Skip to main content

Discrete Grüss type inequality on fractional calculus

Abstract

We give a discrete Grüss type inequality on fractional calculus.

Introduction

Motivated by Grüss [1], our purpose is to prove more general versions of Grüss type inequalities for delta discrete fractional calculus. It is well known that Grüss type inequalities in continuous and discrete cases play a crucial role in studying the qualitative behavior of differential and difference equations, respectively, as well as many other areas of mathematics [29]. For the background and a summary on these particular subjects, we refer the interested reader to the excellent references [2, 1018].

The study of discrete fractional calculus was pioneered by Diaz and Osler [19]. In the mentioned work, the authors used an infinite sum to give a definition of discrete fractional sum, whereas Gray and Zhang used a finite sum in [20]. In the last decade, new results in this area have been established [2124], as well as importance has been gained by inequalities on discrete fractional calculus in [10, 2427]

Preliminaries

We begin with basic definitions and results from [10].

Definition 1

The vth fractional sum of f is defined by

$$ \Delta^{-v}f(t,a)=\frac{1}{\Gamma(v)}\sum_{s=a}^{t-v}(t-s-1)^{\underline{v-1}}f(s), $$

where f and \(\Delta^{-v}f \) are defined for \(s=a \operatorname{mod}(1)\) and \(t=(a+v)\operatorname{mod}(1)\), respectively. In particular, \(\Delta^{-v}\) maps functions defined on \(\mathbb{N} _{a} \) to functions defined on \(\mathbb{N} _{a+v}\), where \(\mathbb{N} _{t}=\{t,t+1,t+2,\ldots\}\).

Here,

$$ t^{\underline{v}}:=\frac{\Gamma(t+1)}{\Gamma(t-v+1)}. $$

From now on in this context for convenience we set \(\Delta ^{-v}f(t,a)=\Delta^{-v}f(t)\).

Theorem 1

[28]

Let f be a real valued function defined on \(\mathbb{N} _{a} \) and let \(\mu,v>0\). Then

$$ \Delta^{-v}\bigl(\Delta^{-\mu}f(t)\bigr)=\Delta^{-(\mu+v)}f(t)= \Delta^{-\mu }\bigl(\Delta^{-v}f(t)\bigr) \quad\textit{for all }t\in \mathbb{N} _{a+\mu+v}. $$

Theorem 2

[21]

For \(v>0 \) and p a positive integer we have

$$ \Delta^{-v}\Delta^{p}f(t)=\Delta^{p}\Delta ^{-v}f(t)-\sum_{k=0}^{v-1} \frac{(t-a)^{\underline{v-p+k}}}{\Gamma (v+k-p+1)}\Delta^{k}f(a), $$

where f is defined on \(\mathbb{N} _{a}\).

Remark 1

Let \(\mu>0 \) and \(m-1<\mu<m\), \(m= \lceil\mu \rceil\), where m is a positive integer, and set \(v=m-\mu>0\). Then by Theorem 2 we have

$$ \Delta^{-v}\Delta^{m}f(t)=\Delta^{m}\Delta ^{-v}f(t)-\sum_{k=0}^{m-1} \frac{(t-a)^{\underline{v-m+k}}}{\Gamma (v+k-m+1)}\Delta^{k}f(a), $$

where f is defined on \(\mathbb{N} _{a} \) and hence

$$ \Delta^{m}\Delta^{-v}f(t)=\Delta_{\ast}^{\mu }f(t)+ \sum_{k=0}^{m-1}\frac{(t-a)^{\underline{v-m+k}}}{\Gamma (v+k-m+1)} \Delta^{k}f(a). $$
(2.1)

Definition 2

[21]

The μth fractional Riemann-Liouville type difference is defined by

$$ \Delta^{\mu}f(t):=\Delta^{m-v}f(t):=\Delta^{m} \bigl(\Delta^{-v}f(t)\bigr), $$

where \(\mu>0\), \(m-1<\mu<m\), and \(v=m-\mu>0\).

So from (2.1) we get

$$ \Delta^{\mu}f(t)=\Delta_{\ast}^{\mu}f(t)+\sum _{k=0}^{m-1}\frac{(t-a)^{\underline{v-m+k}}}{\Gamma(v+k-m+1)}\Delta^{k}f(a), $$
(2.2)

where f is defined on \(\mathbb{N} _{a}\).

Theorem 3

[10]

For \(\mu>0\), μ noninteger, \(m= \lceil\mu \rceil\), \(v=m-\mu\), the following holds:

$$ f(t)=\sum_{k=0}^{m-1}\frac{(t-a)^{\underline{k}}}{k!}\Delta ^{k}f(a)+\frac{1}{\Gamma(\mu)}\sum _{s=a+v}^{t-\mu}(t-s-1)^{\underline {\mu-1}}\Delta_{\ast}^{\mu}f(s) $$
(2.3)

for all \(t\in \mathbb{N} _{a+m}\), where f is defined on \(\mathbb{N} _{a} \) with \(a\in \mathbb{Z} ^{+}:=\{0,1,2,\ldots\}\).

Remark 2

Here \([ a,b ] \) denotes the discrete interval \([ a,b ] =[a,a+1,a+2,\ldots,b]\), where \(a< b \) and \(a,b\in\{0,1,\ldots\}\). Let \(\mu >0 \) be noninteger such that \(m-1<\mu<m\), i.e. \(m= \lceil \mu \rceil \). Consider a function f defined on \([ a,b ] \). Then clearly the fractional discrete Taylor formula (2.3) is valid only for \(t\in [ a+m,b]\), \(a+m< b\).

We now give a discrete Caputo type fractional extended Taylor formula.

Theorem 4

[10]

Let \(\mu>p\), \(p\in \mathbb{N} \), μ not integer, \(m= \lceil\mu \rceil\), \(v=m-\mu\). Then

$$ \Delta^{p}f(t)=\sum_{k=p}^{m-1} \frac{(t-a)^{\underline {k-p}}}{(k-p)!}\Delta^{k}f(a)+\frac{1}{\Gamma(\mu-p)}\sum _{s=a+v}^{t-\mu +p}(t-s-1)^{\underline{\mu-p-1}} \Delta_{\ast}^{\mu}f(s) $$
(2.4)

for all \(t\in \mathbb{N} _{a+m-p}\), where f is defined on \(\mathbb{N} _{a}\), \(a\in \mathbb{Z} ^{+}\).

Remark 3

We assume that f is defined on \([ a,b ] \). Then (2.4) is valid only for \([ a+m-p,b ] \) with \(a+m-p< b\). Notice \(p=0 \) applied to (2.4) yields (2.3).

Remark 4

For \(\mu>0\), μ not an integer, \(m= \lceil\mu \rceil\), \(v=m-\mu\), f defined on \(\mathbb{N} _{a}\), \(a\in \mathbb{Z} ^{+} \) and \(\Delta^{k}f(a) \) for \(k=0,\ldots,m-1\), we get

$$ f(t)=\frac{1}{\Gamma(\mu)}\sum_{s=a+v}^{t-\mu }(t-s-1)^{\underline{\mu-1}} \Delta_{\ast}^{\mu}f(s) \quad\mbox{for all }t\in \mathbb{N} _{a+m}. $$
(2.5)

Remark 5

For \(\mu>p\), \(p\in \mathbb{N} \), μ noninteger, \(m= \lceil\mu \rceil\), \(v=m-\mu\); f defined on \(\mathbb{N} _{a}\), \(a\in \mathbb{Z} ^{+}\), if we assume that \(\Delta^{k}f(a)=0\), \(k=p,\ldots,m-1\), then we obtain

$$ \Delta^{p}f(t)=\frac{1}{\Gamma(\mu-p)}\sum_{s=a+v}^{t-\mu +p}(t-s-1)^{\underline{\mu-p-1}} \Delta_{\ast}^{\mu}f(s) \quad\mbox{for all }t\in \mathbb{N} _{a+m-p}. $$
(2.6)

Main results

We present the following discrete delta Grüss type inequality.

Theorem 5

Let \(\mu>p\), \(p\in \mathbb{Z} ^{+}\), μ not an integer, \(m= \lceil\mu \rceil\), \(v=m-\mu\), f, g be defined on \(\mathbb{N} _{a}\), \(a\in \mathbb{Z} ^{+} \) and \(a+m-p< b\), \(b\in \mathbb{N} \). Assume that

$$ \Delta^{k}f(a)=\Delta^{k}g(a)=0 \quad\textit{for }k=p+1, \ldots,m-1, p< m-2, $$

and

$$ m_{1}\leq\Delta_{\ast}^{\mu}f(s)\leq M_{1},\qquad m_{2}\leq\Delta_{\ast }^{\mu}g(s) \leq M_{2} $$

for \(s=a+1,\ldots,b\), where \(m_{1}\), \(m_{2}\), \(M_{1}\), and \(M_{2} \) are positive constants. Then

$$\begin{aligned} &\frac{1}{b-a-m+p}\sum_{j=a+m-p+1}^{b} \bigl[ \bigl( \Delta ^{p}f ( j ) \bigr) \bigl( \Delta^{p}g ( j ) \bigr) \bigr] \\ &\qquad{}-\frac{1}{ ( b-a-m+p ) ^{2}} \Biggl[ \sum_{j=a+m-p+1}^{b}\Delta^{p}f ( j ) \Biggr] \Biggl[ \sum_{j=a+m-p+1}^{b} \Delta ^{p}g ( j ) \Biggr] \\ &\quad\leq\frac{M_{1}M_{2}C_{1}-m_{1}m_{2}C_{2}}{ [ \Gamma ( \mu -p+1 ) ] ^{2}}, \end{aligned}$$

where

$$ C_{1}:= ( b-a-m+p ) \sum_{j=a+m-p+1}^{b} \bigl[ ( j-a-v ) ^{\underline{\mu-p}} \bigr] ^{2} $$

and

$$ C_{2}:=\frac{ [ ( b-a-v+1 ) ^{\underline{\mu-p+1}}- ( m-p-v+1 ) ^{\underline{\mu-p+1}} ] ^{2}}{ ( \mu -p+1 ) ^{2}}. $$

Proof

By (2.4), we have

$$ \Delta^{p}f(j)=\sum_{k=p}^{m-1} \frac{(j-a)^{\underline {k-p}}}{(k-p)!}\Delta^{k}f(a)+\frac{1}{\Gamma(\mu-p)}\sum _{s=a+v}^{j-\mu +p}(j-s-1)^{\underline{\mu-p-1}} \Delta_{\ast}^{\mu}f(s). $$

By hypothesis \(\Delta^{k}f(a)=0\), \(k=p+1,\ldots,m-1\), \(p< m-2\). So we have

$$ \Delta^{p}f(j)=\Delta^{p}f(a)+\frac{1}{\Gamma(\mu-p)} \sum _{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta _{\ast }^{\mu}f(s) \bigr) $$
(3.1)

and

$$ \Delta^{p}g(j)=\Delta^{p}g(a)+\frac{1}{\Gamma(\mu-p)} \sum _{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta _{\ast }^{\mu}g(s) \bigr) $$
(3.2)

for all \(j\in [ a+m-p+1,\ldots,b ] \). Multiplying (3.1) and (3.2) gives us

$$\begin{aligned} \bigl( \Delta^{p}f(j) \bigr) \bigl( \Delta^{p}g(j) \bigr) =&\bigl( \Delta^{p}f(a) \bigr) \bigl( \Delta^{p}g(a) \bigr)+ \frac{1}{ [ \Gamma(\mu-p) ] ^{2}} \\ &{}\times \Biggl[ \sum_{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu }f(s) \bigr) \Biggr]\Biggl[ \sum _{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}g(s) \bigr) \Biggr] \\ &{}+\frac{\Delta^{p}f(a)}{\Gamma(\mu-p)} \Biggl[ \sum_{s=a+v}^{j-\mu +p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}g(s) \bigr) \Biggr] \\ &{}+\frac{\Delta^{p}g(a)}{\Gamma(\mu-p)} \Biggl[ \sum_{s=a+v}^{j-\mu +p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}f(s) \bigr) \Biggr] . \end{aligned}$$

Summing from \(a+m-p+1 \) to b yields

$$\begin{aligned} &\sum_{j=a+m-p+1}^{b} \bigl( \Delta^{p}f(j) \bigr) \bigl( \Delta ^{p}g(j) \bigr) \\ &\quad=\sum _{j=a+m-p+1}^{b} \bigl( \Delta^{p}f(a) \bigr) \bigl( \Delta^{p}g(a) \bigr) +\frac{1}{ [ \Gamma(\mu-p) ] ^{2}}\\ &\qquad{}\times\sum_{j=a+m-p+1}^{b} \Biggl\{ \Biggl[ \sum_{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu -p-1}} \bigl( \Delta_{\ast}^{\mu}f(s) \bigr) \Biggr] \Biggl[ \sum_{s=a+v}^{j-\mu+p}(j-s-1)^{\underline {\mu -p-1}} \bigl( \Delta_{\ast}^{\mu}g(s) \bigr) \Biggr] \Biggr\} \\ &\qquad{}+\frac{\Delta^{p}f(a)}{\Gamma(\mu-p)}\sum_{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta _{\ast}^{\mu}g(s) \bigr) \Biggr] \\ &\qquad{}+\frac{\Delta^{p}g(a)}{\Gamma(\mu-p)}\sum_{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta _{\ast}^{\mu}f(s) \bigr) \Biggr] . \end{aligned}$$

Then

$$\begin{aligned} &\frac{1}{b-a-m+p}\sum_{j=a+m-p+1}^{b} \bigl( \Delta ^{p}f(j) \bigr) \bigl( \Delta^{p}g(j) \bigr) \\ &\quad=\frac{1}{b-a-m+p}\sum_{j=a+m-p+1}^{b} \bigl[ \bigl( \Delta ^{p}f(a) \bigr) \bigl( \Delta^{p}g(a) \bigr) \bigr] +\frac{1}{ ( b-a-m+p ) [ \Gamma(\mu-p) ] ^{2}} \\ &\qquad{}\times\sum_{j=a+m-p+1}^{b} \Biggl\{ \Biggl[ \sum_{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu }f(s) \bigr) \Biggr] \Biggl[ \sum _{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}g(s) \bigr) \Biggr] \Biggr\} \\ &\qquad{}+\frac{\Delta^{p}f(a)}{ ( b-a-m+p ) \Gamma(\mu-p)}\sum_{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu +p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}g(s) \bigr) \Biggr] \\ &\qquad{}+\frac{\Delta^{p}g(a)}{ ( b-a-m+p ) \Gamma(\mu-p)}\sum_{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu +p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}f(s) \bigr) \Biggr] . \end{aligned}$$
(3.3)

On the other hand,

$$\begin{aligned} &\frac{1}{b-a-m+p}\sum_{j=a+m-p+1}^{b} \bigl( \Delta ^{p}f(j) \bigr)\\ &\quad=\Delta^{p}f(a) +\frac{1}{ ( b-a-m+p ) \Gamma(\mu-p)}\sum_{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}f(s) \bigr) \Biggr] \end{aligned}$$

and

$$\begin{aligned} &\frac{1}{ ( b-a-m+p ) }\sum_{j=a+m-p+1}^{b} \bigl( \Delta^{p}g(j) \bigr) \\ &\quad=\Delta^{p}g(a) +\frac{1}{ ( b-a-m+p ) \Gamma(\mu-p)}\sum_{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu+p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}g(s) \bigr) \Biggr] . \end{aligned}$$

Multiplying the above two terms yields

$$\begin{aligned} &\frac{1}{ ( b-a-m+p ) ^{2}} \Biggl[ \sum _{j=a+m-p+1}^{b} \bigl( \Delta^{p}f(j) \bigr) \Biggr] \Biggl[ \sum _{j=a+m-p+1}^{b} \bigl( \Delta^{p}g(j) \bigr) \Biggr] \\ &\quad= \bigl( \Delta^{p}f(a) \bigr) \bigl( \Delta^{p}g(a) \bigr) \\ &\qquad{}+\frac{\Delta^{p}f(a)}{ ( b-a-m+p ) \Gamma(\mu-p)}\sum_{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu +p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}g(s) \bigr) \Biggr] \\ &\qquad{}+\frac{\Delta^{p}g(a)}{ ( b-a-m+p ) \Gamma(\mu-p)}\sum_{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu +p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}f(s) \bigr) \Biggr] \\ &\qquad{}+\frac{1}{ ( b-a-m+p ) ^{2}\Gamma(\mu-p)^{2}} \sum_{j=a+m-p+1}^{b} \Biggl[ \sum _{s=a+v}^{j-\mu +p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}f(s) \bigr) \Biggr] \\ &\qquad{}\times\sum _{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu +p}(j-s-1)^{\underline{\mu-p-1}} \bigl( \Delta_{\ast}^{\mu}g(s) \bigr) \Biggr]. \end{aligned}$$
(3.4)

So, using (3.3) and (3.4), we get

$$\begin{aligned} &\frac{1}{b-a-m+p}\sum_{j=a+m-p+1}^{b} \bigl[ \bigl( \Delta ^{p}f(j) \bigr) \bigl( \Delta^{p}g(j) \bigr) \bigr] \\ &\qquad{}-\frac{1}{ ( b-a-m+p ) ^{2}} \Biggl[ \sum_{j=a+m-p+1}^{b} \bigl( \Delta^{p}f(j) \bigr) \Biggr] \Biggl[ \sum _{j=a+m-p+1}^{b} \bigl( \Delta^{p}g(j) \bigr) \Biggr] \\ &\quad\leq\frac{M_{1}M_{2}}{ ( b-a-m+p ) ^{2} [ \Gamma ( \mu -p ) ] ^{2}}\sum_{j=a+m-p+1}^{b} \Biggl[ \sum _{s=a+v}^{j-\mu+p} ( j-s-1 ) ^{\underline{\mu -p-1}} \Biggr] ^{2} \\ &\qquad{}-\frac{m_{1}m_{2}}{ ( b-a-m+p ) ^{2} [ \Gamma ( \mu -p ) ] ^{2}} \Biggl[ \sum_{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu+p} ( j-s-1 ) ^{\underline{\mu -p-1}} \Biggr] \Biggr] ^{2}. \end{aligned}$$

Now, calculating the sums:

$$ \sum_{s=a+v}^{j-\mu+p} ( j-s-1 ) ^{\underline{\mu-p-1} }= \int_{\tau=a+v}^{j-\mu+p+1} \bigl( j-\sigma(\tau) \bigr) ^{\underline{\mu-p-1}}\Delta\tau=\frac{1}{\mu-p} ( j-a-v ) ^{ \underline{\mu-p}}, $$

We get

$$\begin{aligned}& \sum_{j=a+m-p+1}^{b} \Biggl[ \sum _{s=a+v}^{j-\mu+p} ( j-s-1 ) ^{\underline{\mu-p-1}} \Biggr] ^{2}\leq\frac{1}{ ( \mu -p ) ^{2}}\sum_{j=a+m-p+1}^{b} \bigl[ ( j-a-v ) ^{\underline{\mu-p}} \bigr] ^{2}, \\& \begin{aligned}[b] \Biggl[ \sum_{j=a+m-p+1}^{b} \Biggl[ \sum_{s=a+v}^{j-\mu +p} ( j-s-1 ) ^{\underline{\mu-p-1}} \Biggr] \Biggr] ={}&\frac{1}{ ( \mu-p ) ^{2}} \Biggl[ \sum_{j=a+m-p+1}^{b-a-v} ( s ) ^{\underline{\mu-p}} \Biggr] ^{2} \\ ={}&\frac{1}{ ( \mu-p ) ^{2}}\frac{1}{ ( \mu -p+1 ) ^{2}} \\ &{}\times\bigl[ ( b-a-v+1 ) ^{\underline{\mu-p+1}}- ( m-p-v+1 ) ^{\underline{\mu-p+1}} \bigr] ^{2}. \end{aligned} \end{aligned}$$

Consequently, we get

$$\begin{aligned} &\frac{1}{b-a-m+p}\sum_{j=a+m-p+1}^{b} \bigl[ \bigl( \Delta ^{p}f(j) \bigr) \bigl( \Delta^{p}g(j) \bigr) \bigr] \\ &\qquad{}-\frac{1}{ ( b-a-m+p ) ^{2}} \Biggl[ \sum_{j=a+m-p+1}^{b} \bigl( \Delta^{p}f(j) \bigr) \Biggr] \Biggl[ \sum _{j=a+m-p+1}^{b} \bigl( \Delta^{p}g(j) \bigr) \Biggr] \\ &\quad\leq\frac{M_{1}M_{2}}{(b-a-m+p) [ \Gamma ( \mu-p ) ] ^{2}}\frac{1}{ ( \mu-p ) ^{2}}\sum_{j=a+m-p+1}^{b} \bigl[ ( j-a-v ) ^{\underline{\mu-p}} \bigr] ^{2} \\ &\qquad{}-\frac{m_{1}m_{2}}{(b-a-m+p)^{2} [ \Gamma ( \mu-p ) ] ^{2}}\frac{1}{ ( \mu-p ) ^{2} ( \mu-p+1 ) ^{2}} \\ &\qquad{}\times \bigl[ ( b-a-v+1 ) ^{\underline{\mu-p+1}}- ( m-p-v+1 ) ^{\underline{\mu-p+1}} \bigr] ^{2} \\ &\quad=\frac{M_{1}M_{2}C_{1}-m_{1}m_{2}C_{2}}{(b-a-m+p)^{2} [ \Gamma ( \mu-p+1 ) ] ^{2}}. \end{aligned}$$

 □

References

  1. Grüss, G: Über das Maximum des absoluten Betrages von \(\frac{1}{b-a}\int_{a}^{b}f(x)g(x)\,dx-\frac{1}{(b-a)^{2}}\int_{a}^{b}f(x)\,dx\int_{a}^{b}g(x)\,dx\). Math. Z. 39, 215-226 (1935)

    Article  MathSciNet  Google Scholar 

  2. Bohner, M, Peterson, A: Advances in Dynamic Equations on Time Scales. Birkhäuser, Boston (2003)

    Book  MATH  Google Scholar 

  3. Cerone, P, Dragomir, SS: A refinement of the Grüss inequality and applications. Tamkang J. Math. 38, 37-49 (2007)

    MATH  MathSciNet  Google Scholar 

  4. Dragomir, SS: A generalization of Grüss’s inequality in inner product spaces and applications. J. Math. Anal. Appl. 237, 74-82 (1999)

    Article  MATH  MathSciNet  Google Scholar 

  5. Dragomir, SS: A Grüss type discrete inequality in inner product spaces and applications. J. Math. Anal. Appl. 250, 494-511 (2000)

    Article  MATH  MathSciNet  Google Scholar 

  6. Liu, Z: Notes on a Grüss type inequality and its applications. Vietnam J. Math. 35, 121-127 (2007)

    MATH  MathSciNet  Google Scholar 

  7. Perić, I, Rajić, R: Grüss inequality for completely bounded maps. Linear Algebra Appl. 390, 287-292 (2004)

    Article  MATH  MathSciNet  Google Scholar 

  8. Pečarić, JE, Tepeš, B: On the Grüss type inequalities of Dragomir and Fedotov. J. Inequal. Pure Appl. Math. 4(5), 91 (2003)

    MathSciNet  Google Scholar 

  9. Pečarić, JE, Tepeš, B: A note on Grüss type inequality in terms of Δ-seminorms. Pril. - Maked. Akad. Nauk. Umet., Odd. Mat.-Teh. Nauki 23/24 (2002/2003); 29-35 (2004)

  10. Anastassiou, GA: Nabla Discrete fractional calculus and inequalities. arXiv:0911.3374v1 [math.CA] (2009)

  11. Anastassiou, GA: Multivariate Fink type identity and multivariate Ostrowski, comparison of means and Grüss type inequalities. Math. Comput. Model. 46, 351-374 (2007)

    Article  MATH  MathSciNet  Google Scholar 

  12. Bohner, M, Matthews, T: The Grüss inequality on time scales. Commun. Math. Anal. 3, 1-8 (2007) (electronic)

    MATH  MathSciNet  Google Scholar 

  13. Graham, RL, Knuth, DE, Patashnik, O: Concrete Mathematics: A Foundation for Computer Science, 2nd edn. Addison-Wesley, Reading (1994)

    MATH  Google Scholar 

  14. Mercer, AM: An improvement of the Grüss inequality. JIPAM. J. Inequal. Pure Appl. Math. 6(4), 93 (2005)

    MathSciNet  Google Scholar 

  15. Mitrinović, DS: Analytic Inequalities. Springer, New York (1970)

    Book  MATH  Google Scholar 

  16. Mitrinović, DS, Pečarić, JE, Fink, AM: Classical and New Inequalities in Analysis. Kluwer Academic, Dordrecht (1993)

    Book  MATH  Google Scholar 

  17. Pachpatte, BG: Some new Ostrowski and Grüss type inequalities. Tamkang J. Math. 38(2), 11-120 (2007)

    Google Scholar 

  18. Boros, G, Moll, V: Irresistible Integrals: Symbols, Analysis and Experiments in the Evaluation of Integrals. Cambridge University Press, Cambridge (2004)

    Book  Google Scholar 

  19. Diaz, JB, Osler, TJ: Differences of fractional order. Math. Comput. 28, 185-202 (1974)

    Article  MATH  MathSciNet  Google Scholar 

  20. Gray, HL, Zhang, NF: On a new definition of fractional difference. Math. Comput. 50, 513-529 (1988)

    Article  MATH  MathSciNet  Google Scholar 

  21. Atıcı, FM, Eloe, PW: Initial value problems in discrete fractional calculus. Proc. Am. Math. Soc. 137(3), 981-989 (2009)

    MATH  Google Scholar 

  22. Atıcı, FM, Şengül, S: Modeling with fractional difference equations. J. Math. Anal. Appl. 369(1), 1-9 (2010)

    Article  MATH  MathSciNet  Google Scholar 

  23. Goodrich, CS: Continuity of solutions to discrete fractional initial value problems. Comput. Math. Appl. 59(11), 3489-3499 (2010)

    Article  MATH  MathSciNet  Google Scholar 

  24. Atıcı, FM, Eloe, PW: Gronwall’s inequality on discrete fractional calculus. Comput. Math. Appl. 64(10), 3193-3200 (2012)

    Article  MATH  MathSciNet  Google Scholar 

  25. Anastassiou, GA: Nabla discrete fractional calculus and nabla inequalities. Math. Comput. Model. 51(5-6), 562-571 (2010)

    Article  MATH  MathSciNet  Google Scholar 

  26. Güvenilir, AF, Kaymakçalan, B, Peterson, AC, Taş, K: Nabla discrete fractional Grüss type inequality. J. Inequal. Appl. 2014, 86 (2014). doi:10.1186/1029-242X-2014-86

    Article  Google Scholar 

  27. Ferreira, RAC: A discrete fractional Gronwall inequality. Proc. Am. Math. Soc. 140(5), 1605-1612 (2012)

    Article  MATH  Google Scholar 

  28. Atıcı, F, Eloe, P: A transform method in discrete fractional calculus. Int. J. Difference Equ. 2, 165-176 (2007)

    MathSciNet  Google Scholar 

Download references

Acknowledgements

Authors are grateful to the editor and reviewers for their suggestions.

Author information

Affiliations

Authors

Corresponding author

Correspondence to Ayşe Feza Güvenilir.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally in writing this article and collaborated in its design in coordination. All authors read and approved the final paper.

Rights and permissions

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.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Akin, E., Aslıyüce, S., Güvenilir, A.F. et al. Discrete Grüss type inequality on fractional calculus. J Inequal Appl 2015, 174 (2015). https://doi.org/10.1186/s13660-015-0688-2

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/s13660-015-0688-2

MSC

  • 39A12
  • 34A25
  • 26A33
  • 26D15
  • 26D20

Keywords

  • discrete fractional calculus
  • discrete Grüss inequality