ( | f | α 1 | g | β 1 ).

  3. (c)

    q ν ( | f | 2 ) q ν ( 1 α | g | α + 1 β | g | β ) q ν ( | f | 2 / α |g|) q ν ( | f | 2 / β |g|).

Proof As a previous one, the proof is based on the Young inequality with the following appropriate choice of parameters:

  1. (a)

    x=|f(τ)| | g ( ρ ) | 2 / α , y= | f ( ρ ) | 2 / β |g(τ)|.

  2. (b)

    x= | f ( τ ) | 2 / α /|f(ρ)|, y= | g ( τ ) | 2 / β /|g(ρ)|, (f(ρ)g(ρ)0).

  3. (c)

    x= | f ( τ ) | 2 / α /|g(ρ)|, y= | f ( ρ ) | 2 / β /|g(τ)|, (g(τ)g(ρ)0).

 □

Theorem 5 Let f and g be two positive functions defined on T t 0 such that for all t>0,

m= min 0 τ t f ( τ ) g ( τ ) ,M= max 0 τ t f ( τ ) g ( τ ) .
(15)

Then the following inequalities hold:

  1. (a)

    0 q ν ( f 2 ) q ν ( g 2 ) ( m + M ) 2 4 m M ( q ν ( f g ) ) 2 .

  2. (b)

    0 q ν ( f 2 ) q ν ( g 2 ) q ν (fg) ( M m ) 2 2 m M q ν (fg).

  3. (c)

    0 q ν ( f 2 ) q ν ( g 2 ) ( q ν ( f g ) ) 2 ( M m ) 2 4 m M ( q ν ( f g ) ) 2 .

Proof It follows from (15) and

( f ( τ ) g ( τ ) m ) ( M f ( τ ) g ( τ ) ) g 2 (τ)0,0τt.
(16)

Multiplying both sides of (15) by ( t q τ ) ( ν 1 ) ̲ / Γ q (ν) and integrating the resulting identity with respect to τ from 0 to t, we can get

q ν ( f 2 ) +mM q ν ( g 2 ) (m+M) q ν (fg).
(17)

On the other hand, it follows from mM>0 and ( q ν ( f 2 ) m M q ν ( g 2 ) ) 2 0 that

2 q ν ( f 2 ) m M q ν ( g 2 ) q ν ( f 2 ) +mM q ν ( g 2 ) .
(18)

According to (17) and (18), we have

4mM q ν ( f 2 ) q ν ( g 2 ) ( m + M ) 2 ( q ν ( f g ) ) 2 ,

which implies (a). By a few transformations of (a), similarly, we obtain (b) and (c). □

Corollary 1 Under the conditions of Theorem 5, if α,β(0,1), α+β=1, then it follows from the arithmetric-geometric mean inequality that

( 1 α q ν ( f 2 ) ) α ( m M β q ν ( g 2 ) ) β q ν ( f 2 ) +mM q ν ( g 2 ) (m+M) q ν (fg),

which implies that

( q ν ( f 2 ) ) α ( q ν ( g 2 ) ) β α α β β m + M ( m M ) β q ν (fg).

Theorem 6 Let f and g be two positive functions on T t 0 and

0< Φ 1 f(τ) Φ 2 <,0< Ψ 1 g(τ) Ψ 2 <.
(19)

Then the following inequalities hold:

  1. (a)

    0 q ν ( f 2 ) q ν ( g 2 ) ( Φ 1 Ψ 1 + Φ 2 Ψ 2 ) 2 4 Φ 1 Ψ 1 Φ 2 Ψ 2 ( q ν ( f g ) ) 2 .

  2. (b)

    0 q ν ( f 2 ) q ν ( g 2 ) q ν (fg) ( Φ 2 Ψ 2 Φ 1 Ψ 1 ) 2 2 Φ 1 Ψ 1 Φ 2 Ψ 2 q ν (fg).

  3. (c)

    0 q ν ( f 2 ) q ν ( g 2 ) ( q ν ( f g ) ) 2 ( Φ 2 Ψ 2 Φ 1 Ψ 1 ) 2 4 Φ 1 Ψ 1 Φ 2 Ψ 2 ( q ν ( f g ) ) 2 .

Proof Under the conditions satisfied by the functions f and g, we have

Φ 1 Ψ 2 f ( τ ) g ( τ ) Φ 2 Ψ 1 .

Applying Theorem 6, we get the inequality (a) and using it, we have (b) and (c). □

Corollary 2 Let f be a positive function on T t 0 satisfying (19). Then the following inequality holds:

q ν ( f 2 ) Γ q ( ν + 1 ) ( Φ 1 + Φ 2 ) 2 4 t ( ν ) ̲ Φ 1 Φ 2 ( q ν ( f ) ) 2 .

Theorem 7 Let f and g be two positive functions on T t 0 and

0<m g ( τ ) f ( τ ) M<
(20)

and p0 be a real number, then the following inequality holds:

q ν ( f 2 p g p ) + m M ( M p 1 m p 1 ) M m q ν ( f p ) M p m p M m q ν (fg)

for p(0,1), or reverse for p(0,1). Especially, for p=2, we have

q ν ( g 2 ) +mM q ν ( f 2 ) (m+M) q ν (fg).

Proof The inequality is based on the Lah-Ribaric inequality [[18], p.9] and [[19], p.123]. □

Theorem 8 Let f and g be two positive functions on T t 0 and p0 be a real number. Then the following inequality holds:

( q ν ( f g ) ) p ( q ν ( f 2 ) ) p 1 q ν ( f 2 p g p )

for p(0,1), or reverse for p(0,1).

Proof The above inequality is obtained via the Jensen inequality for the convex functions. □

Corollary 3 Let f be a positive function on T t 0 and p0 be a real number. Then the following inequality holds:

( q ν ( f ) ) p ( t ( ν ) ̲ Γ q ( ν + 1 ) ) p 1 q ν ( f p )

for p(0,1), or reverse for p(0,1).

Theorem 9 Let p, f and g be three positive functions on T t 0 satisfying (19). If 0<αβ<1, α+β=1, then the following inequalities hold:

(21)
(22)

Proof Since (βf(τ)α Φ 1 )(f(τ) Φ 2 )0 on T t 0 , we have

β f 2 (τ)(α Φ 1 +β Φ 2 )f(τ)+α Φ 1 Φ 2 0.
(23)

Multiplying both sides of (23) by p(τ)/f(τ), we get

βp(τ)f(τ)+α Φ 1 Φ 2 p ( τ ) f ( τ ) (α Φ 1 +β Φ 2 )p(τ).
(24)

From (24) and arithmetric-geometric mean inequality, we obtain

(25)

which implies (21).

Replacing p and f by pfg and f/g in (25), respectively, and Φ 1 / Ψ 2 f(τ)/g(τ) Φ 2 / Ψ 1 , we get

which implies (22). □

Corollary 4 Let p, f and g be three positive functions on T t 0 satisfying (20). If 0<αβ<1, α+β=1, then the following inequality holds:

α q ν ( p g 2 ) +βmM q ν ( p f 2 ) (αm+βM) q ν (pfg).
(26)

Proof Replacing Φ 1 , Φ 2 and f(τ) by m, M and g(τ)/f(τ) in (24), and multiplying both sides by ( t q τ ) ( ν 1 ) ̲ / Γ q (ν) and integrating the resulting identity with respect to τ from 0 to t, we get (25). □

Theorem 10 Let p, f and g be three functions on T t 0 with p(τ)0.

  1. (a)

    If there exist four constants Φ 1 , Φ 2 , Ψ 1 , Ψ 2 R such that ( Φ 2 g(τ) Ψ 1 f(τ))( Ψ 2 f(τ) Φ 1 g(τ))0 for all τ>0, then

    Φ 1 Φ 2 q ν ( p g 2 ) + Ψ 1 Ψ 2 q ν ( p f 2 ) ( Φ 1 Ψ 1 + Φ 2 Ψ 2 ) q ν ( p f g ) | Φ 1 Ψ 1 + Φ 2 Ψ 2 | ( q ν ( p f 2 ) + q ν ( p g 2 ) ) .
    (27)

Moreover, if Φ 1 Φ 2 Ψ 1 Ψ 2 >0, then

(28)
(29)
  1. (b)

    If there exist four constants Φ 1 , Φ 2 , Ψ 1 , Ψ 2 R such that ( Φ 2 g(τ) Ψ 1 f(ρ))( Ψ 2 f(ρ) Φ 1 g(τ))0 for all τ,ρ>0, then

    (30)
  2. (c)

    If Φ 1 Φ 2 >0 and Ψ 1 Ψ 2 >0, then

    Φ 1 Φ 2 ( q ν ( p g ) ) 2 + Ψ 1 Ψ 2 ( q ν ( p f ) ) 2 ( Φ 1 Ψ 1 + Φ 2 Ψ 2 ) q ν (p) q ν (pfg).
    (31)
  3. (d)

    If Φ 1 Φ 2 >0 and Ψ 1 Ψ 2 >0, then

    Φ 1 Φ 2 ( q ν ( p g ) ) 2 + Ψ 1 Ψ 2 ( q ν ( p f ) ) 2 ( Φ 1 Ψ 1 + Φ 2 Ψ 2 ) q ν (pf) q ν (pg).
    (32)

Proof Case (a). It follows from the assumption that

p(τ) ( Φ 2 g ( τ ) Ψ 1 f ( τ ) ) ( Ψ 2 f ( τ ) Φ 1 g ( τ ) ) 0

for all τ0, which implies that

Φ 1 Φ 2 p(τ) g 2 (τ)+ Ψ 1 Ψ 2 p(τ) f 2 (τ)( Φ 1 Ψ 1 + Φ 2 Ψ 2 )p(τ)f(τ)g(τ).
(33)

Multiplying both sides of (33) by ( t q τ ) ( ν 1 ) ̲ / Γ q (ν) and integrating the resulting identity with respect to τ from 0 to t, we obtain the left-hand side of (27). Furthermore, by Cauchy’s inequality, we get the right-hand side of (27).

Multiplying both sides of the inequality

Φ 1 Φ 2 q ν ( p g 2 ) + Ψ 1 Ψ 2 q ν ( p f 2 ) ( Φ 1 Ψ 1 + Φ 2 Ψ 2 ) q ν (pfg)

by 1/ Φ 1 Φ 2 Ψ 1 Ψ 2 , we get (28).

On the other hand, it follows from Φ 1 Φ 2 Ψ 1 Ψ 2 >0 and ( Φ 1 Φ 2 q ν ( p g 2 ) Ψ 1 Ψ 2 q ν ( p f 2 ) ) 2 0 that

2 Φ 1 Φ 2 q ν ( p g 2 ) Ψ 1 Ψ 2 q ν ( p f 2 ) Φ 1 Φ 2 q ν ( p g 2 ) + Ψ 1 Ψ 2 q ν ( p f 2 ) .
(34)

According to (27) and (34), we have

4 Φ 1 Φ 2 Ψ 1 Ψ 2 q ν ( p g 2 ) q ν ( p f 2 ) ( Φ 1 Ψ 1 + Φ 2 Ψ 2 ) 2 ( q ν ( p f g ) ) 2 ,

which implies (29).

Case (b). It follows from the assumption that

p(τ)p(ρ) ( Φ 2 g ( τ ) Ψ 1 f ( ρ ) ) ( Ψ 2 f ( ρ ) Φ 1 g ( τ ) ) 0

for all τ,ρ>0, which implies that

(35)

Multiplying both sides of (35) by ( t q τ ) ( ν 1 ) ̲ ( t q ρ ) ( ν 1 ) ̲ / Γ q 2 (ν) and integrating the resulting identity with respect to τ and ρ from 0 to t, respectively, we obtain (30).

Case (c) and (d). It follows from Cauchy’s inequality that

( q ν ( p f ) ) 2 q ν (p) q ν ( p f 2 ) , ( q ν ( p g ) ) 2 q ν (p) q ν ( p g 2 ) .

Combining (a), (b) and the preceding two inequalities, we see that

Φ 1 Φ 2 ( q ν ( p g ) ) 2 + Ψ 1 Ψ 2 ( q ν ( p f ) ) 2 Φ 1 Φ 2 q ν ( p ) q ν ( p f 2 ) + Ψ 1 Ψ 2 q ν ( p ) q ν ( p g 2 ) ( Φ 1 Ψ 1 + Φ 2 Ψ 2 ) q ν ( p ) q ν ( p f g ) ,

which implies (31). Furthermore,

Φ 1 Φ 2 ( q ν ( p g ) ) 2 + Ψ 1 Ψ 2 ( q ν ( p f ) ) 2 Φ 1 Φ 2 q ν ( p ) q ν ( p f 2 ) + Ψ 1 Ψ 2 q ν ( p ) q ν ( p g 2 ) ( Φ 1 Ψ 1 + Φ 2 Ψ 2 ) q ν ( p f ) q ν ( p g ) ,

which implies (32). □

Theorem 11 Let p, f and g be three positive functions on T t 0 with p(τ)0. Then we have

( q ν ( p ) q ν ( p f g ) + q ν ( p f ) q ν ( p g ) ) 2 ( q ν ( p ) q ν ( p f 2 ) + ( q ν ( p f ) ) 2 ) × ( q ν ( p ) q ν ( p g 2 ) + ( q ν ( p g ) ) 2 ) .
(36)

Moreover, under the assumptions of (a) and (b) in Theorem 10, the following inequality holds:

(37)

Proof First of all, we give the proof of (36). By Cauchy’s inequality and the element inequality 2xy u v x 2 u+ y 2 v, for all x,y,u,v0, we have

which implies (36).

Next, we prove that (37) holds. It follows from (a) and (b) in Theorem 10 that

( Φ 1 Ψ 1 + Φ 2 Ψ 2 ) q ν ( p ) q ν ( p f g ) Φ 1 Φ 2 q ν ( p ) q ν ( p g 2 ) + Ψ 1 Ψ 2 q ν ( p ) q ν ( p f 2 ) Φ 1 Φ 2 q ν ( p ) q ν ( p g 2 ) + Ψ 1 Ψ 2 ( q ν ( p f ) ) 2 , ( Φ 1 Ψ 1 + Φ 2 Ψ 2 ) q ν ( p f ) q ν ( p g ) Φ 1 Φ 2 q ν ( p ) q ν ( p g 2 ) + Ψ 1 Ψ 2 q ν ( p ) q ν ( p f 2 ) Φ 1 Φ 2 ( q ν ( p g ) ) 2 + Ψ 1 Ψ 2 q ν ( p ) q ν ( p f 2 ) .

Combining the preceding two inequalities and the element inequality ( x + y ) 2 4xy, we see that

which implies (37). □

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Acknowledgements

The authors sincerely thank the editor and reviewers for their valuable suggestions and useful comments that have led to the present improved version of the original manuscript.

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Correspondence to Wengui Yang.

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Zhu, C., Yang, W. & Zhao, Q. Some new fractional q-integral Grüss-type inequalities and other inequalities. J Inequal Appl 2012, 299 (2012). https://doi.org/10.1186/1029-242X-2012-299

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Keywords

  • fractional q-integral
  • integral inequalities
  • Grüss-type inequalities